The fin temperature distribution in the array of fins in Figure 7.9 is considered in Example 7.5 and is governed by the dimensionless energy equation (Equation 7.11b)

Question

[latex]-\mu \frac{d^2\vartheta (X)}{dX^2}+\vartheta ^4(X)=F_{dX-B}+\int_{Z=0}^{1}{\left[-\mu (1-\epsilon )\frac{d^2\vartheta (Z)}{dZ^2}+\vartheta ^4(Z) \right]dF_{dX-dZ} } [/latex] (7.11b)

[latex]-\mu \frac{d^2\vartheta (X)}{dX^2}+\vartheta ^4(X)=F_{dX-B}+\int_{Z-0}^{1}{\left[-\mu (1-\epsilon )\frac{d^2\vartheta (Z)}{dZ^2}+\vartheta ^4(Z) \right]dF_{dX-dZ} }(X,Z) [/latex] (7.37)

where

[latex]F_{dX-B}(X)=\frac{1}{2}\left[1-\frac{X}{(B^2+X^2)^{1/2}} \right] [/latex]

[latex]dF_{dX-dZ}(X,Z)=\frac{1}{2}\frac{B^2}{[B^2+(Z-X)^2]^{3/2}}dZ[/latex]

[latex]μ = ka /2\epsilon σT^3_b W^2[/latex] and [latex]B =b/W[/latex]

The boundary conditions are [latex]ϑ = 1[/latex] at the fin base X = 0, and [latex]dϑ/dX = 0[/latex] at X = 1 as it is assumed for simplicity that the end edge of each fin has negligible energy loss. To evaluate the integral on the right-hand side of Equation 7.37, the entire distributions of temperature and its second derivative must be known. Hence, if this equation is written at each of a set of X values, each equation will involve the unknown [latex]ϑ(X)[/latex] at all of the X values, and all of the equations must be solved simultaneously.

Accepted Answer

To proceed with the solution of Equation 7.37, the fin is divided into N small elements, so that [latex]X_i=i\Delta X [/latex] and [latex]Z_j=j\Delta Z [/latex] , where 0 ≤ i ≤ N and 0 ≤ j ≤ N. The second derivative is approximated by

[latex]\frac{d^2\vartheta }{dX^2}=\frac{\vartheta _{i+1}-2\vartheta _i+\vartheta _{i-1}}{(\Delta X)^2} [/latex]

The integral on the right-hand side of Equation 7.37 can be approximated using the trapezoidal rule (Equation 7.31) as

[latex]\int_{z_0}^{z_N}{f(x,z)}dz\approx \Delta z\left[\frac{1}{2}f_0(x)+\sum\limits_{j=1}^{N-1}{f_j(x)+\frac{1}{2}f_N(x) } ight] [/latex] (7.31)

[latex]\int_{0}^{1}{f(X,Z)}dZ\approx \Delta Z\left[\frac{1}{2}f_0(X)+\sum\limits_{j=1}^{N-1}{f_j(X)+\frac{1}{2}f_N(X) } ight] [/latex]

where

[latex]f(X,Z)=\frac{B^2}{2}\left[-\mu (1-\epsilon )\frac{d^2\vartheta }{dZ^2} +\vartheta ^4(Z) ight]\frac{1}{[B^2+(Z-X)^2]^{3/2} } [/latex]

Using ΔZ = ΔX and substituting the finite-difference form of the second derivative results in

[latex]f_j(X_i)=\frac{B^2}{2}\left[-\mu (1-\epsilon )\frac{\vartheta _{j+1}-2\vartheta _j+\vartheta _{j-1}}{(\Delta X)^2}+\vartheta ^4_j ight]\frac{1}{\left\{B^2+[(j-i)\Delta X]^2 ight\}^{3/2} } [/latex]

The two limits where j = 0 and j = N are evaluated by applying the boundary conditions. For [latex]j = i = 0, ϑ_0 = 1[/latex] and [latex]d^2 ϑ/dZ^2=0.[/latex] The latter condition is needed because the [latex]ϑ_{i−1} [/latex] term would otherwise be undefined; the condition arises because the energy entering the fin at the base is all by conduction, so the temperature gradient is linear at that location. Then, for j = 0,

[latex]f_0(X_i)=\frac{B^2}{2[B^2+(i\Delta X)^2]^{3/2}} [/latex]

For [latex]j = N, ϑ_{N+1} = ϑ_{N− 1}[/latex] from [latex](d\vartheta /dx)_{X=1}=(\vartheta _{N-1}-\vartheta _{N-1})/(2\Delta X)=0,[/latex] where N + 1 is a symmetric image point of N − 1. Then

[latex]f_N(X_i)=\frac{B^2}{2}\left[-\mu (1-\epsilon )\frac{2(\vartheta _{N-1}-\vartheta_N)}{(\Delta X)^2}+\vartheta ^4_N ight]\frac{1}{\left\{B^2+[(N-i)\Delta X]^2 ight\}^{3/2} } [/latex]

The energy equation (Equation 7.37) for element i at [latex]X_i = i ΔX[/latex] can now be written in finitedifference form as

[latex]-\mu \frac{\vartheta _{i+1}-2\vartheta _{i-1}}{(\Delta X)^2}+\vartheta ^4_1=\frac{1}{2}\left\{\frac{1}{\left\{[B/(i\Delta X)]^2+1 ight\}^{1/2} } ight\} +\Delta X\left[\frac{1}{2}f_0(X_i)+\sum\limits_{j=1}^{N-1}{f_j(X_i)+\frac{1}{2}f_N(X_i) } ight] [/latex] (7.38)

