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Source: HeatTransfer Calculations
Part
1 Introductory Calculations
HeatTransfer Calculations opens with eight brief industrial heattransfer calculations. While they are short, especially when compared to the calculations that make up the bulk of this handbook, these opening calculations are not trivial. And they deal with real life, which is the hallmark of the calculations throughout the book:
A chemical processing technology that needs a thinner liquid film coating for better adhesive properties of a new product
A new passive method of cooling an electronic assembly that uses a closedloop thermosyphon
During casting of an industrial metal component, a more efficient gating and risering system that minimizes the amount of metal poured to produce a good casting
A furnace wall in a coalfired power plant
A reinforcedconcrete smokestack that must be lined with a refractory on the inside
The condenser in a large steam power plant
The duct of an air conditioning plant
A concentric tube heat exchanger with specified operating conditions
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Introductory Calculations
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Source: HeatTransfer Calculations
Chapter
1 Multiphase Films and Phase Change
Greg F. Naterer University of Ontario Institute of Technology Oshawa, Ontario, Canada
Problem 1 A chemical processing technology needs a thinner liquid film coating for better adhesive properties of a new product. The film flows down under gravity along an inclined plate. Effective thermal control is needed to achieve a specified level of adhesive quality. This goal is accomplished by changing the wall temperature at a certain point along the plate, which affects the thermal boundary layer within the film. Before evaluating heat transfer based on the Nusselt number, the film velocity is needed to predict thermal convection within the film. Then the growth of the thermal boundary layer can be determined. The spatial profile of temperature remains constant, after the point where the edge of the thermal boundary layer reaches the edge of the film. This point is needed for sizing of the processing equipment. Thus, a thermal analysis is needed to find the film velocity, boundarylayer growth, and Nusselt number for evaluating the temperature within the film. This processing temperature affects the adhesive properties of the film. Consider a liquid film (ethylene glycol at 300 K) during testing of the chemical process. The film slows steadily down along a flat plate in the x direction. The plate is inclined at an angle of with respect to the horizontal direction. The x position is measured along the direction of the plate. It is assumed that the film thickness remains nearly uniform. A reference position of x = 0 is defined at a point where the wall boundary conditions change abruptly. Before this point, the film, wall, 1.3
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Multiphase Films and Phase Change 1.4
Introductory Calculations
and air temperatures are T0 . Then the surface temperature increases to Tn (downstream of x = 0) and a thermal boundary layer T develops and grows in thickness within the film. Part (a) Using a reduced form of the momentum equation in the liquid
film, as well as appropriate interfacial and wall boundary conditions, derive the following velocity distribution in the film: y y 2 u =2 − U where U refers to the velocity at the filmair interface. Neglect changes of film velocity in the x direction along the plate. Part (b) Consider a linear approximation of the temperature profile
within the thermal boundary layer, specifically, T = a + by, where the coefficients a and b can be determined from the boundary conditions. Then perform an integral analysis by integrating the relevant energy equation to obtain the thermal boundarylayer thickness T in terms of x, , U, and . For U = 1 cm/s and a film thickness of 2 mm, estimate the x location where T reaches the edge of the liquid film. Part (c) Outline a solution procedure to find the Nusselt number, based
on results in part (b). Solution Part (a) Start with the following mass and momentum equations in the
film: ∂u ∂v + =0 ∂x ∂y u
∂u ∂u ∂p ∂ 2u + v =− + 2 + g (sin ) ∂x ∂y ∂x ∂y
(1.1)
(1.2)
From the continuity equation, the derivative involving u vanishes since the terminal velocity is attained, and thus v equals zero (since ∂v/∂ y = 0 subject to v = 0 on the plate). As a result, both terms on the left side of Eq. (1.2), as well as the pressure gradient term (flat plate), disappear in the momentum equation. Also, we have the boundary condition with a zero velocity at the wall (y = 0) and an approximately zero velocity gradient at the airwater interface. The gradient condition means that ∂u/∂ y = 0 at y = as a result of matching shear stress values ( ∂u/∂ y) on both air and water sides of the interface and a 0 since the transient derivative was neglected in Eq. (1.23). Part (b) At a specified time ts , the temperature reaches a specified frac
tion of Tf , denoted by aTf , when qw aTf = Tf − ks
qw ts − x L
(1.31)
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Multiphase Films and Phase Change Multiphase Films and Phase Change
1.11
This result can be rearranged in terms of the position x1 as follows: x=
ks Tf qw ts − (1 − a) L qw
(1.32)
Using the thermophysical properties for tin, as well as the given numerical values in the problem statement, we obtain 0.04 =
8000 32.9 × (505) × ts − 0.01 × 6940 × 59,000 8000
(1.33)
A time of 3116 s is required before the temperature at x = 0.04 m reaches 500 K, or 99 percent of the fusion temperature. Reference 1. G. F. Naterer, Heat Transfer in Single and Multiphase Systems, CRC Press, Boca Raton, Fla., 2002.
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Multiphase Films and Phase Change
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Source: HeatTransfer Calculations
Chapter
2 Industrial HeatTransfer Calculations
Mohammad G. Rasul School of Advanced Technologies and Processes Faculty of Engineering and Physical Systems Central Queensland University Rockhampton, Queensland, Australia
Introduction Heat transfer is an important aspect of thermodynamics and energy. It is fundamental to many engineering applications. There is a heat or energy transfer whenever there is a temperature difference. Engineers are often required to calculate the heattransfer rate for the application of process technology. A high heattransfer rate is required in many industrial processes. For example, electrical energy generation in a power plant requires a high heat transfer from the hot combustion gases to the water in the boiler tubes and drums. There are three different modes of heat transfer: conduction, convection, and radiation. In practice, two or more types of heat transfer occur simultaneously, but for ease of calculation and analysis, the modes may be separated initially and combined later on. Details of heattransfer analysis, design, and operation can be found elsewhere in the literature [1–5]. In this chapter basic heattransfer equations with several workedout examples on heattransfer calculations and thermal design of heatexchange equipment, usually used in process industries, are presented.
2.1
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Industrial HeatTransfer Calculations 2.2
Introductory Calculations
Basic HeatTransfer Equations The heattransfer rate is the amount of heat that transfers per unit time (usually per second). If a hot metal bar has a surface temperature of t2 on one side and t1 on the other side, the basic heattransfer rate due to conduction can be given by . Q = UA t (2.1) . where Q = heattransfer rate in watts (W) U = heattransfer coefficient in W/m2 K A = surface area of the hot metal bar in m2 t = t2 − t1 = temperature difference in kelvins If a hot wall at a temperature t2 is exposed to a cool fluid at a temperature t1 on one side, the convective heattransfer rate can be given by . Q = hAt (2.2) where h is the convective heattransfer coefficient in W/m2 K. The convective heattransfer coefficient is usually given a special symbol, h, to distinguish it from the overall heattransfer coefficient U. Because of the many factors that affect the convection heattransfer coefficient, calculation of the coefficient is complex. However, dimensionless numbers are used to calculate h for both free convection and forced convection [1–5]. For forced convection, the formula expressing the relationship between the various dimensionless groups may generally be written in the following form: Nu = A Rea Prb
(2.3)
where Nu = hd/, Re = vd/, Pr = C p/, and A, a, and b are constant for the particular type of flow [1–5]. Here, Nu is the Nusselt number, Re is the Reynolds number, and Pr is the Prandtl number; d is the diameter or characteristic dimension in meters, is the thermal conductivity in W/m K, is the density in kg/m3 , is the dynamic viscosity in pascals, and Cp is the specific heat capacity at constant pressure in kJ/kg K. For composite materials in series as shown in Fig. 2.1, the overall heattransfer coefficient U due to combined conduction and convection heat transfer is given by 1 x 1 1 + + = U hA hB
(2.4)
where hA = heattransfer coefficient of fluid film A hB = heattransfer coefficient of fluid film B
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Industrial HeatTransfer Calculations Industrial HeatTransfer Calculations
Figure 2.1
2.3
Heat transfer through a composite wall [5].
= thermal conductivity of materials x = wall thickness Thermal resistance R is the reciprocal of U for unit area; therefore RA =
1 hA A
, R1 =
x1 xn 1 , . . . , Rn = , RB = 1 A n A hB A
The total resistance to heat flow in series is then given by RT = RA + R1 + R2 + · · · + Rn + RB
or
RT =
1 hA A
+
x 1 + A hB A (2.5)
Similarly, the total resistance in parallel is given by 1 1 1 1 1 1 = + + + ··· + + RT RA R1 R2 Rn RB
(2.6)
The overall heattransfer coefficient U is then given by U = 1/RT A. The total resistance to heat flow through a cylinder, which has crosssectional inner radius of r1 and outer radius of r2 , is given by RT =
1 ln(r2 /r1 ) 1 + + ho Ao 2 hi Ai
(2.7)
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Industrial HeatTransfer Calculations 2.4
Introductory Calculations
The total resistance to heat flow through a sphere, which has inner radius of r1 and outer radius of r2 , is given by RT =
1 r2 − r1 1 + + ho Ao 4r1r2 hi Ai
(2.8)
where subscripts i and o represent inside and outside, respectively. Any material which has a temperature above absolute zero radiates energy in the form of electromagnetic waves. The wavelength and frequency of these waves vary from one end to the other end over the spectral band of radiation. It was discovered experimentally by Stefan and verified mathematically by Boltzmann that the energy radiated is proportional to the fourth power of the absolute temperature. The StefanBoltzmann law of radiation is given by . Q = ε A t24 − t14 (2.9) where t2 t1 ε
= absolute temperature of body in kelvins = absolute temperature of surroundings in kelvins = StefanBoltzmann constant (56.7 × 10−9 W/m2 K4 ) = emissivity (dimensionless)
Equation (2.9) indicates that if the absolute temperature is doubled, the amount of energy radiated increases 16 times. However, over a restricted range of Celsius temperatures, little error is introduced by calculating an equivalent heattransfer coefficient for radiation hR defined by [1] . Q = ε A t24 − t14 = hR A(t2 − t1 ) (2.10) Therefore, hR = [ε(t24 − t14 )/(t2 − t1 )] equivalent radiation heattransfer coefficient. Heat Exchanger The equipment used to implement heat exchange between two flowing fluids that are at different temperatures and separated by a solid wall is called a heat exchanger. Heat exchangers may be found in applications such as space heating and air conditioning, power production, wasteheat recovery, and chemical processing. Different types of heat exchangers are used in process industries [2,7]: Concentric tube Counterflow Parallel flow
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Industrial HeatTransfer Calculations Industrial HeatTransfer Calculations
2.5
Crossflow Single pass C max (mixed), Cmin (unmixed) C min (mixed), Cmax (unmixed)
Shell and tube Singleshell pass and doubletube pass Multipass and so on
There are two basic methods of heatexchanger analysis: logarithmic meantemperaturedifference (LMTD) method and the numberoftransferunits (NTU) method. Details on heatexchanger analysis, design, and operation can be found in the book by Shah and Sekulic [2]. In LMTD method, the heattransfer rate can be calculated by the following equation for a concentric tube heat exchanger as shown in Fig. 2.2: Q = UAtlm
tlm =
where
t1 − t2 ln(t1 /t2 )
and
A = dl (2.11)
In the NTU method, effectiveness (ε) is calculated through NTU as follows: ε=
e−NTU(1−R) 1 − R.e−NTU(1−R)
(2.12)
Figure 2.2 Parallelflow (a) and counterflow (b) concentric heat exchangers and the tem
perature distribution with length [5].
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Industrial HeatTransfer Calculations 2.6
Introductory Calculations
. . . where NTU = UA/Cmin , R = Cmin /Cmax , ε = Q/ Qmax , and Qmax = . Cmin (th,i − tc,i ). Then C = mC p and C p is evaluated at the average temperature of the fluid tavg = (ti + to )/2. Workedout Examples The following assumptions have been made to solve the workedout examples presented below:
Negligible heat transfer between exchanger and surroundings
Negligible kineticenergy (KE) and potentialenergy (PE) change
Tube internal flow and thermal conditions fully developed
Negligible thermal resistance of tube material and fouling effects
All properties constant
Example 1
In a coalfired power plant, a furnace wall consists of a 125mmwide refractory brick and a 125mmwide insulating firebrick separated by an airgap as shown in Fig. 2.3. The outside wall is covered with a 12 mm thickness of plaster. The inner surface of the wall is at 1100◦ C, and the room temperature is 25◦ C. The heattransfer coefficient from the outside wall surface to the air in the room is 17 W/m2 K, and the resistance to heat flow of the airgap is 0.16 K/W. The thermal conductivities of the refractor brick, the insulating firebrick, and the plaster are 1.6, 0.3, and 0.14 W/m K, respectively. Calculate 1. The rate of heat loss per unit area of wall surface. 2. The temperature at each interface throughout the wall. 3. The temperature at the outside surface of the wall.
Figure 2.3
Composite wall for Example 1.
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Industrial HeatTransfer Calculations Industrial HeatTransfer Calculations
2.7
Solution Known. Specified thermal conductivities and thickness of wall materials, the room temperature, and the wall surface temperature. Analysis
1. Consider 1 m2 of surface area of the wall in Fig. 2.3. Then, using equation for resistance, we have Resistance of refractory brick =
x 125 = A 1.6 × 103
= 0.0781 K/W Similarly Resistance of insulating firebrick = Resistance of plaster = Resistance of air film on outside surface =
125 = 0.417 K/W 0.3 × 103 12 = 0.0857 K/W 0.14 × 103 1 1 = = 0.0588 K/W ho A 17
Resistance of airgap = 0.16 K/W Hence, the total resistance is RT = 0.0781 + 0.417 + 0.0857 + 0.0588 . + 0.16 = 0.8 K/W. Using the heattransfer rate equation, we have Q = (tA − tB)/RT = (1100 − 25)/0.8 = 1344 W; thus the rate of heat loss per square meter of surface area is 1.344 kW. 2. Referring to Fig. 2.3, the interface temperatures are t1 , t2 , and t3 ; the outside temperature is t4 . Using the thermal resistance . calculated above, we have Q = 1344 = (1100 − t1 )/0.0781; therefore . ◦ C. Similarly, Q = 1344 = (t − t2 )/0.16; therefore t.2 = 780◦ C. t1 = 995 1 . Also, Q = 1344 = (t2 − t3 )/0.417; therefore t3 = 220◦ C, and Q = 1344 = (t3 − t4 )/0.0857; therefore t4 = 104◦ C. by considering the heat3. The temperature t4 can also be determined . transfer rate through the air film, Q = 1344 = (t4 − 25)/0.0857; therefore t4 = 104.1◦ C. The temperature at the outside surface of the wall is 104.1◦ C. Example 2
In a process industry, a reinforcedconcrete smokestack must be lined with a refractory on the inside. The inner and outer diameters of the smokestack are 90 and 130 cm, respectively. The temperature of the
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Industrial HeatTransfer Calculations 2.8
Introductory Calculations
d2 = 90 cm
t1 t3
t2
d1
Figure 2.4
Figure for Example 2.
d3 = 130 cm
inner surface of the refractory lining is 425◦ C, and the temperature of the inner surface of the reinforced concrete should not exceed 225◦ C. The thermal conductivity of the lining material is 0.5 W/m K, and that of the reinforced concrete is 1.1 W/m K. Determine the thickness of the refractory lining and temperature of the outer surface of the smokestack under the conditions that the heat loss must not exceed 2000 W per meter height. Solution
Referring to Fig. 2.4, r2 = 45 cm, r3 = 65 cm, t1 = 425◦ C, and t2 = 225 C, and thermal conductivities are as specified. Known.
◦
Analysis. The smokestack can be considered as a case of heat transfer through a cylinder. For heat transfer across a cylinder, we have
. Q=
2(t1 − t2 ) 2(t2 − t3 ) = (1/1 ) ln(r2 /r1 ) (1/2 ) ln(r3 /r2 )
Substituting values, we obtain 2000 =
2(425 − 225) (1/0.5) ln(45/r1 )
Therefore, r1 = 32.87 cm or d1 = 65.74 cm. Then the lining thickness is =
d2 − d1 90 − 65.74 = = 12.13 cm 2 2
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2.9
Therefore 2000 =
2(225 − t3 ) (1/1.1) ln(65/45)
or
t3 = 118.6◦ C
The temperature of the outer surface of the smokestack is 118.6◦ C. Example 3
In a large steam power plant, the condenser (where steam is condensed to liquid) can be assumed as a shellandtube heat exchanger. The heat exchanger consists of a single shell and 30,000 tubes, each executing two passes. The tubes are constructed with d = 25 mm and steam condenses on their outer surface with an associated convection coefficient of ho = 11,000 W/m2 K. The heattransfer rate Q is 2 × 109 W, and this is accomplished by passing cooling water through the tubes at a rate of 3 × 104 kg/s (the flow rate per tube is therefore 1 kg/s). The water enters at 20◦ C, and the steam condenses at 50◦ C. What is the temperature of the cooling water emerging from the condenser? What is the required tube length per pass? Solution Known. Specifications of a shellandtube heat exchanger with specified heat flow rate and heattransfer coefficient. Water inlet and outlet temperatures.
Assume coldwater average temperature = 27◦ C, from thermophysical properties of saturated water [3, app. A6], C p = 4179 J/kg K, = 855 × 10−6 N s/m2 , = 0.613 W/m K, and Pr = 5.83. Properties.
Analysis. The coolingwater outlet temperature can be determined from the following energybalance equation: . . Q = mc C p,c (tc,o − tc,i ) or 2 × 109 = 30,000 × 4179(tc,o − 20)
Therefore tc,o = 36◦ C. The heatexchanger design calculation could be based on the LMTD or. NTU method. Assuming the LMTD method for calculation, we have Q = UA tlm, where A = N × 2 × dl and 1 1/hi + 1/ho
U=
Here, hi may be estimated from an internal flow correlation, with Reynolds number (Re) given as . 4m 4×1 Re = = = 59,567 d (0.025) 855 × 10−6
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Industrial HeatTransfer Calculations 2.10
Introductory Calculations
Therefore, the flow is turbulent. Using dimensionless correlation [1,3], we have Nu = 0.023Re0.8 Pr0.4 = 0.023(59,567)0.8 (5.83)0.4 = 308. Hence 0.613 hi = Nu = 308 × = 7552 W/m2 K d 0.025 Therefore 1 = 4478 W/m2 K 1/7552 + 1/11,000
U=
(50 − 36) − (50 − 20) (th,i − tc,o ) − (th,o − tc,i ) = = 21◦ C ln[(th,i − tc,o )/(th,o − tc,i )] ln(14/30) . Q 2 × 109 l= = = 4.51 m U(2Nd)tlm (4478)(30,000 × 2 × 0.025) × 21 tlm =
Example 4
In an airconditioning plant, fluid at 20◦ C flows through a duct which has a wall thickness of 100 mm. The surrounding temperature is 10◦ C. The convective heattransfer coefficient inside the duct is 12 W/m2 K, and that of outside is 5 W/m2 K. The thermal conductivity of the duct is 0.5 W/m K, and the external emissivity is 0.85. Assuming negligible internal radiation, determine the heat flow rate through the duct per square meter of duct surface and external wall surface temperature. Use an initial trial value of 40◦ C for external duct wall temperature. Solution Known. Specifications of an airconditioning duct and specified operating conditions. Inside and outside heattransfer coefficients. Analysis
Trial 1. The external duct wall temperature t = 40◦ C = 313 K. The equivalent heattransfer coefficient for external radiation is given by hR =
ε A(t24 − t14 ) 0.85 × 56.7 × 10−9 (3134 − 2834 ) = = 5.115 W/m2 K t2 − t1 40 − 10
Therefore, the combined convectionradiation heattransfer coefficient outside is ho = 5.115 + 5 = 10.115 W/m2 K. The overall heattransfer coefficient U can be determined by 1 x 1 1 1 0.1 1 + + = = + + = 0.382 U hi ho 12 0.5 10.115
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Industrial HeatTransfer Calculations Industrial HeatTransfer Calculations
2.11
Therefore, the overall heattransfer coefficient U = 2.616 W/m2 K and . Q = UA t = 2.616 × 190 = 497.13 W. The outside temperature difference is t = 497.13/10.115 = 49.15◦ C. Therefore, t = 49.15 + 10 = 59.15◦ C. Trial 2. t = 59.15◦ C = 332.15 K. The equivalent heattransfer coefficient for external radiation is now hR =
ε A(t24 − t14 ) 0.85 × 56.7×10−9 (332.154 − 2834 ) = = 5.645 W/m2 K t2 − t1 49.15
The combined convectionradiation heat transfer coefficient outside is then ho = 5.645 + 5 = 10.645 W/m2 K. Therefore 1 x 1 1 1 0.1 1 + + = = + + = 0.377 U hi ho 12 0.5 10.645 Therefore the overall heattransfer coefficient U = 2.65 W/m2 K and Q = UA t = 2.65 × 190 = 503.6 W. The outside temperature difference is t = 503.6/10.645 = 47.3◦ C; therefore t = 47.3 + 10 = 57.3◦ C. Further iterations could be made to refine the accuracy even more, but these are unnecessary as this result is only a small improvement. Therefore, the heat flow rate is 503.6 W and the wall temperature is 57.3◦ C. Example 5
A concentric tube heat exchanger with an area of 50 m2 operating under the following conditions:
Description
Heat capacity rate Cp , kW/K
Inlet temperature ti , ◦ C
Outlet temperature to , ◦ C
Hot fluid Cold fluid
6 3
70 30
— 60
1. What would be the outlet temperature of hot fluid? 2. Is the heat exchanger operating in counterflow or parallel flow or cannot be determined from the available information? 3. Determine the overall heattransfer coefficient. 4. Calculate the effectiveness of the heat exchanger. 5. What would be the effectiveness of this heat exchanger if its length were greatly increased?
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Industrial HeatTransfer Calculations 2.12
Introductory Calculations
th,i = 70°C ∆t1 = 10
Ch = 6 kW/K
tc,o = 60°C
th,o ∆t2
Cc = 3 kW/K
tc,i = 30°C
Counterflow concentric heat exchanger and its temperature distribution for Example 5.
Figure 2.5
Solution Known. Specified operating conditions of heat exchanger including heat capacity rate. Analysis
1. The outlet temperature of hot fluid .can be determined using the energybalance equation as follows: Q = Cc (tc,o − tc,i ) = Ch(th,i − th,o ) or 3000(60 − 30) = 6000(70 − th,o ). Therefore, th,o = 55◦ C. 2. The heat exchanger must be operating in counterflow, since th,o < tc,o . See Fig. 2.5 for temperature distribution. 3. The rate equation with A = 50 m2 can be written as . Q = Cc (tc,o − tc,i ) = UA tlm tlm =
(70 − 60) − (55 − 30) t1 − t2 = = 16.37◦ C ln(t1 /t2 ) ln(10/25)
Therefore, 3000(60 − 30) = U × 50 × 16.37, so U = 109.96 W/m2 K. 4. The effectiveness with the cold fluid as the minimum fluid Cc = Cmin is . Q 60 − 30 Cc (tc,o − tc,i ) ε= . = = 0.75 = Cmin (th,i − tc,i ) 70 − 30 Qmax 5. For a very long heat exchanger, the outlet of the minimum fluid Cmin = Cc will approach th,i : . . Q = Cmin (tc,o − tc,i ) = Qmax
∴
. Q ε= . =1 Qmax
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Industrial HeatTransfer Calculations Industrial HeatTransfer Calculations
2.13
Example 6
You have been asked to design a heat exchanger that meets the following specifications:
Description
Mass flow rate m, kg/s
Inlet temperature ti , ◦ C
Outlet temperature to , ◦ C
Hot fluid Cold fluid
28 27
90 34
65 60
As in many realworld situations, the customer hasn’t revealed or doesn’t know the additional requirements that would allow you to proceed directly to a final design. At the outset, it is helpful to make a firstcut design based on simplifying assumptions, which may be evaluated later to determine what additional requirements and tradeoffs should be considered by the customer. Your submission must list and explain your assumptions. Evaluate your design by identifying what features and configurations could be explored with your customer in order to develop more complete specifications. Solution Known. Specified operating conditions of heat exchanger including mass flow rate. Properties. The properties of water at mean temperature, determined from thermophysical properties of saturated water [3], are given in Table 2.1. Analysis. Using the rate equation, we can determine the value of the UA product required to satisfy the design criteria. Sizing the heat exchanger involves determining the heattransfer area A (tube diameter, length, and number) and associated overall convection coefficient U such that U × A satisfies the required UA product.
TABLE 2.1
Properties of Saturated Water at Mean Temperature
Description
Units
Hotwater mean temperature, 77.5◦ C
Coldwater mean temperature, 47◦ C
Density Specific heat Cp Viscosity Conductivity Prandtl number Pr
kg/m3 kJ/kg K kg/m · s W/m K —
972.8 4.196 3.65 × 10−04 0.668 2.29
989 4.18 5.77 × 10−04 0.64 3.77
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Industrial HeatTransfer Calculations 2.14
Introductory Calculations
Your design approach should have the following five steps: 1. Calculate the UA product. Select a configuration and calculate the required UA product. 2. Estimate the area A. Assume a range of area for overall coefficient, calculate the area, and consider suitable tube diameter(s). 3. Estimate the overall convection coefficient U. For selected tube diameter(s), use correlations to estimate hot and coldside convection coefficients and overall coefficient. 4. Evaluate firstpass design. Check whether the A and U values (U × A) from steps 2 and 3 satisfy the required UA product; if not, then follow step 5. 5. Repeat the analysis. Iterate on different values for area parameters until a satisfactory match is made U × A = UA. To perform the design and analysis, the effectivenessNTU method may be chosen. 1. Calculate the required UA. For initial design, select a concentric tube, counterflow heat exchanger, as shown in Fig. 2.6. Calculate UA using the following set of equations: e−NTU(1 − R) 1 − R.e−NTU(1 − R) . . . where NTU = UA/Cmin , R = Cmin /Cmax , ε = Q/ Qmax , and Qmax = Cmin . (th,i − tc,i ). Then C = mC p, and C p is evaluated at the average temperature of the fluid, tavg = (ti + to )/2. Substituting values, we find ε = 0.464, NTU = 0.8523, Q = 2.94 × 106 W, and UA = 9.62 × 104 W/K. 2. Estimate the area A. The typical range of U for watertowater exchangers is 850 to 1700 W/m2 K [3]. With UA = 9.62 × 104 W/K, the range for A is 57 to 113 m2 , where A = di lN and l and d are length and diameter of the tube, respectively. Considering different values of di with length l = 10 m to describe the heat exchanger, we have three different cases as given in Table 2.2. ε=
th,i = 90°C ∆t1
mh = 28 kg/s th,o = 65°C ∆t2
tc,o = 60°C
mc = 27 kg/s
tc,i = 34°C
Counterflow concentric heat exchanger and its temperature distribution for Example 6.
Figure 2.6
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Industrial HeatTransfer Calculations Industrial HeatTransfer Calculations
TABLE 2.2
2.15
Result of Analysis for Three Different di Values
Case
di , mm
l, m
N
A, m2
1 2 3
25 50 75
10 10 10
73–146 36–72 24–48
57–113 57–113 (answer) 57–113
3. Estimate the overall coefficient U. With inner (i.e., hot) and outer (i.e., cold) water in the concentric tube arrangement, the overall coefficient is given by 1 1 1 = + U hi ho where hi and ho are estimated using dimensionless correlation assuming fully developed turbulent flow. Coefficient, hot side, hi : For flow in the inner side (hot side), the coefficient hi can be determined using the Reynolds number for. mula Redi = 4mh,i /di h and m ˙h = m ˙ h,i N, and the correlation, Nu = hi di / = 0.037Re0.8 Pr0.3 , where properties are evaluated at average mean temperature th,m = (th,i + th,o )/2. Coefficient, cold side, ho : For flow in the annular space do − di , the relations expressed above apply where the characteristic dimension is the hydraulic diameter given by dh,o = 4Ac,o /Po , Ac,o = [(do2 − di2 )]/4, and Po = (do + di ). To determine the outer diameter do , the inner and outer flow areas must be the same: di2 (do2 − di2 ) = 4 4 The results of the analysis with l = 10 m are summarized in Table 2.3. For all these cases, the Reynolds numbers are above 10,000 and turbulent flow occurs. 4. Evaluate firstpass design. The required UA product value determined in step 1 is 9.62 × 104 W/K. The U × A values for cases 1 and 3 are, respectively, larger and smaller than the required value of 9.62 × 104 W/K. Therefore, case 2 with N = 36 and di = 50 mm could be considered as satisfying the heatexchanger specifications in the firstpass design. A strategy can now be developed in step 5 to iterate the Ac,i = Ac,o
TABLE 2.3
or
Results of hi , ho, U , and UA Product for Three Different di and N Values
Case
di , mm
N
A, m2
hi , W/m2 K
ho , W/m2 K
U, W/m2 K
U × A, W/K
1 2 3
25 50 75
73 36 24
57 57 57
4795 2424 1616
4877 2465 1644
2418 1222 814
1.39 × 105 6.91 × 104 4.61 × 104
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Industrial HeatTransfer Calculations 2.16
Introductory Calculations
analysis on different values for di and N, as well as with different lengths, to identify a combination that will meet complete specifications. References 1. R. Kinskey, Thermodynamics: Advanced Applications, McGrawHill, Sydney, Australia, 2001. 2. R. K. Shah and D. P. Sekulic, Fundamentals of Heat Exchanger Design, Wiley, Hoboken, N.J., 2003. 3. F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 5th ed., Wiley, New York, 2002. 4. S. C. Arora and S. Domkundwar, A Course in Heat & Mass Transfer, 3d ed., Dhanpat Rai & Sons, Delhi, India, 1987. 5. T. D. Eastop and A. McConkey, Applied Thermodynamics for Engineering Technologist, 5th ed., PrenticeHall, UK, 1993. 6. M. J. Moran and H. W. Shapiro, Fundamentals of Engineering Thermodynamics, 5th ed., Wiley, Hoboken, N.J., 2004.
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Source: HeatTransfer Calculations
Part
2 SteadyState Calculations
The collection of calculations in this part deals with a variety of circumstances encountered in industrial settings. They are as follows:
The fabrication of sheet and fiberlike materials Hightemperature gas turbine components Hammerbank coils on a highspeed dotmatrix printer Fuel cells Large turbogenerators Doublepaned windows
Contributors include academics, an engineer working in industry, and a consultant. All of the calculations are steadystate and mostly involve cooling using gases and liquids. Determination of heattransfer coefficients, a subject encountered later in this handbook, is a key factor in several calculations. Thermal resistance plays an important role.
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1
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SteadyState Calculations
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2
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Source: HeatTransfer Calculations
Chapter
3 HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid
John P. Abraham School of Engineering University of St. Thomas St. Paul, Minnesota
Ephraim M. Sparrow Department of Mechanical Engineering University of Minnesota Minneapolis, Minnesota
Abstract A method is developed for determining the heat transfer between a moving sheet that passes through a moving fluid environment. The method also enables the streamwise variation of the sheet temperature to be determined. The numerical information needed to apply the method is presented in tabular form. With this information, the desired results can be calculated by algebraic means.
Introduction In the fabrication of sheet and fiberlike materials, the material customarily is in motion during the manufacturing process. The processing may involve heat transfer between the material and an adjacent 3.3
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid 3.4
SteadyState Calculations
Processing station consisting of a moving sheet situated in a parallel gas flow.
Figure 3.1
fluid which may also be in motion. Typical materials include polymer sheets, paper, linoleum, roofing shingles, and finefiber matts. Illustrative diagrams showing possible processing configurations are presented in Figs. 3.1 and 3.2. In Fig. 3.1, an unfinished or partially finished material is unrolled and becomes a moving sheet. In order to complete the finishing operation, a fluid flows over the sheet either to heat or cool it before rollup. In Fig. 3.2, a polymeric sheet is seen emerging from a die. Air flowing through an opening in the die passes in parallel over the sheet to facilitate the heattransfer process. Another alternative (not illustrated) is to eliminate the blown fluid flows of Figs. 3.1 and 3.2 and instead pass the moving sheet through a confined liquid bath. A necessary prerequisite for the solution of the heattransfer problem for such processing applications is the determination of local and average heattransfer coefficients. The issue of heat transfer and the degree to which it is affected by the simultaneous motion of the participating media will be the topic of the present chapter. From an analysis of laminar fluid flow and convective heat transfer, a complete set
Figure 3.2 A moving sheet extruded or drawn through a die. A parallel fluid stream passes over the sheet.
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid Heat Transfer for Moving Sheet in Moving Fluid
3.5
Figure 3.3 Side view showing the webprocessing station of Fig. 3.1.
of heattransfer results have been obtained. The results are parameterized by the ratio Uf /Us and the Prandtl number. A designoriented example will be worked out to provide explicit instructions about how to use the tabulated and graphed heattransfer coefficients. In practice, it is highly likely that the thermal interaction between the moving sheet and the moving fluid would give rise to a streamwise variation of the sheet temperature. An algebraic method will be presented for dealing with such variations. The model on which the analysis will be based is conveyed by the diagrams shown in Figs. 3.3 and 3.4. These diagrams are, respectively, side views of the pictorial diagrams of Figs. 3.1 and 3.2. Each diagram shows dimensional nomenclature and coordinates. The problems will be treated on a twodimensional basis, so that it will not be necessary to deal further with the z coordinate which is perpendicular to the plane of Figs. 3.3 and 3.4. Similarity Solutions and Their Physical Meanings The velocity problem, solved in Ref. 1, was based on a similarity solution of the conservation equations for mass and momentum. To understand
Side view showing the extrusion process of Fig. 3.2.
Figure 3.4
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid 3.6
SteadyState Calculations
Figure 3.5 Velocity profiles in the boundary layer corresponding to a flat sheet
moving through an otherwise quiescent fluid. The profiles correspond to four streamwise locations x1 < x2 < x3 < x4 .
the nature of a similarity solution, the reader may refer to Fig. 3.5, which shows representative velocity profiles for the case of a moving surface situated in an otherwise stationary fluid and where u is the local streamwise velocity in the fluid and y is the coordinate perpendicular to the surface of the sheet. Each velocity profile corresponds to a streamwise location x. The nomenclature is ordered with respect to position so that x1 < x2 < x3 < x4 . If a more concise presentation of the velocity information were sought, the optimal end result would be a single curve representing all velocity profiles at all values of x. Such a simple presentation is, in fact, possible if the y coordinate of Fig. 3.5 is replaced by a composite coordinate which involves both x and y. Such a coordinate is normally referred to as a similarity variable. For the case of a moving surface in a stationary fluid, the similarity variable is =y
Us x
(3.1)
When the velocity profiles of Fig. 3.5 are replotted with as the abscissa variable, they fall together as shown in Fig. 3.6.
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid Heat Transfer for Moving Sheet in Moving Fluid
3.7
Figure 3.6 Velocity profiles of Fig. 3.5 replotted as a function of the similarity variable of Eq. (3.2). All the profiles fall together on a single curve.
The temperature problem will be solved for parametric values of the ratio = Uf Us in the range between 0 and ∞. The temperatures of the fluid and the surface, respectively Tf and Ts , will not enter the solution as such because they are removed by the introduction of a dimensionless temperature variable. The solutions will encompass both streamwiseuniform and streamwisevarying sheet temperatures. The fluids to be considered are gases (primarily air) and liquid water. The Prandtl numbers of these fluids vary between 0.7 and 10. The workedout example will demonstrate how to deal with sheet surfaces whose temperatures vary in the streamwise direction. Analysis Analysis of the heattransfer problem has, as its starting point, the conservationofenergy principle. The equation expressing conservation of energy for a boundary layer is ∂2T ∂T ∂T f cf u (3.2) +v = kf 2 ∂x ∂y ∂y To continue the analysis, it is necessary to make use of the existing solutions for the velocity components u and v. The expressions for u and
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid 3.8
SteadyState Calculations
v are available from Eqs. (3.A6) and (3.A7). These expressions involve a special similarity variable which is appropriate for the situation in which both of the participating media are in motion. In Eqs. (3.A6) and (3.A7), the quantity Urel appears. Its definition is Urel = Us − U f (3.3) To make Eq. (3.2) compatible with the expressions for u and v, it is necessary to transform the physical coordinates, x and y, to the similarity variable , which is used for the velocity problem as shown in the appendix at the end of this chapter. Furthermore, for compactness, it is useful to define a dimensionless temperature as (x, y) =
T (x, y) − Tf Ts − Tf
(3.4)
so that T (x, y) = Ts − Tf (x, y) + Tf
(3.5)
The fluid temperature Tf that appears in Eq. (3.4) will be a constant for all values of x. On the other hand, the temperature Ts of the moving sheet may vary with x as the sheet either cools down or heats up as a result of heat exchange with the fluid. To keep the analysis sufficiently general to accommodate variations of Ts with x, the temperature difference (Ts − Tf ) that appears in Eq. (3.5) will be assumed to have the form Ts = Tf + ax n
(3.6)
where a is a constant. The exponent n may be zero or any positive integer. The case n = 0 corresponds to a uniform sheet temperature. We will now consider the quantity ∂ T/∂ x appearing in Eq. (3.2). This quantity can be evaluated by applying the operation ∂/∂ x to Eq. (3.5), with the result ∂ ∂ (Ts − Tf ) + Tf ∂ Ts − Tf ∂T = = Ts − Tf + ∂x ∂x ∂x ∂x = (Ts − Tf )
∂ + nax n − 1 ∂x
(3.7)
where the derivative ∂ Ts /∂ x has been evaluated using Eq. (3.6). Similarly, the terms ∂ T/∂ y and ∂ 2 T/∂ y2 follow as ∂ ∂ (Ts − Tf ) + Tf ∂T = = Ts − Tf (3.8) ∂y ∂y ∂y
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3.9
and
∂ − T ) ∂ (T s f ∂2T ∂ 2 ∂y = (T = − T ) s f ∂y ∂ y2 ∂ y2
(3.9)
Although depends on both x and y, it is advantageous to tentatively assume that is a function only of the similarity variable , which is a combination of x and y. Then, with = (), it remains to evaluate the quantities ∂/∂ x, ∂/∂ y, and ∂ 2 /∂ y2 in terms of derivatives involving the similarity variable . For that evaluation, and with the definition of taken from Eq. (3.A3), we obtain ∂ y Urel /x ∂ = =− ∂y ∂y 2x
(3.10)
and ∂ y Urel /x ∂ Urel = = ∂y ∂y x
(3.11)
With these relationships, the quantities ∂/∂ x, ∂/∂ y, and ∂ 2 /∂ y2 follow as ∂ d ∂ d (3.12) = =− = − ∂x d ∂ x 2x d 2x ∂ Urel d Urel d ∂ (3.13) = = = ∂y d ∂ y x d x ∂ Urel /x (d/d) ∂ 2 Urel ∂ (d/d) Urel d 2 ∂ = = = ∂y x ∂y x d2 ∂ y ∂ y2 =
Urel d 2 Urel = x d2 x
(3.14)
When Eqs. (3.7) to (3.9), (3.12) and (3.13), (3.A6), and (3.A7) are substituted into the energyconservation equation [Eq. (3.2)] and if the notation = d/d is used, there results nf −
1 1 f = 2 Pr
(3.15)
Everything in this equation is a function only of . As a final specification of the temperature problem, it is necessary to state the boundary
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid 3.10
SteadyState Calculations
conditions. In physical terms, they are T = Ts
at
y=0
and
T → Tf
as
y→∞
(3.16)
These conditions may be transformed in terms of the new variables and to yield =1
at = 0
and
→0
as
→ ∞
(3.17)
Numerical Solution Scheme A suggested method of numerically solving the dimensionless energy equation [Eq. (3.15)], along with its boundary conditions [Eq. (3.17)] will now be presented. The relevant equations are () = Pr[nf () () − 1/2 f () ()]
(3.18)
( + ) = () + ()
(3.19)
( + ) = () + ()
(3.20)
to which Eqs. (3.A14) to (3.A17) along with Eq. (3.A10) must be added if the velocity problem has not already been solved. This scheme involves simultaneous solution of the velocity and temperature problems. For these equations, the quantities Uf /Us , n, and Pr are prescribed parameters. The solution of these equations, including Eqs. (3.A14) to (3.A17), starts by assembling the values of , , f, f , and f at = 0. Among these, the known values are = 1, f = 0, f = Us /Urel , and f from Table 3.A1. Since the value of at = 0 must be known in order to initiate the numerical integration of Eqs. (3.18) to (3.20), it is necessary to make a tentative guess of the value of (0). Once this guess has been made, the known values of Pr, n, f, f , , and at = 0 are introduced into Eq. (3.18) to yield (0). Then, the righthand sides of Eqs. ( 3.19) and (3.20) are evaluated at = 0, yielding the values of () and (). At the same time, the f equations, Eqs. (3.A14) to (3.A17), are used as described in the text following Table 3.A1 to advance f, f , f , and f from = 0 to = . A similar procedure is followed to advance all the functions from = to = 2. These calculations are repeated successively until takes on a constant value. If the thusobtained value does not equal zero as required by Eq. (3.17), the tentatively estimated value of (0) must be revised. In this case, a new value of (0) is selected, and the foregoing computations are repeated. This process is continued until the behavior of stipulated by Eq. (3.17) is achieved.
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid Heat Transfer for Moving Sheet in Moving Fluid
3.11
Extraction of HeatTransfer Coefficients For later application, heattransfer coefficients are needed. These coefficients can be extracted from the numerical solutions that have just been described. The local heattransfer coefficient at any location x is defined as hn =
q = Ts − Tf
∂ T kf ∂ y y=0 Ts − Tf
(3.21)
where the subscript n corresponds to the exponent n in the sheet surface temperature distribution of Eq. (3.6). The use of the absolutevalue notation is motivated by the fact that heat may flow either from the moving surface to the fluid (Ts > Tf ) or vice versa (Tf > Ts ). The derivative ∂ T/∂ y can be transformed into the variables of the analysis by making use of Eqs. (3.8) and (3.13), with the end result hn = k f
Urel n(0) x
(3.22)
A dimensionless form of the heattransfer coefficient can be expressed in terms of the local Nusselt number Nux whose definition is hn x Nux,n = = kf
Urel · x · n(0) = x
Urel x n(0) = Rex n(0) (3.23)
where Rex represents the local Reynolds number: Rex = Urel x/. Val ues of n(0) /Pr0.45 needed to evaluate hn and Nux,n for three Prandtl numbers (0.7, 5, and 10) are listed in Table 3.1 for n = 0 (uniform sheet temperature), 1, and 2. The factor Pr0.45 is used to diminish the spread of the results with the Prandtl number and to improve the accuracy of the interpolation for Prandtl numbers other than those listed. Inspection of the table reveals that the values of (0) are greater at higher Prandtl numbers. This is because the thermal boundary layer becomes thinner as the Prandtl number increases, and, as a consequence, the temperature n() must decrease more rapidly between its terminal values of 1 and 0. Also noteworthy is the observation that the sensitivity of n(0) to the velocity ratio increases with increasing Prandtl number. For the case of uniform sheet temperature (n = 0), the local Nusselt number can be integrated to obtain a surfaceaveraged heattransfer
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid 3.12
SteadyState Calculations
TABLE 3.1
Listing of Θ (0)/Pr0.45 Values for n = 0 (Uniform Sheet Temperature),
1, and 2 n= 0 Uf = Us 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.11 1.25 1.43 1.67 2.00 2.50 3.33 5.00 10.0 ∞
n= 1
n= 2
0.7
5
10
0.7
5
10
0.7
5
10
0.410 0.453 0.500 0.554 0.618 0.696 0.800 0.946 1.186 1.715 1.700 1.164 0.917 0.766 0.656 0.572 0.504 0.445 0.392 0.344
0.559 0.594 0.635 0.685 0.746 0.825 0.931 1.085 1.342 1.915 1.931 1.299 1.004 0.818 0.683 0.576 0.487 0.410 0.343 0.279
0.596 0.631 0.673 0.723 0.786 0.866 0.974 1.132 1.395 1.987 1.911 1.286 0.994 0.809 0.676 0.569 0.480 0.401 0.328 0.258
0.943 1.014 1.097 1.194 1.311 1.459 1.657 1.942 2.412 3.459 3.373 2.288 1.786 1.471 1.245 1.068 0.920 0.792 0.675 0.564
1.164 1.233 1.313 1.410 1.532 1.685 1.893 2.197 2.705 3.846 3.688 2.478 1.912 1.552 1.291 1.082 0.905 0.747 0.599 0.452
1.224 1.294 1.376 1.476 1.599 1.758 1.972 2.285 2.809 3.986 3.812 2.556 1.966 1.592 1.317 1.098 0.910 0.741 0.580 0.417
1.316 1.407 1.510 1.633 1.786 1.980 2.238 2.612 3.235 4.625 4.484 3.032 2.356 1.933 1.626 1.384 1.182 1.005 0.842 0.686
1.573 1.662 1.769 1.897 2.057 2.261 2.537 2.941 3.616 5.134 4.913 3.294 2.535 2.054 1.701 1.419 1.179 0.962 1.241 0.548
1.647 1.739 1.848 1.980 2.144 2.354 2.639 3.055 3.752 5.321 5.078 3.400 2.612 2.110 1.741 1.445 1.191 0.960 0.737 0.506
coefficient h¯ 0 , defined as
L Q = 1 ¯h0 = q dA A(T − T ) A Ts − Tf 0 s f 1 = WL Ts − Tf
L
0
1 h0 Ts − Tf Wdx = L
L
h0 dx
(3.24)
0
where W is the width of the sheet. Equation (3.24) shows that h¯ 0 can be obtained as the length average of h0 . The indicated integration can be carried out by substituting Eq. (3.22) into the Eq. (3.24), giving
L 2k f 0 (0) Urel L U 1 rel (0) dx = h¯ 0 = (3.25) kf 0 L 0 x L or Nu L,0
h¯ 0 L = = 2 0 (0) kf
Urel L = 2 0 (0) ReL
(3.26)
For situations where the sheet temperature is not uniform (n = 0), the average heattransfer coefficient cannot be obtained by averaging h.
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid Heat Transfer for Moving Sheet in Moving Fluid
3.13
Illustrative Example The purpose of this workedout example is to demonstrate how to deal with generalized streamwise temperature variations of the moving sheet. Problem statement
A linoleum sheet moving at 0.5 ft/s is to be cooled by a parallel stream of air whose velocity is 5 ft/s. The temperature of the sheet emerging from a die is 160◦ F, while the air temperature is 80◦ F. The thickness of the sheet is 1/32 in. (1/384 ft), its width is 3 ft, and the distance between the die and the takeup roll is 5 ft. The thermophysical properties of the participating media are Linoleum sheet
Density ( s ) = 22 lbm /ft3 Specific heat (cs ) = 0.21 Btu/lbm ◦ F Air
Density ( f ) = 0.0684 lbm /ft3 Specific heat (cf ) = 0.241 Btu/lbm ◦ F Thermal conductivity (kf ) = 0.0160 Btu/hft◦ F Kinematic viscosity (f ) = 0.690 ft2 /h Prandtl number (Pr) = 0.7 Find
The temperature decrease of the sheet as it cools between its departure from the die and its arrival at the takeup roll. Discussion
The solution strategy to be used to achieve the required result is as follows: 1. Obtain an estimate of the change in the temperature of the moving sheet between the initiation of cooling (x = 0) and the termination of cooling (x = 5 ft). This estimate will enable you to judge whether the issue of a streamwise temperature variation is worthy of further pursuit from the standpoint of engineering practice. In this regard, it should be noted that the maximum temperature difference between the moving sheet and the adjacent air is 80◦ F. If a 5 percent decrease in this
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid 3.14
SteadyState Calculations
temperature difference is deemed to be allowable, then a temperature drop of 4◦ F along the length of the sheet would be regarded as acceptable. 2. If the estimated overall temperature decrease is judged to be significant, then a refined approach is warranted. The first step in that approach is to determine the variation of the sheet temperature TsI with x by making use of the local heattransfer coefficients that correspond to uniform surface temperature. Here, the superscript I is used to denote the temperature variation obtained from the first iteration. This approach, while seemingly inconsistent, is appropriate as the starting point in the successive refinement of the solution. 3. The temperature TsI (x) is then fitted with a polynomial of the form TsI (x) ≈ a0 + a1 x + a2 x 2 This polynomial approximation to the sheet temperature will be required in the first step in the refinement process. 4. It is noted that the fitted curve TsI (x) contains x0 , x1 , and x2 . This observation, taken together with Eq. (3.6), suggests that heattransfer coefficient information is needed for the cases n = 0, 1, and 2. 5. Use the heattransfer coefficients identified in paragraph 4 to determine the local heat flux q(x) at the surface of the sheet. 6. The q(x) information enables TsII (x) to be determined. 7. Compare TsI (x) with TsII (x) and, if necessary, continue iterating until convergence. Step 1. An upperbound estimate of the streamwise decrease in tem
perature of the moving sheet can be made by utilizing a heattransfer model in which the sheet temperature is constant during the process. This calculation requires the determination of the surfaceaveraged heattransfer coefficient h¯ 0 from Eq. (3.25). The value of 0 (0) can be found in Table 3.1 for the velocity ratio Uf /Us = 10, Pr = 0.7, and n = 0 to be 0.392(0.70.45 ) = 0.334. Next, the Reynolds number Re L is calculated to be Re L =
(4.5 ft/s)5 ft 3600 s Urel L = = 117,000 h 0.690 ft2 /h
(3.27)
With this information, the heattransfer coefficient emerges as 2k f 0 (0) Urel L 2 0.0160 Btu/hft◦ F 0.334 h¯ 0 = 117,000 = L 5 ft = 0.731
Btu hft2 ◦ F
(3.28)
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid Heat Transfer for Moving Sheet in Moving Fluid
3.15
An energy balance on the moving sheet can be used to connect the estimated heat loss due to convection with the change in temperature of the moving sheet. That energy balance is m ˙ s cs Ts = −h¯ 0 A(Ts − Tf )
(3.29)
where m ˙ s is the mass flow rate of the sheet, cs is the specific heat of the sheet, and Ts is the temperature change experienced by the sheet during the cooling process. The convective area A includes both the upper and lower surfaces of the sheet. When the mass flow rate of the sheet and the convective area are expressed in terms of known quantities, Eq. (3.29) becomes sWtUs cs Ts = −2h¯ 0 WL(Ts − Tf )
(3.30)
where W, t, and L represent, respectively, the width, thickness, and length of the sheet. Equation (3.30) can be solved for the sheet temperature change to give Ts = − =−
2h¯ 0 L(Ts − Tf ) s tUs cs 2(0.731 Btu/hft2 ◦ F)(5 ft) (80◦ F)(1 h/3600 s) 1 (22 lbm /ft3 ) 384 ft (0.5 ft/s)(0.21 Btu/lbm ◦ F)
= −27.0◦ F
(3.31)
A temperature decrease of this magnitude warrants a refined approach that accounts for temperature variations all along the sheet. Step 2. The next step in the solution procedure is to determine TsI
based on local heattransfer coefficients that correspond to uniform surface temperature. The symbol TsI represents a preliminary temperature variation of the sheet which will be used as the basis of further refinements. An energy balance can be performed on the sheet to connect the TsI with the local rate of heat loss. This balance leads to 2h0 (x) TsI (x) − Tf dTsI 2q(x) =− =− dx s tcs Us s tcs Us
(3.32)
which can be rearranged and integrated along the sheet length to give
Ts T0
dTsI =− I Ts (x) − Tf
x 0
2h0 ()d 2 =− s tcs Us s tcs Us
x
h0 ()d
(3.33)
0
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid 3.16
SteadyState Calculations
in which is a variable of integration. After Eq. (3.22) is substituted into the righthand side, and the indicated integrations are performed, there follows I
x Ts (x) − Tf Urel 2 ln 0 (0) d =− kf I s tcs Us 0 Ts (0) − Tf
2k f 0 (0) Urel x 1 =− d (3.34) s tcs Us 0 After integrating on the righthand side and inserting the limits of integration, we obtain I 4k f 0 (0) Urel x Ts (x) − Tf ln =− (3.35) s tcs Us TsI (0) − Tf which can be solved explicitly for TsI (x) to give 4k f 0 (0) Urel x I I Ts (x) = Ts (0) − Tf exp − + Tf s tcs Us = TsI (0) − Tf e− + Tf
(3.36)
where , which was used to simplify the expression of Eq. (3.36), is found to be 4 0.0160 Btu/hft◦ F 0.334 4.5 ft/s · x = 0.690 2 1 (22 lbm /ft3 ) 384 ft (0.21 Btu/lbm ◦ F)(1800 ft/h) 3600 ft /s √ = 0.151 x
(3.37)
Equations (3.36) and (3.37) can be combined to give √
TsI (x) − 80(◦ F) = (80)e−0.151
x ◦
( F) ≈ 75.86 − 6.907x + 0.6641x 2 (◦ F) (3.38)
where the polynomial on the righthand side is a leastsquares fit used to approximate TsI . Step 3. A refined calculation can be carried out using the polynomial
expression of Eq. (3.38). The first step in the refinement is the recognition that the firstlaw equation [Eq. (3.2)] is linear with the temperature T; that is, wherever T appears, it is to the power 1. It is also known that for a linear differential equation, solutions for that equation for
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid Heat Transfer for Moving Sheet in Moving Fluid
3.17
different boundary conditions can be added together. This realization suggests that a suitable solution form for T(x, y) in the presence of a quadratic variation of the sheet temperature is T(x, y) − Tf = a0 0 () + a1 1 ()x + a2 2 ()x 2
(3.39)
where the n terms are functions of alone. Note that at the sheet surface, where n(0) = 1 for all n, Eq. (3.39) reduces to Ts (x) − Tf = a0 + a1 x + a2 x 2
(3.40)
which is identical in form to Eq. (3.38). In addition, as y → ∞, it follows that n(0) → 0 for all n. Furthermore, substitution of Eq. (3.39) into the first law [Eq. (3.2)] shows that the differential equation obeyed by n() is that of Eq. (3.15). In light of these facts, it follows that the solutions for n() have already been obtained and tabulated for n = 0, 1, and 2 (Table 3.1). The heat flux at the surface of the moving sheet can be obtained from Eq. (3.39) by application of Fourier’s law, which gives ∂T q(x) = −k f ∂y
y=0
∂ a0 0 () + a1 1 ()x + a2 2 ()x 2 = −k f ∂y
y=0
(3.41)
When the indicated differentiations are carried out, there follows
Urel (3.42) a0 0 (0) + a1 x 1 (0) + a2 x 2 2 (0) x The numerical values of 0 (0), 1 (0), and 2 (0) were determined from Table 3.1 to be 0.334, 0.575, and 0.717, respectively. When the values of a0 , a1 , and a2 are taken from Eq. (3.38) and inserted into Eq. (3.42), the local heat flux emerges as q(x) = k f
q(x) = k f
Urel 25.3 − 3.97x + 0.476x 2 x
(3.43)
Step 4. Equation (3.43) can be used as input to the energy balance of
Eq. (3.32) to give dTsII 2q(x) =− =− dx s tcs Us
2k f
Urel (25.3 − 3.97x + 0.476x 2 ) x s tcs Us
(3.44)
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid 3.18
SteadyState Calculations
or, on integration
Ts (x) Ts (0)
dTsII = −
x 0
2k f
Urel (25.3 − 3.97 + 0.4762 )d s tcs Us
(3.45)
where the variable of integration has been introduced. When the integration of Eq. (3.45) is carried out, we obtain 2k f Urel / − =− (50.6x 1/2 − 2.65x 3/2 + 0.190x 5/2 ) s tcs Us (3.46) When the known quantities are substituted into Eq. (3.46), the seconditeration sheet temperature is found to be TsII (x)
TsII (0)
TsII (x) = −0.226(50.6x 1/2 − 2.65x 3/2 + 0.190x 5/2 ) + 160
(3.47)
A comparison of TsI with TsII will enable a determination of whether further iteration is required. That comparison, set forth graphically in Fig. 3.7, shows satisfactory agreement between TsI (dashed curve) and TsII (solid curve). The close agreement (within 1.7◦ F) obviates the need for further revision. Equation (3.47) can be used to determine the overall temperature decrease experienced by the sheet throughout
Figure 3.7
Comparison between the first and seconditeration results TsI and TsII .
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid Heat Transfer for Moving Sheet in Moving Fluid
3.19
the entire cooling process. That decrease is found to be TsII (L) = −0.226 50.6L1/2 − 2.65L3/2 + 0.190L5/2 + 160 = 138.7◦ F (3.48) Concluding Remarks A method has been developed for calculating the heat exchange between a moving fluid and the surface of a sheet which also moves. Both the fluid and the sheet move in the same direction. The heat transfer is initiated by a temperature difference between the fluid and the sheet which exists at their first point of contact. The initial temperature of the sheet may be set by upstream processes to which the sheet is subjected. Under normal conditions, the sheet temperature will vary in the streamwise direction as a result of thermal interaction with the moving fluid. Determination of the temperature variation of the sheet is another focus of the methodology developed here. The information needed to determine both the heattransfer rate and the sheet temperature variation is provided in tabular form. With this information, the desired results can be obtained by algebraic operations. Nomenclature a
Constant value used in Eq. (3.6)
a 0 , a1 , a2
Coefficients used in the polynomial fit of TsI
A
Surface area of sheet
c
Specific heat
f
Similarity function
h h¯
Local heattransfer coefficient
k
Thermal conductivity
L
Length of sheet
m ˙
Mass flow rate of moving sheet
n
Constant value used in Eq. (3.6)
Nu
Nusselt number
p
Pressure
Pr
Prandtl number
q
Local heat flux
Q
Rate of heat flow
Averaged heattransfer coefficient
Re
Reynolds number
t
Sheet thickness
T
Temperature
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid 3.20
SteadyState Calculations
Ts I Ts
Firstiteration sheet temperature
II
Seconditeration sheet temperature
u, v
x, y velocities
U
Velocity
W
Sheet width
x, y
Coordinates
Greek Similarity variable based on the relative velocity
Dimensionless temperature
Dynamic viscosity
Kinematic viscosity
Similarity variable based on the sheet velocity
Density
Streamfunction 4k f  (0) Urel x = s tcs Us
Subscripts f
Fluid
rel
Relative
s
Surface
0, 1, 2
Corresponding to zeroth, first, and secondorder terms in the temperature polynomial
Reference 1. John P. Abraham and Ephraim M. Sparrow, Friction Drag Resulting from the Simultaneous Imposed Motions of a Freestream and Its Bounding Surface, International Journal of Heat and Fluid Flow 26(2):289–295, 2005.
Appendix: Similarity Solutions for the Velocity Problem The starting point of the analysis is the equations expressing conservation of mass and x momentum for laminar boundarylayer flow of a constantproperty fluid. These are, respectively ∂u ∂v + =0 ∂x ∂y
(3.A1)
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid Heat Transfer for Moving Sheet in Moving Fluid
3.21
and u
∂u ∂u 1 dp ∂ 2u +v =− + 2 ∂x ∂y dx ∂y
(3.A2)
In these equations, u and v are the velocity components that respectively correspond to x and y, with p the pressure, and and , respectively, the density and the viscosity of the fluid. Although both the freestream flow and the bounding surface move, the streamwise pressure gradient dp/dx is essentially zero, as it is for the two limiting cases of a moving fluid in the presence of a stationary surface and a moving surface passing through a stationary fluid. These equations may be transformed from the realm of partial differential equations to that of ordinary differential equations by using the similarity transformation from the x, y plane to the plane. To begin the derivation of the similarity model, it is useful to define Urel = y (3.A3) x and =
Urel x f ()
(3.A4) where Urel = U f − Us . These variables can be employed to transform the conservation equations, Eqs. (3.A1) and (3.A2), to similarity form d 3f 1 d 2 f df 1 = f + f f = 0 + 2 d2 d 2 d3
(3.A5)
In the rightmost member of this equation, the notation = d/d for compactness. The fact that this equation is an ordinary differential equation is testimony to the fact that the magnitude of the relative velocity is a suitable parameter for the similarity transformation. It still remains to demonstrate that the boundary conditions depend only on and are independent of x. To this end, the velocity components are needed. They are expressible in terms of f and f as u = Urel f 1 Urel v= (f − f ) 2 x
(3.A6) (3.A7)
The physical boundary conditions to be applied to Eqs. (3.A6) and (3.A7) are v=0
and u → Uf
u = Us as
at y→∞
y=0
(3.A8) (3.A9)
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid 3.22
SteadyState Calculations
The end result of the application of the boundary conditions is f (0) = 0
f (0) =
Us Urel
f ( → ∞) =
Uf Urel
(3.A10)
The transformed boundary conditions of Eq. (3.A10) are seen to be constants, independent of x. Therefore, the similarity transformation is complete. From the standpoint of calculation, it is necessary to prescribe the values of f at = 0 and as → ∞. To this end, it is convenient to define =
Uf Us
(3.A11)
so that f (0) =
1 Us 1 = = Urel 1 − U f /Us 1 −
(3.A12)
and f ( → ∞) =
Uf U f /Us = = 1 − U f /Us 1 − Urel
(3.A13)
A review of Eqs. (3.A5), (3.A10), (3.A12), and (3.A13) indicates a complete definition of the similaritybased, relativevelocity model. To implement the numerical solutions, values of = Uf /Us were parametrically assigned from 0 (stationary fluid, moving surface) to ∞ (stationary surface, moving fluid). The values of f  = 0 that correspond to the specific values of are listed in Table 3.A1. The knowledge of f  = 0 , together with an algebraic algorithm, enables Eq. (3.A5) to be solved without difficulty. Many algorithms can be used for solving Eq. (3.A5). In the present instance, where f  = 0 is known, the equations which constitute the algorithm are f () = −
1 · f () · f () 2
(3.A14)
f ( + ) = f () + · f ()
(3.A15)
f ( + ) = f () + · f ()
(3.A16)
f ( + ) = f () + · f ()
(3.A17)
The use of the algorithms set forth in Eqs. (3.A14) to (3.A17) will now be illustrated. The first step is to choose a velocity ratio as represented
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid Heat Transfer for Moving Sheet in Moving Fluid
3.23
Listing of f ( = 0) Values for the SimilarityBased, RelativeVelocity Model
TABLE 3.A1
= U f /Us 0.000 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.111 1.250 1.429 1.667 2.000 2.500 3.333 5.000 10.00 ∞
f ( = 0) −0.4439 −0.4832 −0.5279 −0.5797 −0.6421 −0.7204 −0.8238 −0.9713 −1.214 −1.717 1.726 1.177 0.9257 0.7689 0.6573 0.5704 0.4991 0.4377 0.3829 0.3319
by . Then, by substituting the value of into Eq. (3.A12), the value of f  = 0 is determined. With this information, the values of f, f , and f at = 0 are known, the latter from Table 3.A1. In addition, the value f = 0 at = 0 follows directly from the differential equation [Eq. (3.A14)]. The numerical work is initiated by evaluating the righthand sides of Eqs. (3.A14) to (3.A17) at = 0. These calculations yield the numerical values of f (), f (), f (), and f (), which are, in turn, used as new inputs to the righthand sides of Eqs. (3.A14) to (3.A17). The end result of this second step are the values of f (2), f (2), f (2), and f (2). These operations are continued repetitively until an value is reached at which f becomes a constant. For checking purposes, the value of that constant should be equal to that obtained from Eq. (3.A13) for the selected value of .
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HeatTransfer and Temperature Results for a Moving Sheet Situated in a Moving Fluid
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24
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Source: HeatTransfer Calculations
Chapter
4 Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil
Ronald S. Bunker Energy and Propulsion Technologies General Electric Global Research Center Niskayuna, New York
Introduction: Gas Turbine Cooling The technology of cooling gas turbine components via internal convective flows of singlephase gases has developed over the years from simple smooth cooling passages to very complex geometries involving many differing surfaces, architectures, and fluidsurface interactions. The fundamental aim of this technology area is to obtain the highest overall cooling effectiveness with the lowest possible penalty on thermodynamic cycle performance. As a thermodynamic Brayton cycle, the efficiency of the gas turbine engine can be raised substantially by increasing the firing temperature of the turbine. Modern gas turbine systems are fired at temperatures far in excess of the material melting temperature limits. This is made possible by the aggressive cooling of the hotgaspath (HGP) components using a portion of the compressor discharge air, as depicted in Fig. 4.1. The use of 20 to 30 percent of this compressed air to cool the highpressure turbine (HPT) presents a severe penalty on the thermodynamic efficiency unless the firing temperature is sufficiently high for the gains to outweigh the losses. In all properly operating cooled turbine systems, the efficiency gain is significant enough to justify the added complexity and cost of the
4.1
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AFT SHAFT
Cross section of highpressure gas turbine.
COMPRESSOR DISCHARGE PRESSURE (CDP) SEAL
FWD SHAFT
FORWARD OUTER SEAL (FOS)
HP TURBINE DISK
AFT AIR SEAL
Figure 4.1
COMPRESSOR DISCHARGE
COMBUSTION ZONE
HP TURBINE VANE
HP TURBINE BLADE
AFT RETAINER
Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil
4.2
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil HeatTransfer Design of Cooled Gas Turbine Airfoil
4.3
cooling technologies employed. Actively or passively cooled regions in both aircraft engine and powergenerating gas turbines include the stationary vanes or nozzles and the rotating blades or buckets of the HPT stages, the shrouds bounding the rotating blades, and the combustor liners and flameholding segments. Aircraft engines also include cooling in the exhaust nozzles and in some lowpressure turbines (LPTs). All such engines additionally cool the interfaces and secondary flow regions around the immediate hotgas path. Cooling technology, as applied to gas turbine components such as the highpressure vane and the blade, consists of four main elements: (1) internal convective cooling, (2) external surface film cooling, (3) materials selection, and (4) thermomechanical design. Film cooling is the practice of bleeding internal cooling flows onto the exterior skin of the components to provide a heatfluxreducing cooling layer. Film cooling is intimately tied to the internal cooling technique used in that the local internal flow details will influence the flow characteristics of the film jets injected on the surface. Materials most commonly employed in cooled parts include hightemperature, highstrength nickel or cobaltbased superalloys coated with yttriastabilized zirconia oxide ceramics [thermal barrier coating (TBC)]. The protective ceramic coatings are today actively used to enhance the cooling capability of the internal convection mechanisms. The thermomechanical design of the components must marry these first three elements into a package that has acceptable thermal stresses, coating strains, oxidation limits, creep rupture properties, and aeromechanical response. Under the majority of practical system constraints, this means the highest achievable internal convective heattransfer coefficients with the lowest achievable friction coefficient or pressure loss. In some circumstances, pressure loss is not a concern, such as when plenty of head is available and the highest available heattransfer enhancements are sought, while in other applications pressure loss may be so restricted as to dictate a very limited means of heattransfer enhancement. In many respects, the evolution of gas turbine internal cooling technologies began in parallel with heatexchanger and fluidprocessing techniques, “simply” packaged into the constrained designs required of turbine airfoils (aerodynamics, mechanical strength, vibrational response, etc.). Turbine airfoils are, after all, merely highly specialized and complex heat exchangers that release the coldside fluid in a controlled fashion to maximize work extraction. The enhancement of internal convective flow surfaces for the augmentation of heat transfer was initially improved around 1975 to 1980 through the introduction of rib rougheners or turbulators, and also pinbanks or pinfins. Almost all highly cooled regions of the highpressure turbine components involve the use of turbulent convective flows and heat transfer. Very few, if
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil 4.4
SteadyState Calculations
any, cooling flows within the primary hot section are laminar or transitional. The enhancement of heattransfer coefficients for turbine cooling makes full use of the turbulent flow nature by seeking to generate mixing mechanisms in the coolant flows that actively exchange cooler fluid for the heated fluid near the walls. These mechanisms include shear layers, boundarylayer disruption, and vortex generation. In a marked difference from conventional heat exchangers, few turbinecooling means rely on an increase in cooling surface area, since the available surface area–volume ratios are very small. Surface area increases are beneficial, but are not the primary objective of enhancements. The use of various enhancement techniques typically results in at least 50 and as much as 300 percent increase in local heattransfer coefficients over that associated with fully developed turbulent flow in a smooth duct.
Problem Statement Estimate the minimum cooling flow requirements for a prescribed highpressure turbine inlet guide vane (Fig. 4.2) meeting predefined system and component constraints, using basic engineering thermalfluid relationships. Determine the optimal configuration of cooling flow and airfoil wall composition. The hotgas (air) inlet total pressure and
e
X = 4 inches
Pr
es
1
Suct
su
re
ion s ide
sid
Simplify
X
Simplify
X=0 X
Hot Air X
P0 = 500 psia T0 = 3400°F Flat plate external flow
Coolant P0 = 520 psia T0 = 1000°F Air Figure 4.2
22
Internal channel flow
Problem geometry and flow simplification.
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil HeatTransfer Design of Cooled Gas Turbine Airfoil
4.5
temperature are 500 psia (3.446 MPa) and 3400◦ F (1870◦ C), respectively. The air coolant supply total pressure and temperature are 520 psia (3.584 MPa) and 1000◦ F (538◦ C), respectively. The substrate metal of the vane may be assumed to have an isotropic thermal conductivity of 10 Btu/h/ft/◦ F (17.296 W/m/K), while the protective ceramic TBC thermal conductivity is 1 Btu/h/ft/◦ F (1.73 W/m/K). The airfoil chordal length is 4 in. (10.16 cm), and the airfoil is 3 in. (7.62 cm) in height. The wall thickness may be considered a constant value along the length and height of the airfoil. The external hotgas Mach number over the airfoil is represented by a linear variation from 0.005 at x = 0 to 1.0 at x = L, where L is the total length. This distribution represents the average of the pressure and suction sides of the airfoil, as well as the endwalls, and suits the average nature of the present estimation. The airfoil is considered choked aerodynamically. The average surface roughness krms on the airfoil is 400 in. (10 m). The conditions are considered to be steadystate, representing a maximum power operating point. Certain constraints must be placed on the solution as typically dictated by material properties or fluid pressure fields. The substrate metal is a Nibased superalloy composition, but must not exceed a maximum temperature of 2000◦ F (1094◦ C) at any given location. The TBC thickness may not exceed 0.010 in. (0.254 mm) as a result of strain limits of bonding with the metal. Also, the TBC maximum temperature must not be more than 2300◦ F (1260◦ C) to avoid the infiltration of molten particulates from the hot gases. The overall temperature increase of the cooling air due to convective exchange with the airfoil interior surfaces must not exceed 150◦ F (66◦ C). We will utilize impingement jet arrays to cool the airfoil inside, and film cooling to reduce the external incident heat flux. The maximum impingement jet Reynolds number cannot exceed 50,000 without reducing coolant pressure to the point of allowing hotgas ingestion inside the airfoil. The average film cooling effectiveness may be assumed to be 0.4 for the entire airfoil. Coolant temperature rise as the flow passes through the film holes is approximately 50◦ F (10.5◦ C). Solution Outline In the complete design of such a hightemperature component, many interdependent aspects of design must be considered for both the individual component and its integration into the engine system. The design system is an iterative method in which aerothermomechanical requirements and constraints must be balanced with material limitations and highlevel system objectives. We will consider here only the thermal design portion of the total solution. In addition, component design may entail one of several levels of analysis, from preliminary
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil
H rad
Metal T
TBCMetal T
SteadyState Calculations
TBC T
4.6
Gas T(x,y,z) Film exit T
Adiabatic wall T
ηaw
Coolant T(x,y,z) Hgas
(k, t)TBC
(k, t)metal
H coolant
Q Figure 4.3
Onedimensional thermal resistance model for a cooled airfoil.
estimates, to detailed twodimensional analyses, to complete threedimensional computational predictions including the conjugate effects of the convective and radiative environments. Each level of analysis has its use as the design progresses from concept to reality. Preliminary design uses mostly bulk quantities and onedimensional simplified equations to arrive at approximate yet meaningful estimates of temperatures and flow requirements. Twodimensional design incorporates boundarylayer analyses, network flow and energy balances, and some thermal gradient estimates to refine the results for local temperature and flow predictions suitable for use in finiteelement stress modeling. Threedimensional design may use complete computational fluid dynamics and heattransfer modeling of the internal and external flow fields to obtain the most detailed predictions of local thermal effects and flow losses. In the present solution, only the simplified preliminary design method will be used. Figure 4.2 shows the vane airfoil and endwalls reduced first to a constant cross section of the aerodynamic shape, and then again to a basic flat plate representing flow from the leadingedge stagnation point to the trailing edge. While the actual airfoil or endwall shape involves many complexities of accelerating and decelerating flows, secondary flows, and discrete film injection holes, a good estimate may still be obtained using fundamental flatplate relations. The solution may be best understood as a sequence of several steps leading to an overall model that is optimized for material thicknesses, cooling configuration, and cooling flow. Figure 4.3 shows the onedimensional thermal model that applies to any location on the airfoil. These steps include 1. Estimation of the external heattransfer coefficient distribution on the airfoil 2. Inclusion of airfoil surface roughness effects for the external heattransfer coefficients
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil HeatTransfer Design of Cooled Gas Turbine Airfoil
4.7
3. Calculation of the average adiabatic wall temperature due to film cooling 4. Estimation of the internal heattransfer coefficients due to impingement cooling 5. Calculation of the required cooling flow rate 6. Optimization of the solution The solution will be iterative to account for fluid property changes with temperature, both internal and external to the airfoil. Thermal radiation heat load incident on the airfoil will be neglected here, but should be included for any locations that are in direct view of the combustor flame. The required properties for air may be calculated by the following equations: Density: = P/RT, lbm /ft3 Dynamic viscosity: = 28.679{[(T/491.6)1.5 ]/(T + 199)}, lbm /ft/h Thermal conductivity: k = 1.065 × 10−4 (T )0.79 , Btu/h/ft/◦ R Specific heat: C p = 2.58 × 10−5 · T + 0.225, Btu/lbm /◦ R Prandtl number: Pr = 0.7 Temperature T is a static reference value in ◦ R, and pressure P is a static value in psia. The local hotgas static temperature is determined from the isentropic relationship between total temperature and local Mach number: Ttotal /Tstatic = 1 + [( − 1)/2] · M 2
(T ) = specificheat ratio
External Boundary Conditions External heattransfer coefficients
By utilizing a simple flatplate model for the airfoil in which the start (leading edge) is at X = 0 in. and the end (trailing edge) is at X = 4 in., we can employ boundarylayer heattransfer equations for the assumed conditions of constant freestream velocity and zero pressure gradient. The laminar and turbulent heattransfer coefficients (HTCs) at some location x may then be expressed per Kays and Crawford [1] as Laminar Turbulent
k · Re0.5 Pr1/3 x k h(x) = 0.0287 · Re0.8 Pr0.6 x
h(x) = 0.332
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil 4.8
SteadyState Calculations
where the local Reynolds number (Re) is based on the distance as Re = Vx/ and the fluid properties are evaluated at a simple average film temperature Tfilm = 0.5 · (Tgas + Twall ). The film temperature is a function of local wall temperature, and since the local wall temperature will change with the external h(x) distribution, the solution becomes iterative. Moreover, when the internal cooling is altered in a subsequent step, wall temperature will again be changed, leading to iteration throughout the solution. The gas temperature in this case is the static hotgas value. Calculation of the laminar and turbulent heattransfer coefficients is not sufficient; we must also approximate the location of transition from laminar to turbulent flow, and model the transition zone. Typical transition of a simple boundary layer occurs between Reynolds numbers 200,000 and 500,000 [1]. We will model the external heattransfer coefficients as laminar up to Re = 200,000, turbulent after Re = 500,000, and use the local average value of the two in the short transition region. Film cooling
Film cooling of gas turbine components is the practice of bleeding the internal cooling air through discrete holes in the walls to form a layer of protective film on the external surface. Figure 4.4 shows the ideal case of a twodimensional tangential slot of film cooling covering a surface. In this ideal case, the layer of film has the exit temperature of the coolant after having been heated by internal cooling and energy pickup through the slot. The gas “film” temperature driving external heat flux to the surface is then that of the coolant rather than that of the hot gas. As the filmcooling layer progresses downstream,
Mainstream hot gas
T gas
T adiabatic wall
Film supply
Tcoolant Figure 4.4 Ideal tangential slot film cooling (threetemperature problem).
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil HeatTransfer Design of Cooled Gas Turbine Airfoil
4.9
diffusion, shear mixing, and turbulence act to mix the coolant with the hot gas and reduce the effective film temperature. Eventually, the two streams of fluid would be completely mixed, and, because of the overwhelming amount of hot gases, the film temperature would return to that of the hot gas. This is also true for the realistic case of film cooling via many discrete jets, where coolant injection and mixing are highly threedimensional processes. In either case, the film cooling serves to reduce the driving potential for external convective heat flux. This threetemperature problem is characterized by the adiabatic film effectiveness defined as =
Tgas − Tadiabatic wall Tgas − Tcoolant exit
where the adiabatic wall temperature is that which would exist in the absence of heat transfer. This adiabatic wall temperature is the reference mixed gas temperature that drives heat flux to the wall in the presence of film cooling. In the present model, the average adiabatic film effectiveness is 0.4. The local value of adiabatic wall temperature is then determined from the local gas static temperature (from the Mach number distribution) and the coolant exit temperature. The latter temperature is a function of the degree of internal cooling.
Surface roughness effect
In most gas turbines, a significant portion of the airfoil external surfaces become rough as a result of deposits, erosion, or corrosion. Roughness serves to elevate heat transfer coefficients drastically when the roughness elements become large enough to disrupt or protrude through the momentum thickness layer 2 . Roughness is generally characterized using a socalled sandgrain roughness ks that is related to the actual physical average height krms of the rough peaks [2]. Depending on the type of roughness, the relationship is in the range of ks = 2krms to 10krms [3]. For the present estimation purposes we shall use ks = 5krms . A local roughness Reynolds number is then defined on the basis of ks . A surface is said to be smooth if the roughness Reks is less than 5, and fully rough when greater than 70, with a transition region between these values [2]. This roughness Reks also employs the coefficient of friction as Vks C f 1/2 Reks = 2
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil 4.10
SteadyState Calculations
In the specific case of a fully rough surface, the coefficient of friction is then related to the momentum thickness by Cf =
0.336 [ ln (864 2 /ks )]2
2 = 0.036 · x · Re−0.2 Finally, the fully rough surface heattransfer coefficient in an external boundary layer may be expressed with the model due to Dipprey and Sabersky [4] as h rgh =
(k/x) · Re · Pr · (C f /2) 0.9 + (C f /2)1/2 /Re−0.2 · Pr−0.44 ks
We may now estimate the magnitude and form of the external heattransfer coefficient distribution by assuming a constant value of external wall temperature, at least until a specific wall temperature distribution is determined using internal cooling. Knowing that our target maximum value is 2300◦ F (1260◦ C), we will use 2200◦ F (1205◦ C) for now as the surface temperature everywhere. We also know that the coolant is allowed to heat up by as much as 150◦ F (66◦ C) inside the airfoil, plus another 50◦ F (10.5◦ C) for the film hole throughwall heating. We shall therefore use a filmcooling exit temperature of 1200◦ F (650◦ C) as an average value. As an aside, it is generally best to fully utilize the allowable internal cooling before exhausting the air as film cooling. All calculations are performed in a spreadsheet format. Figure 4.5 shows the current intermediate estimates for Re and Reks distributions, the smooth and roughsurface heattransfer coefficient distributions, and for reference, the Mach number. The locations of laminar and turbulent transition are reflected in the discontinuities present in the heattransfer coefficients. The fully rough condition of Reks = 70 occurs at x/L 0.25. The effect of roughness is seen to be nearly a doubling of the local heattransfer coefficients. Overall, it is somewhat startling to note that heattransfer coefficients may vary by an order of magnitude over the entire surface. We should note that the current estimate has not accounted for the airfoil leadingedge stagnation region heat transfer, which actually acts as a cylinder in crossflow with augmentation due to freestream turbulence and roughness. Hence the heattransfer coefficients for x/L < 0.1 should be much higher, but the addition is not large compared to the rest of the airfoil. We should also mention that film injection affects local heattransfer coefficients, sometimes significantly, but the complexity is too much to include in this simple estimation. Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil HeatTransfer Design of Cooled Gas Turbine Airfoil
4.11
Reynolds Number
140 120
e5*Rex (x 105) Rex Reks Re ks
100 80 60 40 20 0 0
0.2
0.4
0.6
0.8
1
X/L
HeatTransfer Coefficient (Btu/h/ft2/°F)
2500 2000 1500 1000
Smooth Surface HTC
500
Rough Surface HTC 0 0
0.2
0.4
0.6
0.8
1
X/L Figure 4.5
External flow Reynolds numbers and heattransfer coefficients.
Internal Cooling Impingement cooling
The use of impingement jets for the cooling of various regions of modern gas turbine engines is widespread, especially within the highpressure turbine. Since the cooling effectiveness of impingement jets is very high, this method of cooling provides an efficient means of component heat load management given sufficient available pressure head and geometric space for implementation. Regular arrays of impingement jets are used within turbine airfoils and endwalls to provide relatively uniform and controlled cooling of fairly open internal surface regions. Such regular impingement arrays are generally directed against the target Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil 4.12
SteadyState Calculations
X Y D
Z
Gj = ρjVj
X
Gc = ρcVc Figure 4.6
Z
General jet array impingement geometry.
surfaces, as shown in Fig. 4.6, by the use of sheetmetal baffle plates, inserts, or covers which are fixed in position relative to the target surface. These arrangements allow for the design of a wide range of impingement geometries, including inline, staggered, or arbitrary patterns of jets. There are several pertinent parameters in such jet array impingement arrangements. The primary fluid parameter is the jet Re D , based on the orifice diameter and exit velocity. The main geometric parameters include the target spacing Z/D, and the jettojet (interjet) spacings x/D and y/D. Also important in an array of jets is the relative crossflow strength ratio of postimpingement mass velocity to jet mass velocity Gc /G j or ( V)c /( V) j . This crossflow effect can in severe cases actually shut down the flow of coolant from subsequent jets. Fortunately, in the design of turbine inlet vanes, the crossflow ratio is relatively weak, usually around Gc /G j = 0.2, as a result of the large number of distributed film holes over the surface that extract the air. Jet array impingement correlations for surfaceaverage heattransfer coefficients exist in the literature for various conditions. The correlation of Bailey and Bunker [5] for square jet arrays is used here: k x Z Z hD = 47.1 − 5.5 + 7.3 − 2.3 D D D D x + Re D 4 × 10−3 − 1.3 × 10−4 · − 1.5 × 10−8 · Re D D Gc x Z + 61.2 − 13.7 − 28 Gj D D
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil HeatTransfer Design of Cooled Gas Turbine Airfoil
4.13
This expression is valid for 1 < x/D < 15, 1.5 < Z/D < 5, and Re D < 100,000. The jet Re D is based on the supply cooling conditions. The coolant properties used in this equation are based on an internal film temperature, the average of the coolant supply temperature, and the local internal wall temperature. Coolant temperature rise
The last element of this process is simply a check of the allowed temperature rise in the coolant. This is obtained by calculating the heat flux through the composite wall, iterating for closure on the wall temperatures, and summing the total heat flux that must be absorbed by the coolant. The total coolant flow rate mc is merely the sum of the impingement jet flows as determined from the jet diameter D, jet Re D , and the number of jets that fit into the airfoil wall chord and height space. The coolant temperature rise is then hD · A · (Twall internal − Tcoolant ) Tc = mc · C p Optimization of Solution Optimization of the current design based on average conditions means determining the minimum coolant usage possible while still satisfying the required constraints. The parameters that are available for variation include the impingement jet array geometry, jet Re D , metal wall thickness, and TBC thickness. Actually, maximum thermal protection is obtained with the maximum TBC thickness of 0.010 in. (0.254 mm), so this parameter is set by common sense. Metal thickness must be set to a value that can be manufactured. Aside from this, a minimum metal thickness will lead to more effective internal heat flux; 0.050 in. (1.27 mm) is selected to meet both needs. In exploring the design space available, the constraints cannot all be exactly satisfied by a single solution. In addition, calculations show that the most limiting constraint is that of the 150◦ F (66◦ C) allowable coolant temperature rise. This means that the coolant impingement heattransfer coefficient will be limited in order to maintain the temperaturerise limit. This solution is represented by the following design: Metal thickness
0.050 in. (1.27 mm)
TBC thickness
0.010 in. (0.254 mm)
Jet diameter
0.031 in. (0.787 mm)
Jet Re D
20,000
Jet x/D
8
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil 4.14
SteadyState Calculations
Jet Z/D
3
Number of jets
195
Cooling HTC
404 Btu/h/ft2 /◦ F (2293 W/m2 /K)
Maximum metal T
1951◦ F (1066◦ C)
Maximum TBC T
2245◦ F (1230◦ C)
Average external T
2079◦ F (1138◦ C)
Bulk metal T
1782◦ F (973◦ C)
Coolant T
149.5◦ F (65.6◦ C)
Coolant flow rate
0.195 lbm /s (88.64 g/s)
The metal and TBC maximum temperatures are each about 50◦ F (10.5◦ C) below their allowable limits. The complete thermal distribution is shown in Fig. 4.7. The external heattransfer coefficient distribution is essentially the same as that of Fig. 4.5, with roughness. Hotgas static temperature decreases significantly as the Mach number increases. The effect of film cooling makes it clear why this cooling method is the best line of defense in design, serving to directly reduce the driving potential for heat flux to the surface. Low external heattransfer coefficients near the airfoil leading edge are balanced by high thermal potentials, while high heattransfer coefficients at the trailing edge encounter lower thermal potentials. Using this coolant flow rate for each of the airfoil sides, and approximating the same amount again for the two endwalls, the entire airfoil cooling flow is 0.78 lbm /s (355 g/s). A turbine airfoil of this size has a count of about 44 in the engine, making the total vane segment cooling 34.32 lbm /s (15.6 kg/s). For an engine 4000
Temperature (°F)
3500 3000 2500 2000 1500
T adiabatic wall T external surface
1000
T external metal T internal metal
500
T hot gas
0 0
0.2
0.4
0.6
0.8
1
X/L Figure 4.7 Optimized wall temperature distribution for constrained problem.
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil HeatTransfer Design of Cooled Gas Turbine Airfoil
4.15
flow rate of approximately 300 lbm /s (136.4 kg/s), the vane cooling flow represents 11.44 percent of the engine compressor airflow, which is very typical of highperformance gas turbine inlet guide vanes. An additional note is appropriate here. Real turbine vane cooling designs make full use of available parameter space in a local manner rather than on an average basis. Film cooling will be varied in strength over the airfoil surface, and internal cooling will also be adjusted locally, to obtain a more uniform distribution of resulting material temperatures leading to lower inplane stresses. References 1. Kays, W. M., and Crawford, M. E., 1980, Convective Heat and Mass Transfer, 2d ed., McGrawHill. 2. Schlichting, H., 1979, Boundary Layer Theory, 7th ed., McGrawHill. 3. Bunker, R. S., 2003, The Effect of Thermal Barrier Coating Roughness Magnitude on Heat Transfer with and without Flowpath Surface Steps, Paper IMECE200341073, International Mechanical Engineering Conference, Washington, D.C. 4. Dipprey, R. B., and Sabersky, R. H., 1963, Heat and Mass Transfer in Smooth and Rough Tubes at Various Prandtl Numbers, International Journal of Heat and Mass Transfer, Vol. 6, pp. 329–353. 5. Bailey, J. C., and Bunker, R. S., 2002, Local Heat Transfer and Flow Distributions for Impinging Jet Arrays of Both Sparse and Dense Extent, GT200230473, IGTI Turbo Expo, Amsterdam.
