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3.5 Analysis of two-dimensional, steady state systems |
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Assume a product solution in the form given in Eq. (3.116).
φ(x, y) = X(x)Y ( y)
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d2 X Y + d2Y X = 0 dx2 dy2
As before, we introduce a constant –l, as follows.
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d2 X + λ2 X = 0 dx2
The roots of the above equation are
D = ±iλ
The general solution to Eq. (3.121) is given by
X = Acos λx + Bsin λx
Consider, Eq. (3.120) which is rewritten below.
d2Y − λ2Y = 0 dy2
The roots for the above equation are
D = ±λ
The general solution to Eq. (3.124) is given by
Y = C cosh λy + Dsinh λy
The general solution to Eq. (3.115) can be written as
φ = ( Acos λx + Bsin λx)(C cosh λy + Dsinh λy)
The boundary condition for the left wall is at x = 0; φ = 0. From Eq. (3.127), we have
A = 0
(3.116)
(3.117)
(3.118)
(3.119)
(3.120)
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(3.123)
(3.124)
(3.125)
(3.126)
(3.127)
(3.128)
The boundary condition for the bottom wall is y = 0; φ = 0.
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CHAPTER 3 Conduction |
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Therefore, from Eq. (3.127), we have |
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C = 0 |
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Hence, the general solution reduces to |
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φ = Bsin(λx)Dsinh(λ y) |
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where E = BD |
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The boundary condition at the right wall is x = L; φ = 0. |
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Eq. (3.133) is satisfied by succession of λ′s given by |
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λ = nπ |
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Here n = 0,1,2,3.... |
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∞ |
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φ = ∑En sin |
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The boundary condition at the top wall is given by y = W; φ = 1. Using Eq. (3.135) and starting the expansion from n = 1, we have
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(3.136) |
n=1
We can now exploit the property of orthogonal functions which we discussed earlier. We have for a general set of orthogonal functions given by g1(x), g2(x), ..... gm(x) and gn(x) to represent a function f(x) as an infinite series with cn’s being the constants of the individual terms in the series.
∫ b |
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∫ b cn gn2 |
(x) dx |
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Now invoking Eq. (3.137) and recognising that f(x) = 1, gn(x) = sin(λnx) and cn = Ensinh(λnx). We have,
∫ L sin(λn x) dx = |
∫ L En sinh(λnW)sin2 (λn x) dx |
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(3.139)
(3.140)
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3.5 Analysis of two-dimensional, steady state systems |
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−1 ((−1)n −1)
Ennπ sinh(λnW) = λn L (3.141)
2
Ennπ sinh(λnW) = |
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nπ sinh(λnW)
(3.142)
(3.143)
(3.144)
Substituting for En in Eq. (3.135), we have the final form of solution for the dimensionless temperature φ.
∞ |
2 (1−(−1)n ) |
nπ x |
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φ = ∑ |
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sin |
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For example consider a square slab where width (W) = length (L), as shown in Fig.3.16A. Please note that now the height is L and the width, W, just to let the readers know that there is no hard and fast rule about the nomenclature. It is important, though, to remember that the periodic part of the solution is along x and the exponential part of the solution is along y. Let the top temperature be 100 °C (403.15 K) and the other three walls be at 30 °C (303.15 K). Isotherms for this example would
look like what are shown in Fig. 3.16B. Temperature symmetry around x = W2 can be clearly seen. The exponential nature of temperature across y is also evident. This
FIGURE 3.16
(A) Schematic representation of a two dimensional plane wall and (B) temperature contours.
98CHAPTER 3 Conduction
explains the reason why we chose “-λ2” in Eq. (3.120) Had we chosen “λ2” instead, we would have landed up with an exponential temperature distribution in the x-direction and a symmetric distribution in the y-direction both of which violate the physics of the problem.
Example 3.4: Consider steady, two-dimensional conduction heat transfer in a square metal plate of a constant thermophysical properties as shown in Fig. 3.17. The temperatures at the boundaries are prescribed to be 273.15 K on all the sides except at the top where the temperature is maintained at 373.15 K. Obtain the temperature from the analytical solution at nodes 1, 2, 3, and 4 with the first two nonzero terms of the series solution.
