Consider a two-dimensional cavity as shown in Fig. 10.27. The cavity surface is maintained at a uniform temperature of Tc while the opening is looking at a background at Tb. The cavity surface may be assumed to be gray with an emissivity of εc = 0.6. Calculate the heat transfer per unit length of

Question

Consider a two-dimensional cavity as shown in Fig. 10.27. The cavity surface is maintained at a uniform temperature of $T_{c}$ while the opening is looking at a background at $T_{b}$ . The cavity surface may be assumed to be gray with an emissivity of $\varepsilon_{c}=0.6$ . Calculate the heat transfer per unit length of cavity to the background. Model the enclosure in both the variants shown in the figure.

Table 10.4 Angle factors for configurations in Fig. 10.27

Surface	Case (a)
Number	1	2
1	0.4	0.6
2	1	0
Surface	Case (b)
Number	1	2	3	4
1	0	0.72076	0.13962	0.13962
2	0.72076	0	0.13962	0.13962
3	0.41886	0.41886	0	0.16228
4	0.41886	0.41886	0.16228	0

Table 10.4 Angle factors for configurations in Fig. 10.27

Surface	Case (a)
Number	1	2
1	0.4	0.6
2	1	0
Surface	Case (b)
Number	1	2	3	4
1	0	0.72076	0.13962	0.13962
2	0.72076	0	0.13962	0.13962
3	0.41886	0.41886	0	0.16228
4	0.41886	0.41886	0.16228	0

Accepted Answer

The analysis is presented for the two cases below. The results will be compared so that the need or otherwise of the non-uniform radiosity analysis becomes clear.

Case (a)

The analysis is very simple in this case. The radiosity of surface 1 is written down as

[latex]J_{1}=\varepsilon_{c} E_{b}\left(T_{c}\right)+\left(1-\varepsilon_{c}\right)\left(F_{11} J_{1}+F_{12} J_{2}\right)[/latex] [latex]=\varepsilon_{c} E_{b}\left(T_{c}\right)+\left(1-\varepsilon_{c}\right)\left(F_{11} J_{1}+\left[1-F_{11}\right] J_{2}\right)[/latex]

where [latex]E_{b}[/latex] represents the emissive power of a black body. This may be rearranged in the form

[latex]J_{1}-J_{2}=\varepsilon_{c}\left[E_{b}\left(T_{c}\right)-J_{2}\right]+\left(1-\varepsilon_{c}\right) F_{11}\left[J_{1}-J_{2}\right][/latex]

Representing the radiosity difference [latex]J_{1}-J_{2}[/latex] as [latex]J_{d, 1}[/latex] . noting that [latex]J_{2}=E_{b}\left(T_{b}\right)[/latex] we have

[latex]J_{d, 1}=\varepsilon_{c}\left[E_{b}\left(T_{c}\right)-E_{b}\left(T_{b}\right)\right]+\left(1-\varepsilon_{c}\right) F_{11} J_{d, 1}[/latex]

The angle factors shown in Table 10.4 will be made use of to get

[latex]J_{d, 1}=0.6\left[E_{b}\left(T_{c}\right)-E_{b}\left(T_{b}\right)\right]+0.16 J_{d, 1}[/latex] or [latex]J_{d, 1}=0.714\left[E_{b}\left(T_{c}\right)-E_{b}\left(T_{b}\right)\right][/latex]

Heat loss from the cavity is easily seen to be given by

[latex]q_{2}=J_{2}-G_{2}=J_{2}-J_{1}=-J_{d, 1}=-0.714\left[E_{b}\left(T_{c}\right)-E_{b}\left(T_{b}\right)\right][/latex]

This agrees with the value predicted by Eq. 10.58 when we note that [latex]A_{s}[/latex] there corresponds to [latex]A_{1}[/latex] and [latex]A_{a}[/latex] there corresponds to [latex]A_{2}[/latex]. The black body emissive power there is to be replaced by the difference of black body emissive powers in the present case.

[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}\right)\left(1-\frac{A_{a}}{A_{s}}\right)}\right][/latex] (10.58)

Case (b): Because of symmetry. [latex]J_{3}=J_{4}[/latex] and hence only three radiosity equations are to be written down. We may again define radiosity differences as [latex]J_{d, i}=J_{i}-J_{2}[/latex] , where i = 1, 3, or 4. Making use of the sum rule for each surface, the radiosity equations may be written down as

[latex]J_{d, 1}=\varepsilon_{c}\left[E_{b}\left(T_{c}\right)-E_{b}\left(T_{b}\right)\right]+\left(1-\varepsilon_{c}\right) \cdot 2 F_{13} J_{d, 3}[/latex]

[latex]J_{d, 3}=\varepsilon_{c}\left[E_{b}\left(T_{c}\right)-E_{b}\left(T_{b}\right)\right]+\left(1-\varepsilon_{c}\right)\left(F_{31} J_{d, 1}+F_{34} J_{d, 3}\right)[/latex]

Introducing the numerical values for the angle factors and the surface emissivity, the above equations become

[latex]J_{d, 1}-0.1117 J_{d, 3}=0.6\left[E_{b}\left(T_{c}\right)-E_{b}\left(T_{b}\right)\right][/latex]

[latex]-0.1675 J_{d, 1}+0.9351 J_{d, 3}=0.6\left[E_{b}\left(T_{c}\right)-E_{b}\left(T_{b}\right)\right][/latex]

The two equations may be solved to get

[latex]J_{d, 1}=0.6854\left[E_{b}\left(T_{c}\right)-E_{b}\left(T_{b}\right)\right][/latex] ; [latex]J_{d, 3}=0.7645\left[E_{b}\left(T_{c}\right)-E_{b}\left(T_{b}\right)\right][/latex]

The radiosities of surfaces 1 and 3 are certainly not the same and hence clubbing these, as in Case (a), is not correct.

The heat loss from the cavity may be calculated as

[latex]q_{2}=J_{2}-G_{2}=-\left(F_{21} J_{d, 1}+2 F_{23} J_{d, 3}\right)=-0.708\left[E_{b}\left(T_{c}\right)-E_{b}\left(T_{b}\right)\right][/latex]

Since the results of Case (a) and Case (b) are very close to each other, there is no need to extend the analysis any further. Since the heat loss from the cavity is our concern, the simple two-surface enclosure analysis may itself be adequate in the present case.

Question 10.13: Consider a two-dimensional cavity as shown in Fig. 10.27. Th...

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