The space between two diffuse-gray spheres (Figure 12.7) is filled with an optically dense stagnant medium having constant attenuation coefficient β. For the limiting condition of radiative equilibrium compute the radiative heat flow

Question

The space between two diffuse-gray spheres (Figure 12.7) is filled with an optically dense stagnant medium having constant attenuation coefficient β. For the limiting condition of radiative equilibrium compute the radiative heat flow [latex]Q_1[/latex] across the gap from sphere 1 to sphere 2 and the temperature distribution T(r) in the medium, using the diffusion method with jump boundary conditions.

Accepted Answer

For a gray medium with constant β, Equation 12.31

[latex]q_{\lambda ,z}(r)d\lambda =-\frac{4\pi }{3\beta _\lambda (r)}\left\lgroup\frac{\partial I_{\lambda b}}{\partial z} \right\rgroup_r d\lambda =-\frac{4}{3\beta _\lambda (r)} \left\lgroup\frac{\partial E_{\lambda b}}{\partial z} \right\rgroup_rd\lambda [/latex] (12.31)

gives the net heat flux in the positive r direction as [latex]q_r = −(4/3β) dE_b/dr.[/latex] From energy conservation, with no internal heat sources, [latex]q_r[/latex] varies with r as [latex]q_r = Q_1/4πr^2.[/latex] Combine these relations and integrate from [latex]R_1[/latex] to [latex]R_2[/latex] to obtain

[latex]\frac{Q_1}{4\pi }\int_{r=R_1}^{R_2}{\frac{dr}{r^2} }=-\frac{4}{3\beta } \int_{E_b=E_{b1}}^{E_{b2}}{dE_b} [/latex] (12.43)

[latex]\frac{Q_1}{4\pi }\left\lgroup\frac{1}{R_2}-\frac{1}{R_1} \right\rgroup =\frac{4}{3\beta } (E_{b,2}-E_{b,1}) [/latex] (12.44)

The [latex]E_{b1}[/latex] and [latex]E_{b2}[/latex] are in the gas adjacent to the boundaries, and jump boundary conditions express these quantities in terms of wall values. The jump boundary conditions are in Equations 12.37

[latex]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 [/latex] (12.37)

and 12.38 and involve second derivatives that are now found. By integrating Equation 12.43 from [latex]R_1[/latex] to r,

[latex]E_{\lambda b,w1} - E_{\lambda ,b1}=\left\lgroup\frac{1}{\epsilon _{\lambda ,w1}}-\frac{1}{2} \right\rgroup(q_{\lambda ,z})_1+\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_1 [/latex] (12.38)

[latex]E_b(r)-E_{b,1} =\frac{3\beta Q_1}{16\pi }\left\lgroup\frac{1}{r} -\frac{1}{R_1} \right\rgroup [/latex] (12.45)

Substitute [latex]r = (x^2 + y^2 + z^2)^{1/2} [/latex] and differentiate twice with respect to x to obtain

[latex]\frac{\partial^2 E_b(r)}{\partial x^2}=-\frac{3\beta Q_1}{16\pi } \frac{(x^2+y^2+z^2)^{3/2}-3x^2(x^2+y^2+z^2)^{1/2}}{(x^2+y^2+z^2)^3} [/latex] (12.46)

and similarly for the y- and z-directions.

In the boundary condition Equation 12.37, point 2 can be conveniently taken in Figure 12.7 at x = y = 0 and [latex]z = R_2.[/latex] This gives

[latex]\left[\frac{\partial^2 E_b(r)}{\partial x^2}\right]_2 =\left[\frac{\partial^2 E_b(r)}{\partial y^2}\right]_2=-\frac{3\beta Q_1}{16\pi } \frac{1}{R_2^3};\left[\frac{\partial^2 E_b(r)}{\partial z^2}\right]_2 =\frac{3\beta Q_1}{8\pi } \frac{1}{R_2^3}[/latex]