Solution:
The analytical solution for two-dimensional heat transfer in a plate is given as
∞ |
2 (1− (−1)n ) |
nπ x |
nπ y |
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sin |
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Temperature at node 1:
W = 0.3 m; L = 0.3 m; x = 0.1 m; y = 0.2 m
φ = |
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FIGURE 3.17
Schematic representation of a two-dimensional plane wall with boundary temperatures considered in example 3.4.
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3.5 Analysis of two-dimensional, steady state systems |
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Temperature at node 2:
W = 0.3 m; L = 0.3 m; x = 0.2 m; y = 0.2 m
φ = |
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W = 0.3 m; L = 0.3 m; x = 0.1 m; y = 0.1 m
φ = |
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W = 0.3 m; L = 0.3 m; x = 0.2 m; y = 0.2 m
φ = |
T( x,y) − TC |
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T( x ,y) = T4 = 285.08 K
The horizontal symmetry of the temperature distribution is confirmed with
T1 = T2 and T3 = T4.
Problems
3.1 An electronic device “(like the processor of a desktop computer) weighing
0.28kg generates 60 W of heat and reaches a temperature of 100 °C in ambient air at 30 °C under steady state conditions. The device is initially at 30 °C. Determine the temperature it will reach 6 min after the power is switched on?
Assume the device to be spatially isothermal. Thermo physical properties of
electronic chip are cp =700 J/kg.K, ρ = 2329 kg/m3, k = 130 W/m.K (Adapted and modified from Incropera et al. (2013)).
3.2 The time constant of a K-type (Chromel-Alumel) thermocouple of diameter
0.70mm has been determined to be 1 s, and the temperature indicated by the thermocouple for a time t that equals its time constant τ value is 74.24 °C and
100 CHAPTER 3 Conduction
the temperature indicated at 0.5 s is 57.54 °C. The specific heat and density of Chromel-Alumel are 420 J/kg-K and 8600 kg/m3, respectively.
Using the lumped heat capacity approach, determine
a.the temperature of the medium whose temperature needs to be measured.
b.the initial temperature of the thermocouple.
c.the rate of initial temperature change (at time t = 0 s) of the thermocouple.
3.3 A cylindrical stainless steel rod 1 cm in diameter and 20 cm in length is initially at 750 °C. It is then submerged in a bath of water at 90 °C. The heat transfer coefficient can be taken as 250 W/m2 K. The density, specific heat, and thermal con-
ductivity of the steel are ρ = 7801kg/m3 , cp = 473J/kgK, and k = 43W/mK, respectively. Determine the time required for the center of the rod to reach 300 °C. What is the key assumption/approximation required to solve the problem with the techniques presented in this chapter?
3.4 In a chocolate industry chocolates are made into spherical balls with a diameter of 2 cm and temperature of 20 °C, and they are kept in a freezer at 2 °C before packing. The chocolates have approximately the same thermophysical properties as that of water (Provided in Chapter 5), and the heat transfer coefficient is around 12 W/m2 K. What will be the temperature of the center of the chocolate after 30 min? What is the time required to bring the center temperature of the chocolate to 6 °C. Also determine the total enthalpy removed by the freezer to bring the chocolate center temperature to 10 °C.
3.5 A wall 15 cm thick, made of clay brick, is initially 100 °C. The surface temperatures of both sides of the brick are suddenly reduced to 27 °C. Find the temperature at a plane 5 cm from the surface after 90 min have passed. How much enthalpy has been lost from the brick wall during that time? (Use Heisler’s charts.)
Properties of the brick are ρ = 1625kg/m3 , cp = 840 J/kgK and k = 0.7 W/mK, α = 5.25×10−7 m2 /s.
3.6 Revisit problem 3.5, in which the two sides of the brick are suddenly exposed to a medium that is at 27 °C with a heat transfer coefficient of 80 W/m2 K. Find the temperature at a point 1.5 cm from the surface after 7 hours have passed. Determine (1) the enthalpy loss from the wall during that time and (2) the average temperature within the wall at the end of 7 hours (Use Heisler’s charts.)
