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

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 Q_1 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.

12.7
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For a gray medium with constant β, Equation 12.31

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           (12.31)

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

\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}                               (12.43)

\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})                 (12.44)

The E_{b1} and E_{b2} 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

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                 (12.37)

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

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                       (12.38)

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

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

\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}                        (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 z = R_2. This gives

\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}

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

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}               (12.47)

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

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\}          (15.38)

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}             (12.48)

Adding Equations 12.47 and 12.48 gives

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]

After substituting this into the right side of Equation 12.44, the result is solved for Q_1 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 Relations^a
\psi =\frac{1}{(3\beta D/4)+\overline{E}_1 +\overline{E}_2+1 } 
\phi (z)=\psi \left[\frac{3\beta }{4}(D-z) +\overline{E}_2+\frac{1}{2} \right]  
Infinitely long concentric
cylinders
\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}  ) } 
\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\} 
Concentric spheres \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}  ) } 
\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\} 
ª Definitions: \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   note, ϕ is valid
only if κ > 0.

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

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}               (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.

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