Also, [latex](q_z )_2= Q_1 /4π R^2_2[/latex] Substituting into Equation 12.37 gives

[latex]E_{b,2} - E_{\lambda ,bw2}=\left\lgroup\frac{1}{\epsilon _{w2}}-\frac{1}{2} \right\rgroup \frac{Q_1}{4\pi R^2_2}-\frac{3Q_1}{32\beta \pi } \frac{1}{R^3_2} [/latex] (12.47)

Similarly, at the inner sphere boundary, from Equation 15.38,

[latex]Q_{s,\lambda } =\frac{8\xi ^4 }{3z^2_1}\left\{[(n^2+k^2)^2+n^2-k^2-2]^2+36n^2k^2\right\} \left\{1+\frac{6}{5z_1}[(n^2+k^2)^2-4]\xi ^2-\frac{8nk}{z_1}\xi ^3 \right\} [/latex] (15.38)

[latex]E_{b,w1} - E_{b,1}=\left\lgroup\frac{1}{\epsilon _{w1}}-\frac{1}{2} \right\rgroup \frac{Q_1}{4\pi R^2_1}+\frac{3Q_1}{32\beta \pi } \frac{1}{R^3_1} [/latex] (12.48)

Adding Equations 12.47 and 12.48 gives

[latex] E_{b,2}- E_{b,1}=E_{b,w2}-E_{b,w1}+ \frac{Q_1}{4\pi}\left[\left\lgroup\frac{1}{\epsilon _{w2}}-\frac{1}{2} \right\rgroup \frac{1}{R^2_2} +\left\lgroup\frac{1}{\epsilon _{w1}}-\frac{1}{2} \right\rgroup \frac{1}{R^2_1}+\frac{3}{8\beta } \left\lgroup\frac{1}{R^3_1}-\frac{1}{R^3_2}\right\rgroup \right] [/latex]

After substituting this into the right side of Equation 12.44, the result is solved for [latex]Q_1 [/latex] to give the ψ in the last entry 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} \right] [/latex]
Infinitely long concentric cylinders	[latex]\psi = \frac{1}{\frac{3}{8} \left[\beta D_1\ln \left(\frac{D_2}{D_1} \right)+\frac{1-(D_1/D_2)^2}{kD_1} \right]+(\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} \right)+\frac{D_1}{kD^2_2} \right]+\left\lgroup \overline{E}_2+\frac{1}{2} \right\rgroup \frac{D_1}{D_2} \right\} [/latex]
Concentric spheres	[latex]\psi = \frac{1}{\frac{3}{8} \left[\beta D_1\ln \left(1-\frac{D_1}{D_2} \right)+2\frac{1-(D_1/D_2)^3}{\beta D_1} \right]+(\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}\right\rgroup +\frac{2D_1^2}{\beta D^3_2} \right]+\left\lgroup \overline{E}_2+\frac{1}{2} \right\rgroup \frac{D_1^2}{D_2^2} \right\} [/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.

To obtain the temperature distribution, integrate Equation 12.43 from [latex]R_2[/latex] to r to obtain [latex]E_b(r)-E_{b,2}=(3\beta Q_1/16\pi )(1/r-1/R_2).[/latex] Add Equation 12.47 to eliminate [latex]E_{b,2}:[/latex]

[latex]E_{b}(r) - E_{b,w2}=\frac{3\beta Q_1}{16\pi } \left\lgroup\frac{1}{r}-\frac{1}{R_2} \right\rgroup +\left\lgroup\frac{1}{\epsilon _{w2}}-\frac{1}{2} \right\rgroup \frac{Q_1}{4\pi R^2_2}-\frac{3Q_1}{32\beta \pi } \frac{1}{R^3_2} [/latex] (12.49)

This gives the last expression for ϕ in Table 12.2. The ϕ is valid only if κ > 0, since temperatures are indeterminate in the limit of pure scattering when there is only radiative transfer.

Question 12.8: The space between two diffuse-gray spheres (Figure 12.7) is ...

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