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## Q. 12.3

What is the temperature distribution in the medium for the conditions in Example 12.2?

## Verified Solution

For radiative equilibrium without internal heat sources (no conduction or convection) $\nabla.q_r=\int_{0}^{\infty }{(dq_{r\lambda }/dx)d\lambda }=0.$ Then, from Equation 12.11,

$\frac{dq_{r\lambda }}{d\tau _\lambda } =-2(J_{\lambda ,1}+J_{\lambda ,2})+4\pi \widehat{I}_\lambda (\tau _\lambda )$
$\frac{dq_{r\lambda }(x)}{dx } =2k_\lambda (x)\left\{4\pi I_{\lambda b}(x)-(J_{\lambda ,1}+J_{\lambda ,2}) \right\}$                      (12.11)

$\int_{\lambda =0}^{\infty }{} k_\lambda (x)\left[2\pi I_{\lambda b}(x)-(J_{\lambda ,1}+J_{\lambda ,2})\right] d\lambda =0$

Substituting Equation 12.12 yields

$J_{\lambda ,1}=\frac{\epsilon _{\lambda ,1}E_{\lambda b,1}+\epsilon _{\lambda ,2}E_{\lambda b,2}(1-\epsilon _{\lambda ,1})}{1-(1-\epsilon _{\lambda ,1})(1-\epsilon _{\lambda ,2})}$                      (12.12a)
$J_{\lambda ,2}=\frac{\epsilon _{\lambda ,2}E_{\lambda b,2}+\epsilon _{\lambda ,1}E_{\lambda b,1}(1-\epsilon _{\lambda ,2})}{1-(1-\epsilon _{\lambda ,1})(1-\epsilon _{\lambda ,2})}$                    (12.12b)

$2\pi \int_{\lambda =0}^{\infty }{} k_\lambda (x)I_{\lambda b}[T(x)]d\lambda =\int_{\lambda =0}^{\infty }{}k_\lambda (x)\frac{2(\epsilon _{\lambda ,1}E_{\lambda b,1}+\epsilon _{\lambda ,2}E_{\lambda b,2})-\epsilon _{\lambda ,1}\epsilon _{\lambda ,2}(E_{\lambda b,1}+E_{\lambda b,2})}{\epsilon _{\lambda ,1}+\epsilon _{\lambda ,2}-\epsilon _{\lambda ,1}\epsilon _{\lambda ,2}} d\lambda$                    (12.14)

This can be solved for T(x) by iteration, noting that $κ_λ$ can be a function of T and x; Equation 12.14 reduces to Equation 12.7 when $ϵ_{λ,1}=ϵ_{λ,2}=1.$

$\int_{\lambda =0}^{\infty }{k_{\lambda }(x)I_{\lambda b}}[T(x)]d\lambda=\frac{1}{2} \int_{\lambda =0}^{\infty }{k_{\lambda }(x)[I_{\lambda b}(T_1)+I_{\lambda b}(T_2)]d\lambda }$              (12.7)

If all properties are independent of both wavelength and temperature, Equation 12.14 reduces to the uniform temperature

$T^4=\frac{1}{2}\frac{2(\epsilon _1T_1^4+\epsilon _2T_2^4)-\epsilon _1\epsilon _2(T_1^4+T_2^4)}{\epsilon _1+\epsilon _2-\epsilon _1\epsilon _2}$