Question 6.5: Two parallel isothermal plates of infinite length and finite...
Two parallel isothermal plates of infinite length and finite width L are arranged as shown in Figure 6.8a. The upper plate is black, while the lower is a highly reflective gray material with parallel deep grooves of open angle 1° in its surface extending along the infinite direction. Such a surface might be constructed by stacking polished razor blades. The surroundings are at zero temperature. Compute the net energy gain by the directional surface if T_2 > T_1, and compare the result to the net energy gain if the directional-gray surface is replaced by a diffuse-gray surface with an emissivity equivalent to the hemispherical emissivity of the directional surface.
In Howell and Perlmutter (1963), the directional emissivity was calculated at the opening of an infinitely long groove with specularly reflecting walls of surface emissivity 0.01. This is given by the dot-dashed line in Figure 6.8b. The angle β_1 is measured from the normal of the opening plane of the grooved surface and is in the cross-sectional plane perpendicular to the length of the groove as in Figure 6.8a. The ϵ_1(β_1) in Howell and Perlmutter (1963) was averaged over all circumferential angles for a fixed β_1. Thus, it is an effective emissivity for radiation from a strip on the grooved surface to a parallel infinitely long strip element on an imaginary semicylinder over the groove and with its axis parallel to the grooves. The angle β_1 is different from the usual cone angle θ_1. The actual emissivity ϵ_1(β_1) of Figure 6.8b is approximated for convenience by the analytical expression ϵ_1(β_1) = 0.830 cos β_1. Using cylindrical coordinates to integrate over all β_1, the hemispherical emissivity of this surface is
\epsilon _1=\frac{\int_{-\frac{\pi}{2} }^{\frac{\pi}{2} }{\epsilon _1(\beta _1)\cos \beta _1d\beta _1} }{\cos \beta _1d\beta _1}=0.0830\int_{0}^{\frac{\pi}{2} }{\cos^2 \beta _1d\beta _1}=0.652
and this is the dashed line shown in Figure 6.8b.
The energy gained by surface 1 will first be determined when surface 2 is black and surface 1 is diffuse with \epsilon _1 = 0.652. The energy emitted by the diffuse surface 1 per unit of the infinite length and per unit time is Q_{e,1}= 0.652σT_1^4L. Since surface 2 is black, none of this energy is reflected back to surface 1.
The energy per unit length and time emitted by black surface 2 that is absorbed by surface 1 is
Q_{a,1}=\alpha _1\sigma T_2^4\int_{A_2}^{}{\int_{A_1}^{}{dF_{d2-d1}dA_2}=\epsilon _1\sigma T^4_2 }\int_{A_1}^{}{\int_{A_2}^{}{dF_{d1-d2}}dA_1 }
The configuration factor between infinite parallel strips, from Example 5.2, is {dF_{d1-d2}}=d(\sin \beta _1)/2 so that
\int_{A_1}^{}{\left\lgroup\int_{A_2}^{}{{dF_{d1-d2}}} \right\rgroup }dA_1=\frac{1}{2}\int_{x=0}^{L}{(\sin \beta _{1,max}-\sin \beta _{1,min})} dx
From Figure 6.8a, \sin β_1=(\xi -x)/[(\xi -x)^2+D^2]^{1/2} , which gives
Q_{a,1}=\epsilon _1\sigma T^4_2\frac{1}{2}\int_{x=0}^{L}{\left[\frac{L-x}{(x^2-2xL+L^2+D^2)^{1/2}}+\frac{x}{(x^2+D^2)^{1/2}} \right]dx }
= 0.652σT^4_2[(L^2+D^2)^{1/2}-D]
The net energy gained by surface 1, Q_{a,1} − Q_{e,1}, divided by the energy emitted by surface 2, is a measure of the efficiency of the surface as a directional absorber. For surface 1, being diffuse, this ratio is (I = L/D)
Eff_{diffuse} =\frac{Q_{a,1} − Q_{e,1}}{\sigma T_2^4L}=\frac{0.652}{I}\left[(1+I^2)^{1/2}-1-\frac{T^4_1}{T_2^4} I\right]
When surface 1 is a directional (grooved) surface, the amount of energy emitted from surface 1 is the same as for a diffuse surface (although it has a different directional distribution) since both have the same hemispherical emissivity. The energy absorbed by the grooved surface is, by using α_1(β_1) = \epsilon_1(β_1) for a gray surface,
Q_{a,1}=\sigma T^4_2\int_{A_2}^{}{\int_{A_1}^{}{\alpha _1}\left(\beta_1\right) =\epsilon _1dF_{d2-d1}dA_2 }=\frac{0.830\sigma T^4_2}{4} \int_{x=0}^{L}{\int_{\beta _{1,min}}^{\beta _{1,max}}{\cos ^2\beta _1d\beta _1dx} }
=\frac{0.830\sigma T^4_2}{4} \int_{x=0}^{L}{\left[\frac{D(L-x)}{x^2-2xL+L^2+D^2}+\tan ^{-1}\left\lgroup\frac{L-x}{D} \right\rgroup +\frac{xD}{x^2+D^2} +\tan ^{-1}\frac{x}{D} \right]dx }
=\frac{0.830\sigma T^4_2}{2}\tan ^{-1}\frac{L}{D}
The absorption efficiency of the directional surface is then
Eff_{diffuse} =\frac{Q_{a,1} − Q_{e,1}}{\sigma T_2^4L}=\frac{0.830}{2}\tan^{-1}I-0.652\left\lgroup\frac{T_1}{T_2} \right\rgroup^4
The absorption efficiencies of the directional-gray and diffuse-gray surfaces are shown in Figure 6.9 as a function of l with ({T_1}/{T_2})^4 as a parameter. The Eff for the directional surface is higher than that for the diffuse surface for all values of l. As l approaches zero, the configuration approaches that of infinite elemental strips, and emission from surface 1 becomes much larger than absorption from surface 2. Thus, Eff_{directional} and Eff_{diffuse} are nearly equal since the surfaces always emit the same amount. As l approaches infinity, the directional effects are lost. At intermediate l a 10% difference in absorption efficiency is attainable.