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Solution for the HeatTransfer Design of a Cooled Gas Turbine Airfoil
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16
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Source: HeatTransfer Calculations
Chapter
5 SteadyState HeatTransfer Problem: Cooling Fin
David R. Dearth Applied Analysis & Technology Huntington Beach, California
Introduction In one of my early engineering consulting projects, finiteelement analysis (FEA) techniques were utilized to investigate extending service life of hammerbank coils on a highspeed dotmatrix printer. One aspect of this overall project was assigned to estimate reductions in coil temperatures when additional cooling fins are added. Figure 5.1 shows the original singlefin and two new dualcoolingfin design concepts selected for investigation. Before beginning development of the FEA idealizations, I recommend doing a few sample “warmup” problems with known textbook solutions. These test problems are extensively investigated prior to tackling the real problem. This sample problem was also a “sanity” check, hand solution created to ensure that I was using the correct units on the heattransfer parameters before I invested time on the actual FEA mathematical idealizations. The representative warmup problem selected for investigating this heattransfer project was a simplified cooling fin idealized as a uniform rod or cylinder. A search through the engineering literature addressing heattransfer problems located various portions of each individual
5.1
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Figure 5.1 Design concepts for cooling fins: (a) original single fin; (b) dualfin concept 1; (c) dualfin concept 2.
(a)
Magnetic Coil
Epoxy
Heat Sink Investment Casting Al Alloy A356
(b)
Magnetic Coil
Epoxy
Heat Sink Investment Casting Al Alloy A356
(c)
Magnetic Coil
Epoxy
Heat Sink Investment Casting Al Alloy A356
SteadyState HeatTransfer Problem: Cooling Fin
5.2
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SteadyState HeatTransfer Problem: Cooling Fin SteadyState HeatTransfer Problem: Cooling Fin
5.3
aspect of these types of heattransfer problems. However, none of the engineering references contained a complete approach to the problem from beginning to end. This sample fin problem contains all the features of any reallife problem as a comprehensive, stepbystep solution to the simplified version of the cooling fins with an FEA model to crosscheck the results. Once we were confident of our methodology, techniques, and procedures from investigating this sample problem, we proceeded to address the reallife problem. Sample Problem: HeatTransfer Solution for a Simplified Cooling Fin Figure 5.2 shows a sample problem of a cylindrical fin with heat source at the base temperature Tb = 250◦ F and surrounding still air at T∞ = 72◦ F. The fin is fabricated from a generic aluminum alloy with thermal conductivity K = 130 Btu/hft◦ F. The first step in the analysis is to determine an average heattransfer film coefficient havg for free (natural) convection to the surrounding ambient air (fluid). For purposes of this sample problem havg = 1.30 Btu/hft2 ◦F. The arithmetic for estimating havg can be found in the detailed hand calculations that follow. For our sample problem, it is desired to estimate the distribution of temperatures ◦ F along the length of the fin and at the tip. C L
D
L
T∞ = 72°F Still Air
L = 6 in. D = 1/2 in.
Free Convection Film Coefficient havg = 1.30 Btu/hft2°F Figure 5.2 Sample heattransfer “fin” model for steadystate temperatures.
Tb = 250°F
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SteadyState HeatTransfer Problem: Cooling Fin 5.4
SteadyState Calculations
To gain confidence in our solution, we will present two approaches: (1) hand solutions using conventional equations found in most engineering textbooks on heat transfer and (2) correlate results with a finiteelement idealization.
Answers An aluminum rod (K = 130 Btu/hft◦ F) having a diameter of 0.500 in. and a length of 6 in. is attached to a surface where the base temperature is 250◦ F. The ambient air is 72◦ F, and the average free convective heattransfer film coefficient along the rod length and end is havg = 1.30 Btu/hft2 ◦ F. Determine the temperature distribution along the length of the cylindrical rod, accounting for the heat transfer at the endface. Step 1: Compute average film coefficient from mean film surface temperature Tfilm
Tfilm =
Tb + T∞ 2
(5.1)
Substituting into Eq. (5.1), we obtain Tfilm =
250 + 72 = 161◦ F 2
Table 5.1 lists properties of air at Tfilm = 161◦ F using liner interpolation from data in table A4 of Ref. 1. Step 2: Compute Grashof number Gr and check for laminar or turbulent free convection
Let Tfilm = Tsurface = Ts : TABLE 5.1
Properties of Air at 161◦ F
Density Specific heat C p Dynamic viscosity Kinematic viscosity Thermal conductivity k Thermal diffusivity Prandtl number Pr
= = = = = = =
0.0634 lb/ft3 0.2409 Btu/lb◦ F 1.3787 × 10−5 lb/sft 21.83 × 10−5 ft2 /s 0.01713 Btu/hft◦ F 1.1263 ft2 /h 0.698
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5.5
For gas (air), Gr = =
g (Ts − T∞ )L3 v2
(5.2)
1 1 = = 1.610 × 10−3 ◦ R−1 T ◦R 161 + 460
(5.3)
Let characteristic length L = Lv = (6/12 ) ft. Substituting into Eq. (5.2), we obtain Gr =
(32.2 ft/s2 )(1.610 × 10−3 ◦ R−1 )(250 − 72) ◦ R (6/12 ft)3 (21.83 × 10−5 ft2 /s)2
= 2.421 × 107 Compute the Rayleigh number Ra: Ra = Gr Pr = (2.421 × 107 ) (0.698) = 1.690 × 107 Since Ra = Gr Pr < 109 , laminar flow conditions may be assumed. Step 3: Compute average film heattransfer coefficient h avg = h
First compute the Nusselt number: Nu =
hL k
(5.4)
The Nusselt number can also be represented using empirical relations depending on the geometry of the free convective system: Nu = C(Gr Pr)a
(5.5)
hL = Nu = C(Gr Pr)a k
(5.6)
Then
For a long cylinder L/D > 10 and laminar flow, select the following coefficients: C = 0.59 and a = 1/4 , from table 7.1 in Ref. 2. Substituting and solving for h, we obtain h=
0.01713 Btu/hft◦ F (0.59)(1.690 × 107 )1/4 6/ 12 ft
= 1.30 Btu/hft2 ◦ F
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SteadyState HeatTransfer Problem: Cooling Fin 5.6
SteadyState Calculations
Step 4: Solve for temperature distribution along the length of the fin
The solution to the differential equations for combined conductionconvection of the fin with heat loss at the endface can be found in Ref. 2: cosh[m(L − x)] + h/mK sinh[m(L − x)] T(x) − T∞ = Tb − T∞ cosh(mL) + h/mK sinh(mL) where
m=
(5.7)
hP ⇐⇒ KA
2h Kr
for a cylinder
(5.8)
Substituting and solving Eq. (5.8), we obtain ◦ 2 × (1.30 Btu/hft2  F) m= = 0.9798 ft−1 (130 Btu/hft◦ F)[(0.50/2)/12 ft] Also ◦
1.30 Btu/hft2  F h = = 0.010206 mK (0.9798 ft−1 )(130 Btu/hft◦ F) Substituting into Eq. (5.7), we obtain T(x) − 72 cosh[0.9798(6/12 − x)] + (0.010206)sinh[0.9798(6/12 − x)] = 250 − 72 cosh[0.9798(6/12 )] + (0.010206)sinh[0.9798(6/12 )] (5.9) Step 5: Evaluate T(x) at 1/2 in. intervals (0.50/12ft intervals)
Table 5.2 lists estimates for external surface temperatures of the fin by using a spreadsheet to minimize roundoff in the arithmetic of Eq. (5.9). Figure 5.3 plots the data of Table 5.2. Step 6: Verify hand calculations using FEA idealization
Table 5.2 estimates the steadystate fin tip temperature Tfin tip = 229.9090◦ F. Figure 5.4 summarizes results for an FEA model idealization in which Fig. 5.2 is processed using MSC/Nastran. Figure 5.4 shows an average tip temperature Tfin tip, avg = 229.9119◦ F. A comparison between the solutions gives a percentage difference equal to ±0.00126 percent.
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SteadyState HeatTransfer Problem: Cooling Fin SteadyState HeatTransfer Problem: Cooling Fin
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TABLE 5.2 Fin External Surface Temperatures
Base
Tip
x, in.
T(x),◦ F
0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.50 6.00
250.0000 246.7976 243.8856 241.2594 238.9144 236.8468 235.0532 233.5305 232.2763 231.2885 230.5654 230.1058 229.9090
Applying the Principles With a high degree of confidence in defining and solving a simplified version of the transient heattransfer problem, the final step is to apply the principles. Because of the complexity of the geometry, FEA methods are better suited for addressing solutions for the actual three fin design concepts shown in Fig. 5.1.
What Is FiniteElement Analysis, and How Does It Work? Finiteelement analysis methods were first introduced in 1943. Finiteelement analysis uses the Ritz method of numerical analysis and 260
Temp. T, °F
250
240 230
220 0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
Length x, in. Figure 5.3
Fin external surface temperatures plot.
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Summary of results of FEA solution: steadystate fin temperatures.
230.0
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Base Temperature Tb = 250°F
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Z
Centerline for Axisymmetric Elements
Ambient Air Temperature = 72°F
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Aluminum Material Properties
Fin Tip Temperature = 230°F (229.9119°F Avg)
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SteadyState HeatTransfer Problem: Cooling Fin
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SteadyState HeatTransfer Problem: Cooling Fin SteadyState HeatTransfer Problem: Cooling Fin
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minimization of variational calculus to obtain approximate solutions to systems. By the early 1970s, FEA was limited to highend companies involved in the aerospace, automotive, and power plant industries that could afford the cost of having mainframe computers. After the introduction of desktop personal computers, FEA was soon made available to the average user. FEA can be used to investigate new product designs and improve or refine existing designs. The FEA approach is a mathematical idealization method that approximates physical systems by idealizing the geometry of a design using a system of grid points called nodes. These node points are connected together to create a “finite” number of regions or mesh. The regions that define the mesh are assigned properties such as the type of material and the type of mesh used. With the mesh properties assigned, the mesh regions are now called “elements.” The types of mesh can be defined as rod elements, beam elements, plate elements, solid elements, and so on. Constraints or reactions are entered to simulate how the idealized geometry will react to the defined loading. Analysis models can represent both linear and nonlinear systems. Linear models use constant parameters and assume that the material will return to its original shape after the loading is removed. Nonlinear models consist of stressing the material past its elastic capabilities, and permanent deformations will remain after the loading is removed. The definitions for the materials in these nonlinear models can be complex. For heat transfer, FEA models can be idealized for both steadystate and transient solutions. The mesh elements are defined using the material conductivity or thermal dynamics to represent the system being analyzed. Steadystate transfer refers to materials with constant thermoproperties and where heat diffusion can be represented by linear equations. Thermal loads for solutions to heattransfer analysis include temperatures, internal heat generation, and convection.
FEA Results: Comparison of Fin Design Concepts With a high degree of confidence in defining and solving the sample heattransfer problem, the final step is to apply the principles. Figure 5.5 shows summary results when the FEA model idealizations of fin designs shown in Fig. 5.1 are analyzed using MSC/Nastran. In Fig. 5.5 magnetic coils for each fin design concept are shown generating internal heat at the rate of 16 Btu/h per coil. Table 5.3 lists a comparison of average coil temperatures for each fin design concept. Table 5.3 shows dualfin concept 1, which produces the largest estimate of a 22 percent reduction in coil temperatures.
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(b)
(c)
Figure 5.5
72.0
77.66
83.32
88.98
94.64
June 20, 2005
Summary of results of FEA solution: fin design concepts. (a) Original single fin; (b) dualfin concept 1; (c) dualfin concept 2.
(a)
Avg Coil Temperature 132°F
Avg Coil Temperature 127°F
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SteadyState HeatTransfer Problem: Cooling Fin
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SteadyState HeatTransfer Problem: Cooling Fin SteadyState HeatTransfer Problem: Cooling Fin
TABLE 5.3
5.11
Summary of Average Coil Temperatures
Average coil temperature Percent reduction
Original singlefin design
Dualfin concept 1
Dualfin concept 2
162◦ F N/A
127◦ F −22%
132◦ F −19%
Dualfin concept 2 comes in at a close second at 19 percent reduction in coil temperatures. As a result of this steadystate heattransfer analysis, an approximate 20 percent reduction in coil operating temperatures could significantly extend the operational life of the magnetic coils. Therefore it is recommended that both dualfin concepts be evaluated by performing actual physical testing before implementation of any production changes. References 1. E. R. G. Eckert and Robert M. Drake, Jr., Heat and Mass Transfer, 2d ed., McGrawHill, 1959. 2. J. P. Holman, “ConductionConvection Systems,” in Heat Transfer, 5th ed., McGrawHill, 1981, chap. 29. 3. W. M. Rohsenow and J. P. Hartnett, Handbook of Heat Transfer, McGrawHill, 1973. 4. Donald R. Pitts and Leighton E. Sissom, Heat Transfer, 2d ed., Schaum’s Outline Series, McGrawHill, 1977.
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12
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Source: HeatTransfer Calculations
Chapter
6 Cooling of a Fuel Cell
Jason M. Keith Department of Chemical Engineering Michigan Technological University Houghton, Michigan
Introduction This problem applies fundamentals of heat transfer to the cooling of a fuel cell, which may have applications in laptop computers, cell phones, and the motor vehicles of the future. In this problem you will predict the electrical current generated within the fuel cell for a certain hydrogen mass flow rate, estimate the efficiency of the fuel cell in generating electricity, estimate the heat loss, and design a cooling system for the fuel cell.
Background: The ProtonExchange Membrane Fuel Cell Each cell within a fuel cell is basically an electrochemical device which converts hydrogen and oxygen gas into water and directcurrent electricity. The byproduct of the reaction is heat, which must be removed to prevent damage to the fuel cell components. A rough schematic of a single cell within a fuel cell is shown in Fig. 6.1. Hydrogen gas flows within channels etched in the bipolar plate (slanted lines in Fig. 6.1). The gas is transferred by convection to the surface of the gas diffusion layer (horizontal lines in Fig. 6.1), which it penetrates to reach the anode (area to left of center with vertical lines in Fig. 6.1). At the interface between the anode and electrolyte is a platinum catalyst, which oxidizes
6.1
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Cooling of a Fuel Cell 6.2
SteadyState Calculations
Figure 6.1 Schematic of one cell of a protonexchange membrane fuel cell. The slanted lines represent the bipolar plates; the horizontal lines, the gas diffusion layer; the vertical lines, the electrodes (left block is the anode; right block is the cathode); and the open block, the electrolyte.
the hydrogen into electrons and protons: 2 H 2 → 4 H+ + 4 e −
(6.1)
The protons are relatively small and can pass through the electrolyte membrane (open block in center of Fig. 6.1). The electrons are relatively large and cannot pass through the electrolyte. They must pass by electrical conduction back through the anode, gas diffusion layer, and bipolar plate, through an electrical load (such as a lightbulb), and around to the other side of the fuel cell. The electrons then conduct through the bipolar plate, gas diffusion layer, and anode, where they recombine with the protons in the presence of oxygen, in a reduction reaction to form water: O2 + 4 H+ + 4 e− → 2 H2 O
(6.2)
The oxygen gets to the cathode via the same convectiondiffusion mechanism that the hydrogen uses to reach the anode. The overall reaction is equal to the sum of the anode and cathode reactions: 2 H2 + O2 → 2 H2 O
(6.3)
The directcurrent electricity produced by the fuel cell will have a current I and a voltage Ecell . It is customary to “stack” n cells together to make a much larger system, called a “fuel cell stack,” with stack voltage E = nEcell and current I. A cartoon of a fuel cell stack is seen in Fig. 6.2. Note that the current is the same regardless of the number of cells in the stack. Additional information on the fuel cell chemistry and construction is available in the textbook by Larminie and Dicks (2003). To predict the heat generation, we need to know the efficiency of the fuel cell. Let us define efficiency as the rate of electrical work within
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Cooling of a Fuel Cell Cooling of a Fuel Cell
6.3
Cathode, Electrolyte, Anode (all very thin)
Conducting Bipolar Plate Figure 6.2
Side view of a fuel cell stack.
the fuel cell divided by the rate of thermal energy generation if the fuel were burned by combustion, such that =
P Ccombustion
(6.4)
The rate of electrical work is also called power and is given as the product of the stack voltage E and current I within the fuel cell, P = IE = InEcell
(6.5)
and the rate of thermal energy generation due to combustion is given as Ccombustion = nMfuel H
(6.6)
where Mfuel is the molar flow rate of hydrogen per cell within the stack and H is the heat of formation, given as H = −241.8 kJ/mol if the water product is steam (lower heating value) or H = −285.8 kJ/mol if the water product is liquid (higher heating value). Fuel cells for commercial and/or automotive uses will often have a condenser to produce liquid product, so the higher heating value will be used here. The electrical current I is also related to the hydrogen molar flow rate Mfuel . Equation (6.1) shows that for each mole of H2 reacted at the anode, two moles of electrons are generated. The charge of a mole of electrons, in units of coulombs per mole, is given as Faraday’s constant F = 96,485 C/mol. Thus, a molar flow rate Mfuel to each cell gives an electrical current of I=
Mfuel mol fuel 2 mol e− FC = 2FMfuel s mol fuel mol e−
A
(6.7)
since one coulomb per second is equal to one ampere of current. Substituting Eq. (6.7) into Eq. (6.5) and dividing by Eq. (6.6) gives the efficiency , =
2FEcell Ecell = H 1.48
(6.8)
if the higher heating value of hydrogen is used.
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Cooling of a Fuel Cell 6.4
SteadyState Calculations
Energy that is not converted into electrical power is released as heat. A simple energy balance on the process gives Ccombustion = P + Q
(6.9)
Combining Eqs. (6.4) and (6.9) to eliminate Ccombustion gives the heat generation within the fuel cell: 1 Q= P −1 (6.10) We will now estimate the electrical energy generation and heat generation in a fuel cell stack. Example calculation: efficiency, fuel usage, and heat generation
Consider a fuel cell stack with five cells of crosssectional area Acell = 10 cm2 . Such a small fuel cell may be used in a cell phone or a laptop computer. Assume that the cell voltage Ecell is related to the current density i = I/Acell in units of mA/cm2 according to the relationship Ecell = Eoc − A ln(i) − ir − m exp(ni)
(6.11)
where, for a certain type of protonexchange membrane fuel cell, the parameters in this equation are given by Laurencelle et al. (2001) as Eoc = 1.031 V, A = 0.03 V, r = 2.45 × 10−4 k cm2 , m = 2.11 × 10−5 V, and n = 8 × 10−3 cm2 /mA. The first term estimates the voltage at zero current density, also called the opencircuit voltage. The second term represents activation losses, the third term represents ohmic losses, and the final term represents masstransfer losses. For additional information, consult the text by Larminie and Dicks (2003). The components of Eq. (6.11) are plotted in Fig. 6.3. It is customary to operate a fuel cell within the “ohmic” region of the graph; thus for a current density of 500 mA/cm2 , the cell voltage Ecell = 0.72 V. The stack voltage is E = nEcell = 5(0.72) = 3.6 V, and the current is given by iAcell = 500 mA/cm2 (10 cm2 )(1 A/1000 mA) = 5 A. The total power is equal to the stack voltage multiplied by the current, P = IE = (5 A)(3.6 V) = 18 W. The molar hydrogen flow rate per cell required to produce this power can be calculated by rearranging Eq. (6.7) in terms of Mfuel : Mfuel =
I 2F
(6.12)
Thus, Mfuel = 5 A/2/(96,485 C/mol) = 2.6 × 10−5 mol/s. The total hydrogen molar flow rate can be calculated by multiplying this value by the number of cells and is equal to 1.3 × 10−4 mol/s. Noting that as the
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Cooling of a Fuel Cell Cooling of a Fuel Cell
6.5
Figure 6.3 Polarization plot for protonexchange mem
brane fuel cell.
molecular weight of hydrogen is 2 g/mol, it would take about one hour to use a gram of hydrogen in this fuel cell. Using Eq. (6.8), the efficiency of the fuel cell is = 0.72/1.48 = 0.487. Thus, the heat generation can be calculated from Eq. (6.10) as Q = 18 W (1/0.487 − 1) = 19.0 W. The heat generation per cell is thus Qcell = 19.0/5 = 3.8 W. This heat will have to be removed by cooling air to maintain the fuel cell at a steadystate temperature, which we will now discuss. Convective Cooling Consider the fuel cell stack shown in Fig. 6.2. Larminie and Dicks (2003) state that fuel cells with P > 100 W require air cooling. To avoid drying out the membrane, separate reactant air and cooling air are required. This is achieved by placing cooling channels within the bipolar plate. A schematic of one cell with such a cooling channel is shown in Fig. 6.4. We assign the following variables, from the perspective taken from a side view of the fuel cell: L is the height of the plate and channel; d is the depth of the plate and channel (so the product Ld is the area Acell ), t is the thickness of the bipolar plate, cathode, electrolyte, and anode; and w is the width of the cooling channel. Heat generation within the fuel cell occurs near the cathode, electrolyte, and anode (left and right sides of the schematic in Fig. 6.4). In essence, a total heat generation rate of Qcell /2 W can be considered to be input at the left and right sides of the cell, as shown in the left portion of Fig. 6.5. Since the cathode, electrolyte, anode, and bipolar plate are made of materials with high thermal conductivity (about 20 W/mK),
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Cooling of a Fuel Cell 6.6
SteadyState Calculations
TOP VIEW L d t
w
Half of cathode, electrolyte & anode
Half of cathode, electrolyte & anode
Conducting bipolar plate Cooling channel SIDE VIEW Figure 6.4
Single cell with cooling channel.
the heat source can be considered to be input directly into the cooling channel, as seen in the right portion of Fig. 6.5. Thus, it is appropriate to model this system as heat transfer to a channel with a constant wall flux at the boundaries (due to a uniform heat generation rate per unit area of the channel surface). An energy balance over the fluid in the channel yields m ˙ coolant C p(Tcoolant,out − Tcoolant,in ) = Qcell
(6.13)
which can be used to estimate the exit temperature of the cooling air. The maximum temperature of the channel wall can be estimated using a heattransfer coefficient. For channels with a constant heat flux at the boundary, fully developed laminar flow, and a very high channel
Figure 6.5 Heat generation within a single cell of a fuel cell stack. The picture on the left shows the heat source within the cathode, electrolyte, and anode of the fuel cell. The picture on the right shows an approximation where the heat is input directly into the cooling channel.
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Cooling of a Fuel Cell Cooling of a Fuel Cell
6.7
aspect ratio, Incropera and DeWitt (1996) report the following Nusselt number: Nu =
hDh = 8.23 kgas
(6.14)
where Dh is the hydraulic diameter, given as Dh =
4Across section Pchannel
(6.15)
where Across section and Pchannel are the crosssectional area and the perimeter of the cooling channel, respectively. For a channel aspect ratio of 1, 2, 4, and 8, the Nusselt number is equal to 3.61, 4.12, 5.33, and 6.49. The correlation for the Nusselt number is valid when the Reynolds number is less than 2300: Re =
4m ˙ coolant gas Pchannel
(6.16)
where gas is the gas viscosity. As the coolant heats up as it travels along the channel, the location of maximum solid surface temperature will be at the exit. An energy balance yields that the temperature difference between the solid and gas is a constant, due to the fact that a uniform heat flux is present: h(Tsurface − Tgas ) =
Qcell LPchannel
(6.17)
Finally, an energy balance within the bipolar plate, cathode, and anode yields the relationship between the surface temperature and that at the cell edge where the heat is supplied ksolid
Tedge − Tsurface Qcell = t LPchannel
(6.18)
Again, this temperature difference is a constant along the entire channel length because of the constantheatflux approximation. Example calculation: cooling the fuel cell
We will apply the principles in the preceding section to remove the 19 W of heat generated within the fuel cell. In general, fuel cells with power output P < 100 W can be cooled by natural convection. As this fuel cell may be placed inside a cell phone or a laptop computer, convective cooling may be justified. The procedure illustrated here may also be extended to a larger stack. Protonexchange membranes are typically operated about 50◦ C. Temperatures above 100◦ C can damage the fuel cell components. If we wish
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Cooling of a Fuel Cell 6.8
SteadyState Calculations
TABLE 6.1
Dimensions and Thermal
Properties Item
Value
L d t w ksolid kgas gas Cp
3.16 cm = 0.0316 m 3.16 cm = 0.0316 m 0.5 cm = 0.005 m 0.1 cm = 0.001 m 20 W/mK 0.0263 W/mK 1.84 × 10−4 g/cms 1.0 J/gK
to maintain the maximum temperature within the fuel cell at 60◦ C, for example, with cooling air at 20◦ C, what air mass flow rate is required? Thermal properties and dimensions of the fuel cell are given along with air thermal properties in Table 6.1. This problem is solved in a procedure reverse as that derived in the previous section. The channel perimeter Pchannel is equal to 2d + 2w = 0.0652 m. Equation (6.18) is then used to estimate the solid surface temperature at the cooling channel exit: Tsurface = Tedge − = 60◦ C −
t Qcell LPchannel ksolid 3.8 W 1 0.005 m 1◦ C = 59.5◦ C 0.0316 m 0.0652 m 20 W/m − K 1 K (6.19)
The aspect ratio of the channel is d/w = 31.6, so it is safe to use the infinite aspect ratio value for the Nusselt number Nu = 8.23. The hydraulic diameter is given by Eq. (6.15): 4Across section 4dw = Pchannel 2(d + w) 4 0.0316 m = 0.001 m = 0.0019 m 2 0.0316 m + 0.001 m
Dh =
(6.20)
which is close to 2w as d w. Equation (6.14) can then be used to calculate the heattransfer coefficient: h = 8.23
kgas = 8.23 Dh
0.0263 W/mK 0.0019 m
= 111 W/m2 K
(6.21)
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Cooling of a Fuel Cell Cooling of a Fuel Cell
6.9
The heattransfer coefficient can now be used to determine the gas exit temperature using Eq. (6.17): Qcell hLPchannel ◦ 1 1 C 3.8 W = 42.8◦ C = 59.5◦ C − 2 111 W/m − K 0.0316 m 0.0652 m K (6.22)
Tgas = Tsurface −
Finally, the air mass flow rate can be determined from Eq. (6.13), by setting Tcoolant,out = Tgas . The resulting expression is Qcell C p(Tcoolant,out − Tcoolant,in ) 3.8 W 1 = = 0.166 g/s 1.0 J/gK 42.8 − 20◦ C
m ˙ coolant =
(6.23)
To prove that this is laminar flow, the Reynolds number must be calculated from Eq. (6.16): Re =
4m ˙ coolant m 0.166 g/s 4 = 550 = −4 gas Pchannel 1.84 × 10 g/cms 0.0652 m 100 cm (6.24)
The total mass flow rate of coolant is equal to the value for one cell multiplied by the number of cells, so m ˙ coolant,total = 0.83 g/s. Discussion and Conclusions Analysis of the sample system presented above shows that the maximum cooling flow rate to maintain laminar flow within the cooling channel is 0.689 g/s per cell or 3.44 g/s total. In this circumstance, the maximum temperature within the fuel cell is Tedge = 42.5◦ C, the maximum surface temperature is Tsurface = 42.0◦ C, and the exit coolant temperature is Tcoolant,out = 25.5◦ C. The minimum cooling flow rate can also be determined by selecting a maximum value for Tedge , say, 90◦ C. The flow rate is 0.072 g/s per cell or 0.36 g/s total. The maximum surface temperature is Tsurface = 89.5◦ C, and the exit coolant temperature is Tcoolant,out = 73.0◦ C. The Reynolds number in this case is 240. In practice, the power required to operate electronic equipment by the fuel cell will vary. During high loads, the fuel cell temperature will likely increase. Using a maximum possible temperature of 90◦ C and a Qcell value of 5.7 W (representing a 50 percent increase over the base case considered here) yields a cooling mass flow rate of 0.128 g/s per cell or 0.64 g/s total.
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Cooling of a Fuel Cell 6.10
SteadyState Calculations
This calculation applies convection and conduction heat transfer to the design of a fuel cell cooling system. The situation studied here can be considered a worstcase scenario as the cooling fluid must remove all the heat generated within the fuel cell. In practical applications, there are additional heat losses to the environment. The total heat generated per cell Qcell can be multiplied by a percentage to account for these effects. References Incropera, F., and DeWitt, D. Fundamentals of Heat and Mass Transfer, 4th ed. Wiley, New York (1996). Larminie, J., and Dicks, A. Fuel Cell Systems Explained, 2d ed. Wiley, Hoboken, N.J. (2003). Laurencelle, F., Chahine, R., Hamelin, J., Fournier, M., Bose, T. K., and Lapperriere, A. “Characterization of a Ballard MK5E Proton Exchange Membrane Stack,” Fuel Cells 1(1), 66–71 (2001).
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Source: HeatTransfer Calculations
Chapter
7 Turbogenerator Rotor Cooling Calculation
James T. McLeskey, Jr. Department of Mechanical Engineering Virginia Commonwealth University Richmond, Virginia
Problem Prior to 1960, large turbogenerator rotors were cooled indirectly and limited in output to approximately 200 MW. Heat generated in the windings was conducted to the surface of the rotor and then removed from the outer diameter by convection to the hydrogen gas. As the capability increased in the late 1950s, it became necessary to find more efficient methods of cooling the rotors. Manufacturers developed cooling systems where the cooling gas was directly in contact with the copper conductors. These directcooled rotors helped make possible the construction of turbogenerator units with capacities as high as 1300 MW. In spite of the improved cooling techniques, rotor heating can still be a limiting factor under lagging power factor conditions. Determining the rotor temperature profile is necessary when designing a new machine. In addition, with improvements in steam turbine design, it is now possible to retrofit older steam turbines and provide power increases of as much as 10 percent over the original rating. It is often necessary to determine the ability of the generator rotor to tolerate this increase in power. In order to better quantify these limits, it is necessary to calculate the cooling gas flow and temperature profile in the generator rotor. The
7.1
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Turbogenerator Rotor Cooling Calculation 7.2
SteadyState Calculations
Gas Outlet at Ro
Retaining Ring
Gas Inlet at Ri
Active Length L Side view of a typical turbogenerator rotor showing cooling gas inlet and outlet regions.
Figure 7.1
following calculation is for a directcooled turbogenerator rotor with axial flow passages in the windings. The dimensions and configuration are not for any particular machine, but are typical. The principles can be applied to other designs with appropriate adjustments. Assumptions 1. All cooling is due to direct cooling of hydrogen in contact with the copper. 2. Radial and axial heat transfer within the copper is ignored.
Cooling principle
In a large turbogenerator, the rotor provides the rotating magnetic field. In order to generate this field, current passes through the windings of the rotor. Naturally, these windings can become quite hot. To prevent damage to the electrical insulation which surrounds the windings, it is necessary to cool the windings. Cooling the winding is often done as shown in Figs. 7.1 to 7.4. Cold gas enters under the retaining ring, travels axially through the copper windings, and exits at the center of the rotor. The pressure difference driving the gas results from the different radii of the inlet and exit points. The inlet radius is less than the exit radius, so the inlet is traveling at a lower linear velocity and, in accordance with Bernoulli’s principle, is at a higher pressure. Calculation Method The calculation consists of five primary steps: 1. Determine the cooling gas properties. 2. Determine the volume flow rate of the cooling gas through the rotor.
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Turbogenerator Rotor Cooling Calculation Turbogenerator Rotor Cooling Calculation
7.3
Rotor Slots (filled with windings)
Retaining Ring
Gas Inlet at Ri
Gas Inlet into Side of Winding
Copper Windings Figure 7.2 Cutaway view of generator winding head region (under the retaining ring) showing the gas inlet and filled rotor slots.
3. Determine the temperature rise in the gas passing through the rotor. 4. Determine the temperature difference between the cooling gas and the copper windings. 5. Determine the absolute temperature of the copper.
Determine the gas properties
Modern turbogenerators are capable of achieving nearly 300 MW of output using air as the cooling medium. However, all units larger than that and most older units greater than 80 MW use hydrogen as the cooling medium because of its low density and viscosity and relatively high heat capacity. Our calculation will assume that hydrogen is used. The first step in calculating the cooling is to determine the gas properties based on the estimated operating conditions. In a computer model, this step might be repeated iteratively after the rest of the calculation is performed, but a generator designer can usually make a good first approximation based on the typical data.
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Turbogenerator Rotor Cooling Calculation 7.4
SteadyState Calculations
Cooling gas outlets from the windings
Cooling gas inlets into windings
L/2
Schematic of a single axially directcooled generator rotor winding turn showing gas inlet and outlet holes.
Figure 7.3
Cooling holes d Copper turn
hturn
Figure 7.4 Crosssectional view of a generator rotor winding slot showing the axial cooling holes in the directcooled rotor windings. This slot contains eight turns.
wturn
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Turbogenerator Rotor Cooling Calculation Turbogenerator Rotor Cooling Calculation
TABLE 7.1
7.5
Operating Conditions
Parameter
Variable
Value
Comment
Cooling medium H2 pressure
— P
Coldgas temperature
Tcold
Hydrogen 75 psig = 517 × 103 Pa 45◦ C = 318 K
H2 purity
H2, pure
98% = 0.98
Estimated maximum temperature rise of rotor Maximum temperature of H2 Estimated average H2 temperature
T max
75◦ C
This is the estimated temperature of the cold gas entering the rotor; it is based on the stator cooler design The hydrogen in the machine will not be 100% pure Limited by insulation
TH2 ,max
120
Tcold + T max
Tavg gas
82.5◦ C = 356 K
(Tcold + TH2 ,max )/2
Gather the operating conditions.
— —
See Table 7.1.
Determine the gas properties at Tavg gas . It is necessary to find the prop
erties for both the hydrogen and the air in the system and then use the hydrogen purity to find a weighted average. Density. Using the idealgas law, we obtain H2 =
P RH2 Tavg gas
where
RH2 = 4.124 × 103 J/kg K
(7.1)
air =
P Rair Tavg gas
where
Rair = 0.287 × 103 J/kg K
(7.2)
for the given conditions, H2 = 0.352 kg/m3 and air = 5.06 kg/m3 . The weighted average is = H2 · H2, pure + air · (1 − H2, pure ) = 0.446 kg/m3
(7.3)
Absolute viscosity
H2 = 1.053 × 10−5 Pas air = 2.155 × 10−5 Pas = H2 · H2, pure + air · (1 − H2, pure ) = 1.075 × 10−5 kg m−1 s−1
(7.4)
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Turbogenerator Rotor Cooling Calculation 7.6
SteadyState Calculations
Kinematic viscosity
k =
= 2.409 × 10−5 m2 s−1
(7.5)
Specific heat
Cp = 10,000 Ws/kgK Thermal conductivity
= 0.2 W/mK Determine the volume flow rate of the cooling gas through the rotor
The primary cooling mode inside a directcooled rotor is forced convection of the heat from the copper to the cooling gas. Therefore, it is necessary to determine the volume flow rate and the velocity of the gas through the copper windings. The pressure driving the flow is generated as a result of the difference in the diameters of the gas inlet and outlet. See Table 7.2. The selfgenerated pressure and the volume flow rate depend on the geometric configuration of the rotor. Although some approximations can be made when the exact dimensions are unknown, more fully defining the problem will lead to more accurate results.
Gather the geometric data.
Because the hydrogen inlet and outlet are at different diameters, the linear velocities are different. The inlet is at a smaller diameter and is therefore traveling at a lower linear velocity and, in accordance with Bernoulli’s principle, is at a higher pressure. This forces the cooling gas through the rotor:
Find the selfgenerated pressure.
Pself =
(2 × rpm/60 s)2 (Ro2 − Ri2 ) 2
= 2.854 × 103 Pa
(7.6)
where rpm is the rotation speed in revolutions per minute (r/min). In some machines, additional pressure for moving the gas through the rotor is provided by external fans.
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Turbogenerator Rotor Cooling Calculation Turbogenerator Rotor Cooling Calculation
TABLE 7.2
7.7
Geometric Dimensions (See Figs. 7.1 to 7.5)
Parameter
Variable
Value
Comment
Rotor body radius H2 inlet radius
Ro Ri
0.5 m 0.4 m
Active length
L
6.0 m
Rotation speed Number of slots Number of Cu turns per slot (layers) Holes per turn Holes per slot Turn thickness Turn width Copper resistivity Cooling passage diameter Excitation current
rpm Nslots Turns
3600 min−1 20 8
Outer radius of rotor body Radius under retaining ring where H2 enters the rotor Length of iron through which the windings pass — — —
Holeturn Holeslot hturn wturn RCu d
2 16 0.0148 m 0.030 m 0.01675 × 10−6 m 8 mm
— Turns · Holeturn — — — —
I
2000 A
—
Determine gas velocity through the windings. The gas enters the rotor
under the retaining rings at each end and moves to the center where it exits. This means that all the calculations are performed for half of the active length of the rotor as shown in Fig. 7.5. The velocity depends on the pressure drop through the rotor. The pressure drop depends on the friction and the geometry: P = Pfriction + Pgeometry
(7.7)
The pressure drop due to friction can be found from Pfriction = f
L/2 v2 d 2
(7.8)
where f = 0.316/Re1/4 = 0.316/(vd/ k)1/4 for smooth copper profiles and turbulent flow. The term L/2 is used because the gas enters at one
Outlet Inlet
d
Velocity L/2 Figure 7.5 Schematic cross section of the side of a single cooling hole in the copper winding.
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Turbogenerator Rotor Cooling Calculation 7.8
SteadyState Calculations
end and exits at the center of the rotor. The dynamic pressure is given by v2 /2. The pressure drop due to geometry is given by Pgeometry = c
v2 2
(7.9)
where c is a geometric coefficient, typically between 5 and 15 depending on the rotor. If we define 1/4 k L/2 a = 0.316 d d = 27.76 (m/s)0.25
(7.10)
and b=2
P
with
P = Pself
= 12,789 Nm/kg
(7.11)
then we can write the equation for the pressure loss as b c = v1.75 + · v2 a a
(7.12)
If we determine a and b using the values given above and assume a value for c (c = 10), the roots of the above equation can be found to give the velocity. Using MathCad to find the roots, the value obtained was v = 23.81 m/s in each cooling hole. The volume flow rate in a single cooling hole can be found by multiplying this velocity by the area of each hole: Vhole = area × velocity =
d 2 ·v 4
= 1.196 × 10−3 m3/s
(7.13)
Then the total volume flow rate in a single slot can be determined by multiplying this value by the number of holes in a single slot: Vslot = Vhole · Holeslot = 19.1 × 10−3 m3/s
(7.14)
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Turbogenerator Rotor Cooling Calculation Turbogenerator Rotor Cooling Calculation
7.9
Determine the temperature rise in the gas passing through the rotor
The rotor is assumed to be operating under steadystate conditions. Therefore, the hydrogen gas must remove any heat generated by the windings. The first step is to determine the amount of heat produced in the winding in each slot (actually in each halfslot since the cooling gas exits at the center). This is due to I 2 R losses from the excitation current. The heat is found by first calculating the crosssectional area of one turn of copper (Fig. 7.4) and multiplying by the resistivity to determine the total resistance. This value is then multiplied by the square of the excitation current. This, however, requires an estimate of the copper temperature rise (assumed here as a first guess to be T = 75◦ C). The crosssectional area of one turn is ACu = (hturn × wturn ) −
d 2 × Holeturn 4
= 343.5 × 10−6 m3
(7.15)
The total resistance of windings in one halfslot is Rslot = RCu (1 + 0.004 · T) · turns · = 1.52 × 10−3
L/2 ACu (7.16)
Power (heat) produced in one halfslot is Pslot = I 2 Rslot = 6085 W
(7.17)
This power (heat) must be removed by the hydrogen. To find the temperature rise of the hydrogen, divide this power by the total heat capacity of the hydrogen flowing through the copper: TH2 =
Pslot Cp Vslot
= 71.25◦ C
(7.18)
The temperature profile of the hydrogen gas is shown in Fig. 7.6. It enters from the end at the cold gas temperature of 45◦ C and reaches a maximum of 116.25◦ C at the center of the rotor.
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Turbogenerator Rotor Cooling Calculation 7.10
SteadyState Calculations
140 120
Temperature (°C)
100 80
Cu temperature H2 temperature
60 40 20 0 0
1
2
3
4
L/2 Axial distance from end of rotor active length Temperature profiles of H2 gas and copper for an axially directcooled generator rotor.
Figure 7.6
Determine the temperature difference between the cooling gas and the copper windings
Under steadystate conditions, all the heat generated in the copper is assumed to be transferred to the hydrogen. However, because there is some thermal resistance between the copper and the hydrogen, the copper is at a higher temperature. This temperature must be determined by finding the heattransfer coefficient between the hydrogen and the copper. Reynolds number of flow in holes
vd = 7907 k
(7.19)
k Cp = 0.538
(7.20)
Re = Prandtl number of flow
Pr =
Heattransfer coefficient for flow in smooth copper pipe
h = 0.023 ·
· Re0.8 · Pr0.4 d
= 589.2 W/m2 ◦ C
(7.21)
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Turbogenerator Rotor Cooling Calculation Turbogenerator Rotor Cooling Calculation
7.11
The temperature difference between the copper and the hydrogen is TCuH2 =
Pslot h · Areasurf
(7.22)
where Areasurf equals surface area of the copper in contact with the hydrogen (the perimeter of the hole times the length of the copper): Areasurf = d ·
L · Holeslot = 1.206 m2 2
TCuH2 = 8.57◦ C
(7.23)
The temperature profile of the copper is shown in Fig. 7.6. The copper temperature remains 8.6◦ C above the hydrogen temperature across the length of the rotor and reaches its maximum at the center of the rotor. Determine maximum temperature of copper
The maximum copper temperature occurs at the center of the rotor for a rotor with this type of cooling. It can be found from TCu = Tcold + TH2 + TCuH2 = 45 + 71.2 + 8.6 = 124.8◦ C
maximum value
(7.24)
In order to protect the insulation, the maximum desired temperature is typically 130◦ C, so this rotor design would be acceptable. Summary and Conclusions The calculation presented here provides a method for determining the temperature profile in a large turbogenerator rotor with direct axial cooling using hydrogen. For the particular parameters chosen, the hydrogen gas reaches a maximum of approximately 116◦ C. This results in a maximum copper winding temperature of approximately 125◦ C. If this were an actual rotor, the maximum temperature would be within the typically quoted class B temperature limit of 130◦ C. The maximum copper temperature occurs at the center of the rotor for this design. A wide variety of directcooled rotor designs are produced by various manufacturers. The principles outlined in this calculation can be applied to other designs with the appropriate adjustments for such things as different cooling flow patterns (Costley and McLeskey, 2003) or substantial indirect cooling (McLeskey et al., 1995). This analysis of the thermal limits of the rotor will allow for maximum utilization under all generator operating conditions.
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Turbogenerator Rotor Cooling Calculation 7.12
SteadyState Calculations
References Costley, J. M., and J. T. McLeskey, Jr., “DiagonalFlow, GapPickup Generator Rotor Heat Transfer Model,” Proceedings of IMECE’03: 2003 ASME International Mechanical Engineering Congress & Exposition, Washington, D.C., vol. 374, no. 1, pp. 389–392, Nov. 16–21, 2003. Incropera, F. P., and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 5th ed., Wiley, New York, 2002. McLeskey, J. T., Jr., A. T. Laforet, D. R. Snow, and J. V. Matthews III, “Generator Rotor Replacement by a Third Party Manufacturer,” Conference Papers of PowerGen Americas ’95, Anaheim, Calif., vol. 4, pp. 391–401, 1995. Munson, B. R., D. F. Young, and T. H. Okiishi, Fundamentals of Fluid Mechanics, 4th ed., Wiley, New York, 2002.
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Source: HeatTransfer Calculations
Chapter
8 Heat Transfer through a DoubleGlazed Window
P. H. Oosthuizen Department of Mechanical Engineering Queen’s University Kingston, Ontario, Canada
David Naylor Department of Mechanical and Industrial Engineering Ryerson University Toronto, Ontario, Canada
Problem Statement Consider heat transfer through a 1mhigh doublepaned window under nighttime conditions, when the effects of solar radiation can be ignored. The heat transfer can be assumed steady and onedimensional, and the effects of the window frame can be ignored. The window glass can be assumed to be 3 mm thick and to have a thermal conductivity of 1.3 W/m◦ C. The gap between the panes is 10 mm. The air temperature inside the building can be assumed to be 20◦ C and the outside air temperature, 5◦ C. On the basis of available results, the heattransfer coefficients on the inner and outer surfaces of the window can be assumed to be 4 and 25 W/m2◦ C, respectively, and the emissivity of the glass can be assumed to be 0.84. Find the heattransfer rate through the window per unit area for the case where the gas between the panes is air and for the case where it is argon. Also express these results in terms of an overall heattransfer coefficient and in terms of the overall heattransfer resistance.
8.1
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Heat Transfer through a DoubleGlazed Window 8.2
SteadyState Calculations
Convection
Convection
Radiation
Radiation Indoors: T = Ti
Outdoors: T = To (<Ti)
Convection and Radiation Conduction
Conduction
Heattransfer situation under consideration.
Figure 8.1
Solution The situation considered here is shown in Fig. 8.1. The following nomenclature is used here: A
Frontal area of window
AR
Aspect ratio; H/w
H
Height of window
hcb
Convective heattransfer coefficient for flow between window panes
hci
Convective heattransfer coefficient for inner surface of window
hco
Convective heattransfer coefficient for outer surface of window
hr
Radiation heattransfer coefficient
hrb
Radiation heattransfer coefficient for flow between window panes
hri
Radiation heattransfer coefficient for inner surface of window
hro
Radiation heattransfer coefficient for outer surface of window
k
Thermal conductivity
Nu
Nusselt number based on w
Nua
First Nusselt number value given by betweenglass convection equations
Nub
Second Nusselt number value given by betweenglass convection equations
Nuc
Third Nusselt number value given by betweenglass convection equations
Q
Heattransfer rate through window
Qb
Heattransfer rate through gap between window panes
Qi
Heattransfer rate to inner surface of window
Qgi
Heattransfer rate through inner pane
Qgo
Heattransfer rate through outer pane
Qo
Heattransfer rate from outer surface of window
Ra
Rayleigh number based on w and T2 − T3
R
Overall heattransfer resistance
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Heat Transfer through a DoubleGlazed Window Heat Transfer through a DoubleGlazed Window
Ri
Inside heattransfer resistance
Ro
Outside heattransfer resistance
Rg
Heattransfer resistance of glass window pane
Rb
Heattransfer resistance of panetopane gap
Ti
Inside air temperature
To
Outside air temperature
T1
Temperature of inside surface of window
T2
Temperature of outer surface of inner window pane
T3
Temperature of inner surface of outer window pane
T4
Temperature of outside surface of window
w
Distance between the two panes of glass
Bulk expansion coefficient
Kinematic viscosity of betweenpanes gas
8.3
Because the effects of solar radiation are ignored and because the glass is essentially opaque to the infrared radiation therefore involved in the situation considered, the glass panes will be assumed to behave as conventional opaque gray bodies. The radiant heat transfer will be treated by introducing radiant heattransfer coefficients hr . Because it is assumed that the heat transfer is steady, it follows that the rate of heat transfer from the inside air to the inner surface of the window is equal to the rate of heat transfer through the inner glass pane, which in turn is equal to the rate of heat transfer across the gap between the panes, which in turn is equal to the rate of heat transfer across the outer glass pane, which in turn is equal to the rate of heat transfer from the outside surface of the outer glass pane to the outside air; thus considering the heattransfer rate per unit window area, we obtain Qgi Qgo Qi Qb Qo = = = = A A A A A The subscripts i, gi, b, go, and o refer to the heattransfer rate from the inside air to the inside glass pane, across the inner glass pane, across the gap between the panes, across the outer glass pane, and from the outside glass pane to the outside air, respectively. The heat transfer from the inside air to the inner glass surface is accomplished by a combination of radiation and convection. These will be expressed in terms of convective and radiative heattransfer coefficients defined by Qci = hci (Ti − T1 ) A
Qri = hri (Ti − T1 ) A
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Heat Transfer through a DoubleGlazed Window 8.4
SteadyState Calculations
w
To
T2
T4
T1
T3
Ti
Temperatures used in obtaining solution.
Figure 8.2
kg, tg
kg, tg
where Ti is the inside air temperature and T1 is the temperature of the inner surface of the inner glass pane as shown in Fig. 8.2. Now it is assumed that the window area is small compared to the size of the room to which it is exposed, so the following can be assumed: Qri = ε Ti 4 − T14 = ε Ti 2 + T12 (Ti + T1 ) (Ti − T1 ) A Hence, since by definition Qri = hri (Ti − T1 ) A it follows that hri = ε Ti 2 + T12 (Ti + T1 ) The total heattransfer rate to the inner glass surface is therefore given by Q Qci Qri = + = (hci + hri ) (Ti − T1 ) A A A The heat is transferred across the inner pane by conduction so that since steady, onedimensional conduction is being considered, it follows that Q T1 − T2 = kg A tg where the temperature T2 is as defined in Fig. 8.2. The heattransfer rate across the gap between the panes is given by Q = (hcb + hrb) (T2 − T3 ) A
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Heat Transfer through a DoubleGlazed Window Heat Transfer through a DoubleGlazed Window
8.5
Because the gap between the panes, 1 cm, is small compared to the height of the window, 1 m, it will be assumed that the radiant heat transfer across the gap is given by the equation for radiant heat between two infinite gray surfaces T24 − T34 T22 + T32 (T2 + T3 ) (T2 − T3 ) Qrb = = A 1/ε + 1/ε − 1 1/ε + 1/ε − 1 2 T2 + T32 (T2 + T3 ) (T2 − T3 ) = 2/ε − 1 where the emissivities of the two surfaces are the same. From this equation it follows that T22 + T32 (T2 + T3 ) T22 + T32 (T2 + T3 ) hrb = = 1/ε + 1/ε − 1 2/ε − 1 For the reasons given when discussing the heat transfer across the inner glass pane, the heattransfer rate across the outer pane is given: Q T3 − T4 = kg A tg Finally, consider heat transfer from the outer surface of the outer glass pane to the outside air. For the same reasons as those discussed when considering the heat transfer from the inside air to the inside glass surface, it follows that Q = (hco + hro ) (T4 − To ) A where hro = ε T42 + To 2 (T4 + To ) The equations above give Q 1 = Ti − T1 A hci + hri
Q 1 = T1 − T2 A kg /tg
Q 1 = T2 − T3 A hcb + hrb
Q 1 = T3 − T4 A kg /tg
Q 1 = T4 − T0 A hco + hro
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Heat Transfer through a DoubleGlazed Window 8.6
SteadyState Calculations
Adding these equations together and rearranging then gives Q = A
Ti − To 1 1 1 1 1 + + + + hci + hri kg /tg hcb + hrb kg /tg hco + hro
The heattransfer rate is commonly expressed in terms of an overall heattransfer coefficient U, which is defined by Q = U (Ti − To ) A From the preceding two equations it follows that 1 1 1 1 1 1 + + + + = U (hci + hri ) kg /tg (hcb + hrb) kg /tg (hco + hro ) Alternatively, the overall heat transfer rate can be expressed in terms of the thermal resistance R of the window, which is defined by Q (Ti − To ) = A R Comparison of these equations then gives R=
1 1 1 1 1 + + + + hci + hri kg /tg hcb + hrb kg /tg hco + hro
= Ri + Rg + Rb + Rg + Ro where Ri =
1 1 1 1 , Rg = , Rb = , Ro = hci + hri kg /tg hcb + hrb hco + hro
The space between the two panes of glass forms a vertical highaspectratio enclosure. In order to find the convective heattransfer rate between the two glass panes, the heattransfer rate across this enclosure is assumed to be expressed in terms of the Rayleigh number Ra, and the aspect ratio AR, which are defined by Ra =
gw3 (T2 − T1 ) 2
AR =
H w
respectively. The following three Nusselt numbers are then defined (see El Sherbiny, S. M., Raithby, G. D., and Hollands, K. G. T., 1982,
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Heat Transfer through a DoubleGlazed Window Heat Transfer through a DoubleGlazed Window
8.7
“Heat Transfer by Natural Convection across Vertical and Inclined Air Layers,” J. Heat Transfer, vol. 104, pp. 94–102) Nua = 0.0605 Ra1/3 3 1/3 0.293 0.104 Ra Nub = 1 + 1.36 1 + 6310/Ra Nuc = 0.242
Ra AR
0.272
and the Nusselt number for the heat transfer across the enclosure is taken as the largest of these values: Nub = maximum (Nua , Nub, Nuc ) The value of the convective heattransfer coefficient for the gap between the glass panes is then given by hci =
Nubw kb
where kb is the thermal conductivity of the gas in the enclosure between the glass panes. The thermal conductivity kb and the kinematic viscosity in the Rayleigh number are evaluated at the mean temperature in the gap, at Tbm =
T2 + T3 2
An iterative solution procedure is required; this procedure involves the following steps: 1. Guess the values of T1 , T2 , T3 , and T4 . 2. Using these values, calculate hri , hrb, hro , and hcb. 3. Calculate Q/A, U, and R. 4. Using this value of Q/A, calculate T1 , T2 , T3 , and T4 using Ti = T1 −
Q 1 A hci + hri
T2 = T1 −
Q 1 A kg /tg
T3 = T2 −
Q 1 A hcb + hrb
T4 = T3 −
Q 1 A kg /tg
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Heat Transfer through a DoubleGlazed Window 8.8
SteadyState Calculations
Figure 8.3 Variation of thermal conductivity k of air and argon with temperature.
5. Using these temperature values, repeat steps 1 to 3. If these new temperature values are essentially the same as the guessed values, the procedure can be stopped. If not, the process is repeated to again give new values for the temperatures. This procedure can be implemented in a number of ways. The results given here were obtained by implementing this procedure in the spreadsheet software Excel. In order to obtain the results, the values of and k for air and argon have to be known over the temperature range covered in the situation considered. The values assumed here are shown in Figs. 8.3 and 8.4.
Figure 8.4 Variation of kinematic viscosity of air and argon with temperature.
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Heat Transfer through a DoubleGlazed Window Heat Transfer through a DoubleGlazed Window
TABLE 8.1
8.9
Thermal Resistance Values, m2 C/W
Gas
Ri
Rg
Rb
Rg
Ro
R
Air Argon
0.1154 0.1152
0.00231 0.00231
0.1605 0.1951
0.00231 0.00231
0.03432 0.03433
0.3148 0.3492
The calculated values of Q/A, U, and R for the two cases are Air Argon
Q/A = 47.65 W/m2 C, U = 3.18 W/m2 C, R = 0.315 m2 C/W Q/A = 42.95 W/m2 C, U = 2.86 W/m2 C, R = 0.349 m2 C/W
It will be seen that the heattransfer rate per unit area with argon between the panes is about 11 percent lower than it is with air between the panes. This is basically due to the fact that argon has a lower thermal conductivity than does air. The R values for the two gases for each component of the window are shown in Table 8.1. From the values given in this table it will be seen that the biggest contributor to the overall thermal resistance is the gas space between the panes. The thermal resistance at the inside surface of the window is the nexthighest contributor. It was also found that the Nusselt number for the betweenpanes heat transfer was 1 in both cases considered, indicating that the heat transfer across the gas layer is effectively by conduction, since if this is true, it follows that Q T2 − T3 = kb A w
that is
(Q/A)w =1 kb (T2 − T3 )
(i.e., Nu = 1). This is because the gap between the panes is small.
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Heat Transfer through a DoubleGlazed Window
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10
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Source: HeatTransfer Calculations
Part
3 Transient and Cyclic Calculations
The contributors of the chapters in this part work in U.S. government defense installations, consulting firms, industry in the United States and Japan, and academia. The calculations they describe come out of real life:
Locating corrosion under paint layers on a metal surface The combustion tube of a prototype pulsed detonation engine Coke deposits in fired heaters processing hydrocarbon liquids, vapors, or gases Tape drives used as backup storage devices of computer data Infrared thermographic (IR) surveys of roofs to detect leaks A motor vehicle painting process that involves a series of paint application steps and paint drying/curing ovens An electronics cooling system Laminates of dissimilar materials, which are commonly found in industrial products, such as printedcircuit boards in electronic equipment Heat and mass transfer with phase change, which is encountered in drying operations, the chemical process industry, air conditioning and refrigeration, and manufacturing operations A method of optimizing vapor compression refrigeration cycles c A Low Temperature, Low Energy Carrier (LoTEC ), a passive thermal carrier designed to permit payload transport
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Transient and Cyclic Calculations 9.2
Transient and Cyclic Calculations
to and from the International Space Station at approximately constant temperature without external power Most of the calculations in this part are timedependent, although there are several in which steadystate calculations do deal adequately with the problem at hand, even though, for example, intermittent heat generation may be involved. Heattransfercoefficient determination again plays a role in several calculations.
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Source: HeatTransfer Calculations
Chapter
9 Use of Green’s Function to Solve for Temperatures in a Bimaterial Slab Exposed to a Periodic Heat Flux Applied to Corrosion Detection
Larry W. Byrd AFRL/VASA WrightPatterson Air Force Base, Ohio
This problem demonstrates the use of separation of variables and Green’s function to determine the transient temperature distribution in a layer of paint covering a corroded metal surface. It simulates a proposed noncontacting method of locating corrosion under paint. The following pages give the details of the analysis and a comparison with a finitedifference approximation. The detection concept is based on the assumption that a corrosion layer between the metal substrate and the paint will have a larger thermal resistance than will areas with no corrosion. A sinusoidally varying heat flux is applied to the painted surface as shown in Fig. 9.1. It is desired to show that this will result in a periodic variation in surface temperature that is at the same frequency as the applied flux. The unique aspect of this approach is that the phase lag between the surface temperature and the incident flux, not a change in amplitude of the resulting sinusoidal variation in surface temperature, is used as the
9.3
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Use of Green’s Function to Solve for Temperatures in a Bimaterial Slab Exposed to a Periodic Heat Flux Applied to Corrosion Detection 9.4
Transient and Cyclic Calculations
Signal Processing
Camera
Detector Signal
Heat Source
Sig. Input
AC Ref. Ampl. Input
Camera Control
Phase Output Output
Computer
Figure 9.1
Corrosion detection system.
indicator of corrosion. The incident flux is chosen to be small enough to cause only a slight rise in the paint temperature. The only way to detect this temperature fluctuation is to filter the sinusoidal signal at the load frequency to eliminate the typical random fluctuations in temperature as shown in Fig. 9.2. This has been claimed to give a resolution as good as 0.001 K in commercial differential thermography systems [1]. A onedimensional heat conduction model of an aluminum plate with corrosion at the interface between the aluminum and the paint is shown in Fig. 9.3. The effect of the corrosion layer is modeled by a change in the thermal conductance h1 between the paint and the substrate. If there is no corrosion and the thermal contact was perfect, the conductance h1 is infinite. This corresponds to zero resistance to heat conduction. In real Typical Noisy Signal T (K) 310 300
Conditioned Signal Removes Components Not at Load Frequency
290
C∆Τ
280 270 260 250 0
Scaled Load Signal Figure 9.2
0.05
0.1
0.15
0.2
Time (s)
Typical signal conditioning used in differential thermography.
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Use of Green’s Function to Solve for Temperatures in a Bimaterial Slab Exposed to a Periodic Heat Flux Applied to Corrosion Detection Use of Green’s Function to Solve for Temperatures
9.5
h1 h2 2
1
q0[1+A sin(ωt)]
qconv L1
Aluminum
2
Paint
x
L Figure 9.3
1
Problem geometry.
systems the conductance is a large, but not infinite, value. As the surface begins to corrode, the thermal conductance decreases and the layer acts as an insulator. This means that less energy will be conducted into the substrate and more will be lost to the environment. The conductance for no corrosion is assumed to be greater than 1000 W/m2 C and is on the order of 100 W/m2 C when corrosion is present. A mathematical model of the system solves the heat conduction equation in each layer with additional boundary conditions at the interface to account for the thermal conductance. The change in volume due to the corrosion is not modeled; thus the interface is shown as having a negligible thickness. The heat conduction equation for each region and the associated boundary conditions are given as Eqs. (9.1a) to (9.1d) and (9.2a) and (9.2b). Equations (9.1b) and (9.1c) specify that the heat flux is continuous across the interface but the temperature is discontinuous, with the difference controlled by the value of h1 . It is assumed that the system and the surroundings are initially at a uniform temperature T0 . At t = 0, the painted surface at x = L is heated by incident radiation such that the absorbed part is given by q0 [1 + A sin(t)] and also loses heat by convection. The back surface at x = 0 is considered to be insulated as shown by Eq. (9.1a), but the results should not be significantly different if heat transfer by convection or radiation was included. Region 1:
1
∂ 2 T1 ∂ T1 = 2 ∂t ∂x
0 < x < L1 ;
t>0
(9.1)
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Transient and Cyclic Calculations
−k1 k1
∂ T1 =0 ∂x
x = 0;
∂ T1 = h1 (T1 − T2 ) ∂x
x = L1 ;
t>0
(9.1b)
∂ T1 ∂ T2 = k2 ∂x ∂x
x = L1 ;
t>0
(9.1c)
T1 = T0
t>0
0 < x < L1 ;
(9.1a)
t=0
(9.1d)
Region 2:
∂ T2 ∂ T2 L1 < x < L; t > 0 = 2 ∂t ∂x ∂ T2 q0 [1 + A sin(t)] k2 + h2 T2 = h2 T0 + ∂x h 2 2
(9.2)
f (t)
x = L; t > 0 T2 = T0
x < L1 < L;
(9.2a) t=0
(9.2b)
SeparationofVariables Technique The separationofvariables method is a wellknown technique used to solve partial differential equations when the solution can be written as a product of functions of a single independent variable [2]. For onedimensional heat conduction in a slab of width L, with no heat generation, the equation is ∂2T 1 ∂T = 2 ∂t ∂x
(9.3)
Substituting T(x, t) = X(x)(t) into Eq. (9.3) gives X (x) (t) = = −2 X(x) (t)
(9.4)
is a constant called an eigenvalue which will be determined by the boundary conditions. (t) can be found in terms of by integrating the second ordinary differential equation in Eq. (9.4): (t) = 0 e−
2
t
(9.5)
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Use of Green’s Function to Solve for Temperatures in a Bimaterial Slab Exposed to a Periodic Heat Flux Applied to Corrosion Detection Use of Green’s Function to Solve for Temperatures
9.7
X(x) is found by solving the first ordinary differential equation in Eq. (9.4) and will have the following form: X(x) = a0 sin
x x x x + a1 cos + b0 sinh + b1 cosh L L L L
(9.6)
X(x) is referred to as an eigenfunction. To satisfy the initial condition, the solution is written as an infinite series of the eigenfunctions multiplied by (t) as T(x, t) =
∞
An X(n, x)(n, t)
(9.7)
n=0
The constants An are found using the initial condition and the orthogonality of the eigenfunctions. Orthogonality means that the integral of the product of any two eigenfunctions X(m,x) and X(n, x) over the domain is zero unless m = n. The constants 0 and those in X(x) have been absorbed into An. The initial condition can be written as T0 =
∞
An X(n, x)
(9.7a)
n=0
Multiplying both sides of Eq. (9.7) by X(m, x) and integrating gives
L
L T0 X(m, x)dx =
x =0
Am X2 (m, x)dx = Am N(m)
(9.8)
x =0
where N(m) is the normalization integral. This is easily rearranged to give Am for any value of m. The present problem cannot be solved by separation of variables in its simplest form because the boundary condition at x = L is not only nonhomogeneous but also timedependent. It can be solved using a Green’s function approach for composite media as given by Ozisik [3]. The problem is first transformed using Ti (x, t) = i (x, t) + i (x) f (t)
(9.9)
to remove the nonhomogeneous boundary condition at x = L. This results in a timedependent volumetric heat source term in the heat conduction equation. The subscript i = 1, 2 refers to the regions 1 and 2. f(t) is defined in Eq. (9.2a), while and are the solutions to the following auxiliary problems.
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Transient and Cyclic Calculations
Region 1:
1
∂1 ∂ 2 1 df (t) = − 1 dt ∂t ∂ x2
−k1 k1
0 < x < L1 ;
t>0
(9.10)
∂1 =0 ∂x
x = 0;
∂1 = h1 (1 − 2 ) ∂x
x = L1 ;
t>0
(9.10b)
∂1 ∂2 = k2 ∂x ∂x
x = L1 ;
t>0
(9.10c)
1 (0) = T0 − f (0)
t>0
0 < x < L1 ;
(9.10a)
t=0
(9.10d)
Region 2:
2
∂2 df (t) ∂2 − 2 = 2 dt ∂t ∂x k2
L1 < x < L;
∂2 + h2 2 = 0 ∂x
x = L;
2 (0) = T0 − f (0)
t>0
t>0
L1 < x < L;
t=0
(9.11) (9.11a) (9.11b)
Region 1:
∂ 2 1 =0 ∂ x2
−k1 k1
0 < x < L1
(9.12)
∂1 =0 ∂x
x=0
(9.12a)
∂1 = h1 (1 − 2 ) ∂x
x = L1
(9.12b)
∂1 ∂2 = k2 ∂x ∂x
x = L1
(9.12c)
Region 2:
∂ 2 2 =0 ∂ x2
k2
∂2 + h2 2 = h2 ∂x
L1 < x < L x = L1
(9.13) (9.13a)
For this problem, 1 = 2 = 1.
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QC: IML/OVY
T1: IML
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July 14, 2005
14:2
Use of Green’s Function to Solve for Temperatures in a Bimaterial Slab Exposed to a Periodic Heat Flux Applied to Corrosion Detection Use of Green’s Function to Solve for Temperatures
9.9
The solution for i (x, t) can be written in terms of Green’s function as i (x, t) =
2 j =1
x j +
1
Gi j (x, tx , )
xj
= 0 Fj (x
)dx
transient
t
x j +
=0
xj
1
+
j Gi j (x, tx , ) g j (x , ) dx d kj
(9.14)
periodic
where Green’s function is defined as kj −n2 (t− ) in(x) jn(x ) ∞ e j Gi j (x, tx , ) = Nn n=1
(9.15)
The normalization integral is k1 Nn = 1
L1 2 1n (x)dx x =0
k2 + 2
−q0 Fj = j (x, 0) = h2
L 2 2n (x)dx x = L1
and
(9.16) j g j (x , ) = −q0 A cos( ) kj
The second term in Eq. (9.14) which is marked periodic will be found to be associated with a sinusoidally varying component, while the first term will show a transient response that goes to a terminal value. The eigenfunctions in are solutions to the following problems. Region 1:
∂ 2 1n n2 + =0 1 ∂ x2
0 < x < L1
(9.17)
∂1n =0 ∂x
x=0
(9.17a)
−
∂1n = H1 (1n − 2n) ∂x
x = L1
(9.17b)
k1
∂1n ∂2n = k2 ∂x ∂x
x = L1
(9.17c)
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GRBT05609
Kutz2103G
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July 14, 2005
14:2
Use of Green’s Function to Solve for Temperatures in a Bimaterial Slab Exposed to a Periodic Heat Flux Applied to Corrosion Detection 9.10
Transient and Cyclic Calculations
Region 2:
∂ 2 2n n2 + =0 2 ∂ x2 ∂2n + H2 2n = 0 ∂x hi Hi = ki
L1 < x < L x=L
Solving these equations gives n x n x 1n(x) = Ain sin √ + Bin cos √ i i
(9.18) (9.18a)
(9.19)
The eigenvalues n and constants Ain and Bin are determined from the boundary conditions to arrive at a solution for i (x, t), which is then substituted into Eq. (9.9) for Ti (x, t). Equation (9.17a) gives A1n = 0. This results in a set of homogeneous algebraic equations: 1 sin(1 L1 ) − cos(1 L1 ) sin(2 L1 ) cos(2 L1 ) 0 H1 B1n k1 2 = 0 A 2n cos(2 L1 ) − sin(2 L1 ) k sin(1 L1 ) 0 B2n 2 1 0 X2 Y2 (9.20) X2 = 2 cos(2 L) + H2 sin(2 L)
(9.20a)
Y2 = −2 sin(2 L) + H2 cos(2 L)
(9.20b)
n i = √ i
(9.20c)
Since the set of equations is homogeneous, the constants can be determined only to a multiplicative constant. Thus, without a loss of generality, B1n is set equal to 1 and A2n and B2n are
1 A2n = sin(2 L1 ) cos(1 L1 ) − sin(1 L1 ) H1 √ k1 2 − √ cos(2 L1 ) sin(1 L1 ) (9.21) k2 1
1 B2n = cos(2 L1 ) cos(1 L1 ) − sin(1 L1 ) H1 √ k1 2 + √ sin(2 L1 ) sin(1 L1 ) (9.22) k2 1
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GRBT05609
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QC: IML/OVY
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GRBT056Kutzv4.cls
July 14, 2005
14:2
Use of Green’s Function to Solve for Temperatures in a Bimaterial Slab Exposed to a Periodic Heat Flux Applied to Corrosion Detection Use of Green’s Function to Solve for Temperatures
9.11
The eigenvalues n are found by using the requirement for a nontrivial solution that the determinant of the coefficient matrix is equal to zero: 1 sin(1 L1 ) − cos(1 L1 ) H1 k1 2 k sin(1 L1 ) 2 1 0
sin(2 L1 ) cos(2 L1 ) X2
− sin(2 L1 ) = 0 Y
cos(2 L1 )
(9.23)
2
The normalization integral can now be written as
k1 Nn = 1 =
k1 1
L1
k2 cos (1n x)dx + 2
L
2
0
2 A2n sin(2n x) + B2n cos(2n x) dx
L1
L1 sin(21n L1 ) + 2 41n
k2 L − L1 sin(22n L) − sin(22n L1 ) + A22n − 2 2 42n
L − L1 sin(22n L) − sin(22n L1 ) 2 + B2n + 2 42n
2A2n B2n cos(22n L1 ) − cos(22n L) + 42n 42n
(9.24)
To demonstrate that the resulting surface temperature will provide a signal at the load frequency, only the periodic part of the solution for region 2 is needed because the rest will be filtered out. From Eq. (9.14) this is
2 p =
∞ t
n=1
e−n (t− ) [−q0 A cos( )]I2n2n(x) 2
d
Nnh2
=0
(9.25)
where k1 I2n = 1
L1 x =0
k2 1n(x) dx + 2
L 2n(x) dx
(9.26)
x = L1
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GRBT05609
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QC: IML/OVY
T1: IML
GRBT056Kutzv4.cls
July 14, 2005
14:2
Use of Green’s Function to Solve for Temperatures in a Bimaterial Slab Exposed to a Periodic Heat Flux Applied to Corrosion Detection 9.12
Transient and Cyclic Calculations
Evaluating the integral over gives
2 n2 en t sin( t + n) −n2 t √ 4 2 − 4 + 2 2n(X ) (−q0 AI2n) ∞ e n n + 2 p(x, t) = Nnh2 n=1
(9.27)
The term with the negative exponential will die out to leave the periodic component: 2 p(x, t) =
∞ I2n2n(x) sin(t + n) −q0 A ! h2 n = 1 Nn n4 + 2
(9.28)
The periodic component of the temperature at x = L is " # ∞ q0 T2 p(L, t) = T∞ + 1 − A Cn sin( t + n) h2 n=0
where C0 = −1/, 0 = 0, and Cn = I2n2n(L)/Nn
$
n4
+
(9.29)
2
for n ≥ 1.
The sum can be rewritten using trigonometric identities to give T2 p(L, t) % & ∞ 2 2 ∞ & q0 = T∞ + 1 − A ' Cn cos(n) + Cn sin(n) h2 n= 0 n= 0 × sin( t + ) ∞ Cn sin(n) n= 0 − = tan−1 ∞ Cn cos(n)
(9.30)
(9.31)
n= 0
The surface temperature can be seen to have the correct form to be used to detect a change in thermal conductance. Specifically, it is proportional to the single load frequency with a phase angle that is a function of the conductance between the paint and substrate. Figure 9.4 shows the surface temperature for a time range of 10 s. Also shown by the blue line is a finitedifference approximation of the surface temperature. This method used an explicit scheme to update the temperature. In Fig. 9.5, the time and temperature scales have been changed to show the temperature profile up to 5000 s. Note that at long times the
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July 14, 2005
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Use of Green’s Function to Solve for Temperatures in a Bimaterial Slab Exposed to a Periodic Heat Flux Applied to Corrosion Detection
Surface Temperature (K)
Use of Green’s Function to Solve for Temperatures
9.13
300.045 300.04 300.035 300.03 300.025 300.02 300.015 300.01 300.005 300 299.995 0
1
2
3
4
5 6 Time (s)
7
8
9
10
Surface temperature for short times: q0 = 10 W/m2 , h1 = 1000 C, h2 = 6 W/m2 C, L1 = 2 mm, L = 2.54 mm.
Figure 9.4
W/m2
exponential part of the solution disappears and the temperature approaches a steadystate temperature with a superimposed ripple. The ripple is not as apparent on this figure because of the scale used to illustrate the total change in temperature, which can be shown to approach q0 / h2 = (10 W/m2 )/(6 W/m2 K) = 1.7 K. The ripple magnitude is small because of the lowlevel incident energy (q0 = 10 W/m2 ) used for this particular simulation. Low levels are desirable because the surface temperature will change by only a few degrees, but this makes it harder to detect the ripple magnitude, so some control is required. Figure 9.6 shows the difference in phase lag between a segment with h1 assumed to be either 1000 or 5000 W/m2 C and the phase lag associated with corrosion (h1 = 100 W/m2 C) for a paint thickness of 0.254 mm. This is convenient for corrosion detection because if a plot of phase lag is
Surface Temperature (K)
301.8 301.6 301.4 301.2 301 300.8 300.6 300.4 300.2 300 0
1000
2000
3000
4000
5000
6000
Time (s) Figure 9.5 Surface temperature for long times: q0 = 10 W/m2 , h1 = 1000 W/m2 C, h2 = 6
W/m2 C, L1 = 2 mm, L = 2.254 mm.
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GRBT05609
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QC: IML/OVY
T1: IML
GRBT056Kutzv4.cls
July 14, 2005
14:2
Use of Green’s Function to Solve for Temperatures in a Bimaterial Slab Exposed to a Periodic Heat Flux Applied to Corrosion Detection 9.14
Transient and Cyclic Calculations
phase difference Φ(h1) − Φ(100) 50 phase difference (degrees)
40 30 h1 = 1000
20
h1 = 5000
10 0 −10
0
1
2
3
4
5
6
7
8
9 10
load frequency (Hz) Figure 9.6 Phase lag referenced to a thermal conductance of 100 W/m2 C: q0 = 10 W/m2 , L1 = 2 mm, L = 2.254 mm, h2 = 6 W/m2 C.
shown over the surface, the areas that are corroded will have a large contrast compared to the pristine regions. The phase lag is dependent on the thermal conductance h1 and the heater frequency. If the heater frequency is too high, the phase lag is the same regardless of h1 , and it is not a good indicator of corrosion. Figure 9.7 shows the effect of the thickness of the paint layer. As can be seen, the thicker the paint is, the lower the frequency needed to detect a change in phase lag. This can be problematic because the filtering technique is faster and has less noise as the load frequency is increased.
phase difference (degrees)
40 freq = 0.25
35
freq = 0.5
30
freq = 0.75
25
freq = 1
20
freq = 2
15 10 5 0 −5 0.2
0.25
0.3
0.35
0.4
0.45
0.5
paint layer thickness (mm) Effect of paint thickness: q0 = 10 W/m2 , h1 = 1000 W/m2 C, h2 = 6 W/m2 C, L1 = 2 mm.
Figure 9.7
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Use of Green’s Function to Solve for Temperatures in a Bimaterial Slab Exposed to a Periodic Heat Flux Applied to Corrosion Detection Use of Green’s Function to Solve for Temperatures
9.15
In summary, this analysis gives the temperature distribution in a twolayer material exposed to a sinusoidally varying heat flux on one face and insulated on the other. The surface temperature shows the typical asymptotic rise of a slab exposed to a hot gas with a superimposed ripple. The phase lag between the incident heat flux and the surface temperature is a function of the material parameters and the thermal conductance between the material layers. References 1. Lesniak, J. R., Differential Thermography for Elevated Temperatures, USAF SBIR Final Report, Contract F3316595C2504, Aug. 25, 1997. 2. Berg, W. B., and McGregor, J. L., Elementary Partial Differential Equations, HoldenDay, San Francisco, 1966. 3. Ozisik, M. N., Heat Conduction, Wiley, New York, 1980.
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Use of Green’s Function to Solve for Temperatures in a Bimaterial Slab Exposed to a Periodic Heat Flux Applied to Corrosion Detection
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16
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14:38
Source: HeatTransfer Calculations
Chapter
10 LumpedCapacitance Model of a Tube Heated by a Periodic Source with Application to a PDE Tube
Larry W. Byrd AFRL/VASA WrightPatterson AFB, Ohio
Fred Schauer Pulsed Detonation Research AFRL/PRTS WrightPatterson AFB, Ohio
John L. Hoke Innovative Scientific Solutions, Inc. WrightPatterson AFB, Ohio
Royce Bradley Innovative Scientific Solutions, Inc. WrightPatterson AFB, Ohio
Introduction This problem considers estimation of temperature as a function of time of a circular tube heated internally by a hot gas with a periodically varying temperature. The lumpedcapacitance method neglects the temperature variation in the tube wall, so it is described as simply T(t). It is also assumed for this problem that the heat transfer is primarily in the radial direction, so axial conduction is neglected. The solution will be 10.1
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14:38
LumpedCapacitance Model of a Tube Heated by a Periodic Source with Application to a PDE Tube 10.2
Transient and Cyclic Calculations
Prototype pulsed detonation engine with four detonation tubes.
Figure 10.1
applied to the combustion tube of a prototype pulsed detonation engine (PDE) shown in Fig. 10.1. Data taken from a watercooled aluminum tube and a freeconvection–radiatively cooled steel tube is analyzed to determine the heattransfer characteristics of the interior and exterior surfaces. A PDE operates cyclically, completing a filldetonationexhaust cycle typically many times per second. Conventional engines chemically release heat through deflagrative combustion, resulting in slow burning and isobaric (constantpressure) heat addition. Detonative combustion propagates at supersonic speeds, resulting in pressure gain combustion and resultant higher thrust and efficiency. The fillburnexhaust cycle of the tube consists of equal time intervals for each phase. The pressure in a PDE is initially low. The detonation processes increases the pressure substantially, and the resultant highmomentum flux from the detonation tube produces thrust and allows the PDE to selfaspirate as a result of the overexpansion observed near 4 ms in Fig. 10.3. In practice, the detonationexhaust portion of the PDE cycle is followed by a purge cycle of cold air which serves as a buffer between hot products and the fresh reactants entering the detonation tube prior to the next detonation and to cool the hot tube walls. The PDE thrust can be controlled by regulating operating various operating parameters such as the frequency of the detonations. Mathematical Formulation Figure 10.2 shows a schematic a section of the tube wall with the convective heat transfer occurring at the inside and outside surfaces Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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14:38
LumpedCapacitance Model of a Tube Heated by a Periodic Source with Application to a PDE Tube LumpedCapacitance Model of Tube Heated by Periodic Source
10.3
T0
Tg(t) qi(t)
q0(t) Figure 10.2 Schematic of the lumpedcapacitance model.
characterized by qi (t) = hi Ai [Tg (t) − T (t)] and qo (t) = ho Ao [T(t) − T0 ], Tg (t) is the gas temperature in the tube, T0 is the ambient temperature outside the tube, and h is the average heattransfer coefficient. The subscripts i and o refer to inner and outer, respectively. An energy balance gives mC p
dT = hi Ai Tg − T − ho Ao (T − T0 ) dt
(10.1)
The gas temperature is modeled as Tg = Tm + T cos(t)
(10.2)
with T = (Tf − T0 )/2 and Tm = (Tf + T0 )/2, and = 2 f is the circular firing frequency. As seen in Fig. 10.3, this is an approximation to the actual temperature, which is a function of position and shows a sharp spike as the detonation occurs. There are several ways to improve the results if needed. One is to include the gas temperature as a Fourier series which could be made to fit any expected time variation. Another method is to break the problem into discrete time increments with constant values of the gas temperature appropriate for the phase of the fillburnpurge cycle and 3500 3000
T (K)
2500 2000
Texp Tg
1500 1000 500 0 0
50
100
150
time (ms) Figure 10.3 Expected gas temperature in PDE tube, Tg,exp ,
and the approximation Tg .
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LumpedCapacitance Model of a Tube Heated by a Periodic Source with Application to a PDE Tube 10.4
Transient and Cyclic Calculations
sum up the changes in tube temperature as time progresses. The heat transfer would also be expected to be a function of the tube temperature, but any difference between the model and the actual apparatus will be absorbed into the values of hi and ho , which will be determined semiemperically. Thus they will be average coefficients over the timespan of interest. Solution Using m = (ro2 − ri2 )L, Ai = 2ri L, and Ao = 2ro L, Eq. (10.1) becomes dT + aT = b + c cos(t) dt
(10.3)
a=
2(hi ri + ho ro ) ro2 − ri2 Cp
(10.3a)
b=
2(hi ri Tm + ho ro T0 ) ro2 − ri2 Cp
(10.3b)
c=
2hi ri T ro2 − ri2 Cp
(10.3c)
The solution to this ordinary differential equation is given by the sum of a particular and homogeneous solution, T = Tp + Th [1]. Temperature Th is the solution to dTh + aTh = 0 dt
(10.4)
This can be solved by separating the variables and integrating to give Th = Th,0 e−at
(10.5)
The particular solution Tp would be expected to have a form similar to the righthand side of Eq. (10.3). It is assumed as Tp = + cos(t) + sin(t)
(10.6)
Substituting into Eq. (10.3) gives [− + a ] sin(t) + [ + a] cos(t) + a = b + c cos(t)
(10.7)
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LumpedCapacitance Model of a Tube Heated by a Periodic Source with Application to a PDE Tube LumpedCapacitance Model of Tube Heated by Periodic Source
10.5
Equating the coefficients of the sin(t) and cos(t) and constant terms gives − + a = 0
(10.7a)
+ a = c
(10.7b)
a = b
(10.7c)
which gives = = =
b a
(10.8a)
2
ca + a2
(10.8b)
2
c + a2
(10.8c)
Thus b a Tp = + c cos(t) + 2 sin(t) a 2 + a2 + a2
(10.9)
This form can be readily transformed into a sinusoidal function with a phase angle by substituting sin() = √
a 2
+
cos() = √
a2
2
+ a2
and using the trigonometric identity: cos(t) sin() + sin(t) cos() = sin(t + ) = tan−1
a
(10.10) (10.10a)
The particular solution is then found to be Tp =
b c sin(t + ) + √ a 2 + a2
(10.11)
The temperature can now be written as T(t) = Th,0 e−at +
b c sin(t + ) + √ a 2 + a2
(10.12)
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14:38
LumpedCapacitance Model of a Tube Heated by a Periodic Source with Application to a PDE Tube 10.6
Transient and Cyclic Calculations
The integration constant Th,0 can be found by using the initial condition T = T0 at t = 0, giving the temperature as b b c sin(t + ) c sin() T(t) = T0 − e−at + + √ (10.13) +√ 2 2 a a +a 2 + a2 This equation will be used to estimate values for a and b so that hi and ho can be determined from experimental data for the special case where c is small. When representative values of the thermal parameters are substituted into the definitions for the constants a, b, and c, it is seen √ that c/ 2 + a2 0.1, one can use chart solutions, called Heisler charts [5]. Figure 12.3 shows a chart solution for an infinite slab at the midplane location (x = x/L = 0 at x = 0). The chart solution method can be somewhat cumbersome to read if one has a lot of points to plot, but it represents an easy way to address Eq. (12.13). An alternate method to chart solutions is to compute values for Eq. (12.13) directly. After some investigation, one can reduce Eq. (12.13) and formulate a oneterm, timedependent approximate solution at the midplane location (x = x/L = 0 at x = 0) with < 2 percent error. The oneterm approximation to Eq. (12.13) is shown as Eq. (12.16) for u(x, t) as u(x, t) = A1 e−1 Fo 2
(12.16)
For the problem at hand, let the characteristic length Lc for Eqs. (12.11) and (12.12) equal onehalf the tape reel width Lc = (0.315/2) in. = 0.1575 in. Substituting into Eq. (12.11) for the Fourier modulus Fo, we obtain Fo =
7.903 × 10−4 ft2 /h · (t, h) 0.1575 in. 2 12 in./ft
= 4.5876 · (t, h) = 0.07646 · (t, min)
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15:24
Heisler chart: centerline temperatures for infinite plate of thickness 2L [5]. Figure 12.3
α Fo = L 2 . t c
Bi =
h . Lc K
Transient HeatTransfer Problem: Tape Pack Cooling
u(x,t) =
T(x,t) − T∞ Ti − T∞ 12.8
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15:24
Transient HeatTransfer Problem: Tape Pack Cooling Transient HeatTransfer Problem: Tape Pack Cooling
12.9
Substituting into Eq. (12.12) for the Biot number Bi, we have Bi =
1.6985 Btu/hft2 ◦ F 0.02168 Btu/hft◦ F
0.1575 in. 12 in./ft
= 1.0283 For Bi = 1.028, solve for the coefficients A1 and 1 using Eqs. (12.14), (12.15a), and (12.15b): A1 = 1.12078
and
1 = 0.86730
Step 5: Estimate time required for midpack to cool down to 99 percent of RT
Substituting into Eq. (12.16) and solving for the time required for the center region of the tape pack (x = x/L = 0 at x = 0) to return to 99 percent of RT (72.72◦ F), t in minutes, we obtain u(x, t) =
T(0, t) − T∞ 72.72 − 72 2 = 1.12078 · e−(0.86730) · (0.07646) · (t, min) = Ti − T∞ 130 − 72 0.012414 = 1.12078 · e−0.05751 · (t, min) 0.012414 ln = −0.05751 · (t, min) 1.12078 −4.50297 = t, min −0.05751 t = 78.29 min
A spreadsheet is used to minimize round off in the arithmetic. Step 6: Evaluate T(0, t) at 5min intervals for comparison to FEA results
Table 12.2 lists estimates for internal temperature of the tape pack at the midplane location (x = x/L = 0 at x = 0) as a function of time t. A spreadsheet is used in this table to minimize roundoff in the arithmetic of Eq. (12.16). Step 7: Verify hand calculations using FEA idealization
Table 12.3 lists a comparison of estimates for temperatures at the tape pack center versus time when the geometry shown in Fig. 12.2 is analyzed using both hand solutions and an FEA mathematical idealization
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Transient HeatTransfer Problem: Tape Pack Cooling 12.10
Transient and Cyclic Calculations
TABLE 12.2
Hand Solution: Tape Pack Cooling Center Temperatures versus Time
Start
End
Time, min
Fo{t}
u(0, t)
T(0, t)
0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 50.00 55.00 60.00 65.00 70.00
0.000 0.382 0.765 1.147 1.529 1.911 2.294 2.676 3.058 3.441 3.823 4.205 4.588 4.970 5.352
1.000 0.841 0.631 0.473 0.355 0.266 0.200 0.150 0.112 0.084 0.063 0.047 0.036 0.027 0.020
130.00 120.76 108.57 99.43 92.58 87.44 83.58 80.68 78.51 76.89 75.66 74.75 74.06 73.55 73.16
75.00 80.00
5.734 6.117
0.015 0.011
72.87 72.65
85.00 90.00 95.00 100.00
6.499 6.881 7.264 7.646
0.008 0.006 0.005 0.004
72.49 72.37 72.28 72.21
Range 99% RT
TABLE 12.3
Tape Pack Cooling Center Temperatures versus Time Comparison of FEA versus Hand Solutions
Start
End
Time, min
FEA
Hand
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70
130.000 120.876 108.692 99.484 92.586 87.419 83.549 80.650 78.479 76.853 75.635 74.723 74.039 73.528 73.144
130.000 120.760 108.574 99.434 92.578 87.435 83.578 80.684 78.514 76.886 75.665 74.749 74.062 73.547 73.160
Percent difference 0.000 −0.001 −0.001 −0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
75 80
72.857 72.642
72.870 72.653
0.000 0.000
85 90 95 100
72.481 72.360 72.270 72.202
72.490 72.367 72.275 72.207
0.000 0.000 0.000 0.000
Range 99% RT
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12.11
140
Center Temp. (°F)
130 120 110 100 90
Hand Solution FEA Solution
80
RT 72F 70
0
10
20
30
40
50
60
70
80
90
100
Time (min) Figure 12.4
Summary comparison results of hand versus FEA solution, transient thermal
response.
processed using MSC/Nastran. A comparison between the solutions estimates a percentage difference no larger then ±0.001 percent during the initial transient response with virtually no difference between the solutions after approximately 15 min of cooling. Figure 12.4 shows a summary comparison between the hand solution and the FEA model solutions as a graph. Applying the Principles With a high degree of confidence in defining and solving a simplified version of the transient heattransfer problem, the final step is to apply the principles. Because of the complexity of the geometry that includes an aluminum hub in addition to the wound tape, FEA methods are better suited for addressing the solutions. Before we discuss the finiteelement solution, a short introduction to FEA follows. What Is FiniteElement Analysis, and How Does It Work? Finiteelement analysis methods were first introduced in 1943. Finiteelement analysis uses the Ritz method of numerical analysis and minimization of variational calculus to obtain approximate solution to systems. By the early 1970s, FEA was limited to highend companies involved in the aerospace, automotive, and power plant industries that
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Transient HeatTransfer Problem: Tape Pack Cooling 12.12
Transient and Cyclic Calculations
could afford the cost of having mainframe computers. After the introduction of desktop personal computers, FEA was soon made available to the average user. FEA can be used to investigate new product designs and improve or refine existing designs. The FEA approach is a mathematical idealization method that approximates physical systems by idealizing the geometry of a design using a system of grid points called nodes. These node points are connected together to create a “finite” number of regions or mesh. The regions that define the mesh are assigned properties such as the type of material and the type of mesh used. With the mesh properties assigned, the mesh regions are now called “elements.” The types of mesh include rod elements, beam elements, plate elements, and solid elements. Constraints or reactions are entered to simulate how the idealized geometry will react to the defined loading. Analysis models can represent both linear and nonlinear systems. Linear models use constant parameters and assume that the material will return to its orignal shape after the loading is removed. Nonlinear models consist of stressing the material past its elastic capabilities, and permanent deformations will remain after the loading is removed. The definitions for the materials in these nonlinear models can be complex. For heat transfer, FEA models can be idealized for both steadystate and transient solutions. The mesh elements are defined using the material conductivity or thermal dynamics to represent the system being analyzed. Steadystate transfer refers to materials with constant thermoproperties and where heat diffusion can be represented by linear equations. Thermal loads for solutions to heattransfer analysis include parameters such as temperature, internal heat generation, and convection.
