Use the emission approximation to find the energy flux emerging from an isothermal gas layer at Tg with an integrated mean absorption coefficient weighted by the blackbody spectrum at Tg

Question

Use the emission approximation to find the energy flux emerging from an isothermal gas layer at [latex]T_g[/latex] with an integrated mean absorption coefficient weighted by the blackbody spectrum at [latex]T_g[/latex] (the “Planck mean” in Section 12.6.1) of [latex]κ_P = 0.010 cm^{−1}[/latex] and thickness D = 1.5 cm, if the layer is bounded by transparent nonradiating walls and cold surroundings (Figure 12.1a).

Accepted Answer

If I(θ) is the total intensity emerging from the layer in direction θ, the emerging flux is [latex]q=2\pi \int_{ heta =0}^{\pi /2}{I( heta )}\cos heta \sin heta d heta . [/latex] The layer is isothermal with constant temperature [latex]T_g[/latex] so, for the inner integral of Equation 12.16, the quantities are independent of S* and the Planck mean of

[latex]I(S)=\int_{S^*=0}^{S}{\left[\int_{\lambda =0}^{\infty }{k_\lambda } (S^*)I_{\lambda b}(S^*)d\lambda ight] }dS^* [/latex] (12.16)

[latex]k_\lambda(T_g)[/latex] gives [latex]\pi \int_{\lambda =0}^{\infty }{\kappa _{\lambda }(T_{g})I_{\lambda b}(T_{g})d\lambda } =\kappa _{p}(T_{g})\sigma T^{4}_{g}[/latex].Then Equation 12.16 can be integrated over any path [latex]S = D/cos θ[/latex] through the layer to yield [latex]I (θ) =(1/\pi )\kappa _P(T_g)\sigma T_g^4D/\cos heta .[/latex] This gives the radiative flux from each boundary as

[latex]q=2\int_{ heta =0}^{\pi /2}{\kappa _P(T_g)\sigma T_g^4D\sin heta d heta} =2\kappa_P(T_g)\sigma T_g^4D[/latex] (12.17)

which gives [latex]q=0.03\sigma T_g^4[/latex] for the specified numerical values. This is not a precise result, even though the layer thickness is optically thin: [latex]κ_PD = 0.015 \ll 1.[/latex] This is because the radiation reaching the layer boundary along each path has passed along a distance D/cosθ. For θ approaching π/2 the optical path length becomes very large, so the emission approximation cannot hold.

A more accurate solution including effects of the proper path lengths gives [latex]q = 1.8κ_P(T_g)\sigma T_g^4D[/latex] (see Section 10.8.4), which is a 10% decrease compared with Equation 12.17.

Question 12.4: Use the emission approximation to find the energy flux emerg...

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