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Chapter 12

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}