FEA Results: Tape Pack with Aluminum Hub Figure 12.5 shows an FEA model idealization of a tape pack with the aluminum hub included as shown in Fig. 12.1. The FEA model shown in Fig. 12.5 is analyzed for the case of a tape pack that is initially at temperature soak of 130◦ F and shows the transient temperature response after 10 min of cooling. Table 12.4 lists numerical values for estimates of tape pack cooling as a function of time when the FEA model shown in Fig. 12.5 is analyzed using MSC/Nastran. Figure 12.6 shows the results graphically. Using linear interpolation of the data shown in Fig. 12.6 from t = 65 min to t = 70 min, one can estimate the time required for the tape pack center to return to 99 percent of RT t99% = 69.1 min.
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Y
Figure 12.5
Z
x
Centerline for Axisymmetric Elements
FEA model for transient heattransfer solution, tape pack with aluminum hub.
Initial "Soak" Temperature Ta = 130°F
w = 0.098 Ib/in.3
K = 1.112 × 10 Btu/minin.°F K = 0.230 Btu/Ib°F
72.0
75.63
79.25
86.5
90.13
93.75
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Hub Reel Aluminum Material Properties
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104.6
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Node 369 Midtape Pack T = 108.6°F After 10 Minutes
122.8
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h = 1.966 × 10−4 Btu/minin.2°F Ambient Air Temperature = 72°F
Average Film Coefficient
126.4
130.0
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Transient HeatTransfer Problem: Tape Pack Cooling
12.13
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Transient HeatTransfer Problem: Tape Pack Cooling 12.14
Transient and Cyclic Calculations
TABLE 12.4
Summary of Tabular Results of FEA Solution: Transient Time History Thermal Response for Tape Pack with Aluminum Hub at Midpack Location (Node 369)
Start
FEA
0 5 10 15 20 25 30 35 40 45 50 55 60
130.000 120.872 108.545 98.960 91.666 86.224 82.227 79.323 77.228 75.725 74.649 73.883 73.337
65 70
72.948 72.673
75 80 85 90 95 100
72.477 72.338 72.240 72.170 72.120 72.085
Range 99% RT
Center Temp. (°F)
End
Time, min
Time (min) Figure 12.6 Summary graphical results of FEA solution, transient time history thermal
response for tape pack with aluminum hub at midpack location (node 369).
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Transient HeatTransfer Problem: Tape Pack Cooling Transient HeatTransfer Problem: Tape Pack Cooling
12.15
As a result of this transient heattransfer analysis, it was conservatively recommended that tape packs suspected of previous exposure to temperature extremes be allowed to stabilize to room temperature for a period of least 90 min before making new backup copies or restoring data from a previous recording. References 1. E. R. G. Eckert and Robert M. Drake, Jr., Heat and Mass Transfer, 2d ed., McGrawHill, 1959. 2. J. P. Holman, “ConductionConvection Systems,” in Heat Transfer, 5th ed., McGrawHill, 1981, chap. 29. 3. W. M. Rohsenow and J. P. Hartnett, Handbook of Heat Transfer, McGrawHill, 1973. 4. Donald R. Pitts and Leighton E. Sissom, Heat Transfer, 2d ed., Schaum’s Outline Series, McGrawHill, 1977. 5. M. P. Heisler, Transactions of the American Society of Mechanical Engineers (Trans. ASME), 69: 227 (1947).
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Transient HeatTransfer Problem: Tape Pack Cooling
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16
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Source: HeatTransfer Calculations
Chapter
13 Calculation of the Transient Response of a Roof to Diurnal Heat Load Variations
Jack M. Kleinfeld Kleinfeld Technical Services, Inc. Bronx, New York
Background The calculations presented here are for the transient heat transfer of a roof under diurnal variation of environmental temperature changes and solar loading. They include an estimate of the impact of water in the roof, due to a leak, on the surface temperature of the roof. This problem is of particular interest for infrared (IR) thermographic surveys of roofs to detect leaks evidenced by the presence of moisture in the roof insulation under the membrane or surface of the roof. Infrared thermography is an established method1 for detecting the presence of wet insulation in roofs that depends on differences in the response of wet and dry roof areas to transient changes of the roof temperature in response to the daily loading cycle. In essence, the wet areas have a higher thermal mass than do the dry areas and lag behind the dry areas in exhibiting temperature changes as the environment changes. In practice, infrared thermographic roof moisture surveys are generally carried out at night while the roof is cooling after the day’s heating. Under those conditions, the wet areas, which are cooling more slowly, will show as relatively warmer than the dry areas.
13.1
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Calculation of the Transient Response of a Roof to Diurnal Heat Load Variations 13.2
Transient and Cyclic Calculations
The calculations as shown are typical of the methodology for generating a transient response to a variable input. The methodology shown is based on using finiteelement analysis (FEA) software to perform the calculations. The presentation is kept as generic as possible, but the work shown used Cosmos/M software and may have some details specific to it. This chapter is not, however, a software tutorial. Problem Statement Determine the temperature history and characteristics of a roof surface for dry and wet conditions of the roof ’s insulation, corresponding to a localized leak into the roof from the surface covering. Conditions selected correspond to August in the New York City area at 40◦ N latitude. Approach A small throughsection of the roof is modeled as representative of the entire roof. Those components which will not have an appreciable effect on the temperature history of the roof are omitted from the model. Where possible, the boundary conditions and properties of the materials are determined from literature sources. Where such are not readily available, estimated values are used for illustration. Separate calculations for the performance of dry and wet roof conditions are carried out, giving the behavioral characteristics of each, but not providing the interaction between wet and dry sections of the roof. The transient boundary conditions are developed as locally linearized approximations of the continuously varying actual conditions. Calculation Details Model
Figure 13.1 shows the model as developed for the analysis. It is a representation of the roof structure. Using an FEA approach means that it is set up on a computer. If only one condition is evaluated per calculation, the problem is, in fact, only a onedimensional (1D) analysis. It is easier, however, to use a twodimensional (2D) model, in terms of the setup in the modeling software. The varying conditions that are run are different moisture contents of the roof, in this case dry and, as an example, 20 percent water by weight in the insulation layer. Additional cases can be run for differing boundary conditions (BCs), such as time of year, or weather conditions. If a different type of roof is of interest, an additional model is needed.
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Calculation of the Transient Response of a Roof to Diurnal Heat Load Variations Transient Response of a Roof to Diurnal Heat Load
13.3
Figure 13.1 Outline
of model used in the FEA software, showing curves 1 to 7 and surfaces or regions 1 and 2.
In some cases, it is desirable to understand the interaction on a roof of wet and dry areas. This would call for a threedimensional (3D) model, since the lateral or horizontal heat transfer between the wet and dry areas, potentially of different sizes, is now of interest. Such a 3D model would be set up with both wet and dry areas designated and run under otherwise similar boundary conditions. The diffusion of the thermal signature or surface temperature profile between the wet and dry areas could then be determined. Since the approach is essentially identical for the 2D and 3D models, only the 2D model, evaluated for two roof conditions, is presented here. The model as configured represents a 2 in. thickness of insulation on a 2in.thick concrete roof deck. The roof membrane is not modeled, since
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Calculation of the Transient Response of a Roof to Diurnal Heat Load Variations 13.4
Transient and Cyclic Calculations
it is assumed to have negligible thermal properties. However, its emissivity is used for the radiational behavior of the top surface of the roof. Assumptions in the model
The interface between the insulation and the roof deck is assumed to have full thermal contact between the materials. This is represented by curve 3 in Fig. 13.1. No air gap between them is considered. If the roof being modeled were known to have air gaps due to unevenness of the deck or of the insulation installation, then either an additional region for the air between the insulation and the deck or a contact gap in the FEA software would be used to represent them. The water in the wet insulation is assumed to be uniformly distributed in the insulation and to not puddle either on top or under the insulation. If puddling were suspected, additional regions in the model would be needed. Boundary conditions
The boundary conditions for the calculation occur at the top and bottom surfaces of the roof. These are represented by curves 2 and 1, respectively, in Fig. 13.1. The sides of the modeled domain, which would be connected to more roof, are treated as adiabatic. These are represented by curves 4 to 7 in Fig. 13.1. This is equivalent to assuming that the adjacent portions of the roof are the same as the modeled portion and that the roof is much larger than 2 in. wide. It also assumes that the portion modeled is not at the edge of the roof or near a structure on the roof. The boundary conditions on the top surface of the roof account for direct solar heating of the roof, radiational transfer between the roof and the sky, and convective transfer between the roof and the surrounding air. While the solar heating is, in fact, a radiation process, it can be described and modeled as a heat flux applied to the roof. This has a significant advantage in FEA modeling, since it removes one level of nonlinearity from the model, making it solve more quickly, and also allows the radiation BC to be used to describe the interchange with the sky. The software used would not allow two radiation BCs at the same location. In addition, the solar loading information needed is available in the literature in the form of heat flux data. The boundary conditions on the bottom surface of the roof are primarily convection to the inside air temperature and radiation to the building interior. Radiation is often approximated as a linear phenomenon over a short temperature span and represented as an additional convectivestyle heattransfer coefficient in the convective equation. This simplifies solutions for hand calculations, but is not necessary or desirable for the system as modeled in an FEA package.
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Calculation of the Transient Response of a Roof to Diurnal Heat Load Variations Transient Response of a Roof to Diurnal Heat Load
TABLE 13.1
Solar Flux
Time, h
Flux from ASHRAE, Btu/h ft2
0 5.5 6 7 8 9 10 11 12 13 14 15 16 17 18 18.5 24
12 62 122 174 214 239 247 239 214 174 122 62 12
13.5
Flux as extended and used with 0.915 absorptivity, Btu/h ft2 0 0 10.98 56.73 111.63 159.21 195.81 218.685 226.005 218.685 195.81 159.21 111.63 56.73 10.98 0 0
The solar flux or solar heat gain can be found in the literature. ASHRAE2 presents it for average cloudless days in tabular form for both vertical and horizontal surfaces, as well as for normal incident radiation. It is tabulated for each month of the year at different latitudes. A single set of values was used here, shown in Table 13.1, for August at 40◦ N latitude. The values were adjusted by a factor of 0.915 representing the average emissivity or absorptivity of roofingtype materials.3 The appropriate value for this and for the degree of cloudiness should be used for a particular situation. In addition, since the software interpolates the flux, and other values, from a linearized tabular curve, it is necessary to extend the table both down to zero flux and out to zero time and 24 h, in order to cover the entire calculational range. This was done by adding points with zero value at 0 and 24 h, representing the end of the day, and at 0530 and 1830 h, representing 1/2 h beyond the end of the data. The extended curve is shown in Fig. 13.2 and its constituent data, in Table 13.1. The roof surface exchanges energy with the surroundings by radiation to the sky, provided there are no significant physical objects in the way, as was assumed. This radiation BC requires a value for the temperature of the sky. Clear skies are actually quite cold, especially at night. In the absence of any hard data, values were assumed for the purposes of this demonstration calculation. They are shown in Table 13.2 and Fig. 13.3. An emissivity value for the roof was used for the radiation BC as it was used for the solar flux.
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Calculation of the Transient Response of a Roof to Diurnal Heat Load Variations 13.6
Transient and Cyclic Calculations
250 August, horizontal surface, 40°N latitude
Flux, Btu/h ft 2
200
150
100
50
0 0
Figure 13.2
2
4
6
8
10
12 14 Time, h
16
18
20
22
24
Solar flux boundary condition.
The interior or underside surface of the roof also needs to have a radiation BC to represent its interchange with the building’s interior. The building was assumed to be at a constant 70◦ F for both this BC and the convective transfer BC. An emissivity of 0.92, corresponding to flatfinish paint, was used for the radiation BC. In many ways the convective BCs, on both the top and bottom surfaces, are the hardest to specify. The heattransfer coefficients are not readily determined. The situation is complicated by the fact, which can be seen from the results, that the convective transport reverses direction over the course of the day. Because of the radiational cooling to the night sky, the roof temperatures drop below the surrounding air temperature. This is true on both the top and bottom surfaces of the roof. The convection can potentially be either free or forced. If free convection pertains to the situation being modeled, determination of the heattransfer coefficient is complicated by the reversal of the heat flow. For a downwardfacing hot surface, ASHRAE4 points out that free convection should not actually take place, since the hotter, lighter layer is above the cooler bulk. They report that some, however, does occur, and
TABLE 13.2
Sky Temperature
Time, h
Sky temperature, ◦ F
0 6 7 18 19 24
25 25 55 55 25 25
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Calculation of the Transient Response of a Roof to Diurnal Heat Load Variations Transient Response of a Roof to Diurnal Heat Load
13.7
90
Temperature, °F
80 70 60
Air temperature, °F Sky temperature, °F
50 40 30 20 0
2
Figure 13.3
4
6
8
10
12 14 Time, h
16
18
20
22
24
Environmental temperatures used for BCs.
suggest that the coefficient is less than half that of the coefficient for the same case but with the temperatures reversed. Perry et al.5 and ASHRAE4 present similar correlations for these cases. A temperature and therefore timedependent formulation of the convective heattransfer coefficient should be used for the most accuracy. This has not been done for this calculation. It becomes difficult to evaluate the various coefficients required. For the top of the roof, where there is the complication of a varying air temperature, it also would be difficult to code the heattransfer coefficient into the FEA software, since the formulations generally use the temperature difference between the surface and the bulk air. Approximate values for the heattransfer coefficients were estimated for free convection. Values of 1 Btu/h ft2◦ F for the exterior and 0.8 Btu/h ft2◦ F for the interior were selected. Were the system considered to be in forced convection, a less surfacetemperaturedependent solution should be applicable. The air temperatures assumed for the outside air are shown in Table 13.3 and Fig. 13.3. If data of hourly air temperatures were available, they would be used here.
TABLE 13.3
Air Temperature
Time, h
Air temperature, ◦ F
0 7 12 15 18 24
70 73 83 85 73 70
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Calculation of the Transient Response of a Roof to Diurnal Heat Load Variations 13.8
Transient and Cyclic Calculations
The 24h cycle temperatures and fluxes were repeated without change for the second 24h cycle in the calculations. This simulates two identical days. The first, which gives results quite similar to those for the second except for some variation in the first few hours, was treated as a breakin period and would not be used for the final results. Properties
For a transient thermal solution, the heat capacity and the density of the materials are needed in addition to the thermal conductivity that is necessary for a steadystate solution. Properties for dry materials are relatively available in the literature. Bench testing can also be used to obtain them if necessary. The properties of wet materials, such as wet insulation, are not readily available. They can be approximated for low to moderate water contents by Wet material property = dry material property + (fraction water) · (water material property) The approach is predicated on the dry material having open pores that are filled with the water as the water content increases, essentially adding the properties of the water in parallel to those of the dry material. For insulation materials, such as mineral wool or fiberglass batts, this should be a good approximation. This approximation fails completely at 100 percent water, where it predicts that water’s properties are equal to the dry material’s plus the water’s property. It also should not be used at high water concentrations. For the current example, with 20 percent water content, the approximation should be quite good. Initial conditions
An approximate, uniform temperature for the entire roof was set before beginning the calculations. In order to reduce the effect of errors in this value, the simulation was run for a 48h period and only the second 24h period would be reported as the result. The full 48h calculation results are generally shown here. Time steps
Selection of the time increments to use in calculating the behavior of the roof will affect the run time of the model and its accuracy. Short time steps will cause long run times, with excessively short steps not necessarily improving the accuracy of the results. Long time steps will cause inaccuracies in the calculations. Another consideration in selecting
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13.9
time steps is the size of the finite elements making up the model. The times must be such that changes in the far boundary of the elements occur over the time interval selected. If this is not the case, that is, if the elements are too large or the time steps too small, the calculations will fail and the software will not converge. One indication of whether the time steps are sufficiently short is the shape and smoothness of the results. An explicit check is to run the simulation with a shorter or longer time step, varying by about 20 percent, to see if the results change significantly. If not, then the time step is sufficiently small. The step used for the calculations here was 900 s (15 min).
Calculation Results General
The primary results of the calculations are the temperature history of the surface of the roof. Temperature histories of other places on the roof, such as the ceiling surface or the top of the concrete deck, are also of interest. The results from a given run of the FEA model provides values for a particular set of roof conditions and environmental conditions. As discussed above, a larger model could incorporate more than one roof condition, for example, by including both wet and dry areas in the same model. The comparison between the wet and dry results in the approach used is made by taking results from two runs. Additional information can also be obtained using the FEA results. For example, once the conditions for a particular time are obtained, the relative importance of the various heattransfer mechanisms, solar flux, convection, and radiation can be compared. This can be done as a hand calculation using the temperature condition at the time selected and the boundary conditions specific for that time. Figures 13.4 and 13.5 show the temperature histories of the top and bottom, or exterior and interior surfaces, of the roof for dry and 20 percent wet conditions. They are plotted on the same scale. The wet roof exterior surface exhibits a peak temperature noticeably lower than that of the dry roof. The exterior surface becomes colder than the interior surface at night and hotter during the day. Both the exterior and interior surfaces drop below their local ambient air temperatures at night, driven by the radiational exchange between the exterior surface and the cold night sky. The temperature history of the top of the concrete deck, under the insulation, is not plotted on these two figures because it is almost identical to the interior surface temperature. The peak temperatures of the exterior and interior surfaces occur at different times. The figures show the full 48h calculation. The first few hours differ from the comparable time period in the second day’s results.
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Calculation of the Transient Response of a Roof to Diurnal Heat Load Variations 13.10
Transient and Cyclic Calculations
150 140 130
Temperature, °F
120 110 T exterior, °F T interior, °F
100 90 80 70 60 50
0
4
8
12
16
20
24 28 Time, h
32
36
40
44
48
Temperature history of dry roof surfaces.
Figure 13.4
In this case, the differences are not large and the effects of the initial conditions are overcome by a single 24h cycle of calculation; this is evident on comparison of the endpoints of the 48h calculation with the beginning of the second 24h cycle. Since they match, additional cycles are not necessary to evaluate the performance. For deeper, or more massive, structures, additional cycles would be needed to overcome the
150 140 130
Temperature, °F
120 110 T exterior, °F T interior, °F
100 90 80 70 60 50 0
Figure 13.5
4
8
12
16
20
24 28 Time, h
32
36
40
44
48
Temperature history of 20 percent wet roof surfaces.
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Calculation of the Transient Response of a Roof to Diurnal Heat Load Variations Transient Response of a Roof to Diurnal Heat Load
13.11
Temperature difference, wet roof − dry roof, °F
6
4
2
0 0
4
8
12
16
20
24
−2 −4 −6 −8 Time, second cycle, h
Figure 13.6
Surface temperature contrast between 20 percent wet and dry roof areas.
impact of the initial conditions selected. If daytoday conditions vary widely, some consideration would have to be given to running a set of calculations with varying conditions over multiple day cycles. For application to IR surveys, a key value is the contrast between wet and dry areas on the exterior surface, which is the one usually surveyed. Figure 13.6 shows the contrast between the wet and dry exterior surfaces. At night the wet areas are warmer than the dry areas. This reverses during the day, when the wet areas are cooler. The peak differences do not occur at the same time as the peaks in the temperature histories, but are offset by several hours. This indicates that the difference is maximized during the steepest transient of the roof. Impact of variations in BCs and conditions
An issue raised above in the discussion of BC selection is the value for the convective heattransfer coefficient. Values were estimated on the basis of natural convection, but the confidence level in the values is not high. The model can be used to evaluate the sensitivity of the results to the heattransfer coefficient values, as well as to other values. Results were generated for an additional case where the heattransfer coefficient was increased by a factor of 1.25 for both the exterior and interior surfaces of a dry roof. The resulting temperature history is shown on Fig. 13.7, which includes the base case as well as the air temperature above the roof. There is a noticeable impact on the results. Where the
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Calculation of the Transient Response of a Roof to Diurnal Heat Load Variations 13.12
Transient and Cyclic Calculations
150 140 130
Temperature, °F
120 110
T exterior, °F T exterior 1.25h, °F T interior, °F T interior 1.25h, °F Air T, °F
100 90 80 70 60 50
0
Figure 13.7
4
8
12
16
20
24 28 Time, h
32
36
40
44
48
Sensitivity to assumed heattransfer coefficient.
primary interest is the ability to find wet versus dry roof areas, the impact will be less important, since the same revised coefficients would be used for both the wet and dry cases. If the primary interest in performing the work is the actual temperatures of the roof, for example, if the material life were being evaluated, these levels of differences could be important. Evaluations of differing conditions, such as time of year, cloud cover, location, roof surface emissivity, interior temperatures, roof construction, and other levels of water content, could be done in exactly the same manner as the work shown. For most of these, the changes would be in the BCs. For changes in the interior temperature, if, for example the building used a programmable thermostat, a time curve could be introduced. For changes in the roof surface, for example, from a black to a white surface, the solar flux would be adjusted by the solar emissivity of the surface. The white surface would probably have the same emissivity as the black in the infrared wavelength region, where the transfer by radiation occurs, so the emissivity for the radiation BC would not be changed. For a silver surface, with lower emissivity at both solar wavelengths and infrared wavelengths, the values of emissivity for the radiation BC would also need to be changed. Changes in the water content of the roof would require changes in the physical properties, as outlined above. Changes in the roof construction would require a new model representing the new roof design, but the BCs would probably not change with the possible exception of emissivity.
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Calculation of the Transient Response of a Roof to Diurnal Heat Load Variations Transient Response of a Roof to Diurnal Heat Load
13.13
References 1. ASTM C115390, Standard Practice for the Location of Wet Insulation in Roofing Systems Using Infrared Imaging, American Society for Testing and Materials, Philadelphia, 1990. 2. 1993 Fundamentals Handbook (IP ed.), American Society of Heating, Refrigerating, and AirConditioning Engineers, Atlanta, 1993, chap. 27, tables 12–18. 3. Ibid., chap. 3, table 3. 4. Ibid., chap. 3. 5. Perry, Robert H., et al., Chemical Engineer’s Handbook, 4th ed., McGrawHill, New York, 1963, chap. 10.
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14
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Source: HeatTransfer Calculations
Chapter
14 Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven
Thomas M. Lawrence Department of Biological and Agricultural Engineering Driftmier Engineering Center University of Georgia Athens, Georgia
Scott Adams Paint Engineering Ford Motor Company Vehicle Operations General Office Allen Park, Michigan
Introduction Production of automobiles, light trucks, and sport utility vehicles currently (as of 2005) averages around 12 million vehicles per year in the United States. Each of these vehicles undergoes a painting process that involves a series of paint application steps and paint dryingcuring ovens. Since automotive assembly plants are automated systems with continuousmotion carriers, the ovens are designed as long enclosed spaces that control the heating process in both duration and temperature. Quality control concerns require strict control of the heating rate, 14.1
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven 14.2
Transient and Cyclic Calculations
maximum temperature exposure, and time duration for exposure above the minimum cure temperature. Since a typical vehicle body is a complex shaped structure with varying part thicknesses and orientations, it is a difficult job to ensure that each portion of the body is exposed to heating conditions that will meet the strict quality control standards. Each assembly plant paint shop will regularly send instrumented vehicle bodies or similar devices to determine the actual air and body temperatures at key points over time while the body travels through an oven. A typical automotive plant paint oven will be 100 to 130 m (300 to 400 ft) in length. The goals of this process heating are first to bring the painted vehicle body up from room temperature to a paint cure point of around 150 to 160◦ C (300 to 325◦ F), and then hold the body at the cure temperature for a period of time ranging from approximately 15 to 25 min. A quality curing job is achieved if the body is exposed to a combination of time at temperature that meets the paint manufacturer’s criteria. Because of the wide range in thickness of a vehicle body part, orientation (horizontal, vertical), distance from the oven walls and ceiling, and the potential for blockage of exposure from other parts of the body, it is a difficult task to ensure that all portions of the painted surfaces reach the proper cure temperature and time. There is potential for dirt being trapped in the wet surface when the freshly painted body first enters the oven. Therefore, a typical oven will first heat the painted vehicle body in zones with minimal air movement to avoid stirring up any dirt in the area. These heatup zones use radiant heating from either infrared heating lamps or radiant panels that are heated from the backside with hot air. Once the body has been in the oven for a while and the outer surface of the paint has dried such that the risk of dirt entrapment is minimal, the oven will typically employ convection—or a combination of both radiant and convection—heating. Temperature
Maximum allowable temperature Minimum cure temperature
Time Radiant heatup zone
Convection hold zone
Body cooler
Plot of typical temperature versus time for representative point on a painted vehicle body.
Figure 14.1
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven Transient Heating of a Painted Vehicle Body Panel
TABLE 14.1
Typical Paint Shop Oven Specifications
Zone∗ Length, m (ft) Temperature, ◦ C (◦ F) Time, min Burner capacity, kW (106 Btu/h) Airflow, m3 /s (cfm) ∗
14.3
1
2
3
4
5
6
12.2 (40) 107 (225) 4 879 (3)
12.2 (40) 171 (340) 4 879 (3)
12.2 (40) 160 (320) 4 879 (3)
15.2 (50) 154 (310) 5 732 (2.5)
18.3 (60) 152 (305) 6 732 (2.5)
21.3 (70) 152 (305) 7 732 (2.5)
14.16 (30,000)
14.16 (30,000)
14.16 (30,000)
16.52 (35,000)
14.16 (30,000)
14.16 (30,000)
Zones 1 to 3 are radiant heatup; zones 4 to 6 are convection hold zones.
A representative plot of the temperature versus time (and hence distance traveled in the oven) is given in Fig. 14.1. Key technical specifications for a typical oven are listed as an example in Table 14.1. A modern car or truck body is a complex structure with a combination of thin and thicker metal elements. The orientation of a particular location on the vehicle body can greatly influence the heating exposure and hence the resulting paint temperature. Thus, there are a myriad of potential variables and combinations that must be accounted for when attempting to analyze the transient temperatures for a typical body as it passes through an oven. The paint shop will also occasionally send a carrier loaded with replacement parts through the painting booth and curing oven; for example, see the parts carrier shown in Fig. 14.2. These
Figure 14.2
Photo of typical spareparts carrier.
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven 14.4
Transient and Cyclic Calculations
parts carriers are exposed to the same temperatures as the regular vehicle bodies since they pass through the oven on the same conveyors. Example Description Metal thicknesses and heattransfer effects vary by location on a vehicle body. As a relatively simple example, this problem will analyze the heatup of a painted door part after it first enters the oven and passes through the initial radiant heatup zones. The problem will consider how a simplified transient heattransfer analysis can be applied to the lower section of a painted part below the window cutout area shown in Fig. 14.2. This section is essentially vertically oriented on the parts carrier. A simplified cross section of the oven in this example is given in Fig. 14.3. Tables 14.2 and 14.3, along with Fig. 14.4, contain a summary of the key parameters and assumptions made for this analysis.
Unheated ceiling High emissivity coating on walls
Radiant panel
Hot air circulates inside panels on oven sidewalls, radiating heat to painted body
Heated floor plenum
Simplified cross section of typical oven zone using radiant heat.
Figure 14.3
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven Transient Heating of a Painted Vehicle Body Panel
TABLE 14.2
14.5
Assumptions and Design Conditions Used in Example Analysis Parameter
Metric
English
Oven air temperature, zone 1 Oven air temperature, zone 2 Oven air temperature, zone 3 Air temperature behind radiant panel, zone 1 Air temperature behind radiant panel, zone 2 Air temperature behind radiant panel, zone 3 Estimated radiant panel wall temperature, zone 1 Estimated radiant panel wall temperature, zone 2 Estimated radiant panel wall temperature, zone 3 Residency time in each zone, min Radiant panel to zone air convection coefficient Radiant panel wall thickness Radiant panel wall conductivity Length of each zone Airflow behind radiant panel Initial body temperature Vehicle body material properties: thermal conductivity k Vehicle body material properties: thermal diffusivity Painted part panel thickness Painted part thermal emissivity Painted part thermal conductivity Painted part specific heat Painted part material density
107◦ C 171◦ C 160◦ C 152◦ C 216◦ C 204◦ C 131◦ C
225◦ F 340◦ F 320◦ F 305◦ F 420◦ F 400◦ F 268◦ F
194◦ C
381◦ F
183◦ C
361◦ F
4 min 5.1 W/m2 · K
4 min 0.9 Btu/h · ft2 · F
3 mm 53.5 W/m · K 12.2 m 14.16 m3 /s 27◦ C 60 W/m · K
1/8 in. 30.9 Btu/h · ft · F 40 ft 30,000 cfm 80◦ F 34.7 Btu/h · ft · F
1.7 × 10−5 m2 /s
0.66 ft2 /h
2.4 mm 0.9 53.5 W/m · K 0.485 kJ/kg · K 7850 kg/m3
3/32 in.
0.9 30.9 Btu/h · ft · F 0.12 Btu/lbm · F 490 lbm /ft3
This example will estimate the transient heating of the painted panel as it proceeds through the first three radiant heating zones of a typical oven. The residency time of the panel in each zone is 4 min, so the total time for this transient simulation will be 12 min. The heattransfer paths for this problem are summarized graphically in Fig. 14.5. The problem analysis consists of two basic steps: 1. Determination of the heattransfer characteristics, rate equations, and reasonable assumptions for the paths shown in Fig. 14.5. TABLE 14.3
Thermal Properties of Heated Air Inside Radiant Panel,
Zone 3 Parameter Kinematic viscosity Prandtl number Pr Thermal conductivity k
Metric 3.55 × 10−5
English m2 /s
0.697 0.03805 W/mK
3.77 × 10−4 ft2 /s 0.697 0.022 Btu/hft◦ R
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven 14.6
Transient and Cyclic Calculations
Outer painted surface panel Thickness = 2.4 mm (3/32 in.) ε = 0.9
Internal air gap between outer painted panel and inner panel (assumed adiabatic)
Doublewall door panel Figure 14.4
Painted door panel dimensions and key parameters.
2. Combining these into a transient analysis model to simulate the painted part temperature during the initial heating. The heattransfer characteristics, coefficients, overall rate equations, and assumptions will be analyzed individually in the following sections. Heat Transfer on Panel Inside Wall Surface Heat transfer on the inside surface of the radiant panel is between the heated air supplied by the oven burner and the inner panel wall. The first step is to estimate the convection heattransfer coefficient. To do so, the airflow velocity must be determined. Radiant wall panel Internal air T
Convection on inside wall
Panel internal air T Figure 14.5
Radiation exchange with ceiling
Painted part
Radiation exchange between wall panel and painted part surface Convection on outside wall Conduction through panel wall
Convection on painted part surface
Backside of painted part assumed adiabatic
Radiation exchange between heated floor and painted part surface
Heattransfer paths analyzed in this example.
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven Transient Heating of a Painted Vehicle Body Panel
14.7
From the information in Table 14.2, a total airflow of 14.16 m3 /s [30,000 ft3 /min (cfm)] is provided to each radiant zone. The zones are 12.2 m (40 ft) in length, with radiant panels (and thus airflow) on both sides of the oven. Therefore, the total airflow per unit panel length along the oven is 3 3 ˙ = total zone flow/2 = 14.16 m /s = 0.58 m /s 375 cfm Vol zone length 2 · (12.2 m) m ft (14.1) Various airflow ducting designs may be employed at different assembly plants. For this example, we will consider a case where the radiant panels are subdivided into sections every 1.5 m (5 ft) of length along the oven path. A diagram of this duct section and dimensions is shown in Fig. 14.6. Therefore, the total airflow in each section is 1.5 × 0.58 = 0.88 m3 /s (1875 cfm). On the basis of the heated air temperature inside the radiant panel in zone 3 of 204◦ C (400◦ F), the air properties listed in Table 14.3 were determined. A similar analysis process can be followed to compute the corresponding heattransfer characteristics for zones 1 and 2. The next step is to determine the Reynolds number for airflow inside the radiant panel. Since the flow path is a rectangular duct, the Reynolds number will be based on the hydraulic diameter, or Dh =
4 · 1.5 m · 0.3 m 4· A = = 0.5 m (1.65 ft) P 3.6 m
(14.2)
1.5 m (5 ft) Figure 14.6
0.3 m (1 ft)
Radiant panel di
mensions.
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven 14.8
Transient and Cyclic Calculations
The airflow velocity inside the radiant panel ducting is V=
total airflow 0.88 m3 /s = 1.96 m/s 6.25 ft/s = A 1.5 m 0.3 m
and the resulting Reynolds number is 1.96 m/s 0.5 m VDh Re Dh = = 27,606 = 3.55 × 10−5 m2 /s
(14.3)
(14.4)
Thus, the heated airflow inside the radiant panel is turbulent since the Reynolds number exceeds the generally accepted transition level of around 2300 for internal flow. The convective heattransfer coefficient on the internal panel walls can be estimated using the DittusBoelter equation [Eq. (8.60) in Incropera and DeWitt (2002)] with the airflow being “cooled.” This equation states that 4/5
Nu Dh = 0.023 Re Dh Prn
where
n = 0.3
(14.5)
Applying Eq. (14.5) to this problem gives an estimate of the Nusselt number of Nu Dh =
hDh = 0.023(27,606)4/5 (0.697)0.3 = 73.7 k
(14.6)
with a resulting convection coefficient 73.7 · 0.03805 W/m · K Nu Dh k hi = = Dh 0.5 m = 5.6 W/m2 · K [1.0 Btu/h · ft2 · ◦ R]
(14.7)
Convective Heat Transfer on Outer Wall Surface Estimating the convection heattransfer coefficient on the outer surface of the radiant panel wall (facing the oven interior) is a difficult process. The air inside the oven, although ideally “still,” will have motion due to the movement of vehicle bodies inside the oven, slight pressure differences between zones, temperature gradients, and other factors. However, one can safely assume that the convection heattransfer coefficient on the outer wall is less than on the inside surface. The convective heattransfer rate on the outer radiant panel wall would also be expected to be slightly higher than for purely free (natural) convection; thus bounds
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven Transient Heating of a Painted Vehicle Body Panel
14.9
for the convection coefficient range are known. This example will assume a convection coefficient value of ho = 5.1 W/m2 K (0.9 Btu/hft2 ◦ R) for the outer panel wall. Heat Transfer through Panel Wall The conduction resistance through the panel wall is computed by R=
L 0.0015 m = 2.8 × 10−5 K/W (3.4 × 10−4 h◦ R/Btu) = kA 53.5 W/m · K 1 m2 (14.8)
Therefore, the resistance to heat flow through the radiant panel wall is negligible compared to the convective heat transfer on both the inner and outer wall surfaces. Estimation of Radiant Panel Outer Wall Surface Temperature The two convection coefficients can be combined to obtain the overall thermal resistance for heat transfer between the heated air inside the radiant panel and the oven air, as shown below based on a unit panel area: Rt =
1 1 = 2 1 1 1 m ·K 1 m2 · K + + hi A ho A (5.6 W)1 m2 (5.1 W)1 m2
= 2.7 K/W (0.47 h◦ R/Btu)
(14.9)
An estimate of the outer panel wall temperature is necessary for use in calculation of the radiation heat transfer to the painted part panel for the transient modeling. Heat transfer on the internal wall from the heated air supply will be lost on the outer wall by both convection and radiation. Computing the energy lost through radiation is a difficult task to simplify, since it depends on the effective surrounding temperature as viewed from the outer panel wall. The effective temperature of the surroundings oscillates as the painted panels or vehicle bodies roll past through the oven and down the length of the oven as the panel heats up. For a good portion of time, the radiant panel views the panel on the opposite side of the oven at essentially the same temperature. Therefore, for this simplified analysis to estimate the panel outer wall temperature, we will first ignore the effect of radiation. The wall temperature is first estimated using the conditions in zone 3, where the oven air temperature is 160◦ C, as noted in Table 14.3. The
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven 14.10
Transient and Cyclic Calculations
outer panel wall temperature can be estimated from the following relation Qto wall = Qfrom wall hi A (Theated air − Twall ) = ho A (Twall − Toven ) 5.6 or
(14.10)
W W (1 m2 ) 204◦ C − Twall = 5.1 2 (1 m2 )(Twall − 160◦ C) 2 m K m K Twall = 183◦ C (361◦ F) (14.11)
Using a similar analysis, the radiant panel wall temperatures in zones 1 and 2 are estimated to be 131◦ C (268◦ F) and 194◦ C (381◦ F), respectively. Table 14.3 includes a listing of the estimated radiant wall panel temperatures for the three zones. View Factors for Radiation Heat Transfer with Vertical Painted Panel From an examination of the diagram in Figs. 14.3 and 14.5, three potential radiation heattransfer paths exist with a vertically mounted painted part. These are with the oven radiant wall panels, plus the floor and ceiling. For the view factor to the floor, the panel “sees” mostly the section of floor that is heated, and it can be assumed that the entire floor viewed is heated without introducing any significant error. The oven is also much longer than one individual painted part; therefore, the view factor relationship between the painted part and the oven wall, ceiling, and floor can be considered to be twodimensional. The typical oven cross section given in Fig. 14.3 shows radiant panel walls that are angled slightly to provide better radiation heattransfer exposure for the painted body surfaces. In computing the radiation view factors, the radiant panel walls can be considered flat vertical plates facing the painted part as it moves through the oven, as shown in Fig. 14.7. The view factor between the vertical section of interest on the painted part and the unheated oven ceiling can be determined from the view factor relations for perpendicular plates with a common edge, as shown in Fig. 14.7b. The view factor relation is as follows: F1→3 = F(1+2)→3 − F2→3
(14.12)
The twodimensional view factor for perpendicular plates with a common edge is found using the method outlined in sec. 13.1 of Incropera and DeWitt (2002): 1 + (wi /w j ) − 1 + (wi /w j )2 Fi→ j = (14.13) 2
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven Transient Heating of a Painted Vehicle Body Panel
14.11
Radiant heating panel
1.8 m (6 ft) Painted part area not studied 0.75 m (2.5 ft)
Painted part area studied: height = 1.2 m (4 ft) 1.5 m (5 ft) (a)
0.75 m Section 3 (2.5 ft) Imaginary section above area studied to ceiling: height = 1.8 m (6 ft) Section 2 Painted part area studied: height = 1.2 m (4 ft) Section 1
Radiation heattransfer view factor estimation (not drawn to scale): (a) dimensions; (b) panel ceiling view factor determination.
Figure 14.7
(b)
In this equation, w j and wi are the lengths of sides i and j, respectively. Using the information in Fig. 14.7b, the view factors are computed as follows: 1 + (3.0/0.75) − 1 + (3.0/0.75)2 F(1+2)→3 = = 0.44 (14.14) 2 1 + (1.8/0.75) − 1 + (1.8/0.75)2 F2→3 = = 0.40 (14.15) 2 Fpanel→ceiling = F1→3 = F(1+2)→3 − F2→3 = 0.44 − 0.40 = 0.04
(14.16)
Radiation HeatTransfer Calculations If the heated floor and sidewall panel surfaces are assumed to be at the same temperature (the wall temperatures computed above), they can be lumped together in the radiation heattransfer analysis, and this is the method chosen for this simple example. Thus, the radiation heat transfer to the painted panel in the example can be summarized by Qradiation = Qrad, wall + Qrad, floor + Qrad, ceiling
(14.17)
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven 14.12
Transient and Cyclic Calculations
One must next decide how to treat the radiation heattransfer exchange between the painted part and the oven wall surfaces. Optimal heat transfer will be obtained when the oven surfaces are as near a blackbody as possible and are designed as such. The walls can be assumed to be essentially a blackbody for the purposes of this example, thus simplifying the analysis so that the net radiation heat transfer to the panel is calculated by 4 4 Qrad, ceiling = Fpanelceiling εpanel Apanel Tceiling (14.18) − Tpanel 4 4 Qrad, wall + Qrad, floor = (1 − Fpanelceiling ) εpanel Apanel Twall − Tpanel (14.19) A more detailed heattransfer analysis would consider this situation as an enclosure with multiple diffuse, gray surfaces, using the known (or estimated) emissivity values for the painted part, walls, ceiling, and floor surfaces to compute the net radiation exchange with the painted part. Convection Heat Transfer to Painted Part The other mode of heat transfer involved is convection between the painted part and the oven air. In the radiant heating zone, air movement is not desired so as to avoid potential dirt contamination in the wet paint. There will be a small amount of air motion across the painted surface due to the movement of the painted part carrier through the oven and various pressure imbalances that may exist. The convective heattransfer coefficient will be relatively small, and the assumed value used for the outer radiant panel wall (h = 5.1 W/m2 K) should also be a reasonable assumption for the painted part surface. The convective heat transfer to be used in the example calculation is given in the following equation: Qconvection = hApanel (Toven − Tpanel )
(14.20)
Transient Panel Temperature Analysis The transient modeling will track the panel temperature starting from ambient during the 12 min it needs to pass through the heatup section. The painted panel is actually a doublewall structure, with the backside of the panel assumed to be an adiabatic surface (see Fig. 14.4). Recognize that there will be some heat transfer through the panel inner wall and into the air gap, but this assumption simplifies greatly the rough level calculations in this example.
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven Transient Heating of a Painted Vehicle Body Panel
14.13
A check should be made to see if the panel can be treated at a uniform temperature through the thickness of the part. Since this panel is composed of relatively thin [2.4 mm (3/32 in.)] material over much of the part, this would be a reasonable expectation. To check this, calculate the Biot number, which is used to determine whether a lumpedcapacitance transient heat transfer is applicable. The Biot number is calculated using Bi =
hLc k
(14.21)
where Lc is the characteristic length. The characteristic length used should be the entire thickness of the painted metal (2.4 mm), since the backside of the painted surface is assumed to be adiabatic in this analysis (see Fig. 14.4). The Biot number is therefore calculated as Bi =
(5.1 W/m2 · K)0.0024 m = 2.3 × 10−4 53.5 W/m · K
(14.22)
Since the Biot number is much less than 0.1, it is proper to assume that the temperature distribution through the thickness of the metal can be considered approximately uniform. Calculation of the transient temperature will be conducted for this example using an explicit (timemarching) process with time steps of 30 s. The choice of time step made was for convenience, but the engineer should always check for potential stability problems in an explicit numerical solution. For example, sec. 5.9 in Ref. 1 describes the stability criteria for a finitedifference method solution. The basic equation used for the transient simulation analysis is net energy into panel = rate of change of panel temperature, or Qnet = mC p
T t
Qrad, wall + Qrad, floor + Qrad, ceiling + Qconvection = mC p Tn+1 = Tn +
(14.23) Tn+1 − Tn time step, s
time step (Qrad, wall + Qrad, floor + Qrad, ceiling + Qconvection ) mC p (14.24)
where Tn+1 = panel temperature at end of time step Tn = panel temperature at beginning of time step m = panel mass = V = A · thickness
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven 14.14
Transient and Cyclic Calculations
180
T = 156.8 °C @ 12 min
Panel temperature (°C)
160 140 120 100
Zone 3
80 Zone 2
60 40
Zone 1
20 0 0
1
2
3
4
5 6 Time (min)
7
8
9
10
11
Predicted painted panel temperatures (metric units).
Figure 14.8
Equations (14.18) to (14.20) and (14.24) were incorporated into a spreadsheet solution of the estimated painted panel temperature over time as it passes through the first three radiant heating zones. Calculation results for the temperatures are given in Figs. 14.8 and 14.9 for the simulation in metric [International System (SI)] and English units, respectively. The temperature values differ slightly because of the rounding differences in the unit conversions for heattransfer coefficients, panel thickness, and other variables. A shift in the temperature trends occurs as the painted part enters the next zone at the 4and 8min marks. Figures 14.10 and 14.11 show a breakdown between
Panel temperature (°F)
350
T = 312 °F @ 12 min
300 250 200
Zone 3
150 Zone 2
100 Zone 1
50 0 0 Figure 14.9
1
2
3
4
5 6 Time (min)
7
8
9
10
11
Predicted painted panel temperatures (English units).
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven Transient Heating of a Painted Vehicle Body Panel
14.15
100 Convection (kJ) Radiation total (kJ)
90
Heat transfer (kJ)
80 70
Zone 1
60 Zone 3
50
Zone 2
40 30 20 10 0 0
Figure 14.10
1
2
3
4
5 6 7 Time (min)
8
9
10
11
Panel heattransfer breakdown (metric units).
convection and total radiation heat transfer to the panel over time for the metric and English units calculations, respectively. Summary This example problem used a typical industrial scenario to demonstrate the application of convection and radiation heat transfer. Several 9 Convection (Btu) Radiation total (Btu)
8 Heat transfer (Btu)
7 6 Zone 1
5 Zone 3
4
Zone 2
3 2 1 0 0
Figure 14.11
1
2
3
4
5 6 Time (min)
7
8
9
10
11
Panel heattransfer breakdown (English units).
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Transient Heating of a Painted Vehicle Body Panel in an Automobile Assembly Plant Paint Shop Oven 14.16
Transient and Cyclic Calculations
simplifying assumptions, and the logic behind these choices, were made to keep the analysis to a reasonable level of effort. The estimated panel temperatures are considered only initial guesses, and a more detailed modeling effort would be required to obtain more accurate results. Nomenclature A
Area
Bi
Biot number
Cp
Specific heat
Dh
Hydraulic diameter
h
Convection heattransfer coefficient
F
View factor
k
Thermal conductivity
Lc
Characteristic length
m
Mass
Nu
Nusselt number
P
Perimeter
Q
Heattransfer rate
R
Thermal resistance
Re
Reynolds number
t
Time
T
Temperature
V ˙ Vol
Air velocity
w
Length of side in view factor calculation
Volumetric airflow per unit length
Greek Delta (change in value) ε
Thermal emissivity
StephanBoltzmann constant
Reference Incropera, F. P., and DeWitt, D. P. Fundamentals of Heat and Mass Transfer, 5th ed. Wiley, New York, 2002.
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Source: HeatTransfer Calculations
Chapter
15 Thermal System Transient Response
James E. Marthinuss, Jr., and George Hall Northrop Grumman Electronics Systems Linthicum, Maryland
Introduction The focus of this chapter will be transient heattransfer calculations of an entire electronicscooling system. It will essentially work from the small to the large aspects in transient calculations, first starting with the heatdissipating parts, then with how the parts are packaged and tied to a heat sink, continuing with how to handle a coldplate transient, and then finishing up with a total system transient response. Topics to be discussed will include when to apply lumped capacitance to a transient calculation; how to use a detailed steadystate analysis to perform a simple transient analysis; and finally when and how to approach transient heattransfer calculations, how to calculate a time constant, and how to properly apply a time constant to obtain a temperature response of a thermal system. Simple approximations will be made to solve complex transients without the aid of a finitedifference solver. Practical solutions to transient thermal problems will be discussed in detail. A typical electronicscooling system consists of electrical components generating heat, cooled by conduction across a board to a rack. The rack in turn is cooled by fluid flowing through it. It is desirable to know not only whether the components will remain cool enough to operate continuously under normal conditions but also if they can operate during transient conditions. Transient thermal events include power surges, coolant flow interruptions, and spikes in coolant temperature. 15.1
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Thermal System Transient Response 15.2
Transient and Cyclic Calculations
Problem Definition Given a thermal system, how to determine the response to a transient condition
The modern, bruteforce approach to analyzing any thermal system is to create a detailed allencompassing automeshed finitedifference analysis/finiteelement analysis (FDA/FEA) thermal model, input the transient event, and let the computer grind away. Depending on the computational resources available and with a little bit of luck in obtaining a proper mesh, a response can be predicted. But as with most computer modeling, a little bit of insight into the problem can easily offset deficiencies in computational power and the need for luck. A more thoughtful approach is to quantify the approximate transient characteristics of a thermal system and then decide how and what to model, considering the transient event. This can be accomplished by determining the approximate transient characteristics of each part of the system and then determining how each part of the system will respond to the transient event. Once the general nature of the system’s response is understood, detailed computer modeling can be devoted to the transientcritical sections of the system as necessary. Typical electronics cooling
An electronics system consisting of numerous parts will be analyzed over several transient temperature profiles. Using this as an example, it will be easy to understand the important factors when performing a transient heattransfer calculation that can be applied to any system. The basic heat path will be described in detail. We will start with the electronics, where the heat is dissipated which consists of commercial
Electrical Component, 6 W
Wedgelock
PWB
Aluminum Heat Sink Air Heat Exchanger Aluminum Chassis
Figure 15.1
25°C 0.01 kg/s Air In
Heat path description.
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Thermal System Transient Response Thermal System Transient Response
TABLE 15.1
15.3
Description of Thermal System Width, cm
Depth, cm
Thermal path length, cm
Density, kg/cm3
Specific heat, Ws/kg◦ C
Conductivity, W/cm◦ C
Electrical component Airgap under component Electrical leads from component to PWB (40 leads) PWB Heat sink Wedgelock
2.5
2.5
0.125
0.0014
1200
jc = 2◦ C/W
2.5
2.5
0.05
1.15×10−6
1090
0.0003
0.025
0.0125
0.1
0.00785
455
2.5 2.5 5
2.5 0.13 0.5
0.15 7.5 N/A
0.0015 0.003 N/A
1100 1050 N/A
Chassis Air heat exchanger
1.0 1.0
2.5 3.0
1.2 5
0.003 N/A
1050 N/A
Description
2.87
0.0078 1.7 R = 3.2 cm2 ◦ C/W 1.7 N/A
electronics mounted to a printed wiring board (PWB). This PWB is then attached to a heat sink; the heat sink is then attached to a chassis sidewall using a wedgelock that provides a pressure contact to the chassis coldplate heat exchanger. This chassis coldplate heat exchanger is then cooled with forced air. A schematic showing the heat path as described above is shown in Fig. 15.1. The details of the system are listed in Table 15.1 and shown in Fig. 15.2.
Transient Characteristics Governing equations
The governing equation in thermal systems is the first law of thermodynamics, which states that at any time the total energy in a system is conserved. For typical electronic cooling systems this can be written as1 . . Qi = Qi
conducted out
. + Qi
stored
. = Qi
impressed
For materials with constant thermal conductivies defined by the constitutive equation1 T1 − T2 R12 = . Q12 conducted
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Thermal System Transient Response 15.4
Transient and Cyclic Calculations
0.1cm Lead Length 0.05cm Airgap
Component Lead and Airgap Information
2.5 cm by 2.5 cm 0.5 cm
5 cm
7.5 cm
0.125 cm 0.13 cm PWB
Aluminum Heat Sink 0.15 cm Figure 15.2
Dimensions.
and constant specific heats defined by1 . Qstored t Cp = VT Then the energy balance can be written as . Ti − Tj . dTi Qi = = Qi + i Vi C pi Ri j dt j
impressed
In order to solve for the temperature of each part of a system over time Ti (t), it is first necessary to determine the resistances between parts and the capacitance of each part.
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Thermal System Transient Response Thermal System Transient Response
T1
Resistances: R1 = Junction to Case R2 = Case to PWB R3 = Through PWB R4 = Across Heat Sink R5 = Through Wedgelock R6 = Through Chassis R7 = Connection to Air
R1 T2 R2 T3 R3 T5 T4
R4
Figure 15.3
T7
T6 R5
R6
15.5
Temperatures: T1 = Junction T2 = Case T3 = PWB Top T4 = Heat Sink Top T5 = At Wedgelock T6 = Chassis T7 = Bottom of Chassis T8 = Air
T8 R7
Resistor network.
Characterizing a Thermal System Determine resistances
In order to calculate the transient response of the system, the thermal resistances must be calculated from the heat source to the boundary sink. To fully understand exactly what needs to be calculated, a simple resistor network is shown in Fig. 15.3. All of these resistances will be expressed in terms of temperature rise per power dissipated or ◦ C/W. The part manufacturer provided the resistance through the electronic part from junction (hottest spot on the part) to the case of the part as 2◦ C/W. This thermal resistance will be used for R1 . Next we must calculate the thermal resistance from the case of the part to the PWB. This part is not bonded to the PWB, but it has 40 electrical leads that connect the case of the part to the board. Knowing that the leads are made of copper and have a cross section of 0.025 × 0.0125 cm and are 0.1 cm long, we can calculate this thermal resistance as1 R=
L kA
where R = resistance from case to PWB, ◦ C/W L = length of the leads 0.1, cm k = conductivity of copper, W/cm◦ C A = area, cm2 Since there are 40 leads, which are 0.025 × 0.0125 cm, the total area of the leads can be calculated as A = 0.025 cm × 0.0125 cm × (40 leads) = 0.0125 cm2
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Thermal System Transient Response 15.6
Transient and Cyclic Calculations
Now that the area is known, it can be used as input to the resistance equation: Ra =
0.1 cm (2.87 W/cm◦ C) × (0.0125 cm2 )
= 2.79◦ C/W This resistance is only for the leads. The thermal resistance for the airgap must also be included. Using the resistance equation and inputting the proper area of the part of 2.5 × 2.5 cm and conductivity of 0.0003 W/cm◦ C for the airgap, the thermal resistance of the 0.05cmlong airgap can be calculated: Rb =
0.05 cm (0.0003 W/cm◦ C) × (6.25 cm2 )
= 26.7◦ C/W The resistance through the leads and the resistance through the air are parallel resistances, so these resistances need to be added up in parallel using the following equation: R2 =
1 1 = 1 1 1 1 + + Ra Rb 2.79◦ C/ W 26.7◦ C/ W
= 2.5◦ C/ W Even though air has an extremely low conductivity, it still lowers the overall resistance from case to PWB by roughly 10 percent. This is due mainly to the short distance between the case and the PWB. The thermal resistance calculation that was just done is for R2 . This same process needs to be done for the thermal resistance through the printed wiring board (PWB), which is R3 . This resistance is calculated by using the length of the heat path, which is the thickness of the PWB at 0.15 cm. The area used is the footprint of the electronic part, which is 6.25 cm2 , and the conductivity is 0.0078 W/cm◦ C. Using the resistance equation once again, the thermal resistances can be calculated: R3 =
0.15 cm (0.0078 W/cm◦ C) × (6.25 cm2 )
= 3.08◦ C/ W The next thermal resistance that must be calculated is the resistance across the aluminum heat sink, which is R4 . This resistance is calculated by using the length of the heat path, from the center of the heat
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Thermal System Transient Response Thermal System Transient Response
15.7
sink to the edge of the heat sink, which is 7.5 cm. The area used is just the thickness of the aluminum, which is 0.13 cm, and this is multiplied by the electronic part width of 2.5 cm. The conductivity of aluminum at 1.7 W/cm◦ C is also needed in order to solve the thermal resistance equation: R4 =
(1.7
W/cm◦ C)
7.5 cm × (0.13 cm × 2.5 cm)
= 13.57◦ C/ W Next the thermal resistance through the wedgelock interface will be calculated. This thermal resistance is calculated by using an interface thermal resistance, which is 3.2◦ Ccm2 / W. The interface resistance is based on measured results and can be obtained from many different sources. This area resistance is divided by the interface area, which is 5 cm long by 0.5 cm wide: R5 = (3.2◦ Ccm2 /W)/(5.0 cm × 0.5 cm) = 1.28◦ C/W The conduction path to the air must be calculated, which is R6 . This resistance is calculated by using the length of the heat path, from the wedgelock interface to the air heat exchanger face, which is 1.2 cm. The area used is just the crosssectional area of the aluminum, which is 1.0 × 2.5 cm. Using the resistance equation and substituting the proper conductivity, length, and area, this resistance can be calculated: R6 =
1.2 cm (1.7 W/cm◦ C) × (1.0 cm × 2.5 cm)
= 0.28◦ C/ W The last thermal resistance to be calculated is for the convective heat transfer to the air in the heat exchanger, which is R7 (Ref. 1): R7 =
1 hA
where h = heattransfer coefficient, W/cm2 ◦ C A = heat exchanger surface area, cm2 In order to calculate the heattransfer coefficient in the heat exchanger, the Reynolds number must be calculated to determine whether the flow is turbulent or laminar. To calculate the Reynolds number, the hydraulic diameter must first be calculated on the 1.0 × 3.0 cm heat exchanger
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Thermal System Transient Response 15.8
Transient and Cyclic Calculations
duct as follows1 : Dh =
4A P
where Dh = hydraulic diameter, cm A = area of passage, cm2 P = wetted perimeter of passage, cm Then Dh =
4 × (1 cm × 3 cm) 2 × (1 cm + 3 cm)
= 1.5 cm The airflow velocity must be calculated on the basis of a mass flow rate of 0.01 kg/s and using air fluid properties at a temperature of 25◦ C. The density of air at 25◦ C is 1.15×10−6 kg/cm3 . The velocity can be calculated from the duct size of 1 × 3 cm by using the following equation1 : m ˙ A
Va = where Va = air velocity, cm/s m ˙ = mass flow rate, kg/s = density of air, kg/cm3 A = area of air passage, cm2 Then Va =
0.01 kg/s (1.15 × 10−6 kg/cm3 ) × (1 cm × 3 cm)
= 2898.6 cm/s Now the Reynolds number can be calculated from the following equation1 by inputting the velocity of 2898.6 cm/s, the hydraulic diameter of 1.5 cm, and the kinematic viscosity of air of 0.156 cm2 /s: Re =
Va Dh
where Re = Reynolds number Va = air velocity, cm/s Dh = hydraulic diameter, cm = kinematic viscosity of air, cm2 /s
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Thermal System Transient Response Thermal System Transient Response
15.9
Then Re =
(2898.6 cm/s) × (1.5 cm) = 27,871 0.156 cm2/s
This Reynolds number is turbulent because it is well above the transition Reynolds number of 2300, so we can now calculate the Nusselt number using the turbulent flow equation below. The Reynolds number of 27,871 and the Prandtl number for air at 25◦ C of 0.7 are input into the following equation1 : Nu = 0.023 Re4/5 Pr1/3 where Nu = Nusselt number Re = Reynolds number Pr = Prandtl number for air Then Nu = 0.023 × (27,871)4/5 (0.7)1/3 = 73.5 Once the Nusselt number is calculated, the heattransfer coefficient can be calculated from the following equation,1 using the Nusselt number of 73.5, the hydraulic diameter of 1.5 cm, and the conductivity of 0.0003 W/cm◦ C: h= where
Nuk Dh
h = heattransfer coefficient, W/cm2 ◦ C Nu = Nusselt number k = conductivity for air 0.0003, W/cm◦ C Dh = hydraulic diameter, cm
Then, h=
73.5 × 0.0003 W/cm◦ C = 0.0147 W/cm2 ◦ C 1.5 cm
This is the heattransfer coefficient, which can now be used as input to the equation below to calculate the resistance of the heat exchanger R7 . The active heatexchanger area is 3 × 5 cm. This resistance is the connection between the air and chassis: R7 =
1 1 = 4.54◦ C/W = 2 ◦ hA (0.0147 W/cm  C) × (3 cm × 5 cm)
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Thermal System Transient Response 15.10
Transient and Cyclic Calculations
We have now calculated the thermal resistances in the entire heat path. These resistances are shown below: R1 = 2◦ C/W R2 = 2.5◦ C/W R3 = 3.08◦ C/W R4 = 13.57◦ C/W R5 = 1.28◦ C/W R6 = 0.28◦ C/W R7 = 4.54◦ C/W Calculate steadystate temperatures
Using the thermal resistances calculated above, the steadystate temperatures along the heat path can be calculated. Knowing that the power dissipated is 6.0 W, the temperatures are shown below under steadystate conditions: T = R × Q T8 − T7 = R7Q = (4.54◦ C/W) × (6.0 W) = 27.2◦ C T7 − T6 = R6Q = (0.28◦ C/W) × (6.0 W) = 1.7◦ C T6 − T5 = R5Q = (12.8◦ C/W) × (6.0 W) = 7.7◦ C T5 − T4 = R4Q = (13.57◦ C/W) × (6.0 W) = 81.4◦ C T4 − T3 = R3Q = (3.08◦ C/W) × (6.0 W) = 18.5◦ C T3 − T2 = R2Q = (2.5◦ C/W) × (6.0 W) = 15.0◦ C T2 − T1 = R1Q = (2◦ C/W) × (6.0 W) = 12.0◦ C The temperature differences calculated above are used to fill in the list below (temperature values for T7 down to T1 ) to obtain temperatures T1 through T7 . Temperature T8 is the boundary air node, which is fixed at 25◦ C. Starting from this boundary node we can calculate the temperature for T7 , then T6 , and so on until we reach the temperature of T1 . This calculation is shown below: T7 = T8 + (T8 − T7 ) = 25◦ C + 13.6◦ C = 52.2◦ C T6 = T7 + (T7 − T6 ) = 52.2◦ C + 1.7◦ C = 53.9◦ C
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Thermal System Transient Response Thermal System Transient Response
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T5 = T6 + (T6 − T5 ) = 53.9◦ C + 7.7◦ C = 61.6◦ C T4 = T5 + (T5 − T4 ) = 61.6◦ C + 81.4◦ C = 143.0◦ C T3 = T4 + (T4 − T3 ) = 143.0◦ C + 18.5◦ C = 161.5◦ C T2 = T3 + (T3 − T2 ) = 161.5◦ C + 15.0◦ C = 176.5◦ C T1 = T2 + (T2 − T1 ) = 176.5◦ C + 12.0◦ C = 188.5◦ C If this unit were allowed to run until steady state is reached, then the resulting temperature would be 189◦ C, which is too hot for most electronics. A common junction temperature limit is 150◦ C. Distributed thermal mass
The thermal capacitance or thermal mass of an object is the product of the specific heat and the mass of the object. In general, the larger the thermal mass, the slower the response of an object to thermal events such as heat inputs and changes to the boundary conditions. The simple lumpedcapacitance approach assumes that the entire object is at an approximately uniform temperature and that all temperature gradients are outside the object.1 Heat conducting through a material causes temperature gradients in the material. If these gradients are significant, then portions of the thermal mass of the material are at significantly different temperatures. Where internal gradients are significant, the lumpedcapacitance assumption is not valid.1 The transient response of an object with significant internal temperature gradients can be approximated by treating the thermal mass as distributed throughout the object. By dividing up the object into multiple parts, each with its own thermal mass and thermal resistance to adjacent parts, the thermal mass of the object can be distributed and the correct thermal response can be approximated. Lump masses where possible
Once the thermal resistances are known and the steadystate temperature gradients are determined, the thermal mass distribution can be considered. It is easy to lump thermal mass when the predominant temperature gradients are not caused by conduction but by convection, radiation, contact resistance, or thin interface layers (bonds). Thermal mass should be distributed where a relatively large conduction temperature gradient exists across a part that has significant thermal mass, such as a heat sink on a PWB. For the thermal system shown in Fig. 15.4, there are several temperature gradients that occur across relatively short distances, indicating
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Thermal System Transient Response 15.12
Transient and Cyclic Calculations
T1 R1 T2 R2 T3 R3 T5 T41
R4/8
R5
T8
T7
T6 R6
R7
Resistance Temperature Temp Gradient Q1= 6 W T1 = Junction R1 = Junction to Case 2.00°C/W 188.50°C 12.000°C T2 = Case R2 = Case to PWB 2.50°C/W 176.50°C 15.000°C T3 = PWB Top R3 = Through PWB 3.08°C/W 161.50°C 18.480°C T41 = Heat Sink Top R41 = Across Heat Sink 1.70°C/W 143.02°C 10.178°C T42 R42 1.70°C/W 132.84°C 10.178°C T43 R43 1.70°C/W 122.67°C 10.178°C T44 R44 1.70°C/W 112.49°C 10.178°C T45 R45 1.70°C/W 102.31°C 10.178°C T46 R46 1.70°C/W 92.13°C 10.178°C T47 R47 1.70°C/W 81.96°C 10.178°C T48 R48 1.70°C/W 71.78°C 10.178°C T5 = At Wedgelock R5 = Through Wedgelock 1.28°C/W 61.60°C 7.680°C T6 = Chassis R6 = Through Chassis 0.28°C/W 53.92°C 1.680°C T7 = Bottom of Chassis R7 = Connection to Air 4.54°C/W 52.24°C 27.240°C T8 = Air 25.00°C Figure 15.4
Distributed resistor network.
an appropriate location to divide thermal mass. These include R2 , the component leads; R5 , the wedgelock; R6 , the resistance through the chassis; and R7 , the fluid convection. These gradients allow the thermal mass of several parts to be lumped. Several other conduction gradients occur across parts with significant thermal mass and require the mass to be distributed along the resistance. These include R1 , the junction to case resistance; R3 , the resistance through the PWB; and R4 , the resistance across the heat sink. The thermal mass and resistance of each of these sections should be divided up into subsections such that the temperature gradient across each subsection is relatively small. As with general meshing practices, it is good practice to divide up large temperature gradients into relatively insignificant ones in order to reduce the error associated with the discrete thermal model. In the case of this example, the gradient across the heat sink (T4 − T5 = 82◦ C) should be divided into less significant gradients, say, 10◦ C. The heat sink should be divided into eight roughly
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Thermal System Transient Response Thermal System Transient Response
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10◦ C parts, each with oneeighth the total heatsink thermal mass. So the thermal model would now be as shown in Fig. 15.4. Calculate capacitances
The capacitance must now be calculated. By using the areas already calculated in the thermal resistance calculation section, a volume may be easily generated, and therefore the capacitance can be paired up to the thermal resistance. By pairing up the thermal resistance to the proper capacitance, a thermal time constant can be calculated for each segment of the system. The first capacitance that we can calculate is for the electronic part. This part is a plastic component with a specific heat of 1200 Ws/kg◦ C and a density of 0.0014 kg/cm3 , which is 2.5 cm wide by 2.5 cm long by 0.125 cm high. By inputting this information into the equation below, the capacitance can be calculated for the electronic part C1 (Ref. 1): C1 = VCp where C1 = capacitance, Ws/◦ C Cp = specific heat, Ws/kg◦ C = density, kg/cm3 V = volume, cm3 Then, C1 = (0.0014 kg/cm3 ) × (2.5 × 2.5 × 0.125 cm) × (1200 Ws/kg◦ C) = 1.3125 Ws/◦ C Next we must calculate the capacitance associated with the connection between the electronic part and the PWB. This consists of the electronic part leads and the airgap. The density of the copper leads is 0.00785 kg/cm3 , and the specific heat is 455 Ws/kg◦ C. Using the area calculated under the resistance section for R2 of 0.03125 cm2 and the lead length of 0.1 cm, the volume can be calculated. The thermal capacitance of the leads can then be calculated as shown below: Ca = VCp = (0.00785 kg/cm3 ) × (0.03125 cm2 × 0.1 cm) × (455 Ws/kg◦ C) = 0.0112 Ws/◦ C The capacitance of the airgap must also be calculated. Using the dimensions of the airgap—2.5 cm long, 2.5 cm wide, and 0.05 cm thick—the volume can be calculated. Using the air density of 1.15 × 10−6 kg/cm3
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Thermal System Transient Response 15.14
Transient and Cyclic Calculations
and the specific heat of 1090 Ws/kg◦ C, the capacitance of the airgap can be calculated as shown below: Cb = VCp = (1.15 × 10−6 kg/cm3 ) × (2.5 × 2.5 × 0.05 cm) × (1090 Ws/kg◦ C) = 0.000392 Ws/◦ C We need to then add together the capacitance of the leads and the airgap, because the thermal path is treated as a single resistance. Adding the two capacitances Ca and Cb together, we can obtain the capacitance C2 as follows: C2 = Ca + Cb = 0.0112 Ws/◦ C + 0.000392 Ws/◦ C = 0.0116 Ws/◦ C The next capacitance that needs to be calculated is for the PWB or C3 . The density of the PWB is 0.0015 kg/cm3 , and the specific heat is 1100 Ws/kg◦ C. Using the area of the component of 6.25 cm2 with a thickness of 0.15 cm, the capacitance can be calculated as shown below: C3 = VCp = (0.0015 kg/cm3 ) × (6.25 cm2 × 0.15 cm) × (1100 Ws/kg◦ C) = 1.547 Ws/kg◦ C The capacitance of the aluminum heat sink must also be calculated, which is C4 . Using the previously calculated area of the heat sink of 0.325 cm2 in the thermal resistance section along with the length of 7.5 cm used in the same section for R4 , the volume of the heat sink is known. Using the aluminum density of 0.003 kg/cm3 and the specific heat of 1050 Ws/kg◦ C, the capacitance of the aluminum heat sink can be calculated as shown below: C4 = VC p = (0.003 kg/cm3 ) × (0.325 cm2 × 7.5 cm) × (1050 Ws/kg◦ C) = 7.68 Ws/◦ C The hightemperature gradients in the heat sink require that its capacitance be distributed. Dividing C4 among the eight heatsink nodes gives C41 = C4 /8 =
7.68 Ws/◦ C = 0.96 Ws/◦ C 8
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Thermal System Transient Response Thermal System Transient Response
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The capacitance of the wedgelock C5 must also be calculated. The area of the wedgelock interface calculated in the thermal resistance section for R5 is 6.5 cm2 . Some assumptions must be made in order to calculate the volume, because at this interface there is no specific thickness. An interface resistance of 3.2 cm2 ◦ C/W was used, so in order to back out what thickness to use to calculate the volume, an airgap thickness will be calculated as shown below on the basis of the interface resistance per unit area: L = (3.2 cm2 ◦ C/W) × k = (3.2 cm2 ◦ C/W) × (0.0003 W/cm◦ C) = 0.00096 cm This calculated length of 0.00096 cm and the area of 6.5 cm2 are used to calculate the volume of the wedgelock, which will be used to calculate the capacitance. Using the air density of 1.15 × 10−6 kg/cm3 and the specific heat of 1090 Ws/kg◦ C, the capacitance of the wedgelock can be calculated as follows: C5 = VCp = (1.15 × 10−6 kg/cm3 ) × (6.5 cm2 × 0.00096 cm) × (1090 Ws/◦ C) = 7.82 × 10−6 Ws/◦ C The secondtolast (penultimate) capacitance that needs to be calculated is the conductive path between the wedgelock and the heatexchanger face, which is C6 . Using the chassis area of 2.5 cm2 , along with the length of 1.2 cm for R6 , the volume of the chassis conductive path is known. Using the aluminum density of 0.003 kg/cm3 and the specific heat of 1050 Ws/kg◦ C, the capacitance of the aluminum chassis can be calculated as shown below: C6 = VC p = (0.003 kg/cm3 ) × (2.5 cm2 × 1.2 cm) × (1050 Ws/kg◦ C) = 9.45 Ws/kg◦ C The final capacitance to be calculated is for the convective path to the air in the heat exchanger: C7 = VC p = (1.15 × 10−6 kg/cm3 ) × (1 × 3 × 5 cm) × (1090 Ws/kg◦ C) = 0.019 Ws/◦ C
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Thermal System Transient Response 15.16
Transient and Cyclic Calculations
T1 R1 T2 R2 T3 R3 T5 T41
R4/8
R5
T8
T7
T6 R6
R7
Resistance Temperature Temp Gradient Capacitance Q1= 6 W 2.00°C/W 188.50°C 12°C 1.313 Ws/°C T1 = Junction R1 = Junction to Case 2.50°C/W 177°C 15°C 0.012 Ws/°C T2 = Case R2 = Case to PWB 3.08°C/W 162°C 18°C 1.547 Ws/°C T3 = PWB Top R3 = Through PWB 1.70°C/W 143°C 10°C 0.96 Ws/°C T41 = Heat Sink Top R41 = Across Heat Sink 1.70°C/W 133°C 10°C 0.96 Ws/°C T42 R42 1.70°C/W 123°C 10°C 0.96 Ws/°C T43 R43 1.70°C/W 112°C 10°C 0.96 Ws/°C T44 R44 1.70°C/W 102°C 10°C 0.96 Ws/°C T45 R45 1.70°C/W 92°C 10°C 0.96 Ws/°C T46 R46 1.70°C/W 82°C 10°C 0.96 Ws/°C T47 R47 1.70°C/W 72°C 10°C 0.96 Ws/°C T48 R48 62°C 8°C 8E06 Ws/°C T5 = At Wedgelock R5 = Through Wedgelock 1.28°C/W 0.28°C/W 54°C 2°C 9.45 Ws/°C T6 = Chassis R6 = Through Chassis 4.54°C/W 52°C 27°C 0.019 Ws/°C T7 = Bottom of Chassis R7 = Connection to Air 25°C T8 = Air Figure 15.5
Distributed capacitances.
All the capacitances have been calculated for the system. These values are listed below and are included in Fig. 15.5: C1 = 1.3125 Ws/◦ C C2 = 0.0116 Ws/◦ C C3 = 1.547 Ws/◦ C C41–48 = 0.96 Ws/◦ C C5 = 7.82 × 10−6 Ws/◦ C C6 = 9.45 Ws/◦ C C7 = 0.019 Ws/◦ C
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Thermal System Transient Response Thermal System Transient Response
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How to Evaluate Response Once the resistances and capacitances are determined, there are several methods for determining the transient response based on the governing equation1 : . Ti − Tj dTi ˙ i impressed Qi = + i Vi C pi =Q R dt ij j It would appear that since the governing equation for each part is a firstorder ordinary differential equation, the solution would be an exponential function of the form1 Ti (t) ≈ e−t/ Although a single mass resistor system does have an exponential response it is a common misconception that a thermal system with distributed mass can be characterized with a single time constant. Singletimeconstant fallacy
An electrical circuit analogy is often used to aid in understanding transient thermal characteristics. Treating a thermal system as a discrete electrical component, with a lumped capacitance, connected with a simple resistor to a known temperature, the transient thermal characteristics of the system can be reduced to a single time constant given by = RC.1 The transient thermal response of a lumpedcapacitance system is a simple exponential. Given a simple system without internal temperature gradients, this may be a valid characterization. Most actual thermal systems have significant internal temperature gradients, that effectively distribute the thermal capacitance. There is not a single time constant that can be associated with the response because of the large temperature gradients through the system. Many electrical component manufacturers will try to use a single time constant. This is not valid and will give erroneous results. Multipletransient response
Treating the sample as a single lumpedmass/singleresistor thermal system is characterized by a classic RC response to a thermal event as shown in Fig. 15.6. The response of this system is the same exponential response regardless of the nature of the thermal event. Figure 15.6 shows the response of the single lumpedmass system to the 6 W input shown as percent of response. The classic exponential response is
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Thermal System Transient Response 15.18
Transient and Cyclic Calculations
100 90 RC=91
Final Response (%)
80 70
62.8
60 50 40 30 20 10 0 0
1
2
3
4
5
Time (min) Figure 15.6
Lumped mass response.
1/e = 63.2 percent in one time constant. This response can be readily curvefit with an exponential form and extrapolated. A distributed mass system is made up of multiple parts with different time constants and thus does not respond with the same exponential decay response to every thermal event. Figure 15.7 shows the response 185
165 T1 = Junction
T1
T2 = Case
T2
145
T3 = PWB Top T41 = Heat Sink Top
T3
T42
Temperature (°C)
125
T41 T42 T43 T44 T45 T46 T47 T48 T5 T6
105
85
65
45
T43 T44 T45 T46 T47 T48 T5 = At Wedgelock T6 = Chassis T7 = Bottom of Chassis T8 = Air
T7 T8
25 0
1
2
3
4
5
Time (min)
Figure 15.7
System response to 6 W over 5 min.
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Thermal System Transient Response Thermal System Transient Response
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of the system, but in this case the thermal mass is distributed. The response of the distributed mass system is significantly different. The response of the outer part is only 40.8 percent, while the response of the more inner parts is significantly slower. When the thermal mass of a system is distributed, the portions close to the transient event respond quickly while those farther away have to wait until the closer portions respond. This characteristic results in an overall response that appears to have a rapid partial response to fast inputs and yet has a slower full response to longer events. In a short transient the adjacent parts will respond, but the more remote ones will have an insignificant response. In a long transient the adjacent parts will form stable temperature gradients with their adjacent parts while the more remote parts continue to slowly respond until they also come to equilibrium. Figure 15.8 shows how close the internal temperature gradients, the temperature difference from one part to the next, reach their final values during the transient. The adjacent parts quickly form constant temperature gradients, while the temperature gradients between remote parts take longer to form. The resulting longterm response, even for the parts near the thermal event, is the sum of their transient response plus the response of all the parts in the system. In general the partial transient response of a distributed mass system cannot be curvefit and cannot be extrapolated. To characterize the transient response of a distributed mass system, thermal modeling must be used with particular detail to the portions partially responding to the transient. 100%
T1
90%
Temperature Difference Relative to Final Difference
80%
T2 T1 = Junction
T3
T2 = Case TT3 = PWB Top T41 = Heat Sink Top
T41
70%
T42 60%
T42 T43
T43 T44 T45 T46 T47 T48 T5
50% 40% 30%
T44 T45 T46 T47 T48 T5 = At Wedgelock T6 = Chassis T7 = Bottom of Chassis
T6 T7
20% 10% 0% 0
1
2
3
4
5
Time (min)
Figure 15.8
Percentage response to 6 W over 5 min.
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Thermal System Transient Response 15.20
Transient and Cyclic Calculations
Iterative solution
One method for solving for temperatures is to input the resistances, capacitances, powers, and boundary temperatures into a finitedifference solver and let the solver iterate through the time steps to solve for the temperatures. At each time step the solver will calculate the temperature rise of each node on the basis of the net heat on the node. This method can also be solved using a spreadsheet with a solver function. The energy imbalances at each node and at each time step are summed up. The solver is used to iterate the temperatures in order to minimize the energy imbalance: Ti − Tj dTi ˙i = ˙ i impressed Q + i Vi C pi =Q R dt i j j Lumpedmass approximation
Another solution is to approximate the energy balance within each part by scaling the net energy flow down to the next part. The advantage to this method is that no iterations are required and the results are generally faster than with the actual response. In this approximation the response of each part over a short time is approximated by an exponential function. To determine the approximate exponential response of each part, a local time constant is calculated for each part relative to its neighboring parts. Determine time constants. Although the longterm response of each
part of the system cannot be described with a single time constant, the shortterm response can be approximated as exponential with a time constant. We have calculated the thermal resistances as well as the thermal capacitances of the entire system, which can now be used to generate the thermal time constants of the distributed thermal system. The time constant for each part is different depending on how the temperature of the part is being changed. By pairing up the thermal resistance to the proper capacitance, a thermal time constant can be calculated for each segment of the system. The equation shown below1 will be used to calculate the time constant for each section: Ci i = (1/Ri j ) j
R1 = 2◦ C/W R2 = 1.1◦ C/W R3 = 3.08◦ C/W
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Thermal System Transient Response Thermal System Transient Response
15.21
R41–48 = 1.7◦ C/W R5 = 0.49◦ C/W R6 = 0.28◦ C/W R7 = 4.54◦ C/W C1 = 1.3125 Ws/◦ C C2 = 0.0116 Ws/◦ C C3 = 1.547 Ws/◦ C C41–48 = 0.96 Ws/◦ C C5 = 7.82 × 10−6 Ws/◦ C C6 = 9.45 Ws/◦ C C7 = 0.019 Ws/◦ C The thermal time constant calculations and results are shown below: 1 = C1 × R1 = (1.3125 Ws/◦ C) × (2◦ C/W) = 2.625 s 2 = C2 ×
1 = (0.0116 Ws/◦ C) 1/R1 + 1/R2 1 = 0.013 s /2 ◦ C/W + 1/2.5◦ C/W
× 3 = C3 ×
1
1 = (1.547 Ws/◦ C) 1/R2 + 1/R3
× 41 = C41 × × 42−48 = C42 × ×
1 = 2.135 s (1/2.5◦ C/W + 1/3.08◦ C/W) 1 = (0.96 Ws/◦ C) 1/R3 + 1/R41 1 = 1.050 s 1/3.08◦ C/W + 1/1.7◦ C/W 1 = (0.96 Ws/◦ C) 1/R41 + 1/R42 1 = 0.814 s 1/1.7 C/W + 1/1.7◦ C/W ◦
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Thermal System Transient Response 15.22
Transient and Cyclic Calculations
5 = C5 × × 6 = C6 × × 7 = C7 × ×
1 = (7.82 × 10−6 Ws/◦ C) 1/R48 + 1/R5 1 = 5.7 × 10−6 s 1/1.7◦ C/W + 1/1.28◦ C/W 1 = (9.45 Ws/◦ C) 1/R5 + 1/R6 1 = 2.17 s 1/1.28 C/W + 1/0.28◦ C/W ◦
1 = (0.019 Ws/◦ C) 1/R6 + 1/R7 1/0.28◦ C/W
1 = 0.005 s + 1/4.54◦ C/W
These time constants indicate the order of magnitude of the response of each section. A given part will generally reach equilibrium with its adjacent parts within three time constants. The overall system response, however, depends on much more than the individual time constants. The system’s thermal characteristics are summarized in Fig. 15.9. Approximating transient response. Using the thermal capacitances and
resistances that were calculated above, an approximation of the thermal system response can be performed. For this example the temperature at the end of 30 s of operation is to be calculated, but this calculation may be applied at any time in the transient. By knowing that the temperature gradients in the layers above affect the temperature response of the layers below, we can approximate this interaction by first calculating the response of the first layer in the thermal mass chain1 : −t T = 1 − exp (Tf ) RC where T = timerelated temperature change, ◦ C = T(t) − T(0) t = time, s Tf = final or steadystate temperature difference, ◦ C = T(t → ∞) − T(0)
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Thermal System Transient Response Thermal System Transient Response
15.23
T1 R1 T2 R2 T3 R3 T5 T41
R4/8
R5
T8
T7
T6 R6
R7
Resistance Temperature Temp Gradient Q1= 6 W T1 = Junction R1 = Junction to Case 2.00°C/W 188.50°C 12°C T2 = Case R2 = Case to PWB 2.50°C/W 177°C 15°C T3 = PWB Top R3 = Through PWB 3.08°C/W 162°C 18°C T41 = Heat Sink Top R41 = Across Heat Sink 1.70°C/W 143°C 10°C 1.70°C/W 133°C 10°C T42 R42 T43 R43 1.70°C/W 123°C 10°C 1.70°C/W 112°C 10°C T44 R44 T45 R45 1.70°C/W 102°C 10°C 1.70°C/W 92°C 10°C T46 R46 T47 R47 1.70°C/W 82°C 10°C 1.70°C/W 72°C 10°C T48 R48 T5 = At Wedgelock R5 = Through Wedgelock 1.28°C/W 62°C 8°C 0.28°C/W 54°C 2°C T6 = Chassis R6 = Through Chassis T7 = Bottom of Chassis R7 = Connection to Air 4.54°C/W 52°C 27°C T8 = Air 25°C
Figure 15.9
Capacitance 1.313 Ws/°C 0.012 Ws/°C 1.547 Ws/°C 0.96 Ws/°C 0.96 Ws/°C 0.96 Ws/°C 0.96 Ws/°C 0.96 Ws/°C 0.96 Ws/°C 0.96 Ws/°C 0.96 Ws/°C 8E06 Ws/°C 9.45 Ws/°C 0.019 Ws/°C
Tau 2.625 s 0.013 s 2.135 s 1.050 s 0.814 s 0.814 s 0.814 s 0.814 s 0.814 s 0.814 s 0.814 s 5.70E6 s 2.171 s 0.005 s
Resistance network summary.
