Consider a wedge cavity as shown in Fig. 10.28 with a wedge angle of 60◦ such that the enclosure has the shape of an equilateral triangle. Consider two cases:

Question

1. Each surface of the wedge is at a uniform temperature of 500 K
2. The temperature of each surface of the wedge varies from 500 K at the vertex to a temperature of 400 K at the tip.
The opening of the wedge may be treated as a black body at 300 K. The wedge surfaces are gray having equal emissivities of [latex]\varepsilon=0.6[/latex]. Obtain the heat loss from the wedge per unit wedge length with uniform and non-uniform radiosity assumptions. Draw conclusions from the results. Solve the integral equation by a a numerical method such as the Gauss–Seidel method (refer to Chap. 7) after writing the integral as a sum over strips. Obtain all the required angle factors by the use of the triangle rule.

Accepted Answer

(1) Uniform wedge temperature case:

Comparisons may be based on the improvement made with reference to the two-surface enclosure model familiar to us already, applied to the case where the wedge surfaces are at a uniform temperature of 500 K. The opening is represented as surface 0 while the wedge surfaces are represented by a single surface 1. In this model, the angle factors are obtained by angle factor algebra and are given by

[latex]F_{11}=0.5[/latex] ; [latex]F_{10}=0.5[/latex] ; [latex]F_{01}=1[/latex] and [latex]F_{00}=0[/latex]

The radiosity of the wedge surface is given by

[latex]J_{1}=\frac{0.6 imes 5.67 imes 10^{-8} imes 500^{4}+(1-0.6) 5.67 imes 10^{-8} imes 300^{4}}{1-(1-0.6) 0.5}=2428.2 W / m ^{2}[/latex]

Based on Eq. 10.58, the heat flux leaving the cavity is given by

[latex]Q_{a}=q_{a} A_{a}=-A_{a} J_{s}=-A_{a} \sigma T_{s}^{4}\left[\frac{\varepsilon_{s}}{1-\left(1-\varepsilon_{s} ight)\left(1-\frac{A_{a}}{A_{s}} ight)} ight][/latex] (10.58)

[latex]q=\frac{0.6 imes 5.67 imes 10^{-8} imes\left(500^{4}-300^{4} ight)}{1-(1-0.6) imes 0.5}=2313.4 W / m ^{2}[/latex]

(2) Variable wedge temperature case:

If we apply the two-surface model to this case, the temperature of each surface of the wedge is taken as the mean between the apex and the tip of the wedge and is given by [latex]T_{1}=450 K[/latex] . The radiosity of the wedge surface is given by

[latex]J_{1}=\frac{0.6 imes 5.67 imes 10^{-8} imes 450^{4}+(1-0.6) 5.67 imes 10^{-8} imes 300^{4}}{1-(1-0.6) 0.5}=1514.2 W / m ^{2}[/latex]

The heat loss from the wedge is then given by

[latex]q=\frac{0.6 imes 5.67 imes 10^{-8} imes\left(450^{4}-300^{4} ight)}{1-(1-0.6) imes 0.5}=1399.3 W / m ^{2}[/latex]

The case where the wedge temperature varies significantly along its length cannot be solved with the two-surface enclosure model since it is a gross approximation. The solution may be improved by considering the radiosity variation by dividing the wedge to a number of strips each of which is considered isothermal. For example, each wedge is divided in to two strips of equal width, as an improvement over the twosurface analysis. This case is shown in Fig. 10.29 and the appropriate calculations are shown below.

All the angle factors may be calculated by the use of the triangle formula or the crossed string method. The reader is encouraged to verify the angle factors given in Table 10.5. Since [latex]J_{1}=J_{3}[/latex] and [latex]J_{2}=J_{4}[/latex] , we have only two radiosity equations that may be written down as

[latex]J_{1}=\varepsilon E_{b 1}+(1-\varepsilon)\left(F_{10} E_{b 0}+F_{13} J_{1}+F_{14} J_{2} ight)[/latex]

[latex]J_{2}=\varepsilon E_{b 2}+(1-\varepsilon)\left(F_{20} E_{b 0}+F_{23} J_{1}+F_{24} J_{2} ight)[/latex]

Table 10.5 Angle factors for configurations in Fig. 10.29

Surface No.	0	1	2	3	4
0	0	0.1830	0.3170	0.1830	0.3170
1	0.3660	0	0	0.5	0.1340
2	0.6340	0	0	0.1340	0.2321
3	0.3660	0.5	0.1340	0	0
4	0.6340	0.1340	0.2321	0	0

Introducing the numerical values and rearranging, we get

[latex]0.8 J_{1}-0.05359 J_{2}=2193.5[/latex]

[latex]-0.05359 J_{1}+0.90718 J_{2}=2242.7[/latex]

These two simultaneous equations are solved to get [latex]J_{1}=2348.7[/latex] and [latex]J_{2}=1490.6[/latex].The radiosity indeed varies significantly along the wedge length! The heat loss fromthe wedge cavity may then be obtained as

[latex]q=2\left(F_{01} J_{1}+F_{02} J_{2} ight)-E_{b 0}[/latex]

[latex]=2(0.183 imes 2348.7+0.317 imes 1490.6)-459.27[/latex]

[latex]=1345.4 W / m ^{2}[/latex]

The heat loss from the cavity has also changed significantly between the two- and the four-surface models!

Question 10.14: Consider a wedge cavity as shown in Fig. 10.28 with a wedge ...

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