3.7 In a plastic welding process, a cylindrical polypropylene rod (filler material with k = 0.5 W/mK) of diameter 3 mm is used and is initially 25 °C. The rod is suddenly exposed to a hot air jet of diameter 2 cm at 560 °C. The effective convective and radiative heat transfer coefficient (total) is 60 W/m2 K. How much time will it take for the polypropylene rod to reach its melting temperature of 160 °C? The density and specific heat of polypropylene are 920 kg/m3 and 1800 J/kg K respectively.
3.8 Stainless steel bearings (ρ = 7900 kg/m3 , cp = 477 J/kgK, and k = 15 W/mK) that have been uniformly heated to 840 °C are hardened by quenching them in an oil bath that is maintained at 40 °C. The ball diameter is 20 mm, and the associated convection coefficient is 1000 W/m2 K. If quenching is to occur until the surface temperature of the balls reaches 100 °C, how long must the balls be kept in oil? What is the center temperature at the end of the cooling period?
3.5 Analysis of two-dimensional, steady state systems 101
3.9 An iron sphere of diameter 80 mm is initially at a uniform temperature of 200 °C. It is suddenly exposed to ambient air at 30 °C with a convection coefficient of 510 W/m2 K.
a.Determine the temperature at the center of the sphere and at a depth of 5 mm from the surface at t = 1 min after the sphere is exposed to air. Also determine the average temperature of the slab at t = 1 min.
b.Calculate the enthalpy removed from the sphere in this duration.
Assume the density, specific heat, thermal conductivity, and thermal diffusivity for the sphere as 8000 kg/m3, 460 J/kgK, 60 W/mK, and 1.6 ×10−5 m2 /s, respectively.
3.10In northern India, the highest temperature of a summer day can go up to 45 °C. In places where refrigeration facilities are not available, drinking such warm water or taking baths is very unpleasant. What minimum burial depth would you recommend to the company laying water pipelines so that even in summer one can get water at a temperature not exceeding 25 °C? Assume that initially
the soil is at 20 °C, and then it is subjected to a constant surface temperature of 40 °C for 60 days. The thermal diffusivity of soil at 20 °C is 0.138 ×10−6 m2 /s.
3.11A semi-infinite slab is initially at a temperature of 100 °C. Suddenly one end of the slab is exposed to boiling water at 100 °C. An arrangement is made with a thermocouple to measure the temperature at a location 12 mm from one end. For the slab under consideration, the temperature at this location at t = 2 min is 65 °C. Determine the thermal conductivity of the material. The density and specific heat of the solid are known to be 2300 kg/m3 and 750 J/kg K, respectively.
3.12A square slab of dimensions 10 cm × 10 cm is very deep in the direction perpendicular to the plane of the paper.The slab is made of material with a thermal conductivity of k = 15 W/mK. Steady state prevails in the slab, there is no heat generation, and all the properties are assumed to be constant. The boundary conditions are given in accompanying Fig. 3.18.
a.Suggest a strategy of making use of the analytical solution presented in this chapter to determine the temperature distribution in the slab.
b.Using the strategy obtained in (a), determine the center temperature, that is, the temperature at (0.05, 0.05) using the first two non-zero terms in the series solution.
3.13Consider the problem of conduction in a square slab as given in problem 3.12. However, the boundary conditions are now different and are given in accompanying Fig. 3.19. The slab is made of material with a thermal conductivity of k = 15 W/mK. Steady state prevails in the slab, there is no heat generation, and all the properties are assumed to be constant.
a.Develop a strategy to obtain a solution to the problem of determination of temperature distribution for this problem using the analytical solution presented earlier in this chapter.
b.Hence, determine the temperature at (0.05, 0.05) using the first two nonzero terms in the series solution.
102 CHAPTER 3 Conduction
FIGURE 3.18
Two-dimensional plane wall with boundary temperatures considered in problem 3.12.
FIGURE 3.19
Two-dimensional plane wall with boundary conditions considered in problem 3.13.
References
Bergman, T.L., Incropera, F.P., DeWitt, D.P. and Lavine, A.S., 2011. Fundamentals of Heat and Mass Transfer. John Wiley and Sons, New York.