Inputting the resistance of 2◦ C/W (R1 ) and the capacitance of 1.313 W/s◦ C (C1 ) along with the steadystate temperature difference of 12◦ C at the 30 s, the equation reduces as shown below: T1 = 1 − exp
−30 s ◦ (2 C/W) × (1.313 W/s◦ C)
(12◦ C)
= 12◦ C This first layer has obviously already reached its steadystate temperature gradient in 30 s. The next layer down or the case of the component will see all the power propagate to itself. Knowing that, in order to change the temperature of the component junction as well as the case temperature, the entire thermal capacitance of the layers above the case must be considered. The capacitance of the junction and the case are added together as shown below, along with the time constant or RC associated with the temperature response of the junction and the case of the component combined. This calculation will be continued until all the temperature gradients through each section are calculated at the
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Thermal System Transient Response 15.24
Transient and Cyclic Calculations
30s time period: RC2 = R2 (C1 + C2 ) + R1 C1 = 2.5◦ C/W (1.313 Ws/◦ C + 0.012 Ws/◦ C) + (2.0◦ C/W × 1.313 Ws/◦ C) = 5.9 s −t −30 s T2 = 1 − exp (15◦ C) (T) = 1 − exp RC 5.9 s = 15◦ C RC3 = R3 (C1 + C2 + C3 ) + RC2 = 3.08◦ C/W (1.313 Ws/◦ C + 0.012 Ws/◦ C + 1.55 Ws/◦ C) + (5.9 s) = 14.8 s −30 s −t T3 = 1 − exp 18.5◦ C (T) = 1 − exp RC 14.8 s = 16.1◦ C Now in layer 3, which is the PWB, the heat has not propagated completely through but just a certain percentage of heat has traveled through the PWB to the heat sink. Knowing this, we can adjust the temperature gradient of the next layer according to the lower percentage of power dissipation, which propagates to it. With the calculated temperature gradient of only 16.1◦ C as compared to the steadystate temperature gradient of 18.5◦ C, only 87 percent of the dissipated power has moved to the heat sink from the PWB. In other words, the PWB has stored 13 percent of the power dissipated by the component. Accordingly, the gradient through the heat sink layer will be adjusted directly by this percentage as shown below: RC41 = R41 (C1 + C2 + C3 + C41 ) + RC3 = 1.7◦ C/W (1.313 Ws/◦ C + 0.012 Ws/◦ C + 1.55 Ws/◦ C + 0.96 Ws/◦ C) + (14.8 s) = 21.3 s −t −30 s T41 = 1 − exp (T) × (% of power) = 1 − exp RC 21.3 s × 10.2◦ C × (0.87) = 6.7◦ C
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Thermal System Transient Response Thermal System Transient Response
15.25
The steadystate temperature gradient through the first section of the heat sink is 10.2◦ C, while after 30 s of operation the temperature gradient is only 6.7◦ C, therefore a 65 percent power dissipation (energy storage) factor will be applied to the calculated temperature gradient for the next part of the heat sink (T42 ): RC42 = R42 (C1 + C2 + C3 + C41 + C42 ) + RC41 = 1.7◦ C/W (1.313 Ws/◦ C + 0.012 Ws/◦ C + 1.55 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C) + (21.3 s) = 29.4 s −t T42 = 1 − exp (T) × (% of power) RC −30 s = 1 − exp (10.2◦ C) × (0.65) 29.4 s = 4.2◦ C The percentage of heat which travels through this section of the heat sink is 4.2◦ C divided by 10.2◦ C, so for the next section of the heat sink (T43 ) a power factor of 41 percent will be used: RC43 = R43 (C1 + C2 + C3 + C41 + C42 + C43 ) + RC42 = 1.7◦ C/W (1.313 Ws/◦ C + 0.012 Ws/◦ C + 1.55 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C) + (29.4 s) = 39.2 s −t T43 = 1 − exp (T) × (% of power) RC −30 s = 1 − exp (10.2◦ C) × (0.41) 39.2 s = 2.2◦ C This calculation will continue until all the temperature gradients through each layer are calculated. These calculations are shown below:
Percent of power =
T43 2.2◦ C = = 0.22 T43 f 10.2◦ C
RC44 = R44 (C1 + C2 + C3 + C41 + C42 + C43 + C44 ) + RC43
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Thermal System Transient Response 15.26
Transient and Cyclic Calculations
= 1.7◦ C/W (1.313 Ws/◦ C + 0.012 Ws/◦ C + 1.55 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C) + (39.2 s) = 50.6 s −t T44 = 1 − exp (T) × (% of power) RC −30 s = 1 − exp (10.2◦ C) × (0.22) 50.6 s = 1.0◦ C Percent of power =
T44 1.0◦ C = = 0.10 T44 f 10.2◦ C
RC45 = R45 (C1 + C2 + C3 + C41 + C42 + C43 + C44 + C45 ) + RC44 = 1.7◦ C/W (1.313 Ws/◦ C + 0.012 Ws/◦ C + 1.55 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C) + (50.6 s) = 63.6 s −t T45 = 1 − exp (T) × (% of power) RC −30 s = 1 − exp (10.2◦ C) × (0.10) 63.6 s = 0.4◦ C Percent of power =
T45 0.4◦ C = = 0.04 T45 f 10.2◦ C
RC46 = R46 (C1 + C2 + C3 + C41 + C42 + C43 + C44 + C45 + C46 ) + RC45 = 1.7◦ C/W (1.313 Ws/◦ C + 0.012 Ws/◦ C + 1.55 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C) + (63.6 s) = 78.2 s
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Thermal System Transient Response Thermal System Transient Response
15.27
−t T46 = 1 − exp (T) × (% of power) RC −30 s = 1 − exp 10.2◦ C × (0.04) 78.2 s = 0.13◦ C It can be seen that farther away from the power dissipation source the thermal gradients are less established. Close to the end on the heat sink the temperature rise is only a fraction of the total steadystate temperature rise, or only about 1.5 percent. We will continue to calculate the temperature rise through the rest of the thermal transient model: Percent of power =
T46 0.13◦ C = = 0.013 T46 f 10.2◦ C
RC47 = R47 (C1 + C2 + C3 + C41 + C42 + C43 + C44 + C45 + C46 + C47 ) + RC46 = 1.7◦ C/W (1.313 Ws/◦ C + 0.012 Ws/◦ C + 1.55 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C) + (78.2 s) = 94.5 s −t T47 = 1 − exp (T) × (% of power) RC −30 s = 1 − exp 10.2◦ C × (0.013) 94.5 s = 0.04◦ C Percent of power =
T47 0.04◦ C = 0.004 = T47 f 10.2◦ C
RC48 = R48 (C1 + C2 + C3 + C41 + C42 + C43 + C44 + C45 + C46 + C47 + C48 ) + RC47
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Thermal System Transient Response 15.28
Transient and Cyclic Calculations
= 1.7◦ C/W (1.313 Ws/◦ C + 0.012 Ws/◦ C + 1.55 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C) + (78.2 s) = 112.4 s −t T48 = 1 − exp (T) × (% of power) RC −30 s = 1 − exp (10.2◦ C) × (0.004) 112.4 s = 0.01◦ C Percent of power =
T48 0.01◦ C = = 0.001 T48 f 10.2◦ C
RC5 = R5 (C1 + C2 + C3 + C41 + C42 + C43 + C44 + C45 + C46 + C47 + C48 + C5 ) + RC48 = 1.28◦ C/W (1.313 Ws/◦ C + 0.012 Ws/◦ C + 1.55 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 8 × 10−6 Ws/◦ C) + (112.4 s) = 125.9 s −t T5 = 1 − exp (T) × (% of power) RC −30 s = 1 − exp (10.2◦ C) × (0.001) 125.9 s = 0.002◦ C Percent of power =
T5 0.002◦ C = = 0.0002 T5 f 10.2◦ C
RC6 = R6 (C1 + C2 + C3 + C41 + C42 + C43 + C44 + C45 + C46 + C47 + C48 + C5 + C6 ) + RC5
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Thermal System Transient Response Thermal System Transient Response
15.29
= 0.28◦ C/W (1.313 Ws/◦ C + 0.012 Ws/◦ C + 1.55 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 8 × 10−6 Ws/◦ C + 9.45 Ws/◦ C) + (125.9 s) = 131.5 s −t T6 = 1 − exp (T) × (% of power) RC −30 s = 1 − exp (1.7◦ C) × (0.0002) 131.5 s T6 < 0.0001◦ C Percent of power =
T6 0.0001◦ C = < 0.0002 T6 f 10.2◦ C
RC7 = R7 (C1 + C2 + C3 + C41 + C42 + C43 + C44 + C45 + C46 + C47 + C48 + C5 + C6 + C7 ) + RC6 = 4.54◦ C/W (1.313 Ws/◦ C + 0.012 Ws/◦ C + 1.55 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 0.96 Ws/◦ C + 8 × 10−6 Ws/◦ C + 9.45 Ws/◦ C + 0.019 Ws/◦ C) + (131.5 s) = 222.4 s −t T7 = 1 − exp (T) × (% of power) RC −30 s = 1 − exp (27.2◦ C) × (0.0002) 222.4 s T7 < 0.0001◦ C The temperature gradients for all the sections have been calculated at a time of 30 s. During this 30s transient only the parts close to the
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Thermal System Transient Response 15.30
Transient and Cyclic Calculations
TABLE 15.2
Approximate Solution Comparison
Time, s
Approximation temperature, ◦ C
Detailed model temperature, ◦ C
30 60 120 180 240 360 480 600
83.1 103.9 131.1 149.2 161.9 176.4 182.9 185.7
76.4 98.9 127.7 146.4 159.2 174.3 181.6 185.1
Percent difference 8.8 5.1 2.7 1.9 1.7 1.2 0.7 0.3
heat source respond. Adding the temperature gradients through each layer and adding them to the boundary node temperature of 25◦ C, we can calculate the temperature of the component at 30 s after power is applied to the component. This calculation is shown below: T1 = T1 + T2 + T3 + T41 + T42 + T43 + T44 + T45 + T46 + T47 + T48 + T5 + T6 + T7 + T8 = 12◦ C + 15◦ C + 16.1◦ C + 6.7◦ C + 4.2◦ C + 2.2◦ C + 1.2◦ C + 0.46◦ C + 0.15◦ C + 0.04◦ C + 0.01◦ C + 0.002◦ C + 0◦ C + 0◦ C + 25◦ C = 83.1◦ C Using this approximation, we calculated a component temperature of 83.1◦ C at the end of the 30s transient. At the same time period the detailed model using finitedifference software yields a temperature of 76.4◦ C, which is very close to this calculated temperature of 83.1◦ C. This approximation may be used for any time period. Because this calculation is timeconsuming, a spreadsheet could be used to calculate the temperature at any point in time very easily. Table 15.2 lists the resulting temperatures in a spreadsheet form at different times to show the temperature response and the exact solution compared to this simple approximation calculation. In most cases the approximation will be within 10 percent of the exact solution. System Response The response to a thermal event depends on the duration and frequency of the event and the distribution of thermal mass within the thermal system. In general only the adjacent parts will respond to a short transient event. An iterative finitedifference solver was used to evaluate the system’s response to these transients.
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Thermal System Transient Response Thermal System Transient Response
15.31
To a short transient
Short transients significantly affect only the adjacent parts of the system. The rapid responses of the junction and case in these transients indicate that more resolution might be needed to accurately simulate their response. As with meshing in general, it is good engineering practice to further divide a model to establish mesh sensitivity. In transient modeling the thermal mass distribution as well as the resistances are meshsensitive. The more remote parts, on the other hand, have little to no response and need no further resolution for this transient. Single event. Figure 15.10 shows the transient response from a 25◦ C
soak to 6 W applied to the junction (T1 ). In this transient the parts near T1 change quickly at first while the chassis barely changes at all. It should be noted that the response of T1 slows from an exponential response as the remote parts begin to respond. Figure 15.11 presents the same data as Fig. 15.10 except that the temperature difference between adjacent parts is plotted as a fraction of its final steadystate value. This figure helps illustrate how the parts near T1 tend to form relatively stable temperature gradients quickly. The remaining temperature fluctuations near T1 are due to that fact that the gradients in the remote parts form slowly as the net heat load finally conducts to them.
85
75
T1
T1 = Junction T2 = Case T3 = PWB Top
T2
65
T41 = Heat Sink Top
Temperature (°C)
T42 T43 T44
55
T3
T45 T46 T47
45
T41 T42
35
25 0
5
10
15
20
25
T48 T5 = At Wedgelock T6 = Chassis
T7 = Bottom of Chassis T43 T8 = Air T44 T45 T46 T47 T48 T5 T6 T7 T8 30
Time (s)
Figure 15.10
System response to 6 W over 30 s.
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Thermal System Transient Response 15.32
Transient and Cyclic Calculations
100% 90% 80%
Temperature Difference Relative to Final Difference
T1 = Junction T2 = Case T3 = PWB Top T41 = Heat Sink Top T42 T43 T44 T45 T46 T47 T48 T5 = At Wedgelock T6 = Chassis T7 = Bottom of Chassis
T1 T2
70% 60%
T3 50%
T41 40%
T42 30%
T43
20%
T44
10%
T45 T46 T47 T48
T5 T6 T7 T8
0% 0
5
10
15
20
25
30
Time (s)
Figure 15.11
Percentage response to 6 W over 30 s.
Repeated event. Figure 15.12 shows the transient response of repeated
applications of 6 W for 10 s to the junction (T1 ) 25 percent of the time (10 s on and 30 s off). This analysis shows more clearly how only the nearby parts respond to short transients while the remote parts stay at relatively constant temperatures. The temperature gradients in the remote parts are equivalent to those resulting from the average power flowing through the parts, in this case 6 W × 25 percent = 1.5 W. To a long transient
During longer transients the adjacent parts form stable temperature gradients while distant parts continue to respond. Single event. Figure 15.13 shows the transient response from a 25◦ C
soak to 6 W applied to the junction (T1 ). Notice in this plot that the response of the remote parts (T45 ) is not at all exponential over the entire transient. The fact that the nearby gradients form almost completely indicates that no further resolution of the component should be needed for this transient. The more remote parts, on the other hand, respond significantly, indicating that more resolution might be needed to accurately simulate their response. As with meshing in general, it is good engineering practice to further divide a model to establish mesh sensitivity. In transient modeling the thermal mass distribution as well as the resistances are sensitive to meshing.
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Thermal System Transient Response Thermal System Transient Response
15.33
85
75
T1
T1 = Junction T2 = Case
T2
Temperature (°C)
65
T3 = PWB Top T41 = Heat Sink Top
T3
T42 T43 55
45
35
T41
T44
T42
T45
T43 T44
T46
T45 T46 T47 T48 T5 T6 T7
T48
T47 T5 = At Wedgelock T6 = Chassis T7 = Bottom of Chassis T8 = Air
T8
25 0
10
20
30
40
50
60
70
80
Time (s)
Figure 15.12
System response to 6 W pulsed for 10 s every 40 s.
185
165 T1 = Junction
T1
T2 = Case
T2
145
T3 = PWB Top T41 = Heat Sink Top
T3
T42
Temperature (°C)
125
T41 T42 T43 T44 T45 T46 T47 T48 T5 T6
105
85
65
45
T43 T44 T45 T46 T47 T48 T5 = At Wedgelock T6 = Chassis T7 = Bottom of Chassis T8 = Air
T7 T8
25 0
1
2
3
4
5
Time (min)
Figure 15.13
System response to 6 W over 5 min.
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Thermal System Transient Response 15.34
Transient and Cyclic Calculations
125 115
T1
T1 = Junction
105
T2 = Case
T2
T3 = PWB Top
Temperature (°C)
95
T41 = Heat Sink Top T42
85
T3
75
T41
T45
T42
T46
T43
T47
T43 T44
65
T48
T44
55
T5 = At Wedgelock
T45 T46 T47 T48 T5 T6 T7
45 35
T6 = Chassis T7 = Bottom of Chassis T8 = Air
T8
25 0
60
120
180
240
300
360
Time (s)
Figure 15.14
System response to 6 W pulsed for 1 min every 3 min.
Repeated event. Figure 15.14 shows the transient response of repeated
applications of 6 W for 1 min to the junction (T1 ) onethird of the time (1 min on and 2 min off ). In this case the power lasts long enough for the remote parts to respond. Notice how their response is progressively smaller and more delayed. Figure 15.15 shows the transient response to repeated 25◦ C steps in air temperature for 1 min out of every 3 min with a constant 3 W applied to T1 . This analysis shows that it is not the inherent nature of the junction to respond quickly or the chassis to respond slowly. What is inherent in a distributed mass thermal system is that the parts near the event respond quickly while the remote parts, in this case T1 , respond slowly and to a lesser degree. Conclusion The transient response of a thermal system with internal temperature gradients is the combined response of the individual exponential responses of its parts. In general the temperature of any part changes exponentially in the short term but more slowly in the long term because of the effects of more remote parts. Over longer transients a distributed system responds more slowly than does a single mass system. The parts closer to the thermal event respond quickly while the remote
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Thermal System Transient Response Thermal System Transient Response
15.35
125
T1
115
T2
T1 = Junction
105
T2 = Case
T3
T3 = PWB Top
Temperature (°C)
95 85 75
T41
T41 = Heat Sink Top
T42
T42
T43
T43
T44
T44 T45
T45
T46 65
T46
T47
T47 55
T48
T48
45
T5 = At Wedgelock
T8
T5 T6 T7
T6 = Chassis T7 = Bottom of Chassis T8 = Air
35 25 0
60
120
180
240
300
360
Time (s)
Figure 15.15
System response to 25◦ C spike for 1 min every 3 min with 3 W.
parts have a delayed, slower response. An iterative solution is possible using a solver. A noniterative approximation is also quite practical. However, a singletimeconstant solution does not accurately represent the true transient response of a real thermal system. Acknowledgments The authors wish to acknowledge the support of Richard Rinick and Richard Porter for their expeditious review of this section. The authors would also be remiss in not acknowledging the steadfast support of their wives. Nomenclature A
Area of thermal path, cm2
C
Lumped thermal capacitance of solid, Ws/◦ C
Cp
Specific heat, Ws/kg◦ C
Dh
Hydraulic diameter, cm
h
Heattransfer coefficient, W/cm2 ◦ C
k
Conductivity of material, W/cm◦ C
L
Thermal path length, cm
m ˙
Mass flow rate, kg/s
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Thermal System Transient Response 15.36
Nu
Transient and Cyclic Calculations
Nusselt number
P
Wetted perimeter of flow passage, cm
Pr . Q
Prandtl number
R
Resistance to conduction heat transfer, ◦ C/W
Re
Reynolds number
T
Temperature, ◦ C
V
Volume, cm3
Va
Flow velocity, cm/s
Heattransfer rate, W
Greek T
Temperature difference, ◦ C
Kinematic viscosity of air, cm2 /s
Density, kg/cm3
Time constant, s
Reference 1. From Frank P. Incropera and David P. DeWitt, Fundamentals of Heat and Mass Transfer, Wiley, New York, 1985, pp. 5, 8, 64, 65, 68, 175–177, 369, 386, 397, 404. This material used by permission of John Wiley & Sons, Inc.
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Source: HeatTransfer Calculations
Chapter
16 Thermal Response of Laminates to Cyclic Heat Input from the Edge
Wataru Nakayama ThermTech International Kanagawa, Japan
Introductory Note Laminates of dissimilar materials are commonly found in industrial products. Heat conduction in laminates is a wellstudied subject, so there is a sizable volume of literature on it. However, actual calculation of temperature and heat flow in laminates is by no means simple, particularly when one deals with transient heat conduction problems. The difficulty amplifies where the laminate structure lacks regularity in the geometry and dimensions of component materials. For instance, printedcircuit boards (PCBs) in electronic equipment have diverse layouts of copper strips embedded in resin matrix. To solve heat conduction problems in such actual laminates, we resort to numerical analysis using the finitedifference or finiteelement method. In more recent years, the art of numerical analysis (code) and the computer on which the code is run have been made powerful enough to deal with complex heat conduction problems. However, in many instances, it is still prohibitively timeconsuming to find the full details of transient heat conduction in laminate structures. We still need to exercise the art of reducing the factors and variables involved in the problem in order to perform the analysis in a reasonable timeframe.
16.1
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Thermal Response of Laminates to Cyclic Heat Input from the Edge 16.2
Transient and Cyclic Calculations
In this section we are concerned with one of the issues involved in the art of modeling. Suppose that a physical volume in that heat is generated intermittently from embedded heat sources, as is the case in almost all electronic devices and equipment. Transient heat generation produces a temperature distribution in the volume that is basically nonsteady (transient). However, when the heat generation occurs at a very high frequency, as on the microprocessor chip, the temporal variation of temperature is restricted to a very narrow region around the heat source, and the rest of the zone in the volume has a quasisteady temperature distribution. So, for practical purposes, the assumption of steadystate temperature is applied, obviating the need to perform transient analysis. Consider another extreme end where the frequency of heat generation is very low. Again, the steadystate assumption makes sense, and the analysis for a situation of peak heat generation suffices to find whether the heatsource temperature can be held below a tolerable level in a given thermal environment. In between these extremes, very high and very low frequency ends, there are intermediate situations where the transient analysis cannot be bypassed altogether. But still, the transient analysis may be applied to a region near the heat source, and a quasisteady temperature field may be assumed in the rest of the region. We suppose a boundary around the heat source that separates the zone of transient temperature and that of quasisteady temperature. If we know such boundaries beforehand, we can program our task in a timesaving way, that is, the timeconsuming transient analysis is performed only where the temperature varies perceptively with time. But how do we know the possible location of boundaries before embarking on fullscale numerical analysis? The size of the region of large temperature variation around the heat source depends on the heat generation frequency, the material composition and dimensions of the laminate, and the heat transfer from the surface(s) of the laminate to the environment. We need a guide that suggests a spatial extent of temporal variation of the temperature, and such a guide should involve relatively simple calculations. A heat conduction problem is devised here to develop a guide. Problem Definition and Formal Solutions
Figure 16.1 explains the problem to be solved. A laminate is composed of low and highthermalconductivity layers, alternately stacked and perfectly bonded rendering interfacial thermal resistance to zero. The laminate is semiinfinite in the direction of the layers (x), and cyclic heat input is applied to the edge (x = 0) of one of the highconductivity layers. The edge of the laminate is adiabatic except for the edge of the heat source layer. The upper and lower surfaces of the laminate are convectively cooled. Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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Infinitely long in the x direction
Figure 16.1 Laminate of alternate low and highconductivity layers with cyclic heat input at the edge of one of the highconductivity layers.
Adiabatic except at the edge of the heatsource layer
q
Heat flux
t
x
Heat dissipation to environment
Heatsource layer
Highconductivity layer
Lowconductivity layer
Heat dissipation to environment
Thermal Response of Laminates to Cyclic Heat Input from the Edge
16.3
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Thermal Response of Laminates to Cyclic Heat Input from the Edge 16.4
Transient and Cyclic Calculations
In terms of the heat flux, q (W/m2 ), the heat input at the edge (x = 0) of the heat source layer is given as q = q0 [1 + cos(2 f t)]
(16.1)
where q0 is the amplitude, f is the frequency in hertz, and t is the time in seconds. Note that the heat flux becomes maximum (2q0 ) at t = (n – 1)/ f , and zero at t = (2n− 1)/2 f , where n is an integer. We assume that t = 0 is taken at an instant long after the start of cyclic heat application. Then, the thermal response of the laminate to the heat input includes the steady component corresponding to the first term (q0 ) on the righthand side of Eq. (16.1) and the cyclic component corresponding to the second term [q0 cos(2 f t)]. Figure 16.2 shows conceptual sketches of the responses of the heatsource layer temperature to cyclic heat input. The temperature here is that averaged across the thickness of the heatsource layer. Also, the environment temperature is set as zero, so is the temperature rise from the environment. When the frequency is low, the cyclic component dominates the temperature distribution along x (Fig. 16.2a). For this quasisteady situation the analysis can be focused on the temperature distribution produced by the peak heat input 2q0 . When the frequency is high, the transient component is diffused in a short distance from the edge, and the rest of the region is covered by the steady component (Fig. 16.2b). We introduce the notion of penetration distance. The penetration distance of the transient component (lt ) is the distance from the edge to a point where the amplitude of cyclic temperature variation is reduced to ( max – min ) × 1/e =0.368( max – min ), where max and min are the peak and the minimum temperatures at x = 0, respectively. The penetration distance of the steady component (ls ) is the distance from the edge to a point where the timeaveraged temperature mid is reduced to mid × 1/e = 0.368mid . These penetration distances are indicative of the extent of the respective regions. Once we know the order of magnitude of penetration distances, we apply the knowledge to the numerical analysis on actual laminates that may have more complex planar patterns of highconductivity layers, but retain essential features such as the number of layers, the layer thicknesses, the component materials, and the heattransfer coefficients on the laminate surfaces. In other words, we define the region of analysis in a distance around the heat source by ls × (a multiplication factor of the analyst’s choice), and also program to perform transient heat conduction analysis only in a region around the heat source whose expanse is set as lt × (a multiplication factor of the analyst’s choice). Such an approach will produce considerable saving in computer memory usage and computing time compared to allout analysis tracking temporal temperature variations in the whole spatial domain. Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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(b) (a)
Figure 16.2 Responses of heatsource layer temperature to cyclic heat input from the edge q = q0 (1 + cos(2f )): (a) lowfrequency input; (b) highfrequency input. In (b), the penetration distances are illustrated; ls is the penetration distance of the steady component, and lt , that of the transient component.
π θ min
π/2 θmid
θ
θmax
t=0
x
θ min
θ mid
θ
θ max
lt
ls
π/2
t=0
π
x
Thermal Response of Laminates to Cyclic Heat Input from the Edge
16.5
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Thermal Response of Laminates to Cyclic Heat Input from the Edge 16.6
Transient and Cyclic Calculations
Besides the penetration distances, knowledge of the amplitude of temperature swing at the heat source, as indicated by max and min in Fig. 16.2, is useful, as the heatsource temperature is of primary interest in thermal design. So, our ultimate interest is in the penetration distances and the temperature swing at the heat source. Knowledge of these characteristic quantities is derived from the solution of the set of equations governing heat conduction in the laminate. However, as the literature testifies, for example, in Refs. 1 and 2, the rigorous analysis involves complex mathematics that deter practicing engineers from even trying to understand it. Also, the rigor of mathematics makes the analysis practically tractable only in those cases where the laminate has a regular composition, that is, a pair of lowconductivity and highconductivity layers, each having a respectively fixed thickness, repeated in the direction of layer stack. In what follows we introduce an approximation that relaxes the constraint on the laminate composition. The approximation is based on the observation of laminates commonly found in electronic equipment. Thus, thin metals (high conductivity layer) are embedded in resin matrix (low conductivity layer), and the following disparity relations hold: kM >> k (M >> )
dM 0
sinh(q · L) sinh(q · L) = q q
If q < 0
sinh(q · L) −1 · sinh(q · L) = q −q
(19.16)
So, use the positive root in the expressions: s q = positive
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Transient Analysis of Low Temperature, Low Energy Carrier (LoTEC©) 19.6
Transient and Cyclic Calculations
Substituting for q into Eq. (19.14) yields d2 i − q2 · i = 0 dx 2
(19.17)
which has the following solution: i = A · cosh(q · x) + B · sinh(q · x) di = A · q · sinh(q · x) + B · q · cosh(q · x) dx
(19.18) (19.19)
Applying the boundary condition at x = 0 yields i = A = I
(19.20)
and the differential equation [Eq. (19.10)] for I gives the following equation for B: Tc −s·t0 sI = · B · q + · − · I · 1 − e (19.21) t0 · s2 Thus, B=
(s + ) · I −
· Tc · 1 − e−s·t0 2 t0 · s ·q
(19.22)
So i = I · cosh(q · x) + sinh(q · x) (s + ) · I Tc −s·t0 · − · · 1 − e ·q t0 · s2 · · q
(19.23)
Applying the boundary condition at x = L yields Tc · 1 − e−s·t0 = I · cosh(q · L) 2 t0 · s s+ · Tc −s·t0 + · sinh(qL) · I − · 1−e ·q t0 · s2 · · q (19.24) or
Tc −s·t0 · 1−e · 1+ · sinh(q · L) ·q t0 · s2 s+ = I · cosh(q · L) + · sinh(q · L) ·q
(19.25)
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Transient Analysis of Low Temperature, Low Energy Carrier (LoTEC©) Transient Analysis of LoTEC
19.7
Solving (19.25) for m yields Tc −s·t0 · 1−e · 1+ · sinh(q · L) ·q t · s2 I = 0 s+ · sinh(q · L) cosh(q · L) + ·q
(19.26)
Now, return to the time domain. Use the Laplace transform inversion theorem: R+ i · ∞ TI = I · es·t dt (19.27) R− i · ∞
Because the integral is in complex space, let s = z, then q = and √ dq 1 1 = = √ √ = dz 2 · q 2 ·z 2 z
√
z/
Then TI =
1 − e−z·t0 · 1 + · sinh(q · L) Tc ·q · · e z·t dt z+ t0 · z2 · sinh(q · L) cosh(q · L) + ·q (19.28)
R+ i · ∞ R− i · ∞
Now let Tc −z·t0 · sinh(q · L) · e z·t (19.29) · 1 − e · 1 + ·q t0 · z2 z+ h(z) = cosh(q · L) + · sinh(q · L) (19.30) ·q g(z) =
There is a pole of order 2 at z = 0 in Eq. (19.28), so take the derivative to get the residue; then take the limit as z approaches zero. The limit turns out to be Tc . For the many zeros of Eq. (19.30), the residue is evaluated as Tc −z·t0 · 1+ · 1−e · sinh(q · L) · e z·t d ·q t0 Residue = lim z + z →0 dz · sinh(q · L) cosh(q · L) + ·q (19.31)
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Transient Analysis of Low Temperature, Low Energy Carrier (LoTEC©) 19.8
Transient and Cyclic Calculations
There are many poles of order 1 for cosh(q · L) +
z+ · sinh(q · L) = 0 ·q
(19.32)
So Residue =
g(z) h (z)
(19.33)
This leads to an infinite series solution because many values of z satisfy Eq. (19.32). Thus, the solution for the temperature is TI (t) = Tc +
∞
Tm (, , , z j , L, t)
(19.34)
j =0
where Tm(, , , L, z, t) √ z Tc − T0 · exp(z · t) · (1 − exp(−t · z)) · 1 + · z · sinh · · L 0 ·z t0 · z2 = √ √ √ (z + )/z 1 1/ · sinh[L · z/] · √ L 1 + ·√z/ · sinh[L · z/]− /2 · · √z/ · sinh[L · z/] 2 z/ · √ + 1/2 · z+· z · cosh[L · z/] · L
(19.35) The results of this equation compare favorably with measurements of the transient temperature (see Fig. 19.4). The values of variables , , , and L are 4.80 ×10−8 m2 /s, 3.15 × 10−6 m/s, 5.25 × 10−6 /s, and 0.0381 m. The sumofsquares error is 0.0764K 2 . The analytical temperatures are calculated using only five terms of the infinite series. The measured values are determined from inserting LoTEC into an environmental chamber and applying the ramp temperature of Fig. 19.3. The time of the ramp t0 is 1 h. The temperature Tc is 20◦ C, and the temperature TE is 60◦ C. The temperatures measured are an average of six temperatures on the inner liner walls. The measured temperatures differ from each other by less than 1◦ C, with an uncertainty of 0.2◦ C. Note that the square root of the sumofsquares error is 0.276◦ C, which is slightly greater than the uncertainty in the measurement. The agreement between the measurements and the analytical results shows that even though the LoTEC has an intricate geometry, the onedimensional simplification is amazingly accurate.
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Transient Analysis of Low Temperature, Low Energy Carrier (LoTEC©) Transient Analysis of LoTEC
19.9
45
40
35 Tanalytical Texperimental 30
25
20
1
1.5
2
2.5
3
3.5
4.5 4 Time (h)
5
5.5
6
6.5
7
Comparison of analytical and measured temperatures using five terms of the summation. Times shorter than 1 h require more than five terms in the infinite series with these values of properties. The sumofsquares error between the two temperatures is 0.0764K 2 .
Figure 19.4
Reference 1. F. C. Wessling, L. S. Stodleck, A. Hoehn, S. Woodard, S. O’Brien, and S. Thomas, “Low c ) and Phase Change Materials (PCMs) Temperature, Low Energy Carrier (LoTEC for Biological Samples,” 30th Proceedings of the International Conference on Environmental Systems, paper no. 2000012280, Toulouse, France, July 2000.
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Transient Analysis of Low Temperature, Low Energy Carrier (LoTEC©)
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10
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Source: HeatTransfer Calculations
Chapter
20 Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc )
Francis C. Wessling Department of Mechanical and Aerospace Engineering University of Alabama in Huntsville Huntsville, Alabama
James J. Swain Department of Industrial and Systems Engineering and Engineering Management University of Alabama in Huntsville Huntsville, Alabama
c A Low Temperature, Low Energy Carrier (LoTEC ; see Fig. 20.1) is a passive thermal carrier designed to permit payload transport to and from the International Space Station at approximately constant temperature without external power [1]. Further details on LoTEC are given in a companion calculation in Chap. 19. An analysis of the transient response of LoTEC with onedimensional assumptions (see Fig. 20.2) is presented in the companion calculation. This solution is for the case of a ramp temperature as shown in Fig. 20.3. As shown in the figure, the temperature of the environment TE varies over time, starting at the initial temperature of the LoTEC and rising to a final temperature of Tc over a period t0 . The solution of the transient temperature response TI (t) for the inner liner of the LoTEC is given in Eq. (19.35) of Chap. 19 (above) and is
20.1
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Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc©) 20.2
Transient and Cyclic Calculations
Figure 20.1 Low Temperature, Low Energy Carrier (LoTEC).
reproduced here as Eq. (20.1): TI (t) = Tc +
∞
Tm(, , , zi , L, t)
(20.1)
i=1
where Tm(, , , L, z, t) = Tc −T0 (zt) e (1 t0 z2
L √ 2 z/
√ sinh[L z/] +
1 √ L z/
√ √ − exp(−t0 z)) 1 + z z sinh[L z/] √ √ √ (z+ √)/z sinh[L z/]+ L(z + ) cosh[L z/] sinh[L z/]− 2L 2 z/
The solution is given in terms of five variables: , , , L, and t. The roots zi , i = 1, . . ., are dependent on the other variables through a
Figure 20.2
Onedimensional representation of LoTEC.
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Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc©) Parameter Estimation of LoTEC
Figure 20.3
20.3
Applied environmental temperature
ramp.
transcendental equation [Eq. (20.2)], and are independent of time t: z z+ F (, , , L, z) = cosh[L z/] + √ sinh L z/
(20.2)
The variables , , and contain eight primitive variables: = ki / i C p,i , = ki Ai / mVmC p,m, and = U As / mVmC p,m. Measurements exist for some of these primitive variables, but not all of them. For example, the effective area of the aerogel is unknown because of the chamfered corners of LoTEC. The area of the aerogel Ai is originally taken to be that of the sides, top, bottom, front, and back of the inner liner. This is adjusted later according to experimental data to give a proper area of the aerogel. The thickness of the aerogel is designated by L and equal to 1.5 in. (0.0381 m). The thermal conductivity of the insulation is ki . Although it is measured separately in the laboratory, its actual value in this application is unknown. The inner liner is taken as a lumped mass mVm and is obtained directly by weighing. The specific heat of the liner is taken to be C p,m, and this value is not well known. The effective conductance of the structural heat gain or loss to the interior not due to aerogel is given by UAs . Each of the three variables , , and have at least one unknown. These are ki , Ai , C p,m, and UAs . A nonlinear leastsquares parameter estimation technique using Gauss linearization [2] is used to determine the variables , , and rather than the primitive variables. The variables Ai , C p,m, and UAs should be repeatable from one LoTEC to another. Their values are dependent on the geometric construction of LoTEC. The LoTEC are assembled using jigs to hold the dimensions to close tolerance. The approach used here is to test two LoTECs to
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Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc©) 20.4
Transient and Cyclic Calculations
estimate unique values of for each LoTEC and estimate the common values of and from both LoTECs. The two unique values of (1 and 2 ), , and can be estimated together using the results of both experiments in combination as will be described. The temperature for each LoTEC is measured every 6 min from the end of the ramp time (t0 is one hour) to 7 h for a total of 61 data points per LoTEC and 122 data points overall. The first 61 points are from LoTEC 1 and the second 61 are from LoTEC 2. The nonlinear estimation problem is used to solve for the values of ˆ 1 , ˆ and ˆ that most closely match the 122 observed temperatures to ˆ 2 , , the values predicted in Eq. (20.1). A Gauss approximation algorithm is used to iteratively solve for the values of the parameters that achieve this best fit. The Gauss algorithm is based on a linearization of the response TI (t) using the derivatives of TI (t) with respect to the parameters 1 , 2 , , and . These derivatives are contained in the sensitivity matrix X consisting of 122 rows and 4 columns, in which each row of the matrix represents the derivative of the TI (t) for a particular observation of t and the value of the parameters at that particular iteration of the algorithm. Note that for the first 61 rows, TI (t) depends on 1 and the final 61 depend on 2 as the only terms that differ in the two experiments. The derivatives are evaluated next. As noted, the variable z are the roots of the transcendental Eq. (20.2). Derivatives must be taken with respect to , , , and L. These are needed in the formulation of the derivatives of Tm(t), which are also functions of z and are given, for example, by DTm/D = ∂ Tm/∂ + ∂ Tm/∂z ∂z/∂, where DTm/D is the derivative of Tm with respect to considering both the explicit derivative ∂ Tm/∂ and the derivative ∂ Tm/∂z from the variable z in Eq. (20.2). Similar derivatives exist for the other three variables. Because z is an implicit function of , , , and L as defined by F of Eq. (20.2), its derivative with respect to is given by ∂z/∂ = −(∂ F/∂)/(∂ F/∂z). Similar expressions obtain for the other derivatives. The actual equations follow. Derivatives of F The transcendental function F is given in Eq. (20.2). The partial derivatives of F with respect to the variables , , , L, and z are now developed in the following equations using the shorthand notation dF d to represent ∂ F/∂: −Lz z(z + ) z dF d(, , , L, z) = sinh L + √ 22 (z/)3/2 22 z/ z z (z + )L × sinh L + cosh L (20.3) 2
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Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc©) Parameter Estimation of LoTEC
20.5
z −(z + ) sinh L (20.4) √ 2 z/ 1 z dF d (, , , L, z) = √ (20.5) sinh L z/ z z z z+ dF dL (, , , L, z) = sinh L + cosh L dF d(, , , L, z) =
(20.6) L z z 1 dF dz (, , , L, z) = sinh L sinh L + √ √ 2 z/ z/ z+ z z (z + )L − sinh L + cosh L √ 2z 2z z/ (20.7) The derivatives of z with respect to the four variables are given in the following equations:
dz d(, , , L, z) =
−dF d(, , , L, z) dF dz(, , , L, z)
(20.8)
dz d(, , , L, z) =
−dF d(, , , L, z) dF dz(, , , L, z)
(20.9)
dz d (, , , L, z) =
−dF d (, , , L, z) dF dz(, , , L, z)
(20.10)
dz dL (, , , L, z) =
−dF dL (, , , L, z) dF dz(, , , L, z)
(20.11)
Derivatives of Temperature We now develop derivatives of temperature. In order to keep the equations within a reasonable length, the temperature terms are separated into differentiable pieces that are then reassembled. The numerator of Eq. (20.1) is evaluated first. Call it N (, , , L, z, t). The partial derivatives will be designated in the form dN d(, , , L, z, t). Similar nomenclature is used for the other variables. Note that since observations are taken after t ≥ t0 , TE is a constant equal to Tc − T0
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Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc©) 20.6
Transient and Cyclic Calculations
as used in Eq. (20.1): TE (zt) z √ N (, , , L, z, t) = e (1−exp (−t0 z)) 1+ z sinh L 2 z t0 z
dN d (, , , L, z, t) =
dN d (, , , L, z, t) =
dN d (, , , L, z, t) = dN dL (, , , L, z, t) =
(20.12) TE zt z e (1 − exp (−t z)) sinh L √ 0 t0 z2 2 z L z − cosh L (20.13) 2 √ − TE z (zt) z e (1 − exp (−t0 z)) sinh L t0 2 z3 (20.14) √ TE z (zt) z (20.15) e (1 − exp (−t0 z)) sinh L 3 t0 z TE (zt) z e (1 − exp (−t0 z)) cosh L (20.16) t0 z2
The derivative of the numerator with respect to z is considered next. The numerator is separated into two parts, N1 and N2 , to simplify the expressions: TE (zt) e (1 − exp (−t0 z)) t0 z2 √ z N2 (, , , L, z, t) = 1 + z sinh L z N1 (, , , L, z, t) =
(20.17) (20.18)
Taking derivatives of these expressions yields dN1 dz (, , , L, z, t) =
−2TE (zt) TE t e (1− exp (−t0 z)) + 2 e(zt) (1− exp (−t0 z)) t0 z3 t0 z
TE (zt) e exp(−t0 z) (20.19) z2 √ − z z z dN2 dz (, , , L, z, t) = sinh L + √ sinh L 2 z 2z z L z + cosh L (20.20) 2z +
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Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc©) Parameter Estimation of LoTEC
20.7
These terms are then combined to yield the expression dN dz(, , , L, z, t) = N1 (, , , L, z, t)dN2 dz(, , , L, z, t) + N2 (, , , L, z, t)dN1 dz(, , , L, z, t)
(20.21)
The denominator D(, , , L, z, t) is approached in the same manner as the numerator by separation into four parts Di (, , , L, z, t), for i = 1, 2, 3, 4: L z z 1 D (, , , L, z, t) = sinh L sinh L + √ √ 2 z/ z/ z+ z z (z + )L − sinh L + cosh L √ 2z 2z z/ (20.22) Only the nonzero denominator derivatives are presented below (e.g., dD1 /d = 0): z z −L2 L cosh L + sinh L √ 42 4 z L z − √ sinh L (20.23) 2 z z z L 1 dD1 dL (, , , L, z, t) = cosh L + √ sinh L 2 2 z dD1 d (, , , L, z, t) =
(20.24) z z L2 L dD1 dz (, , , L, z, t) = cosh L − √ sinh L 4z 4z z (20.25) The derivatives for the second denominator term D2 (, , , L, z, t) are provided next: L z z + sinh L cosh L 2 2 z (20.26) z −1 dD2 d (, , , L, z, t) = sinh L (20.27) √ 22 z/ dD2 d (, , , L, z, t) =
1 √
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Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc©) 20.8
Transient and Cyclic Calculations
z 1 cosh L (20.28) L z z −1 dD2 dz (, , , L, z, t) = + cosh L sinh L √ 2z 2z z/
dD2 dL (, , , L, z, t) =
(20.29) The derivatives for the third denominator term D3 (, , , L, z, t) are provided next: z z −(z + ) L(z + ) dD3 d(, , , L, z, t) = + cosh L √ sinh L 4z 4z z z z+ dD3 d(, , , L, z, t) = sinh L √ 2 2 z z/
(20.30) (20.31)
z 1 sinh L √ 2z z/
(20.32)
z −(z + ) dD3 dL (, , , L, z, t) = cosh L 2z
(20.33)
dD3 d (, , , L, z, t) =
z+ z −1 dD3 dz (, , , L, z, t) = + sinh L √ √ 2z z/ 2z2 z/ z z z+ × sinh L sinh L + √ 2 4z z/ L(z + ) z − cosh L (20.34) 2 4z The final denominator term D4 has the following derivatives: z −L2 (z + ) dD4 d (, , , L, z, t) = sinh L √ 2 4 z/ z −L(z + ) dD4 d (, , , L, z, t) = cosh L 2 2 z L z dD4 d (, , , L, z, t) = cosh L 2z
(20.35)
(20.36)
(20.37)
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Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc©) Parameter Estimation of LoTEC
20.9
z z+ √ sinh L 2 z z+ z + cosh L (20.38) 2z L(z + ) z z 1 dD4 dz (, , , L, z, t) = cosh L − cosh L 2 2z 2z L2 (z + ) z + sinh L (20.39) √ 4z z
dD4 dL (, , , L, z, t) =
The derivatives of the denominator can be obtained by summation over the four terms: dD d(, , , L, z, t) =
4
dDi d(, , , L, z, t)
(20.40)
dDi d (, , , L, z, t)
(20.41)
dDi d (, , , L, z, t)
(20.42)
dDi dL (, , , L, z, t)
(20.43)
dDi dz (, , , L, z, t)
(20.44)
i=1
dD d (, , , L, z, t) =
4 i=1
dD d (, , , L, z, t) =
4 i=1
dD dL (, , , L, z, t) =
4 i=1
dD dz (, , , L, z, t) =
4 i=1
The partial derivatives of the temperature expression Tm with respect to the various variables can be expressed in terms of the following numerator and denominator terms: dTm d(, , , L, z, t) =
D(, , , L, z, t)dN d(, , , L, z, t) − N(, , , L, z, t)dD d(, , , L, z, t) D2 (, , , L, z, t) (20.45)
dTm d(, , , L, z, t) =
D(, , , L, z, t)dN d (, , , L, z, t) − N(, , , L, z, t)dD d (, , , L, z, t) D2 (, , , L, z, t) (20.46)
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Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc©) 20.10
Transient and Cyclic Calculations
dTm d (, , , L, z, t) D(, , , L, z, t)dN d (, , , L, z, t) − N(, , , L, z, t)dD d (, , , L, z, t) D2 (, , , L, z, t) (20.47) dTm dL(, , , L, z, t) =
D(, , , L, z, t)dN dL(, , , L, z, t) − N(, , , L, z, t)dD dL(, , , L, z, t) D2 (, , , L, z, t) (20.48) dTm dz(, , , L, z, t) =
=
D(, , , L, z, t)dN d z(, , , L, z, t) − N(, , , L, z, t)dD d z(, , , L, z, t) D2 (, , , L, z, t) (20.49)
The total derivatives of the temperature Tm with respect to the variables can be found with an implicit z as a function of the variables: DTm D (, , , L, z, t) = dTm d(, , , L, z, t) + dTm dz(, , , L, z, t) × dz d(, , , L, z) (20.50) DTm D (, , , L, z, t) = dTm d(, , , L, z, t) + dTm dz(, , , L, z, t) × dz d (, , , L, z) (20.51) DTm D (, , , L, z, t) = dTm d (, , , L, z, t) + dTm dz (, , , L, z, t) × dz d (, , , L, z) (20.52) DTm DL(, , , L, z, t) = dTm dL(, , , L, z, t) + dTm dz(, , , L, z, t) × dz dL (, , , L, z) (20.53) Having determined the form of the derivative of Tm (, , , L, z, t), we can obtain the derivative of TI (t) through summation over the roots zi from Eq. (20.2): DTI D(, , , L, z, t) =
∞
dTm d (, , , L, zi , t)
(20.54)
dTm d (, , , L, zi , t)
(20.55)
dTm d (, , , L, zi , t)
(20.56)
dTm dL (, , , L, zi , t)
(20.57)
i=1
DTI D(, , , L, z, t) =
∞ i=1
DTI D (, , , L, z, t) =
∞ i=1
DTI DL(, , , L, z, t) =
∞ i=1
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Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc©) Parameter Estimation of LoTEC
20.11
Parametric Estimation Using Gauss Linearization The Gauss algorithm is used to solve the nonlinear estimation problem, using the derivatives DTI D, DTI D, and DTI D in Eqs. (20.54) through (20.56). These expressions are also known as the “sensitivity coefficients.” To simplify the description of the algorithm, let the parameters be summarized using the column vector as the transpose of (1 , 2 , , ). The initial value of the parameters are denoted by ˆ 0 , values of the parameters during iterations u = 1, 2, . . . are denoted by ˆ u, and the final estimate will be denoted ˆ . The sensitivity matrix X contains DTI (t j )/Dk, for row j = 1, . . . , 122 and column k = 1, 2, 3, 4. The sensitivity matrix Xu is computed using ˆ u. The sensitivity coefficients can be plotted as functions of time t and compared. The independence of these functions is a condition for identifiability in the estimation problem. To reduce the variability in the magnitudes among the sensitivity coefficients, normalized sensitivity coefficients are plotted: V = DTI D, V = DTI D, V = DTI D, V L = LDTI DL. As shown in Fig. 20.4, the sensitivity functions are independent over the timeframe of the experiments, from 1 to 7 h, so this timeframe is sufficient for determining the parameters , , and . Define the actual temperature observations y = (y1 , y2 , . . . , y122 )T , where the first 61 observations are from the first LoTEC and the remaining 61 observations are from the second LoTEC. The residual error between observations and the predicted temperatures is given by eu = (e1u, e2u, . . . , e122u)T , where e ju = y j − TI (t j ), where TI (t j ) is evaluated
Figure 20.4
Sensitivity coefficients for calculated temperatures.
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Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc©) 20.12
Transient and Cyclic Calculations
TABLE 20.1
Summary of Parameter Estimation by Gauss Method
Trial no.
ˆ 1 × 108
ˆ 1 × 108
ˆ × 106
ˆ × 106
SSe
0 1 2 3 4 Uncertainty
5.0339 5.109 5.1061 5.1061 5.1061 0.1168
5.0004 5.0754 5.0725 5.0725 5.0725 0.1157
3.0805 3.1905 3.1943 3.1943 3.1943 0.0438
3.905 1.6825 1.6428 1.6425 1.6425 0.9117
1.311 0.4299 0.4268 0.4268 0.4268
using ˆ u. The estimates are chosen to minimize the sum of squared residuals: SS e = eT e =
122
e2j
(20.58)
j=1
In each iteration an improved estimate of the parameters can be obtained using ˆ u+1 = ˆ u + (XuT Xu)−1 Xueu
(20.59)
The initial estimates ˆ 0 were calculated from measurements and/or estimates of the eight primitive variables. Poor initial estimates result in imaginary roots to Eq. (20.2). Approximately four iterations were required to give the minimum least squares of the errors. As shown in Table 20.1, convergence occurred rapidly using these starting values. For the experimental data used, the values of ˆ 1 , ˆ 2 , ˆ 2 , and ˆ 1 are 5.1061 × 10−8 m2 /s, 5.0725 × 10−8 m2 /s, and 3.1943 × 10−6 m/s, 1.6425 × 10−6 Hz. The total sum of squared residuals SSe at this solution is 0.4268K 2 . This is a good fit to the data, as shown in Fig. 20.5. An estimate of the uncertainties in the estimated parameters can be obtained using the SSe and the sensitivity matrix Xu using ˆ . An estimate of the error variance is given by s2 = SSe /(n − p), where n is the number of observations (122) and p is the number of estimated parameters (4). Then
−1 Var[ˆ ] = s2 XuT Xu
(20.60)
The square roots of the diagonal terms of this expression are the standard deviations of the individual estimated coefficients. Using twice the standard deviations provides an uncertainty amount for each parameter. The uncertainties for ˆ 1 , ˆ 2 , ˆ 2 , and ˆ 1 are, respectively, 2.1, 2.3, 1.4,
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Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc©) Parameter Estimation of LoTEC
20.13
45
40 Temperature (°C) y
35
TI 30
25
20 0
Figure 20.5
1.5
2
2.5
3
3.5
4.5 4 Time (h)
5
5.5
6
6.5
7
Comparison of measured and calculated temperatures for two LoTECs.
and 55 percent. Fortunately, the temperature is only weakly dependent on the value of . References 1. F. C. Wessling, L. S. Stoieck, A. Hoehn, S. Woodard, S. O’Brien, and S. Thomas, “Low c ) and Phase Change Materials (PCMs) Temperature, Low Energy Carrier (LoTEC for Biological Samples,” 30th Proceedings of the International Conference on Environmental Systems, paper no. 2000012280, Toulouse, France, July 2000. 2. J. V. Beck and K. J. Arnold, Parameter Estimation in Engineering and Science, Wiley, 1977.
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T1: IML July 21, 2005
13:25
Parameter Estimation of Low Temperature, Low Energy Carrier (LoTECc©)
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14
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Source: HeatTransfer Calculations
Part
4 HeatTransferCoefficient Determination
As can be seen in a number of calculations presented earlier in this handbook, heattransfer coefficients play a key role. Their determination can be difficult in instances in which surface geometries are complex or surfaces are fouled or corroded. This brief part consists of two chapters, both by academics. The first confronts geometrically complex surfaces exposed to convection by using liquid crystal or infrared thermography in conjunction with a semiinfinite solid assumption. Where there is buildup of scale or contamination around the tubes of a heat exchanger, or corrosion of internal baffles alters the individual film coefficient for heat transfer on the side affected, determining individual film coefficients experimentally is simplified by a method called the Wilson analysis; this is described in the second chapter.
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1
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HeatTransferCoefficient Determination
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Source: HeatTransfer Calculations
Chapter
21 Calculation of Convective HeatTransfer Coefficient Using Semiinfinite Solid Assumption
Srinath V. Ekkad Mechanical Engineering Department Louisiana State University Baton Rouge, Louisiana
Introduction Twodimensional surface heattransfer distributions on convective surfaces provide a complete picture for complex flowsolid interactions in heat transfer. The semiinfinite solid assumption provides an opportunity to obtain detailed surface heattransfer behavior using thermography. Typically, liquid crystal or infrared thermography is used in conjunction with the semiinfinite solid assumption. Calculation of HeatTransfer Coefficient Using Semiinfinite Solid Assumption The local heattransfer coefficient h over a surface coated with liquid crystals can be obtained by using a onedimensional (1D) semiinfinite solid assumption for the test surface. The 1D transient conduction 21.3
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Calculation of Convective HeatTransfer Coefficient Using Semiinfinite Solid Assumption 21.4
HeatTransferCoefficient Determination
equation, the initial condition, and the convective boundary condition on the liquid crystal coated surface are k
∂2T ∂T = Cp ∂t ∂ x2
(21.1)
Boundary conditions: At t = 0, T = Ti At x = 0, −k
∂T = h(Tw − Tm) ∂x
As x → ∞, T = Ti Solving Eq. (21.1) with prescribed initial and boundary conditions, the solution thus obtained is 2 T(x, t) − Ti x h t hx = erfc + − exp √ k 2 t Tm − Ti k2 √ h t x × erfc (21.2) + √ k 2 t Since the surface heattransfer coefficient is required, one obtains the nondimensional temperature at the convective boundary surface (at x = 0) [1]: √ 2 h t h t Tw − Ti erfc = 1 − exp (21.3) k Tm − Ti k2 The colorchange temperature or the prescribed wall temperature Tw is obtained either using a liquid crystal coating or through infrared thermography. The initial temperature Ti of the test surface and the mainstream temperature Tm are measured before and during the test, respectively. The time of the wall temperature change from the initial transient test temperature is determined. The local heattransfer coefficient h can be calculated from Eq. (21.3). Use of a lowconductivity material such as plexiglass or polycarbonate and maintaining test conditions such that the time of temperature imposition on the surface is short enables one to maintain the validity of the semiinfinite solid assumption on the test surface as the test duration does not allow for temperature penetration into the test surface. The theory allows for a shortduration experiment, and complex geometries can be tested for obtaining heattransfer coefficients. Other variations of these experiments involve the use of a secondary flow that is at a temperature different from that of the primary flow. In
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Calculation of Convective HeatTransfer Coefficient Using Semiinfinite Solid Assumption Semiinfinite Solid Assumption
21.5
that situation, the heattransfer coefficient is dependent on the mixing temperature of both flows. There are two unknowns in Eq. (21.3). Film cooling over a surface is a threetemperature problem involving the mainstream temperature Tm, the coolant temperature Tc , and the wall temperature Tw . In filmcooling situations, the mainstream temperature Tm in Eq. (21.3) must be replaced by a film temperature Tf , which is a mixed temperature between the mainstream and coolant temperatures that governs the convection from the liquidcrystalcoated surface. To find the unknown Tf in terms of known quantities Tm and Tc , a nondimensional temperature is defined as the film cooling effectiveness (): =
Tf − Tm Tc − Tm
or
Tf = (Tc − Tm) + Tm = Tc + (1 − ) Tm
(21.4)
Replacing Tm in Eq. (21.3) by Tf from Eq. (21.4), we get the following equation with two unknowns, h and : √ 2 h t h t erfc [ Tc + (1 − ) Tm − Ti ] Tw − Ti = 1 − exp 2 k k (21.5) To obtain both the heattransfer coefficient h and the film effectiveness , it is necessary to obtain two equations with two unknowns (h and ) and solve for h and . Therefore, two similar transient tests are run to obtain two different sets of conditions. Concluding Remarks A method for calculating heattransfer coefficients on complex geometries exposed to convection is explained. The semiinfinite solid assumption is used to determine the heattransfer coefficient in a shortduration transient test. The advantage of obtaining full surface results for complex geometries is extremely significant in improving quality of heat transfer experiment. Several results are presented in Ref. 2. References 1. Incropera, F. P., and Dewitt, D. P., Introduction to Heat Transfer, 4th ed., Wiley, New York, 2002. 2. Ekkad, S. V., and Han, J. C., “A Transient Liquid Crystal Thermography Technique for Gas Turbine Heat Transfer Measurements,” Measurement Science & Technology, special edition on gas turbine measurements, vol. 11, pp. 957–968, July 2000.
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T1: IML July 8, 2005
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Calculation of Convective HeatTransfer Coefficient Using Semiinfinite Solid Assumption
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Source: HeatTransfer Calculations
Chapter
22 Determination of HeatTransfer Film Coefficients by the Wilson Analysis
Ronald J. Willey Department of Chemical Engineering Northeastern University Boston, Massachusetts
Introduction Over time, the performance of shellandtube heat exchangers has changed. The common explanation is “fouling,” the buildup of scale or contamination around the tubes. Another possibility is the corrosion of internal baffles or related components that alter the individual film coefficient for heat transfer on the side affected. Determination of individual film coefficients experimentally is simplified by a method called the Wilson analysis. This method is most successful for singlephase (with no boiling or condensation on the side of interest) heat exchangers. It requires the ability to vary the flow rate of the fluid on the side of interest, while holding the flow rate of the opposite fluid constant. The overall heattransfer coefficient U is determined for each flow rate. By plotting the inverse of U versus the inverse of the velocity or the flow rate, normally raised to a fractional power between 0.6 and 0.8, the inverse of the film coefficient hi for the side of interest can be experimentally determined. The film coefficient can be compared to values predicted by correlations in the literature. If the experimentally determined
22.1
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Determination of HeatTransfer Film Coefficients by the Wilson Analysis 22.2
HeatTransferCoefficient Determination
film coefficient is 25 percent or more below that predicted, then most likely the exchanger internals are not functioning as designed. The results assist in troubleshooting heat exchangers, and pinpointing what service may be needed to repair (e.g., cleaning versus replacement). My personal experience is in the academic laboratory. Students analyze several types of heat exchangers. They determine the experimental film coefficients by the Wilson analysis, and then compare the results to the empirical correlations as offered in textbooks and handbooks. Amazingly, the tube side matches predictions by the DittusBoelter correlation (or related tube correlations) within ± 20 percent. For shellandtube heat exchangers, the shellside film coefficient often varies from that predicted by correlations. The closest empirical correlation is the Donohue equation [1]. It predicts results well for new shellandtube (S&T) exchangers that have no internal leakage between baffles. For plate heat exchangers, experimental work at Northeastern University in the early 1960s provided some of the first empirical correlations for film coefficients with these exchangers. The Wilson analysis method helped the researchers obtain film coefficients that later led to Nusselt number correlations for plate heat exchangers operating in the turbulent flow regime [2]. Elemental Tubular HeatExchanger Design Equations Heat exchangers are sized and designed by using an overall heattransfer coefficient U (for further background about heat transfer, see, for example, the text by McCabe et al. [1]). The inverse of the overall heattransfer coefficient, in its simplest form, is a sum of resistances due to heat transfer attributed to convection resistance on both sides of the wall that separates the two fluids, conduction resistance across the wall, and fouling resistances: 1 1 = + other resistances Ui hi
(22.1)
where Ui is the overall heattransfer coefficient for the design equation Q = Ui Ai T lm where
(22.2)
Q = heating or cooling duty required A = area required for heat transfer Tlm = logmean driving force temperature between fluids (definition varies depending on flow patterns and geometry) hi = [in Eq. (22.1)] film coefficient for convective heat transfer on the fluid side of interest
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Determination of HeatTransfer Film Coefficients by the Wilson Analysis HeatTransfer Film Coefficients and Wilson Analysis
22.3
Individual Film Coefficients Estimated by Empirical Equations Film coefficients are defined by Newton’s law of cooling hi =
Q (Tb − Tw ) A
(22.3)
where Q/A = heat flux flowing from the hot fluid to the cold fluid Tb = local temperature of the bulk fluid (the local hot stream temperature as expressed above) Tw = local temperature at the heatexchanger wall Over the years, engineers and scientists have developed empirical equations that allow the estimation of film coefficients depending on the flow regime and whether a phase change is involved. For a singlephase fluid flowing in the turbulent flow regime, a classical empirical equation is the DittusBoelter relation [3] 0.8 n NNu = 0.023NRe N Pr
(22.4)
where n = 0.4 for heating and 0.3 for cooling, and NNu =
h kDi
vDi cp NPr = k
NRe =
(Nusselt number for tubeside fluid)
(22.5)
(Reynolds number for tubeside fluid)
(22.6)
(Prandtl number for tubeside fluid)
(22.7)
where cp = fluid’s heat capacity Di = inside diameter in tube k = fluid’s thermal conductivity v = average velocity of fluid inside tube = fluid’s viscosity = fluid’s density For shellside correlations, see chap. 11 in Ref. 4 or the Donohue correlation as presented in Ref. 1, chap. 15. Individual Film Coefficients by Experimental Means The Wilson analysis (originally described by Wilson in 1915 [5]), provides researchers and engineers with an experimental method for determining individual heattransfer film coefficients in convective heat
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Determination of HeatTransfer Film Coefficients by the Wilson Analysis 22.4
HeatTransferCoefficient Determination
transfer in which no phase change occurs. The objective is to maintain constant flow rate on one side of the heat exchanger while varying the flow rate on the opposite side (the side of interest). The overall heattransfer coefficient U is determined for each point on the basis of the flow rates and temperatures acquired for the exchanger under investigation. The area used in the calculation is the area normal to the heat transfer on the side that flow rate is being varied. Since flow rate for only one side is varied, the change in the overall heat transfer coefficient is directly related to a change in the individual convective film coefficient hi for the side on which the flow rate is being varied. By plotting 1/U versus 1/flow ratex using the proper exponent x (detailed below), a straight line results. This plot is sometimes called a Wilson plot. The intercept is equaled to the sum of all resistances to heat transfer except the side that is being varied. The resistance on the variable flow side, 1/ hi for each point, is given as the difference of 1/Ui minus the intercept found from in the Wilson plot. The exponent x depends on the flow regime, and is exchangerdependent. Typically, 0.8 is used for turbulent flow inside tubes, 0.33 for laminar flow inside tubes, and 0.6 for crossflow for tube bundles. Problem Statement Determine the individual tubeside film coefficients and corresponding Nusselt (NNu ) numbers for the five data points listed in Table 22.1 for a 12ftlong, sixtube (i.d. = 0.425 in.), singlepass shellandtube heat exchanger. Water is the working fluid used in the experiment on both sides. The cold fluid is on the tube side. The streams are flowing countercurrently to each other. Compare the results to the NNu predicted by an empirical formula available in the literature. Solution Outline 1. Determination of Ai , area normal to heat transfer for the heat exchanger.
TABLE 22.1
Data Acquired in a ShellandTube HeatExchanger Experiment Flow rate
Point
Cold flow, L/min
1 2 3 4 5
4.7 8.5 16.3 20.2 35.7
Stream temperatures Cold in, 12.2 11.7 11.1 11.1 10.6
◦C
Cold out, ◦ C
Hot in, ◦ C
Hot out, ◦ C
40.6 37.8 33.9 32.2 27.1
45.0 45.0 45.0 45.0 45.0
40.6 37.6 32.7 30.9 25.4
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Determination of HeatTransfer Film Coefficients by the Wilson Analysis HeatTransfer Film Coefficients and Wilson Analysis
22.5
2. Determination of the heating load for each point acquired. 3. Determination of Tlm for each data point acquired. 4. Determination of Ui for each data point. 5. Determination of vi for each point acquired. 6. Formation of Wilson plot by plotting 1/Ui against 1/vix for each point; see “Introduction” above for suggested values of x to use. 7. Determination of the underlying resistances to heat transfer for all resistances except the film coefficient of interest. This is the intercept given on the Wilson plot. 8. Determination of the film coefficient for each point (1/Ui —the intercept found above). 9. Comparison of each film coefficient to its counterpart empirical values. Huge differences indicate fouling or other problems. Determination of area
Ai = Di LNt where Di = 0.425 in. = 0.425 in. · 0.0254 m/in. = 0.0108 m L = 12 ft = 12 ft · 0.3049 m/ft = 3.66 m Ai = · 0.0108 m · 3.66 m · 6 = 0.745 m2 Determination of Qc for point 1
Given a cold stream average temperature of (40.6 + 12.2)/2 = 26.4◦ C from the tables, find the following properties of water: = 996 kg/m3 Cp = 4176 J/kg C = 0.866 · 10−3 kg/m s k = 0.622 W/m C NPr = 5.81 Qc = m Cp(Tc out − Tc in ) where m is the mass flow rate, kg/s. Then m = flow rate times = 4.7 L/min · 1 m3 /1000 L · 1 min/60 s · 996 kg/m3 = 0.078 kg/s Qc = 0.078 kg/s · (40.6◦ C − 12.2◦ C) · 4176 J/kg C = 9250 W
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Determination of HeatTransfer Film Coefficients by the Wilson Analysis 22.6
HeatTransferCoefficient Determination
Few details are provided for the hot stream for comparison of the heat balance; therefore, Qh cannot be computed and compared. Normally, this value should be within 5 percent of Qc . Determination of ∆Tlm for point 1
For countercurrent flow, Tlm is as follows: Tlm =
=
(Th in −Tc out ) − (Thout −Tc in ) Thin −Tc out ln Thout −Tc in (45 − 40.6) − (40.6 − 12.2) = 12.9◦ C 45 − 40.6 ln 40.6 − 12.2
Determination of Ui for point 1
Ui =
Q/A = 9250 W/0.745 m2 /12.9◦ C = 962 W/m2 C Tlm
Determination of vi for point 1
vi = flow rate/total crosssectional area Flow rate = 4.7 L/min · 1 m3 /1000 L · 1 min/60 s = 7.83 × 10−5 m3 /s Crosssectional area for flow = Nt · · Di 2 /4 = 6 tubes · · (0.0108 m)2 / 4 = 5.5 × 10−4 m2 vi = (7.83 × 10−5 m3 /s)/(5.5 × 10−4 m2 ) = 0.142 m/s Formation of Wilson plot
See Table 22.2 and Fig. 22.1 for the completed calculations. The plot is of 1/Ui versus 1/vi for each point. Determination of the underlying resistances
From Fig. 22.1, the data are linear with the exponent on the velocity equal to 0.8 (turbulent flow throughout). In this case linear regression can be used to determine the intercept of 1.16 × 10−4 m2 ◦ C/W. Determination of film coefficient for each point
From Fig. 22.1 1 1 = + intercept U1 h1
(22.8)
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40.6 37.8 33.9 32.2 27.1
45.0 45.0 45.0 45.0 45.0
40.6 37.6 32.7 30.9 25.4
Hot out, ◦C 970 1410 2200 2480 3370
Ui , W/m2◦ C
1100 1690 2970 3510 5600
Experimental hi , W/m2◦ C
1060 1680 2760 3240 4960
Empirical hi , W/m2◦ C
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12.2 11.7 11.1 11.1 10.6
4.7 8.5 16.3 20.2 35.7
1 2 3 4 5
Hot in, ◦C
Cold out, ◦C
Cold flow, L/min
Point
Cold in, ◦C
Stream temperatures
Completed Calculations for Experimental hi and Resultant Nusselt Number Comparisons
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Flow rate
TABLE 22.2
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Determination of HeatTransfer Film Coefficients by the Wilson Analysis
22.7
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Determination of HeatTransfer Film Coefficients by the Wilson Analysis 22.8
HeatTransferCoefficient Determination
0.0012
1/Ui m2 C/W
0.001 0.0008 0.0006 1/hi = 1/Ui − intercept 0.0004 0.0002 Other resistances’ contribution
0 0
1
2
Intercept = 1.16 X 10−4 Figure 22.1
1/vi0.8
3
4
5
(s/m)0.8
Wilson plot for data given in problem.
1 1 = − intercept = 1.038 × 10−3 − 0.116 × 10−3 = 0.922 m2 C/W h1 U1 h1 =
1 1 = (m2 C/W)−1 = 1090 W/m2 C h1 0.922
NNu1 =
1090 W/m2 C · 0.0108 m h1 D = = 18.8 k 0.622 W m C
Comparison of each film coefficient to its counterpart empirical values 0.4 NNu = 0.023N 0.8 Re N Pr
(22.4)
NRe =
vDi 0.142 m/s · 0.0108 m · 996 kg/m3 = 1770 = 0.866 · 10−3 kg/m s
NPr =
cp = 5.81 k
NNu empir = 0.023 · 17700.8 · 5.810.4 = 18.4 hi = Nu ·
k 18.4 · 0.622 W/m C = = 1060 W/m2 C D 0.0108 m
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Determination of HeatTransfer Film Coefficients by the Wilson Analysis HeatTransfer Film Coefficients and Wilson Analysis
22.9
The percentage difference is (18.8 − 18.4)/18.4 · 100% = 2 percent, which is considered excellent agreement in this case; the exchanger is functioning as expected on the tube side. References 1. W. McCabe, J. Smith, and P. Harriott, Unit Operations of Chemical Engineering, McGrawHill, New York, 2001, chap. 12. 2. R. A. Troupe, J. C. Morgan, and J. Prifti, Chemical Engineering Progress, 56 (1):124– 128 (1960). 3. F. W. Dittus and L. M. K. Boelter, Publication on Engineering (Univ. Calif. Berkeley), 2:443 (1930). 4. R. H. Perry and D. W. Green, Perry’s Chemical Engineers’ Handbook, McGrawHill, New York, 1997. 5. R. E. Wilson, Transactions ASME, 37:47 (1915).
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Determination of HeatTransfer Film Coefficients by the Wilson Analysis
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10
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Source: HeatTransfer Calculations
Part
5 Tubes, Pipes, and Ducts
Even when a fluid flows through a circular pipe, heattransfer calculations are not as simple as a practicing engineer would like them to be. The following calculations, which were contributed by academics, a consultant, and an engineer working in industry, are useful for dealing with reallife situations:
Assessment of performance of heatexchanging equipment and development of heattransfer correlations The case of a pumped fluid entering a heated pipe as a slightly subcooled or saturated liquid, being heated sufficiently to cause vaporization, and leaving the pipe as liquidvapor mixture or totally vapor; or the inverse, which is related, in which the vessel pressure and the exit pressure fix the driving force, and the flow rate from the vessel is required When a “waxy” crude oil containing highmolecularweight alkanes or paraffin waxes flowing through a pipeline is exposed to a cold environment that is below its freezingpoint temperature (or solubility temperature), and solids deposition on the pipe wall is likely to occur The combustor of a hypersonic vehicle over a critical portion of its flight trajectory
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1
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Tubes, Pipes, and Ducts
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Source: HeatTransfer Calculations
Chapter
23 Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube
Afshin J. Ghajar and Jaeyong Kim School of Mechanical and Aerospace Engineering Oklahoma State University Stillwater, Oklahoma
Introduction
Heattransfer measurements in pipe flows are essential for assessment of performance of heatexchanging equipment and development of heattransfer correlations. Usually, the experimental procedure for a uniform wall heat flux boundary condition consists of measuring the tube outsidewall temperatures at discrete locations and the inlet and outlet bulk temperatures in addition to other measurements such as the flow rate, voltage drop across the test section, and current carried by the test section. Calculations of the local peripheral heattransfer coefficients and local Nusselt numbers thereafter are based on knowledge
23.3
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Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube 23.4
Tubes, Pipes, and Ducts
of the tube insidewall temperatures. Although measurement of the insidewall temperature is difficult, it can be accurately calculated from the measurements of the outsidewall temperature, the heat generation within the tube, and the thermophysical properties of the pipe material (electrical resistivity and thermal conductivity). This chapter presents the general finitedifference formulations used for this type of heattransfer experiment, provides specific applications of the formulations to single and twophase convective heat transfer experiments, gives details of implementation of the calculation procedure (finitedifference method) in a computer program, and shows representative reduced heattransfer results. FiniteDifference Formulations The numerical solution of the conduction equation with internal heat generation and nonuniform thermal conductivity and electrical resistivity was originally developed by Farukhi (1973) and introduced by Ghajar and Zurigat (1991) in detail. The numerical solution is based on the following assumptions: 1. Steadystate conditions exist. 2. Peripheral and radial wall conduction exists. 3. Axial conduction is negligible. 4. The electrical resistivity and thermal conductivity of the tube wall are functions of temperature. On the basis of these assumptions, expressions for calculation of local insidewall temperature and heat flux and local and average peripheral heattransfer coefficients will be presented next. The heat balance on a control volume of the tube at a node P (refer to Fig. 23.1) is given by q˙ g = q˙n + q˙e + q˙s + q˙w
(23.1)
From Fourier’s law of heat conduction in a given direction n, we have q˙ = −kA
dT dn
(23.2)
Now, substituting Fourier’s law and applying the finitedifference formulation for a control volume on a segment (slice) of the tube with nonuniform thermal conductivity in Eq. (23.1), we obtain n− n+ −1 q˙n = (23.3) + An (TP − TN) kP kN
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Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube InsideWall Parameters from OutsideWall Temperatures
23.5
N
Control Volume Interface Node Point
δn+
qn
δn
P qe
δeδ w δs δ s+ qs
qw

+
δw

δe +
S
W
x
E
r z
θ
y Figure 23.1
Finitedifference node arrangement on a segment (slice).
q˙e =
−1
e− e+ + kP kE
−1
s− s+ q˙s = + kP kS
w− w+ q˙w = + kP kW
Ae (TP − TE )
(23.4)
As (TP − TS)
(23.5)
−1
Aw (TP − TW )
(23.6)
Note that, in order to deal with the nonuniform thermal conductivity, the thermal conductivity at each control volume interface is evaluated as the sum of the thermal conductivities of the neighboring node points based on the concept that the thermal conductance is the reciprocal of the resistance (Patankar, 1991). The heat generated at the control volume is given by q˙ g = I 2 R
(23.7)
Substituting R = l/Ac into Eq. (23.7) gives q˙ g = I 2
l Ac
(23.8)
Substituting Eqs. (23.3) to (23.6) and (23.8) into Eq. (23.1) and solving for TS gives
TS = TP − q˙ g − q˙n − q˙e − q˙w
s− s+ + kP kS
−1 As
(23.9)
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Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube 23.6
Tubes, Pipes, and Ducts
Equation (23.9) is used to calculate the temperature of the interior nodes. Once the local insidewall temperatures are calculated from Eq. (23.9), the local peripheral insidewall heat flux can be calculated from the heatbalance equation, Eq. (23.1). From the local insidewall temperature, the local peripheral insidewall heat flux, and the local bulk fluid temperature, the local peripheral heattransfer coefficient can be calculated as follows: h=
q˙ Tw − Tb
(23.10)
Note that, in these analyses, it is assumed that the bulk fluid temperature increases linearly from the inlet to the outlet according to the following equation: Tb = Tin +
(Tout − Tin )z L
(23.11)
The local average heattransfer coefficient at each segment can be calculated by the following equation: h¯ =
q˙ T¯w − Tb
(23.12)
In this section, we have presented the basic formulations of the local insidewall temperature, the local peripheral heattransfer coefficient, the local average heattransfer coefficient, and the overall heattransfer coefficient from the given local outsidewall temperature at a particular segment (slice) of an electrically heated circular tube. Next, we will show specific applications of the formulations to single and twophase convective heattransfer experiments. Application of the FiniteDifference Formulations In this section, the finitedifference formulations developed in the previous section are applied to actual heattransfer experiments. The experiments were performed to study single and twophase heat transfer in an electrically heated tube under a variety of flow conditions. The obtained data were then used to develop robust single and twophase heattransfer correlations. For this purpose, accurate heattransfer measurement is critical; therefore, finitedifference formulations were used as a key tool in obtaining accurate heattransfer coefficients from the measured outsidewall temperatures.
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Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube InsideWall Parameters from OutsideWall Temperatures
23.7
Experimental setup
A brief description of the experimental setup is presented to help the readers understand how the finitedifference formulations are used in actual experimental work. A schematic diagram of the overall experimental setup is shown in Fig. 23.2. The experimental setup shown in the figure is designed to systemically collect pressure drop and heattransfer data for singleand twophase flows for various flow conditions and flow patterns (in case of twophase flow) and different inclination angles. The test section is a 27.9mm i.d. (inside diameter) straight standard stainlesssteel 316 schedule 10S pipe with a length : diameter ratio of 100. The uniform wall heat flux boundary condition is maintained by a welder which is a power supply to the test section. The entire length of the test section is insulated using fiberglass pipe wrap insulation, which provides the adiabatic boundary condition for the outside wall. Ttype thermocouples are cemented with an epoxy adhesive having high thermal conductivity and electrical resistivity to the outside wall of the test section at uniform intervals of 254 mm (refer to Fig. 23.3). There are 10 thermocouple stations with four thermocouples in 90◦ peripheral intervals at each station in the test section. The inlet and exit bulk temperatures are measured by Ttype thermocouple probes. To ensure a uniform fluid bulk temperature at the inlet and exit of the test section, a mixing well is utilized. More details of the experimental setup may be found in Ghajar et al. (2004). Heattransfer measurements at uniform wall heat flux boundary condition are carried out by measuring the outside wall temperatures at the 10 thermocouple stations along the test section and the inlet and outlet bulk temperatures in addition to other measurements such as the flow rates of test fluids, system pressure, voltage drop across the test section, and current carried by the test section. A National Instruments data acquisition system is used to acquire the data measured during the experiments. The computer interface used to monitor and record the data is a LabVIEW Virtual Instrument written for this specific application.
Finitedifference formulations for the experimental setup
In order to apply the finitedifference formulations developed in the previous section to the experimental setup, the grid configuration of the test section is designed as shown in Fig. 23.4 according to the basic node configuration shown in Fig. 23.1. The computation domain is divided into control volumes of uniform thickness except at the inside
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Figure 23.2
Regulator
Drain
Shell and Tube Heat Exchanger
Drain
Schematic of the experimental setup.
Coriolis Water Flow Meter
Transmitter
Filter & Separator
Filter Pump
Air Compressor
Drain
Water Storage Tank
Air Out
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Main Control Valve for Water Flow
Coriolis Air Flow Meter
Air Side Heat Exchanger
SCXI 1125 Module
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Main Control Valve for Air Flow
SCXI1000
SCXI
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NIDAQ Card CPU with LabVIEW Virtual Instrument
SCXI 1102/B/C Modules
Thermocouple Probe
Mixing Well
Insulation
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Clear Polycarbonate Observation Section
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Electric Screw Jack
StainlessSteel Test Section
Lincoln SA750 Welder
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Clear Polycarbonate Calming & Observation Section
Aluminum IBeam
Mixing Well
Thermocouple Probe
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Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube
23.8
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25.4 cm 25.4 cm
C Figure 23.3
Test section.
D
A
Thermocouple
B 2.7864 cm 3.3401 cm
25.4 cm 25.4 cm 25.4 cm 17.78 cm
Thermocouple Station Number
25.4 cm
Copper Plate
Flow Direction
25.4 cm
25.4 cm
264.16 cm
Stainless Steel 316 2.54 cm (1 in.) Schedule 10S Test Section
25.4 cm
Tail of Flow Direction
17.78 cm
Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube
23.9
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Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube 23.10
Tubes, Pipes, and Ducts
ZeroThickness Control Volume at Boundary
Control Volume Interface
(and also
of of
(and also
of of
(a)
Node for Thermocouple Station L
(b)
Grid configuration for the experimental setup: (a) at a segment (slice); (b) at axial direction. Figure 23.4
and outside walls of the tube in the radial direction, where the control volumes have a zero thickness. The node points are placed in the centerpoint of the corresponding control volumes, and the node points along the circumferential and axial directions are placed in accordance with the actual thermocouple locations at the test section. According to the assumption that the axial conduction is negligible, each thermocouple station (segment) is solved independently of the others. The calculation in a segment is marched from the outsidewall node
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23.11
points, the temperatures of these nodes are experimentally measured, to the insidewall node points based on the heat balance at each control volume. First, determination of the geometric variables at a node point (i, j) is as follows. The distances from a grid point to the interface of the control volume are (Do − Di ) NCVL r n+ = n− = s+ = s− = = (23.13) 2 2 for the radial direction and 2 ri
[email protected] (23.14) (e+ )i = (e− )i = (w+ )i = (w− )i = 2 for the circumferential direction. Because of the zerothickness control volume layer at the boundaries (i = 0 and NCVL + 1), we have n+ = 0 at i = 1 and s+ = 0 at i = NCVL . The areas of all faces for a control volume are
2 ri + (n− )i zk (23.15) ( An)i =
[email protected] (23.16) ( Ae )i = ( Aw )i = (n− )i + (s− )i zk
2 ri − (s− )i zk (23.17) ( As )i =
[email protected] The length and the crosssectional area of the control volume in the axial direction are set as l = zk ( Ac )i =
2 ri r
[email protected] (23.18) (23.19)
Now we can calculate the heat generation and the heat flux at a control volume by applying these geometric variables. Note that because of the zerothickness control volume layer at the boundaries (i = 0 and NCVL + 1), all heat fluxes (qn, qe , qs , and qw ) and the heat generation (qg ) terms are set to zero at i = 0 and NCVL + 1. Therefore, Eqs. (23.3), (23.4), (23.5), (23.6), and (23.8) for i = 1, 2, . . . , NCVL and j = 1, 2, . . . ,
[email protected] become as follows: ki, j for i=1 − ( An)i Ti, j − Ti−1, j n (q˙n)i, j = − + −1 n + n for i = 2, 3, . . . , NCVL ( An)i Ti, j − Ti−1, j ki, j ki−1, j (23.20)
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Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube 23.12
(q˙e )i, j
(q˙s )i, j
Tubes, Pipes, and Ducts
−1 (e− )i (e+ )i = + ( Ae )i Ti, j − Ti, j+1 for ki, j ki, j+1
i = 1, 2, . . . , NCVL
(23.21) −1 + − s + s ( As )i Ti, j − Ti+1, j for i = 1, 2, . . . , N CVL − 1 ki, j ki+1, j = ki, j for i = N CVL ( As )i Ti, j − Ti+1, j s− (23.22)
(q˙w )i, j
−1 (w− )i (w+ )i = + ( Aw )i Ti, j − Ti, j − 1 for ki, j ki, j−1
i = 1, 2, . . . , N CVL (23.23)
The heat generated at the control volume is given by zk q˙ g i, j =Ii,2 j i, j ( AC )i
i = 1, 2, . . . , NCVL ; j = 1, 2, . . . ,
[email protected] for
(23.24)
Substituting Eqs. (23.20) to (23.23) into Eq. (23.1) and solving for Ti+1, j gives Ti+1, j =
for
Ti, j −
(q˙ g )i, j − (q˙n)i, j − (q˙e )i, j − (q˙w )i, j s− s+ −1 + ( As )i ki, j ki+1, j
(23.25)
i = 0, 1, 2, . . . , NCVL ; j = 1, 2, . . . ,
[email protected] With Ti = 0, j [= (Tow )k, j ] given, Eq. (23.25) for the kth thermocouple station will be forced to converge after some iterations since the equation is developed on the basis of heat balance in a control volume. Then, the local insidewall temperature at the kth thermocouple station is (Tiw )k, j = TNCVL +1, j
for
j = 1, 2, . . . ,
[email protected] (23.26)
The local insidewall heat flux at the kth thermocouple station is TN , j − TNCVL +1, j q˙in k, j = kNCVL , j CVL r 2
for
j = 1, 2, . . . ,
[email protected] (23.27)
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23.13
The bulk temperature at the kth thermocouple station is (Tb)k = Tin +
(Tout − Tin )zk L
(23.28)
The local heattransfer coefficient at the kth thermocouple station is q˙in k, j hk, j = for j = 1, 2, . . . ,
[email protected] (23.29) (Tiw )k, j − (Tb)k Finally, the local average heattransfer coefficient at the kth thermocouple station is
h¯ k =
1
N
[email protected] [email protected] j=1
1
N
[email protected] [email protected] j=1
q˙in k, j (23.30)
(Tiw )k, j − (Tb)k
At times it is necessary to calculate the overall heattransfer coefficient for certain test runs. In these cases, the overall heattransfer coefficient can be calculated as follows: TCST 1 1 N h˜ = (23.31) h¯ k zk h¯ dz = L L k= 1 Note that, to execute these calculations, the thermophysical properties of the pipe material (thermal conductivity and electrical resistivity) are required. These and other thermophysical properties needed for these type of calculations will be presented next. Thermophysical properties
In order to implement the formulations above in a computer program and represent the heat transfer and flow data in a dimensionless form, knowledge of thermophysical properties of pipe material and the working fluids are required. For the heattransfer experiments presented in this chapter to demonstrate the applications of the finitedifference formulations, the pipe material was made of stainless steel 316 and the working fluids were air, water, ethylene glycol, and mixtures of ethylene glycol and water. The equations used in Tables 23.1 to 23.4 present the necessary equations for the calculation of the pertinent thermophysical properties. The equations presented were mostly curve fitted from the available data in the literature for the range of experimental temperatures.