This equation is written for each element i for the range 1 ≤ i ≤ N, giving N equations for the N unknown temperatures [latex]ϑ_i[/latex] . Each equation contains every unknown [latex]ϑ_i[/latex] , which appear in the [latex]f_j[/latex] terms. If the resulting set of equations is written in matrix form, the coefficient matrix is full. This is in contrast to one-dimensional pure conduction problems where the coefficient matrix is usually tridiagonal. This illustrates that in radiative transfer problems the temperature of every element can be influenced by the temperature of all of the surrounding elements. Equation 7.38 is written for element i = 1 to give

[latex]-\mu \frac{\vartheta _{2}-2\vartheta _{1}+1}{(\Delta X)^2}+\vartheta ^4_1=\frac{1}{2}\left\{1-\frac{1}{\left\{[B/(\Delta X)]^2+1 ight\}^{1/2} } ight\} +\frac{\Delta XB^2}{2} \left\lgroup\frac{1}{2}\frac{1}{[B^2+(\Delta X)^2]^{3/2}} ight group -\sum\limits_{j=1}^{N-1}{\left[\mu (1-\epsilon )\frac{\vartheta _{j+1}-2\vartheta _j+\Theta _{j-1}}{(\Delta X)^2} -\Theta _j^4 ight] } \frac{1}{\left\{B^2+[(j-1)\Delta X]^2 ight\}^{3/2} } +\frac{1}{2}\left[-\mu (1-\epsilon )\frac{2(\vartheta _{N-1}-\vartheta_N)}{(\Delta X)^2}+\Theta ^4_N ight]\frac{1}{\left\{B^2+[(N-1)\Delta X]^2 ight\}^{3/2} } [/latex] (7.39)

For the infinite array of fins, [latex] ϑ_i = ϑ_{j =1} [/latex] so that, gathering terms that have [latex]ϑ_j [/latex] and [latex]ϑ_ j^ 4 [/latex] and using [latex]ϑ_0 = 1, [/latex] Equation 7.39 is written as (a repeated subscript denotes a summation over the values of that subscript, e.g., [latex]A _{1j} ϑ_j=\sum\limits_{j=1}^{N}{A _{1j} ϑ_j}) [/latex]

[latex]A _{1j} ϑ_j+{B _{1j} ϑ_j^4}=C_1 [/latex] (7.40)

where [latex][let \ \widetilde{B}\equiv B^2\mu (1-\epsilon ) /2\Delta X ] [/latex]

[latex]A_{11}=\widetilde{B}\left\{\frac{4}{(1-\epsilon )\Delta XB^2}-\frac{2}{B^3}+\frac{1}{[B^2+(\Delta X)^2]^{3/2}} ight\} [/latex]

[latex]A_{12}=\widetilde{B}\left\{\frac{-2}{(1-\epsilon )\Delta XB^2}+\frac{1}{B^3}-\frac{2}{[B^2+(\Delta X)^2]^{3/2}}+ \frac{1}{[B^2+(2\Delta X)^2]^{3/2}} ight\} [/latex]

.

[latex]A_{1j}=\widetilde{B} \left\lgroup \frac{1}{\left\{B^2+[(j-2)\Delta X]^2 ight\}^ {3/2}}-\frac{2}{\left\{B^2+[(j-1)\Delta X]^2 ight\}^ {3/2}}+\frac{1}{\left[B^2+(j\Delta X)^2 ight]^{3/2} } ight group [/latex] 2< j< N-1

.

[latex]A_{1(N-1)}=\widetilde{B} \left\lgroup \frac{1}{\left\{B^2+[(N-3)\Delta X]^2 ight\}^ {3/2}}-\frac{2}{\left\{B^2+[(N-2)\Delta X]^2 ight\}^ {3/2}}+\frac{1}{\left\{B^2+[(N-1)\Delta X]^2 ight\}^{3/2} } ight group [/latex]

[latex]A_{1N}=\widetilde{B} \left\lgroup \frac{1}{\left\{B^2+[(N-2)\Delta X]^2 ight\}^ {3/2}}-\frac{2}{\left\{B^2+[(N-1)\Delta X]^2 ight\}^ {3/2}} ight group [/latex]

[latex]B_{1j}=\delta _{1j}-\frac{B^2\Delta X}{\beta _j\left\{B^2+[(j-1)\Delta X]^2 ight\}^{3/2} } [/latex] ; [latex]\begin{matrix} \beta _j=2, \ 1\leq j\lt N \ \beta _j=4, j=N \end{matrix} [/latex]

[latex]C_1=\frac{1}{2} \left\{1-\frac{\Delta X}{[B^2+(\Delta X)^2]^{1/2}} ight\} +\frac{\mu }{(\Delta X)^2}+\frac{B^2(\Delta X)}{4[B^2+(\Delta X)^2]^{3/2}} -\frac{\mu (1-\epsilon )}{2B(\Delta X)}[/latex]

This is done for each element i, 1 ≤ i ≤ N, and a matrix equation is generated of the form

[latex][A_{ij}][\vartheta _j]+[B_{_{ij}}][\vartheta ^4_j]=[C_i][/latex] (7.41)

This is a set of nonlinear algebraic equations for the unknown temperatures [latex]ϑ_i = ϑ_{j=i}[/latex]. Solution methods are in Section 7.8.

Question 7.11: The fin temperature distribution in the array of fins in Fig...

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