Question 11.5: A plane layer with thickness from x = 0 to D is initially at...
A plane layer with thickness from x = 0 to D is initially at T_i(x) and is in vacuum. The layer is gray and is composed of a translucent porous solid material that emits, absorbs, and scatters radiation. Scattering is isotropic, and the thermal conductivity of the layer is a constant, k. The absorption and scattering coefficients, κ and σs are uniform within the layer. The surfaces of the porous layer are assumed nonreflecting. For time t > 0 an internal heat generation q′′′(x,t) is applied. The surrounding environment for x < 0 is at T_{e,1}, and for x > D it is at T_{e,2}. Provide the energy relations to solve for the transient temperature distribution within the layer, T(x,t).
From Equation 10.1 the energy equation is
\rho c_p\frac{DT}{Dt} =\beta T\frac{DP}{Dt} +∇⋅(k∇T-q_r)+\dot{q}+\Phi _d (10.1)
\rho c\frac{\partial T}{\partial t} =k\frac{\partial^2T}{\partial x^2} -\frac{\partial q_r}{\partial x} +\dot{q} (x,t)
The ∂q_r /∂x is found from Equation 10.50 for isotropic scattering and, for a gray material,
-\nabla q_r(S)=4\pi \frac{k(\tau )}{\omega (\tau )}[\widehat{I}(\tau )-I_b(\tau ) ] =4\frac{k(S)}{\omega (S)}[\pi \widehat{I}(S)-\sigma T^4(S) ] (10.50)
\frac{{\partial q_r}}{{\partial x}} =4\pi \frac{k}{\omega }\left[\frac{\sigma T^4}{\pi } (x,t)-\widehat{I}(x,t) \right]
so the energy equation becomes
\rho c\frac{\partial T}{\partial t} =k\frac{\partial^2T}{\partial x^2} -4\frac{k}{\omega }[\sigma T^4(x,t)-\pi \widehat{I}(x,t) ] +\dot{q} (x,t)
Let T_{i,0} be an arbitrary reference temperature such as T_i(0, 0) and ϑ = T/T_{i,0},\tau =(k+\sigma _s)x and \tau_D =(k+\sigma _s)D. The energy equation in dimensionless form is then
\frac{\partial \vartheta }{\partial \widetilde{t} } =4N\frac{\partial^2 \vartheta }{\partial \tau ^2}-\frac{4(1-\omega )}{\omega } [\vartheta ^4(\tau ,\widetilde{t} )-\widetilde{I}(\tau ,\widetilde{t} )+\widetilde{q}(\tau ,\widetilde{t} ) ] (11.110)
where
\widetilde{t}=\frac{(k+\sigma _s)\sigma T_{i,0}^3}{\rho c} t ; N=\frac{k(k+\sigma _s)}{4\sigma T_{i,0}^3} ; \widetilde{I}=\frac{\pi \widehat{I} }{\sigma T_{i,0}^4}; \widetilde{q}=\frac{\dot{q} }{(k+\sigma _s)\sigma T_{i,0}^4}
The equation for \widetilde{I} is obtained from Equation 11.31, where for nonreflecting boundaries I^+(0,μ)σT^4_{e,1}/π and I^-(k_D,-μ)=σT^4_{e,2}/π, with the surroundings assumed to act as black radiation sources; this gives
\widehat{I}(\tau)=(1-\omega ) \frac{\sigma T^4(\tau )}{\pi } +\frac{\omega }{2} \left[\int_{\mu =0}^{1}{I^+}(0,\mu )e^{-\tau /\mu }d\mu +\int_{\mu =0}^{1}{I^-}(\tau _D,-\mu )e^{-(\tau _D-\tau )/\mu } d\mu +\int_{\tau ^*=0}^{\tau _D}{\widehat{I} (\tau ^*)E_1(\left|\tau ^*-\tau \right| )}d\tau ^* \right] (11.31a)
\widehat{I}(\tau)=(1-\omega ) \frac{\sigma T^4(\tau )}{\pi } +\frac{\omega }{2\pi } \left[J_1E_2(\tau ) +J_2E_2(\tau _D-\tau )+\pi \int_{\tau ^*=0}^{\tau _D}{\widehat{I}(\tau ^*)E_1(\left|\tau ^*-\tau \right| )d\tau ^* } \right] (11.31b)
\widetilde{I}(\tau,\widetilde{t} )=(1-\omega ) \vartheta ^4(\tau ,\widetilde{t} ) +\frac{\omega }{2} \left[\vartheta _{e,1}^4E_2(\tau ) +\vartheta _{e,2}^4E_2(\tau _D-\tau )+ \int_{0}^{\tau _D}{\widetilde{I} (\tau ^*,\widetilde{t} )E_1(\left|\tau ^*-\tau \right| )d\tau ^* } \right] (11.111)
Equations 11.110 and 11.111 are solved numerically subject to the boundary conditions ∂ϑ(0,\widetilde{t})/∂τ = 0,∂ϑ(τ_D,\widetilde{t})/∂τ=0 (no heat conduction into surrounding vacuum), and the specified initial condition ϑ(τ,0)=T_i(x)/T_{i,0}. Numerical solution methods are in Chapter 13.