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Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube 23.14
Tubes, Pipes, and Ducts
TABLE 23.1
Equations for Thermophysical Properties of Stainless Steel 316 Range and accuracy
Data source
Thermal conductivity: kss = 13.0 + 1.6966 ×10−2 T − 2.1768 × 10−6 T2 , with T in ◦ C and kss in W/mK
300–1000 K; R2 = 0.99985
Incropera and Dewitt (2002)
Electrical resistivity: ss = 73.152 + 6.7682 ×10−2 T − 2.6091 × 10−6 T2 − 2.2713 × 10−8 T3 , with T in ◦ C and ss in cm
−196 to 600◦ C; R2 = 0.99955
Davis (1994)
Fitted equation
TABLE 23.2
Equations for Thermophysical Properties of Air Fitted equation
Range and accuracy
Data source
Density: air = p/[T (R/Mair )], where T is in kelvins, air in kg/m3 , p is absolute pressure in Pa, R is the universal gas constant (= 8314.34 J/kmol · K), and Mair is the molecular weight of air (= 28.966 kg/kmol) Viscosity: air = 1.7211 × 10−5 + 4.8837 × 10−8 T − 2.9967 × 10−11 T2 , with T in ◦ C and air in Pa · s
−10 to 120◦ C; R2 = 0.99994
Kays and Crawford (1993)
Thermal conductivity: kair = 2.4095 × 10−2 + 7.6997 × 10−5 T − 5.189 × 10−8 T2 , with T in ◦ C and k air in W/m · K Specific heat: c p air = 1003.6 + 3.1088−2 T + 3.4967 × 10x −4 T2 , with T in ◦ C and (c p )air in J/kg · K
−10 to 120◦ C; R2 = 0.99996
Kays and Crawford (1993)
−10 to 330◦ C; R2 = 0.99956
Kays and Crawford (1993)
TABLE 23.3
Equations for Thermophysical Properties of Water Fitted equation 999.96 + 1.7158 × 10−2 T
Range and accuracy
Data source
Density: water = − 5.8699 × 10−3 T2 + 1.5487 × 10−5 T3 , with T in ◦ C and water in kg/m3
0–100◦ C; R2 = 0.99997
Linstrom and Mallard (2003)
Viscosity: water = 1.7888 × 10−3 − 5.9458 × 10−5 T + 1.3096 × 10−6 T2 − 1.8035 × 10−8 T3 + 1.3446 × 10−10 T4 − 4.0698 × 10−13 T5 , with T in ◦ C and water in Pa · s
0–100◦ C; R2 = 0.99998
Linstrom and Mallard (2003)
Thermal conductivity: kwater = 5.6026 × 10−1 − 2.1056 × 10−3 T − 8.6806 × 10−6 T2 − 5.4451 × 10−9 T3 , with T in ◦ C and kwater in W/m · K Specific heat: c p water = 4219.8728 − 3.3863T + 0.11411T2 − 2.1013 × 10−3 T3 + 2.3529 × 10−5 T4 − 1.4167 × 10−7 T5 + 3.58520 × 10−10 T6 , with T in ◦ C and (c p )water in J/kg · K
0–100◦ C; R2 = 0.99991
Linstrom and Mallard (2003)
0–100◦ C; R2 = 0.99992
Linstrom and Mallard (2003)
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i=1 j=1
3 3
Ci j X( j−1) T(i−1)
i=1 j=1
2 3
Ci j
+ j=1
3
C3 j
X( j−1)
T2 0.82935 4.8136 × 10−3 −5.8790 × 10−16
1/4 −10 to 100◦ C within ±5%
Ghajar and Zurigat (1991)
July 20, 2005
with T in ◦ C and mix in mPa · s
2.6492 −3.1496 × 10−2 2.2389 × 10−15
X( j−1) T (i−1)
0.55164 C = −2.7633 × 10−2 −6.0629 × 10−17
where
ln mix =
Viscosity:
−0.049214 4.1024 × 10−4 −9.5278 × 10−8
Ghajar and Zurigat (1991)
GRBT056Kutzv4.cls
0.17659 −9.9189 × 10−4 2.4264 × 10−6
0–150◦ C within ±0.25%
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with T in ◦ C and mix in g/cm3
1.0004 C = −1.2379 × 10−4 −2.9837 × 10−6
where
mix =
Data source
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Fitted equation
Equations for Thermophysical Properties of Mixture of Ethylene Glycol and Water
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Density:
TABLE 23.4
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23.15
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Fitted equation
0.60237 3.7454 × 10−3 2.3777 × 10−17
0–150◦ C within ±0.25%
Ghajar and Zurigat (1991)
Ghajar and Zurigat (1991)
July 20, 2005
Thermal expansion coefficient: −4 1 −1.2379 × 10 − 9.9189 × 10−4 X + 4.1024 × 10−4 X2 mix = − −6 −6 −8 2 mix +2 −2.9837 × 10 + 2.4614 × 10 X − 9.5278 × 10 X T with T in ◦ C, mix in g/cm3 , and mix in 1/◦ C
0–150◦ C within ±1%
GRBT056Kutzv4.cls
Thermal conductivity: kmix = 1 − X kwater + Xkeg − kF kwater − keg 1 − X X where keg = 0.24511 + 1.755 × 10−4 T − 8.52 × 10−7 T2 , kF = 0.6635 − 0.3698X − 8.85 × 10−4 T, with T in ◦ C and kwater , keg , kF , and kmix in W/m · K
3.0411 −2.5424 × 10−2 2.5283 × 10−15
Ghajar and Zurigat (1991)
QC: IML/OVY
with T in ◦ C
2.5735 C = −3.1169 × 102 1.1605 × 10−16
where
−10 to 100◦ C within ±5%
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j=1
Data source
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i=1 j=1
Range and accuracy
Equations for Thermophysical Properties of Mixture of Ethylene Glycol and Water (Continued )
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Prandtl number: 1/4 3 2 3 ( j−1) (i−1) ( j−1) ln Prmix = Ci j X T + C3 j X T2
TABLE 23.4
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Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube
23.16
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23.17
Stainless steel 316. In the finitedifference equations, the thermal con
ductivity and electrical resistivity of each node are determined as a function of temperature from the equations shown in Table 23.1 for stainless steel 316. Test fluids. The test fluids used in the experiments were air, distilled
water, ethylene glycol, and different mixtures of distilled water and ethylene glycol. The equations for thermophysical properties of these fluids are presented in Tables 23.2 to 23.4. Computer program
The finitedifference formulations presented in the previous section are implemented into a computer program written for the experimental works described in this chapter. The computer program consists of five parts: reading and reducing input data, executing the finitedifference formulations, providing thermophysical properties, evaluating all the heattransfer and flow parameters, and printing outputs. Since the finitedifference formulations and thermophysical properties have already been discussed in detail, the inputs and outputs from the computer program will be presented next. Input data. All the necessary inputs for the computer program are pro
vided by a database file managing a data set and each raw data file for the data set. The raw data file is produced through recordings from the experimental data acquisition process. A sample of the input files is given as Fig. 23.5a and b. The database file shown in Fig. 23.5a includes the preset information for an experimental run:
GroupNo
Data group number (e.g., the first digit indicates 1→singlephase, 2→twophase)
RunNo Phase Liquid MC_EG Pattern IncDeg
Test run number 1→singlephase flow test, 2→twophase flow test Test liquid (W→water, G→ethylene glycol) Mass fraction of ethylene glycol in the liquid mixture Flow pattern information of the twophase flow test Inclination angle of the test section
With the information provided in the database file, the computer program opens and reads data from each individual raw data file for an experimental run shown in Fig. 23.5b and proceeds to all the required data reductions and calculations and then saves the results in an output file for each test run. Output. A sample of the outputs from the computer program is shown
in Figs. 23.6 and 23.7 for a singlephase flow experimental run and a Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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format for experimental run.
(b)
Figure 23.5 Input data formats. (a) Database format for experimental run; (b) raw data
(a)
Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube
23.18
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23.19
(a)
(b)
Output for singlephase flow heattransfer test run. (a) Part 1: summary of a test run. (b) Part 2: detailed information at each thermocouple location. (c) Part 3: detailed information at each thermocouple station. Figure 23.6
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Tubes, Pipes, and Ducts
(c) Figure 23.6
(Continued )
twophase flow experimental run, respectively. As shown in the figures [Fig. 23.6 is in English units and Fig. 23.7 is in SI (metric) units], the user has the option of specifying SI or English units for the output. The output file starts with a summary of some of the important information about the experimental run for a quick reference as shown in Figs. 23.6a and 23.7a. The program then lists (see Figs. 23.6b and 23.7b) the measured outsidewall temperatures, the calculated insidewall temperatures, the calculated Reynolds numbers (superficial Reynolds numbers for each phase of the twophase flow run) of the flow based on the insidewall temperature, the calculated heat flux, and the calculated peripheral heattransfer coefficients for each thermocouple location. In the last part of the output (refer to Figs. 23.6c and 23.7c), the tabulated summary of the local averaged heattransfer results at each thermocouple station, such as its location from the tube entrance, bulk Reynolds numbers, bulk Prandtl numbers, viscosities at the wall and bulk, local average heattransfer coefficient, and Nusselt numbers are displayed. In the case of twophase flow runs, some auxiliary information for the twophase flow parameters is listed as shown in Fig. 23.7d. More details about the information that appears in Fig. 23.7d may be found in Ghajar (2004). Utilization of the finitedifference formulations
The outputs presented in Figs. 23.6 and 23.7 contain all the necessary information for an indepth analysis of the experiments. One way
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(a)
(b) Figure 23.7 Output for twophase flow heattransfer experimental run. (a) Part 1: sum
mary of a test run. (b) Part 2: detailed information at each thermocouple location. (c) Part 3: detailed information at each thermocouple station. (d) Part 4: auxiliary information.
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21
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Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube 23.22
Tubes, Pipes, and Ducts
(c)
(d)
Figure 23.7
(Continued )
to verify the reliability of the finitedifference formulations is to compare the experimental heattransfer coefficients with the predictions from the wellknown empirical heattransfer correlations. Figure 23.8 shows the comparison of the experimental Nusselt numbers (dimensionless heattransfer coefficients) obtained from the computer program with those calculated from selected wellknown singlephase heattransfer correlations of Colburn (1933), Sieder and Tate (1936), Gnielinski (1976), and Ghajar and Tam (1994). As shown in the figure, the majority of the experimental data is excellently matched with the calculated values from the correlations, and the maximum deviation between the calculated and the experimental values is within ±20 percent. Note that in Fig. 23.8 Gnielinski’s (1976) correlations, labeled as [1] and [3], refer to the first and the third correlations proposed in his work. The finitedifference formulations presented here have been successfully applied to develop correlation in single and twophase flow convective heattransfer studies. Ghajar and Tam (1994) utilized the
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23.23
Nucal
200
100 90 80 70 60 50 40 30 +20%
Sieder & Tate (1936) Colburn (1933) Ghajar & Tam (1994) Gnielinski [1] (1976) Gnielinski [3] (1976)
20 20% 10 10
20
30
40
50 60 70 80 90 100
200
Nuexp Comparison of experimental Nusselt numbers with predictions from selected singlephase heattransfer correlations at thermocouple station 6.
Figure 23.8
finitedifference formulations for heattransfer measurements in a horizontal circular straight tube with three different inlet configurations (reentrant, squareedged, and bellmouth inlets) under uniform wall heat flux boundary condition. From the measurements, they successfully developed a correlation for prediction of the developing and fully developed forced and mixed singlephase convective heattransfer coefficients in the transition region for each inlet. They also proposed heattransfer correlations for laminar, transition, and turbulent regions for the three inlets. In the case of the gasliquid twophase flow heat transfer, Kim and Ghajar (2002) applied the finitedifference formulations in order to measure heattransfer coefficients and developed correlations for the overall heattransfer coefficients with different flow patterns in a horizontal tube. A total of 150 twophase heattransfer experimental data were used to develop the heattransfer correlations. Ghajar et al. (2004) extended the work of Kim and Ghajar (2002) and studied the effect of slightly upward inclination (2◦ , 5◦ , and 7◦ ) on heat transfer in twophase flow. In the research works mentioned here, a computer program based on the presented finitedifference formulations was used as the key tool
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Tubes, Pipes, and Ducts
to analyze the experimental data. As demonstrated, the computational procedure (computer program) presented in this chapter can be utilized as an effective design tool to perform parametric studies or to develop heattransfer correlations. It can also be used as a demonstration tool for the basic principles of heat conduction and convection.
Summary A detailed computational procedure has been developed to calculate the local insidewall temperatures and the local peripheral convective heattransfer coefficients from the local outsidewall temperatures measured at different axial locations along an electrically heated circular tube (uniform wall heat flux boundary condition). The computational procedure is based on finitedifference formulation and the knowledge of heat generation within the pipe wall and the thermophysical properties of the pipe material and the working fluids. The method has applications in a variety industrial heatexchanging equipment and can be used to reduce the heattransfer experimental data to a form suitable for development of forced and mixedconvection flow heattransfer correlations in an electrically heated circular tube for different flow regimes.
Nomenclature A
Area, m2
Ac
Crosssectional area of a control volume in the axial direction, m2
C
Coefficient; refer to Table 23.4
cp
Specific heat at constant pressure, J/(kg · K)
Di
Circular tube inside diameter, m
Do
Circular tube outside diameter, m
E
Neighboring node point to a given node point in the eastern direction; refer to Fig. 23.1
h h¯
Heattransfer coefficient, W/(m2 · K) Local average heattransfer coefficient, W/(m2 · K); refer to Eqs. (23.12) and (23.30)
h˜
Overall heattransfer coefficient, W/(m2 · K); refer to Eq. (23.31)
I
Electrical current, A
i
Index in the radial direction
j
Index in the circumferential direction
k
Thermal conductivity, W/(m · K), or index in the axial direction
L
Length of the test section, m
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23.25
l
Length of a control volume in the axial direction, m
N
Neighboring node point to a given node point in the northern direction; refer to Fig. 23.1
NCVL
Number of control volume layers in the radial direction except the boundary layers
[email protected] Number of thermocouples at a thermocouple station
NTCST
Number of thermocouple stations
Nu
Nusselt number, hDi /k, dimensionless
n
Normal distance, m
P
A given node point P; refer to Fig. 23.1
Pr
Prandtl number, c p/k, dimensionless
p
Pressure, Pa
q˙
Heattransfer rate, W Heat generation rate, W
q˙ g q˙
Heat flux, W/m2
R
Electrical resistance, or the universal gas constant (= 8314.34 J/kmol · K)
R2
Correlation coefficient, dimensionless
r
Radius, m
T
Temperature, ◦ C or K
S
Neighboring node point to a given node point in the southern direction; refer to Fig. 23.1
W
Neighboring node point to a given node point in the western direction; refer to Fig. 23.1
X
Mass fraction of ethylene glycol in the mixture of ethylene glycol and water, dimensionless
x
A spatial coordinate in a cartesian system, m
y
A spatial coordinate in a cartesian system, m
z
A spatial coordinate in a cartesian system; axial direction, m
Greek
Thermal expansion coefficient, 1/◦ C
Electrical resistivity, · m
Distance from a node point to a control volume face at a given direction, m
Designates a difference
Angular coordinate, ◦
Dynamic viscosity, Pa · s
Density, kg/m3
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Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube 23.26
Tubes, Pipes, and Ducts
Subscripts b
Bulk
cal
Calculated
E
Evaluated at neighboring node point to a given node point in eastern direction; refer to Fig. 23.1
e
Evaluated at eastern control volume interface of a given node point; refer to Fig. 23.1
e+
Evaluated at eastern control volume interface of a given node point in side of neighboring node point; refer to Fig. 23.1
e−
Evaluated at eastern control volume interface of a given node point in side of given node point; refer to Fig. 23.1
eg
Ethylene glycol
exp
Experimental
i
Index in radial direction
in
Evaluated at inlet
iw
Evaluated at inside wall of a tube
j
Index in circumferential direction
k
Index in axial direction
mix
Mixture of ethylene glycol and water
N
Evaluated at neighboring node point to a given node point in northern direction; refer to Fig. 23.1
n
Evaluated at northern control volume interface of a given node point; refer to Fig. 23.1
n+
Evaluated at northern control volume interface of a given node point in side of neighboring node point; refer to Fig. 23.1
n−
Evaluated at northern control volume interface of a given node point in side of given node point; refer to Fig. 23.1
out
Evaluated at outlet
ow
Evaluated at outside wall of a tube
P
Evaluated at a given node; refer to Fig. 23.1
S
Evaluated at neighboring node point to a given node point in southern direction; refer to Fig. 23.1
s
Evaluated at southern control volume interface of a given node point; refer to Fig. 23.1
s+
Evaluated at southern control volume interface of a given node point in side of neighboring node point; refer to Fig. 23.1
s−
Evaluated at southern control volume interface of a given node point in side of given node point; refer to Fig. 23.1
ss
Stainless steel
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Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube InsideWall Parameters from OutsideWall Temperatures
23.27
W
Evaluated at neighboring node point to a given node point in western direction; refer to Fig. 23.1
w
Evaluated at western control volume interface of a given node point (refer to Fig. 23.1); evaluated at inside wall of a tube
w+
Evaluated at western control volume interface of a given node point in side of neighboring node point; refer to Fig. 23.1
w−
Evaluated at western control volume interface of a given node point in side of given node point; refer to Fig. 23.1
Superscript −
Average
References Colburn, A. P. (1933), “A Method of Correlating Forced Convective Heat Transfer Data and a Comparison with Liquid Friction,” Trans. Am. Inst. Chem. Eng., vol. 29, pp. 174–210. Davis J. R. (ed.) (1994), Stainless Steels, in series of ASM Specialty Handbook, ASM International, Materials Park, Ohio. Farukhi M. N. (1973), An Experimental Investigation of Forced Convective Boiling at High Qualities inside Tubes Preceded by 180 Degree Bend, Ph.D. thesis, Oklahoma State University, Stillwater. Ghajar, A. J. (2004), “NonBoiling Heat Transfer in GasLiquid Flow in Pipes—a Tutorial,” invited tutorial, Proceedings of the 10th Brazilian Congress of Thermal Sciences and Engineering—ENCIT 2004, Rio de Janeiro, Brazil, Nov. 29–Dec. 3. Ghajar, A. J., Kim, J., Malhotra, K., and Trimble, S. A. (2004), “Systematic Heat Transfer Measurements for AirWater TwoPhase Flow in a Horizontal and Slightly Upward Inclined Pipe,” Proceedings of the 10th Brazilian Congress of Thermal Sciences and Engineering—ENCIT 2004, Rio de Janeiro, Brazil, Nov. 29–Dec. 3. Ghajar, A. J., and Tam, T. M. (1994), “Heat Transfer Measurements and Correlations in the Transitional Region for a Circular Tube with Three Different Inlet Configurations,” Experimental Thermal and Fluid Science, vol. 8, no. 1, pp. 79–90. Ghajar, A. J., and Zurigat, Y. H. (1991), “MicrocomputerAssisted Heat Transfer Measurement/Analysis in a Circular Tube,” Int. J. Appl. Eng. Educ., vol. 7, no. 2, pp. 125–134. Gnielinski, V. (1976), “New Equations for Heat and Mass Transfer in Turbulent Pipe and Channel Flow,” Int. Chem. Eng., vol. 16, no. 2, pp. 359–368. Incropera, F. P., and DeWitt, D. P. (2002), Introduction to Heat Transfer, 4th ed., Wiley, New York. Kays, W. M., and Crawford, M. E. (1993), Convective Heat and Mass Transfer, 3d ed., McGrawHill, New York. Kim, D., and Ghajar, A. J. (2002), “Heat Transfer Measurements and Correlations for AirWater Flow of Different Flow Patterns in a Horizontal Pipe,” Experimental Thermal and Fluid Science, vol. 25, no. 8, pp. 659–676. Linstrom, P. J., and Mallard, W. G. (eds.) (2003), NIST Chemistry WebBook, NIST Standard Reference Database no. 69, National Institute of Standards and Technology, http://webbook.nist.gov. Patankar, S. V. (1991), Computation of Conduction and Duct Flow Heat Transfer, Innovative Research, Inc., Minn. Sieder, E. N., and Tate, G. E. (1936), “Heat Transfer and Pressure Drop in Liquids in Tube,” Ind. Eng. Chem., vol. 29, pp. 1429–1435.
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Calculation of Local InsideWall Convective HeatTransfer Parameters from Measurements of Local OutsideWall Temperatures along an Electrically Heated Circular Tube
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Source: HeatTransfer Calculations
Chapter
24 TwoPhase Pressure Drop in Pipes
S. Dyer Harris Equipment Engineering Services, P.A. Wilmington, Delaware
Problem Calculating the head pressure required to force a given flow rate through a length of pipe is a common problem in engineered systems, and is straightforward when the fluid is a liquid far from its saturation state. A more complex problem is the case of the pumped fluid entering a heated pipe as a slightly subcooled or saturated liquid, being heated sufficiently to cause vaporization, and leaving the pipe as liquidvapor mixture or totally vapor. This is called a singlecomponent, twophase flow (an example of a multicomponent twophase flow would be a mixture of air and water). The inverse case is related, in which the vessel pressure and the exit pressure fix the driving force, and the flow rate from the vessel is required. In either case the change in specific volume requires a higher local velocity for a given mass flow, and consequently there is an increased frictional pressure drop. Additional losses from liquidvapor interfaces, elevation changes, and other factors can combine to make the twophase pressure drop considerably higher than that of the singlephase. To properly design a flowing system, derive an algorithm to calculate the pressure drop expected when a singlephase liquid at saturation conditions is pumped through a heated tube, considering twophase flow effects. 24.1
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TwoPhase Pressure Drop in Pipes 24.2
Tubes, Pipes, and Ducts
Calculate conditions over a range of pressures, heater powers, and tube diameters to illustrate the effect of each parameter on pressure change. Given information
Geometry Heated horizontal tube Singlephase liquid water at inlet Smooth internal pipe surface Conditions Tube length (heated) = 3 m Tube inside diameter = 19 mm, 25 mm Inlet pressure = 10 bar, 35 bar, 70 bar Mass flow = 1 kg/s Heater power = 15 kW, 18 kW
Assumptions
Saturated liquid conditions at pipe inlet No noncondensible gases present Steadystate operation Uniform wall heat flux over heated length Neglect inlet/outletfitting effects Coolant state may be represented by mean bulk properties (this important assumption is called the homogeneous model, and although commonly used, there are others; the homogeneous model is used here for illustration, and alternatives are discussed) Nomenclature A Flow area D
Tube diameter
f
Friction factor
G
Mass flux (mass flow rate per unit cross section); acceleration due to gravity
gc
Unit system factor
hf
Enthalpy of liquid phase at saturation
hfg
Enthalpy change from liquid to vapor (latent heat)
L
Heated length of tube
M
Mass flow rate
N
Number of subdivisions of heated tube for analysis
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TwoPhase Pressure Drop in Pipes TwoPhase Pressure Drop in Pipes
n
Subdivision number
P
Pressure (absolute)
x
Thermodynamic quality (vapor mass fraction)
V
Fluid velocity
v
Fluid specific volume
vf
Specific volume of saturated liquid
v fg
Specific volume change from liquid to vapor
Wf
Mass of liquid
Wg
Mass of vapor
z
Distance along tube from inlet
Angle of elevation, such that increase in height = z sin
Dynamic viscosity
Fluid density
24.3
Use of a consistent set of units, such as SI or English, is always required. TwoPhase Flow Pressure Drop Fundamentals With incompressible singlephase flow, coupling between flow rate and pressure loss is usually through property variation with temperature and is relatively weak. Bulk temperatures and averages suffice. In contrast, for heated twophase flow the heat transfer and the pressure drop are very closely coupled. Heat addition causes a phase change and therefore significant changes in bulk properties, vapor bubbles, and flow patterns along the tube length. In turn these affect the heattransfer rates and the pressure drop. In many cases the two phases may not be in thermodynamic equilibrium at a particular point. Furthermore, particularly at high heattransfer rates, the combination of proposed flow, heat transfer, and phase change may not converge to a stable operating point, a situation called flow instability. There is significant literature on this subject. Much of this literature treats special cases meaningful to a particular author or employer. For instance, there is a lot of data and discussion of conditions at high pressure and temperature, driven primarily by the nuclear power industry of the 1970s and 1980s. If your need is for a process design with steam and water in tube bundles above 70 bar (at 1000 psia), someone probably has solved your problem (e.g., see Refs. 1 and 5). In general, the pressure drop through a tube will be higher for a twophase flow than for the same mass flow of the liquid. Much of the research has focused on finding or computing a multiplier for the singlephase calculation to determine how much higher. Ideally, it would be a
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TwoPhase Pressure Drop in Pipes 24.4
Tubes, Pipes, and Ducts
Figure 24.1
Flow patterns in boiling twophase flow in a heated tube.
simple number, like 2.5. A little reflection immediately squashes that idea because there are so many variables. Among these are
Flow regime changes with level of boiling (Fig. 24.1).
There is an acceleration term, often significant, because of the major density changes. (Lowering density requires a higher velocity to maintain constant total mass flow.)
There is tight coupling of the flow regimes to the heat transfer.
There are more fluids than just water used in twophase flow systems. For many of them detailed flow and thermodynamic property data are not available.
TwoPhase ForcedConvection Flow Patterns It is useful to visualize the flow patterns in a heated pipe with twophase flow (please refer to Fig. 24.1). Conceptually, it shows consecutive sections of a heated tube, with wall temperature above Tsat . Subcooled liquid enters at the bottom, is heated to saturation over some length, then vapor bubbles appear and are swept into the flow (bubble pattern). With more heating the pattern changes as shown from left to right. The sequence of patterns illustrated is fuzzily defined and tends to move from one to another gradually. There is usually no sharp demarcation between them. A simple way to visualize the process is of a liquid flow, initially containing trace vapor, transitioning over the length of the tube to a vapor containing trace liquid. Many applications will not completely vaporize the liquid, and only one or another of the regimes will be present over most of the length.
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TwoPhase Pressure Drop in Pipes TwoPhase Pressure Drop in Pipes
24.5
For practical reasons predicting flow P and heat transfer in heated tubes will be an estimate at best. Efforts to refine the estimate through additional decimal places or shorter integration steps in time and space may be disproportionate to value. But the estimate can be indicative and certainly an important consideration in a process design. The Momentum Equation for TwoPhase Flow A basic equation will be derived for modeling flow suitable for practical calculation, albeit with the aid of a computer. This expression forms the starting point for many analyses. Increased accuracy for specific situations is obtained by substitution of appropriate friction term models and assumptions. It should also not be surprising that the many correlations and methods are rooted in empirical data, not developed from theory. The general method will be shown, and examples of the simplest given. A control volume momentum analysis of a fluid in many fluid mechanics texts [2,3] produces an expression similar to the following by summing the forces on an element of fluid in a stream tube: v dP +
VdV gc
+
g f V2 dz = 0 sin dz + gc D 2gc
Applying this equation in one dimension to an internal pipe flow implies that bulk or mean values of the variables are used. Boundarylayer and velocity profile effects are not explicit. A boiling, twophase flow is compressible, and local mean velocity varies with density and phase. It is simpler to replace the velocity by the mass flow rate per unit crosssectional area, using a form of the continuity equation: M = AV = V=
VA v
M v = Gv A
where G = (M/A) is the mass flux (kg/sm2 ). Substituting for V and dV, and dividing by v, we obtain G 2 dv dP + + gc
g sin gc v
dz +
f G 2v dz = 0 D 2gc
The first term is the pressure differential, the second term describes acceleration from volume change, the third term is static head, and the fourth is a model for pressure losses due to friction.
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TwoPhase Pressure Drop in Pipes 24.6
Tubes, Pipes, and Ducts
For flow between two points along the streamtube or pipe, we obtain P2 P1
G2 dP + gc
v2 v1
g dv + gc
G2 g ( P1 − P2 ) + (v2 − v1 ) + gc gc
z2 z1
sin 1 G2 dz + v 2gc D
z2 − z1 v¯
1 G2 sin + 2gc D
z2 f v dz = 0 z1
z2 f v dz = 0 z1
For a singlephase and incompressible flow, both specific volume and the friction factor f are constants. The second term becomes zero, the third term is static pressure change from elevation, and the fourth term becomes the familiar expression for friction loss in a pipe. In this problem, the fluid is initially a liquid at or near saturation. Increasing elevation (flow upward) and friction pressure drop along the wall will cause flashing of the liquid to vapor. If there is also heat addition from the walls fluid, enthalpy increases and vaporization is intensified. Because the quality varies along the length of the pipe, the effective friction factor also must vary. The twophase challenge is evaluation of the integrals by suitable definitions or models for the friction factor– specific volume relationship. An iteration scheme is required. The value of the friction factor depends on fluid properties, and the fluid properties depend on the pressure and enthalpy at each point along the pipe, which in turn depend on the friction loss. An added complication is the variation in the flow patterns as shown in Fig. 24.1. Intuitively, each pattern should require a different treatment of the relation between liquid and vapor phases, and the resulting fluid state. A simplifying solution scheme is to divide the line into small segments over which values of these properties might be assumed nearly constant, compute the segment P, and then sum the whole. Iteration is required until the changes in state and the property values are in balance. Alternatively, if the overall pressure difference is given and G is to be found (as in a relief line), the scheme might involve a double iteration wherein a G is assumed, then the P calculated, tested against the available P, and an adjustment made to G. The example problem illustrates use of the equation with the same common assumptions, and illustrates the solution procedure and the general nature of a twophase flow. Procedure The following series of steps may be programmed in Basic, FORTRAN, or one of the math packages, according to one’s preference:
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TwoPhase Pressure Drop in Pipes TwoPhase Pressure Drop in Pipes
24.7
1. Obtain tables or correlations of properties for the fluid used. 2. Divide the pipe into a number of segments and assign the appropriate heat addition to each segment. 3. For the given pipe geometry and flow rate, calculate the singlephase, isothermal pressure profile as the first approximation. 4. Beginning with the singlephase isothermal profile, calculate the fluid state properties in each segment, then revise the pressure in each using the equation derived above. The change in enthalpy from the heat flux over the segment is added to the saturation enthalpy at each step. This produces a new pressure profile. 5. Iterate the pressureproperty solutions until the calculated pressure profile converges. Each of these steps will now be discussed and illustrated by calculated results for the problem posed.
Solution Details 1. Obtaining fluid property values can be the most vexing part of the computation. Tabulations with detail equivalent to the steam tables for water are rare for most other fluids of interest. The twophase flow phenomena being modeled depend on incremental changes in properties from segment to segment. Whatever the source, the property data should be fitted with polynomial or powerlaw regressions for use by the computer, so that property change is continuous computationally. 2. When dividing the pipe into segments, the guiding principle is that segments should be short enough that pressure and hence the properties can be reasonably represented with average values in the segment. However, the segments should not be so small that coarseness in the property tables or accuracy of regression fit will affect the solution. While the computer is willing work with a large number of small subdivisions, this may give a false optimism about the accuracy of results. 3. A reasonable initial estimate of the pressure profile in the tube is the singlephase isothermal pressure profile, calculated in the traditional way. Various correlations for the friction factor have been proposed. Any will do that fit the physical circumstances. In this example the Blasius [6] smooth pipe correlation is used. Note that this and any other friction factor correlation will contain a Reynolds number as an independent variable. Calculating the Reynolds number in turn requires the fluid properties density and viscosity.
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TwoPhase Pressure Drop in Pipes 24.8
Tubes, Pipes, and Ducts
4. This section will comment on evaluation of each term in the equation for each segment. The key parameters defining fluid state are the pressure and the thermodynamic quality. The pressure at beginning and end of a segment is calculated in the previous iterative step. From classsical thermodynamics, quality is defined as the mass of vapor divided by the total mass of fluid: x=
Wg W f + Wg
This definition is strictly only true for thermodynamic equilibrium and in addition for a flowing system in which the liquid and vapor velocities are equal. These assumptions are tacit in the homogeneous model, but may not be for other models. The classical thermodynamics definition of quality should be noted when evaluating or using other models. In this example, the mass quality will be evaluated from the enthalpy in the segment. Initially, the enthalpy is the saturated liquid enthalpy h f (1) commensurate with the inlet pressure P(1). The enthalpy, or energy content, will be increased by addition of heat h over the segment. At any segment n (n = 1, 2, 3, . . . , N ) the energy content will be h(n) = h(1) + (n − 1)h (If the heat flux were not uniform, then a h defined for each segment could be used.) Because the local pressure has changed from friction losses, elevation changes, and vapor acceleration, the thermodynamic state values of h f and h f g have changed slightly, so that a new quality can be calculated from x=
h(n) − h f hfg
All the other thermodynamic properties can be obtained from this revised segment quality:
The acceleration term represents the change in pressure related to the change in velocity between two points, in this case the beginning and the end of each segment. Evaluation is straightforward, requiring calculation of the specific volume at beginning and end using the quality.
The elevation term is essentially the vertical component of the weight per unit mass of fluid in the segment. For a reasonably short segment in which conditions are not changing rapidly, a mean value of the specific volume in a segment is convenient to use.
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TwoPhase Pressure Drop in Pipes TwoPhase Pressure Drop in Pipes
24.9
The friction loss term is the most difficult term to evaluate and the subject of many studies and models. Clearly there should be differences in irreversible losses depending on the flow pattern extant. The homogeneous model is reasonable for a liquid flow with trace vapor (low quality, bubbly), or a vapor flow with trace liquid (high quality, annular dispersed). A pattern such as semiannular or annular, with a liquid film on the wall surface and a mostly vapor core, does not fit the homogeneous concept at all. Alternatives are referenced and mentioned in the concluding section. Nearly all analyses begin with the concept of a multiplier, generally defined by the ratio of the twophase pressure gradient at a point to the singlephase gradient with liquid at the same total mass flow [1]: 2 =
(dP/dz)tp (dP/dz)sp
Under the assumptions of homogeneous flow, this is effectively defining a twophase friction factor as the singlephase factor multiplied by the specific volume. However, recall that the friction factor correlates as a function not only of the density but also as the viscosity through the Reynolds number. The viscosity of the twophase mixture therefore needs better definition. One possibility is to evaluate viscosity in the same manner as the formal thermodynamic properties as a linear function of quality: = f + (1 − x) f g Another viscosity model proposed [1] is 1 x 1−x = + g f The latter gives more weight to the vapor phase at increasing quality, reflecting the observation that the vapor void fraction, especially at lower pressures, is physically dominant in the tube. Pick one or the other depending on knowledge of or belief about the dominating pattern in the system being analyzed. In this example the latter definition is used. Collier and Thome [1] present an expression for the friction multiplier that combines this definition with the Blasius form for smooth pipe friction factor: f g −1/4 vf g 2 = 1 + x 1+x vf g Figure 24.2 is a plot of this expression for steam and water over a range of pressures. Note that it can be a very large number at low pressures
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TwoPhase Pressure Drop in Pipes 24.10
Tubes, Pipes, and Ducts
Multiplier value
1000
100
10
1 0.00
0.20
0.40
0.60
0.80
1.00
Quality 1 bar Figure 24.2
7 bar
35 bar
70 bar
Homogeneous multiplier for water as a function of quality for a range of
pressures.
and quality. This expression is effectively the integrand of the friction term in the equation. The procedure used here for evaluation of each segment is effectively the trapezoidal rule for numerical integration, using the inlet and outlet property values for a segment. Results Calculations made for the problem conditions specified and the assumptions and models discussed above are exhibited in graphical form in Figs. 24.3 through 24.6.
Absolute pressure, bar
12 10 8 6 4 Pin = 10 bar 2 Diam = 19 mm Power = 18 kW 0 0
0.5
1
1.5
2
2.5
3
Location in tube, m Figure 24.3
Pressure profile in a heated 19mm tube at low pressure.
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TwoPhase Pressure Drop in Pipes TwoPhase Pressure Drop in Pipes
24.11
80
Inlet pressure, bar
70 60 50 40 30 20 10 0 0
0.5
1
1.5
2
2.5
3
3.5
Length along tube, m Pin = 10 bar Figure 24.4
Pin = 35 bar
Pin = 70 bar
Effect of increasing system pressure on the pressure profile.
Figure 24.3 shows the calculated pressure profile in the 19mm tube at 18 kW total heat at 10bar inlet pressure. As energy is added, quality increases and the P per segment increases. The vapor specific volume of water at this pressure is relatively large for a given quality. The effect is reminiscent of compounding interest.
12
Absolute pressure, bar
10 8 6 4 Pin = 10 bar Power = 18 kW
2 0 0
0.5
1
1.5
2
2.5
3
Location in tube, m Diam = 19 mm Figure 24.5
Diam = 25 mm
Effect of increasing tube diameter on the pressure profile.
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TwoPhase Pressure Drop in Pipes 24.12
Tubes, Pipes, and Ducts
11
Absolute pressure, bar
10 9 8 7 6
Tube diameter = 19 mm
5 4 0
0.5
1
1.5
2
2.5
3
Location along tube, m 15 kW Figure 24.6
18 kW
18.5 kW
18.6 kW
Effect of increasing tube power on the pressure profile.
Figure 24.4 shows the same tube and power, but compares inlet pressure. Consistent with intuition, at the higher pressures, 35 and 70 bar (500 and 1000 psia), the total P values are much less because the qualities and hence void fractions are much lower. Of course, this also means that to increase permissible heat load for a given available pumping head, increasing overall system pressure is an option. Figure 24.5 shows P profiles for the same flow rate, total power, and lower inlet pressure, but an increased tube diameter. Increasing the tube size for the same power decreases the wall heat flux just from geometry. It also decreases the mean velocity (or equivalently the mass flux), so that frictional and acceleration pressure losses are less for any quality. This means that excessive losses can be avoided by increasing the pipe size, just as in singlephase flow. As before, the additional size would allow higher power loads for the same losses. Figure 24.6 is a study of limiting flows. Twophase flow is compressible, and the effects are nonlinear as quality and void fraction increase. In singlephase compressible (gas) flow, phenomena such as supersonic flow, shock waves, and choking (mass flow limit) occur, all related to the speed of sound in the gas. Twophase flow does not exhibit limits that are closely related to its sonic velocity, but there are still limits that are analogous to choking. Looking at the 18kW line in Fig. 24.3, one can surmise that increasing the power added will continue to increase the pressure drop. The practical limit of the pressure drop is the difference between the supply pressure and ambient pressure to which it is discharged. Obviously, absolute outlet pressure cannot be a negative number, even if discharging into a vacuum. Leung [7] has derived Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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TwoPhase Pressure Drop in Pipes TwoPhase Pressure Drop in Pipes
24.13
expressions for the limiting rate of discharge of an adiabatic (the tube is not heated) twophase flow from a relief line. The algorithm just illustrated can be used to find such limits by repeatedly running the problem at incrementally higher input powers as shown in Fig. 24.6. If modeling an adiabatic flow, one can estimate the discharge flow rate by incrementing flow rate until the calculated outlet pressure equals the available discharge pressure.
Other Models The homogeneous model, while simple in concept, does not reflect the true physical characteristics of all twophase flows encountered. For example, situations typified by an annular flow (Fig. 24.1) are not modeled well. Therefore, a lot of effort has been made to improve the flow model for this regime, often at the expense of increasing complexity. The most frequently referenced model originated with Martinelli and Nelson [4]. Their approach was to treat the liquid film and the core vapor separately, with some recognition for the fluid vapor interface as a source of friction loss. Part of the method included empirical fits to available data. Since their initial paper, many improvements have been made both in the formulation and in the data supporting it [1]. The more recent trend toward miniaturization of electronics has increased interest and development of cooling methods in small channels with high heat flux. Intentionally or unintentionally, boiling of the coolant occurs. However, the dimensions of the flow channels drastically affect the phenomena. A nucleating bubble, small relative to a 15mm tube, may completely span a 0.01mm passage. High mass flux levels may still be a laminar flow because of the small dimension. Some of the methods and models traditionally used, including the ones discussed here, may not apply. A paper by Kandlikar [8] reviews these issues and suggests possible approaches to analysis. The analysis and example results shown used the homogeneous model. Twophase models that treat each phase separately with some form of coupling between them to account for energy and momentum exchange can be found and substituted in the basic equation. Here is a brief summary of the major properties of each approach. Homogeneous model Easiest to use, but simplistic in assumptions Expect to work best at Very lowquality (lowvoidfraction) flows Highpressure systems where liquid and vaporspecific volumes are converging High mass flow rates, implying turbulent mixing of the phases
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TwoPhase Pressure Drop in Pipes 24.14
Tubes, Pipes, and Ducts
Adiabatic flow where the vapor phase appears uniformly from reduction in pressure
Annular models Harder to use, but takes flow patterns into better account Basis has roots in empirical data Expect to work best at Annular or separated flow occurs over most of tube length Lower mass flow rates
References 1. Collier, J. G., and J. R. Thome, 1996, Convective Boiling and Condensation, 3d ed., Oxford University Press. 2. Rohsenhow, W. M., and J. P. Hartnett, 1973, Handbook of Heat Transfer, McGrawHill. 3. ASHRAE Handbook—Fundamentals, 2001. 4. Martinelli, R. C., and D. B. Nelson, 1948, “Prediction of Pressure Drop during Forced Circulation Boiling of Water,” Trans. ASME 70:165. 5. Tong, L. S., 1975, Boiling Heat Transfer and TwoPhase Flow, Kriegler Publishing. 6. Pitts, Donald R., and L. E. Sissom, 1998, Theory and Problems of Heat Transfer, 2d ed., Schaum’s Outline Series. 7. Leung, J. C., 1986, “Simplified Vent Sizing Equations for Emergency Relief Requirements in Reactors and Storage Vessels,” J. AIChE, 32(10):1622. 8. Kandlikar, S. G., 2004, “Heat Transfer Mechanisms during Flow Boiling in Microchannels,” ASME J. Heat Transfer, 126:8.
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Source: HeatTransfer Calculations
Chapter
25 HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils
Anil K. Mehrotra and Hamid O. Bidmus Department of Chemical and Petroleum Engineering University of Calgary Calgary, Alberta, Canada
When a liquid flowing through a pipeline is exposed to a cold environment that is below its freezingpoint temperature (or solubility temperature), solids deposition on the pipe wall is likely to occur. This phenomenon takes place frequently during the transportation of “waxy” crude oils that contain highmolecularweight alkanes or paraffin waxes. Paraffin waxes have a reduced solubility in crude oils at lower temperatures, causing their crystallization and deposition on cooler surfaces. The adverse effects of wax deposition are encountered in all sectors of the petroleum industry, ranging from oil reservoir formations to blockage of pipelines and process equipment. The deposited wax impedes the flow of oil through the pipeline, causing an increase in the pumping power. If not controlled adequately, the deposited solids may block the pipeline, resulting in high costs for its cleaning. Remediation can be carried out using mechanical cleaning methods, chemical cleaning methods, or by heating to melt the deposit.
25.1
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils 25.2
Tubes, Pipes, and Ducts
The highest temperature at which the first crystals of paraffin wax start to appear on cooling of paraffinic mixtures, such as waxy crude oils, is called the wax appearance temperature (WAT) or the cloudpoint temperature (CPT). Wax molecules start to crystallize out of the liquid mixture when the crudeoil temperature falls below the WAT, which leads eventually to solid deposition on the pipe wall. Holder and Winkler [1] observed solid wax deposits using crosspolarized microscopy and found that the wax crystallites have structures of platelets that overlap and interlock. A solid network structure is formed when sufficient quantities of solid paraffin crystals are formed, which leads to the formation of a gellike structure with entrapped liquid oil. The wax deposit is thus not entirely solid but rather consists of two phases: liquid oil and solid wax. The waxoil gel is formed as a result of the flocculation of orthorhombic wax crystallites that appear in the solution during cooling [2]. It has been reported that as little as 2 percent of precipitated wax can be sufficient to form a gellike deposit [3]. Solids deposition from waxy crude oils is a complex engineering problem, which involves the consideration of thermodynamics, solidliquid multiphase equilibria, crystallization kinetics, fluid dynamics, heat transfer, mass transfer, rheology, and thermophysical and transport properties. Modeling of solids deposition has been attempted via molecular diffusion and mass transfer, shear dispersion, Brownian motion, and heat transfer. In this chapter, however, we will treat the solids deposition from waxy crude oils mainly as a heattransfer problem; that is, solids deposition at steady state is assumed to occur primarily because of the presence of a thermal driving force, and momentum and masstransfer effects are neglected. EnergyBalance and HeatTransfer Equations for Solids Deposition at Steady State Consider a long pipeline carrying a “waxy” mixture that is completely immersed in water at a constant temperature Tc . This is similar to the transportation of waxy crude oils through subsea pipelines from offshore wells. Heat transfer to the surrounding “colder” seawater would decrease the crudeoil temperature Th, which could lead to the precipitation and deposition of paraffin solids on the pipe wall. As mentioned above, wax crystals would form a deposit layer comprising a solid (wax) phase with immobile liquid oil trapped within it. With time, the deposit layer would grow in thickness until such time that its growth does not occur any more. At this point, the rates of heat transfer across the flowing oil, the wax deposit layer, and the pipe wall would be the same and remain constant with time. When this happens, the thickness
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils Solids Deposition with “Waxy” Crude Oils in Pipeline
25.3
deposit
C
Th
pipe wall Td
Twi
Two
Tc
xd ri ro Radial temperature profile through the various thermal resistances.
Figure 25.1
of the deposit as well as the oil and wall temperature would each attain a constant value. Once the deposit layer achieves a constant layer thickness, the rate of heat transfer can be assumed to be under a thermal steady state because of the stable temperatures. As shown schematically with radial temperature profiles in Fig. 25.1, the transfer of thermal energy from the “hot” fluid (the flowing crude oil) to the “cold” fluid (the surrounding seawater) at steady state involves four thermal resistances in series: two convective resistances due to the flowing crude oil and the seawater and two conductive resistances offered by the pipe wall and the deposited layer. Assuming onedimensional heat transfer in the radial direction of the pipe, the rate of heat transfer could be equated to the rate of thermal energy lost by the hot waxy crude oil, the rate of thermal energy gained by the cold seawater, and the rate of heat exchange between hot and cold fluids, as follows: q=m ˙ h Ch(Th in − Th out ) = m ˙ c Cc (Tc out − Tc in ) = Ui Ai (Th − Tc ) where
(25.1)
q = rate of heat transfer m ˙ h, m ˙ c = mass flow rates of “hot” and “cold” streams Ch, Cc = average specificheat capacities of “hot” and “cold” streams Th in , Th out = inlet and outlet “hot” stream temperatures Tc out , Tc in = outlet and inlet “cold” stream temperatures
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils 25.4
Tubes, Pipes, and Ducts
Ui = overall heattransfer coefficient based on inside pipe surface area Ai Th, Tc = average temperatures of “hot” and “cold” streams, respectively The rate of heat transfer can also be equated to the heat flow across the two convective thermal resistances, as follows: q = hc Ac (Two − Tc ) = hh Ah(Th − Td)
(25.2)
where hc , hh = convective heattransfer coefficients on outside and inside of the pipe, respectively Ac = pipe outside surface area in contact with cold stream Ah = inside surface area at depositoil interface Td = temperature at this interface Note that, for a clean pipe with no deposit, Ah would be replaced by Ai and Td would become equal to Twi . Next, the overall or combined thermal resistance is expressed as the sum of four individual thermal resistances: 1 1 ln(ro /ri ) ln(ri /(ri − xd)) 1 = + + + Ui Ai hh Ah 2km L 2kd L hc Ac
(25.3)
At steady state, the heat flux (i.e., the rate of heat transfer per unit inside pipe wall area) through each of the four thermal resistances included in Ui is as follows: q hh(Th − Td) kd(Td − Twi ) km(Twi − Two ) hc (Two − Tc ) = = = = Ai ri /(ri − xd) ri ln(ri /(ri − xd)) ri ln(ro /ri ) ri /ro (25.4) where xd = deposit thickness (assumed to be uniform along pipe length) Twi , Two = average inside and outside pipe wall temperatures kd, km = thermal conductivities of deposit layer and pipe wall, respectively When dealing with the solidification or melting of a pure substance, the liquiddeposit interface temperature Td would be the melting or freezing temperature; however, it would be the liquidus or saturation temperature for a multicomponent mixture. From a modeling study, Singh et al. [4] estimated Td to approach the WAT of waxy mixtures when the deposit layer thickness stops growing, while Bidmus and Mehrotra [5] verified experimentally that Td and WAT are the same temperature at pseudosteadystate conditions.
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils Solids Deposition with “Waxy” Crude Oils in Pipeline
25.5
The mathematical relationships for heat transfer and energy balance presented above can be utilized for calculations dealing with solids deposition in a pipeline. In the following sections, we will illustrate the use of this mathematical framework by solving eight example problems. Example 1: Determination of Deposit Thickness xd A “waxy” crude oil is produced and transported via a pipeline from an offshore oil production platform to an onshore refinery for processing. The crude oil has a WAT of 26◦ C. At the beginning of the operation, about 10,000 barrels of oil per day (bopd) at a temperature of 40◦ C leaves the offshore platform into the pipeline. After a few months of pipeline operation, it is observed that the pressure drop across the pipeline has increased considerably and that the crude oil arrives at the refinery at a temperature of 28◦ C, which has been found to be more or less constant for several days. The engineer managing the operation believes that the increased pressure drop is a result of solids deposition, and the deposit thickness needs to be estimated before deciding on a suitable remedial action. It can be assumed that the seawater at an average temperature of 10◦ C flows across (or normal to) the pipeline at an average velocity of 0.1 m/s. How could the deposit thickness be estimated from heattransfer considerations? Pertinent pipeline data as well as average crudeoil and seawater properties are listed in Table 25.1. TABLE 25.1 Data and Average Properties Used in Example Calculations
Property Crude oil Flow rate F Specificheat capacity Ch Thermal conductivity kh Viscosity h Density h Seawater Specificheat capacity Cc Thermal conductivity kc Viscosity c Density c Deposit Thermal conductivity kd Pipeline Wall thermal conductivity km Inside diameter Di Wall thickness (ro − ri ) Outside diameter Do
Value 10,000 bopd or 0.0184 m3 /s 2400 J/kg K 0.15 W/m K 10 cP or 0.010 Pa s 750 kg/m3 4200 J/kg K 0.65 W/m K 1 cP or 0.001 Pa s 1020 kg/m3 0.24 W/m K 24 W/m K 10 in. or 0.254 m 0.625 in. or 0.0159 m 11.25 in. or 0.286 m
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils 25.6
Tubes, Pipes, and Ducts
Solution
Since the temperature of the crude oil arriving at the plant has been observed to be constant for some time, it can be assumed that the pipelineseawater system has attained a thermal steady state. Under these conditions, the deposit layer would have a constant thickness and the temperature at the depositoil interface would be at the crudeoil WAT. Although the deposit thickness will most likely not be uniform along the pipeline, due to the varying temperature of the crude oil from the inlet to the outlet, it will be assumed to be uniform for these calculations. Equation (25.4) can thus be used to solve for the three unknowns: Twi , Two , and xd. Estimation of inside heattransfer coefficient hh.
The average velocity of
crude oil in the pipeline is uh =
F 4F 4 × 0.0184 = = = 0.36 m/s Ai × 0.2542 Di2
The Reynolds number of crude oil in the pipeline is Re =
h uh Di 750 × 0.36 × 0.254 = = 6915 i 0.010
The Prandtl number is Pr =
Chh 2400 × 0.010 = = 160 kh 0.15
Given that the flow is turbulent (Re > 4000) and assuming L/Di to be large, the DittusBoelter correlation [6] can be used to estimate the average heattransfer coefficient in the pipeline hh: Nu =
hh Di 0.023 × 69150.8 × 1600.3 × 0.15 = 0.023 Re0.8 Pr0.3 ⇒ hh = kh 0.254
= 73 W/m2 K Estimation of outside heattransfer coefficient hc
Re =
c uc Do 1020 × 0.1 × 0.286 = = 29,150 c 0.001
Pr =
Cc c 4200 × 0.001 = = 6.46 kc 0.65
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils Solids Deposition with “Waxy” Crude Oils in Pipeline
25.7
The ChurchillBernstein equation [7] can be used to obtain the average heattransfer coefficient for crossflow across a cylindrical surface: 5/8 4/5 0.62 Re1/2 Pr1/3 Re Nu = 0.3 + 1/4 1 + 282,000 2/3 1 + (0.4/Pr) 0.62 × 29,150 × 6.46 ⇒ hc = 0.3 + 1/4 1 + (0.4/6.46)2/3 1/2
×
1/3
1+
29,150 282,000
5/8 4/5
0.65 = 515 W/m2 K 0.286
Estimation of deposit thickness xd . The average oil temperature is T h = (40 + 28)/2 = 34◦ C. As already explained, the depositoil interface temperature Td will be assumed to be equal to the WAT of 26◦ C. Therefore, using Eq. (25.4), we obtain
73(34 − 26) 0.24(26 − Twi ) = 0.127/(0.127 − xd) 0.127 ln(0.127/(0.127 − xd)) =
24(Twi − Two ) 502(Two − 10) = 0.127 ln(0.143/0.127) 0.127/0.143
Solving these equalities simultaneously for Twi , Two , and xd would yield Twi = 11.3◦ C, Two = 11.0◦ C, and xd = 6.1 × 10−3 m or 6.1 mm. Thus, the deposit layer in the pipeline is estimated to be approximately 6 mm thick, which is about 5 percent of the pipeline inner radius. It should be noted that the inside heattransfer coefficient for the pipeline with deposit has been assumed to be the same as that for the clean pipeline. Including the effect of deposit thickness would increase the value of hh from 73 to 77 W/m2 K. Also, while a radial temperature gradient would exist across each section of the pipeline, the inlet and outlet crudeoil temperatures are bulk values and the calculated crudeoil temperatures are average values for the entire length of the pipeline.
Example 2: The Case of Insulated Pipeline If the pipeline in Example 1 were insulated over its entire length with a waterproof plastic material having a thermal conductivity of 0.15 W/m K, what minimum thickness of insulation material would be required to prevent solids deposition?
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils 25.8
Tubes, Pipes, and Ducts
Solution
With an additional thermal resistance (due to insulation), Eq. (25.4) becomes hh(Th − Td) kd(Td − Twi ) km(Twi − Two ) kins (Two − Tins ) = = = ri /(ri − xd) ri ln(ri /(ri − xd)) ri ln(ro /ri ) ri ln((ro + xins )/ro ) =
hc (Tins − Tc ) ri /(ro + xins )
(25.5)
where kins and xins are the thermal conductivity and the thickness of the insulation material, respectively, and Tins is the temperature of the insulation material surface in contact with the seawater. Without any solids deposition (i.e., xd = 0), Twi ≥ Td. Equation (25.5), therefore, becomes hh(Th − Twi ) =
km(Twi − Two ) kins (Two − Tins ) hc (Tins − Tc ) = = ri ln(ro /ri ) ri ln((ro + xins )/ro ) ri /(ro + xins ) (25.6)
The minimum thickness of the insulation material required can be obtained for Twi = Td = 26◦ C: 73(34 − 26) = =
24(26 − Two ) 0.15(Two − Tins ) = 0.127 ln(0.143/0.127) 0.127 ln((0.143 + xins )/0.143) 502(Tins − 10) 0.127/(0.143 + xins )
Solving these equalities gives Two = 25.6◦ C, Tins = 11.0◦ C, and xins = 4.3 × 10−3 m. The minimum insulation thickness required is, therefore, 4.3 mm. Note that this thickness is not much less than the deposit layer thickness obtained in Example 1, because the thermal conductivity of insulation material is not much lower than that of the deposit layer. During the solids deposition process, the deposit layer acts as an insulation to heat transfer, thereby preventing further solids deposition. The importance of the deposit thermal conductivity is illustrated in Example 5. Also note the much larger temperature difference across the insulation material (25.6 − 11.0 = 14.6◦ C) compared to that across the pipe wall (26.0 − 25.6 = 0.4◦ C). Example 3: Effect of CrudeOil Composition (or WAT) A change in the crudeoil composition will affect its WAT. Basically, the waxier a crude oil is, the higher is its WAT. The effect of wax composition will, therefore, be investigated by varying the WAT. For the operating
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils Solids Deposition with “Waxy” Crude Oils in Pipeline
25.9
conditions and crudeoil properties used in Example 1, what would be the deposit thickness for the crudeoil WAT values of (a) 27◦ C and (b) 25◦ C? Solution
Since crudeoil properties and other data (listed in Table 25.1) as well as operating conditions are the same, the individual heattransfer coefficients obtained in Example 1 can be used. Equation (25.4) can be solved for xd by substituting the given WAT values for the interface temperature Td: Part a
73 (34 − 27) 0.24 (27 − Twi ) = 0.127/(0.127 − xd) 0.127 ln(0.127/(0.127 − xd)) =
24(Twi − Two ) 515(Two − 15) = 0.127 ln(0.143/0.127) 0.127/0.143
Solving these three equalities gives xd = 7.6 × 10−3 m or 7.6 mm. Part b
73(34 − 25) 0.24(25 − Twi ) = 0.127/(0.127 − xd) 0.127 ln(0.127/(0.127 − xd)) =
24(Twi − Two ) 515(Two − 15) = 0.127 ln(0.143/0.127) 0.127/0.143
Again, solving these three equalities gives xd = 5.0 × 10−3 m or 5.0 mm. The higher WAT of 27◦ C gave a larger deposit thickness of 7.6 mm, and a lower WAT of 25◦ C gave a smaller deposit thickness of 5.0 mm. These results imply that an increase in the crudeoil wax concentration would lead to an increase in the amount of deposit in the pipeline. This effect is further illustrated in Fig. 25.2, where the deposit thickness has been obtained for WAT values ranging from 20 to 27◦ C. It should be noted that, in these calculations, it was assumed that all crudeoil properties remain constant even though the concentration of wax was varied. Properties such as the viscosity and density of the crude oil can be altered by an increase in wax concentration, which would alter the value of inside heattransfer coefficient. However, even with such crudeoil property changes, an increase in wax concentration would lead to increased amount of deposit in the pipeline.
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils 25.10
Tubes, Pipes, and Ducts
Deposit Thickness (xd /ri)
0.07 0.06 0.05 0.04 0.03 0.02 0.01 20
22
24
26
28
WAT, °C Figure 25.2 Effect of crudeoil WAT on deposit
thickness under similar operating conditions.
Example 4: Effect of CrudeOil Flow Rate The pipeline throughput is increased from 10,000 to 15,000 bopd (0.0276 m3 /s). What would be the thickness of the deposit layer in the pipeline at steady state? Solution Estimation of inside heattransfer coefficient hh. The inside heattransfer coefficient would be altered by an increase in the production rate. The new velocity of crude oil in the pipeline would be uh = F/Ai = 4F/( Di2 ) = (4 × 0.0276)/(3.142 × 0.2542 ) = 0.54 m/s. Then
Re =
huh Di 750 × 0.54 × 0.254 = 10,370 = h 0.01
Pr = 160
The inside heattransfer coefficient is hh =
0.023 × 10,3700.8 × 1600.3 × 0.15 = 102 W/m2 K 0.254
Estimation of deposit thickness xd .
Using Eq. (25.4), we obtain
102(34 − 26) 0.24(26 − Twi ) = 0.127/(0.143 − xd) 0.127 ln(0.127/(0.127 − xd)) =
24(Twi − Two ) 515(Two − 10) = 0.127 ln(0.143/0.127) 0.127/0.143
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils Solids Deposition with “Waxy” Crude Oils in Pipeline
25.11
Solving these equations would give xd = 4.3 × 10−3 m or 4.3 mm. Thus, when compared to Example 1, the deposit thickness is reduced as a result of an increase in the crudeoil flow rate. This is caused by the increased inside heattransfer coefficient due to the increased flow rate. The resulting higher wall temperature then yields a reduction in the amount of deposition. Example 5: Deposit Thermal Conductivity As was shown in Example 2, when solids deposition occurs in a pipeline, the layer of deposit already formed in the pipeline acts as insulation to further reduce heat transfer between the crude oil and the seawater. The deposit thickness at which this happens would depend on the thermal conductivity of the deposit layer. Suppose that a section of the pipeline in Example 1 was removed for inspection. It was found that the actual deposit thickness was higher than the estimated value by 50 percent. An analysis suggested that this error was perhaps due to the use of an incorrect deposit thermal conductivity value in the calculations. What is a more appropriate estimate for the thermal conductivity of the deposit layer? Solution
Once again, all other conditions will be assumed to be unchanged. Although Eq. (25.4) can still be applied, a different approach will be used to solve this problem. Deposit thickness. The deposit thickness is 50 percent higher: xd = 1.5 × 6.1 × 10−3 = 9.2 × 10−3 m or 9.2 mm. Rate of heat transfer. With the estimated value of the inside heattransfer coefficient, the rate of heat transfer can be obtained from Eq. (25.2):
q = hh Ah(Th − Td) = hh 2(ri − xd)L(Th − Td) ⇒ q = 73 × 2 × L (0.127 − 9.2 × 10−3 )(34 − 26) = 435L W Thermal conductivity. The revised rate of heat transfer is the same as that for the temperature difference (Td – Tc )
q=
Td − Tc ln(ro /ri ) ln(ri /(ri − xd)) 1 + + 2km L 2kd L hc 2ro L
(25.7)
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils 25.12
Tubes, Pipes, and Ducts
where kd is the only unknown, which can be obtained by solving 435L =
⇒ kd =
26 − 15 ln(0.143/(0.127 − 9.2 × 10−3 )) ln(0.143/0.127) 1 + + 2 × × 24 × L 2 × × kd × L 515 × 2 × × 0.143 × L 5.20 = 0.353 W/m K 14.72
So the actual thermal conductivity of the deposit layer in the pipeline is 0.353 W/m K. Considering that the wax deposit comprises a gellike mixture of liquid oil entrapped in a solid wax matrix, it is expected that the thermal conductivity value of the deposit would be between the values of the crude oil and the solid wax. Since the thermal conductivities of crude oil and solid wax are in the range of 0.10 to 0.35 W/m K, the value of 0.353 W/m K for the deposit obtained above is somewhat higher. However, high thermal conductivity values for deposited solids have been reported experimentally [8,9]. When the deposit was removed and the thermal conductivity was measured, they were generally lower than those calculated from the heattransfer consideration [8]. The actual conditions of the deposit during the deposition process are complex and difficult to reproduce in a thermal conductivity measurement. It is pointed out that the deposit consists of a network of solid wax with liquid oil entrapped in it, which is subjected to a temperature difference during the deposition process. Therefore, the liquid trapped in the deposit may provide convective effects due to the temperature difference across the deposit layer. While these conditions cannot be reproduced in actual measurements, the calculated thermal conductivity values will be the apparent or effective values for the whole deposit layer and will include any convective effects in the deposit. Example 6: Effect of CrudeOil and Seawater Temperatures What will be the deposit thickness if (a) the seawater temperature in Example 1 is 15◦ C and (b) the pipeline inlet and outlet temperatures are 45 and 28◦ C, respectively? Solution
Using Eq. (25.4) with the same properties and heattransfer coefficients, we get the following.
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils Solids Deposition with “Waxy” Crude Oils in Pipeline
25.13
Part a Deposit thickness
73(34 − 26) 0.24(26 − Twi ) = 0.127/(0.127 − xd) 0.127 ln(0.127/(0.127 − xd)) =
24(Twi − Two ) 515(Two − 15) = 0.127 ln(0.143/0.127) 0.127/0.143
Solving these equations gives xd = 5.2 mm. Part b Deposit thickness. The average oil temperature Th = (45 + 28)/2 = 36.5◦ C. Using this value in Eq. (25.4), we obtain
73(36.5 − 26) 0.24(26 − Twi ) = 0.127/(0.127 − xd) 0.127 ln(0.127/(0.127 − xd)) =
24(Twi − Two ) 515(Two − 10) = 0.127 ln(0.143/0.127) 0.127/0.143
which gives xd = 3.8 mm. The calculations presented above were repeated for several values of the average crude oil and seawater temperatures. The predictions are shown in Figs. 25.3 and 25.4, which indicate that there is a reduction in the amount of deposited solids with an increase in either the
Deposit Thickness (xd /ri)
0.07 0.06 Tc = 10°C
0.05
Tc = 15°C
0.04 0.03 0.02 Tc = 20°C
0.01 0.00 0
3
6
9
12
15
18
(Th − WAT), °C Predicted variations in deposit thickness with crudeoil temperature at different seawater temperatures.
Figure 25.3
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils 25.14
Tubes, Pipes, and Ducts
Deposit Thickness (xd /ri)
0.07 Th = 30°C
0.06 0.05
Th = 35°C
0.04 0.03 0.02
Th = 40°C
0.01 0.00 0
3
6
9
12
15
18
(WAT − Tc ), °C Predicted variations in deposit thickness with seawater temperature at different crudeoil temperatures.
Figure 25.4
crudeoil temperature or the seawater temperature. Conversely, a reduction in either the crudeoil or the seawater temperature would yield a larger amount of deposited solids. These predictions are in good agreement with experimental results reported by Bidmus and Mehrotra [5], who had performed wax deposition experiments over a range of operating conditions using a concentric draft tube assembly with a “hot” waxsolvent mixture flowing in the inner tube and cold water flowing through the annular region. Example 7: Minimum CrudeOil Temperature for Preventing Solids Deposition in Pipeline As was shown in Example 6 as well as Figs. 25.3 and 25.4, an increase in either the crudeoil temperature or the seawater temperature would decrease the amount of deposited solids. The maximum seawater temperature above which solids deposition would not occur is the crudeoil WAT. What is the minimum average crudeoil temperature for which there would be no solids deposition in the pipeline? Solution
Equating the two expressions for the rate of heat transfer given by Eqs. (25.2) and (25.7), we obtain hh Ah(Th − Td) =
Td − Tc ln(ro /ri ) ln(ri /(ri − xd)) 1 + + 2km L 2kd L hc 2ro L
(25.8)
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils Solids Deposition with “Waxy” Crude Oils in Pipeline
Equation (25.8) is rearranged as follows:
ln(ri /(ri − xd)) Td − Tc ln(ro /ri ) 1 1 − = + 2kd L hh Ah Th − Td 2km L hc 2ro L
25.15
(25.9)
Note that, when there is no deposit layer, Ah = 2 ri L and xd = 0:
1 Td − Tc ln(ro /ri ) 1 = (25.10) + hhri Th − Td km hc ro The minimum average crudeoil temperature to avoid any solids deposition can be expressed as Td − Tc Th = Td + hhri
ln(ro /ri ) 1 + km hc ro
−1 (25.11)
Furthermore, because of the relatively high thermal conductivity of the pipeline material, its thermal resistance can be neglected, which would simplify Eq. (25.11) as follows: Th = Td +
hc ro (Td − Tc ) hhri
(25.12)
Accordingly, the average crudeoil temperature should be higher than the crudeoil WAT (or Td ) by [(hc ro / hhri )(Td − Tc )], in order to avoid any solids deposition in the pipeline. For the case considered in Example 1, Th = 26 +
515 × 0.143 (26 − 10) = 152.2◦ C 73 × 0.127
that is, the average or bulk crudeoil temperature would have to be maintained at about 152◦ C to prevent any solids deposition in the pipeline. Since this value is the average temperature in the pipeline, the inlet crudeoil temperature would have to be higher than 152◦ C. The technical and economic feasibilities of preheating the crude oil (at the offshore platform) to a temperature higher than 152◦ C, prior to its pumping, would need to be investigated. Importance of Fractional Thermal Resistance Offered by the Deposited Solids As was shown in Example 6, when either the crudeoil temperature or the seawater temperature is increased, the amount of deposited solids decreases. The difference between these two temperatures is the driving force for the deposition process, and it can be related to the amount of solids deposited in the pipeline. This would mean that the larger the
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils 25.16
Tubes, Pipes, and Ducts
difference between the crudeoil and the seawater temperatures, the greater would be the amount of deposit in the pipeline. However, the results in Fig. 25.3 indicate that, at a constant seawater temperature, the amount of deposit actually would decrease when the crudeoil temperature is increased (thereby increasing the temperature differential). Whereas the results in Fig. 25.4 show that when the temperature difference is increased, while keeping the crude oil temperature constant and decreasing the seawater temperature, the amount of deposit would increase. This reasoning suggests that the overall temperature difference alone is not sufficient for determining the amount of deposited solids. A more appropriate quantity for describing the amount of deposit is the ratio of the thermal resistance due to the deposit layer and the total thermal resistance; this ratio of thermal resistances is denoted by d [5]: d =
Rd Rh + Rd + Rm + Rc
(25.13)
Fractional Thermal Resistance (θ)
The deposit fractional thermal resistance is a useful quantity for estimating the amount of deposited solids [5]. For the case considered in Example 1, a plot between the individual fractional thermal resistances and the amount of deposit for various conditions of temperature is shown in Fig. 25.5. While all the other fractional thermal resistances decrease with an increase in the deposit thickness, the fractional thermal resistance for the deposit layer increases. However, the use of Eq. (25.13) would require a priori knowledge of the deposit thickness and thermal conductivity. 0.8
θd 0.6
0.4
θh
0.2
θm
θc
0.0 0.00 0.02 0.04 0.06 0.08 0.10 0.12
Deposit Thickness (xd /ri ) Relationship between fractional resistances and deposit thickness.
Figure 25.5
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils Solids Deposition with “Waxy” Crude Oils in Pipeline
25.17
Bidmus and Mehrotra [5] showed that d is also related to the ratio of the temperature difference across the deposit layer and the overall temperature difference at steady state: d =
Td − Twi Th − Tc
(25.14)
For a pipeline made of metal with a relatively high thermal conductivity and small wall thickness, the average wall temperature can be assumed to be equal to the seawater temperature; hence, it can be shown that Td − Tc d ∼ = Th − Tc
(25.15)
This equation can be used to estimate the extent of solids deposition at steady state with the knowledge of only the bulk “hot” and “cold” fluid temperatures and the crudeoil WAT (or Td). Example 8: Application of d A waxy crude oil with an average temperature of 36◦ C and a WAT of 25◦ C is transported in a pipeline submerged in seawater at 10◦ C. If the same crude oil with a lower average temperature of 34◦ C is transported in the same pipeline submerged in warmer seawater at 15◦ C, which scenario is likely to result in a higher solids deposition? Assume that the pipeline is made of a metal having a high thermal conductivity and with a small wall thickness such that its thermal resistance can be neglected. Solution
For both cases, the fractional thermal resistance of the deposit layer d can be estimated from Eq. (25.15) as follows: For scenario 1: d = (25 − 10)/(36 − 10) = 0.58 For scenario 2: d = (25 − 15)/(34 − 15) = 0.53 Scenario 1 has a slightly higher value of d and is, therefore, likely to have more deposit under the same operating conditions. It should be noted that this type of analysis involving a comparison of d is appropriate for similar values of WAT and hydrodynamic conditions in the pipeline. Even though further study is required to establish and validate a relationship between d and other operating conditions, d is an important parameter that could be useful for scaling up of solids deposition in pipelines [5].
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HeatTransfer Calculations for Predicting Solids Deposition in Pipeline Transportation of “Waxy” Crude Oils 25.18
Tubes, Pipes, and Ducts
Concluding Remarks This chapter presented a mathematical framework for estimating solids deposition in a pipeline for transporting “waxy” crude oils under steady state. It is based entirely on heattransfer considerations, involving relationships for energy balance and heat transfer. The mathematical model is relatively simple, and its use was illustrated through a number of example calculations for predicting the amount (or thickness) of deposited solids. It is emphasized that this heattransfer model does not deal with the effects of wax concentration on the WAT value as well as the composition of the deposited solids; these considerations would require appropriate thermodynamic calculations involving solidliquid phase equilibria along with the estimation of thermophysical and transport properties. Similarly, fluid dynamics (or momentum transfer) considerations would determine the pressure drop across the pipeline with or without solids deposition. Moreover, waxy crude oils and waxsolvent mixtures display complex rheological (thixotropic) behavior, especially at temperatures below the WAT [10,11]. The liquidsolid phase ratio as well as the composition of deposited solids have been found to change somewhat with time through a process referred to as “deposit aging” [4,5,8,9], which may be explained in terms of diffusional mass transfer as well as the shear stress between the flowing crude oil and the deposited solid layer. References 1. Holder, G. A., and Winkler, J., “Wax Crystallization from Distillate Fuels. I. Cloud and Pour Phenomena Exhibited by Solutions of Binary nParaffin Mixtures,” J. Inst. Petrol. 51, 228, 1965. 2. Dirand, M., Chevallier, V., Provost, E., Bouroukba, M., and Petitjean, D., “Multicomponent Paraffin Waxes and Petroleum Solid Deposits: Structural and Thermodynamic State,” Fuel 77, 1253, 1998. 3. Holder, G. A., and Winkler, J., “Wax Crystallization from Distillate Fuels. II. Mechanism of Pour Point Depression,” J. Inst. Petrol. 51, 235, 1965. 4. Singh, P., Venkatesan, R., Fogler, H. S., and Nagarajan, N., “Formation and Aging of Incipient Thin Film WaxOil Gels,” AIChE J. 46, 1059, 2000. 5. Bidmus, H. O., and Mehrotra, A. K., “HeatTransfer Analogy for Wax Deposition from Paraffinic Mixtures,” Ind. Eng. Chem. Res. 43, 791, 2004. 6. Holman, J. P., Heat Transfer, 9th ed., McGrawHill, New York, 2002. 7. Churchill, S. W., and Bernstein, M. A., “Correlating Equation for Forced Convection from Gases and Liquids to a Circular Cylinder in Crossflow,” J. Heat Transfer 99, 300, 1977. 8. Bidmus, H. O., A Thermal Study of Wax Deposition from Paraffinic Mixtures, M.Sc. thesis, University of Calgary, Calgary, Canada, 2003. 9. Parthasarathi, P., Deposition and Aging of Waxy Solids from Paraffinic Mixtures, M.Sc. thesis, University of Calgary, Calgary, Canada, 2004. 10. Wardhaugh, L. T., and Boger, D. V., “Measurement of the Unique Flow Properties of Waxy Crude Oils,” Chem. Eng. Res. Des. 65, 74, 1987. 11. Tiwary, D., and Mehrotra, A. K., “Phase Transformation and Rheological Behaviour of Highly Paraffinic ‘Waxy’ Mixtures,” Can. J. Chem. Eng. 82, 162, 2004.
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Source: HeatTransfer Calculations
Chapter
26 Heat Transfer in a Circular Duct
Nicholas Tiliakos Technology Development Dept. ATKGASL Ronkonkoma, New York
Introduction The following calculations were performed to assess the thermal loads (i.e., heat flux q, convective heattransfer coefficient hconv , etc.) for the combustor of a hypersonic vehicle over a critical portion of its flight trajectory. Calculation 1 provides the relevant information to describe the combustor thermal environment (i.e., convective and radiative heat flux, recovery temperature, convective heattransfer coefficient, radiative equilibrium temperature), which is then used in calculation 3 to determine how a section of the combustor, namely, a test specimen, can be instrumented with thermocouples which are then used to determine the heat flux into the specimen, from experimental data obtained during aeropropulsion wind tunnel testing. Aeropropulsion tunnels are used to simulate the thermal and aerodynamic environment that the hypersonic vehicle will encounter at a critical point in its trajectory. Calculation 2 was required to assess the survivability of a fuel injector embedded inside a duct upstream of the combustor. Our concern was with the integrity of a brazed component internal to this fuel injector.
26.1
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Heat Transfer in a Circular Duct 26.2
Tubes, Pipes, and Ducts
Calculation 1: Convective and Radiative Heat Transfer in a Circular Duct Determine the convective and radiative heat transfer of a hightemperature combustible mixture flowing in a circular duct; then determine the shear stress acting on the duct wall. Calculation procedure
The circular duct could represent the combustor section of an airbreathing engine, or the exhaust stack of a steam power plant; in the case presented below it shall represent the combustor of a hypersonic engine. The working fluid is a hightemperature combustible mixture whose properties can be determined using standard chemical equilibrium codes [1]. The conditions in the combustor, obtained from the ramjet performance analysis (RJPA) chemical equilibrium code [2], are presented in Table 26.1. 1. Determine the recovery temperature inside the combustor. The recov
ery temperature is given by Trecovery = r [Tt − Ts ] + Ts
(26.1)
where r is the recovery factor and Tt and Ts are the total and static temperatures of the flowfield, respectively. The recovery factor is defined as r = (Pr)1/n
(26.2)
where n = 1/2 for laminar flow and n = 1/3 for turbulent flow. 1a. To assess whether the flow in the duct is laminar or turbulent, calculate the Reynolds number on the basis of the duct diameter. (Note: If this were TABLE 26.1
Geometric, Thermodynamic, and Transport Properties of the Combustor (Calculation 1)
Combustor diameter D Combustor length L Static pressure Ps Total pressure Pt Static temperature Ts Total temperature Tt Density Equilibrium specific heat C p Equilibrium specificheat ratio Mach number M Velocity V Prandtl number Pr Absolute viscosity Thermal conductivity
5 in. 36 in. 32 psia 90 psia 4200◦ R 4800◦ R 0.0211 lbm /ft3 0.444 Btu/lbm ◦ R 1.21 1.3 3700 ft/s 0.68 5.1 × 10−5 lbm /ft · s 1.02 × 10−4 Btu/ft · s · ◦ R
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Heat Transfer in a Circular Duct Heat Transfer in a Circular Duct
26.3
not a circular duct, the Reynolds number would have to be calculated from the hydraulic diameter, defined as Dh = 4Ac /P, where Ac is the flow crosssectional area and P is the wetted perimeter [3].) The Reynolds number based on the combustor (duct) diameter is Re D =
VD
(26.3)
where , V, and are the density (lbm /ft3 ), velocity (ft/s), and absolute viscosity (lbm /ft · s) of the working fluid in the combustor, respectively, and D is the combustor (duct) diameter (ft). The viscosity is a transport property that can be obtained from standard chemical equilibrium codes, such as NASA’s Chemical Equilibrium Analysis (CEA) code [1]. If the mixture is predominately nitrogen, then Sutherland’s viscosity power law can be used to estimate the mixture viscosity [4]: ≈ 0
T T0
3/2
T0 + S T+S
(26.4)
where S is an effective temperature, called the Sutherland constant, which is characteristic of the gas [table 12 of Ref. 4]; T0 and 0 are a reference temperature and reference viscosity, respectively; 0 is evaluated at T0 . Using Eq. (26.3), and the combustor (i.e., duct) properties in Table 26.1, the Reynolds number is calculated to be Re D =
(0.0211 lbm /ft3 )(3700 ft/s)(5 in. · 1 ft/12 in.) = 6.378 × 105 5.1 × 10−5 lbm /ft · s
Since Re D > 2300, the transition Reynolds number in a circular pipe [3], it follows that flow in the combustor is turbulent. Now that we know that the flow is turbulent, the recovery temperature is calculated using n = 1/3 in Eq. (26.1), to get Trecovery = (0.681/3 )[4800◦ R − 4200◦ R] + 4200◦ R = 4728◦ R 2. Determine the hydrodynamic length in the circular duct. Internal flows
are usually characterized as either developing or fully developed. For external flow there are no constraints to the flow, and the only consideration is whether the flow is laminar or turbulent. In contrast, for internal flow the fluid is confined by the surface, so that the boundary layer is unable to develop without eventually being constrained by the duct walls [3]. The fluid flowing over the internal duct wall will form a boundary layer, which, after a certain distance, will merge at the centerline. Beyond this merger, viscous effects dominate throughout the duct
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Heat Transfer in a Circular Duct 26.4
Tubes, Pipes, and Ducts
cross section and the velocity profile no longer changes with increasing distance from the duct entrance. At this point the flow is then said to be fully developed, and the distance from the duct entrance to the merger point is called the hydrodynamic entry length xfd,h [3]. For turbulent flow in a duct, the hydrodynamic length xfd,h is approximately independent of Reynolds number, and is given, as a first approximation, as [3,5] x fd,h 10 ≤ ≤ 60 (26.5) D turb For the combustor, xfd,h is between = 10 · (D = 5 in.) = 50 in. and 60 · (D = 5 in.) = 300 in. Since both the lower and upper limits of xfd,h > L (= 36 in.), it follows that the flow in the combustor is developing, turbulent flow. Note that for laminar flow in a duct, the hydrodynamic length xfd,h is defined as follows [3]: x fd,h ≈ 0.05Re D (26.6) D laminar 3. Determine the thermal entry length in the circular duct. For turbulent
flow in a duct, the thermal entry length xfd,t is defined similar to the hydrodynamic length, except that in this case beyond the merger of the wall’s thermal boundary layer, viscous effects dominate throughout the duct cross section and the temperature profile no longer changes with increasing distance from the duct entrance. At this point the flow is then said to be thermally fully developed, and the distance from the duct entrance to the merger point is called the thermal entry length xfd,t [3]. For turbulent flow in a duct, the thermal entry length xfd,t is approximately independent of Reynolds number and Prandtl number, and is given, as a first approximation, as [3,5] x fd,t = 10 (26.7) D turb For the combustor, xfd,t = 10 · (D = 5 in.) = 50 in. and since xfd,t > L (= 36 in.), it follows that the flow in the combustor is thermally developing, turbulent flow. Note that for laminar flow in a duct, the thermal entry length xfd,t is defined as [3] x fd,t ≈ 0.05Re D Pr (26.8) D laminar 4. Determine the Nusselt number in the circular duct. We have determined
that the flow in the combustor is still developing (both hydrodynamically and thermally), so that the bottom surface of the duct has no
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Heat Transfer in a Circular Duct Heat Transfer in a Circular Duct
26.5
“communication” with the top surface; therefore, for all practical purposes, flow over the bottom and top surfaces is similar to external flow over a flat plate. As a result we shall analyze the flow inside the combustor (duct) as external flow over a flat plate. For such a configuration, many closedform correlations for determining the Nusselt number exist [3]. For a critical Reynolds number = Rex,c = 5 × 105 , Re L ≤ 108 , and 0.60 < Pr < 60, the average Nusselt number for external flow over a flat plate is given as [3] 1/3 Nu L = 0.037 Re0.8 L − 871Pr
(26.9)
Substituting in the previously determined values, we obtain (Nu) L = 0.037(4.592 × 106 )0.8 − 871(0.68)1/3 = 6184.3 Note that in the calculation above we used the Re number based on the duct length L, since, as mentioned above, the flow can be analyzed as if it were external flow over a flat plate. If, however, the flow were fully developed, then we could use the following Nusselt number correlation for turbulent, fully developed flow in a circular duct [3]: (Nu) D =
( f/8)(Re D − 1000) Pr 1 + 12.7( f/8)1/2 (Pr2/3 −1)
(26.10)
where f is the friction factor, given as [3] f = (0.790 ln Re D − 1.64)−2
(26.10a)
Note that Eq. (26.10) applies for turbulent, fully developed flow in a circular duct, for 0.5 < Pr < 2000, 3000 ≤ Re D ≤ 5 × 106 , (L/D) ≥ 10, and Eq. (26.10a) applies for turbulent, fully developed flow in a circular duct, for 3000 ≤ Re D ≤ 5 × 106 . 5. Determine the convective heattransfer coefficient in the circular duct.
Since the flow in the combustor is still developing (both hydrodynamically and thermally), we can analyze the flow inside the combustor as external flow over a flat plate. Using the Nusselt number for external flow over a flat plate [Eq. (26.9) in step 4], the convective heattransfer coefficient is determined as follows: h=
Nu L L
(26.11)
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Heat Transfer in a Circular Duct 26.6
Tubes, Pipes, and Ducts
Substituting in numbers from above and from Table 26.1, the convective heattransfer coefficient is h=
(6184.3)(1.02 × 10−4 Btu/ft · s · ◦ R) = 0.21 Btu/ft2 · s · ◦ R 3 ft
= 1.458 × 10−3 Btu/in.2 · s · ◦ R If, however, the flow were fully developed, then we could still use Eq. (26.11), but with the Nusselt number correlation for flow in a circular duct [Eqs. (26.10) and (26.10a)] [3], as presented at the end of step 4. 6. Determine the coldwall convective heat transfer in the circular duct, for a duct coldwall temperature equal to Twall = 520◦ R. The convective heat
flux is given by Newton’s law of cooling [3]: qconv = h(Trecovery − Twall )
(26.12)
where h was calculated in step 5, Trecovery in step 1, and Twall is given as 520◦ R. Substituting the appropriate numbers from steps 1 and 5, the coldwall convective heattransfer coefficient is qconv, cold = (0.21 Btu/ft2 s ◦ R)(4728◦ R − 520◦ R) = 883.7 Btu/ft2 s = 6.14 Btu/in.2 s 7. Determine the hotwall convective heat transfer in the circular duct, for a duct hotwall temperature equal to the radiation equilibrium temperature.
To determine the duct wall temperature for radiation equilibrium, neglecting conduction into the wall as small compared to the other modes of heat transfer (especially at steady state, where temperature gradients in the wall go to zero), we balance the convective and radiative heattransfer rates qconv = qrad , with qconv given in Eq. (26.12) and qrad given as 4 4 qrad = ε F Twall (26.13) − Tsurr where is the StefanBoltzmann constant, 3.3063 × 10−15 Btu/in.2 · s · ◦ R4 ; ε is the duct wall emissivity, which may be a function of the duct wall temperature; F is the view factor, which depends on geometry [3]; and Tsurr is the temperature of the fluid surrounding the duct. We shall assume ε = 1, F = 1 for this calculation. Setting Eq. (26.12) equal to Eq. (26.13), we obtain 4 4 FεTwall − FεTsurr + hTwall − hTrecovery = 0
(26.14)
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Heat Transfer in a Circular Duct Heat Transfer in a Circular Duct
26.7
and noting that for radiation equilibrium Twall = Trad,equil , Eq. (26.14) becomes 4 4 FεTrad,equil − FεTsurr + hTrad,equil − hTrecovery = 0
(26.15)
Substituting F = 1, = 3.3063 × 10−15 Btu/in.2 · s · ◦ R4 , ε = 1, Tsurr = 520◦ R, we can guess a Trad,equil and iterate until it satisfies Eq. (26.15). Doing this, we obtain Trad,equil = 4092◦ R. Knowing the duct hotwall temperature, Twall,hot = Trad,equil = 4092◦ R, we can calculate the hotwall convective heattransfer coefficient qconv,hot using Eq. (26.12) in step 6, with Trecovery = 4728◦ R, given in step 1, and Twall = Twall, hot = Trad, equil = 4092◦ R, given above: qconv, hot = (1.458 × 10−3 Btu/in.2 s ◦ R) (4728 − 4092◦ R) = 0.93 Btu/in.2 s 8. Determine the shear stress acting on the interior of the duct walls. The
duct wall shear stress is given by w = cf Qdyn = cf
1 Ps M 2 2
(26.16)
where is the specificheat ratio, Ps is the static pressure in the duct (combustor), M is the Mach number in the duct, and cf is the skin friction coefficient, which can be found from the ChiltonColburn analogy [3,6,7]: cf = StPr2/3 2
(26.17)
where St is the Stanton number, St = Nu/(RePr), and Pr is the Prandtl number; Eq. (26.17) applies for 0.6 < Pr < 60. For laminar flows, Eq. (26.17) is applicable only when dP/dx ∼ 0; however, in turbulent flow, conditions are less sensitive to pressure gradients, so that Eq. (26.17) is approximately valid [3]. Substituting in the values for Re L (step 4), Nu (step 4), and Pr (Table 26.1), cf = [2(6184.3)/(4.592 × 106 )(0.68)](0.68)2/3 = 0.0031. The values for , Ps , and M (see Table 26.1) should be local values (as close to the duct wall as possible). Substituting these values into Eq. (26.16), we find that the shear stress acting on the duct wall is wall = (0.0031)(0.5)(1.21)(32 psia)(1.3)2 = 14.6 lb f /ft2 . Calculation 2: Convective and Conductive Heat Transfer in a Circular Duct A liquid fuel injector, made of stainless steel 440 (SS 440) shown in R Fig. 26.1, is embedded inside a combustor wall, made of Poco Graphite ,
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Heat Transfer in a Circular Duct 26.8
Tubes, Pipes, and Ducts
Brazed Component
Liquid Fuel Injector
Hot Gas Flow
T s2
T s1
L
Combustor Wall JP10 Fuel r1 r2
Figure 26.1 Liquid fuel injector embedded inside a combustor wall, showing the internal
brazed component.
which reaches 2700◦ F (3160◦ R) at steady state. Type JP10 fuel, initially at TJP10 = 70◦ F, flows through the fuel injector, at 0.1 lbm /s. Material properties for SS 440 and Poco Graphite, as well as JP10 fuel properties, are presented in Table 26.2. Inside the fuel injector, there is a brazed component (assume same material properties as TABLE 26.2
Material and JP10 Fuel Properties for Liquid Fuel Injector (Calculation 2) JP10 Fuel
Specific heat C p
0.365 Btu/lbm ◦ R
Fuel Injector and Components: Stainless Steel 440 [8] Density Specific heat C p Thermal conductivity Melting temperature Tmelt
0.28 lbm /in.3 0.11 Btu/lbm ◦ F at 32–212◦ F 14 Btu/h · ft · ◦ F at 212◦ F = 3.241 × 10−4 Btu/in. · s · ◦ R 2500–2750◦ F
Combustor Wall: Poco Graphite [8,9] Density Thermal conductivity
0.0641 lbm /in.3 55 Btu/h · ft · ◦ F = 1.273 × 10−3 Btu/in. · s · ◦ R
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Heat Transfer in a Circular Duct Heat Transfer in a Circular Duct
26.9
SS 440). Note that the fuel injector is L = 0.4 in. long, with inner radius r1 = 0.0545 in. and outer radius r2 = 0.0885 in. At steady state 1. What temperature does the brazed component reach, and will it melt? 2. What is the temperature distribution inside the fuel injector wall (in the region r1 ≤ r ≤ r2 ? Calculation procedure 1. Determine the steadystate temperature of the braze component Ts1 . The
fuel injector, shown in Fig. 26.1, can be modeled as a hollow cylinder, where the brazed component extends to radius r1 at temperature Ts1 and the outside wall of the fuel injector extends to radius r2 at temperature Ts2 . At steadystate conditions the combustor wall temperature, 2700◦ F, will equal the fuel injector outerwall temperature Ts2 . We shall assume onedimensional (1D) radial heat transfer through a cylinder (i.e., the fuel injector) with an inner and outer radius of r1 and r2 , respectively, as shown in Fig. 26.1. The fuel injector outer wall and the combustor wall will get to the steadystate temperature more quickly than will the fuel injector body since the surrounding combustor wall material has higher thermal conductivity than does the fuel injector body (compare SS 440 and Poco Graphite thermal conductivity values in Table 26.2). The fuel injector material will soak up this heat via radial heat conduction, which will be balanced by the heat transfer to the JP10 fuel. The JP10 fuel, flowing at m ˙ = 0.1 lbm /s through the injector, will have heat transferred to it at the rate of qconv = m ˙ JP10 C p, JP10 (TJP10 − Ts1 )
(26.18)
Radial heat conduction through a hollow cylinder is given as [3] qcond =
2L (Ts1 − Ts2 ) ln(r2 /r1 )
(26.19)
where Ts1 and Ts2 are the braze component and combustor wall steadystate temperatures, respectively (Fig. 26.1). Equating Eqs. (26.18) and (26.19) and solving for Ts1 , we get Ts1 =
m ˙ JP10 C p, JP10 ln(r2 /r1 )TJP10 + 2LTs2 m ˙ JP10 C p, JP10 ln (r2 /r1 ) + 2L
(26.20)
where the thermal conductivity is for SS 440, presented in Table 26.2. Substituting in the appropriate numbers in Eq. (26.20) and solving for Ts1 , we get Ts1 = 645◦ R = 185◦ F; this temperature is less than the melting temperature of the SS 440 injector braze material, given as
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Heat Transfer in a Circular Duct 26.10
Tubes, Pipes, and Ducts
2500◦ F in Table 26.2. Therefore for the combustor conditions outlined above, the fuel injector braze component will not melt. 2. Determine the temperature distribution in the cylindrical fuel injector.
