Question 12.11: Using the diffusion method, find the steady temperature prof...
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 T_1 and T_2 spaced D apart with the lower wall 1 at x = 0. What is the heat transfer across the layer?
For this geometry Equation 12.52 becomes, in dimensionless form,
q=q_r+q_c=-\left\lgroup\frac{16\sigma T^3}{3\beta _R}+k \right\rgroup \nabla T (12.52)
\frac{q}{\sigma T_1^4} =-\left\lgroup\frac{4}{3} \frac{d\vartheta ^4}{d\tau } +4N_1\frac{d\vartheta }{d\tau } \right\rgroup (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 τ_D to yield
\frac{q}{\sigma T_1^4}\tau _D =-\left\{\frac{4}{3}[\vartheta ^4(\tau _D)-\vartheta ^4(0)]+4N_1[\vartheta (\tau _D)-\vartheta (0)] \right\} (12.57)
where ϑ(0) and ϑ(τ_D) 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 T_1 and T_2.
At wall 1, for the particular N_1 of the problem, the ψ_1 is found from Figure 12.11 and set equal to \psi _1=\sigma [T_1^4-T^4(0)] /q_{r,1} . From Equations 12.51 and 12.52 the radiative flux q_{r,1} at the wall is
q_r=-\frac{4}{3\beta _R} \nabla E_b=-\frac{16\sigma T^3}{3\beta _R}\nabla T (12.51)
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)]}
Then ψ_1 from Equation 12.54 becomes
\psi _1=\frac{\sigma [T_1^4-T^4(x\longrightarrow 0)]}{q_{r,1}} (12.54)
\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)]}
This is rearranged into
\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)] \right\} (12.58)
As shown in the derivation of ψ1, the conditions for which the diffusion solution is valid lead to the jump T_1 – T(0) being small. For convenience, a portion of Equation 12.58 can then be linearized. With T_1 – T(0) = δ, where δ is small,
\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)
Then Equations 12.58 becomes
\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)] \right\}
or in dimensionless form,
\frac{4}{3} \psi _1\frac{q_1}{\sigma T_1^4}=\frac{4}{3} [1-\vartheta ^4(0)]+4N_1[1-\vartheta (0)] (12.59a)
Similarly, at wall 2 (note that ψ_2 is found by using the value of N_2 = kβ_R /4σ T_2^3 as the abscissa in Figure 12.11),
\frac{1}{3} \psi _2\frac{q_1}{\sigma T_1^4}=\frac{1}{3} [\vartheta ^4(\tau _D)-\vartheta _2^4]+N_1[\vartheta (\tau _D)-\vartheta _2] (12.59b)
Now add Equations 12.57, 12.59a, and 12.59b to eliminate the unknown temperatures in the medium ϑ(0) and ϑ(τ_D). This yields the energy flux transferred across the layer as
\frac{q_1}{\sigma T_1^4}=\frac{[1-\vartheta _2^4]+3N_1[1-\vartheta _2]}{3\tau _D/4+\psi _1+\psi _2} (12.60)
The results of Equation 12.60 are plotted in Figure 12.12 and compared with the exact and additive solutions. For τ_D = 1, the results compare very well with the exact solution, and Equation 12.60 will provide good results for larger τ_D . For τ_D = 0.1 a simple additive solution provides better energy transfer values.
