Question 12.10: By using the additive approximation, obtain a relation for t...
By using the additive approximation, obtain a relation for the energy transfer from a gray infinite wall at T_1 with emissivity ϵ_1 to a parallel infinite gray wall at T_2 with emissivity ϵ_2. 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.
The energy flux by only conduction from surface 1 to surface 2 is q_c = k(T_1 −T_2 )/ D . 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 | 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. |
|
as q_r=\sigma (T_1^4-T_1^4) /(3\beta D/4+1/\epsilon _1+1/\epsilon _2-1). Since the two energy transfers are assumed independent, the additive solution gives q = qc + qr. Using the dimensionless quantities N_1=k\beta /4\sigma T_1^2,\vartheta =T/T_1 and τ_D = (κ + σ_s)D = βD gives the combined flux as
\frac{q}{\sigma T_1^4} =\frac{4N_1(1-\vartheta _2)}{\tau _D}+\frac{1-\vartheta _2^4}{3\tau _D/4+1/\epsilon _1+1/\epsilon _2-1} (12.50)
Equation 12.50 is correct for N_1 = 0 (radiation only) within the accuracy of the diffusion solution, and for N_1 → ∞ (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.