The temperature distribution inside the fuel injector, between r2 and r1 as shown in Fig. 26.1, is given by [3] T (r ) =
Ts1 − Ts2 r1 + Ts,2 ln ln (r1 /r2 ) r2
(26.21)
Calculation 3: Conductive Heat Transfer inside a Duct Wall A SS 440 material test piece, of thickness t = 0.75 in., is embedded inside a combustor engine wall, subject to the same flow conditions discussed in calculation 1 (presented in Table 26.1) and is to be tested inside an aeropropulsion wind tunnel, simulating Mach 1.3 hotgas flow, as shown in Fig. 26.2. The material test piece will require instrumentation with type K thermocouples. You are asked to determine (1) whether the test piece can be modeled as a lumped capacitance; (2) the theoretical temperature distribution within the test piece, without the zirconia; (3) how many thermocouples should be embedded inside the material, and where they should be located; (4) the steadystate heat flux into the specimen deduced from experimental (i.e., thermocouple temperature) data; and (5) the SS 440 material wall temperature (below the zirconia coating).
Thermocouple 2 SS 440 Thermocouple 1
TT / C2 TT / C1 .25"
Thermocouple Bead
+x
.025" qconv Hot Gas Flow
Tcomb,wall 0.025in.Thick Zirconia
Thermocouples embedded in a test section made of SS 440 and covered with 0.025in. zirconia, exposed to a hotgas flow.
Figure 26.2
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Heat Transfer in a Circular Duct Heat Transfer in a Circular Duct
26.11
Calculation procedure 1. Determine whether the SS 440 material test piece can be modeled as a lumped capacitance. A material’s temperature distribution can be mod
eled as a lumped capacitance if its Biot number Bi < 0.1 [3]. The Biot number is defined as Bi =
resistance to conduction hLc = resistance to convection
(26.22)
where h is the convective heattransfer coefficient and Lc is a characteristic length in the direction of heat transfer, defined as volume V = surface area As
Lc =
(26.23)
For this calculation, let’s take Lc = material thickness, equal to t = 0.75 in. The thermal conductivity of SS 440 is 3.241 × 10−4 Btu/in. · s · ◦ R (Table 26.2), and the convective heat transfer for these flow conditions was determined in calculation 1, step 5 and h = 1.458 × 10−3 Btu/in.2 · s · ◦ R. Substituting these numbers into Eq. (26.23), we determine that the Bi = 3.37. Since Bi 0.1, we cannot use the lumpedcapacitance model for the SS 440 material test piece, and as such more than 1 thermocouple is required to assess the heat transfer into the test piece. 2. Determine the theoretical temperature distribution within the test specimen (without the zirconia). In order to calculate the temperature dis
tribution inside the SS 440 test piece, we need to assume that the SS 440 specimen can be modeled as a 1D, semiinfinite solid, with a constant heat flux boundary condition and with constant thermal properties. Note that with respect to the thermocouple bead, the 0.75in.thick specimen does appear as a semiinfinite solid. The 1D, timedependent heattransfer equation is ∂ 2 T(x, t) 1 ∂ T(x, t) = ∂t ∂ x2
(26.24)
with the boundary conditions (BCs) −
∂ T(0, t) = q = constant ∂x T(∞, t) = Ti
as
for
x = 0, t > 0
x → ∞, t > 0
(26.25) (26.26)
and the initial condition (IC) T(x, 0) = Ti
at
t = 0 for all x
(26.27)
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Heat Transfer in a Circular Duct 26.12
Tubes, Pipes, and Ducts
The boundary condition is a nonhomogenous BC [Eq. (26.26)]. To readily obtain an analytical solution to Eq. (26.24), we let T∗ = T − Ti , which will transform the nonhomogenous BC [Eq. (26.26)] into a homogenous BC, as shown below, where Eqs. (26.24) to (26.27) have been recast for the “new, homogenous” problem ∂ 2 T∗ (x, t) 1 ∂ T∗ (x, t) = ∂t ∂ x2
(26.28)
with the BCs −
∂ T∗ (0, t) = q = constant ∂x T∗ (∞, t) = 0
as
for
x = 0, t > 0
x → ∞, t > 0
(26.29) (26.30)
and the IC T ∗ (x, 0) = 0
at
t=0
for all x
(26.31)
The integral method can be used to solve the homogenous set of Eqs. (26.28) to (26.31) [10]. The temperature distribution can be derived to give √ q t −x ∗ T (x, t) = exp √ (26.32) t or, noting that T ∗ = T − Ti
√ q t −x T (x, t) = Ti + exp √ t
(26.33)
where Ti = initial temperature of the specimen at t = 0 (◦ R), q is the heat flux into the specimen (Btu/in.2 · s), is the specimen thermal diffusivity (in.2 /s), is the specimen thermal conductivity (Btu/in. · s · ◦ R), and x is the location of the temperature T(x,t) (provided by an embedded thermocouple) into the specimen, at depth x (in). Equation (26.33) can be used to assess the temperature distribution into the test specimen (without zirconia) and allow us to determine where to embed the thermocouples into the test specimen. Assuming that Ti = 530◦ R, and, for steadystate conditions, the hotwall convective heat flux into the combustor wall was calculated in calculation 1, step 7, as qconv,hot = 0.93 Btu/in.2 s. Using the thermal diffusivity, = / C p = 0.0105 in.2 /s, for SS 440 (Table 26.2) and the thermal conductivity (Table 26.2), the theoretical temperature distribution inside the test specimen, for several steadystate times, is presented in Fig. 26.3. Note that when deciding where to locate thermocouples inside a test piece, Eq. (26.33) can be used to determine T (x,t) for a range of q values [recall that Eq. (26.33) was derived for a semiinfinite solid]. Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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Heat Transfer in a Circular Duct Heat Transfer in a Circular Duct
26.13
2000 T (°R) tSS = 20 s
1800
T (°R) tSS = 10 s 1600 T (°R) tSS = 5 s
Temperature (°R)
1400
1200
1000
800
600 400
200
0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Depth into Specimen x (in.)
Theoretical temperature distribution inside SS 440 test specimen (without the zirconia coating), for several steadystate times, and for a specimen q = 0.93 Btu/in.2 s (calculation 1). Specimen initial temperature Ti = 530◦ R.
Figure 26.3
3. How many thermocouples should be embedded inside the material, and where should they be located? We saw in step 1 that the test specimen
temperature distribution cannot be modeled as a lumped capacitance because its Biot number Bi 0.1. In step 2 we determined the theoretical temperature distribution, also shown in Fig. 26.3, where we see that at least three thermocouples should be embedded inside the material thermocouples to adequately determine the temperature distribution inside the test specimen, as well as the heat flux to the test specimen. These locations could be at the specimen surface (x = 0 in.), at x = 0.35 in., and x = 0.75 in. (backside of test piece). If the temperature distribution calculation is repeated for a different material, such as Poco Graphite, which has a thermal diffusivity and thermal conductivity approximately 10 and 13 times that of SS 440, respectively, the thermal gradients are much less, thus requiring at most two thermocouples inside that material. 4. Determine the SS heat flux into the specimen from experimental temperature data. Equation (26.33), step 2, can be rearranged and solved for
the steady state q, as shown below: q (x, t)SS =
[TSS − Ti ] −x exp √ √ tSS tSS
(26.34)
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Heat Transfer in a Circular Duct 26.14
Tubes, Pipes, and Ducts
T T/C1 (°R)
Thermocouple Temperature
TSS
Ti
Time t (s)
tSS
Figure 26.4 Schematic of experimental data from a thermocouple embedded inside a test specimen, described in calculation 3, and shown in Fig. 26.2.
If we use the experimental temperature data, shown in Fig. 26.4, from one of the thermocouples embedded to a known depth, x = 0.25 in. (Fig. 26.2), we can determine the steadystate heat flux into the SS 440 test specimen using Eq. (26.34). The experimental temperature data are tSS = 11 s, TSS = 2530◦ R, Ti = 530◦ R, x = 0.25 in., and thermal diffusivity (SS 440) = 0.0105 in.2 /s; therefore qSS = 0.913 Btu/in.2 s at x = 0.25 in. Recall that Eq. (26.34) was derived neglecting the zirconia layer and applies only for the homogenous SS 440 solid beneath. 5. Determine the test specimen wall temperature Tcomb,wall as shown in Fig. 26.2. The heat flux across the zirconia coating is given by Fourier’s law
of heat conduction, for steadystate conditions only [3]: qSS =
zirc [Tcomb,wall − TT/C1 ] x
(26.35)
The thermal conductivity of the zirconia coating is zirc = 5.087 × 10−4 Btu/in. · s · C, and Tcomb, wall and TT/C1 are the steadystate temperatures (shown labeled in Fig. 26.2); x is the separation distance, in the direction of heat flow, between TT/C1 and Tcomb, wall , and in this case this is the thickness of the zirconia coating, 0.025 in. The heat flux into the test specimen was determined in step 4, Eq. (26.34); therefore,
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Heat Transfer in a Circular Duct Heat Transfer in a Circular Duct
26.15
Eq. (26.35) can be rearranged to solve for the steadystate Tcomb,wall Tcomb,wall =
q x + TT/C1 zirc
(26.36)
where q = 0.913 Btu/in.2 s is the steadystate heat flux into the test specimen, calculated in step 4 using Eq. (26.34); TT/C1 is the steadystate thermocouple measurement [TT/C1 = 2530◦ R at tSS = 11 s (shown in Figs. 26.2 and 26.4)]; and x = 0.025 in. Substituting these numbers into Eq. (26.36), Tcomb, wall = 3103◦ R = 2643◦ F. References 1. McBride, B. J., and Gordon, S., Chemical Equilibrium Analysis (CEA) Code, NASA Reference Publication 1311, June 1996, Computer Program for Calculation of Complex Chemical Equilibrium Compositions & Applications, vol. I, Analysis; vol II, User’s Manual & Program Description. 2. Pandolfini, P. P., and Friedman, M. A., Instructions for Using Ramjet Performance Analysis (RJPA), IBM PC version 1.24, document JHU/APL AL92P175, June 1992. 3. Incropera, F. P., and DeWitt, D. P., Fundamentals of Heat and Mass Transfer, 5th ed., Wiley, New York, 2002. 4. White, F. M., Viscous Fluid Flow, McGrawHill, New York, 1974. 5. Kays, W. M., and Crawford, M. E., Convective Heat and Mass Transfer, McGrawHill, New York, 1980. 6. Colburn, A. P., Trans. Am. Inst. Chem. Eng., 29, 174, 1933. 7. Chilton, T. H., and Colburn, A. P., Ind. Eng. Chem., 26, 1183, 1934. 8. Materials Selector, published by Materials Engineering, a Penton publication, issued Dec. 1992, p. 45. 9. www.poco.com/graphite/igrade.asp. 10. Ozisik, M. N., Heat Conduction, Wiley, New York, 1993.
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Heat Transfer in a Circular Duct
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Source: HeatTransfer Calculations
Part
6 Heat Exchangers
Heatexchanger analyses and design calculations, including those involving compact heat exchangers, are a staple of many heattransfer books. This handbook is no exception. The calculations and analyses included in this part are all by academics in the United States and Canada. Three of them deal with the real world:
Air cooling of a highvoltage power supply using a compact heat exchanger A heat recovery system for an industrial clothes dryer “Passive” condensers (heat exchangers driven by natural forces and not requiring any power supplies, moving parts, or operator actions) in nuclear reactors
The remaining three chapters present general methodologies:
A rigorous, stepbystep methodology for calculating core dimensions of a compact heat exchanger with the most intricate basic flow arrangement situation in a singlepass configuration—a crossflow in which fluids do not mix orthogonal to the respective flow directions An analysis methodology for an aircooled heat exchanger with singlephase natural circulation loops A design methodology based on optimization via an evolutionary algorithm that manipulates populations of solutions by mating the best members of the current population (natural selection) and by randomly introducing perturbations into the newly created population (mutation)
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1
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Heat Exchangers
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Source: HeatTransfer Calculations
Chapter
27 Air Cooling of a HighVoltage Power Supply Using a Compact Heat Exchanger
Kirk D. Hagen Weber State University Ogden, Utah
Introduction A heat exchanger with a heattransfer surface areatovolume ratio of at least 700 m2 /m3 is classified as a compact heat exchanger. Compact heat exchangers are typically used in applications that require a heat exchanger with a high heattransfer capacity and a small volume and when one of the fluids is a gas. Consequently, this type of heat exchanger is usually found in aerospace and other applications where the size and weight of the cooling unit must be minimized. Of all the applications for which compact heat exchangers find effective use, the thermal management of electronics is probably the most common. The following heattransfer analysis pertains to the air cooling of a highvoltage power supply using a single stream compact heat exchanger, commonly called a “coldplate.”1 A coldplate typically consists of a layer of folded metal fins sandwiched between two metal baseplates. Coldplates are usually constructed by dip brazing the fins to the baseplates to ensure intimate thermal contact. The heatproducing
27.3
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Air Cooling of a HighVoltage Power Supply Using a Compact Heat Exchanger 27.4
Heat Exchangers
Figure 27.1
Aircooling system.
component, a power supply in this case, is mounted to one of the baseplates. A fan forces ambient air through the passages formed by the folded fins and baseplates, which make up the coldplate core. The air receives the heat dissipated by the power supply and transports that heat to the surroundings, thereby maintaining the power supply temperature below a maximum allowable value specified by the manufacturer. This basic cooling system is illustrated in Fig. 27.1. Statement of the Problem The mounting plate of a highvoltage power supply in a groundbased communications system is to be held below a temperature of 70◦ C while operating in an environment where the air temperature is 27◦ C and the atmospheric pressure is 1 atm. Under normal operating conditions, the power supply dissipates 550 W. The mounting plate of the power supply measures 18.0 × 12.0 cm. Because the unit in which the power
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Air Cooling of a HighVoltage Power Supply Using a Compact Heat Exchanger HighVoltage Power Supply and Compact Heat Exchanger
27.5
supply operates is a groundbased system, a 110V alternatingcurrent (ac) power source is available. The most convenient and costeffective means of cooling the highvoltage power supply is to utilize a compact coldplate with a small ac axial fan. Because of packaging constraints, the coldplate cannot extend more than 3.0 cm beyond the perimeter of the power supply baseplate, and it cannot extend more than 3.5 cm normal to the power supply baseplate surface. Hence, the coldplate cannot exceed the overall size of 24.0 × 18.0 × 3.5 cm. There is no constraint, however, on a reasonably sized fan and a transition duct from the fan to the coldplate core. To meet these constraints, we choose a coldplate that measures 22.0 × 16.0 cm, with the smaller dimension corresponding to the direction of airflow. The third dimension is governed by the required fin height, and will be determined as part of the analysis. The analysis is performed using SI (International System; metric) units.
Analysis Assumptions To simplify the analysis, the following assumptions are made: 1. Heat transfer and airflow rate are steady. 2. Conduction in the coldplate is onedimensional (normal to the coldplate surface). 3. Radiation is neglected. 4. All the heat dissipated by the power supply is transferred into the coldplate. 5. Heat dissipated by the power supply is uniformly distributed over the coldplate baseplate. 6. The flow rate of air is equal in all ducts of the coldplate core. 7. The coldplate surface is isothermal. 8. The thermal resistance of the coldplate baseplate is negligible. 9. Thermal resistance at the power supply–coldplate interface is negligible. 10. The tips of the fins are adiabatic. Thus, even though only one baseplate receives heat, the other baseplate is not considered as an extension of the fins. 11. The heattransfer coefficient is constant over the entire surface of the fins. 12. All thermal properties are constant, and are based on a temperature of 27◦ C (300 K).
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Air Cooling of a HighVoltage Power Supply Using a Compact Heat Exchanger 27.6
Heat Exchangers
Thermal Properties Aluminum 1100 (folded fins): k = 222 W/m · ◦ C Air at 1 atm: = 1.1614 kg/m3 , cp = 1007 J/kg · ◦ C, = 15.89 × 10−6 m2 /s, Pr = 0.707 Calculations We begin by considering the dimensions of the coldplate and the folded fins. The overall length and width, respectively, of the coldplate are L = 16.0 cm and W = 22.0 cm, where the smaller dimension (L) is chosen to correspond to the direction of airflow to minimize the pressure drop. Plain folded fins are commercially available in a wide range of fin thicknesses, heights, and pitches. Fin pitch is a spacing parameter that defines the number of fins per inch. Folded fins are usually manufactured with either a rounded or a flat crest. The flatcrest type is used here. Illustrated in Fig. 27.2 is an end view of a portion of the coldplate core showing the fin thickness , fin height b, and fin pitch Nf . At the beginning of the calculations, we do not know the heattransfer surface area or airflow rate required to maintain the power supply temperature below its allowable value. Recognizing the potential iterative nature of the analysis, we choose a reasonable set of fin dimensions and a flow rate to initiate the calculations. For the fin thickness and pitch, we let = 0.010 in. = 0.0254 cm Nf = 8 fins/in. = 3.1496 fins/cm Using the duct geometry shown in Fig. 27.2, the duct aspect ratio ∗ is given by the relation ∗ =
Figure 27.2
b− 1/Nf −
(27.1)
Coldplate fin configuration.
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Air Cooling of a HighVoltage Power Supply Using a Compact Heat Exchanger HighVoltage Power Supply and Compact Heat Exchanger
27.7
Based on the work of Kays and London,2 a common duct aspect ratio for which heat transfer and friction data exist is ∗ = 8. Solving Eq. (27.1) for the fin height b, we obtain b = ∗
1 − + Nf
1 =8 − 0.0254 + 0.0254 3.1496 = 2.362 cm = 0.02362 m Recall that the overall thickness of the coldplate cannot exceed 3.5 cm. Using the calculated fin height and assuming that the baseplates have a thickness of 0.3 cm, the overall coldplate thickness t is t = 2(0.3) + 2.362 = 2.962 cm, which is below the allowable value. The wetted perimeter P of a duct is 1 P = 2 b+ − 2 Nf 1 = 2 2.362 + − 2(0.0254) 3.1496 = 5.257 cm = 0.05257 m
(27.2)
and the crosssectional area for flow A of a single duct is
1 − Nf = (2.362 − 0.0254)
A = (b − )
1 − 0.0254 3.1496
= 0.6825 cm2 = 6.825 × 10−5 m2
(27.3)
The hydraulic diameter Dh of a duct is Dh = =
4A P 4(6.825 × 10−5 ) 0.05257
= 5.193 × 10−3 m
(27.4)
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Air Cooling of a HighVoltage Power Supply Using a Compact Heat Exchanger 27.8
Heat Exchangers
The number of ducts Nd in the coldplate is the product of the fin pitch and coldplate width Nd = Nf W = (3.1496)(22.0) = 69.3
(27.5)
which we round down to 69 to avoid a fractional number of ducts. The total crosssectional area for flow At is A t = Nd A = (69)(6.825 × 10−5 ) = 4.709 × 10−3 m2
(27.6)
Now that the coldplate dimensions have been determined, we need to estimate a flow rate. A reasonable volume flow rate for a typical axial fan suitable for this type of application is 65 cfm (ft3 /min). Thus, we have Q = 65 ft3 /min × 1 m3 /35.3147 ft3 × 1 min/60 s = 0.03068 m3 /s The air velocity v based on this assumed flow rate is Q v= At =
0.03068 4.709 × 10−3
= 6.515 m/s
(27.7)
which yields a mass flow rate m ˙ of m ˙ = At v = (1.1614)(4.709 × 10−3 )(6.515) = 0.0356 kg/s
(27.8)
and a mass velocity G of G = v = (1.1614)(6.515) = 7.567 kg/s · m2
(27.9)
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Air Cooling of a HighVoltage Power Supply Using a Compact Heat Exchanger HighVoltage Power Supply and Compact Heat Exchanger
27.9
The Reynolds number Re is Re = =
vDh (6.515)(5.193 × 10−3 ) 15.89 × 10−6
= 2129
(27.10)
which indicates laminar flow. For laminar flow, the hydrodynamic entry length xfd,h may be obtained from the approximation xfd,h ≈ 0.05 Re Dh lam = (0.05)(2129) = 107
(27.11)
Similarly, the thermal entry length xfd,t may be obtained from the approximation xfd,t ≈ 0.05 Re Pr Dh lam = (0.05)(2129)(0.707) = 75
(27.12)
The actual duct lengthtodiameter ratio for our coldplate is L 0.16 = = 31 Dh 5.193 × 10−3 which indicates that the velocity and thermal boundary layers are not fully developed. The graph of heat transfer and friction data from Kays and London2 for a duct with an aspect ratio of ∗ = 8 includes curves for the friction factor f and the Colburn j factor jH , for L/Dh = 40. These curves can be used in this analysis with little error. A version of this graph is shown in Fig. 27.3. From Fig. 27.3, knowing that Re = 2129, the friction factor and Colburn j factor are f = 0.013 and jH = 0.004, respectively. The Colburn j factor is defined by the relation jH = St Pr2/3
(27.13)
where St is the Stanton number and Pr is the Prandtl number. The Stanton number may be expressed in terms of the heattransfer
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Air Cooling of a HighVoltage Power Supply Using a Compact Heat Exchanger 27.10
Heat Exchangers
Figure 27.3
Friction factor and Colburn j factor data. (Redrawn from Ref. 2.)
coefficient h and the mass velocity G as St =
h Gcp
(27.14)
where cp is the specific heat of air. Combining Eqs. (27.9), (27.13), and (27.14), and solving for the heattransfer coefficient, we obtain h = jH cp v Pr−2/3 = (0.004)(1.1614)(1007)(6.515)(0.707)−2/3 = 38.4 W/m2 · ◦ C
(27.15)
The next step in the analysis is to determine the fin efficiency. From Kern and Kraus,3 the relation for fin efficiency of a longitudinal fin is = where
m= =
2h k
tanh(mb) mb
(27.16)
1/2
2(38.4) (222)(2.54 × 10−4 )
= 36.905 m−1
1/2
(27.17)
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Air Cooling of a HighVoltage Power Supply Using a Compact Heat Exchanger HighVoltage Power Supply and Compact Heat Exchanger
27.11
Thus =
tanh[(36.905)(0.02362)] (36.905)(0.02362)
= 0.7022/0.8717 = 0.8056 Now, we find the heatexchanger surface efficiency. The fin heattransfer surface area Sf for one duct is Sf = 2(b − )L = 2(2.362 − 0.0254)(16.0) = 74.771 cm2 = 7.477 × 10−3 m2
(27.18)
and the total heattransfer surface area St for one duct (fin area plus interfin area) is 1 St = S f + − L Nf 1 = 74.771 + − 0.0254 (16.0) 3.1496 = 79.445 cm2 = 7.945 × 10−3 m2
(27.19)
Note that this result does not include the interfin area near the fin tips because, as stated in assumption 10 in the “Analysis Assumptions” section, the baseplate on the other side of the coldplate is not considered as an extension of the fins and therefore does not participate in the heat transfer. From Kraus and BarCohen,4 the heatexchanger surface efficiency is = 1−
Sf (1 − ) St
= 1 − (74.771/79.445)(1 − 0.8056) = 0.8170
(27.20)
We are now ready to calculate the coldplate surface temperature. Neglecting changes in potential and kinetic energies, a simple energy balance on the air in the coldplate core yields the relation q=m ˙ cp(To − Ti )
(27.21)
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Air Cooling of a HighVoltage Power Supply Using a Compact Heat Exchanger 27.12
Heat Exchangers
where q is the heat dissipated by the power supply and To and Ti are the outlet and inlet air temperatures, respectively. Solving Eq. (27.21) for To , we obtain q To = + Ti m ˙ cp =
550 + 27 (0.0356)(1007)
= 42.3◦ C For an isothermal coldplate surface, Incropera and DeWitt5 derive a relation for the surface temperature Ts as Ts − To = e Ts − Ti
(27.22)
PLh m ˙ cp
(27.23)
where =−
Rearranging Eq. (27.22) and solving for Ts , we obtain Ts =
To − Ti e 1 − e
(27.24)
Because the wetted perimeter P in Eq. (27.23) pertains to a single duct, the mass flow rate m ˙ must also pertain to a single duct. Thus, the value for is =
−(0.05257)(0.16)(0.8170)(38.4) (0.0356/69)(1007)
= −0.5079 Finally, solving for Ts from Eq. (27.24), we obtain Ts =
42.3 − 27e−0.5079 1 − e−0.5079
= 65.4◦ C which is below the maximum allowable temperature of 70◦ C. Recognizing that this result is based on some assumed fin dimensions and an estimated flow rate of air, we now need to do a pressure drop analysis for the coldplate in order to select a fan for this application. This is accomplished by generating the system curve for the coldplate
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Air Cooling of a HighVoltage Power Supply Using a Compact Heat Exchanger HighVoltage Power Supply and Compact Heat Exchanger
27.13
and superimposing that curve on some candidate fan curves until the operating point coincides with the assumed flow rate in the foregoing heattransfer analysis. If this process does not lead to an acceptable operating point, the coldplate dimensions and/or the fin dimensions will have to be modified. Using information given by Mott6 and referring to Fig. 27.1, we see that there are three energy losses in the system. There is an entrance loss, a friction loss in the coldplate core, and a gradual enlargement loss in the fan transition. The total loss hL is the sum of these three losses. Hence, we have 2 v L hL = K1 + f + K2 (27.25) Dh 2g where K1 and K2 are the loss or resistance coefficients for the entrance and fan transition, respectively, f is the friction factor obtained from Fig. 27.3, and v is the air velocity in the coldplate core. For the entrance loss coefficient, we let K1 = 0.5. The gradual enlargement loss coefficient is a function of the cone angle and the ratio of the hydraulic diameter of the fan to the hydraulic diameter of the coldplate exit. To be conservative, we will assume an infinite hydraulic diameter ratio and a cone angle of 60◦ . These two assumptions lead to a loss coefficient of K2 = 0.72. The pressure drop p corresponding to the total loss is 2 v L p = hL = K1 + f (27.26) + K2 Dh 2 where is the specific weight of air. Because fan curves are expressed in terms of pressure as a function of volume flow rate, it is more convenient to express Eq. (27.26) in terms of Q rather than v. Substituting Eq. (27.7) into Eq. (27.26), we obtain Q2 L p = K1 + f + K2 (27.27) Dh 2A2t The units on p and Q in Eq. (27.27) are Pa and m3 /s, respectively. At this point in the analysis, we generate a system curve for the coldplate using Eq. (27.27). This step is best facilitated by creating a table of pressure drops and corresponding volume flow rates and recognizing that the friction factor f is a function of Reynolds number obtained from Fig. 27.3. The results are shown in Table 27.1. We then select a candidate axial fan that runs on 110 V ac power and that has a fan curve that appears to supply the required flow rate of air for our application. Shown in Fig. 27.4 is a graph of such a fan curve and the system curve
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Air Cooling of a HighVoltage Power Supply Using a Compact Heat Exchanger 27.14
Heat Exchangers
TABLE 27.1
System Curve Data Generated from Eq. (27.27) p
Q m3 /s
ft3 /min
Re
f
Pa
in. H2 O
0.00 0.01 0.02 0.03 0.03068 0.04
0.0 21.9 42.4 63.6 65.0 84.8
0 694 1388 2082 2129 2776
— 0.033 0.018 0.014 0.013 0.010
0.0 5.86 18.6 38.9 39.9 64.0
0.0 0.0235 0.0743 0.156 0.160 0.257
that was obtained from the data in Table 27.1. Note that one of the entries in Table 27.1 is for the estimated volume flow rate of 65 ft3 /min. The point at which the fan and system curves cross is the operating point. The operating point defines the volume flow rate and pressure drop for the system when the fan in question is used. We see that the operating point is approximately 65 ft3 /min and 0.16 in. H2 O. This volume flow rate matches the value that was estimated at the beginning of the analysis, so no modifications of coldplate or fin dimensions are necessary. However, if we desired a larger margin between the calculated baseplate temperature and the allowable temperature, we would have to increase the fin pitch and/or fin height and perhaps use a fan with a greater flow capacity. As a final calculation, we determine the heattransfer surface areatovolume ratio to verify that our coldplate can be classified as a compact
Figure 27.4
System curve and fan curve.
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27.15
heat exchanger. The total heattransfer surface area Stotal is 1 Stotal = 2 (b − ) + − LNd Nf 1 = 2 (2.362 − 0.0254) + − 0.0254 (16.0)(69) 3.1496 = 5804 cm2 = 0.5804 m2
(27.28)
For consistency with the standard definition of heattransfer surface area in heat exchangers, Eq. (27.28) includes the interfin area near the fin tips. The volume of the coldplate V, excluding the baseplates, is V = WLb = (22.0)(16.0)(2.362) = 831 cm3 = 8.31 × 10−4 m3
(27.29)
Hence, the heattransfer surface areatovolume ratio is = =
Stotal V 0.5804 8.31 × 10−4
= 698 m2 /m3
(27.30)
Strictly speaking, must be at least 700 m2 /m3 to classify the heat exchanger as “compact,” but since our result is extremely close to this value, we consider our coldplate to be a compact heat exchanger. A value of much less than 700 m2 /m3 would not have invalidated the heattransfer analysis, however. Concluding Remarks This chapter presents a heattransfer analysis of a coldplate for air cooling a highvoltage power supply. The coldplate consists of a standard configuration of folded aluminum fins brazed between two baseplates. A small axial fan, selected by means of a pressure drop analysis, draws ambient air through the fin passages, carrying the dissipated heat to the surroundings. The analysis shows that the power supply baseplate temperature is lower than the allowable value, thereby assuring reliable power supply operation with respect to thermal considerations.
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Air Cooling of a HighVoltage Power Supply Using a Compact Heat Exchanger 27.16
Heat Exchangers
While this analysis did not require modifications of coldplate or fin dimensions to find a suitable operating point, most analyses of this type involve an iterative process, which is most efficiently facilitated by the use of a computer. In order to meet the temperature specification, the thermal analyst or designer must iterate on the primary dimensional parameters while simultaneously selecting a fan that supplies sufficient airflow. This chapter outlines the basic analytical steps required to successfully do this. References 1. Hagen, K. D., Heat Transfer with Applications, PrenticeHall, Upper Saddle River, N.J., 1999, pp. 388, 410. 2. Kays, W. M., and A. L. London, Compact Heat Exchangers, 3d ed., McGrawHill, New York, 1984, p. 146. 3. Kern, D. Q., and A. D. Kraus, Extended Surface Heat Transfer, McGrawHill, New York, 1972, p. 90. 4. Kraus, A. D., and A. BarCohen, Thermal Analysis and Control of Electronic Equipment, McGrawHill, New York, 1983, p. 234. 5. Incropera, F. P., and D. D. DeWitt, Introduction to Heat Transfer, 4th ed., Wiley, New York, 2002, pp. 449, 450. 6. Mott, R. L., Applied Fluid Mechanics, 5th ed., PrenticeHall, Upper Saddle River, N.J., 2000, pp. 239–283.
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Source: HeatTransfer Calculations
Chapter
28 Energy Recovery from an Industrial Clothes Dryer Using a Condensing Heat Exchanger
David Naylor Department of Mechanical and Industrial Engineering Ryerson University Toronto, Ontario, Canada
P. H. Oosthuizen Department of Mechanical Engineering Queen’s University Kingston, Ontario, Canada
Description of the Problem A heat recovery system for an industrial clothes dryer is shown in Fig. 28.1. Warm moist air at a temperature of 45◦ C and relative humidity of 90 percent is exhausted from the clothes dryer at flow rate of 0.13 m3 /s. The total pressure is assumed to be 100 kPa. The building coldwater supply enters a thinwalled copper tube (D = 1.0 cm) at 8◦ C and at a flow rate of 6 L/min. The water supply makes nine passes across the exhaust duct, before going to the building’s hotwater heater. The total length of piping inside the duct is L = 1.7 m. The intent of the system is to recover latent heat by the formation of condensation on the outer surface of the tube. This will serve to preheat the cold water and reduce the energy demand at the water heater.
28.1
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Figure 28.1
Side view and front view of the heat recovery system.
Energy Recovery from an Industrial Clothes Dryer Using a Condensing Heat Exchanger
28.2
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Energy Recovery from an Industrial Clothes Dryer Using a Condensing Heat Exchanger Energy Recovery from Industrial Clothes Dryer
28.3
The following calculation illustrates a method to estimate the amount of energy that can be recovered with this design. The heat exchanger is analyzed as a combined heat/masstransfer problem using the log mean temperaturedifference (LMTD) method.
Solution Procedure The thermal resistance diagram for this problem, shown in Fig. 28.2, has three resistances: Ri is the thermal resistance associated with the convection inside the pipe, Rcond is the thermal resistance associated with the condensate film, and Rmass is the effective resistance associated with the mass transfer of vapor to the pipe wall. The thermal resistance across the pipe wall is neglected. The resistance Rmass requires further discussion. When noncondensable gases (air in the present case) are present in the flow, the concentration of these “noncondensables” builds up at the pipe surface. This effect reduces the partial pressure of vapor at the vaporliquid interface. As a result, there can be a substantial temperature difference between the freestream dewpoint temperature (T1,dp ) and the interface temperature (Ti ). It should be kept in mind that Rmass is not actually a thermal resistance, but it will be treated as such for the purpose of this iterative method. For a thinwalled pipe (Ai ≈ Ao ), the overall heattransfer coefficient U can be calculated as U=
1 (Rmass + Rcond + Ri )−1 = A
A Rmass +
1 hcond
+
1 hi
−1 (28.1)
where hcond and hi are the average convective heattransfer coefficients produced by condensation on the outside of the pipe and forced convection on inside of the pipe.
Figure 28.2
Thermal resistance diagram.
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Energy Recovery from an Industrial Clothes Dryer Using a Condensing Heat Exchanger 28.4
Heat Exchangers
Masstransfer resistance
The mass transfer of water vapor to the surface of the pipe is driven by the vapor concentration gradient. This gradient can be estimated using a heat/masstransfer analogy, combined with an empirical correlation for forced convection over a cylinder. Forced convection from a cylinder in crossflow can be calculated from the correlation by Hilpert [1]: Nu =
hD 1/3 = CRem D Pr k
(28.2)
The ChiltonColburn analogy [1] can be used to get an expression for the average Sherwood number. This analogy is applied by replacing the Nusselt number with the Sherwood number and the Prandtl number with the Schmidt number: Sh =
gm D 1/3 = CRem D Sc
(28.3)
where gm is the masstransfer conductance, is the mean density, is the binary diffusion coefficient (water vapor diffusing into air), and Sc is the Schmidt number (Sc = /). To initiate the iterative solution procedure, it is necessary to assume a value for the mean temperature on the outside of the condensate film. In the first iteration, the mean vaporcondensate interface temperature will be arbitrarily taken to be Ti = 15◦ C. This is just an initial guess, which will be corrected as the calculation proceeds. The air and masstransfer properties are obtained (from Ref. 1) at an approximate film temperature Tf = (T1 + Ti )/2 = 30◦ C: = 2.667 × 10−5 m2 /s, = 1.861 × 10−5 N s/m2 , Sc = 0.61. Since the concentration of vapor is low, the dynamic viscosity is approximated as that of dry air. The density of the vapor and air can now be calculated at the freestream conditions and at the interface conditions. For a relative humidity of 1 = 90 percent at T1 = 45◦ C, the partial pressure of water vapor at the inlet is Pv,1 = 1 Psat,1 = 0.9(9593 Pa) = 8634 Pa
(28.4)
where the saturation pressure is obtained from steam tables. Using this vapor pressure, the vapor density in the freestream can be calculated from the idealgas equation of state: v,1 =
Pv,1 Mv (8634)(18) = = 0.0587 kg/m3 RT1 (8314)(318)
(28.5)
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Energy Recovery from an Industrial Clothes Dryer Using a Condensing Heat Exchanger Energy Recovery from Industrial Clothes Dryer
28.5
The air density in the freestream is A,1 =
PA,1 MA (100,000 − 8634)(29) = = 1.002 kg/m3 RT1 (8314)(318)
(28.6)
At the vaporcondensate interface, the vapor pressure will correspond to the saturation pressure at the assumed interface temperature (for Ti = 15◦ C, Pv,i = 1705 Pa). So, the vapor density at the interface is v,i =
Pv,i Mv (1705)(18) = = 0.0128 kg/m3 (8314)(288) RTi
(28.7)
Similarly, the air density at the interface is A,i =
PA,i MA (100,000 − 1705)(29) = = 1.191 kg/m3 (8314)(288) RTi
(28.8)
Now, the mass fraction of vapor can be calculated in the freestream m1 =
v,1 0.0587 = = 0.0553 v,1 + A,1 0.0587 + 1.002
(28.9)
and at the interface mi =
v,i 0.0128 = = 0.0106 v,i + A,i 0.0128 + 1.191
(28.10)
In the present problem the mean freestream velocity is Q/Ac = (0.13 m3 /s)/(0.2)2 m2 = 3.25 m/s. So, the Reynolds number for the external flow is Re D =
V D (1.13)(3.25)(0.01) = 1973 = 1.861 × 10−5
(28.11)
where the density in the Reynolds number has been evaluated as the mean of the total density ( v + A) in the freestream and at the interface. At this Reynolds number the constant and exponent for Hilpert’s correlation are C = 0.683 and m = 0.466. Evaluating the average Sherwood number, we obtain Sh = 0.683 (1973)0.466 (0.61)1/3 = 19.88
(28.12)
Using this result, the convective masstransfer conductance gm is gm =
(1.13)(19.88)(2.667 × 10−5 ) Sh = = 0.0599 kg/m2 s D 0.01
(28.13)
The convective mass transfer is driven by the difference in the mass fraction of vapor in the freestream and at the vaporcondensate
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Energy Recovery from an Industrial Clothes Dryer Using a Condensing Heat Exchanger 28.6
Heat Exchangers
interface. The masstransfer rate, which is equal to the condensation rate, is given by m ˙ v = gm A(mv,1 − mv,i )
(28.14)
The total surface area of the pipe is A = DL = (0.01)(1.7) = 0.0534 m2 . Applying Eq. (28.14) and the results from Eqs. (28.9), (28.10), and (28.13) gives the condensation rate: m ˙ v = (0.0599)(0.0534)(0.0553 − 0.0106) = 1.43 × 10−4 kg/s
(28.15)
The latent heattransfer rate required to condense vapor at this rate is q=m ˙ v h f g = (1.43 × 10−4 )(2,466,000) = 353 W
(28.16)
where h f g is the latent heat of vaporization, evaluated at the interface temperature. The dewpoint temperature corresponding to Pv,1 = 8.364 kPa is T1,dp = 43.0◦ C. For the assumed vaporcondensate interface temperature, the effective thermal resistance due to mass transfer can be calculated as Rmass =
T1,dp − Ti 43.0 − 15.0 = = 0.0793◦ C/W q 353
(28.17)
Condensation heat transfer
The heattransfer coefficient on the outer surface of the condenser coil (hcond ) can be estimated using relations for laminar film condensation on vertical tube bundles. For the first iteration it is necessary to make an initial guess for the temperature drop across the vapor film. Initially, it will be assumed that Ti − T S = 2◦ C, where T S is the mean surface temperature of the pipe. Again, an improved estimate will be made for this quantity, later in the calculation. The outer heattransfer coefficient can be estimated from a theoretical relation for laminar film condensation on a bank of horizontal cylinders [2]. For n cylinders of diameter D, which are stacked vertically, the average Nusselt number is Nucond
1/4 gh f g ( − v )(nD)3 hcond (nD) = = 0.729 k k (Ti − T S)
(28.18)
In this equation, the properties of the liquid film are evaluated at mean temperature of the condensate film (T S + Ti )/2. For the first iteration we will evaluate the liquid water properties at 14◦ C: k = 0.592 W/m K,
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Energy Recovery from an Industrial Clothes Dryer Using a Condensing Heat Exchanger Energy Recovery from Industrial Clothes Dryer
28.7
= 1000 kg/m3 , and = 1.18 × 10−3 N s/m2 . The latent heat is evaluated at the interface temperature, h f g = 2466 kJ/kg K. The density of the airvapor mixture is about 1000 times lower than the density of the liquid film. So, in Eq. (28.18), the term ( − v ) will be approximated as 2 . In the current design, there are nine stacked pipes (n = 9). Evaluating Eq. (28.18), we obtain Nucond = 0.729
(9.81)(2466 × 103 )(1000)2 (9 × 0.01)3 (1.18 × 10−3 ) (0.592) (2.0)
1/4 = 1374 (28.19)
The average condensation heattransfer coefficient is then hcond =
Nucond k (1374)(0.592) = = 9037 W/m2 K nD 9(0.01)
(28.20)
Internal pipe flow (forced convection)
The properties of the water inside the pipe are evaluated at the mean bulk temperature. For the first iteration, the properties are evaluated at Tw,i = 8◦ C: kw = 0.584 W/m K, c p,w = 4196 J/kg K, w = 1000 kg/m3 , w = 1.377 × 10−3 N s/m2 , and Prw = 9.93. The mass flow rate in the pipe is 6 L/min or m ˙ w = 0.1 kg/s. Now the internal Reynolds number can be calculated: Re D =
wV D 4m ˙w 4(0.1) = = 9246 (turbulent) = w D (0.01)(1.377 × 10−3 ) (28.21)
There are many correlations in the literature for turbulent fully developed forced convection in circular pipes. In a review of empirical correlations, the most recent edition of the Handbook of Heat Transfer [3] recommends the equation by Gnielinski [4]: 0.4 Nui = 0.012 Re0.87 D − 280 Pr
1.5 ≤ Pr ≤ 500
3 × 103 ≤ Re D ≤ 106 (28.22)
Evaluating Eq. (28.22), we obtain Nui = 0.012 [(9246)0.87 − 280] (9.93)0.4 = 76.4
(28.23)
Evaluating the average heattransfer coefficient, we obtain hi =
Nui kw 76.4(0.584) = = 4462 W/m2 K D 0.01
(28.24)
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Energy Recovery from an Industrial Clothes Dryer Using a Condensing Heat Exchanger 28.8
Heat Exchangers
Overall heattransfer rate
The overall heattransfer coefficient is now calculated using Eq. (28.1): U = [(0.0534)(0.0793) + 1/9037 + 1/4462 ]−1 = 219 W/m2 K
(28.25)
Using an LMTD analysis, the total heattransfer rate q the heat exchanger can be expressed as (T1,dp − Tw,i ) − (T1,dp − Tw,o ) q = FUATlm = FUA T1,dp − Tw,i ln T1,dp − Tw,o =m ˙ w c p,w (Tw,o − Tw,o )
(28.26)
where Tw,i and Tw,o are the inlet and outlet temperatures of the domestic cold water. Note that this situation can be treated as a pure counterflow, since the temperature on the outside of the condensate film is essentially constant. So, the correction factor is unity (F ≈ 1). Solving Eq. (28.26) for the domestic water outlet temperature Tw,o , we have UA Tw,o = T1,dp − (T1,dp − Tw,i ) exp − (28.27) m ˙ w c p,w Evaluating Eq. (28.27) yields (219)(0.0534) Tw,o = 43.0 − (43.0 − 8.0) exp − = 8.96◦ C (0.1)(4196)
(28.28)
Using this result, we can now get an improved estimate of the total heattransfer rate to the condenser coils: q=m ˙ w c p,w (Tw,o − Tw,i ) = (0.1)4196(8.96 − 8.0) = 403 W
(28.29)
Now a better estimate of the interface temperature Ti can be made. Referring to the thermal resistance diagram in Fig. 28.2, the improved interface temperature is Ti = Tdp,1 − q Rmass = 43.0 − (403)(0.0793) = 11.0◦ C
(28.30)
This is significantly different from the initial guess of Ti = 15◦ C. So, at least one more iteration will be required. To proceed with the next iteration, an improved estimate of the temperature drop across the condensate film is also needed. This quantity can be obtained using the improved heattransfer rate estimate and the
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Energy Recovery from an Industrial Clothes Dryer Using a Condensing Heat Exchanger Energy Recovery from Industrial Clothes Dryer
TABLE 28.1
28.9
Various Quantities after First Three Iterations and for Converged
Solution Iterations Quantity Heattransfer rate q, W Vaporcondensate interface temperature Ti , ◦ C∗ Temperature difference across condensate Ti − T S, ◦ C† Condensation heattransfer coefficient hcond , W/m2 K Internal convection coefficient hi , W/m2 K Condensation rate m ˙ v , kg/s ∗ Initial † Initial
First
Second
Third
Converged solution
404 11.0
380 10.7
378 10.6
378 10.6
0.84
0.65
0.61
0.59
9037
10,978
11,675
11,932
4462
4195
4195
4195
1.43 × 10−4
1.52 × 10−4
1.53 × 10−4
1.53 × 10−4
assumed value Ti = 15◦ C. assumed value Ti − TS = 2◦ C.
condensation heattransfer coefficient (hcond ), as follows: Ti − Ts =
q 403 = = 0.835◦ C hcond A (9037)(0.0534)
(28.31)
Improved values have now been calculated for all the parameters that were “guessed” in the first iteration. So, the procedure can be repeated until convergence. Of course, an alternate method for solving this set of equations is to use commercial equation solving software which has built in functions for vapor properties (e.g., Engineering Equation Solver [5]). Table 28.1 shows some of the key parameters after the first three iterations and the final converged solution. The converged value for the total heattransfer rate is q = 378 W. It can be seen from the table that the solution converges quickly, in about three iterations. Conditions after the coil
The flow rate and inlet conditions for this problem are based on an industrial electrical clothes dryer with a rated power of 20 kW. So, for this application, 378 W is a relatively low quantity of heat to recover. To be practical, more coils would likely have to be added. In this case, the analysis would follow along the same lines, with the outlet conditions of the first coil becoming the inlet conditions for the second coil, and so on. The conditions at state 2 (after the coil) can be calculated using the condensation rate m ˙ v on the first coil, as follows: w2 = w1 −
m ˙v m ˙A
(28.32)
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Energy Recovery from an Industrial Clothes Dryer Using a Condensing Heat Exchanger 28.10
Heat Exchangers
where w1 and w2 are the values of specific humidity before and after the coil. Specific humidity w is the mass fraction of vapor per kilogram of dry air. The specific humidity before the coil can be calculated, assuming idealgas behavior as w1 =
Pv,1 V Mv /RT1 Mv Pv,1 Pv,1 = = 0.622 PA,1 V MA/RT1 MA PA,1 PA,1
(28.33)
Evaluating Eq. (28.33), the specific humidity at the inlet is w1 = 0.622
8.634 = 0.05878 100 − 8.634
(28.34)
The mass flow rate of dry air is m ˙ A = AQ =
(100 − 8.634)29 PA,1 MA Q= 0.13 = 0.13 kg/s RT (8.314)(318)
(28.35)
Evaluating Eq. (28.32), using the condensation rate for the converged solution, we obtain w2 = w1 −
m ˙v 1.53 × 10−4 = 0.0576 = 0.05878 − m ˙A 0.13
(28.36)
Equation (28.33) can be written at state 2 (after the coil) as follows: w2 = 0.622
Pv,2 Patm − Pv,2
(28.37)
Solving Eq. (28.37) for Pv,2 gives Pv,2 =
Patm w2 (100)(0.0576) = = 8.47 kPa 0.622 + w2 0.622 + 0.0576
(28.38)
Since the heat transfer to the pipe is assumed to be all sensible heat, the temperature of the flow can be assumed to remain at 45◦ C. So, relative humidity after the first coil is 2 =
Pv,2 8.47 = = 88.3% Psat,2 9.593
(28.39)
This outlet condition could be used as the inlet condition for the next condenser coil, if one existed. Concluding Remarks It should be noted that the current analysis probably underestimates the total heattransfer rate. The main reason is the conservative assumption that the heat transfer to the coil occurs by condensation only. Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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Energy Recovery from an Industrial Clothes Dryer Using a Condensing Heat Exchanger Energy Recovery from Industrial Clothes Dryer
28.11
In reality, there will also be a component of sensibleheat transfer, due to convection and radiation. These effects could be added quite easily to the current method. In addition, the effects of vapor shear have been neglected. The flow in the duct can be expected to thin the condensate film on the upstream side of the tube, thereby enhancing the heattransfer rate. However, in the present case, masstransfer effects are the main “resistance” to condensation. So, improvements to the condensation model would not be expected to greatly improve the estimated heattransfer rate. As a final comment, one should take note of the need to include the effects of noncondensable vapors in this type of calculation. As shown in Table 28.1, the high concentration of air suppresses the temperature of the vaporcondensate interface far below the freestream dewpoint temperature. In fact, if this analysis is repeated with the interface temperature at the saturation temperature corresponding to the freestream vapor pressure (i.e., a standard Nusselttype analysis), the heattransfer rate is overpredicted by a factor of more than 10. This general result has been shown before. For example, the reader is referred to the finitevolume numerical study of the forced convection on a flat plate with condensation by Chin et al. [6]. References 1. Incropera, F. P., and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 4th ed., Wiley, New York, 1996. 2. Oosthuizen, P. H., and D. Naylor, Introduction to Convective Heat Transfer Analysis, WCB/McGrawHill, New York, 1999. 3. Ebadian, M. A., and Z. F. Dong, “Forced Convection, Internal Flow in Ducts,” in Handbook of Heat Transfer, 3d ed., McGrawHill, New York, 1999, chap. 5. 4. Gnielinksi, V., “New Equations for Heat and Mass Transfer in Turbulent Pipe and Channel Flow,” Int. Chem. Eng., 16, 359–368, 1976. 5. Engineering Equation Solver, FChart Software, Madison, Wis., 2004. 6. Chin, Y. S, Ormiston, S. J., and Soliman, H. M., “Numerical Solution of the Complete Twophase Model for Laminar Film Condensation with a Noncondensable Gas,” Proceedings of 10th International Heat Transfer Conference, Brighton, U.K., 1994, vol. 3, pp. 287–292.
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Energy Recovery from an Industrial Clothes Dryer Using a Condensing Heat Exchanger
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12
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Source: HeatTransfer Calculations
29 Chapter
Sizing of a Crossflow Compact Heat Exchanger
Dusan P. Sekulic Department of Mechanical Engineering University of Kentucky Lexington, Kentucky
Summary This text offers a rigorous, stepbystep methodology for calculating core dimensions of a compact heat exchanger. Considering the analytical complexity of implemented calculations, the most intricate basic flow arrangement situation in a singlepass configuration would be a crossflow in which fluids do not mix orthogonal to the respective flow directions. Such a flow arrangement is selected to be considered in this calculation. The set of known input data is provided in the problem formulation. Subsequently, calculations are executed using an explicit stepbystep routine. The procedure follows a somewhat modified thermal design (sizing) procedure derived from the routine advocated in Shah and Sekulic (2003). The modification is related to the fact that iterative steps are listed and executed for both intermediate and overall refinements of all assumed and/or estimated entities. The overall pressure drop constraints are ultimately satisfied without further optimization of the design. The main purpose of the calculation sequence is to illustrate the procedure, usually hidden behind a userfriendly, but contentnonrevealing, platform of any existing commercial software package. Such a blackbox approach executed by a computer, although very convenient for handling by a not necessarily sophisticated user, is utterly nontransparent for a deeper insight. Consequently, this example
29.1
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Sizing of a Crossflow Compact Heat Exchanger 29.2
Heat Exchangers
Lgas
COLDAIR FLOW
Lstack
Secondary heattransfer surface Primary heattransfer surface
Lair HOTAIR FLOW Figure 29.1
Schematic of a crossflow heat exchanger with both fluids unmixed.
calculation is not intended to discuss a particular design; rather, it illustrates the procedure. Therefore, the numerical values calculated do not have importance for any particular design.
Problem Formulation A task at hand is to design (“to size”) a heat exchanger, specifically, to determine principal heatexchanger core dimensions (width, length, and height of the core having specified heattransfer surfaces) (see Fig. 29.1). The heat exchanger has to cool a hotair gas stream, available at an elevated temperature, with a coldair stream, available at a significantly lower temperature. Terminal states of both fluid streams are known, except for the outlet of the hot stream as follows:
Fluid Property Inlet temperature Outlet temperature Inlet pressure Mass flow rate Pressure drop Fluid type
Cold fluid
Hot fluid
Symbol
Unit
Value
Symbol
Unit
Value
Tc,i
K
500
Th,i
K
700
Tc,o
K
620
Th,o
K
—
pc,i
kPa
500
ph,i
kPa
100
m ˙c
kg/s
20
m ˙h
kg/s
20
pc
kPa
5
pc
kPa
4.2
Air
—
Air
—
—
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Sizing of a Crossflow Compact Heat Exchanger Sizing of a Crossflow Compact Heat Exchanger
29.3
Given data
The following information is provided: cold fluid—air; hot fluid—air, Tc,i , Th,i , Tc,o , pc,i , ph,i , m ˙ c, m ˙ h, pc , ph; heat exchanger flow arrangement type—singlepass, crossflow unmixedunmixed flow arrangement. Find
The principal dimensions of the core must be determined: (1) fluid flow lengths (core dimensions) in directions of hot and cold fluid flows and (2) the dimension of the stack of alternate layers of flow passages in the direction orthogonal to crossflow planes. Assumptions
Determination of the core dimensions assumes an a priori decision regarding selection of heattransfer surface types on both sides of a heat exchanger. This selection is, as a rule, within the realm of an engineer’s decisions for any sizing problem; thus, it is not necessarily given in the problem formulation. In the present calculation, a decision regarding the surface selection will be made at a point when geometric and heattransfer and/or hydraulic characteristics of the core need to be assessed for the first calculation iteration. That decision may always be modified and calculation repeated. The types of heattransfer surfaces will be selected, and data involving geometric, heattransfer, and hydraulic properties will be obtained from a database given in Kays and London (1998). The assumptions on which the calculation procedure is based are listed and discussed in detail in Shah and Sekulic (2003, chap. 3, p. 100) and will not be repeated here (standard assumptions for designing a compact heat exchanger). Calculation Procedure Design procedure for a sizing problem features two distinct segments of calculation. The first one delivers the magnitude of the thermal size of the core, expressed as an overall heattransfer area A, and/or formulated as a product of the overall heattransfer coefficient and the heattransfer area UA. Determination of this quantity should be based on an application of thermal energy balance [i.e., the heattransfer rate delivered by one fluid is received by the other; no losses (gains) to (from) the surroundings are present]. Formulation of this balance involves a fundamental analysis of heattransfer phenomena within the heat exchanger core, which can be summarized through a concept of heat exchanger effectiveness, Shah and Sekulic (2003). The resulting design procedure is the “effectiveness number of heattransfer units” method. The effectiveness is expressed in terms of known (or to be
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Sizing of a Crossflow Compact Heat Exchanger 29.4
Heat Exchangers
determined) inlet/outlet temperatures, and mass flow rates (for known fluids). The unknown temperatures (for some problem formulations) must be determined (i.e., they may not be known a priori—such as the outlet temperature of the hot fluid in this problem), and any assumed thermophysical properties should be recalculated multiple times (i.e., an iterative procedure is inherent). This feature of the calculation is only one aspect of the design methodology that ultimately leads to an iterative calculation sequence. The second reason for an iterative nature of this procedure is, as a rule, inherently transcendent structure of the effectivenessnumberofheattransfer correlation (for the crossflow unmixedunmixed arrangement, as in the case that will be revealed in step TDG8 in the tabular list below). Finally, the third and main reason for an iterative procedure is a constraint imposed on pressure drops. The magnitudes of pressure drops must be obtained from the hydraulic part of the design procedure. The hydraulic design part of the procedure cannot be decoupled from the thermal part, which leads to the calculation of pressure drops after thermal calculations are completed, and hence is followed by a comparison of calculated pressure drops with the imposed limits. As a rule, these limits are not necessarily satisfied after the first iteration, and subsequent iterations are needed. In this routine calculation presentation, determination of the thermal size of the heat exchanger will be termed the “targeting the design goal” (TDG) procedure. Each step will be separately marked for the purpose of crossreferencing. The second segment of the calculation is devoted to the determination of actual overall dimensions of the core, in a manner to satisfy the required overall heattransfer area and to achieve the overall heattransfer coefficient, that is, to satisfy the required thermal size. This segment is inherently iterative because it requires a satisfaction of pressure drop constraints as emphasized above. This segment of calculation will be termed “matching (the design goal and) geometric characteristics” (MGC) procedure. Both procedures will be organized as a continuous sequence of calculations and presented in a tabular format for the sake of compactness and easy access to various steps. The most important comments will be given as the notes to the respective calculation steps immediately after the equation(s) defining the step. A detailed discussion of numerous aspects of these calculations, and the issues involving relaxation of the assumptions, are provided in Shah and Sekulic (2003). The reader is advised to consult that source while following the stepbystep calculation procedure presented here. Some numerical values of the derivative variables presented may differ from the calculated values because of rounding for use elsewhere within the routinely determined data.
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Tc,i + Tc,o 500 + 620 = 2 2
TDG5
TDG4
C∗ =
C1 20.82 = C2 21.50
At this point, it is convenient to determine which of the two fluid streams has a larger heat capacity (for a nonbalanced case). Note that the designator 1 denotes the “weaker” fluid and the designator 2, the “stronger” fluid.
Designation of fluid streams Ch = C2 > Cc = C1
0.9685
—
21.50
—
—
kW/K
kW/K
July 27, 2005
Heat capacity rates of the fluid streams represent the products of respective mass flow rates and corresponding specific heats, calculated at the estimated referent temperatures.
20.82
Cc = (mc ˙ p )c = 20 × 1.041 ˙ p )h = 20 × 1.075 Ch = (mc
TDG3b
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TDG3a
kJ/kg K kJ/kg K
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The specific heat of either of the two fluids (for singlephase flows) is determined at the calculated referent temperatures. Since both fluids are gases in this case, and since both are considered as air, an ideal gas (see any idealgas thermodynamic properties data source) properties will be assumed.
1.041 1.075
c p,c = c p,air (Tc,ref ) = c p,air (560) c p,h = c p,air (Th,ref ) = c p,air (700)
K
K
Units
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700
560
Value
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To initiate the iterative procedure for a sizing problem like the one given in this problem formulation, a determination of referent temperatures of both fluids is needed. As a first guess, either an arithmetic mean of temperature terminal values or a given temperature value (if single) for each fluid may be selected.
Th,ref = Th,i = 700
Tc,ref =
Calculation
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TDG1b
TDG1a
Step
Targeting the design goal (TDG) procedure
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29.5
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T2,o = Th,o
Tc,in + Tc,out 500 + 620 = 2 2
Th,ref =
Th,in + Th,out 700 + 583.8 = 2 2
The second iteration for Tc,ref is identical to the first because both terminal temperatures were already provided in the problem formulation and an arithmetic average of the two was used (i.e., not, say, an integral mean value for which the distribution throughout the core would be needed).
Tc,ref =
641.9
560
583.8
K
K
K
—
July 27, 2005
T1,ref = Tc,ref
Note that the relationship used for determining the outlet temperature of the hot fluid is a straightforward consequence of adopted definitions of the heatexchanger effectiveness and heat capacity rate ratio, both expressed as functions of terminal temperatures. Now, having this originally unknown temperature estimated, a new value of the referent temperature for the hot fluid can be determined; that is, steps TDG1 through TDG7 should be repeated.
= 500 + (1 − 0.9685 × 0.6)(700 − 500)
Th,o = Tc,i + (1 − C ∗ ε)(Th,i − Tc,i )
0.600
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TDG1b/2
T1,o − T1,i 620 − 500 = T2,i − T1,i 700 − 500
Units
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Heatexchanger effectiveness represents the dimensionless temperature of the weaker fluid (the one with C1 = Cc ) (Sekulic, 1990, 2000). Note that the current decision on which fluid is weaker was based on the rough estimate, namely, a first iteration of referent temperatures. These are not necessarily the best assumptions, in particular for the hot fluid in this case. So, the outlet temperature of the hot fluid must be determined with more precision (see TDG7), and steps TDG1 through TDG7 should be repeated.
ε=
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TDG1a/2
Calculation The heat capacity rate ratio is not equal to 1; therefore, the heat exchanger operates with nonbalanced fluid streams.
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TDG7
TDG6
Step
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29.6
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TDG7/2
C1 20.82 = C2 21.22
T1,o − T1,i 620 − 500 = T2,i − T1,i 700 − 500
= 500 + (1 − 0.9814 × 0.6)(700 − 500)
Th,o = Tc,i + (1 − C ∗ ε)(Th,i − Tc,i )
This represents a refined value of the heatexchanger effectiveness; there is no change in this value because all the values involved were already known (i.e., all were given in the problem formulation), and are not affected by intermediate calculations.
ε=
This is a new, refined value of the heat capacity rate. The new value is slightly larger than the value in the first iteration.
C∗ =
582.2
0.600
0.9814
—
K
—
—
—
kW/K
July 27, 2005
TDG6/2
TDG5/2
Note that the implemented refinement of the value of the heat capacity rate on the hotfluid side did not change the designations of the fluids.
21.22
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Designation of fluid streams Ch = C2 > Cc = C1
This is a new value for the heat capacity rate of the hot fluid.
Ch = (mc ˙ p )h = 20 × 1.061
There is no change in this iteration for the cold fluid.
kW/K
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TDG4/2
TDG3b/2
20.82
kJ/kg K
kJ/kg K
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Cc = (mc ˙ p )c = 20 × 1.041
1.061
1.041
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The result obtained represents a new value for the specific heat of the hot fluid.
c p,h = c p,air (Th,ref ) = c p,air (641.9)
There is no change in this iteration for the coldfluid specific heat.
c p,c = c p,air (Tc,ref ) = c p,air (560)
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TDG2b/2
TDG2a/2
Note that the second iteration of Th,ref differs significantly from the first iteration: 641.9 K instead of 700 K.
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29.7
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kW/K
July 27, 2005
The product U A, also termed the “thermal size,” is a compounded thermal and physical size of the unit. This size involves the physical size (area of the heattransfer surface A) and heattransfer size (U is the overall heattransfer coefficient). The subsequent procedure (i.e., MGC) untangles intricate relations between these entities and physical size of the heat exchanger, leading to explicit values of all core dimensions for selected types of both (hotand coldside) heattransfer surfaces.
37.71
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U A = NTU · C1 = 1.811 × 20.82
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TDG9
n=1
C ∗ NTU Therefore, the exact expression for the heatexchanger effectiveness of an unmixedunmixed crossflow arrangement is algebraically very complex, regardless of the fact that it may be represented in different forms (Baclic and Heggs, 1985). See Shah and Sekulic (2003) for an approximate equation and Sekulic et al. (1999) for methods of determining the effectiveness relationships for other (not necessarily crossflow) flow arrangements. Graphical representation, as well as tabular data for the crossflow unmixedunmixed flow arrangement can be found in Baclic and Heggs (1985). It is important to realize that regardless of which correlation is used, the determination of NTU versus effectiveness is algebraically an intractable task without some kind of iterative procedure. In the correlation presented, In, n = 1, 2, . . . , ∞, represent modified Bessel functions of the integer n order.
ε =1−
—
Units
Kutz2103G
For the crossflow unmixedunmixed arrangement the relationship between effectiveness and the number of units (NTU) (explicit in terms of effectiveness but not explicit in terms of NTU) is as follows (Baclic and Heggs, 1985): ∞ √ ∗ n ∗ n/2 I (2NTU C ∗ ) e−NTU(1+C ) n 1 (C )
1.811
Value
GRBT05629
NTU = NTU(ε = 0.600; C ∗ = 0.9814; unmixedunmixed)
We obtained a new value of originally unknown outlet temperature of the hot fluid. The criterion for a termination of the iterative procedure may involve either a sufficiently small change of two successive values of this temperature, or a change of the successive values for heatexchanger effectiveness. In this case, these comparisons indicate that either no change or a very small change takes place, and the iterative procedure is terminated at this point.
Calculation
P2: IML/OVY
TDG8
Step
P1: IML/OVY T1: IML 17:33
Sizing of a Crossflow Compact Heat Exchanger
29.8
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Distribution of the total dimensionless thermal size between two fluid sides is determined in this step. Since both fluids are gases, both thermal resistances are to be assumed as equal in the first iteration. That leads to the given distribution of NTU1 and NTU2 versus NTU (Shah and Sekulic, 2003).
3.555
—
—
July 27, 2005
3.622
GRBT056Kutzv4.cls
NTU1 = NTUc = 2NTU = 2 × 1.811
K kPa kPa kJ/kg K 106 Pa s 102 W/m K — kg/m3 kg/m3 kg/m3
Kutz2103G
NTU2 = NTUh = C ∗ NTU1 = 0.9814 × 3.622
642 100 >96 1.061 31.67 4.80 0.699 0.498 0.574 0.533
Units
GRBT05629
QC: IML/OVY
The fluid properties are usually determined at arithmetic (or integral) mean values of fluid temperatures. In this calculation, we will adopt the arithmetic mean values from the second iteration. Certain data (temperatures, pressures) are provided in the problem formulation (see above), and/or devised from the inlet data and known pressure drops. Specific heats and viscosities are based on the mean temperatures. Densities are calculated assuming the idealgas assumption. The mean density is based on the following relationship: −1 1 1 1 i,m = + 2 i o
560 500 >495 1.041 28.95 4.32 0.698 3.484 2.782 3.093
Fluids 1 and 2
Value
P2: IML/OVY
MGC2
Determination of fluids’ thermophysical properties
Referent temperature Tc,ref = T1,ref , Th,ref = T2,ref Inlet pressures p , p 1,i 2,i Outlet pressures p 1,o , p2,o Specific heats c p,c = c p,1 , c p,h = c p,2 Viscosities , 1 2 Thermal conductivities k , k 1 2 Prandtl numbers Pr , Pr 1 2 Densities (inlet) 1,i , 2,i Densities (outlet) 1,o , 2,o Densities (bulk mean) 1,m, 2,m
MGC1
Calculation
Step
Matching geometric characteristics (MGC) procedure
P1: IML/OVY T1: IML 17:33
Sizing of a Crossflow Compact Heat Exchanger
29.9
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0.29
—
July 27, 2005
Surface characteristics involve the ratio of Colburn and Fanning factors j/f for plane plate fin surface 19.86 (Kays and London, 1998). For this surface, the ratio of j and f over the wide range of Reynolds numbers is approximately constant: j ≈ const = 0.29 f c,h
m m−1 m m m m−1 — m
GRBT056Kutzv4.cls
The sizes and shapes of heattransfer surfaces are correlated with the heattransfer and hydraulic characteristics. However, these characteristics in turn are needed to determine the sizes and shapes of the heattransfer surfaces! This interrelation renders the calculation procedure iterative. A selection of the surface geometry (i.e., selection of both fluid flow area geometries) should be done first a priori. Subsequently, calculation of heattransfer and fluid flow characteristics may be conducted to establish whether the surfaces selection fits the thermal size distribution and the overall thermal size (but in a manner to satisfy the pressure drop constraints). We will select the surface designated as 19.86 (Kays and London, 1998) for both fluid sides. See also Webb (1994) for further discussion of issues involving enhanced heattransfer surfaces.
6.350 × 10−3 782 1.875 × 10−3 0.152 × 10−3 63.75 × 10−3 1841 0.849 2 × 10−3
QC: IML/OVY
MGC4
Plate spacing b Number of fins n f Hydraulic diameter Dh Fin thickness Uninterrupted flow length lf Heattransfer area per volume between passes Fin area per total area A f /A Plate thickness a (under designer’s control)
Units
Kutz2103G
Value
GRBT05629
Selection of heat surface types: plane plate fin surface 19.86 (Kays and London, 1998)
Calculation
P2: IML/OVY
MGC3
Step
P1: IML/OVY T1: IML 17:33
Sizing of a Crossflow Compact Heat Exchanger
29.10
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July 27, 2005
In this step, the first estimates of core mass velocities are based on the estimates of j/f and parameters as discussed in MGC4 and MGC5. This estimate, as given above, is based on a simplified expression for G that takes into account the assumptions as follows: (1) only friction contributes to the pressure drop, (2) fouling resistances are neglected, (3) thermal resistance of the heattransfer wall is neglected, and (4) thermal resistances caused by formations of convective boundary layers on both fluid sides are equal.
kg/m2 s 19.94
GRBT056Kutzv4.cls
kg/m2 s
—
Kutz2103G
QC: IML/OVY
53.22
0.9
GRBT05629
An initial guess of the first estimate for the total surface temperature effectiveness for both sides should be made at this point (i.e., 0,c = 0,h). It is generally assumed that the total surface temperature effectiveness for a compact heattransfer surface (for a good design) must be within the range of 0.7 to 0.9. Let us assume the value to be at the high end of this range for both fluid sides (note that the same geometry was suggested for both surfaces). 1/2 p 2 pin 0 m 1/2 j 5 × 103 2 × 500 × 103 × 0.9 × 3.093 = 0.29 Gc ≈ f pin NTU Pr2/3 c 500 × 103 3.622 × 0.6982/3 1/2 1/2 p 2 pin 0 m j 4 × 103 2 × 100 × 103 × 0.9 × 0.533 = 0.29 Gh ≈ f pin NTU Pr2/3 h 100 × 103 3.555 × 0.6992/3
P2: IML/OVY
MGC6
MGC5
Although selection of surface types leads to the known heattransfer geometries on both fluid sides, the calculation of j(Re) and f (Re) parameters (i.e., heattransfer and friction factors in dimensionless form) cannot be performed straightforwardly at this point. This is because the Reynolds numbers for fluid flows are still unknown. This hurdle can be resolved by first guessing at the magnitude of the ratio of j to f (i.e., j/f ). Since this ratio is nearly constant in a wide range of Reynolds numbers, a unique value can be suggested as indicated above. In the first iteration (which will follow) only the value of j/f (rather than separate j and f values) would be needed for calculation of both fluid core mass velocities. Subsequently, these core mass velocities (see step MGC6) will be used to determine the first iteration of Reynolds numbers, leading to the corresponding values of j and f . Subsequently, the second iteration for j/f can be calculated from known j and f values. In this case, j/f ∈ (0.25, 0.37) for Re ∈(500, 4000), so an educated guess would lead to j/f ≈ 0.29.
P1: IML/OVY T1: IML 17:33
Sizing of a Crossflow Compact Heat Exchanger
29.11
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—
0.01006 0.01707 8.145 7.074
fc = exp(b0 + rc (b1 + rc (b2 + rc (b3 + rc (b4 + b5 rc ))))) fh = exp(b0 + rh (b1 + rh (b2 + rh (b3 + rh (b4 + b5 rh))))) rc = ln(Rec ) = ln(3446) rh = ln(Reh) = ln(1181)
—
—
July 27, 2005
This explicit calculation of the refined values for j and f is conducted by using j(Re) and f (Re) correlations for the selected geometry as listed in MGC4 (note that both fluid sides have the same geometry, based on a decision elaborated in MGC3). The values are calculated for Reynolds numbers as determined in step MGC7. The form of the correlation is dictated by the curve fitting of the experimentally obtained data listed in Kays and London (1998).
—
0.00506
jh = exp(a0 + rh (a1 + rh (a2 + rh (a3 + rh (a4 + a5 rh)))))
—
0.003718
GRBT056Kutzv4.cls
jc = exp(a0 + rc (a1 + rc (a2 + rc (a3 + rc (a4 + a5 rc )))))
—
—
QC: IML/OVY
MGC8
Note that uncertainties involved with an experimental determination of the Reynolds values, and subsequently j and f , are ± 2, ± 14, and ± 3 percent, respectively. So, the first estimates for Reynolds numbers must be refined later (in subsequent iterations) up to the margin of ±2 percent. One iteration would likely suffice.
1181
—
Units
Kutz2103G
Gh Dh,h 19.94 × 1.875 × 10−3 = h 31.67 × 10−6
Value
GRBT05629
3446
P2: IML/OVY
Reh =
Calculation Gc Dh,c 53.22 × 1.875 × 10−3 Rec = = c 28.95 × 10−6
Step
MGC7
P1: IML/OVY T1: IML 17:33
Sizing of a Crossflow Compact Heat Exchanger
29.12
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=
=
h
c
0.00506 0.01707
0.003718 0.01006
p pin
NTU Pr2/3
2 pin 0 m h
1/2 4 × 103 2 × 100 × 103 × 0.9 × 0.533 = 0.2965 100 × 103 3.555 × 0.6992/3
1/2
Gh Dh,h 20.16 × 1.875 × 10−3 = h 31.67 × 10−6
Reh =
1194
3890
—
—
kg/m2 s
July 27, 2005
Refined values of Reynolds numbers can now be calculated again. Note that Rec has increased in this iteration; that is, it differs from the previous iteration by 13 percent—this is obviously far more than a standard uncertainty margin of 2 percent allowed for determining the Reynolds number. Although Reh differs from the previous iteration for less than 2 percent, we will continue iterations until both Reynolds number changes reduce to below that margin.
Gc Dh,c 60.08 × 1.875 × 10−3 = c 28.95 × 10−6
Rec =
20.16
GRBT056Kutzv4.cls
These are the refined values for core mass velocities, but still based on an approximate G–( j/f ) relationship.
Gh ≈
j f
kg/m2 s
—
—
Kutz2103G
60.08
0.2965
0.3696
GRBT05629
QC: IML/OVY
MGC7/2
j f
j f
Refined values of j/f ratios are now based on newly calculated values of j and f factors for the estimated Reynolds numbers. 1/2 j 5 × 103 2 × 500 × 103 × 0.9 × 3.093 p 2 pin 0 m 1/2 Gc ≈ = 0.3696 f pin NTU Pr2/3 c 500 × 103 3.622 × 0.6982/3
P2: IML/OVY
MGC6/2
MGC9
P1: IML/OVY T1: IML 17:33
Sizing of a Crossflow Compact Heat Exchanger
29.13
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—
0.01692 8.266 7.085
rc = ln(Rec ) = ln(3890) rh = ln(Reh) = ln(1194)
MGC6/3
j f
p pin
NTU Pr2/3
2 pin 0 m h
1/2
= 0.2972
4 × 103 2 × 100 × 103 × 0.9 × 0.533 100 × 103 3.555 × 0.6992/3
These are refined values for the core mass velocities.
Gh ≈
1/2 20.19
60.36
kg/m2 s
kg/m2 s
—
0.2972
July 27, 2005
This is the next iteration for j/f ratios. 1/2 p 2 pin 0 m 1/2 j 5 × 103 2 × 500 × 103 × 0.9 × 3.093 = 0.3731 Gc ≈ f pin NTU Pr2/3 c 500 × 103 3.622 × 0.6982/3
—
GRBT056Kutzv4.cls
0.3731
Kutz2103G
—
GRBT05629
QC: IML/OVY
MGC9/2
— —
0.009779
fc = exp(b0 + rc (b1 + rc (b2 + rc (b3 + rc (b4 + b5 rc ))))) fh = exp(b0 + rh (b1 + rh (b2 + rh (b3 + rh (b4 + b5 rh)))))
P2: IML/OVY
In this step, the next estimation of the j and f factors is executed. Note that j factors differ from the ones determined in the previous iteration for less than 2 percent, respectively, for both fluids the f factors differ a bit more. As indicated in the MGC7 step, the actual values of j factors (determined experimentally) usually have a margin of error for an order of magnitude larger than calculated here in two successive calculations. The f factors usually have the experimental margin of error at the calculated level. So, an additional iteration would probably be sufficient. 0.003649 j = f c 0.009779 0.005029 j = f h 0.01692
—
—
0.005029
jh = exp(a0 + rh (a1 + rh (a2 + rh (a3 + rh (a4 + a5 rh)))))
Units
Value 0.003649
Calculation jc = exp(a0 + rc (a1 + rc (a2 + rc (a3 + rc (a4 + a5 rc )))))
Step
MGC8/2
P1: IML/OVY T1: IML 17:33
Sizing of a Crossflow Compact Heat Exchanger
29.14
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MGC10 Tw =
Th, ref +
NTUc C ∗ Tc,ref Tc,ref + Th,ref 560 + 642 NTUh = = ∗ NTUc C 2 2 1+ NTUh
These are the last estimations for j/f ratios. 601
K
—
—
July 27, 2005
0.2972
0.3732
—
—
—
—
—
—
GRBT056Kutzv4.cls
This is the last estimation of j and f factors. Calculated values are practically identical to the ones calculated in the previous iteration. j 0.003646 = f c 0.009769 0.005026 j = f h 0.01691
7.086
0.01691
fh = exp(b0 + rh (b1 + rh (b2 + rh (b3 + rh (b4 + b5 rh))))) rh = ln(Reh) = ln(1195)
0.009769
fc = exp(b0 + rc (b1 + rc (b2 + rc (b3 + rc (b4 + b5 rc )))))
8.271
0.005026
rc = ln(Rec ) = ln(3909)
0.003646
jc = exp(a0 + rc (a1 + rc (a2 + rc (a3 + rc (a4 + a5 rc )))))
—
—
Kutz2103G
jh = exp(a0 + rh (a1 + rh (a2 + rh (a3 + rh (a4 + a5 rh)))))
1195
3909
GRBT05629
QC: IML/OVY
MGC9/3
Gh Dh,h 20.19 × 1.875 × 10−3 = h 31.67 × 10−6
Reh = These are the new values of Reynolds numbers. They differ well below the margin of 2 percent from the previously calculated values (adopted here as a criterion for termination of the iterative procedure). Therefore, this should be considered as the last iteration.
Gc Dh,c 60.36 × 1.875 × 10−3 = c 28.95 × 10−6
Rec =
P2: IML/OVY
MGC8/3
MGC7/3
P1: IML/OVY T1: IML 17:33
Sizing of a Crossflow Compact Heat Exchanger
29.15
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MGC13 m = −0.1
0.00970
—
—
July 27, 2005
The hot fluid experiences cooling conditions. The flow regime is in the laminar region. Therefore, the exponent n = 0 in Shah and Sekulic (2003, table 7.12, p. 531). Tw m 601 −0.1 = 0.009769 fc,corr = fc Tc,ref 560
0.005026
GRBT056Kutzv4.cls
MGC12
—
−0.1185
QC: IML/OVY
Note that cold air (i.e., gas) is exposed to heating, and that its flow regime is turbulent. For details of the alternate exponent determination, see Shah and Sekulic (2003, table 7.13, p. 531). The conditions to be satisfied are 1 < Tw,ref /Tc,ref < 5; 0.6 < Pr < 0.9. Tw n = jh = 0.005026 jh,corr = jh Th,ref
—
Units
Kutz2103G
0.003616
Value
GRBT05629
The temperature of the heattransfer surface wall between the fluids is calculated from a balance equation that relates heattransfer rates delivered from one fluid to those received by the other. These heattransfer rates are expressed in terms of fluidtowall and walltofluid temperature differences and the respective thermal resistances on both fluid sides. The wall temperature is needed to perform a correction of thermophysical properties. This correction is due to temperature gradients between the fluids and the heattransfer surface wall across the respective boundary layers on both fluid sides. Tw n 601 −0.1185 = 0.003646 jc, corr = jc Tc,ref 560 1/4 Tw 601 1/4 = 0.3 − log10 n = 0.3 − log10 Tc,ref 560
Calculation
P2: IML/OVY
MGC11
Step
P1: IML/OVY T1: IML 17:33
Sizing of a Crossflow Compact Heat Exchanger
29.16
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60.36 × 1041 0.6982/3
2/3 Prh
Ghc p,h = 0.005026
20.19 × 1061 0.6992/3
lc =
b −= 2
f,c =
2
6.35 × 10−3
− 0.152 × 10−3
tanh(ml)c tanh(137.8 × 0.00302) = (ml)c 137.8 × 0.00302
mc = 2hc /k = 2 × 288.7/(200 × 0.152 × 10−3 )
The heattransfer coefficient for the hot fluid is determined from definition of the Colburn factor.
hh = jh,corr
m
m−1 137.8 0.00302
—
W/m2 K
W/m2 K
0.9460
136.7
288.7
July 27, 2005
MGC17
= 0.003616
GRBT056Kutzv4.cls
MGC16
Pr2/3 c
Gc c p,c
QC: IML/OVY
The heattransfer coefficient for the cold fluid is determined from the definition of the Colburn factor.
hc = jc,corr
Kutz2103G
For a fluid cooling case and laminar flow, the exponent is equal to 0.81. The conditions to be satisfied are: 0.5 < Tw,ref /Th,ref = 0.94 < 1; 0.6 < Pr = 0.699 < 0.9.
—
GRBT05629
m = 0.81
0.01603
P2: IML/OVY
MGC15
MGC14
Cold fluid is heated, and the flow regime is turbulent. The suggested calculation of the exponent in the correction term is valid for the range of temperature ratios as follows: Tw,ref 3.1 kPa. Therefore, this requirement is satisfied. However, since the pressure drop for the cold fluid was not satisfied, the new iteration is needed, as emphasized in step MGC34.
kPa
—
0.031
Kutz2103G
3.1
kPa
7.9
GRBT05629
The hot fluid side pressure drop divided by the inlet fluid pressure at the hot fluid side can now be calculated in the same manner as for the cold fluid. Of course, one may decide to calculate the pressure drop right away (step MGC36). The analytical expression for this calculation is given in the step MGC32. p ph = pi,h = 100 × 103 × 0.031 pi h
p = 500 × 103 × 0.0159 pi c From the input data, an allowed pressure drop is 5 kPa < 7.9 kPa. This indicates that the imposed condition is not satisfied! This prompts a need to reiterate the calculation with a new value of the mass velocity (in the first iteration, the mass velocity was calculated in step MGC6 by using the first approximation based on a weak dependence of j/f on the Reynolds number). 2 p 20.19 (1 − 0.32812 + 1.22) + 0.01603 0.198 × 0.498 = pi h 2(100 × 103 × 0.498) 1.875 × 10−3 0.5 4 0.498 0.498 − 1 − (1 − 0.32812 − 0.2) +2 0.574 0.574 pc = pi,c
P2: IML/OVY
MGC36
MGC35
MGC34
P1: IML/OVY T1: IML 17:33
Sizing of a Crossflow Compact Heat Exchanger
29.23
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1/2
0.498 0.498 − 1 − (1 − 0.32812 − 0.2) 0.574 0.574
1/2
Gh Dh,h 22.93 × 1.875 × 10−3 = h 31.67 × 10−6
Reh =
The new iteration cycle requires determination of new Reynolds numbers (with changed G values, including all the earlier local iterations; this is the fourth iteration of Reynolds numbers). This iteration repeats step MGC7.
Gc Dh,c 47.91 × 1.875 × 10−3 = c 28.95 × 10−6
Rec =
1357
3103
—
—
July 27, 2005
The new value of the mass velocity is increased from 20.19 to 27.24 kg/m2 s, that is, for roughly 7 percent.
(1 − 0.32812 + 1.22) + 0.01603 0.198 × 0.498 + 2 1.875 × 10−3 0.5 4
=
4 × 103 × 0.498 × 100 × 103
GRBT056Kutzv4.cls
2 × 100 × 103
kg/m2 s
Units Kutz2103G
22.93
Value
GRBT05629
QC: IML/OVY
MGC7/4
Calculation The new iteration loop starts with the determination of the set of new mass velocities. These values will be used to calculate the refined values of Reynolds numbers in step MGC7. Subsequently, steps MGC8 through MGC36 should be revisited. The new mass velocities G should be calculated from the exact expression for the pressure drop, as given in steps MGC33 and MGC35, assuming G values as unknown and the other numerical values in these equations as given. The convergence would be very fast. The new value of the mass velocity of 53.61 kg/m2 s is significantly smaller than the assessed value in the first iteration (60.37 kg/m2 s); thus, a reduction of almost 10 percent is achieved. p 1/2 2( pin in ) pi h Gh = L i 1/2 i i +2 − 1 − (1 − 2 − Ke ) (1 − 2 + Kc ) + f rh m o o h
P2: IML/OVY
MGC38
Step
P1: IML/OVY T1: IML 17:33
Sizing of a Crossflow Compact Heat Exchanger
29.24
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1/(0.9775 × 136.7 × 419.3) 1.785 × 10−5 560 560 642 + 1/(0.9542 × 288.7 × 419.3) 8.657 × 10−6 = −5 1/(0.9775 × 136.7 × 419.3) 1.785 × 10 1+ 1+ 1/(0.9542 × 288.7 × 419.3) 8.657 × 10−6
The wall temperature is calculated this time by taking into account the difference in thermal resistances on both fluid sides (compare this with step MGC10). With a difference of slightly more than 10 K (vs. the last iteration) for air, the correction of Colburn and Fanning friction factors will still remain relatively small.
642 +
586.7
K
July 27, 2005
=
Tw =
1/( 0 hA)h Rh Tc,ref Th,ref + Tc,ref 1/( 0 hA)c Rc = Rh 1/( 0 hA)h 1+ 1+ Rc 1/( 0 hA)c
Th,ref +
GRBT056Kutzv4.cls
MGC10/2
—
0.306
QC: IML/OVY
The ratios of Colburn and Fanning factors are calculated exactly for new Reynolds numbers. Note that these values did change, but as in any other iteration for determining this ratio, these changes are not significant beyond the first two decimal places.
—
Kutz2103G
0.3651
GRBT05629
The new iteration for the MGC8 estimation of j and f factors takes the values for Reynolds numbers from MGC7/4. j 0.003777 = f c 0.01034 0.004709 j = f h 0.01539
—
8.040 7.213
rc = ln(Rec ) = ln(3103) rh = ln(Reh) = ln(1357)
—
— — — —
0.003777 0.004709 0.01034 0.01539
jc = exp(a0 + rc (a1 + rc (a2 + rc (a3 + rc (a4 + a5 rc ))))) jh = exp(a0 + rh(a1 + rh(a2 + rh(a3 + rh(a4 + a5 rh))))) fc = exp(b0 + rc (b1 + rc (b2 + rc (b3 + rc (b4 + b5 rc ))))) fh = exp(b0 + rh(b1 + rh(b2 + rh(b3 + rh(b4 + b5 rh)))))
P2: IML/OVY
MGC9/4
MGC8/4
P1: IML/OVY T1: IML 17:33
Sizing of a Crossflow Compact Heat Exchanger
29.25
Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
m = 0.81 For the fluid cooling case and laminar flow, the exponent is equal to 0.81. The conditions to be satisfied are 0.5 < Tw,ref /Th,ref = 0.91 < 1; 0.6 < Pr = 0.699 < 0.9.
