Question 12.9: A plane layer of gray medium with constant properties is ori...
A plane layer of gray medium with constant properties is originally at a uniform temperature T_0. The attenuation coefficient is β, and the layer thickness is D. The heat capacity of the medium is c_υ and its density is ρ. At time t = 0 the layer is placed in surroundings at zero temperature. Neglecting conduction and convection, discuss the solution for the transient temperature profiles for cooling only by radiation when β is very large. Contrast this with the solution for β being very small as in Section 12.2.
At the layer center, x = D/2 (boundaries are at x = 0 and D), the symmetry condition provides that at any time ∂T/∂x = 0 for x = D/2. At t = 0 for any x, T = T_0. For only radiation being included, there is a temperature jump at the boundaries x = 0, D, so the boundary temperatures are finite rather than being equal to the zero outside temperature.
For large β the diffusion approximation can be employed, and from Equation 12.30 the heat flux in the x-direction is
q(x)=-\frac{4\sigma }{3\beta (x)}\frac{d(T^4)}{dx}d\lambda =-\frac{16}{3\beta (x)} \sigma T^3\frac{dT}{dx} (12.30)
q(x,t)=-\frac{4}{3\beta }\frac{\partial E_b(x,t)}{\partial x } =-\frac{4\sigma}{3\beta } \frac{\partial T^4(x,t)}{\partial x }
Energy conservation gives −∂q(x,t)/∂x = ρc_υ ∂T /∂t. Combining these two equations to eliminate q gives the transient energy diffusion equation for the temperature distribution in the layer with uniform attenuation coefficient:
\rho c_v\frac{\partial T}{\partial t} =\frac{4\sigma}{3\beta }\frac{\partial^2 T^4(x,t)}{\partial x^2}
Defining dimensionless variables as \widetilde{t}=\beta \sigma T_0^3t/\rho c_v,\tau =\beta x, and \vartheta =T/T_0 gives
\frac{\partial \vartheta }{\partial \widetilde{t} } =\frac{4}{3}\frac{\partial^2 \vartheta ^4(\tau ,\widetilde{t} ) }{\partial\tau ^2 }
The initial condition and the symmetry condition at x = D/2 are ϑ(τ,0) = 1 and \partial \vartheta /\partial \widetilde{t}\mid _{\tau =kD,\widetilde{t} } =0. At the boundary τ = βD, a slip condition must be used. For surroundings that are empty space at zero temperature, E_{bw} = 0 and \epsilon_{w}=1, so that at the exposed boundary of the medium for any time Equation 12.37 gives
E_{\lambda ,b2} – E_{\lambda ,bw2}=\left\lgroup\frac{1}{\epsilon _{\lambda ,w2}}-\frac{1}{2} \right\rgroup(q_{\lambda ,z})_2-\frac{1}{2\beta ^2_\lambda } \left\lgroup\frac{\partial^2E_{\lambda b}}{\partial z^2}+\frac{1}{2} \frac{\partial^2E_{\lambda b}}{\partial y^2}+\frac{1}{2} \frac{\partial^2E_{\lambda b}}{\partial x^2} \right\rgroup_2 (12.37)
\sigma T^4\mid _{x=D}=\frac{1}{2} \left\lgroup-\frac{4\sigma }{3\beta }\frac{\partial T^4 }{\partial x} \right\rgroup_{x=D}-\frac{\sigma }{2\beta ^2} \frac{\partial^2 T^4 }{\partial x^2} \mid _{x=D}
or
0=\left\lgroup2\vartheta ^4+\frac{4}{3}\frac{\partial \vartheta ^4}{\partial \tau }+\frac{\partial^2 \vartheta ^4}{\partial \tau^2} \right\rgroup_{\tau =\beta D}
Similar relations apply at x = 0. For these conditions, a numerical solution is necessary. For comparison, for the limit of a small β an analytical solution is obtained in Example 12.5.
Viskanta and Bathla (1967) obtained numerical solutions to the transient energy equation, along with limiting solutions. Results for an optical thickness of βD = 2 are in Figure 12.8 for transient profiles in one-half of the symmetric layer. The results in this section have been for the important limit where radiation is the dominant mode of energy transfer; conduction and convection have been neglected. Combining heat conduction with radiation is now considered; convection combined with radiation is treated later.
