By using the additive approximation, obtain a relation for the energy transfer from a gray infinite wall at T1 with emissivity ϵ1 to a parallel infinite gray wall at T2 with emissivity ϵ2. The spacing between the walls is D, and the region between the walls is filled with gray medium having

Question

By using the additive approximation, obtain a relation for the energy transfer from a gray infinite wall at [latex]T_1[/latex] with emissivity [latex]ϵ_1[/latex] to a parallel infinite gray wall at [latex]T_2[/latex] with emissivity [latex]ϵ_2.[/latex] The spacing between the walls is D, and the region between the walls is filled with gray medium having a constant absorption coefficient κ, isotropic scattering coefficient σs, and thermal conductivity k. Use the diffusion approximation for the radiative transfer.

Accepted Answer

The energy flux by only conduction from surface 1 to surface 2 is [latex]q_c = k(T_1 −T_2 )/ D .[/latex] The diffusion solution for only radiation from 1 to 2 is in Table 12.2

TABLE 12.2 Diffusion-Theory Predictions of Energy Transfer and Temperature Distribution for an Absorbing-Emitting Gray Gas with Isotropic Scattering in Radiative Equilibrium between Gray Walls, and without Internal Heat Sources
Geometry	[latex]Relations^a[/latex]
	[latex]\psi =\frac{1}{(3\beta D/4)+\overline{E}_1 +\overline{E}_2+1 } [/latex] [latex]\phi (z)=\psi \left[\frac{3\beta }{4}(D-z) +\overline{E}_2+\frac{1}{2} ight] [/latex]
Infinitely long concentric cylinders	[latex]\psi = \frac{1}{\frac{3}{8} \left[\beta D_1\ln \left(\frac{D_2}{D_1} ight)+\frac{1-(D_1/D_2)^2}{kD_1} ight]+(\overline{E}_1+\frac{1}{2} )+\frac{D_1}{D_2} (\overline{E}_2+\frac{1}{2} ) } [/latex] [latex]\phi (r)=\psi \left\{-\frac{3}{8} \left[\beta D_1\ln \left(\frac{D}{D_2} ight)+\frac{D_1}{kD^2_2} ight]+\left\lgroup \overline{E}_2+\frac{1}{2} ight group \frac{D_1}{D_2} ight\} [/latex]
Concentric spheres	[latex]\psi = \frac{1}{\frac{3}{8} \left[\beta D_1\ln \left(1-\frac{D_1}{D_2} ight)+2\frac{1-(D_1/D_2)^3}{\beta D_1} ight]+(\overline{E}_1+\frac{1}{2} )+\frac{D_1^2}{D_2^2} (\overline{E}_2+\frac{1}{2} ) } [/latex] [latex]\phi (r)=\psi \left\{-\frac{3}{8} \left[\beta D_1 \left\lgroup\frac{D_1}{D_2} -\frac{D_1}{D} ight group +\frac{2D_1^2}{\beta D^3_2} ight]+\left\lgroup \overline{E}_2+\frac{1}{2} ight group \frac{D_1^2}{D_2^2} ight\} [/latex]
ª Definitions: [latex]\overline{E} _N=(1-\epsilon _{wN})/\epsilon _{wN} ,\psi= Q_1/\sigma (T^{4}_{w1}- T^{4}_{w2}),\phi (\xi )=[T^4(\xi )-T^4_{w2}]/(T^4_{w1}-T^4_{w2}),D=2r,\beta =k+\sigma _s[/latex] note, ϕ is valid only if κ > 0.

as [latex]q_r=\sigma (T_1^4-T_1^4) /(3\beta D/4+1/\epsilon _1+1/\epsilon _2-1).[/latex] Since the two energy transfers are assumed independent, the additive solution gives q = qc + qr. Using the dimensionless quantities [latex]N_1=k\beta /4\sigma T_1^2,\vartheta =T/T_1[/latex] and [latex]τ_D = (κ + σ_s)D = βD[/latex] gives the combined flux as

[latex]\frac{q}{\sigma T_1^4} =\frac{4N_1(1-\vartheta _2)}{ au _D}+\frac{1-\vartheta _2^4}{3 au _D/4+1/\epsilon _1+1/\epsilon _2-1} [/latex] (12.50)

Equation 12.50 is correct for [latex]N_1 = 0[/latex] (radiation only) within the accuracy of the diffusion solution, and for [latex]N_1 → ∞[/latex] (conduction only), because the solution adds these two limiting results. Examples 11.1 and 11.2 show that superposition provides the exact solution for the limits of a medium with only scattering or an optically thin medium.

Question 12.10: By using the additive approximation, obtain a relation for t...

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