0.01431
—
July 27, 2005
MGC14/2
—
GRBT056Kutzv4.cls
Cold fluid is heated, and the flow regime is turbulent. The suggested calculation of the exponent in the correction term is valid for the range of temperature ratios as follows: Tw,ref 4.3 kPa. This result indicates that this condition is now safely satisfied. Consequently, because of this pressure drop, there is no need for further iterations. 22.932 p = pi h 2(100 × 103 × 0.498)
pc = pi,c
P2: IML/OVY
MGC36/2
MGC35/2
MGC34/2
P1: IML/OVY T1: IML 17:33
Sizing of a Crossflow Compact Heat Exchanger
29.31
Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
P1: IML/OVY
P2: IML/OVY
GRBT05629
Kutz2103G
QC: IML/OVY GRBT056Kutzv4.cls
T1: IML July 27, 2005
17:33
Sizing of a Crossflow Compact Heat Exchanger 29.32
Heat Exchangers
Conclusion The stepbystep heatexchanger design procedure clearly demonstrates how intricate the sizing of a compact heat exchanger may be. However, an inevitably iterative routine converges very rapidly. The main dimensions of this heatexchanger core are determined to be Lc = 48 cm, Lh = 23 cm, and Lstack = 550 cm. The core is made of plane triangular plate fin surfaces [surface designation 19.86 (Kays and London, 1998)]. Imposed limitations on the pressure drops are both satisfied. If, because of say, space considerations, the core dimensions must satisfy certain a priori imposed aspect ratios (fluid flow vs. stack length), further iterations would be needed. Calculation is presented in a most explicit manner, by listing each step (regardless of whether it may consist merely of a repeated calculation, already exercised in a previous step). All algebraic operations are, as a rule, included. This is done keeping in mind a need for full transparency of the calculation algorithm. Such design is conducted, as a rule, in practice by using a computer routine (what would never be fully transparent but eliminates any calculation errors that may often be present in a calculation as given here). Still, following the procedure as presented here, one can easily devise such a routine and execute the calculation. References Baclic, B. S., and P. J. Heggs, 1985, “On the Search for New Solutions of the SinglePass Crossflow Heat Exchanger Problem, Int. J. Heat Mass Transfer, vol. 28, no. 10, pp. 1965–1976. Kays, W. M., and A. L. London, 1998, Compact Heat Exchangers, reprint 3d ed., Krieger, Malabar, Fla. Sekulic, D. P., 1990, “A Reconsideration of the Definition of a Heat Exchanger,” Int. J. Heat Mass Transfer, vol. 33, pp. 2748–2750. Sekulic, D. P., 2000, “A Unified Approach to the Analysis of Unidirectional and Bidirectional Parallel Flow Heat Exchangers,” Int. J. Mech. Eng. Educa., vol. 28, pp. 307–320. Sekulic, D. P., R. K. Shah, and A. Pignotti, 1999, “A Review of Solution Methods for Determining EffectivenessNTU Relationships for Heat Exchangers with Complex Flow Arrangements,” Appl. Mech. Reviews, vol. 52, no. 3, pp. 97–117. Shah, R. K., and D. P. Sekulic, 2003, Fundamentals of Heat Exchanger Design, Wiley, Hoboken, N.J. Webb, R. L., 1994, Principles of Enhanced Heat Transfer, Wiley, New York. Zhao, H., A. J. Salazar, and D. P. Sekulic, 2003, “Influence of Topological Characteristics of a Brazed Joint Formation on Joint Thermal Integrity,” Int. Mech. Eng. Congress, vol. 1, ASME paper IMECE200343885, Washington, D.C.
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Source: HeatTransfer Calculations
Chapter
30 SinglePhase Natural Circulation Loops An Analysis Methodology with an Example (Numerical) Calculation for an AirCooled Heat Exchanger
Daniel M. Speyer Mechanical Engineering Department The Cooper Union for the Advancement of Science and Art New York, New York
Introduction When heat transfer in a fluid (flow) circuit results in a vertical temperature variation (a nonsymmetric variation), the resulting density differences and effects of friction will establish a circulation flow. In effect, the density differences will “pump” the fluid. This is called natural circulation (not to be confused with natural convection), and it is used by design, or is present, in many systems or situations involving heat transfer and a singlephase gas, singlephase liquid, or twophase liquid and gas; in these situations the usual forcedconvection heattransfer equations are applicable. Engineering examples of these are aircooled heat exchangers used to cool the dielectric oil in electric transformers, hotwater circulation heating systems in residential buildings and steamliquid water
30.1
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SinglePhase Natural Circulation Loops 30.2
Heat Exchangers
circulation (recirculation) in steam boilers in electricitygenerating power stations. In the case of singlephase gas or liquid systems, typically the analysis is not difficult, conceptually or in practice; and the calculations are not timeconsuming. However, the possibility of, applications of, and analysis of such natural circulation systems are not discussed in many general heattransfer and fluid flow texts, and are seldom realized by engineers. The present analysis develops the equations and applies them to an example problem. In addition, the following text also discusses methods to obtain optimum solutions—and as such represents a methodology for singlephase natural circulation calculations. Description of SinglePhase Natural Circulation A singlephase natural circulation fluid flow circuit is shown in Fig. 30.1. The system is distributed vertically, with fluid heated in the riser and cooled in the downcomer. As a consequence of density differences and gravity, the fluid is subject to (initially) an unbalanced buoyant force. As the fluid velocity increases, the temperature and resulting density differences will decrease, and friction effects in the form of minor losses and viscous shear will increase. These effects, with regard to minor losses and density effects, reduce the buoyant force, while the viscous ELEV. P2
P2 T2
d
T2
(L3) dncmr
(L3) riser
C
(L2) dncmr T2
(L2) riser
HE
b
AT
IN
c
f
a
T1
Figure 30.1
d e
T1
f
(DOWNCOMER)
b
T1
a
c
(L1) dncmr
(RISER)
P1
d e
G
T1 (L1) riser
ING OOL
a
P1
T2
(TEMP.)
ρ1
(DENSITY)
T2 ρ2
Natural circulation circuit, including modeling and notation used.
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SinglePhase Natural Circulation Loops SinglePhase Natural Circulation Loops
30.3
drag force is increased, and a steady state is reached when these forces are numerically equal. By itself the heating and cooling of the fluid does not establish a net buoyant force. A symmetric (temperature) distribution would create no unbalanced force. Basically it is the existence of isothermal (adiabatic) sections in the upper portion of the riser (the chimney) and lower portion of the downcomer that give rise to the net force—as their respective fluid densities are at a minimum and maximum.1 Major Analysis Assumptions The following restrictions and assumptions are made in order to develop relatively simple equations, which are nevertheless broadly applicable. The fluid is singlephase (liquid or gas), with a steady onedimensional flow; and the (spatial) differences in pressure, temperature, and velocity (and the velocity itself) are small. As a result, the differences in density are small and the fluid flow can be considered to be incompressible. The fluid density is assumed to be a linear function of temperature, and other fluid properties, including specific heat cp and viscosity , are constant, or the use of an average value is satisfactory. The rate of heat transfer per length is assumed to be a symmetric function (in elevation), symmetric with respect to the vertical midpoint of the heated and cooled regions. The resulting equations (for flow rate and inlet or outlet temperature) are identical to those obtained for the simpler case of a constant heat transfer (per unit length). If the heat transfer is constant, the temperature and density will be linear functions of elevation. Figure 30.1 exhibits this trend; however, this behavior is shown for simplicity—it is not a requirement. In twophase liquid vapor systems the basic approach is similar, although the details can be considerably more complex. Effects of compressibility (changes in momentum, critical flow, etc.), flow regime, differing gas and liquid (phase) velocity in a mixture, flooding, and other phenomena may be significant. These may even preclude a steady state.
1 In Fig. 30.1 the area b c e f in the densityversuselevation plot, multiplied by acceleration due to gravity g, is the total buoyant force—without the local pressure losses due to friction. This can be compared to the larger rectangular area which represents the maximum buoyant force, if the heating could be accomplished over negligible height at the bottom of the riser, and cooling over negligible height at the top of the downcomer. The figures and description are consistent with a fluid that expands when heated (i.e., ∂ /∂ T < 0).
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SinglePhase Natural Circulation Loops 30.4
Heat Exchangers
Development of Natural Circulation Equations for Flow Rate and Temperature The change in pressure between two locations is the same for any path; thus (from Fig. 30.1) the pressure difference (P2 − P1 ) will be identical for fluid flowing up the riser from P1 to P2 and fluid flowing down the downcomer from P2 to P1 . The pressure varies in the riser and downcomer as a result of hydrostatic head, viscous shear, and minor losses: dP = −g → dz
Hydrostatic head
dP = −g
dz
(30.1)
For the viscous shear and minor losses, it is convenient to use the mass flux, calculated at a single convenient area, as the flow variable: . G0 = m/A0 . Since loss coefficients (Klocal ) and friction factors ( f ) are in general based on different areas, a ratio of the flow area A0 to the local flow area is included:
dPi = −
Viscous shear
4 fi Di
Li
A0 Ai
2
G0 dz → Pi 2 i
A0 2 G20 Ai 2 i 2 2 A0 G0 Pj = −( Klocal ) j A j 2 j =−
Local minor losses
4 fi Li Di
(30.2) (30.3)
Integrating along the height of the riser and downcomer, we obtain Riser
P2 = P1 −
g dz −
G20
riser
+ (Klocal ) j
1 2 j
4 fi Li 1 A0 Di 2 i Ai riser A0 Aj
Downcomer
P2 = P1 − dncmr
+ (Klocal ) j
g dz + G20
1 2 j
2 (30.4)
4 fi Li 1 A0 Di 2 i Ai dncmr
A0 Aj
2
(30.5)
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SinglePhase Natural Circulation Loops SinglePhase Natural Circulation Loops
30.5
The energy equation (first law of thermodynamics) provides the fluid temperature, and thus the density, versus height: q˙ . dz = mcp dT → L
q˙ . dz = mcp L
dT = G0 A0 cp
dT
(30.6)
If the heat transfer per length is a function of height z and is a symmetric function with respect to the midpoint of a heated or cooled section, then the average temperature Tavg is the mean of the inlet and outlet temperatures (T1 and T2 for the riser) and the average density is the mean of the inlet and outlet densities, 1 and 2 . For T(z) symmetric about z = 0, we obtain L/2 −L/2
T(z)dz L
≡ Tavg =
T1 + T2 1 + 2 → avg = 2 2
(30.7)
We will divide the riser and downcomer (each) into three vertically stacked sections (regions) as shown in Fig. 30.1: two isothermal (adiabatic) sections, one above and one below the heated (and cooled) sections. These are isothermal sections of height (L1 )riser and (L1 )dncmr with fluid at density 1 ; using Eq. (30.7), heated and cooled sections (L2 )riser and (L2 )dncmr at avg ; and isothermal sections (L3 )riser and (L3 )dncmr at 2 . Substituting these heights and densities in Eqs. (30.4) and (30.5) results in expression (30.8) for the mass flow per unit area G0 , and integration of the energy equation [Eq. (30.6)] provides a second equation [Eq. (30.9)] in the common variables T1 , T2 , and G0 : G20 =
1 g[(L1 )dncmr − (L1 )riser ] + avg g[(L2 )dncmr − (L2 )riser ] + 2 g[(L3 )dncmr − (L3 )riser ]
(4 fi Li /Di ) (1/2 i ) (A0 /Ai )2 + (Klocal ) j (1/2 j ) (A0/A j )2 all
(30.8) q˙riser = −q˙dncmr = mc ˙ p(T2 − T1 ) = G0 A0 c p(T2 − T1 )
(30.9)
These two equations contain the heat transfer q, ˙ the riser or downcomer outlet temperature (T1 or T2 ), and the mass flux G0 ; and can be (numerically) solved since the fluid properties versus temperature, friction factors, and loss coefficients are readily available—often with correlations in the form of equations, tables, or graphs. However, the solution to Eqs. (30.8) and (30.9) presents some difficulties, particularly for a “casual analysis.” It is typically iterative, numerical accuracy issues are present (in particular, obtaining accurate density differences), and tables and graphs are timeconsuming and difficult to implement in computer programs.
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SinglePhase Natural Circulation Loops 30.6
Heat Exchangers
These problems can be resolved, or significantly lessened, if the density, friction factors, and loss coefficients are expressed by equations of a particular functional form. The remainder of this analysis is directed largely toward the particular equations chosen to address these problems and the equations obtained from Eqs. (30.8) and (30.9). Particular Equations (Correlations) for Density, Loss Coefficients, and Friction Factors It should be noted that the buoyant (force) terms, the numerator in Eq. (30.8), are equivalent to differences in density. Consider a typical case: the configuration used to analyze an aircooled heat exchanger in the following example calculation. Assume a threesection riser with unheated height (L1 )riser at (fluid density) 1 , heated height (L2 )riser at avg , and unheated height (L3 )riser at 2 and a downcomer with its entire length unheated at density 1 ,2 resulting in Eq. (30.8) with the numerator replaced by ( 1 − avg )gL2 + ( 1 − 2 )gL3 . This expression is typical; in general, the expression for the buoyant force will include density differences. In a typical problem (e.g., the following example calculation) the density values ( 1 , avg , and 2 ) are the same to two significant digits; thus to obtain an accuracy of only 10 percent, it would be necessary to use density values accurate to four digits. Most property tables only provide three or four significant digits! However, if the density differences are replaced by temperature differences, the large errors in evaluating density differences are avoided, and the resulting equation will be more easily solved. Such a linear dependence (of density) on temperature is satisfactory for gases, as well as liquids, since the pressure and temperature differences are generally small. This linear dependence is conveniently expressed by the coefficient of thermal expansion ()—the fractional change in volume per unit change in temperature: 1 ∂ 1 − 1 ∼ =− → − 1 = − 1 (T − T1 ) =− ∂T P 1 T − T1 (30.10) 2 − 1 = − 1 (T2 − T1 ) avg − 1 = −1/2 1 (T2 − T1 ) Values of for common liquids are often tabulated; however, is easily calculated from the values of density at two temperatures. For 2 This corresponds to air heated, rising, and mixing at the top of the riser or downcomer with cooler ambient air, which then flows down the downcomer. Mixing is equivalent to heat removal over a negligible height: (L2 )dncmr = (L3 )dncmr = 0, and (L1 )dncmr = (L1 + L2 + L3 )riser .
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SinglePhase Natural Circulation Loops SinglePhase Natural Circulation Loops
30.7
the ubiquitous liquid water and ideal gas (at ambient temperature), we obtain Liquid water at 25◦ C
= 2.1 × 10−4 K−1
Ideal gas at 25◦ C
=
1 1 = = 0.00336 K−1 T 298 K
With regard to the loss coefficients (and friction factors) in prior analysis—with equations such as Eqs. (30.4) and (30.5), or Eq. (30.8)— Speyer [9] found it useful to describe loss coefficients as the sum of the laminar equation (varies as Re−1 ) plus a constant turbulent value.3 For several of the loss coefficients, such an equation was realistic,4 and the expectation was it would be applicable to friction factors as well. In the present application this would lead to polynomials which can easily be solved5 : Klocal = f =
KRe + CCK Re f Re + CC f Re
where where
KRe = const, CCK = const
(30.11)
f Re = const, CC f = const
(30.12)
With regard to loss coefficients (Klocal ), although the minor losses frequently represent the majority of the pressure drop, laminar coefficients (KRe) are seldom required—the fully turbulent value, which is equivalent to the CCK value in Eq. (30.11),6 is sufficient. Also, a large number of geometries are involved, and laminar data are scarce. In any
3 As this calculation was being completed, three references in Idelchik [5] were found that apparently describe the use of an equation [e.g., Eq. (30.11)] for loss coefficients. Two references are for A. D. Altshul, and one for Idelchik. These were not available and are not included in the references list. 4 Orifices (particularly tangent orifices) and tube banks with a staggered square or triangular pitch are some examples where the form of Eq. (30.11) works well, as are screens and flow normal to a circular disk; however, flow normal to a sphere, a single tube, and multiple tubes, all in a line, are some examples where it seems to be a poor model (due to boundarylayer separation, etc.). 5 Laminar flow in pipes, and similar uniform interior flow geometries, are common in natural circulation calculations, and for these conditions, a constant friction factor is a particularly poor representation. For transitional or fully turbulent Reynolds numbers, if Eq. (30.11) or (30.12) and the associated constants are not sufficiently accurate for a particular flow regime, the constants can be revised (once an approximate solution for flow rate and Reynolds number is obtained). Thus it is an improvement over simply using a constant value for friction factor and loss coefficient—or at least no worse, as a constant value is obtained if the laminarflowbased constants (KRe and f Re) are set to zero: Klocal = CCK, f = CC f . 6 The local Reynolds number is based on a velocity and characteristic dimension appropriate to the flow geometry. It is unrelated to that used for the friction factor.
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SinglePhase Natural Circulation Loops 30.8
Heat Exchangers
event the laminar component to the loss coefficients (KRe) were not required for the example problem (based on a calculation of the Reynolds numbers—developed as part of that calculation), and are not reviewed herein. In laminar flow the friction factor is inversely proportional to the Reynolds number ( f = f Re/Re). For two common cases, flow inside a circular duct (pipe flow), and flow in the narrow passageway between parallel surfaces (slit flow), the constants ( f Re) are Laminar flow, f =
f Re : Re
f Re = 16
(pipe flow)
f Re = 24
(slit flow)
(30.13)
For other geometries the values are readily available; for instance, Holman [4] summarizes the results of Shah and London [8], for values of f Re (his f Re DH /4), as well as Nusselt number Nu, for various geometries.7 For turbulent flow in smooth and commercial piping, a number of authors have suggested a constant value of (about) 0.005 Turbulent flow
f ∼ = 0.005
(30.14)
which is used for smooth tubing, and for commercial piping, f ∼ = 0.0075 is used.8 The laminar and turbulent equations [Eqs. (30.13) and (30.14)] are now combined, and to assess how realistic this combined equation is, the laminar flow [Eq. (30.13)] is used, and for turbulent flow in smooth and rough pipes the equation [Eq. (30.15)] due to Colebrook [2] is used: 1 2.523 ε/D Turbulent flow + = − 0.869 ln (30.15) 3.7 4f Re 4 f This equation, with a surface roughnesstodiameter of zero (ε/D = 0), reproduces Prandtl’s universal law of friction for smooth pipes; and (for ε/D = 0.01), the equation of Drew and Genereaux [3], which McAdams [6] recommends as being applicable for “commercial pipes, steel, cast iron, etc., ±10 percent.”
7 If using an earlier edition of Holman [4], the values should be checked. There are errors in the fifth edition (1981). 8 Although the difference is rarely large, prior experience with a particular application, known or estimated surface roughness, and known diameter(s), may suggest another turbulent friction factor constant (CC f ). For large surface roughness and small diameter an order of magnitude larger value is possible, in which case Eq. (30.12) would be particularly inaccurate for laminar and intermediate Reynolds numbers (2300 > Re > ∼ 100).
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SinglePhase Natural Circulation Loops SinglePhase Natural Circulation Loops
30.9
Equation (30.13) is found to be realistic, with an error of 25 percent or less, for most Reynolds numbers of interest. In particular, as a general guide (for pipe and slit flow), f = ( f Re/Re) + CC f : f Re ≈ 16 to 24 and CC f ≈ 0.005 to 0.0075, for which 1. Generally error in f < 25 percent 2. f value generally conservative 3. Large overestimates likely for ≈1000 ≤ Re ≤ 2300 Incorporating Density, Friction Factor, and Loss Coefficient Correlations in Natural Circulation Equations Equation (30.8) is now modified to include temperature differences rather than density differences, and the friction factor and loss coefficient equations. Equations (30.8) and (30.9) are iterative if solving for flow (G0 ) and outlet temperature (T1 or T2 ); however, the revised equations are usually not iterative,9 unless there are large changes (with temperature) in fluid properties (principally viscosity)—which is not typical of natural circulation in air or liquid water. Combining Eqs. (30.8) and (30.10) to (30.12) results in a secondorder polynomial: [(L2 )riser + 2(L3 )riser − (L2 )dncmr − 2(L3 )dncmr ] /2 1 g(T2 − T1 ) 2 1 2 2 × 2 1 KRe j j A j A0 1 4Li f Rei i Ai A0 1 + CC f i + + CCK j Ai 2 i Di G0 Di A0 Aj 2 j G0 D j A0 1
G20 =
all
(30.16) The solution for G0 is Eq. (30.17), and the sign of the coefficients are a ≥ 0, b ≥ 0, and c ≤ 0. √ −b + b2 − 4ac 2 aG0 + bG0 + c = 0 → G0 = 2a 2 c = − (L2 )riser − 2(L3 )riser + (L2 )dncmr + 2(L3 )dncmr 1 g(T2 −T1 ) 4Li f Rei i 1 KRe j j 1 A0 A0 (30.17) b= + 2 A D A D i i j j j i all 2 2 4Li CC f i 1 A0 1 A0 a= + CCK j D A A i i i j j all 9 When the natural circulation equations are applied to a heat exchanger, as in the following example, the solution is often iterative. But, this is due to the additional heattransfer rate equation.
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SinglePhase Natural Circulation Loops 30.10
Heat Exchangers
If the Reynolds number is large, coefficient b is negligible, resulting in the following equation, where a and c are as defined in (30.17): aG20
+ c = 0 → G0 =
−
c a
(30.18)
Although unlikely, if the flow is laminar (for all the important friction factors and loss coefficients), the coefficient a is negligible, resulting in the following equation, where b and c are as defined in Eq. (30.17): bG0 + c = 0 → G0 = −
c b
(30.19)
If the riser outlet temperature T2 and heat transfer are both unknown, then T2 − T1 can be replaced by the heat transferred, by combining Eqs. (30.9) and (30.16). This results in a thirdorder polynomial in G0 , Eq. (30.20).10 As in Eqs. (30.18) and (30.19), if applicable, the solutions may be simplified by setting a = 0 or b = 0 [and b and a are as defined in Eq. (30.17)]:
aG30 + bG20 + c = 0 q˙riser c = c A0 c p(T2 − T1 )
= [−(L2 )riser − 2(L3 )riser + (L2 )dncmr + 2(L3 )dncmr ]
12 gq˙riser
(30.20)
A0 c p
The equations (to this point) are written assuming that the riser inlet temperature T1 is known. If it is the downcomer inlet temperature T2 that is known (and not the riser inlet), a more convenient equation, equivalent to Eq. (30.16), is as follows: [−(L2 )riser −2(L1 )riser +(L2 )dncmr + 2(L1 )dncmr ] /2 2 g(T2 − T1 ) 2 2 2 2 × 2 2 KRe j j A j A0 1 4Li f Rei i Ai A0 1 +CC f i + +CCK j Ai 2 i Di G0 Di A0 Aj 2 j G0 D j A0 1
G20 =
all
(30.21)
10 A number of the commercial computer programs will easily find the roots of a cubic polynomial.
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SinglePhase Natural Circulation Loops SinglePhase Natural Circulation Loops
30.11
The solutions to Eq. (30.21) are similar to those for Eq. (30.16).11 Equation (30.22) is analogous to Eq. (30.17): −b + b 2 − 4a c 2 a G0 + b G0 + c = 0 → G0 = 2a 2 c = [(L2 )riser + 2(L1 )riser − (L2 )dncmr − 2(L1 )dncmr ] 2 g(T2 − T1 ) 4Li f Rei i 2 (30.22) KRe A A j j 0 2 0 b = + 2 i Ai Dj j Aj Di all 2 2 4Li CC f i 2 A0 2 A0 a = + CCK j D A A i i i j j all Similarly, Eq. (30.23) is analogous to Eq. (30.19): a G30 + b G20 + c = 0 c = c
q˙riser A0 c p(T2 − T1 )
2 gq˙riser = [(L2 )riser + 2(L1 )riser − (L2 )dncmr − 2(L1 )dncmr ] 2 A0 c p
(30.23)
and b and a are as given in Eq. (30.22). Example Calculation for an AirCooled Heat Exchanger Aircooled heat exchangers are used to cool the oil in large electric transformers, as shown in Fig. 30.2. This calculation develops the air natural circulation flow, its outlet temperature, and the heat transfer.12 The source of the heat is resistance heating due principally to current in the transformer’s metal windings. This is generally referred to as load loss. To calculate the heat transfer, the air natural circulation flow and air outlet temperature must be calculated. It is assumed that the oil pumps are operating, the airside heat transfer is controlling, and the oil and ambient air temperatures are known. The natural circulation process shown in Fig. 30.2 differs from the earlier configuration in Fig. 30.1, in which the entire natural circulation 11 In both Eqs. (30.20) and (30.23), q ˙riser is used. If the standard thermodynamic convention is observed (with reference to a system, heat flow in is numerically positive, and heat flow out is negative), the equation would be correct with the heat transfer q˙riser replaced by −q˙dncmr . 12 The calculation uses approximate (made up, but reasonable) numerical values to describe the geometry.
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SinglePhase Natural Circulation Loops 30.12
Heat Exchangers
T1
MIXING REGION
•
q TRANSFORMER TANK T2
LOW
LOAD
RISER
HIGH
RISER
DOWNCOMER
IRON CORE
•
q HIGH (VOLTAGE) T1 FANS IDLE (OPTIONAL)
Figure 30.2
PUMPS OPERATING (OPTIONAL)
Schematic view of electric transformer aircooled heat exchanger for example
calculation.
flow was contained by a physical boundary. In this case, except for the inlet at the bottom of the heat exchanger, the airflow is (often) not enclosed by actual physical boundaries. Because of this difference it may not be immediately evident that a natural circulation analysis applies, or it may not be evident how it is to be applied. The following description should clarify matters. A mass (per unit time) of air enters the bottom of the heat exchanger, and as it rises, it is heated. It is assumed that the air does not mix with ambient air at the periphery of the riser section. At the top of the heat exchanger the heated air mixes with the ambient cooler air. (If the imagined enclosure were made real, this would be the location of a second heat exchanger.) To satisfy steadystate conservation of mass, the identical mass of air flows down external to the heat exchanger, and it enters the bottom, thus completing the flow circuit. As the air flows, it encounters several locations where pressure loss will occur. These include contractions and expansions in flow area and viscous drag. With a solid vertical boundary, and assuming radial variation in heat transfer is small, the riser flow can be described by a onedimensional (1D) model, as assumed herein. Without the vertical boundary, an infinitely wide riser (compared to its height) remains in 1D flow; however, reducing the width increases lateral flow. This warmer fluid (air) can mix with cooler ambient fluid at the periphery—resulting in increased heat transfer. In the following numerical example (and typical of electric transformers) the riser width and height are similar, and lateral flow resistance Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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Kutz2103G
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T1: IML
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SinglePhase Natural Circulation Loops SinglePhase Natural Circulation Loops
30.13
is not large (the geometry is rather “open”). The 1D assumption is not realistic and an underprediction (in heat transfer) by a factor of 2 would not be unexpected; however, if a solid vertical boundary were present the 1D calculation would be expected to be realistic. The heat exchanger comprises parallel flattened tubes, roughly rectangular in shape: 10 cm long, 1.1 cm wide, and 2 m high. Above and below the 2m tubes are structural members and a manifold. These entry and exit regions are each an additional 0.25 m high. Hot oil from the transformer enters a manifold at the top of the heat exchanger, splitting the oil flow between the (flattened) tubes (3400 tubes). The oil flows down the interior of the tubes, and a manifold at the bottom collects the cooled oil, which is returned to the transformer. The spacing between these tubes provides a partially enclosed flow area for the cooler air—in counterflow to the hotter oil. The tubes are arranged in rows; the distance between the ends of successive tubes is 1.5 cm, and the distance between the sides of adjacent tubes is 3.5 cm. In the horizontal plane a repeating unit (unit cell) can be assumed to be the rectangular region connecting the centers of four neighboring tubes. The heat exchanger has a 3 × 6m footprint and a 2.5m height. The long sides of the flattened tubes are parallel to the 6m dimension. For riser: (10 + 1.5) × 3.5 + (1.1 × 1.5) cm2 = 41.9 cm2
Flow area in unit cell
4 × 52.9 cm2 × 10−2 m/cm = 0.0755 m 2 × (10 + 1.1) cm
Equivalent diameter of unit cell (see below) Height of isothermal sections Height of heated section
(L1 )riser = (L3 )riser = 0.25 m (L2 )riser = 2.00 m 49.1 cm2 = 14.2 m2 52.9 cm2 3 × 6 m2 Aht = 2 × (10 + 1.1 cm) × 2 m × 52.9 cm2 100 cm 2 = 1510 m × m A0 = 3 × 6 m2 ×
Flow area Heattransfer (ht) area
For heat exchanger overall: Projected area of unit cell
(10 + 1.5) × (1.1 + 3.5) cm2 = 52.9 cm2
The mixing of air with ambient (as was discussed earlier) is modeled by removing all the heat at the top of the downcomer. For downcomer: Height of cooling section
( L3 )dncmr = 0 ( L2 )dncmr = 0
Height of bottom isothermal section
( L1 )dncmr = 2.5 m
Height of upper isothermal section
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Kutz2103G
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T1: IML
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July 27, 2005
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SinglePhase Natural Circulation Loops 30.14
Heat Exchangers
The bottom entry is modeled as an orifice with an area that is 20 percent of the projected horizontal area. There is a sudden (area) contraction at a 0.6m clearance between the bottom of the riser and the concrete pad. In addition, the presence of the manifold results in a reduced area, representing 34 percent of the total projected area, at the top and bottom of the riser. For heat exchanger overall: Total projected (horizontal) area (top area)
A = 3 m × 6 m = 18 m2
Minimum flow area (orifice) at inlet
Aorifice = 0.2 × 18 m2 = 3.6 m3
For the calculation of loss coefficients, the permanent pressure loss for an orifice and sudden contraction and sudden expansion (in flow area) are required. The following (for turbulent flow) are adapted from a summary in Bird, Stewart, and Lightfoot [1]. For a sudden contraction, the coefficient is increased from 0.45 to 0.5. Defining an area ratio , the ratio of smaller cross sectional area to larger cross sectional area, we obtain
Porifice
Korifice =− 2 local
= PSE PSC
Aorifice
m ˙ Adownstream
2
2.7(1 − 2 )(1 − ) =
Korifice Adownstream 2 m ˙ KSE =− 2 local Aupstream 2 m ˙ KSC =− 2 local Adownstream
KSE = (1 − )2 KSC = 0.5(1 − )
(30.24)
(30.25) (30.26)
(where subscripts SC and SE denote sudden compression and contraction, respectively). For the natural circulation flow G0 , the riser flow area is the most convenient area to use (for A0 ) in Eq. (30.17). The loss coefficients are often the major part of a fluid flow calculation. Typically, drawings and other sources provide the detail, and then a large number of individual resistances must be developed and documented. Thus the calculations should be developed in a clear, compact manner—in this case a tabular format, such as Table 30.1. The effect of the density ratio ( 1 / local ) will be small, and thus it will be simplest if the losses are combined into a single sum (this simplifies the bookkeeping). About twothirds of the losses occur at the inlet conditions, and 1 / local = 1 / avg is used (this is slightly conservative). Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
SC = sudden compression. SE = sudden expansion.
(14.2/10.8)2 = 1.73 (14.2/18)2 = 0.622 (14.2/6.12)2 = 5.38
(A0 /Ai )2
0.2 × 1.73 = 0.35 4.56 × 0.789 = 2.84 0.33 × 5.38 = 1.78 0.32 × 5.38 = 1.74 0.29 × 5.38 = 1.53 0.44 × 5.38 = 2.35
K(A0 /Ai )2
GRBT056Kutzv4.cls
†
0.5(1 − ) = 0.20 (1 − )2 (1 − )/ = 4.56 0.5(1 − ) = 0.33 (1 − )2 = 0.32 0.5(1 − ) = 0.29 (1 − )2 = 0.44
Ki
QC: IML/OVY
∗
Clearance at entry 0.6 × 2(3 + 6) = 10.8 SC∗ 10.8/18 = 0.6 Minimum entry area 0.25 × 18 = 4.5 Orifice 0.25 Bottom manifold 0.34 × 18 = 6.12 SC 0.34 6.12/14.2 = 0.43 SE† Top manifold SC SE 0.34
i
Kutz2103G
Ki for
GRBT05630
Ai , m2
Calculation of Loss Coefficients for Sample Calculation of AirCooled Heat Exchanger
P2: IML/OVY
Description
TABLE 30.1
P1: IML/OVY T1: IML July 27, 2005 19:21
SinglePhase Natural Circulation Loops
30.15
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SinglePhase Natural Circulation Loops 30.16
Heat Exchangers
All loss coefficients are adjusted by the area ratio (A0 /Alocal ) and then summed. The resulting sum of the loss coefficients (10.6) is rounded to 11: 2 2 1 A0 1 A0 1 ∼ ∼ CCK i K 11 = = i i Ai avg Ai avg Since the laminar loss coefficient contribution was not included, it is a good idea to develop a justification particularly when a laminar result (for friction is obtained and/or in new situations), although a viscous oil, or the equivalent, is typically required to obtain laminar flow for the minor losses. The following provides the justification. The Reynolds number for flow in the downcomer (Re0 ) can be related to the local Reynolds number (Relocal ) for the evaluation of the Klocal values: Relocal =
A0 D local A0 mD ˙ local D0 mD ˙ 0 D local A0 × × = = Re0 A local A0 D0 A0 D0 A local D0 A local (30.27)
From Table 30.1, the principal losses are at the manifold and the inlet orifice. The equivalent diameter of the manifold will be similar to the tubes (Dmanifold ≈ D0 ), and the equivalent diameter of the orifice is assumed to be 1.5 m. Looking ahead, the result of this example calculation is Re0 =1430, and we obtain D orifice A0 0.15 m √ = 1430 × 0.622 ≈ 2400 D0 A orifice 0.07 m D manifold A0 0.07 m √ = Re0 = 1430 × 5.38 ≈ 3340 D0 A manifold 0.07 m
Refan = Re0 Remanifold
The manifold regions are essentially orifices,13 and the orifice K values (for < 0.5) vary by about ±15 percent for Reorifice > 40.14 Thus the use of constant turbulent K values is justified.
13 The values of are 0.25, 0.33, and 0.43, from Table 30.1. The average is ∼ 0.35, and the orifice K is 4.4; however, to compare this to the sudden expansion and sudden compression, the areas need to be put on an equal basis [see Eqs. (30.24) to (30.26)]. If all are based on the smaller area, the orifice K value is K = 4.4 × 0.352 = 0.54. This compares reasonably well to the sum of the sudden compression and sudden expansion values, 0.33 + 0.42 = 0.75. 14 Perry and Chilton [7, figure 518] has discharge coefficient varying from 0.62 to 0.72, for < 0.5 and Reorifice > 40 (Reynolds number calculated with D = orifice diameter). The equivalent K values are 2.60 (0.62−2 ) and 1.93 (0.72−2 ), or K ∼ 2.3 ± 15 percent.
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30.17
Now returning to our (main) calculation for the natural circulation, in order to calculate the laminar flow friction and heat transfer, it is necessary to adjust the flow geometry to a case where a laminar flow solution is readily available. The unit cell is similar to slit flow (D = 2 × gap width), and this is used. Note that the equivalent diameter calculated for the unit cell, and the value for a slit, are quite similar (0.0755 and 0.07 m). From Holman [4], f Re DH /4 = f Re = 24; and for constant temperature NuT = 7.541 and constant heat flux Nu H =8.235. Constant heat flux is consistent with expected poor heat transfer and a large, and thus relatively constant, tube walltoair temperature difference: Equivalent diameter (for slit flow)
Driser = D0 = 2 × 3.5 cm × 10−2 m/cm = 0.07 m
Laminar friction factor
f Re = 24
Laminar Nusselt number (constant heat flux)
(Nuriser )laminar = Nu H = 8.235
For turbulent flow the friction factor is independent of geometry, but it depends on surface roughness. Heat exchangers such as this are placed in the outdoors and the exterior surfaces are painted, and (thus) not hydraulically smooth. Friction factor (turbulent and not smooth)
(ε/D ≈ 0.01) → CC f = 0.0075
The unknowns are the natural circulation flow and the outlet air temperature—and assuming a large oil flow rate the inlet and outlet oil temperatures will be essentially the same. For illustrative purposes the oil was assumed to change by 2◦ C; and air inlet temperature = T1 = 25◦ C, oil inlet temperature = (Toil )in = 61◦ C, and oil outlet temperature = (Toil )out = 59◦ C. In order to calculate the airflow and outlet temperature, Eq. (30.17) is used. As the friction (flow area) is chosen to be A0 and the loss coefficients are calculated, including the adjustment to the area A0 [i.e., (A0 /Alocal )2 → (A0 /Ai )2 ], the calculation will proceed with ( A0 /Ai )2 = 1. To start the calculation, a value of the air outlet temperature (T2 ) is assumed. The first estimate (guess) is T2 = 40◦ C: Tavg =
25 + 40◦ C = 32.5◦ C 2
Air properties are 1 = 1.168 kg/m3 , avg = 1.872 × 10−5 kg/(m s), avg = 1.139 kg/m3 , and kavg = 2.672 × 10−5 kW/m/◦ C, with = (32.5 + 273.15 K)−1 = 3.272 × 10−3 K−1
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Kutz2103G
QC: IML/OVY GRBT056Kutzv4.cls
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SinglePhase Natural Circulation Loops 30.18
Heat Exchangers
Then c = [−2.0 − 2 × 0.25 m + 0 + 0]3.272 × 10−3 K−1 (1.139 kg/m3 )2 × 9.8 m/s2 (40 − 25◦ C) = −1.640[kg/(m2 s)]2 1 1.168 kg/m3 = = 1.025 avg 1.139 kg/m3 b=
A0 =1 Ai
(see discussion in text)
4 × 2.5 m × 24 × 1.872 × 10−5 kg/(m2 s) × 1.025 × 1 (0.07 m)2 all
= 0.9402 kg/(m2 s) 4 × 2.5 m × 0.0075 a= × 1.025 + 11 × 1.025 × 12 = 12.37 0.07 m all −0.9402 kg/(m2 s) + (0.94022 [kg/(m2 s)]2 − 4 × 12.37 × (−1.640) G0 = 2 × 12.37 0.3281 kg = m2 s A review of this calculation reveals that an incorrect tube length of 2.5 m, not 2 m, was used in the calculation of the coefficients a and b, immediately above. However, the effect of this minor error is not significant (and it is conservative), and the calculation was not revised: (Velocity0 )avg = (Re0 )avg =
G0 0.3281 kg/(m2 s) m = = 0.2809 3 avg s 1.168 kg/m G0 D0 0.3281 kg/(m2 s) 0.07 m = 1227 = 1.872 × 10−5 kg/(m s)
One version of the heat transfer is obtained from conservation of energy [Eq. (30.9)]: q˙riser = G0 A0 c p(T2 − T1 ) = ×
0.3316 kg 1.008 kJ × 14.2 m2 × 2 kg ◦ C m s
kW × (40 − 25◦ C) = 71.2 kW 1 kJ/s
The other version of the heat transfer is obtained from the (heat exchanger) rate equation. As the Reynolds number is less than 2300,
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July 27, 2005
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SinglePhase Natural Circulation Loops SinglePhase Natural Circulation Loops
30.19
the laminar flow convective heattransfer coefficient is used: (Nuriser )laminar = Nu = =
Nu k hD = 8.235 → h = k D
8.235 × 2.672 × 10−5 kW/(m ◦ C) kW = 0.003143 2 ◦ 0.07 m m C
The oil and air are in counterflow: q˙ = U × A × LMTD ∼ =
×
(Aht )air 1 1 (Aht )air 1 xwall + + + hair hoil (Aht )oil hfouling k
[(Toil )in − T2 ] − [(Toil )out − T1 ] (Toil )in − T2 ln (Toil )out − T1
(30.28)
For a very small airside heattransfer coefficient, the fouling factor, wall conductance, and oilside heattransfer coefficient will be negligible: q˙ =
0.003143 kW m2 ◦ C
× 1510 m2
(61 − 40◦ C) − (59 − 25◦ C) = 128.0 kW 61 − 40◦ C ln 59 − 25◦ C
The two values of the heat transfer (71.2 and 128.0 kW) are in poor agreement, and a new value of the air outlet temperature is estimated (guessed). Once the second calculation is completed, the difference in heat transfer for each calculation can be used to extrapolate to a better estimate of heat transfer. The results for the first three values of air outlet temperature are summarized in Table 30.2.
TABLE 30.2
Summary of Outlet Temperature T2 and Heat Transfer q˙ for Three Iterations for Sample Calculation of AirCooled Heat Exchanger q, kW
Air outlet temperature T2 , ◦ C
Basis of T2
G0 , kg/(m2 s)
Eq. (30.9)
Eq. (30.27)
q˙
40 45 45.56
Guess Guess q˙ → 0
0.3281 0.3805 0.3859
70.44 108.9 113.6
128.0 114.1 112.4
−45.0 −4.5 1.0
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Kutz2103G
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T1: IML
GRBT056Kutzv4.cls
July 27, 2005
19:21
SinglePhase Natural Circulation Loops 30.20
Heat Exchangers
The final (third iteration) results are as follows: Riser outlet temperature
T2 = 45.56◦ C
Mass flux of air in riser
G0 = 0.3846 kg/(m2 s) q˙ = 1/2 (113.0 + 112.9 kW) = 113.0 kW
Total forcedconvection heat transfer
Summarizing some of the other numerical results, we obtain riser air velocity (Velocityavg ) of 0.3418 m/s; outside heattransfer coefficient (hconv ) of 0.003167 kW/(m2 ◦ C); LMTD of 23.51◦ C; and skin friction (Pfric ) and minor loss (Pε ) pressure drop of 0.2285 and 0.7253 Pa, respectively. Finally, the radiation heat transfer, and natural convection heat transfer from the peripheral tubes, should be calculated. As the manifold is also at the oil temperature, the total height (2.5 m) is used. The radiation heat transfer is from the four sides and top. The graybody radiation is given by 4 4 q˙radiation = Aradiation ε Twall − Tambient (30.29) The tube wall temperature is the average oil temperature, and paints typically have values of emissivity of 0.8 to about 0.95, and a value of 0.90 is used. The outside (envelope) area is applicable—four sides and the top: Aradiation = 2 × (6 m × 2.5 m + 3 m × 2.5 m) + 6 m × 3 m = 63 m2 q˙ =
5.67 × 10−11 kW 63 m2 0.90[(60 + 273.15 K)4 m2 K4 −(25 + 273.15 K)4 ] = 14.20 kW
Assuming turbulent free convection at the outer surface of the peripheral tubes (i.e., vertical plates), McAdams [6] recommends (Nu)film =
hL = 0.13(Gr Prfilm )1/3 kfilm
109 < Gr Prfilm < 1012
(30.30)
and the Grashof and Prandtl numbers are evaluated at the film temperature (average of wall and ambient temperatures): Gr Prfilm = (Gr × Pr)film = gL3 (Twall − Tambient )( 2 /2 )film Prfilm 1 25 + 45.42 ◦ Tfilm = + 60 C = 47.61◦ C 2 2 Air properties are film = 1.086 kg/m3 , film = 1.944 × 10−5 kg/(m s), Prfilm = 0.0703, kfilm = 2.783 × 10−5 kW/(m ◦ C), and = (47.61 + 273.15 K)−1 = 0.003118 K−1 .
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Kutz2103G
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July 27, 2005
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SinglePhase Natural Circulation Loops SinglePhase Natural Circulation Loops
30.21
Then
Anat conv = 2 6 m × 2.5 m ×
1.1 cm 10 cm + 3 m × 2.5 m × 10 + 1.5 cm 3.5 + 1.1 cm
= 29.67 m2 9.8 m/s2 2.53 m3 3.118 × 10−3 K−1 (60 − 25◦ C)(1.086 kg/m3 )2
Gr Prfilm =
3.118 × 10−3 K[1.944 × 10−5 kg/(m s)]2 × 0.703 = 3.665 × 1010
h = 0.13 × (3.665 × 1010 )1/3 × = 0.004807
2.783 × 10−5 kW/(m ◦ C) 2.5 m
kW m2 s
This heattransfer coefficient is similar in magnitude to the airside natural convection value of 0.003167 kW/(m2 ◦ C)—thus the contribution of the peripheral tube heat transfer will be small. The additional heat transfer and the total are q˙ = [0.004807 − 0.003167 kW/(m2 ◦ C)] 23.7 m2 (60 − 25◦ C) = 1.361 kW The total heat transfer is q˙total = 113.0 + 14.20 + 1.361 kW = 128.6 kW. References 1. Bird, R. B., W. E. Steward, and E. N. Lightfoot, Transport Phenomena, Wiley, p. 217 (1960). 2. Colebrook, C. F., “Turbulent Flow in Pipes, with Particular Reference to the Transition Region between Smooth Rough Pipe Laws,” J. Inst. Civil Eng. (London), 11, 133–156 (1938–1939). 3. Drew, T. B., and R. P. Genereaux, Trans. Am. Inst. Chem. Eng., 32, 17–19 (1936). 4. Holman, J. P., Heat Transfer, 9th ed., McGrawHill, p. 273 (2002). 5. Idelchik, I. E., Handbook of Hydraulic Resistance, 2d ed., Hemisphere Publishing, p. 3 (1986). 6. McAdams, W. H., Heat Transmission, 3d ed., McGrawHill, pp. 56–157 (1954). 7. Perry, R. H., and C. H. Chilton, eds., Chemical Engineers’ Handbook, 5th ed., pp. 5–13 (1973). 8. Shah, R. K., and A. L. London, Laminar Flow: Forced Convection in Ducts, Academic Press (1978). 9. Speyer, D. M., ShellSide Pressure Drop in Baffled ShellandTube Heat Exchangers, Ch.E. Ph.D. thesis, New York University (1973). 10. Streeter, V. L., and E. B. Wylie, Fluid Mechanics, 8th ed., McGrawHill, p. 215 (1985).
References Colebrook [2] and Drew [3] were not available. The information was taken from secondary sources: Streeter [10] and McAdams [6], respectively.
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SinglePhase Natural Circulation Loops
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22
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10:22
Source: HeatTransfer Calculations
Chapter
31 Evaluation of Condensation Heat Transfer in a Vertical Tube Heat Exchanger
Karen Vierow School of Nuclear Engineering Purdue University West Lafayette, Indiana
Problem Some nuclear reactors are equipped with “passive” condensers to remove heat when the reactor is isolated from the turbine and other normal operation systems. In this context “passive” means that the heat exchangers are driven by natural forces and do not require any power supplies, moving parts, or operator actions. The condensers are vertical tube bundles that sit in a pool of water and condense steam on the tube side. The condensers are designed to take steam generated in the reactor and return condensate back to the reactor vessel. Noncondensable gases such as air may enter the system and seriously degrade heattransfer performance. To design one of these heat exchangers, evaluate the local condensation heattransfer coefficient in a test loop from experimental data provided in Table 31.1. Estimate how many tubes would be needed if the total heat removal rate were 1 MW and the tube length were reduced by a factor of 2.
31.1
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Evaluation of Condensation Heat Transfer in a Vertical Tube Heat Exchanger 31.2
Heat Exchangers
TABLE 31.1
Temperature Measurements
Distance from tube inlet z, m
Condenser tube wall outer surface temperature, ◦ C
Coolant bulk temperature, ◦C
0.03 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
57.0 52.9 41.9 22.9 12.3 11.8 10.9 11.5 11.3
21.0 — 12.6 — 10.7 — 10.1 — 10.0
Given information
Geometry Doublepipe, concentric tube vertical heat exchanger Steamnoncondensable gas mixture flowing into center tube from top Condensate draining out the bottom of center tube Coolant water flowing upward in the coolant annulus around condenser tube Stainlesssteel condenser tube 2 m in length, 1 in. o.d., 1.5mm tube wall thickness Wellinsulated outer pipe
Conditions Steam inlet mass flux = 6.75 kg/(m2 s) Noncondensable gas inlet mass flux = 0.75 kg/(m2 s) Primary side inlet pressure = 220 kPa Coolant flow rate = 0.13 kg/s Temperature measurements as in Table 31.1
Assumptions
Saturated conditions at condenser tube inlet Noncondensable gas and steam in thermal equilibrium Coolant temperature well represented by bulk temperature due to turbulent mixing Steadystate operation Each tube in prototype tube bundle carrying an equal heat load Nomenclature A Heattransfer area C1 , C2
Constants in coolant axial temperature profile fit
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10:22
Evaluation of Condensation Heat Transfer in a Vertical Tube Heat Exchanger Condensation Heat Transfer in Vertical Tube Heat Exchanger
cp
Specific heat
Din
Condenser tube inner diameter
Dout
Condenser tube outer diameter
hc
Condensation heattransfer coefficient
hfg
Heat of vaporization
hf g
Modified heat of vaporization
kwall
Condenser tube wall thermal conductivity
m ˙
Mass flow rate
P
Pressure
Pinlet
Total pressure at condenser tube inlet
Ptotal
Total pressure
Pv
Vapor partial pressure
q
Heat rate
q
31.3
Heat flux
Rc
Condensation thermal resistance
Rcoolant
Coolant thermal resistance
Rwall
Condenser tube wall thermal resistance
T
Temperature
TCL
Condenser tube centerline temperature
Tcoolant
Coolant temperature
Tsat
Saturation temperature
Twall,in
Condenser tube wall inner surface temperature
Twall,out
Condenser tube wall outer surface temperature
X
Mole fraction
z
Axial distance downstream from condenser tube inlet
Calculation Method The heattransfer coefficient is obtained from experimental data because there are no widely applicable mechanistic analysis models for the condensation heattransfer coefficient in tubes under forced convection with noncondensable gases. Empirical correlations are available for given ranges of steam flow rates, noncondensable gas flow rates, pressures, and tube diameters against which the calculation results could be compared; for example, see the paper by Vierow and Schrock (1991). Definition of the heattransfer coefficient
The condensation heattransfer coefficient must first be defined by identifying the thermal resistances. The steam–noncondensable gas
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Evaluation of Condensation Heat Transfer in a Vertical Tube Heat Exchanger 31.4
Heat Exchangers
mixture enters the condenser tube from the top. Steam directly contacts the cold tube wall and immediately begins to condense, forming a thin condensate film along the inner surface of the condenser tube. This film thickens with distance from the inlet and represents an increasing thermal resistance. As the steam–noncondensable gas mixture is drawn to the condensate surface, a noncondensable gas boundary layer develops and provides additional thermal resistance. There is also a thermal resistance in the tube wall and the coolant. Noting that the coolant is assumed to be well mixed, the radial profile for the coolant temperature at any axial location may be taken as a constant temperature over the cross section. The condensation heattransfer coefficient is theoretically defined as the heattransfer coefficient at the vaporliquid interface. The heattransfer coefficient between the condenser tube centerline and the condenser tube inner surface is more practical to calculate because information on the condensate thickness is seldom available and the heattransfer resistance radially within the steamgas mixture is small enough to be neglected. The heattransfer resistances considered in the calculation are as in Fig. 31.1. If the local heat flux and the temperatures are known, the thermal resistances and therefore the heattransfer coefficients may be calculated. The condensation heattransfer coefficient can then be obtained as below, where z is the axial distance from the condenser tube inlet and Tsat (z) is the local saturation temperature in the condenser
Condenser Tube Wall
Condensate Film Outer Pipe Wall
. .. .
Steam/Gas TCL Twall,in
Tcoolant
Twall,out
Coolant Annulus
.
TCL Figure 31.1
Rc
. . Rwall
Twall,in
Twall,out
Rcoolant
.
Tcoolant
Thermal resistances.
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Evaluation of Condensation Heat Transfer in a Vertical Tube Heat Exchanger Condensation Heat Transfer in Vertical Tube Heat Exchanger
31.5
tube at z: Dout Din hc (z) = Tsat (z) − Twall, in (z) q (z)
(31.1)
The diameter ratio factor is necessary because the heat flux will be calculated at the condenser tube outer surface and must be corrected to correspond to the tube inner surface. Estimation of local heat flux from axial coolant temperature profile
An energy balance on the coolant provides the local heat flux: q (z)dA = −m ˙ coolant c pdT(z)
(31.2)
Note that the righthand side is negative since the coolant is in countercurrent flow with the primary side, from bottom to top: q (z) = −m ˙ coolant c p
m ˙ coolant c p dT(z) dT(z) =− Dout dz Dout dz
(31.3)
The axial temperature gradient can be obtained by fitting the coolant temperature as a function of z. The top four points of the coolant data provided in Table 31.1 can be fit by an exponential. Because the temperature change is minimal between 2.0 and 1.5 m, the heat flux is smaller than can be expected from condensation heat transfer and the heattransfer modes are most likely forced convection and conduction. Since the axial temperature rise rather than the actual temperature is of interest, an equation for the temperature rise is calculated. For 0.03 ≤ z ≤ 1.5 m, we obtain T(z) = T(z) − T (1.5 m) = C1 × e−z/C2
(31.4)
For the current data, C1 = 12.37, C2 = 0.32, and T(z) = 12.37e−z/0.32 : q (z) = −
m ˙ coolant c p dT(z) m ˙ coolant c p C1 e−z/C2 m ˙ coolant c p dT(z) =− = Dout dz Dout dz Dout C2 (31.5)
Determination of local saturation temperature
The local saturation temperature is obtained from knowledge of the local steam partial pressure. To determine local steam partial pressure, the local steam flow rate and noncondensable gas flow rate must be known.
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Evaluation of Condensation Heat Transfer in a Vertical Tube Heat Exchanger 31.6
Heat Exchangers
The local steam flow rate is the difference between the inlet steam flow rate and the amount of steam that has been condensed. The latter is found from the heat removal rate between the tube inlet and the desired distance from the tube inlet z: qi→i+1 =
zi+1
q (z)dA = −m ˙ coolant c p12.37 e−zi+1 /0.32 − e−zi /0.32
(31.6)
zi
m ˙ condensate (z) =
q0→z hf g (z)
(31.7)
To account for condensate subcooling, the modified heat of vaporization suggested by Rosenhow is employed: hf g (z) = h f g (z) + 0.68cp,l (Tsat (z) − Twall, in (z))
(31.8)
The local saturation temperature is obtained in later steps and must be iterated on: m ˙ steam (z) = m ˙ steam, inlet − m ˙ condensate (z)
(31.9)
The local noncondensable gas flow rate is not a function of z. It is vented out the bottom of the condenser tube under steadystate conditions, and the local mass flow rate may be considered constant throughout the tube. The local steam partial pressure is obtained as Pv (z) = Ptotal (z) × Xsteam (z) = Pinlet × Xsteam (z)
(31.10)
where Xsteam (z) is the local steam mole fraction. The total pressure at any location in the tube can be assumed the same as the inlet pressure. This assumption arises from the requirement that the pressure drop through the condenser tube be very small so that steam can easily enter the condenser in a passive mode. Let the noncondensable gas be air. The local steam fraction is approximately m ˙ steam (z) 18.02 Xsteam (z) = m ˙ steam (z) m ˙ air + 28.97 18.02
(31.11)
The local saturation temperature is obtained from Incropera and DeWitt (2002): Tsat (z) = Tsat (Pv (z))
(31.12)
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Evaluation of Condensation Heat Transfer in a Vertical Tube Heat Exchanger Condensation Heat Transfer in Vertical Tube Heat Exchanger
31.7
This saturation temperature and hf g at Pv (z) are substituted back into Eq. (31.8). Calculation of tube wall inner surface temperature
The axial profile of the tube wall outer surface temperature has been measured and provided in the given information. The inner surface temperature is much more difficult to measure because the instrumentation can easily affect the condensation phenomena. Hence, the inner surface temperature is deduced from the outer surface temperature by a simple conduction heattransfer calculation: Dout ln Twall, in (z) − Twall, out (z) D in q(z) = Rwall = (31.13) Rwall 2kwall dz Rearranging, we obtain Twall, in (z) = q (z) × Rwall dA + Twall, out where
ln Rwall dA =
Dout Dout D in 2kwall
(31.14)
(31.15)
Note that the wall temperature data do not monotonically decrease. This is due to the uncertainty in the thermocouple measurement of approximately ±0.5◦ C. The parameters necessary to calculate the local condensation heattransfer coefficient are now available. Example calculation at z = 0.03 m
From Eq. (31.5), the local heat flux on the condenser tube outer surface is (0.13 kg/s) 4.18 kJ/(kg K) 12.37e−(0.03 m)/0.32 q (z) = 1m 0.32 (1 in.) 39.3 in. = 239 kW/m2 From Eq. (31.8), the modified heat of vaporization is hf g (0.03 m) = 2202 kJ/kg + 0.68 4.18 kJ/(kg K) (121◦ C − 57◦ C) = 2384 kJ/kg K
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Evaluation of Condensation Heat Transfer in a Vertical Tube Heat Exchanger 31.8
Heat Exchangers
Note that hf g and Tsat must be taken at the vapor partial pressure calculated below. This must be iterated on once the partial pressure is known. The procedure is simple to do in software such as an Excel spreadsheet. The local condensate flow rate is obtained from Eqs. (31.6) and (31.7): m ˙ cond (0.03 m) = −
(0.13 kg/s)(4.18 kJ/kg K)12.37 K[e−(0.03 m)/0.32 − e−(0 m)/0.32 ] 2384 kJ/kg
= 2.5 × 10−4 kg/s
Equation (31.9) provides the local steam flow rate: 2 1 in. m ˙ steam (0.03 m) = (6.75 kg/m s) − 2(0.0015 m) 4 39.3 in./m 2
− 2.5 × 10−4 kg/s = 0.0027 kg/s − 0.00025 kg/s = 0.0024 kg/s To obtain the steam mole fraction, the noncondensable gas mass flow rate is first calculated. 2 1 in. 2 m ˙ air = (0.75 kg/m s) − 2(0.0015 m) = 0.00030 kg/s 4 39.3 in./m From Eq. (31.11), we obtain 0.0024 kg/s 18.02 kg/kg mol Xsteam (0.03 m) = 0.00030 kg/s 0.0024 kg/s + 28.97 kg/kg mol 18.02 kg/kg mol = 0.93 The local steam partial pressure can now be estimated from Eq. (31.10): Pv (0.03 m) = (220 kPa) × (0.9242) = 204 kPa From ASME steam tables, we have Tsat (0.03 m) = 121◦ C The solution is iterated on between hf g and Tsat .
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Evaluation of Condensation Heat Transfer in a Vertical Tube Heat Exchanger Condensation Heat Transfer in Vertical Tube Heat Exchanger
31.9
The condenser tube inner wall temperature is as follows, from Eqs. (31.14) and (31.15): ln Rwall dA =
1 in. 39.3 in. 39.3 in. 1 in. − 2(0.0015 m) 1m 1m 2(15.5 W/m K) 1 in.
= 1.0 × 10−4
K m2 W
Twall, in (0.03 m) = (239 W/m2 K) × (1.0 × 10−4 K m2 /W) + 57◦ C = 57.03◦ C From Eq. (31.1), the local condensation heattransfer coefficient becomes 0.0254 m 239 kW/m2 0.0224 m hc (0.03 m) = 121◦ C − 57◦ C = 4.23 kW/m2 K The local heattransfer coefficient and related parameters are tabulated in Table 31.2 at each tube wall temperature measurement location. At 1.5 m, the condensate mass flow rate was calculated to be greater than the inlet mass flow rate. From this point on, the calculation procedure becomes invalid because the condensation rate is very small. To estimate the number of tubes needed for a heat removal rate of 1 MW at half the tube length, note that essentially all the condensation TABLE 31.2
HeatTransfer Calculation Results z, m
Parameter
0.03
0.25
0.5
0.75
1.0
1.25
1.5
q (z)
239
120
55
25
12
5
2
121 2384 0.0024
119 2394 0.0011
112 2422 0.00048
104 2476 0.00022
94 2505 0.00011
74 2500 3.61E05
— — —
0.93
0.86
0.72
0.54
0.36
0.16
—
204 4.23
189 2.06
159 0.89
119 0.35
80 0.16
36 0.098
— —
on tube outer surface, kW/m2 Tsat (z), ◦ C hf g (z), kJ/kg Steam flow rate (z), kg/s Steam mole fraction (z) Psteam (z), kPa hc (z), kW/m2 K
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Evaluation of Condensation Heat Transfer in a Vertical Tube Heat Exchanger 31.10
Heat Exchangers
takes place within the first meter from the tube inlet. The local heattransfer coefficients and heat fluxes from about 1.0 m on are too low to be due primarily to condensation; the tubes in this example appear to be oversized. The number of tubes required is Number of tubes =
1 MW heat removal/tube
Each tube condenses the total steam flow coming in within 1 m from the tube inlet. Heat removal per tube is found from Eq. (31.6) to be 6.42 kW per tube. Thus, 162 tubes are required. References Incropera, F. P., and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 5th ed., Wiley, New York, 2002. Vierow, K., and V. E. Schrock, “Condensation in a Natural Circulation Loop with Noncondensable Gases, Part I—Heat Transfer,” Proceedings of International Conference on Multiphase Flow ’91 (Japanese Society of Multiphase Flow, ANS), Tsukuba, Japan, Sept. 1991, pp. 183–186.
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Source: HeatTransfer Calculations
Chapter
32 HeatExchanger Design Using an Evolutionary Algorithm
Keith A. Woodbury Mechanical Engineering Department University of Alabama Tuscaloosa, Alabama
Abstract Heatexchanger design is achieved by satisfying a required heat transfer while not exceeding an allowable pressure drop. Iterative procedures for hand or calculator operation have been used for many years, but the common availability of powerful desktop computers admits a wider class of design approaches. In this chapter, a design methodology based on optimization via an evolutionary algorithm is detailed. The evolutionary algorithm manipulates populations of solutions by mating the best members of the current population (natural selection) and randomly introducing perturbations into the newly created population (mutation). Heatexchanger design by optimization allows inclusion of additional constraints such as minimum volume or pressure drop.
Introduction Heatexchanger design
Hodge and Taylor [1] state that “heat exchanger design requirements can present multifaceted heat transfer and fluiddynamic requirements
32.1
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HeatExchanger Design Using an Evolutionary Algorithm 32.2
Heat Exchangers
on both the hot and cold fluid sides” and that these “Heat exchanger specifications are normally centered about the [heattransfer] rating and the allowable pressure drop.” Heatexchanger design requires satisfaction of the heattransfer (rating) specification while not exceeding the allowable pressure drop. Note that allowable pressure drop specifications for both hot and cold streams could be specified. Shah [2] outlines a procedure suitable for hand calculations to design heat exchangers. In Shah’s procedure applied to a shellandtube exchanger, a number of tubes of a chosen diameter are assumed implicitly through the mass velocity G, and the length of these tubes is chosen to satisfy the heattransfer requirement. If this geometry combination exceeds the allowable pressure drop, then a new number of tubes is chosen in accordance with the allowable pressure drop and the process is repeated until convergence is attained. Hodge and Taylor [1] provide a synopsis of Shah’s method. Design procedures such as Shah’s are simple and effective, but cannot be extended to include other design constraints, such as total volume. Procedures based on optimization methods are easily extensible and can obtain results quickly with a desktop computer. Genetic and evolutionary algorithms
Goldberg [3] offers the following definition: “Genetic algorithms are search algorithms based on the mechanics of natural selection and natural genetics.” Most genetic algorithm (GA) purists hold that a GA must contain all the elements of binary encoding, natural selection, crossover breeding, and mutation. However, binary encoding imposes a priori precision limitations and an extra layer of coding/decoding when the unknown parameters are continuous real numbers. A slightly different approach is to use strings of real numbers (arrays) to directly represent the parameters in an optimization problem. An evolutionary algorithm (EA) manipulates a population of solutions through the essential elements of natural selection and mutation, but does not utilize binary encoding to represent the realvalued parameters of the process. This is the approach used in the present application. Overview
In the remainder of this chapter the EA is first described in detail. Next the spiral plate heat exchanger (SPHE) is briefly described, along with the design constraints and properties for a particular application. Then the EA is applied to the design of a spiral plate heat exchanger. An ensemble design is performed, and some useful statistical information
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HeatExchanger Design Using an Evolutionary Algorithm HeatExchanger Design Using an Evolutionary Algorithm
32.3
about the design parameters is extracted from these results. Finally, the fitness function is modified to allow minimization of the exchanger volume and the SPHE is redesigned. Some conclusions close out the chapter. Evolutionary Optimization Algorithm The EA used here is the same as the realnumber encoding algorithm described by Woodbury [4]. However, a substantial change has been made in the selection of parents for the creation of a new generation [5]. A summary of the basic algorithm is presented below. Outline of the algorithm
The evolutionary algorithm incorporates the following basic steps: 1. A population of possible solutions is initialized randomly. 2. The fitness of each solution is determined according to an appropriate performance index (also known as the fitness function). 3. Pairs of the population are selected as parents of the new generation. This selection of parents is biased toward the mostfit members of the current population. 4. The selected parents mate to form children, which constitute the new population. 5. Variation is introduced into the new population through mutation, crossover, and creep. 6. Steps 2 to 5 are repeated until a specified number of generations are processed. These basic steps are amplified in the following sections. Initial population
The number of unknown parameters (nparm ) and the allowable range for each must first be specified, as well as a number of solutions to be contained in the population (npop ). The “population” is a collection of npop vectors, and each vector contains nparm values of the parameters. A fitness value that indicates the “goodness” of each member of the population is calculated from the performance index for this initial generation. Selection of parents
The process of selection is implemented to allow better members of the population to contribute the most to the new generation. This
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HeatExchanger Design Using an Evolutionary Algorithm 32.4
Heat Exchangers
helps ensure that the characteristics of the better solutions persist through future generations, and that poor solutions do not exert influence over subsequent solutions. During selection, parents are chosen from the current population to reproduce children for the next generation. Tournament selection is utilized to create a mating pool for each generation, where two or more members of the population are randomly selected to compete. The winner of the tournament is declared as the population member with the best fitness value, and this population member is copied into the mating pool for subsequent reproduction [6]. In the present algorithm, three members of the current population are randomly selected from the entire population, and the member with the best fitness is added to the mating pool. This process is repeated until the mating pool is the same size as the initial population (i.e., contains npop members). Note that it is possible, and actually very likely, that the mating pool will have repeat members. After the pool has been created, two parents will be randomly chosen from the pool for reproduction to create a new offspring for the next generation. Reproduction
Reproduction in a traditional GA is accomplished by crossover recombination of binary strings in the population. However, crossover of the realvalued arrays to produce children may not be a good approach if the elements of the vector represent drastically different quantities. An alternative method of mating the parents is through a weighted average [4]. A random weight w between 0 and 1 is chosen for each child. The child is then produced as follows: Offspring = w ∗ Parent1 + (1 − w) ∗ Parent2
(32.1)
where Parent is the vector array of parameters constituting a single individual. Enough children are created to fill the new population with as many members as the original population. Crossover
It is also possible to use randomly triggered crossover in realnumber encoding to introduce variation into a new generation [4]. This is done by generating a random number between zero and one for each member of the new population. If that number is below a userspecified crossover threshold ( pcross ), the population members are recombined analogous to crossover of binary strings. Thus, a crossover site is selected within the vector array, and another member of the new population is selected as a
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HeatExchanger Design Using an Evolutionary Algorithm HeatExchanger Design Using an Evolutionary Algorithm
32.5
crossover partner, and the two segments of the corresponding arrays are switched at the crossover site. Crossover is meant to introduce largescale variation in the each generation. Mutation
After a new generation of children is produced, mutations may occur. Mutations are important because they widen the search space of the genetic algorithm by introducing new values not contained in the current population. A random number between zero and one is generated for each population member, and if that number is below a userspecified mutation threshold ( pmutation ), a mutation will occur to that member. In the realnumber encoding of the present algorithm, mutations occur by changing the value of the population member to a random number that is within the allowable (specified) range of that parameter. Creep
The notion of creep [4] can be applied only to realnumber encoding. Unlike mutation, which generally introduces large changes into the population, creep introduces small changes into the population. The parameters that control the creep process are a specified maximum creep value (Creepmax ) and the probability that creep will occur ( pcreep ). As in mutation, a random number between zero and one is selected for each population member, and if the random number is below the specified threshold pcreep , the member will creep according to the following formula
New_member = (1 + c)∗ Original_member
(32.2)
where c is a random number generated in the range of [−Creepmax , Creepmax ]. Elitism
After creating a new generation and applying the techniques of crossover, mutation, and creep, one may find that the best member in the new population is not as good as the best member in the previous generation. The technique of elitism preserves the best member of the population and lets it persist into the new generation, ensuring that the best fitness for the new generation is never worse than that of the previous generation. Elitism enhances the performance of the evolutionary algorithm by guaranteeing that the solutions will not diverge.
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HeatExchanger Design Using an Evolutionary Algorithm 32.6
Heat Exchangers
The present EA utilizes elitism by preserving the best member of the population. At each generation the population is sorted, besttoworst by fitness value, and during reproduction the first member of the population is not replaced. Therefore, this elite population member will persist unchanged throughout generations until a better member is created. Termination
The EA is terminated after a specified number of generations have been created. However, an alternate stopping criterion is to continue until a specific fitness level is achieved. Design of a Spiral Plate Heat Exchanger Reference 8 depicts a spiral plate heat exchanger (SPHE), which is a counterflow heat exchanger with rectangular flow passages created by wrapping separation plates into a spiral. The assembly process is described by Martin [7, p. 74]: “Each spiral channel is sealed by welding on one edge, the other side being accessible for inspection and cleaning.” Martin reports that this type of exchanger has advantage in applications of energy recovery and is especially advantageous in handling solid suspensions. SPHEs are also claimed to be more resistant to fouling [8]. As mentioned previously, any heatexchanger design requires satisfaction of both a heattransfer requirement and a pressure drop requirement. Thus, suitable phenomenological models relating these quantities to the conditions of the flow are needed. Martin [7] provides the following heattransfer and pressure drop relations for a SPHE. Heat transfer
The heattransfer coefficient for flow in the rectangular channel in the SPHE can be found from the correlation Nu = 0.04 Re0.74 Pr0.4
(32.3)
which is valid for 400 < Re < 30, 000. Here the Reynolds and Nusselt numbers are based on the hydraulic diameter of the rectangular crosssectional area channel. The performance of a heat exchanger is typically characterized either by its effectiveness or using the log mean temperature difference (LMTD). Martin gives the correction factor for use with the LMTD for
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HeatExchanger Design Using an Evolutionary Algorithm HeatExchanger Design Using an Evolutionary Algorithm
SPHEs as Fc =
n NTU tanh NTU n
32.7
(32.4)
where n is the number of turns (wraps) in the exchanger and NTU is the number of transfer units: NTU =
UA Cmin
(32.5)
Here U is the overall heattransfer coefficient, A is the heatexchanger surface area, and Cmin is the lesser of mc ˙ p for the hot and cold fluids. Pressure drop
The pressure drop in the SPHE is computed as p =
f L V2 Dh 2
(32.6)
where L is the length of the flow passage, Dh is the hydraulic diameter, V is the velocity of the flow, and f is the friction factor, which is given by Martin as follows: f = 1.5
64 + 0.2 Re−0.1 Re
(32.7)
Properties and design data
Property and design data are summarized in Table 32.1. The target application requires heating of m ˙ c = 16 kg/h of a highheatcapacity fluid using m ˙ h = 50 kg/h of hot waste exhaust gas. The cold fluid enters 32.1 Properties and Data for SPHE Design
TABLE
Ti ,◦ C To ,◦ C m, ˙ kg/h c p , J/kgK , kg/m3 Pr , Ns/m2 k, W/m◦ C
Cold fluid
Hot fluid
20 225 16 6960 0.3774 0.96 1.3 × 10−5 0.0270
500 156.7 50 1330 0.4526 0.70 2.5 × 10−5 0.0346
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HeatExchanger Design Using an Evolutionary Algorithm 32.8
Heat Exchangers
at 20◦ C and is to be heated to 225◦ C, resulting in a heating load of 6342 W. The hot fluid enters at 500◦ C and, accounting for the heating load, must exit at 157.6◦ F in this counterflow exchanger. Design parameters
The spiral heat exchanger is characterized by the following geometric parameters: Width W
Width of rectangular channel
Hotspace thickness th
Height of hotflow passage
Coldspace thickness tc
Height of coldflow passage
Core diameter Dc
Diameter of inner core of spiral
Number of turns n
Number of wraps of spiral
Thickness t
Thickness of metal sheet separating passages
All other geometric parameters (such as flow path length and total heat exchanger volume) can be computed from this set of input parameters. Design fitness function
There are already two constraints that must be met for a successful design: the heat rating and tobespecified pressure drop. Additional constraints on the design might be desirable, for example, a specified volume for the exchanger. A general fitness function for the EA optimization is utilized in the form F=
N i=1
Wi
desiredi − actuali desiredi
2 (32.8)
where “desired” is the required value of the constraint and “actual” is the value resulting from the present set of design parameters. Here Wi is an optional relative weight attached to a particular constraint. For example, it might be very important that the heattransfer requirement be met, so a larger weighing could be applied to that term. For the design performed here, a specific form of Eq. (32.8) is employed: F = 100 +
Qdesired − Qact Qdesired
2
Pdesired − Pcold Pdesired
+
2
Pdesired − P hot Pdesired
+
2
Voldesired − Volact Voldesired
2 (32.9)
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HeatExchanger Design Using an Evolutionary Algorithm HeatExchanger Design Using an Evolutionary Algorithm
32.9
The larger weight is applied to the heattransfer term to ensure that this requirement is met. A single design
Because the EA relies heavily on random processes, different searches of the same parameter space will result in different, but similar, designs. A design was obtained with the Matlab routine EAez.m [10] with the following inputs and search ranges: Qdesired = 6341.5 W Pdesired = 25 Pa Voldesired = 35,000 cm3 1.0 cm ≤ W ≤ 10 cm 1 ≤ n ≤ 50 0.1 cm ≤ tc ≤ 5.0 cm 0.1 cm ≤ th ≤ 5.0 cm 1.0 cm ≤ Dc ≤ 10.0 cm 0.1 cm ≤ t ≤ 1.0 cm The convergence history from this design search is shown in Fig. 32.1 as the best fitness value from the population [computed from Eq. (32.9)] at each generation of the EA. After 500 generations, the best fitness value has reached almost 10−5 , indicating good satisfaction of the constraints in Eq. (32.9). The results from the EA for this single design are W = 4.16 cm, n = 12, tc = 0.857 cm, th = 1.791 cm, Dc = 5.911 cm, and t = 0.6947 cm. Ensemble design
Because the EA relies heavily on random procedures during the search, different designs are achieved for each execution of the search. This fact may seem unsettling, but it can be exploited to gain some insight about the design. Table 32.2 lists the results from a series of 10 designs obtained from the EA starting with the parameters from the previous section. The first column gives only the design (sequence) number, while the second column gives the best fitness value from the population [from Eq. (32.9)] for each design. The next four columns report the actual values of the constraints Q, pc , ph, and Vol corresponding to the best member of each population. The last six columns give the values of the design parameters obtained for each design.
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HeatExchanger Design Using an Evolutionary Algorithm 32.10
Heat Exchangers
102 101 100
Fitness
10−1 10−2 10−3 10−4 10−5 10−6
0
Figure 32.1
50
100
150
200
250 300 Generation
350
400
450
500
Typical convergence history for evolutionary algorithm SPHE design.
The last four lines of Table 32.2 are statistics computed from the 10 values in each column. The first is the mean or expected value of the 10 values; the second is the sample standard deviation. The third line is the 95 percent confidence interval for the mean value reported on the first of the last four lines. This confidence interval is computed TABLE 32.2 Design 1 2 3 4 5 6 7 8 9 10
Results from 10 EA Designs F 4.27×10−5 1.13×10−5 2.52×10−5 1.18×10−5 2.38×10−5 1.79×10−6 3.20×10−5 8.70×10−6 2.92×10−6 2.52×10−6
Q 6342.8 6340.5 6339 6340.8 6340.7 6342.1 6343.2 6341.8 6342.0 6341.9
Pc
Ph
Vol
W
24.9 25.0 25.0 25.1 25.1 25.0 25.0 25.0 25.0 25.0
25.0 25.1 25.0 25.0 25.1 25.0 25.0 24.9 25.0 25.0
34,841 35,032 35,074 34,956 34,899 34,990 35,158 35,003 34,961 35,030
3.85 4.15 4.05 3.84 3.84 3.57 3.74 3.83 3.83 3.81
34,994 89
3.85 12.9 0.944 1.999 0.16 0.57 0.047 0.112
Mean 1.63×10−5 6341.5 25.0 25.0 1.24 0.05 0.04 Standard 1.40×10−5 deviation 95% confidence ± 1.2 0.9 0.04 ± percent, % 0.01 0.1 0.1
0.03 0.2
63.6 2.9
n 13 12 12 13 13 14 13 13 13 13
tc
th
Dc
t
0.943 0.860 0.885 0.945 0.945 1.034 0.979 0.947 0.950 0.955
1.993 1.799 1.860 2.001 2.001 2.212 2.079 2.009 2.011 2.022
4.703 6.136 8.318 4.615 4.705 3.084 7.202 4.859 5.045 5.325
0.495 0.689 0.629 0.500 0.497 0.311 0.429 0.493 0.488 0.481
0.11 0.41 3.1 3.6
5.399 0.501 1.48 0.102
0.034 0.080 1.06 4.0 19.6 14.6
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HeatExchanger Design Using an Evolutionary Algorithm HeatExchanger Design Using an Evolutionary Algorithm
32.11
for small samples (< 30) with unknown standard deviation [9] using Student’s t distribution. Note that this confidence interval is the “±” uncertainty associated with the mean value reported, in the units of the parameter. Thus, the average heattransfer rating of the 10 designs is 6341.5 W ± 1.2 W. Finally, the last line of the table shows the percentage uncertainty in the corresponding column, which is simply the plus/minus variation amount divided by the mean. Looking at the last line of Table 32.2, it can be seen that the design constraints are met to a high degree, as the 10 different designs vary by 0.2 percent or less for all the specified constraints. One can also observe that some of the design parameters have less variation than do others. Specifically, W, tc , th, and n vary by less than 5 percent over the 10 designs, while Dc and t vary by 15 percent or more. These results demonstrate that the design of the SPHE is relatively insensitive to the values of Dc and t; that is, similar values of the fitness function F result for relatively wide variations in these two parameters. This means that the design is very forgiving with regard to these parameters, and that some convenient values of these two parameters could be selected and the design process repeated excluding them. “Asking for too much . . . ”
It’s possible for the designer to ask too much of the EA, and it is important to be able to recognize this when it happens. Figure 32.2 shows a convergence history from a single execution of the EAez.m [10] routine with all the same input parameters as before except that Voldesired = 3500 cm3 (an order of magnitude less than before). Note in Fig. 32.2 that the fitness function decreases exponentially (linearly on the semilog arithmic plot) until about generation 200; then the F value has “hit the wall” and cannot be reduced further. The values of the design constraints after 500 generations for this case are Q = 6114.6 W, pc = 31.59 Pa, ph = 31.12 Pa, and Vol = 5165.3 cm3 , which do not agree well with the corresponding constraint values. As can be seen, the algorithm is now trying to minimize the volume to satisfy the unrealistic constraint, but is unable to reduce the fitness F any further without increasing the penalty added to F by Q, pc , and ph. This is because the Vol term of the fitness function now dominates the value of F, and no combination of design parameters can lower F. Physically, this is because an exchanger with the desired volume cannot also satisfy the heattransfer and pressure drop requirements for the process. Inequality constraints
A reasonable request would be to design the heat exchanger to satisfy the heattransfer requirement and the pressure drop constraint, but to
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HeatExchanger Design Using an Evolutionary Algorithm 32.12
Heat Exchangers
10 2
Fitness
101
100
10−1
0
50
100
150
200
250
300
350
400
450
500
Generation Figure 32.2
Convergence history when “asking for too much.”
make the total volume as small as possible. This can be accomplished by modifying the fitness function [Eq. (32.9)] as F = 100 +
Qdesired − Qact Qdesired
2
Pdesired − Pcold Pdesired
+
2
Pdesired − P hot Pdesired
+
Volact Volnominal
2
2 (32.10)
where Volnominal is chosen so that the last term will be on the same order as the others when converged. This may require a tryitandsee approach. Figure 32.3 shows the convergence history for 500 generations of a design search using Eq. (32.10) with Volnominal = 1 × 105 cm3 . It is not obvious from the convergence history in Fig. 32.3 that the best solution has been obtained. Table 32.3 shows results from this search after 500, 1000, and 1500 generations. Note that only modest improvement in the fitness function F [Eq. (32.10)] is made after the original 500 generations,
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HeatExchanger Design Using an Evolutionary Algorithm HeatExchanger Design Using an Evolutionary Algorithm
32.13
101
Fitness
100
10−1
10−2 0 Figure 32.3
50
100
150
200
250 300 Generation
350
400
450
500
Convergence history for first 500 generations using Eq. (32.9).
but that further refinement does result in a smallervolume exchanger (9825 cm3 vs. 12,300 cm3 ). Execution times
The EA search for the best design can be very quick. However, in the present design, a nonlinear subsearch [involving the ln() function] must be performed to find the outlet temperatures for each flow stream for each member of the population at each generation. This subsearch requirement is due entirely to the LMTD approach used by Martin [7] to characterize heatexchanger performance. If an effectivenessNTU TABLE 32.3
Ngen
Results for Design Search Using Inequality Constraint [Eq. (32.10)] F
Q
Pc
Ph
Vol
W
n
tc
th
Dc
t
1.56 × 10−2
500 6332.9 25.1 24.6 12,297 6.6 13 0.498 1.006 3.675 0.112 1000 1.02 × 10−2 6339 25.0 25.0 10,095 7.56 13 0.430 0.855 2.465 0.101 1500 9.72 × 10−3 6344.5 24.9 24.9 9,824 7.77 13 0.419 0.832 2.247 0.101
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HeatExchanger Design Using an Evolutionary Algorithm 32.14
Heat Exchangers
correlation were available for the SPHE, this subsearch could be eliminated and the design time significantly shortened. As it stands, a single design using EAez.m [10] requires about 55 s of CPU time on a 2.0GHz Pentium 4 computer with 1 Gbyte of RAM, so that the ensemble design requires about 10 min to achieve.
Conclusions A spiral plate heat exchanger has been designed for a particular application. By performing an ensemble design, statistical information about the design parameters can be recovered. Inequality constraints on total exchanger volume can be incorporated into the fitness function to minimize the size of the designed exchanger. Designs can be obtained in about 1 min on a 2.0GHz Pentium 4 computer.
Nomenclature A
Area, m2
Creepmax
Maximum amount of creep for a parameter
Cmin
Smaller of values of mc ˙ p for the two streams of an exchanger
cp
Constantpressure specific heat
Dh
Hydraulic diameter
Dc
SPHE core diameter
F
Fitness value
Fc
Correction factor for LMTD method
f
Friction factor
G
Mass velocity G = m/A, ˙ kg/m2 s
k
Thermal conductivity
L
Length of flow passage
LMTD
log mean temperature difference
m ˙
Mass flow rate, kg/s
nparm
Number of parameters in solution
npop
Number of solutions in population
n
Number of turns in SPHE
N
Number of design constraints
NTU
Number of transfer units, UA/Cmin
pcross
Probability of crossover
pcreep
Probability of creep mutation
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HeatExchanger Design Using an Evolutionary Algorithm HeatExchanger Design Using an Evolutionary Algorithm
pmutation
Probability of mutation
p
Pressure drop
Pr
Prandtl number = /
Re
Reynold’s number VD/
t
Thickness of separating plate
tc
Thickness of cold space
th
Thickness of hot space
U
Overall heattransfer coefficient for exchanger
V
Velocity
w
Weight for reproduction
W
Width of SPHE flow passage
32.15
Greek
Thermal diffusivity k/ c p
Dynamic viscosity
Kinematic viscosity /
Density
Subscripts and superscripts c
Cold fluid
h
Hot fluid
References 1. Hodge, B. K., and Taylor, R. P., Analysis and Design of Energy Systems, 3d ed., Upper Saddle River, N.J.: PrenticeHall, 1999. 2. Shah, R. K., “Compact Heat Exchanger Design Procedures,” in Heat Exchangers: ThermalHydraulic Fundamentals and Design, Kakac, S., Bergles, A. E., and Mayinger, F. (eds.), New York: Hemisphere, 1981, pp. 495–536. 3. Goldberg, D. E., Genetic Algorithms in Search, Optimization, and Machine Learning, Boston: AddisonWesley, 1989. 4. Woodbury, K. A., “Application of Genetic Algorithms and Neural Networks to the Solution of Inverse Heat Conduction Problems: A Tutorial,” in Inverse Problems in Engineering, Proceedings of the 4th International Conference on Inverse Problems in Engineering, Angra dos Reis, Brazil, May 2002, pp. 73–88. 5. Woodbury, K. A., Graham, C., Baker, J., and Karr, C., “An Inverse Method Using a Genetic Algorithm to Determine Spatial Temperature Distribution from Infrared Tranmissivity Measurements in a Gas,” 2004 ASME Heat Transfer/Fluids Engineering Summer Conference, paper HTFED0456779, Charlotte, N.C., July 2004. ¨ 6. Back, T., Fogel, D. B., and Michalewicz, Z., Evolutionary Computation 1: Basic Algorithms and Operators, Philadelphia, Pa.: Institute of Physics Publishing, 2000. 7. Martin, H., Heat Exchangers, Washington, D.C.: Hemisphere, 1992.
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HeatExchanger Design Using an Evolutionary Algorithm 32.16
Heat Exchangers
8. Crutcher, M., “No More Fouling: The Spiral Heat Exchanger,” Process Heating, 1999 (http://www.processheating.com/CDA/ArticleInformation/coverstory/ BNPCoverStoryItem/0,3154,18383,00.html). 9. Bluman, A. G., Elementary Statistics, 5th ed., Boston: McGrawHill, 2004. 10. Woodbury, K. A., SPHE Matlab Functions, www.me.ua.edu/inverse/SPHEdesign, 2005.
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Source: HeatTransfer Calculations
Part
7 Fluidized Beds
Gasfluidized beds are addressed in many heattransfer books because they are a common reactor scheme for a wide range of applications where intimate gassolid contact is required. Gasfluidized beds are used for such diverse applications as coating, pharmaceutical production, coal combustion, and petroleum cracking. The extensive surface area of particles, coupled with the steady particle mixing induced by bubbles, ensures nearly isothermal operation and impressive convective heattransfer coefficients. This part is composed of a single chapter, by two U.S. academics, that presents two typical design problems, one in the bubbling regime and the other for a more dilute circulating fluidized bed. Both illustrate very common gassolid contacting methods with quite different hydrodynamics, so each is described briefly.
