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

Question

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 [latex]T_2 > T_1[/latex], 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.

Accepted Answer

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 [latex]β_1[/latex] 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 [latex]ϵ_1(β_1)[/latex] in Howell and Perlmutter (1963) was averaged over all circumferential angles for a fixed [latex]β_1[/latex]. 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 [latex]β_1[/latex] is different from the usual cone angle [latex]θ_1[/latex]. The actual emissivity [latex]ϵ_1(β_1)[/latex] of Figure 6.8b is approximated for convenience by the analytical expression [latex]ϵ_1(β_1) = 0.830 cos β_1[/latex]. Using cylindrical coordinates to integrate over all [latex]β_1[/latex], the hemispherical emissivity of this surface is

[latex]\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 [/latex]

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 [latex]\epsilon _1 = 0.652[/latex]. The energy emitted by the diffuse surface 1 per unit of the infinite length and per unit time is [latex]Q_{e,1}= 0.652σT_1^4L[/latex]. 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

[latex]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 } [/latex]

The configuration factor between infinite parallel strips, from Example 5.2, is [latex]{dF_{d1-d2}}=d(\sin \beta _1)/2 [/latex] so that

[latex]\int_{A_1}^{}{\left\lgroup\int_{A_2}^{}{{dF_{d1-d2}}} ight group }dA_1=\frac{1}{2}\int_{x=0}^{L}{(\sin \beta _{1,max}-\sin \beta _{1,min})} dx [/latex]

From Figure 6.8a, [latex]\sin β_1=(\xi -x)/[(\xi -x)^2+D^2]^{1/2} [/latex], which gives

[latex]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}} ight]dx } [/latex]

[latex]= 0.652σT^4_2[(L^2+D^2)^{1/2}-D] [/latex]

The net energy gained by surface 1, [latex]Q_{a,1} − Q_{e,1},[/latex] 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)

[latex]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 ight] [/latex]

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 [latex]α_1(β_1) = \epsilon_1(β_1) [/latex] for a gray surface,

[latex]Q_{a,1}=\sigma T^4_2\int_{A_2}^{}{\int_{A_1}^{}{\alpha _1}\left(\beta_1 ight) =\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} } [/latex]

[latex]=\frac{0.830\sigma T^4_2}{4} \int_{x=0}^{L}{\left[\frac{D(L-x)}{x^2-2xL+L^2+D^2}+ an ^{-1}\left\lgroup\frac{L-x}{D} ight group +\frac{xD}{x^2+D^2} + an ^{-1}\frac{x}{D} ight]dx } [/latex]

[latex]=\frac{0.830\sigma T^4_2}{2} an ^{-1}\frac{L}{D} [/latex]

The absorption efficiency of the directional surface is then

[latex]Eff_{diffuse} =\frac{Q_{a,1} − Q_{e,1}}{\sigma T_2^4L}=\frac{0.830}{2} an^{-1}I-0.652\left\lgroup\frac{T_1}{T_2} ight group^4 [/latex]

The absorption efficiencies of the directional-gray and diffuse-gray surfaces are shown in Figure 6.9 as a function of l with [latex]({T_1}/{T_2})^4[/latex] 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, [latex]Eff_{directional}[/latex] and [latex]Eff_{diffuse}[/latex] 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.

Question 6.5: Two parallel isothermal plates of infinite length and finite...

Related Answered Questions

A cylindrical cavity open at one end has a specularly reflecting cylindrical wall and base (Figure 6.28a). Determine the fraction of radiation from ring element dX1 that reaches dX by means of one reflection from the base with reflectivity ρs,1 and one reflection from the cylindrical wall with ...

Verified Answer:

An enclosure with three sides is shown in Figure 6.24b. The length L is long enough that the triangular ends can be neglected in the radiative heat balances. Two surfaces are black, and the third is a gray diffuse emitter with ϵ1 = 0.05. ...

Verified Answer:

A black surface A1 faces a smaller parallel mirror A2 as shown in Figure 6.22. Compute the configuration factor F1−1(2) between A1 and the image of A1 formed by one specular reflection in A2. The surfaces are infinitely long in the direction normal to the plane of the drawing. ...

Verified Answer:

Two infinite parallel plates as shown in Figure 6.15a are specular reflectors, but have emissivities (and hence absorptivities) that depend on angle θ (but not on Φ) such as for polished metals in Section 3.5.3. What is the equation for radiative exchange between the surfaces? ...

Verified Answer:

The infinitely long (normal to the cross section) groove shown in Figure 6.19 has specularly reflecting sides that emit diffusely. What fraction of the energy emitted from A2 reaches the black receiver surface element dA3 (an infinitely long differential strip)? Express the result in terms ...

Verified Answer:

For the previous example, to illustrate the insulating ability of a vacuum bottle, how long will it take for the coffee to cool from 368 to 322 K if the only heat loss is by radiation? ...

Verified Answer:

A spherical vacuum bottle consists of two silvered, concentric glass spheres, the inner being 15 cm in diameter and the evacuated gap between the spheres being 0.65 cm. The emissivity of the silver coating is 0.02. If hot coffee at 368 K is in the bottle and the outside temperature is 294 K, ...

Verified Answer:

An enclosure consists of three diffuse surfaces of finite width and infinite length normal to the cross section shown in Figure 6.5. One surface is concave. The radiative properties of each surface ...

Verified Answer:

For an example that can be carried out in analytical form, a small area element dA1 is considered on the axis of, and parallel to, a black circular disk as shown in Figure 6.14. The element is at T1, the disk at T2, and the environment ...

Verified Answer:

Consider the geometry of Figure 6.5. Total energy flux is supplied to two of the three surfaces at rates q1, q2, and T3 is known. Determine q3 and the unknown surface temperatures. ...

Verified Answer: