Using the diffusion method, find the steady temperature profile in a medium with constant attenuation coefficient β and thermal conductivity k, with q″ ′ = 0, and contained between infinite parallel black walls at T1 and T2 spaced D apart

Question

Using the diffusion method, find the steady temperature profile in a medium with constant attenuation coefficient β and thermal conductivity k, with q″′ = 0, and contained between infinite parallel black walls at [latex]T_1[/latex] and [latex]T_2[/latex] spaced D apart with the lower wall 1 at x = 0. What is the heat transfer across the layer?

Accepted Answer

For this geometry Equation 12.52 becomes, in dimensionless form,

[latex]q=q_r+q_c=-\left\lgroup\frac{16\sigma T^3}{3\beta _R}+k ight group abla T [/latex] (12.52)

[latex]\frac{q}{\sigma T_1^4} =-\left\lgroup\frac{4}{3} \frac{d\vartheta ^4}{d au } +4N_1\frac{d\vartheta }{d au } ight group [/latex] (12.56)

From energy conservation, with no internal heat sources, the q is constant across the layer between the walls. Equation 12.56 is then integrated from 0 to [latex]τ_D[/latex] to yield

[latex]\frac{q}{\sigma T_1^4} au _D =-\left\{\frac{4}{3}[\vartheta ^4( au _D)-\vartheta ^4(0)]+4N_1[\vartheta ( au _D)-\vartheta (0)] ight\} [/latex] (12.57)

where [latex]ϑ(0)[/latex] and [latex]ϑ(τ_D)[/latex] are in the medium at the boundaries. These two temperatures are eliminated by using the jump boundary conditions to relate them to the specified wall temperatures [latex]T_1[/latex] and [latex]T_2.[/latex]

At wall 1, for the particular [latex]N_1[/latex] of the problem, the [latex]ψ_1[/latex] is found from Figure 12.11 and set equal to [latex]\psi _1=\sigma [T_1^4-T^4(0)] /q_{r,1} [/latex]. From Equations 12.51 and 12.52 the radiative flux [latex]q_{r,1} [/latex] at the wall is

[latex]q_r=-\frac{4}{3\beta _R} abla E_b=-\frac{16\sigma T^3}{3\beta _R} abla T [/latex] (12.51)

[latex]q_{r,1}=-\frac{16\sigma T^3}{3\beta _R}\frac{dT}{dx} \mid _1=\frac{16\sigma T_1^3}{3\beta _R}\frac{q}{(16\sigma T_1^3\beta _R)+k} =\frac{4q}{3\beta _R[(4\sigma /3\beta _R)+(k/4T_1^3)]} [/latex]

Then [latex]ψ_1[/latex] from Equation 12.54 becomes

[latex]\psi _1=\frac{\sigma [T_1^4-T^4(x\longrightarrow 0)]}{q_{r,1}} [/latex] (12.54)

[latex]\psi _1=\frac{\sigma [T_1^4-T^4(0)]}{(4\sigma /3\beta _R)q/[(4\sigma /3\beta _R)+(k/4T_1^3)]} [/latex]

This is rearranged into

[latex]\frac{4}{3\beta _R} q_1=\frac{1}{\psi _1} \left\{\frac{4\sigma}{ 3\beta _R}[T_1^4-T^4(0)]+\frac{k}{4T_1^3}[T_1^4-T^4(0)] ight\} [/latex] (12.58)

As shown in the derivation of ψ1, the conditions for which the diffusion solution is valid lead to the jump [latex] T_1 – T(0)[/latex] being small. For convenience, a portion of Equation 12.58 can then be linearized. With [latex] T_1 – T(0) = δ,[/latex] where δ is small,

[latex]\frac{T_1^4-T^4(0)}{4T_1^3} =\frac{T_1^4-(T_1-\delta )^4}{4T_1^3}\approx \frac{T_1^4-T_1^4+4T_1^3\delta }{4T_1^3} =\delta =T_1-T(0) [/latex]

Then Equations 12.58 becomes

[latex]\frac{4}{3\beta _R} q_{1}\simeq \frac{1}{\psi _1} \left\{\frac{4\sigma}{ 3\beta _R}[T_1^4-T^4(0)]+k[T_1-T(0)] ight\} [/latex]

or in dimensionless form,

[latex]\frac{4}{3} \psi _1\frac{q_1}{\sigma T_1^4}=\frac{4}{3} [1-\vartheta ^4(0)]+4N_1[1-\vartheta (0)] [/latex] (12.59a)

Similarly, at wall 2 (note that [latex]ψ_2[/latex] is found by using the value of [latex]N_2 = kβ_R /4σ T_2^3 [/latex] as the abscissa in Figure 12.11),

[latex]\frac{1}{3} \psi _2\frac{q_1}{\sigma T_1^4}=\frac{1}{3} [\vartheta ^4( au _D)-\vartheta _2^4]+N_1[\vartheta ( au _D)-\vartheta _2] [/latex] (12.59b)

Now add Equations 12.57, 12.59a, and 12.59b to eliminate the unknown temperatures in the medium [latex]ϑ(0)[/latex] and [latex]ϑ(τ_D).[/latex] This yields the energy flux transferred across the layer as

[latex]\frac{q_1}{\sigma T_1^4}=\frac{[1-\vartheta _2^4]+3N_1[1-\vartheta _2]}{3 au _D/4+\psi _1+\psi _2} [/latex] (12.60)

The results of Equation 12.60 are plotted in Figure 12.12 and compared with the exact and additive solutions. For [latex] τ_D = 1,[/latex] the results compare very well with the exact solution, and Equation 12.60 will provide good results for larger [latex] τ_D[/latex] . For [latex] τ_D = 0.1[/latex] a simple additive solution provides better energy transfer values.

Question 12.11: Using the diffusion method, find the steady temperature prof...

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