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Fluidized Beds
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Source: HeatTransfer Calculations
Chapter
33 FluidizedBed Heat Transfer
Charles J. Coronella and Scott A. Cooper Chemical and Metallurgical Engineering Dept. University of Nevada Reno, Nevada
Background Gasfluidized beds are a common reactor scheme for a wide range of applications where intimate gassolid contact is required, and are used for such diverse applications as coating, pharmaceutical production, coal combustion, and petroleum cracking. The extensive surface area of particles, coupled with the steady particle mixing induced by bubbles, ensures nearly isothermal operation and impressive convective heattransfer coefficients. In order to predict heattransfer rates in fluidized beds, it is necessary to have a basic understanding of fluidization hydrodynamics, so a brief summary is given here. In the design of an actual reactor, the engineer is advised to consult additional publications, starting with the excellent text by Kunii and Levenspiel (1991) or the text edited by Geldart (1986). Fluidized beds are composed of a large collection (a bed) of solid particles, as small as 40 m, up to a maximum of about 5 mm, with a gas rising through them. You can imagine that at a low gas throughput, the particles are still, in the mode of a packed bed, but as the gas flow is increased, eventually a point is reached such that the drag force on the particles is sufficient to overcome the gravitational force. This gas flow corresponds to the minimum fluidization velocity, and the particles are lifted and become mobile. The fluidized particles are not carried out of the reactor, but move around, resembling a boiling liquid. As the gas flow is increased further, the particles mix quite rapidly, with gas bubbles rising through the bed. If the gas flow is increased still further, 33.3
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FluidizedBed Heat Transfer 33.4
Fluidized Beds
particles are slowly ejected from the bed, at first slowly, until a gas flow is reached where all the particles in the bed are carried up (elutriated) and removed from the bed. In this mode of operation, the chamber has a low solids volume fraction, perhaps only 1 to 3 percent, and solids are continuously injected to make up a bed. Two typical design problems are given below, one in the bubbling regime and the other for a more dilute circulating fluidized bed (CFB). Both illustrate very common gassolid contacting methods with quite different hydrodynamics, so each is described briefly. In the bubbling regime, all important aspects are dominated by the flow of gas bubbles, including heat transfer. The phase map given by Geldart (1973) is useful in characterizing this phenomenon. Small and light particles, typically catalysts, are classified as group A, in the range of 40 m up to approximately 100 m. These particles fluidize smoothly at low gas flow rates, and bubbles appear only in beds fluidized at a gas flow rate somewhat higher than the minimum required for fluidization. Group B particles are those larger than 100 m and smaller than 800 m; bubbles are always present when these particles are fluidized. Fluidization of the largest particles, group D, is characterized by slowmoving bubbles, and much of the gas flow bypasses the bubbles, unlike in groups A and B, and flows through the interstices between particles. The smallest particles, group C, are not fluidized easily, due to strong interparticle forces. A bubbling fluidized bed is often characterized as consisting of two phases: a bubble phase, consisting primarily of gas, and a dense “emulsion” phase, consisting of both gas and solids. The dense phase is often said to have a void fraction εmf identical to that of a bed fluidized at the minimum flow necessary for fluidization Umf and is independent of gas flow. The volume fraction of the bed occupied by bubbles εb is a function of the size of bubbles and the speed at which they travel through the bed. It is quite common to control the temperature and remove heat from a fluidized bed by contact between the bed and either the reactor walls themselves or heattransfer tubes immersed within the bed horizontally or vertically. Note that tubes in a fluidized bed are subject to substantial abrasion, and if the environment is also corrosive, tube failure is likely without special care. At any moment, a tube may be surrounded by a rising gas bubble, therefore undergoing very slow heat transfer, since the gas is characterized by a relatively small thermal conductivity kg . When a tube is in contact with the dense phase, heat is transferred into an emulsion consisting of both gas and solid. The relatively high thermal conductivity and heat capacity of the solid substantially enhances heat transfer. A “packet” of particles may be in contact with a surface briefly, and heats (or cools) relatively quickly from the bed temperature to the tube surface temperature. The passage of the next bubble replaces that packet with Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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FluidizedBed Heat Transfer FluidizedBed Heat Transfer
33.5
another packet of particles, this one, like the previous packet, initially at the reactor temperature, reinvigorating the heattransfer rate. After a packet of solids is removed from a surface, it mixes into the dense emulsion phase, and rapidly changes temperature to that of the bed itself, since each particle is surrounded by many others at the reactor temperature. Thus, the overall rate of heat transfer between a fluidized bed and a surface is a function of the relative size of each phase, and also a function of the rate of replacement of packets. Starting at a relatively low gas flow, increasing gas flow causes greater bubble flow, which in turn causes an increase in εb and in the frequency of packet movement. As packets of solids are replaced more frequently, the rate of heat transfer should increase. Simultaneously, however, the fraction of heattransfer surface surrounded by gas increases with εb, with a slower rate of heat transfer relative to the rate to particle packets. Typically, a plot of heattransfer rate versus gas flow shows a rapid increase from the minimum bubbling gas flow and then a broad maximum, so that the rate of heat transfer is near the maximum over a fairly broad range of gas flow, and finally decreases with further increased gas flow. Circulating fluidized beds (CFBs) are characterized by very high gas flow rates, causing the particles within the bed to be elutriated out of the bed. The gassolid flow leaving the bed goes through a cyclone separator, so that the solids may be returned to the base of the bed, as illustrated in Fig. 33.1. There is significant variation in the distribution of solids between the bottom gas distributor and the top exit of the CFB riser. Typically, fresh and recycled solids are introduced into the riser near the bottom. Here, the solids concentration is the greatest, resembling a turbulently bubbling bed. The rapidly moving gas carries particles out of this dense region and up into the riser, lifting the solids up through the riser to the exit at the top. Werther and Hirschberg (1997) report that the region immediately above this is dilute, having an average crosssectional concentration of solids typically on the order of 1 to 2 percent. Along the length of the dilute zone, particles tend to loosely accumulate into clusters. These clusters tend to regularly settle out of the gas suspension, falling downward along the riser walls. After descending along the walls for some distance, the clusters break apart and are reentrained into the upwardflowing gas stream. The rapid mixing of solids and gas helps maintain a relatively even temperature profile through the CFB riser, even for highly exothermic reactions. The hydrodynamics within the upper region of a CFB is complex, and is well described by Werther and Hirschberg (1997). To summarize, the gas rapidly moves upward through the center of the riser, carrying with it a dilute concentration of solids. Along the wall there exists a higher concentration of solids streaming downward in clusters. This downward flow is downward relative to the upward gas flow, and may be upward Downloaded from Digital Engineering Library @ McGrawHill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGrawHill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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FluidizedBed Heat Transfer 33.6
Fluidized Beds
Gas
Exhaust Gas + Solids
Cyclone
CFB Riser
Solids
Solids Return
Charge
Gas Feed This schematic depicts the arrangement of a typical circulating fluidized bed. Solids are fed into the bottom of the CFB riser, where they are carried upward by the rapidly moving gas stream. On exiting the riser, the solids are separated from the gas by a cyclone and are eventually returned to the bottom of the riser.
Figure 33.1
or downward relative to the riser wall. A timeaveraged cross section of the riser would show a relatively solidladen annulus that surrounds a relatively solidfree core. Accordingly, this description is known as the coreannulus model. Heat is transported from the bed to the wall by three mechanisms following this coreannulus flow description: 1. Particles move from the hot core of the combustor to the downwardmoving annulus, exchanging heat with the nearby wall before returning to the hot core. Some correlations describe this particle convection as a singleparticle mechanism, while others model the movement of particle clusters.
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FluidizedBed Heat Transfer FluidizedBed Heat Transfer
33.7
2. Heat transfer occurs by gas convection along the exposed walls. 3. Radiation is an important mechanism of heat transfer when the CFB is operated at high temperatures, as are all combustors. Glicksman (1997) describes how the contributions of these three heattransfer mechanisms are summed according to the amount of contact that each mechanism has with the riser wall. Problem Statement I: Bubbling Fluidized Bed A fluidized bed is being designed to heat and volatilize bitumen from tar sands. The bed will operate in the bubbling regime, the fluidizing gas consists primarily of nitrogen at 450◦ C, and the particles are essentially round sand of 200 m diameter. Heat is supplied to the reactor through 3/4in. tubes immersed vertically in the fluidized bed. The gas flow is 200 standard cubic feet per minute (SCFM). The reactor is 1 m in diameter, is fluidized by gas coming through eight distinct nozzles, and is filled with sand fluidized to a height 0.8 m. There are 20 heattransfer tubes pushing up through the gas distributor at the base of the reactor to a height of 0.3 m. Calculate the convective heattransfer coefficient between the tubes and the bed. Solution
Sand in this size range fluidized by a gas is typical of Geldart group B fluidization, and therefore we can be sure that the bed is bubbling (provided the gas flow is below the particle terminal velocity). Thus, we should expect a fairly large heattransfer coefficient. First, we should calculate some preliminary variables that will be needed in the calculation of the convective coefficient. We’ll need the minimum fluidization velocity, which is calculated from the (dimensionless) Ergun equation, set equal to the weight of solids Ar = 150
1 − εmf 1.75 · Remf + 3 · Re2mf 2 εmf εmf
(33.1)
where Ar is the Archimedes number and Remf is the particle Reynolds number at minimum fluidization conditions: Ar =
g ( s − g ) · g · d3p 2
Remf =
dp · Umf · g
The viscosity of N2 at 450◦ C is = 33.5 × 10−6 Pa · s, and the gas density is calculated from the idealgas law g = 0.472 kg/m3 . Sand density is s = 2600 kg/m3 , the particle diameter is dp = 200 m, and
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FluidizedBed Heat Transfer 33.8
Fluidized Beds
so the Archimedes number is calculated as Ar = 85.8. In practice, the void fraction at minimum fluidization εmf should always be measured, but here it is estimated as εmf = 0.45, a typical value for relatively round, small particles. Thus Remf is calculated from Eq. (33.1), Remf = 0.0946, from which the minimum fluidization velocity can be found, Umf = 0.0336 m/s. The actual superficial gas velocity U is also required: ft3 200 F 450 + 273 m U= = min · = 0.297 A 293 s · (1 m)2 4 The second term accounts for the raised temperature, and unit conversions aren’t shown. The gas flow is nearly 9 times that required for fluidization, which means that the bed is indeed fluidized, and bubbling will be vigorous. An empirical correlation presented by Zabrodsky (1966) is quite good at predicting hmax , the maximum heattransfer coefficient to be expected at some gas velocity: hmax = 35.8 s0.2 · kg0.6 · d−0.36 p
(33.2)
All variables are given in SI (International System; metric) units in this correlation. For this particular application, it predicts hmax = 627 W/(m2 · K). It is tempting to stop here, since we know that the overall heat transfer is near the maximum rate over a broad range of U. However, this gas flow is fairly small, so we might suspect that the actual h is somewhat below the maximum value. There is no uniformly accepted correlation for predicting h in bubbling beds. We will use the correlation given by Molerus et al. (1995). However, the interested reader should consult one or more of the excellent published reviews (Kunii and Levenspiel, 1991, chapter 13; Botterill, 1986; Xavier and Davidson, 1985). The correlation presented by Molerus et al. (1995) is given below. Despite its complex appearance, it is not difficult to use, and is reasonably accurate for bubbling fluidization systems over a wide range of operating conditions: Nu ≡
h· = Nupc + Nugc kg
(33.3)
The dimensionless convective coefficient is given as the sum of a component from particle convection and a component from gas convection. The former accounts for heat transfer to moving packets, and the latter accounts for heat transfer directly to the gas phase, which becomes
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FluidizedBed Heat Transfer FluidizedBed Heat Transfer
33.9
more important at high gas velocities and in beds composed of larger particles: √ 3 0.165 Pr · R · V Nugc = (33.4) V + 0.05 −1 33.3 0.125(1 − εmf ) · 1 + √ 3 V · Ue (33.5) Nupc = √ kg 0.28 1+ · 1+ · (1 − εmf )2 · R ·Ue 2 2 · Cs · V Intermediate parameters are defined as follows, g U − Umf s · Cs V≡ R≡ Ue ≡ 3 · (U − Umf ) Umf s − g kg · g
(33.6)
and the length constant is defined as
≡ √ g · ( s − g )
2/3 (33.7)
Both V and Ue are two ways to represent dimensionless excess gas velocity and are evaluated as follows: m m − 0.034 0.297 U − Umf s s = 7.74 V= = m Umf 0.034 s s · Cs Ue = 3 · (U − Umf ) kg · g 2600 kg · 740 J m
m kg · K m3 3 = − 0.034 = 41.0 · 0.297 W m
s s 0.0518 · 9.81 2 m·K s The term R is simply a density ratio, and is evaluated easily: g R= = s − g
0.472
kg m3
kg kg 2600 3 − 0.472 3 m m
= 1.816 × 10−4
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FluidizedBed Heat Transfer 33.10
Fluidized Beds
We also need the gasphase Prandtl number: kg J · 3.35 × 10−5 1104 Cg · kg · K m·s Pr ≡ = = 0.714 W kg 0.0518 m·K Now we can evaluate the contribution of heat transfer directly to the gas phase from Eq. (33.4): √ 3 0.165 Pr · R · V Nugc = V + 0.05 0.165 3 (0.714) · (1.81610−4 ) · (7.74) = = 8.298 × 10−3 7.74 + 0.05 We can also evaluate the contribution of particle packets from Eq. (33.5):
Nupc
Nupc
−1 33.3 0.125(1 − εmf ) · 1 + √ 3 ( 7.74)(41.0) = √ kg 0.28 · 1+ · (1 − εmf )2 · R · Ue2 1+ 2 · Cs · V −1 33.3 0.125(1 − 0.45) · 1 + √ 3 ( 7.74)(41.0) = W 0.0518 √ 0.28 m·K · 1+ 1+ · (1 − 0.45)2 · 1.816 × 10−4 · (41.0)2 J kg 7.74 · 3.35 × 10−5 2 · 740 kg · K m·s
Nupc = 0.02116
The Nusselt number is evaluated from Eq. (33.3): Nu = Nugc + Nupc = 8.298 × 10−3 + 0.02116 = 0.0295 To find the convective coefficient, we need to evaluate from Eq. (33.7): = √
g · ( s − g )
2/3
=
3.35 × 10
−5
kg m ·s
2/3
m kg 9.81 2 · (2600 − 0.472) 3 s m
= 2.57 m
Finally, we obtain W (0.0295) · 0.0518 Nu · kg W m·K h= = = 595 2 2.57 m m ·K
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FluidizedBed Heat Transfer FluidizedBed Heat Transfer
33.11
This coefficient, then, is the one to use to predict the rate of heat transfer between the fluidized bed and the tubes in the bed. The prediction indicates that the rate is near the maximum rate predicted by Eq. (33.2). In fact, the maximum coefficient predicted by this correlation is 635 W/(m2 K) at a gas flow rate 85 percent greater than that specified in this design, remarkably similar to the maximum coefficient predicted by Eq. (33.2). Economics will determine the preferred gas flow rate. Note that, according to this correlation, the predicted convective coefficient is independent of the size, location, and number of tubes. Of course, the total rate of heat transfer to the bed is proportional to the tube area, using q = hAt T. The effect of tubes on bubble flow in a fluidized bed is well documented, causing bubbles to be smaller, rise more slowly, and therefore to reduce the flow of particle “packets.” Like all correlations for convective coefficients, this prediction should be used cautiously. In this calculation, we’ve neglected any radiation effects. For these conditions, that is reasonable, but for higher temperatures (>800◦ C) radiation might make a substantial contribution to overall heat transfer. Problem Statement II: Circulating Fluidized Bed Circulating fluidized beds are often used as coal combustion furnaces. One such unit consists of a 24mtall riser with a 4.0 × 4.0 m2 cross section. The bottom 10 m of the riser is covered in a refractory material. The walls of the upper 14 m of the riser are made up of membrane tubes through which steam is heated. On average, the fluidized particles in the riser consist of round coal particles with a diameter of 200 m. From pressure drop measurements, it is estimated that solids represent approximately cb = 1.8 percent of the volume in the dilute zone of the riser where the membrane tubes are located. The fluidizing medium is air with a superficial velocity of 6.5 m/s. If the bed temperature is uniform at 850◦ C, what is the heattransfer coefficient from the bed to the membrane wall? The coal properties include the density specified as s = 1350 kg/m3 and the heat capacity of Cs = 1260 J/(kg·K). At the given temperature, the air has a density of g = 0.31 kg/m3 , a viscosity of = 4.545 × 10−5 kg/(m·s), and a heat capacity of Cg = 1163 J/(kg·K). Solution
We will give two solution techniques, noting that predicting heat transfer in CFB risers is still not well developed. The first method is a very simple empirical method: h = a bn
(33.8)
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FluidizedBed Heat Transfer 33.12
Fluidized Beds
On the basis of data taken from commercial boilers, Basu and Nag (1996) suggest a = 40 and n = 0.5 when b is given in kg/m3 and h is predicted in W/(m2 K). These parameter values are restricted to dp ≈ 200 m, 750◦ C < Tb < 850◦ C, and 5 kg/m3 < b < 25 kg/m3 . First, we calculate the average bed density, to ensure that the correlation applies: b = cb s + (1 − cb) g = 24.6 kg/m3 Thus, the correlation is valid, and we predict an average heattransfer coefficient: h = a bn = 40 (24.6)0.5 = 198
W m2 K
Another far more detailed, mechanistic method is also given here. Most such CFB heattransfer correlations are derived from experimental studies of relatively small, laboratoryscale units. Care must be taken when applying these correlations to large CFB combustors as differences in hydrodynamics between small and large units are considerable and not yet well predicted. The heattransfer coefficient from the bed suspension to the wall is calculated by summing the contributions from particle convective heat transfer, gas convective heat transfer, and radiation. The heattransfer contributions from the particle and gas convection are dependent on the fraction of the wall covered by the cluster phase f h = f hc + (1 − f )hg + hr
(33.9)
where hc is the contribution of particle clusters, hg is the effect of the gas phase, and hr is the radiation contribution, assumed to apply to the entire surface. Particle cluster convective term. Lints and Glicksman (1994) present a
model for calculation of the particle convective heattransfer coefficient. According to this model, particles coalesce into clusters as they exit the upwardmoving core into the dense annular region within the riser. As the cluster approaches the wall and travels downward, heat is transferred from the cluster to the wall. After descending along the wall for some distance, the cluster is reentrained into the hot riser core: · dp 1 · tc = + (33.10) hc kg kc · Cs · s · cc where is a dimensionless cluster thickness, predicted empirically: = 0.0287(cb)−0.581 = 0.0287(0.018)−0.581 = 0.296
(33.11)
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FluidizedBed Heat Transfer FluidizedBed Heat Transfer
33.13
The fraction of solids in the cluster is also predicted empirically from the average solids concentration: cc = 1.23(cb)0.54 = 1.23(0.018)0.54 = 0.141
(33.12)
The thermal conductivity of a cluster is predicted as a linear average of its components: kc = cc ks + (1 − cc )kg (33.13) W W W = (0.141) 0.26 + (1 − 0.141) 0.073 = 0.099 mK mK mK The final parameter required for evaluating hc from Eq. (33.10) is tc , the time of contact between the cluster and the wall. It is evaluated from the ratio of the distance traveled to the speed at which a cluster travels tc =
Lc Uc
(33.14)
where Lc is the distance that a cluster travels along a riser wall before being entrained back into the upward bulk suspension, and is correlated empirically: Lc = 0.0178( b)0.596 = 0.0178(24.6)0.596 = 0.120 m
(33.15)
The cluster speed is assumed to always equal 1 m/s. Thus tc =
0.120 m = 0.120 s 1 m/s
and hc can now be evaluated from Eq. (33.10): · dp 1 · tc (0.296) · (0.0002 m) = + = W hc kg kc · Cs · s · cc 0.073 m·K · (0.120 s) + W J kg 0.099 · 1260 · 1350 3 · (0.141) m·K kg · K m hc = 209
W m2 K
The resistance to conduction through the cluster (dp/kg ) is somewhat less than the resistance in the transient term, and has only a slight effect on the cluster convective coefficient.
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FluidizedBed Heat Transfer 33.14
Fluidized Beds
Gas convection term. One simple way to determine the gas convective
heattransfer coefficient is to use the correlation presented by Martin (1984), applicable when gas velocities are greater than the minimum fluidization velocity: Nu = 0.009 Pr1/3 Ar1/2
(33.16)
The Prandtl number is evaluated from its definition: J −5 kg · 1163 4.545 × 10 · Cg m·s kg · K = 0.724 Pr = = W kg 0.073 m·K (33.17) Additionally, the Archimedes number must be calculated: g · ( s − g ) · g · (dp)3 2 kg m
kg kg 3 0.3105 · 1350 · 9.81 · (0.0002 m) − 0.3105 m3 m3 m3 s2 = = 15.92 2 kg −5 4.545 × 10 m·s
Ar =
We calculate the Nusselt number from Eq. (33.16): Nu = 0.009 Pr1/3 Ar1/2 = (0.009)(0.724)1/3 (15.92)1/2 = 0.032 The gas convective heattransfer coefficient is calculated from the Nusselt number: W (0.032) · 0.073 Nu · kg m·K = 11.68 W hg = = dp 0.0002 m m2 · K
Radiation term. To calculate the radiation heattransfer coefficient, one
must first determine the effective emissivity of the bed. This is related to the bed volume where heat transfer occurs and to the surface area of the suspended particles, the mean beam length, and the mean extinction coefficient as described by Glicksman (1997). Looking at the region of the CFB riser whose walls are composed of membrane tubes, the volume of this region of the bed is Vbed = (24 m − 10 m) × (4 m) (4 m) = 224 m3
(33.18)
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FluidizedBed Heat Transfer FluidizedBed Heat Transfer
33.15
Recalling the volume fraction of solids in the riser cb = 0.018, the volume of particles in this region of the riser can be found: VTS = Vbed × cb = (224 m3 ) × (0.018) = 4.032 m3
(33.19)
If the particles are assumed to be smooth spheres, then the volume of a single particle is Vp =
3 d = 4.19 × 10−12 m3 6 p
(33.20)
and the surface area of a single particle is Sp = d2p = 1.26 × 10−7 m2
(33.21)
The total number of particles in this region of the riser is found from the total volume occupied by the particles and the volume of a single particle: Np =
VTS 4.032 m3 = = 9.63 × 1011 particles Vp 4.189 × 10−12 m3
(33.22)
The surface area of all of the particles can then be calculated: Sbed = Np · Sp = (9.63 × 1011 ) · (1.257 × 10−7 m2 ) = 1.21 × 105 m2 (33.23) From this information, an estimation of the mean beam length Lm, can be made. 224 m3 Vbed Lm = 3.5 · = (3.5) · = 0.00648 m (33.24) Sbed 1.21 × 105 m2 If the particle emissivity is estimated to be εrp = 0.5, as suggested by Glicksman (1997), then the mean extinction coefficient K can be found: 3 3 cb 0.018 K= = · εrp · · (0.5) · = 67.5 m−1 2 dp 2 0.0002 m (33.25) If the bed is sufficiently dilute and the effective bed emissivity εrb is less than approximately 0.5 to 0.8, then the following correlation, suggested by Glicksman (1997), can be used: εrb = 1 − exp(−K · Lm) = 1 − exp[(−67.5 m−1 ) · (0.00648 m)] = 0.354 (33.26) If the wall emissivity εrw is estimated to be 0.3, and the wall temperature is estimated at 600◦ C, then the radiation heattransfer coefficient can
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FluidizedBed Heat Transfer 33.16
Fluidized Beds
be calculated as reported by Basu and Nag (1996): 4 Tb − Tw4 · hr = 1 1 + − 1 · (Tb − Tw ) εrb εw W 4 4 −8 [(1123 K) − (873 K) ] · 5.67 × 10 m 2 · K4 = 1 1 + − 1 · (1123 K − 873 K) 0.354 0.3 = 44.4
W m2 K
(33.27)
Total bedtowall heattransfer coefficient. At this point, the heattransfer
coefficient for each of the three mechanisms has been calculated. All that is remaining is to determine the fractional wall coverage of particle clusters f , as required in Eq. (33.9). Lints and Glicksman (1994) present the following correlation: f = 3.5 · (cb)0.37 = 3.5 · (0.018)0.37 = 0.79
(33.28)
Along this fraction of surface area f , particle convective heat transfer is occurring. Along the remainder of the surface area (1 − f ), gas convective heat transfer occurs. Radiation acts over the entire surface. Recalling the results calculated previously, we have hc = 209
W m2 · K
hg = 11.68
W m2 · K
hr = 44.4
W m2 · K
The total bedtowall heattransfer coefficient is found from Eq. (33.9): h = f · hc + (1 − f ) · hg + hr = 0.79(209) + (1 − 0.79) · (11.68) + 44.4 = 212
W ·K
m2
The simple empirical solution for this case is similar to the more rigorous solution, differing by less than 7 percent. Such close agreement is unusual. The total rate of heat transfer from the CFB to the membrane walls is found from q = hAt T, with At consisting of the area of the membrane wall = (4 × 4 m) × (14 m) = 224 m2 . Nomenclature a
Empirical constant in Eq. (33.8)
A
Fluidizedbed crosssectional area
At
Area of all heattransfer surfaces
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FluidizedBed Heat Transfer FluidizedBed Heat Transfer
Ar
Dimensionless Archimedes number
cb
Volume fraction of solids in the riser
cc
Volume fraction of solids in clusters
Cg
Heat capacity of gas
Cs
Heat capacity of solids
dp
Harmonic mean particle diameter
f
Fraction of riser wall covered by particle clusters
33.17
F
Gas volumetric flow rate
g
Acceleration due to gravity
h
Mean convective coefficient between a fluidized bed and a surface
hc
Convective heattransfer coefficient through particle clusters
hg
Convective heattransfer coefficient through fluidizing gas
hr
Radiation heattransfer coefficient
kc
Thermal conductivity of a particle cluster
kg
Thermal conductivity of fluidizing gas
ks
Thermal conductivity of solid particles
K
Mean extinction coefficient
Lc
Distance that a cluster travels along the wall
Lm
Mean beam length
n
Empirical constant in Eq. (33.8)
Np
Number of particles
Nu
Dimensionless Nusselt number
Pr
Dimensionless Prandtl number of the gas
q
Total heattransfer rate between surface and a fluidized bed
R
Dimensionless density defined in Eq. (33.6)
Remf
Particle Reynolds number at minimum fluidization gas velocity
S bed
Surface area of all particles in riser section
Sp
Surface area of a single particle
tc
Cluster contact time with wall
Tb
Bed temperature
Tw
Wall temperature
U
Superficial gas velocity
Uc
Speed at which a cluster travels along a wall
Ue
Dimensionless excess gas velocity defined in Eq. (33.6)
Umf
Minimum superficial gas velocity necessary for fluidization
V
Dimensionless excess gas velocity defined in Eq. (33.6)
Vbed
Volume of circulating fluidizedbed riser subject to heat transfer
Vp
Volume of a single particle
VTS
Volume of total number of solid particles in riser
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FluidizedBed Heat Transfer 33.18
Fluidized Beds
Greek
Dimensionless gaslayer thickness at riser wall
T
Temperature difference between fluidized bed and heattransfer surface
εmf
Void fraction of a bed at minimum fluidization conditions
εb
Volume fraction of a fluidized bed occupied by gas bubbles
εpb
Radiation emissivity of single particle
εrb
Radiation emissivity of particle bed
εwb
Radiation emissivity of riser wall
Characteristic length in a bubbling fluidized bed
Gas viscosity
b
Bulk suspension density in a fluidized bed
g
Gas density
s
Particle density
StefanBoltzmann constant
References P. Basu and P. K. Nag, “Heat Transfer to Walls of a Circulating FluidizedBed Furnace,” Chem. Eng. Sci., 51(1), 1–26 (1996). J. S. M. Botterill, “Fluid Bed Heat Transfer,” in Gas Fluidization Technology, D. Geldart (ed.), Wiley, New York, 1986. D. Geldart (ed.), Gas Fluidization Technology, Wiley, New York, 1986. D. Geldart, “Types of Gas Fluidisation,” Powder Technol., 7, 285–292 (1973). L. R. Glicksman, “Heat Transfer in Circulating Fluidized Beds,” in Circulating Fluidized Beds, J. R. Grace, A. A. Avidan, and T. M. Knowlton (eds.), Blackie Academic and Professional, New York, 1997, pp. 261–311. D. Kunii and O. Levenspiel, Fluidization Engineering, 2d ed., ButterworthHeinemann, Boston, 1991. M. C. Lints, and L. R. Glicksman, “Parameters Governing ParticletoWall Heat Transfer in a Circulating Fluidized Bed,” in Circulating Fluidized Bed Technology IV, A. Avidan, (ed.), American Institute of Chemical Engineers (AIChE), New York, 1994, pp. 297– 304. H. Martin, “Heat Transfer between Gas Fluidized Beds of Solid Particles and the Surfaces of Immersed Heat Exchanger Elements, Part I,” Chem. Eng. Process., 18, 157–169 (1984). O. Molerus, A. Burschka, and S. Dietz, “Particle Migration at Solid Surfaces and Heat Transfer in Bubbling Fluidized Beds—II. Prediction of Heat Transfer in Bubbling Fluidized Beds,” Chem. Eng. Sci., 50, 879–885 (1995). J. Werther and B. Hirschberg, “Solids Motion and Mixing,” in Circulating Fluidized Beds, J. R. Grace, A. A. Avidan, and T. M. Knowlton (eds.), Blackie Academic and Professional, New York, 1997, pp. 119–148. A. M. Xavier and J. F. Davidson, “Heat Transfer in Fluidized Beds: Convective Heat Transfer in Fluidized Beds,” in Fluidization, 2d ed., J. F. Davidson, R. Clift, and D. Harrison (eds.), Academic Press, New York, 1985. S. S. Zabrodsky, Hydrodynamics and Heat Transfer in Fluidized Beds, MIT Press, Cambridge, Mass., 1966.
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Source: HeatTransfer Calculations
Part
8 Parameter and Boundary Estimation
The three chapters in this part, two by U.S. academics and one by an engineer employed in industry in Japan, present calculations that vary from being conjectural to representing an industrial situation. The first chapter, which covers parameter estimation, is arguably the most theoretical in the entire handbook. The second chapter deals with the thermal design of boxes whose dimensions do not exceed 1 m and whose function is to house heatgenerating components that must be prevented from overheating. Examples include desktop computers, laptop computers, and cell phones. Relevant calculations involve the upper bounds of heat transfer from sealed boxes, from boxes having vent ports for a natural draft, and from boxes cooled by fans. The third chapter in this part presents calculations that can be used to predict the freezing time of strawberries being frozen in a fluidizedbed freezer.
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Parameter and Boundary Estimation
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Source: HeatTransfer Calculations
Chapter
34 Estimation of Parameters in Models
Ashley F. Emery Mechanical Engineering Department University of Washington Seattle, Washington
Consider the onedimensional heat transfer through a plane layer whose temperature satisfies c
∂T ∂2T =k 2 ∂t ∂x
(34.1)
with boundary conditions x = 0: x = L:
h(T0 − T (0, t)) = −k −k
∂T ∂x
∂T = h(T(L, t) − TL) ∂x
(34.2a) (34.2b)
and initial conditions t = 0:
T(x, 0) = 0
(34.2c)
We wish to estimate the thermal diffusivity = k/ c. We will do this by measuring the temperature at a selected location x = x ∗ as a function of time. The idea is to choose a value of the conductivity for which the predictions of the model [Eq. (34.1)] match the measurements as closely as possible. 34.3
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Estimation of Parameters in Models 34.4
Parameter and Boundary Estimation
We will do this by minimizing the leastsquares error E: E=
M N j
(Te (x j , ti ) − Tp(x j , ti , ))2 =
i
M N j
ei,2 j ()
(34.3)
i
where Te and Tp represent the experimental and predicted temperatures, respectively; M is the number of measurement locations; N is the number of measurements at each location; and ei, j () are called the residuals, which are functions of . We could simply choose different values of the diffusivity until we find a value that minimizes E by plotting E() as a function of and choosing that value of for which E() is a minimum. Note that in Eq. (34.3) all the residuals are given equal weight. We could associate a different weight with each residual, but for simplicity in this problem we will treat each measurement as being of equal importance. Readers interested in a more detailed treatment should consult one of the references listed. Estimation and Sensitivity Unfortunately, although plotting E() works, it doesn’t tell us much about where and when to make the measurements. A better way is to investigate the sensitivity of our estimate of to the measurements. We do this as follows. Since Tp(x, t, ) is a nonlinear function of , we expand T(x, t, ) about some initial estimate 0 , giving a residual of ei, j () = Te (xi , t j ) − Tp(xi , t j , 0 ) −
∂ T ( − 0 ) ∂ 0
(34.4a)
Substituting Eq. (34.4a) into Eq. (34.3) and differentiating with respect to and setting the derivative equal to 0 gives N M i j (Te (xi , t j ) − Tp(xi , t j , 0 ))Ai, j (0 ) − 0 = (34.4b) N M 2 i j Ai, j () where Ai, j = ∂ T()/∂ evaluated at xi , t j , 0 and is called the sensitivity of the model. Now this will give us an improved value of , but since Tp is a function of , we will have to iterate. Our procedure is then N M i j (Te (xi , t j ) − Tp(xi , t j , p))Ai, j (p) p+1 − p = (34.5a) N M 2 i j Ai, j (p) where Ai, j (p) =
∂ T () ∂
(34.5b)
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Estimation of Parameters in Models Estimation of Parameters in Models
34.5
evaluated at xi , t j , p, and we iterate until we have achieved convergence. From Eq. (34.5a) it is apparent that the rate of convergence will depend strongly on the denominator that is a function of the sensitivities. The higher the sensitivity (i.e., the larger the absolute value of A), the faster our method will converge. In fact, it is known that (), ˆ the standard deviation of our final estimate, is given by (Te ) () ˆ = N M 2 ˆ i j Ai, j ()
(34.6)
where (Te ) is the noise associated with our experimental measurements. If there is no noise, then ˆ will be the exact value. Thus the first thing that we want to do before making the measurements is to compute the sensitivities Ai, j and to choose measurement locations and times that give us large absolute values. Another way to look at it is to avoid measurements for which the sensitivity is low. The Matlab program Slab_Sensitivity computes and plots the sensitivities for a slab subjected to convective boundary conditions. The call is
[T,S] = Slab_Sensitivity(Fo,kappa,bc,Nt) where Fo is the time duration in terms of the Fourier number, kappa is the true thermal diffusivity, bc defines the boundary conditions, and Nt is the number of time steps; bc is in the form of a vector with components in terms of the Nusselt number and nondimensional temperature, bc =[h0 L/k;T0 ;hL L/k;TL ]. Figure 34.1 depicts the temperatures and the sensitivities for Fo = 1, kappa = 1, and bc = [10;0;100;1] at x/L = 0, 0.25, 0.50, 0.75, and 1.00. In Fig. 34.1a and b, note how the sensitivities change with time and position. The sensitivity at x/L = 1 is essentially zero because the high value of the convective heattransfer coefficient hL causes the temperature to reach a constant value very quickly and to be insensitive to the diffusivity. On the other hand, the temperature at x/L = 0, where h0 is low, continuously changes and the sensitivity is significantly larger than that at x/L = 1, and eventually becomes the greatest of all positions. For moderate times, the best location to measure the temperature is near the center, someplace between 0.25 and 0.75. We define information about our estimate of to be Information about ˆ =
2 () ˆ 2 (noise)
(34.7)
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Estimation of Parameters in Models 34.6
Parameter and Boundary Estimation
h0, T0 = 10, 0
1
hL, TL = 100, 1
0.9
x/L = 0.00 0.25 0.50 0.75 1.00
Nondimensional Temperature
0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (Fo) (a)
h0, T0 = 10, 0
0.3
hL, TL = 100, 1 x/L = 0.00 0.25 0.50 0.75 1.00
Nondimensional Sensitivity
0.25
0.2
0.15
0.1
0.05
0 −0.05
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (Fo) (b) Figure 34.1
(a) Temperature histories; (b) sensitivity histories.
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Estimation of Parameters in Models Estimation of Parameters in Models
34.7
Figure 34.2a illustrates the instantaneous and cumulative information versus time. While it might appear from Fig. 34.1b that a measurement at x/L = 0.5 is the most sensitive, the contributions of the different measurements to the information are shown in Fig. 34.2b and it is easy to see how each contributes differently to the total information. At early times, the measurement made near the boundary where the heat is added is the most informative. As time goes on, the measurements made toward the boundary at x/L = 0 become more important and finally at long times, the sensor at x/L = 0 is the most valuable, but contributes little to overall accuracy of the estimation. Estimation We simulate an actual experiment by adding noise to the temperature computed using the exact value of . The diffusivity is estimated by comparing the measured (simulated) temperatures to that predicted from the model using the program Estimate_kappa. The call is
[kappahat,sigma_khat] = Estimate_kappa(Fo,kappa,bc,Nt,Ixm, noise) where Ixm is the node number of the measured temperature. The noise variable can be entered in one of two ways: (1) noise > 0, where the noise is in terms of the nondimensional value of temperature (T/TL); or (2) noise < 0, where the noise is a percentage of the nondimensional temperature. Using the same values for bc as in the sensitivity calculation, measuring the temperature at the midpoint Ixm = 11, and using a noise of 0.02 gives the simulated measured temperature shown in Fig. 34.3a. The estimated thermal diffusivity is 1.0215, and its standard deviation is 0.0410. Figure 34.3b shows the convergence as a function of the iterations. Noise The noise is generated using a randomnumber generator. Each time the program is run, a different set of random numbers is generated, simulating different experiments. If you want the measurement noise to be constant, you should modify the program as shown in the code. Programs Pseudocode and Matlab programs for these calculations are available at www.me.washington.edu/people/faculty/emery/Handbook Programs.
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Estimation of Parameters in Models 34.8
Parameter and Boundary Estimation
h0, T0 = 10, 0
2
hL, TL = 100, 1
1.8 1.6
Information
1.4
instantaneous cumulative
1.2 1 0.8 0.6 0.4 0.2 0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (Fo) (a)
h0, T0 = 10, 0
0.9
hL, TL = 100, 1 x/L = 0.00 0.25 0.50 0.75 1.00
0.8
Relative Contribution
0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (Fo) (b) Figure 34.2
(a) History of the information; (b) contributions to the infor
mation.
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Estimation of Parameters in Models Estimation of Parameters in Models
h0, T0 = 10, 0
0.6
hL, TL = 100, 1
x* = 11
34.9
Noise = 0.02
0.5 exact measured
Temperature
0.4
0.3
0.2
0.1
0
−0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (Fo) (a)
0.12
h0, T0 = 10, 0
hL, TL = 100, 1
x* = 11
Noise = 0.02
0.1
Residuals
0.08
0.06
0.04
0.02
0 0.5
1
1.5
2
Estimate of κ (b) Figure 34.3
of .
(a) Temperature histories; (b) convergence of estimated value
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Estimation of Parameters in Models 34.10
Parameter and Boundary Estimation
References Beck, J. V., and Arnold, K. J., 1977, Parameter Estimation in Engineering and Science, Wiley, New York. Norton, J. P., 1986, An Introduction to Identification, Academic Press, New York. Ross, Gavin J. S., 1990, Nonlinear Estimation, SpringerVerlag, New York. Sorenson, H. W., 1980, Parameter Estimation: Principles and Problems, Marcel Dekker, New York. Strang, G., 1993, Introduction to Linear Algebra, WellesleyCambridge Press, Wellesley, Mass.
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Source: HeatTransfer Calculations
Chapter
35 Upper Bounds of Heat Transfer from Boxes
Wataru Nakayama ThermTech International Kanagawa, Japan
Introduction In this chapter we consider boxes that house heatgenerating components. We confine our attention to those boxes that we can readily handle. So, their dimensions are of the order of centimeters and do not exceed approximately one meter. The items of our primary interest are electronic boxes and gadgets. The most familiar examples are desktop computers, laptop computers, and cell phones. Many other electronic boxes are integrated in various machines and systems. Common to all those electronic boxes is the need for thermal design to prevent the components in the box from overheating. The first step of thermal design is the determination of cooling mode. The designer faces a question as to whether a particular future product can be encased in a sealed box, or a box having vent ports to allow cooling air through it, or to use one or more fans to drive air mechanically out of the box. In this first phase of the design analysis the ruleofthumb calculation suffices, and this chapter explains such calculations. The upper bounds of heat transfer from sealed boxes, boxes having vent ports for natural draft, and those cooled by the fan(s) are studied here. Knowledge about the upperbound heat transfer serves as a useful guide in assessment of different cooling modes.
35.1
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Upper Bounds of Heat Transfer from Boxes 35.2
Parameter and Boundary Estimation
The calculation of upperbound heat transfer is performed using simplified models. Internal organization of the box, often complex in actual equipment, is omitted from consideration here. We focus our attention on the exterior envelope of the box. The “envelope” is the exterior surface in the case of a sealed box, and includes the vent ports in the case of a ventilated box. The upperbound heat transfer is determined by several factors. In the case of sealed boxes, the laws of natural convection heat transfer and radiation heat transfer, and the surface area of the box are the determinants. To calculate the upperbound heat transfer from ventilated boxes, we need only the enthalpy transport equation applied to the airflow from the vent port. To calculate upperbound values we also need to specify the allowable temperature potential to drive heat flow from the box. In the thermal design of consumer electronic equipment, 30◦ C is commonly assumed as a maximum environmental temperature. Meanwhile, the surface temperature of those commercial products that have accidental contact with or are worn by human users needs to be held below 45◦ C. This criterion also applies to the temperature of exhaust air from the vent port, which may also come in contact with human users. So, the temperature potential available for heat dissipation from the box to the environment is assumed to range from 10 to 25 K.
Passive Cooling of Sealed Box It is assumed that the materials filling the box have infinite thermal conductivity and hence that heat generated from any parts of the interior instantaneously diffuse to the surface of the box. The model is an isothermal lump (Fig. 35.1). The heat flow from the box surface Q [in watts (W)] is the sum of heat flow by natural convection QNC (W), and
Gravity
H
L
W Figure 35.1
Isothermal solid lump.
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Upper Bounds of Heat Transfer from Boxes Upper Bounds of Heat Transfer from Boxes
35.3
radiation heat flow Q R (W): Q = QNC + Q R
(35.1)
The convective heat flow QNC is the sum of heat flows from different surfaces QNC = QV + QU + Q L
(35.2)
where QV , QU , and Q L are the heat flow from the vertical, the upper, and the lower surfaces, respectively. Each term on the righthand side of Eq. (35.2) is written in the form Qi = hi Ai T
(35.3)
where hi is the heattransfer coefficient (W/m2 K), Ai is the surface area (m2 ), T is the difference between the surface temperature and the environment temperature (K), and the subscript i is V, U, or L. The sum of vertical surface areas is AV = 2(W + L)H. The horizontal surface area is AU = AL = WL (≡ AH ). The heattransfer coefficient is found from the Nusselt number– Rayleigh number correlations. These numbers are defined as Nui ≡ Ra ≡
hi LB k
(35.4)
gL3BT · Pr 2
(35.5)
where the overbar on the Nusselt number means the average value over the surface, LB is the characteristic length (m) [LB = H for vertical surface; LB = AH /2(W + L) for horizontal surface], and g is the natural gravity, 9.807 m/s2 . The physical properties of air in these equations are as follows: k is the thermal conductivity (W/m K), is the volumetric expansion coefficient (K−1 ), is the kinematic viscosity (m2 /s), and Pr is the Prandtl number. The NuRa formulas used to estimate the natural convection heat transfer from the isothermal surface are as follows. For the sake of simplicity, the subscript i (V, U, or L) will be omitted in the following:
Vertical surface (Refs. 1 and 2) 0.670 Ra1/4
Nu = 0.68 +
1+
4/9 (0.492/Pr)9/16
Ra ≤ 109
(35.6)
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Upper Bounds of Heat Transfer from Boxes 35.4
Parameter and Boundary Estimation
Upper horizontal surface (Ref. 1) Nu = 0.54 Ra1/4
104 ≤ Ra ≤ 107
(35.7)
Nu = 0.15 Ra1/3
107 ≤ Ra ≤ 1011
(35.8)
Lower horizontal surface (Ref. 1) Nu = 0.27 Ra1/4
105 ≤ Ra ≤ 1010
(35.9)
The radiation heat transfer is computed from Q R = As hRT
(35.10)
where As is the total surface area exposed to the environment (m2 ), hR is the radiation heattransfer coefficient (W/m2 K) given as (Ref. 1) hR = 4ε T 03
(35.11)
where ε is the emissivity of the box surface, is the StefanBoltzmann constant (5.670 × 10−8 W/m2 K4 ), and T0 is the temperature in absolute scale (K). The T0 is, by definition, the mean temperature of the surface and the environment; however, for most practical purposes and ease of calculation, it may be equated to the environment temperature. Using hR and hi (restoring the subscript i), we may write Eq. (35.1) as Q = T (35.12) (hi + hR)Ai i
where the summation extends over the heattransfer areas. Calculations of heat flow were carried out for a wide range of box dimensions. Table 35.1 shows the dimensions of model boxes. The model designation consists of two numbers; one before a hyphen signifies the model, and that after the hyphen, the orientation of the box. In this dimension range are included those from cell phones, handheld calculators, laptop computers, desktop computers, and mediumsized server computers. It is assumed that the box is placed on a large adiabatic floor or attached to an adiabatic ceiling; hence, five surfaces are available for heat dissipation. This means that, for convection heattransfer estimation, either QU or Q L is set to zero in Eq. (35.2). For convenience of explanation, we first calculate the radiation heattransfer coefficient using Eq. (35.11). Except for shiny metal surfaces, the emissivity of surfaces of commercial and industrial products is in the range 0.8 to 0.9. For upperbound estimation we set ε = 1 and T0 = 300 K. Then hR = 6.12 W/m2 K
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Upper Bounds of Heat Transfer from Boxes Upper Bounds of Heat Transfer from Boxes
TABLE 35.1
35.5
Dimensions of Model Boxes (m)
Model no.
W
L
H
11 12 13
0.6 0.6 1.0
0.2 1.0 0.2
1.0 0.2 0.6
21 22
0.5 0.5
0.2 0.5
0.5 0.2
31 32 33
0.5 0.5 0.2
0.2 0.1 0.1
0.1 0.2 0.5
41 42 43
0.3 0.3 0.2
0.2 0.01 0.01
0.01 0.2 0.3
51 52 53
0.1 0.1 0.05
0.05 0.02 0.02
0.02 0.05 0.1
61 62 63
0.1 0.1 0.05
0.05 0.01 0.01
0.01 0.05 0.1
In calculation of convective heattransfer coefficient, the properties of air listed in Table 35.2 are used. Substituting these values into Eq. (35.5), we have Ra = 9.063 × 107 L3B T
(35.13)
The range of T is practically 5 to 25 K. For the 17 dimensions listed in Table 35.1, the Rayleigh number takes a wide range of values. On the vertical surfaces, Ra = 453 (H = 0.01 m, T = 25 K) ∼ 2.27 × 109 (H = 1.0 m and T = 25 K). On the horizontal surfaces, Ra = 33 (LB = 0.0042 m, T = 5 K) ∼ 1.49 × 107 (LB = 0.188 m, T = 25 K). These extreme values are outside the applicability range of some of the correlations [Eqs. (35.6) to (35.9)]. However, those cases of Rayleigh numbers TABLE 35.2 Physical Properties of Air at 300 K and Other Constants Used in Calculations
Property Density Specific heat at constant pressure Kinematic viscosity Thermal conductivity Volumetric thermal expansion coefficient Prandtl number Surface emissivity StefanBoltzmann constant Gravity acceleration
Symbol
Value
cp k Pr ε g
1.161 kg/m3 1007 J/kg K 1.589 × 10−5 m2 /s 0.0263 W/m K 0.00333 K−1 0.707 1 5.670 × 10−8 W/m2 K4 9.807 m/s2
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Upper Bounds of Heat Transfer from Boxes 35.6
Parameter and Boundary Estimation
outside the applicability range occupy only small corners in the matrix of dimensions and temperature difference. From Eq. (35.6) we compute the heattransfer coefficient on the vertical surface hV , which is reduced to hV = 0.0179 + 1.317H−1/4 T1/4
(35.14)
On the small horizontal surfaces the Rayleigh number is much smaller than the lower bound of Eqs. (35.7) and (35.9). Fortunately, the contribution from small areas to the overall heat transfer is small; the following formulas from Eqs. (35.7) and (35.9), respectively, serve the purpose of estimating the upperbound heat transfer: −1/4
hU = 1.39LB
T1/4
−1/4
hL = 0.693LB
(35.15)
T1/4
(35.16)
Equation (35.12) is now rewritten to calculate the lowerbound thermal resistance min (K/W): min ≡
T 1 = Q (hi + hR) Ai
(35.17)
i
Substituting the dimensional data of Table 35.1 into Eqs. (35.14) to (35.16) and setting hR = 6.12 W/m2 K, we find the overall thermal resistance as shown in Fig. 35.2. The horizontal axis (abscissa) is the total heattransfer area As . The heattransfer area depends on the Cell phone
Laptop
Desktop
Office computer server
Θmin (K/W)
100 10
U05 U25 L05 L25 Correlation
1 0.1 0.01 0.001
0.01
0.1
1
10
Surface area As (m2) Figure 35.2
Lowerbound overall thermal resistance from sealed boxes.
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Upper Bounds of Heat Transfer from Boxes Upper Bounds of Heat Transfer from Boxes
35.7
orientation of the box; it is large when the box is erected with its narrow area on the adiabatic floor. Hence, the results for the same model box spread over a certain range of the area in Fig. 35.2. The areas of typical electronic equipment are indicated assuming their usual orientations. In the data designations, U and L indicate the surface exposed to the environment in addition to the vertical surfaces that are exposed in all cases; the subscripts U and L denote the upper and lower surfaces, respectively. The numerals indicate T; 05 is 5 K, and 25 is 25 K. The calculated thermal resistances fall in a narrow band. The line in Fig. 35.2 represents the following correlation min =
1 1 = As heq As (9.18 − 0.6 ln As )
K/W
(35.18)
where heq is the equivalent heattransfer coefficient (W/m2 K) obtained from line fitting of the calculated results: heq = 9.18 − 0.6 ln As . Equation (35.18) correlates the calculated results in a range −23 to +30 percent, and provides a means to perform the ruleofthumb calculation of lowerbound thermal resistance. Some examples are as follows. 1. For a portable gadget having a 0.01 m2 heattransfer area, the lowerbound thermal resistance is calculated from Eq. (35.18) as 8.4 K/W. If we allow a 15 K (= T) temperature rise on the surface, the upperbound heat flow is 1.8 W. 2. For a mediumsized computer having a 1 m2 heattransfer area, the lowerbound thermal resistance is 0.11 K/W. For T = 15 K, the upperbound heat flow is 138 W. Enthalpy Transport from Vent Ports Natural draft ventilation
Figure 35.3 shows a box having top and bottom sides available as vent ports. The following derivation of equations is commonly employed in the analysis of natural draft systems (e.g., see Ref. 3). The static pressure immediately below the bottom of the box is pi = p0 − 12 0 U2
(35.19)
where p0 is the pressure of the environment at the level of the box bottom and U is the velocity that is assumed to be uniform over the area W × L. The static pressure immediately above the top of the box is pe = pi − pH
(35.20)
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Upper Bounds of Heat Transfer from Boxes 35.8
Parameter and Boundary Estimation
pe
T0
p0
Te
U
W
Natural draft through a box.
Figure 35.3
pi
where p H is the pressure difference between the bottom and the top of the box. From Eqs. (35.19) and (35.20), we have pe = p0 − pH − 12 0 U2
(35.21)
In the atmosphere above the box the dynamic pressure given to the plume is lost by viscous diffusion. Hence, considering the force balance outside the box, we have pe = p0 − 0 gH
(35.22)
This boundary condition at the exit, together with the inlet boundary condition (35.19), is commonly employed in the numerical analysis of ventilated systems. Canceling out p0 and pe from Eqs. (35.21) and (35.22), we have pH = − 12 0 U2 + 0 gH
(35.23)
Meanwhile, pH is the sum of the frictional pressure drop p f and the staticpressure head gH, where is the air density that varies along the box height: pH = p f + gH
(35.24)
The staticpressure head depends on the temperature distribution in the block. However, for the sake of simple formulation, we suppose
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Upper Bounds of Heat Transfer from Boxes Upper Bounds of Heat Transfer from Boxes
35.9
a fictitious situation; the air in the box is uniformly heated up to the temperature Te . Then, the density in Eq. (35.24) is written as e = 0 {1 − (Te − T0 )}. Eliminating pH from Eqs. (35.23) and (35.24), we have p f = − 12 0 U2 + 0 gH (Te − T0 )
(35.25)
This is a nonlinear equation, since the terms on the lefthand side are functions of the velocity U. Suppose that we solved Eq. (35.25) for U; then, the enthalpy equation would give the heat flow Q from the box: Q = 0 c p UAe (Te − T0 )
W
(35.26)
where Ae is the crosssectional area of the exit vent (= W × L) (m2 ). Denoting the volumetric flow rate as V = 60UAe (m3 /min), we write the thermal resistance as =
Te − T0 60 = Q 0c pV
K/W
(35.27)
Note that the variation of air density is taken into account only in the buoyancy term (the Boussinesq approximation); hence, the density in Eqs. (35.26) and (35.27) is 0 . Also, note that Eqs. (35.26) and (35.27) are valid for mechanical draft systems as well; thus, they will be used in the subsequent sections. The upper bound of heat transfer is now derived assuming the flow resistance along the path of airflow to be negligibly small. Then, from Eq. (35.25) we have Umax = 2gH (Te − T0 ) (35.28) Using Umax to compute V in Eq. (35.27), we obtain the lower bound of thermal resistance as min =
1
0 c p Ae 2gH (Te − T0 )
(35.29)
Substituting the numerical values of Table 35.2 and setting Te − T0 = 20 K in Eq. (35.29), the ruleofthumb estimation of the lowerbound thermal resistance is obtained as min =
1
√ 1337Ae H
K/W
(35.30)
The dimensions of the models of Table 35.1 are now substituted in Eq. (35.30). Some examples are plotted in Fig. 35.4 by solid triangles,
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Upper Bounds of Heat Transfer from Boxes 35.10
Parameter and Boundary Estimation
10
63
Θmin (K/W)
53
1 43
0.1
Sealed Models Vent Opt.
33 21
0.01 11
0.001 0.001
0.01
0.1
1
10
Surface area As (m2) Figure 35.4 Comparison of lowerbound thermal resistances in natural draft
cooling: Sealed box (upper line); some models of Table 35.1 with vents (triangles); boxes of optimum dimensions for natural draft cooling (lower line). Note that As is the surface area of five sides of the sealed box.
where the horizontal axis is the external heattransfer area of the sealed box As , defined in the previous section. These results belong to the boxes having their longest dimensions as the heights. Also shown in Fig. 35.4 is a line representing the correlation in Eq. (35.18) for sealed boxes. The min for case 43 is very close to that for the sealed box, indicating that, for a thin, flat box, the ventilation effect is comparable to the natural cooling from the larger vertical sides. The min for a given As is actually lower than the results calculated for the model boxes. In order to find the lower bound for a given As , we allow the variation of the dimensions, H, W, and L, to minimize min . Geometric consideration defines the maximum height as given by Eq. (35.31), so min of Eq. (35.30) can be written in terms of Ae or H alone. The optimization condition is as follows: As − Ae √ 4 Ae W = L = Ae H=
Ae =
3 5
As
(35.31) (35.32) (35.33)
Substitution of optimum Ae and H into Eq. (35.30) yields the following equation: min = 0.00347A−5/4 s
(35.34)
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Upper Bounds of Heat Transfer from Boxes Upper Bounds of Heat Transfer from Boxes
35.11
The lower line in Fig. 35.4 represents Eq. (35.34). The vertical separation between the sealed box correlation (the upper line) and the ventilated box correlation indicates the order of magnitude of maximum benefits obtained by natural ventilation. For a box having a 0.01m2 heattransfer area, ventilation reduces the lowerbound thermal resistance from 8.4 to 1.1 K/W. For a 1m2 box, the corresponding figures are 0.11 and 0.0035 K/W. However, as we will see in the next section, the presence of flow resistance inside the actual box seriously compromises the benefit of ventilation. In terms of the volume of the box having the optimum dimensional proportion [VB (m3 )], Eq. (35.34) is rewritten as −5/6
min = 4.12 × 10−4 V B
(35.35)
Mechanical draft ventilation
We suppose appropriate fans for the models of Table 35.1. First, consider the largest box, model 1. The fans used in most electronic boxes of interest here are propeller fans driven by brushless directcurrent (dc) motors. Among the commercially readily available fans, the largest one has the dimensions 12 × 12 cm. In order to maximize the mechanical draft, four fans are accommodated in the top plane of the model 11 box, as shown in Fig. 35.5. Figure 35.6 shows a typical characteristic curve of a fan of this size (curve pFAN); the vertical axis (ordinate) is the staticpressure rise p (Pa) produced by the fan, and the horizontal axis (abscissa) is the collective volumetric flow rate of four fans. Following the definition set for natural draft cooling, the upperbound heat transfer is again determined assuming negligible flow resistance in the box. Then, the upperbound heat flow is determined by the flow rate at
20 cm
12 cm
12 cm
60 cm Figure 35.5
Top view of model 11 with four propeller fans.
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Upper Bounds of Heat Transfer from Boxes
∆p (Pa)
35.12
Parameter and Boundary Estimation
45 40 35 30 25 20 15 10 5 0
pFAN pRESIST pNATURAL
0
2 Natural draft 1.8 m3/min
4
6
V
(m3/min)
8
10
Mechanical draft 6.5 m3/min
Figure 35.6 Fan curve (pFAN; V = collective flow rate of four fans), resistance curve (pRESIST), and natural draft potential (pNATURAL) for model 11.
the freedelivery point of the fan curve (p = 0), VF (m3 /min): Q F = c p VF (Te − T0 )/60
W
(35.36)
where Te is the temperature of exhaust air and T0 is the environment temperature. From Fig. 35.6 we find VF = 8 m3 /min. The corresponding lowerbound thermal resistance [Eq. (35.27)] is min =
60 = 0.0064 1.161 × 1007 × 8
K/W
(35.37)
This value almost coincides with the lower bound of natural draft thermal resistance for model 11 (0.0062 K/W). As we will see in the next section, the benefit of mechanical draft lies in its ability to overcome flow resistance in the box. For the smallest box among the models of Table 35.1, that is, model 6, a fan having a 4cm side length is commercially available. The fan is designed to be accommodated in a thin space; the air is sucked in the direction of a rotor axis and driven out through a slot provided on the side of the fan frame. Figure 35.7 shows such a fan accommodated in the model 6 box, with a sketch of a stack of cards to be cooled. A typical characteristic curve is shown in Fig. 35.8 (pFAN). The freedelivery
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Upper Bounds of Heat Transfer from Boxes Upper Bounds of Heat Transfer from Boxes
5 mm
5 cm
4 cm
4 cm
1 cm
10 cm Air inlet
Air exit Cards stack
Figure 35.7
∆p (Pa)
35.13
90 80 70 60 50 40 30 20 10 0
Sideblow fan
Model 6 box equipped with a sideblow fan.
pFAN pRESIST 0.1 0.08 0.06 0.04 0.02
.0
0 0
0
0.0001
0.0002
0.01
0.20
0.03
0.04
V (m3/min) Natural draft 0.00022 m3/min
Mechanical draft 0.0285 m3/min
Fan curve (pFAN), resistance curve (pRESIST), and natural draft potential (horizontal line in the inset) for model 63.
Figure 35.8
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Upper Bounds of Heat Transfer from Boxes 35.14
Parameter and Boundary Estimation
flow rate is VF = 0.033 m3 /min. The corresponding lowerbound thermal resistance is min =
60 = 1.56 1.161 × 1007 × 0.033
K/W
(35.38)
This value is almost onethird the lowerbound thermal resistance of natural draft cooling for model 63 (4.73 K/W). The Effects of Flow Resistance on Thermal Resistance In this section we consider situations that are a little closer to reality. Again, we take the largest model (model 1) and the smallest (model 6) as our illustrative examples. The resistance to airflow is defined in terms of the staticpressure drop from the inlet to the exit of the box and the volumetric flow rate. Large flow resistance means a large pressure drop for a given flow rate, or a low flow rate under a given pressure difference. In Figs. 35.6 and 35.8, typical resistance curves are shown (pRESIST). The actual flow resistance is a function of the dimensions and configurations of internal airflow paths. To determine the flow resistance curve of actual electronic equipment, we need to conduct an experiment. In the experiment, the test object is placed in a wind tunnel. The wind tunnel provides a means to accurately measure the airflow rate and the pressures upstream and downstream of the test object. The intersection of the resistance curve and the fan curve gives an actual operating point of the flow system; that is, the fan provides a driving potential in the form of pressure rise that matches the pressure drop caused in the box. From Fig. 35.6, we read the actual volumetric flow rate in the model 1 box as V = 6.5 m3 /min. The corresponding thermal resistance is = 0.0079 K/W, representing an approximately 20 percent increase from the lower bound calculated earlier in the “Mechanical Draft Ventilation” section. In Fig. 35.6 a line showing the maximum driving potential of natural draft is also included (for the posture of model 11, see Table 35.1); that is, for H = 1 m and (Te − T0 ) = 20 K, we have 0 gH(Te − T0 ) = 1.161 × 9.807 × 0.00333 × 20 = 0.759 Pa
(35.39)
The intersection of this line with the resistance curve gives the actual natural draft flow rate as V = 1.8 m3 /min. The corresponding thermal resistance is = 0.029 K/W, almost fivefold the lower bound calculated earlier in the “Natural Draft Ventilation” section. As illustrated above, the natural draft has very low driving potential. Low potential of natural draft is amplified as the box size reduces. Figure 35.8 shows a typical resistance curve for equipment of model 6
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Upper Bounds of Heat Transfer from Boxes Upper Bounds of Heat Transfer from Boxes
TABLE 35.3
35.15
Comparison of Thermal Resistance Values (K/W) Natural draft
Model no.
Completely sealed
11 63
0.066 6.8
Mechanical draft
Lower bound
With flow resistance
Lower bound
With flow resistance
0.0062 4.73
0.0286 233
0.0064 1.56
0.0079 1.80
size. The intersection of the resistance curve with the fan curve gives V = 0.0285 m3 /min; hence, = 1.8 K/W. The natural draft potential for H = 0.1 m and (Te − T0 ) = 20 K is 0.0759 Pa. From the inset of Fig. 35.8 we read the volumetric flow rate as V = 0.00022 m3 /min. The thermal resistance of model 63 in natural draft cooling with flow resistance is calculated as = 233 K/W. Table 35.3 compares the thermal resistance values of models 1 and 6. The values for sealed boxes are calculated using Eq. (35.18). Table 35.3 provides some insight into the cooling technique for small electronic gadgets. For small boxes of the size of model 6 or smaller, if natural convection cooling is favored, heat dissipation from the surfaces is the realistic mode of cooling. Heat generated in the box has to be spread in the box and conducted to the surface. Vent holes bring only marginal enhancement of cooling. To benefit from mechanical draft, the fan must be made smaller than currently available models; occupation of almost 40 percent of the box space as illustrated in Fig. 35.7 is unacceptable. With the reduction of fan size, the rotational speed of the fan needs to be raised to provide a sufficiently high fan performance. This has to be accompanied by the efforts to mitigate acoustic noise from the fan. References 1. F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 4th ed., Wiley, New York, 1996, pp. 493–498. 2. S. W. Churchill and H. H. S. Chu, “Correlating Equations for Laminar and Turbulent Free Convection from a Vertical Plate,” International Journal of Heat and Mass Transfer, 18, 1323–1329, 1975. 3. T. S. Fisher and K. E. Torrance, “Free Convection Cooling Limits for PinFin Cooling,” ASME Journal of Heat Transfer, 120, 633–640, 1998.
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Upper Bounds of Heat Transfer from Boxes
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Source: HeatTransfer Calculations
Chapter
36 Estimating Freezing Time of Foods
R. Paul Singh Department of Biological and Agricultural Engineering University of California Davis, California
Freezing of foods is a common unit operation employed in the food industry. Many fruits and vegetables, such as peas, strawberries, diced carrots, and green beans, are frozen in fluidizedbed freezers where the product comes into direct contact with air at subfreezing temperatures. As the food undergoes freezing, there is a change in phase of water into ice; this complicates the heattransfer computations required to estimate freezing times. In this example, we will predict freezing time of strawberries being frozen in a fluidizedbed freezer. The initial temperature of a strawberry is 15◦ C, and it has a moisture content of 75 percent (wet basis). The shape of the strawberry is assumed to be a sphere with a diameter of 2.5 cm. The final desired center temperature of the strawberry is −18◦ C. The air temperature in the fluidizedbed freezer is −40◦ C. The convective heattransfer coefficient is measured to be 80 W/(m2 K). The properties of the strawberry are assumed as follows. The density of an unfrozen strawberry is 1130 kg/m3 , the density of a frozen strawberry is 950 kg/m3 , the specific heat of an unfrozen strawberry is 3.55 kJ/(kg ◦ C), and the specific heat of a frozen strawberry is 1.5 kJ/(kg ◦ C). The thermal conductivity of the frozen
36.1
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Estimating Freezing Time of Foods 36.2
Parameter and Boundary Estimation
strawberry is 1.5 W/(m K). Calculate how long it will take to reduce the center temperature of the strawberry from 15 to −18◦ C.
Approach To solve this problem, we will use a method proposed by Pham (1986). Pham’s method is useful in calculating freezing time of foods with high moisture content (>55 percent). The following assumptions were made by Pham: (1) the temperature of the air used in the freezer is constant, (2) the initial temperature of the product is constant, (3) the value of the final temperature is fixed, and (4) the convective heat transfer on the surface of the food is described by Newton’s law of cooling. Pham divides the total heat removed during freezing from some initial temperature to a desired final temperature into two components. As seen in Fig. 36.1, a plot of center temperature versus heat removal is divided into two parts using a mean freezing temperature Tm. The first part of this curve represents heat removal in precooling and some initial change of phase as the product begins to freeze; the second part represents the remaining heat removal during phase change and additional cooling to reach the final desired temperature. The mean freezing temperature Tm is obtained using the following equation, which has been empirically obtained for foods with moisture content exceeding 55 percent Tm = 1.8 + 0.263Tc + 0.105Ta
(36.1)
where Tc is the final center temperature of the product (◦ C) and Ta is the air temperature used in freezing (◦ C).
Temperature (°C)
Ti
Tm
Tc
∆H1
∆H2
Heat Removal (kJ/kg) Figure 36.1 A plot of temperature at the product center versus heat removal during freezing of foods.
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Estimating Freezing Time of Foods Estimating Freezing Time of Foods
36.3
The time to freeze the product from an initial temperature Ti (◦ C) to a final temperature of Tc is obtained using the following equation: t=
d NBi H1 H2 1+ + Eh 2 T1 T2
(36.2)
where d is the characteristic dimension, which is the shortest dimension from the surface to the center [for a sphere it is the radius (m)]; h is the convective heat transfer [W/(m2 K)]; and E is the dimensional shape factor (it is 1 for an infinite slab, 2 for an infinite cylinder, and 3 for a sphere). The term H1 represents the change in the volumetric enthalpy (J/m3 ) for the precooling and initial periods of phase change: H1 = ucu(Ti − Tm)
(36.3)
where u is the density of unfrozen material, cu is the specific heat of the unfrozen material [J/(kg K)], and Ti is the initial temperature of the product (◦ C). The term H2 is the change in the volumetric enthalpy (J/m3 ) for the period involving the remaining phase change and postcooling of the product to the final center temperature: H2 = f [L f + c f (Tm − Tc )]
(36.4)
where f is the density of the frozen product, c f is the specific heat of the frozen product [kJ/(kg K)], and L f is the latent heat of fusion of the food undergoing freezing [J/(kg K)]. The Biot number NBi is defined as NBi =
hd kf
(36.5)
where h is the convective heattransfer coefficient [W/(m2 K)], d is the characteristic dimension (m), and k f is the thermal conductivity of the frozen material [W/(m K)]. The temperature gradient T1 is obtained from T1 =
Ti + Tm − Ta 2
(36.6)
where Ti is the initial temperature (◦ C), Ta is the temperature of the air (◦ C), and Tm is the mean temperature as defined in Eq. (36.1).
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Estimating Freezing Time of Foods 36.4
Parameter and Boundary Estimation
The temperature gradient T2 is obtained from T2 = Tm − Ta
(36.7)
For more information on estimating thermal properties of foods required in Eqs. (36.3) to (36.5), refer to Singh and Heldman (2001) and Singh (2004). Next, we will use the preceding equations to calculate freezing time for the given problem. Solution Given
Initial temperature of strawberry = 15◦ C Moisture content of strawberry = 75 percent (wet basis) Diameter of strawberry = 2.5 cm Final desired center temperature of strawberry = −18◦ C Air temperature in fluidizedbed freezer = −40◦ C Convective heattransfer coefficient = 80 W/(m2 K) Density of unfrozen strawberry = 1130 kg/m3 Density of frozen strawberry = 950 kg/m3 Specific heat of unfrozen strawberry = 3.55 kJ/(kg ◦ C) Specific heat of frozen strawberry = 1.5 kJ/(kg ◦ C) Thermal conductivity of frozen strawberry = 1.5 W/(m K) Procedure
1. Using Eq. (36.1), calculate Tm: Tm = 1.8 + 0.263 × (−18) + 0.105 × (−40) = −7.134◦ C 2. Using Eq. (36.3), calculate H1 : H1 = 1130 kg/m3 × 3.55 kJ/kg K × 1000 J/kJ × (15 − (−7.134)) ◦ C = 88,790,541 J/m3 3. The latent heat of fusion of strawberries is obtained as a product of moisture content and latent heat of fusion of water (333.2 kJ/kg): Lf = 0.75 × 333.2 kJ/kg × 1000 J/kJ = 249,900 J/kg
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Estimating Freezing Time of Foods Estimating Freezing Time of Foods
36.5
4. Using Eq. (36.4), calculate H2 : 249,900 J/kg + 1.5 kJ/kg K × 1000 J/kJ 3 H2 = 950 kg/m × × (−7.134 − (−18))◦ C = 252,889,050 J/m3 5. Using Eq. (36.6), calculate T1 : 15 + (−7.134) T1 = − (−40) 2 = 43.93◦ C 6. Using Eq. (36.7), calculate T2 : T2 = [−7.134 − (−40)] = 32.87◦ C 7. The Biot number is calculated using Eq. (36.5) as follows: NBi =
80 W/m2 K × 0.0125 m 1.5 W/m2 K
= 0.667 8. Substituting results of steps 1 through 7 in Eq. (36.2), noting that for a sphere E f = 3, we have 0.0125 m 88790541 J/m3 252889050 J/m3 t= + 43.93◦ C 32.87◦ C 3 × 80 W/m2 K 0.667 × 1+ 2 Time = 674.7 s = 11.2 min Result
The estimated time to freeze strawberries in a fluidizedbed freezer is 11.2 min. References Pham, Q. T., 1986. “Simplified Equation for Predicting the Freezing Time of Foodstuffs,” J. Food Technol. 21:209–219. Singh, R. P., and Heldman, D. R., 2001. Introduction to Food Engineering, Elsevier, Amsterdam, The Netherlands. Singh R. P., 2004. Food Properties Database, RAR Press, Davis, Calif.
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Estimating Freezing Time of Foods
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Source: HeatTransfer Calculations
Part
9 Temperature Control
In the single chapter in this part, the author, who works in industry in the United States, describes how precise temperature control of an electronic device can be accomplished by mounting it on a thermoelectric module (TEM). Depending on the direction of the current flowing through a TEM, heat is released or absorbed at the interface between it and the electronic device by virtue of the Peltier effect. Because the rate of heat released or absorbed by the Peltier effect is proportional to the magnitude of the current, TEMs are effective temperature controllers of devices such as lasers and optical routers.
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Temperature Control
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Source: HeatTransfer Calculations
Chapter
37 Precision Temperature Control Using a Thermoelectric Module
Marc Hodes Bell Laboratories Lucent Technologies Murray Hill, New Jersey
Introduction Depending on the direction of the current flowing through a thermoelectric module (TEM), heat is absorbed or released at the interface between it and an electronic device mounted on it by virtue of the Peltier effect. Moreover, the rate of heat absorbed or released by the Peltier effect is directly proportional to the magnitude of the current. Hence, TEMs are well suited to precision temperature control of electronic devices. Here the equations governing the operation of a TEM are presented and used to calculate parameters relevant to precision temperature control of an optical router.
Background Thermal energy may be reversibly converted to electrical energy and vice versa in electrically conducting materials (including semiconductors) by thermoelectric effects (see, e.g., Foiles [1]). Thermoelectric effects are small in metals but moderate in certain types of semiconductors. The moving charge carriers constituting an electric current I carry different amounts of electrical energy in different conductors. 37.3
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Precision Temperature Control Using a Thermoelectric Module 37.4
Temperature Control
Thus heat is evolved or absorbed when current flows through the interface between two conductors. This is known as the Peltier effect. The rate of (reversible) heat absorption at the interface between two conductors q due to the Peltier effect is q = I(B − A)T
(37.1)
where I is current, is the Seebeck coefficient (a material property), and T is absolute temperature. Current is positive when positive charge carriers flow from conductor A to conductor B. The irreversible (bulk) effects of Ohmic (I 2 R) heating and heat conduction must also be considered in the analysis of thermoelectricity. Thermoelectric modules are solidstate devices that exploit the Peltier effect in order to cool, heat, or generate electrical power. They are commonly used to cool electronics or to precisely control their temperature through cooling and heating. A singlestage TEM is shown in Fig. 37.1. It consists of an array of positively doped (ptype) and negatively doped (ntype) semiconductor pellets connected electrically in series and thermally in parallel between ceramic substrates. Each adjacent pair of ptype and ntype pellets is referred to as a thermocouple, and there are N thermocouples in a TEM. Here it is assumed that the controlled side of the TEM is that which does not connect to the external leads according to Fig. 37.1. The other side is defined as the uncontrolled ControlledSide Substrate
n=3
n=N
n=2
n=1
UncontrolledSide Substrate VN Figure 37.1
V0
Cutaway view of a TEM.
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Precision Temperature Control Using a Thermoelectric Module Precision Temperature Control Using a TEM
Controlled Side
{
+I: Peltiercooled csi −I: Peltierheated csi
Cooled and/or Heated Object or Medium
Ceramic Substrate
Cu Electrical Interconnect
I
n
Semiconductor Pellet Uncontrolled Side
ControlledSide Interface (csi) x = 0, T = Tc
+
p
37.5
x
{ VN
V0
UncontrolledSide Interface (usi) x = L, T = Tu
VTEM = V0 − VN +
−
−
+
DC Source Figure 37.2
Schematic representation of a singlethermocouple TEM.
side. The controlledside interface (csi) of a TEM is defined as that between the thermocouples inside it and the controlled side substrate and the uncontrolledside interface (usi) is defined analogously according to Fig. 37.2, which shows a single thermocouple (N = 1) TEM. Currents are taken to be positive when positive charge flows from ntype pellets to ptype pellets at the controlled side of a TEM, that is, counterclockwise according to Fig. 37.2. Since p and n are positive and negative quantities, respectively, heat is absorbed by the Peltier effect at the csi and released by the Peltier effect at the usi when current is positive as per Eq. (37.1). Analogously, when current is negative, heat is absorbed and released by the Peltier effect at the usi and the csi, respectively. When a TEM operates in cooling or heating modes, dc power must be supplied to it. Moreover, for precision temperature control applications, the polarity of the power source must be reversible in order to switch between cooling and heating modes. Thermoelectric modules posses a number of highly advantageous attributes. For example, they have no moving parts and are of very simple construction, acoustically silent, compact, rugged, lightweight, moderatecost, and modular. The major disadvantage of TEMs is that their thermodynamic efficiencies are smaller than those of competing technologies, such as vapor compression refrigerators.
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Precision Temperature Control Using a Thermoelectric Module 37.6
Temperature Control
Governing Equations The equations resulting from the onedimensional analysis of a singlestage TEM by Hodes [2] are presented in this section. The assumptions made by Hodes [2] are commonplace in the onedimensional analysis of TEMs. Seifert et al. [3] and Yao et al. [4] attained good agreement between equivalent onedimensional models and numerical solutions to more rigorous formulations. Hence, solutions to the present formulation provide sufficiently accurate results for many TEM applications such as the one described below. The net rate of heat conducted out of a differential control volume normal to the x direction defined in Fig. 37.2 equals the rate of Ohmic heat generation inside it. Hence the form of the heat equation governing conduction inside the pellets is d2 T I2 + =0 2 dx kA2P
(37.2)
where x is distance from the csi, k is thermal conductivity of a pellet, and AP is the crosssectional area of a pellet. The boundary conditions on the heat equation at the csi and the usi are T = Tc at x = 0 and T = Tu at x = L, respectively, where L is the height of a pellet, as per Fig. 37.2. It follows that the (parabolic) temperature distribution in the pellets is: 2 −I 2 2 I L Tu − Tc T= x + + (37.3) x + Tc L 2kA2P 2kA2P The rates of heat transfer from the controlledside substrate into the csi qc and from the usi into the uncontrolledside substrate qu may be determined from surface energy balances. The sign convention adopted in the surface energy balances is that qc and qu are positive in the positive xdirection defined in Fig. 37.2. Physically, the rate of heat transfer from the controlledside substrate into the csi qc equals the rate of energy absorbed by the Peltier effect at the csi of [I( p − n)Tc ] per thermocouple plus the rate of heat transfer by conduction into the pellets at the csi of (−2kAP dT/dxx=0 ) per thermocouple. The quantity ( p − n) is subsequently denoted by p,n. To accommodate appreciable amounts of heat at realistic supply currents and voltages, a TEM must contain an array of N thermocouples as shown in Fig. 37.1. For an Nthermocouple TEM, the surface energy balances at the csi and the usi are I2 R qc = N I p,nTc − K(Tu − Tc ) − (37.4) 2 I2 R qu = N I p,nTu − K(Tu − Tc ) + (37.5) 2
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Precision Temperature Control Using a Thermoelectric Module Precision Temperature Control Using a TEM
37.7
where the Ohmic resistance R and the thermal conductance K of a thermocouple are defined conventionally as R=
2 L AP
(37.6)
K=
2kAP L
(37.7)
An energy balance on a control volume around the interconnected pellets . forming a TEM shows that the rate of electrical work done on it (WTEM ) is equal to (qu − qc ). Hence . WTEM = N I p,n(Tu − Tc ) + I 2 R
(37.8)
. WTEM is positive when a dc power supply does work on a TEM in cooling and heating modes. An energy balance on a control volume around the . power supply in Fig. 37.2 shows that (IV) N + WTEM = (IV)0 , where the voltages V0 and VN in a TEM are defined according to Fig. 37.2. Hence, the voltage across a TEM is VTEM = V0 − VN = N [ p,n(Tu − Tc ) + IR ]
for
I = 0
(37.9)
Despite the utility of the results developed thus far, the boundary conditions at the csi and the usi in a TEM are usually more complicated in practice; that is, Tc and Tu are not specified but must be computed (see, e.g., the analysis by Yamanashi [5]). This necessitates accounting for the three thermal resistances shown in Fig. 37.3. First, the thermal resistance between the location on the object or in the medium where constant temperature is being maintained, subsequently referred to as the control point (cp), and its local ambient (Rcp−∞,c ) must be accounted for. This necessitates the addition of a surface energy balance at the control point to the balance equations. Also, the thermal resistances between the control point and the csi (Rcp−c ) and between the usi and its local ambient (Ru−∞,u) must be accounted for. Constriction/spreading resistances in the controlled and uncontrolled substrates of a TEM are embedded in Rcp−c and Ru−∞,u, respectively. The temperatures of the local ambients around the control point (T∞,c ) and the usi (T∞,u) may or may not be the same. Specifying all the aforementioned thermal resistances by their corresponding overall conductances (K), the surface energy balances at the control point, csi and usi are given by Eqs. (37.10) through (37.12). The overall energy balance given by Eq. (37.8)
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Precision Temperature Control Using a Thermoelectric Module 37.8
Temperature Control
T∞,c Rcp−∞,c Tcp
qcp Control Point Surface Energy Balance Rcp−c
Tc Thermocouple N
p
n
csi Surface Energy Balance Thermocouple 1
qc n
p
p
n +
qu Tu
I
Energy Balance on Control Volume around TM Pellets
usi Surface Energy Balance
Ru−∞,u T∞,u VN
V0
VTEM = V0 − Vn Figure 37.3
Representation of onedimensional analysis of a TEM.
remains valid: qcp = Kcp−c (Tcp − Tc ) + Kcp−∞,c (Tcp − T∞,c ) I2 R Kcp−c (Tcp − Tc ) = N I p,nTc − K(Tu − Tc ) − 2 I2 R Ku−∞,u(Tu −T∞,u) = N I p,nTu − K(Tu −Tc ) + 2
(37.10) (37.11) (37.12)
Calculational Example Problem
Reconfigurable optical add/drop modules (ROADMs) are analog devices that route light containing information (voice, video, and/or data) among optical fibers in telecommunications networks. Light from
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Precision Temperature Control Using a Thermoelectric Module Precision Temperature Control Using a TEM
37.9
incoming optical fibers travels along waveguides (often silica) fabricated on a large die (often silicon) and is routed to the appropriate outgoing fiber by various optical components. ROADMs use thermooptic phase shifters to locally heat the waveguides in order to change their temperature. Since the index of refraction of the waveguides is temperaturedependent, the phase of the light traveling through them may be controlled. This phase control in combination with interferometers enables light to be switched between different paths. Integrated on the same die as the heatdissipating thermooptic phase shifters, however, are temperaturesensitive optical components, such as optical filters, which must be maintained within narrow operating temperature ranges (e.g., ±1◦ C) to function properly. In order to maintain the die at constant temperature except in the vicinity of the thermooptic phase shifters as, for example, the ambient temperature changes, it may be mounted on a heat spreader which, in turn, is mounted on a TEM. A TEM has been selected to maintain the temperature of the die in a ROADM at 75◦ C for ambient temperatures ranging from −5 to 65◦ C. The TEM consists of a uniform array of 142 Bi2 Te3 pellets (71 thermocouples), and the pellet properties are p = 2.0 × 10−4 V/K, n = −2.0 × 10−4 V/K, = 1.25 × 10−5 m, and k = 1.5 W/(m K). Pellet height L equals 1.14 mm and the pellet footprint is 1.40 × 1.40 mm. The electrical interconnects for the pellets and the external leads of the TEM are copper, which may be assumed to contribute no electrical or thermal contact resistances and have a Seebeck coefficient of 0 V/K. The pellets are located between 30mm × 30mm × 0.75mmthick alumina [k = 36.0 W/(m K)] substrates. The die has the same footprint as the TEM on which it is mounted, and its power dissipation may be approximated as uniform. Assume that the die is thermally insulated from the environment except through a layer of thermally conductive epoxy of interfacial resistance Rti = 1.70 × 10−5 m2 W/K coupling it to the controlside substrate of the TEM and that the uncontrolledside substrate of the TEM transfers heat to the ambient by conduction through a layer of thermal grease (Rti = 2.58 × 10−5 m2 W/K) in series with a 1.0◦ C/W thermal resistance representing a heat sink. (A). Assuming that the die dissipates 10 W of power, calculate the ambient temperatures at which it is unnecessary to supply power to the TEM when its external leads terminate in an open circuit, when they are electrically shortcircuited together and when +4 V is applied across them. In each case quantitatively describe the temperature distribution in the pellets inside the TEM. (B). What are the minimum and maximum powers that the die can dissipate when the power available to the TEM is limited to 10 W and when it is unlimited?
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Precision Temperature Control Using a Thermoelectric Module 37.10
Temperature Control
(C). Assume that the size of the heat sink and, hence, its thermal resistance, may be varied. When the die dissipates 10 W, what is the optimum thermal resistance of the heat sink insofar as minimizing the power consumed by the TEM? (D). Assume, for the sake of this example, that the objective is to bring the die to 200◦ C. Is it more efficient (i.e., is less electrical power required) when the die is used as a resistive heater or when the TEM is used to heat the die? Assume that all electrical power is used to drive either the die or the TEM. Perform the calculations assuming that T∞,u = 65.0◦ C and RHS = 1.0◦ C/W.
Solution Known
Thermocouple properties, geometry, and number: p = 2.0 × 10−4 V/K, n = −2.0 × 10−4 V/K, p,n = 4.0 × 10−4 V/K, = 1.25 × 10−5 m, k = 1.5 W/(m K), L = 1.14 mm, AP = 30 × 30 mm, N = 71. 30 × 30mm footprint by 0.75mmthick TEM substrates with k = 36.0 W/(m K). Controlpoint temperature and uncontrolledside ambient temperature range: Tcp = 75◦ C, −5◦ C ≤ T∞,u ≤ 65◦ C. Data required to compute thermal resistances between control point and csi of the TEM and the usi of the TEM and its local ambient. Schematic. A schematic representation of the problem is shown in
Fig. 37.4. Assumptions
Only steadystate conditions are relevant. All of the assumptions in the onedimensional analysis of a TEM by Hodes [2] are valid. No heat transfer between the control point and its local ambient (Kcp−∞,c = 0). Analysis. Surface energy balances at the control point (the silicon die),
the csi of the TEM, and the usi of the TEM [Eqs. (37.10) through (37.12), respectively] and the overall energy balance on a control volume around the interconnected pellets within the TEM [Eq. (37.8)] are necessary to solve this problem. [Note that the surface energy balance at the control point reduces to qcp = Kcp−c (Tcp − Tc ) because it has been assumed that heat transfer from the control point to its local ambient is negligible,
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Precision Temperature Control Using a Thermoelectric Module Precision Temperature Control Using a TEM
37.11
qcp
Thermal Epoxy n=2 to n = 70
p
n = 71
n
csi
p
n=1
n
s
x
V = V0
V = VN
Grease & Heat Sink in Series
Open Circuit
VTEM
Short Circuit Applied Voltage (Electrical Work Done on TEM) Figure 37.4
Schematic representation of the problem.
i.e., Kcp−∞,c = 0.] It is convenient to compute the parameters required to solve the foregoing equations before proceeding with the analysis. The electrical resistance Rand thermal conductance K of each thermocouple in the TEM are given by Eqs. (37.6) and (37.7), respectively. The result is that R = 1.45 × 10−2
and
K = 5.16 × 10−3 W/K
The overall conductance between the die (control point) and the csi (Kcp−c ) is the reciprocal of the sum of the thermal resistance across the thermal epoxy between the die and the alumina substrate (Repoxy = Rti /Asubstrate = 1.89 × 10−2 K/W) and that due to conduction through the alumina substrate. The thermal resistance to conduction through the substrate equals the onedimensional resistance through the substrate [R1d = t/(kA)] plus the constriction resistance associated with conduction into the pellets. Each pellet is surrounded by four lines of symmetry forming a corresponding unit cell from which the constriction resistance can be calculated. Approximating the square unit cell
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Precision Temperature Control Using a Thermoelectric Module 37.12
Temperature Control
and square pellets as concentric circles with equivalent crosssectional areas, it follows that the pellet radius a and the unit cell radius b are 0.79 and 1.42 mm, respectively. Then the constriction resistance associated with conduction through the alumina in each unit cell may be calculated from the results of Yovanovich et al. [6]. The result is that the constriction resistance equals 2.30◦ C/W.1 The corresponding onedimensional resistance is 3.29◦ C/W. Since the 142 pellets are thermally in parallel, it follows that the overall conductance between the control point and control side of the TEM (Kcp−c ) is Kcp−c =
1 = 17.17 W/K Rep + (R1d + Rc )/142
The overall conductance between the usi and its local ambient (Ku−∞,u) may be calculated analogously. The thermal resistance to conduction through the uncontrolledside substrate (including spreading) is identical to that through the controlledside substrate (including constriction). This conduction resistance is in series with the thermal resistances of the grease layer between the uncontrolledside substrate and the heat sink and the heat sink itself. The result is that Ku−∞,u = 0.936 W/K (A). When the external leads of the TEM are opencircuited, the current I through the TEM equals 0. Setting I = 0 and simultaneously solving the surface energy balances [Eqs. (37.10) through (37.12)] results in Tc , Tu, and T∞,u values of 74.42, 47.11, and 36.43◦ C, respectively. When the external leads of the TEM are shortcircuited together but no external power is supplied to the TEM, the voltage across the TEM as given by Eq. (37.9) must equal 0. The values of I, Tc , Tu, and T∞,u which result from the simultaneous solution of the surface energybalance equations subject to this constraint are 0.436 A, 74.42◦ C, 58.58◦ C, and 47.90◦ C, respectively. It is noted that when the external leads of the TEM are shortcircuited together, both the voltage across them and the . rate of external electrical work done on the TEM ( W TEM ) equal 0. However, the voltage constraint must. be used to determine the state of the TEM rather than the fact that W TEM = 0. Otherwise, the opencircuit
1 The boundary condition at the thermal epoxyalumina interface was taken to be the heat flux distribution corresponding to an almost isothermal boundary condition ( = −1/2 in the Yovanovich et al. [6] nomenclature), and the aluminapellet interface was assumed to be isothermal. The constriction resistance calculation was done by a spreadingresistance calculator available on the University of Waterloo’s Microelectronics Heat Transfer Laboratory (MHTL) Website [7]. It is noted that Yovanovich [8] recomme