A typical example of an enclosure with gray and diffuse walls is shown in Fig. 10.20. The enclosure is a rectangular parallelepiped of length 1 m, width 0.5 m, and height 0.5 m. The top face identified as surface 1 is supplied with a heat input of 50 kW. The bottom face identified as surface 2 is m

Question

A typical example of an enclosure with gray and diffuse walls is shown in Fig. 10.20. The enclosure is a rectangular parallelepiped of length 1 m, width 0.5 m, and height 0.5 m. The top face identified as surface 1 is supplied with a heat input of 50 kW. The bottom face identified as surface 2 is maintained at a temperature of 500 K. The
right face identified as surface 3 is maintained at a temperature of 600 K. The other three sides (the front, the left, and the back faces) are specified as reradiating surfaces and hence are clubbed together as a single surface 4 (it is concave). All surfaces are specified to have a common emissivity of 0.8. Determine the temperatures of those surfaces for which the heat fluxes are specified and the heat fluxes at the surfaces for which the temperatures are specified.

Accepted Answer

The solution is in two parts. In the first geometric part, the required angle factors are evaluated. The second part uses the radiosity irradiation formulation to obtain surface radiosities. The third part uses these radiosities to get the required answers.

1. Geometric part:

All the surfaces are either rectangles or consist of rectangles and hence the areas are easily calculated and tabulated as shown in Table 10.2. There are 4 surfaces and hence the total number of angle factors is [latex]4^{2} = 16[/latex]. However, there are [latex]4_{C_{2} } =\frac{4! }{2!2!} = 6[/latex] reciprocity relations. Four expressions due to the sum rule are available. Out of 4 surfaces, 3 are flat and hence self-angle factors for these are zero. That is, there are 3 angle factors, which are known. Because of symmetry [latex]F_{13} = F_{23}[/latex]. Thus, the number of angle factors that have to be independently determined is 16 − 6 − 4 − 3 − 1 = 2. [latex]F_{12} and F_{31}[/latex] are identified as the two angle factors that need to be independently

Table 10.2 Areas in Example 10.8

surface	[latex]Area m^{2}[/latex]
1	1×0.5=0.5
2	1×0.5=0.5
3	0.5×0.5=0.25
4	1×0.5+1×0.5+0.5×0.5=1.25

determined. Surfaces 1 and 2 are two parallel and equal rectangles. Using the notation of Fig. G.1, we have a = 1 m, b = 0.5 m, and c = 0.5 m. Hence, we have

[latex]X=\frac{a}{c}=\frac{1}{0.5}=2; Y=\frac{b}{c}=\frac{0.5}{0.5} =1[/latex]

From Fig. G.2, we read off the required angle factor as [latex]F_{12} = 0.28[/latex]. However, if we use the results of Table G.1, the value is [latex]F_{12} = 0.28588 ≈ 0.286[/latex]. Surfaces 3 and 1 are two perpendicular rectangles sharing a common edge. Using the notation of Fig. G.1, we have a = 1 m, b = 0.5 m, and c = 0.5 m. Hence, we have

[latex]Z=\frac{a}{b}=\frac{1}{0.5}=2; Y=\frac{c}{b}=\frac{0.5}{0.5} =1 or \acute{Y}=\frac{1}{Y}=1[/latex]

Table G.1 Angle factors for diffuse radiation interchange between equal parallel rectangles (continued)

X	Y
X	0.1	2	3	4	5	6	7	8	9	10
0.1	0.03119	0.03514	0.03965	0.04209	0.0436	0.04463	0.04537	0.04592	0.04636	0.04671
0.2	0.06186	0.06971	0.07868	0.08353	0.08654	0.08858	0.09005	0.09116	0.09202	0.09272
0.3	0.09152	0.10318	0.1165	0.12372	0.1282	0.13123	0.13342	0.13507	0.13636	0.13739
0.4	0.11979	0.13513	0.15268	0.16219	0.16809	0.17209	0.17497	0.17715	0.17885	0.18021
0.5	0.14641	0.16527	0.18687	0.19859	0.20586	0.21079	0.21435	0.21703	0.21913	0.22081
0.6	0.17123	0.19342	0.21888	0.23271	0.2413	0.24712	0.25132	0.25449	0.25697	0.25896
0.7	0.19417	0.21951	0.24864	0.26447	0.27431	0.28098	0.2858	0.28943	0.29227	0.29455
0.8	0.21527	0.24356	0.27614	0.29387	0.3049	0.31238	0.31778	0.32186	0.32504	0.32759
0.9	0.2346	0.26564	0.30147	0.321	0.33314	0.34139	0.34734	0.35184	0.35535	0.35816
1	0.25226	0.28588	0.32474	0.34596	0.35917	0.36813	0.37461	0.3795	0.38332	0.38638
1.5	0.32006	0.36405	0.41541	0.44366	0.46131	0.47331	0.48199	0.48855	0.49367	0.49779
2	0.36405	0.41525	0.47558	0.50899	0.52993	0.54421	0.55454	0.56236	0.56847	0.57338
3	0.41541	0.47558	0.54738	0.58762	0.613	0.63037	0.64297	0.65251	0.65998	0.66599
4	0.44366	0.50899	0.58762	0.63204	0.6602	0.67954	0.69359	0.70426	0.71261	0.71933
5	0.46131	0.52993	0.613	0.6602	0.69024	0.71093	0.72599	0.73743	0.7464	0.75363
6	0.47331	0.54421	0.63037	0.67954	0.71093	0.73258	0.74838	0.76039	0.76982	0.77741
7	0.48199	0.55454	0.64297	0.69359	0.72599	0.74838	0.74673	0.77717	0.78695	0.79482
8	0.48855	0.56236	0.65251	0.70426	0.73743	0.76039	0.77717	0.78995	0.8	0.80811
10	0.49779	0.57338	0.66599	0.71933	0.75363	0.77741	0.79482	0.80811	0.81856	0.82699

From Table G.4, we read off the required angle factor as [latex]F_{31} = 0.23285 ≈ 0.233[/latex]. All other angle factors are obtained by any one of the following:
1. Symmetry condition
2. Reciprocity relation
3. Sum rule

Table G.4 Angle factors for diffuse radiation interchange between perpendicular rectangles sharing a common edgeTable G.4 Angle factors for diffuse radiation interchange between perpendicular rectangles sharing a common edge

[latex]\acute{Y}[/latex]	Z
[latex]\acute{Y}[/latex]	1.5	2	3	4	5	6	7	8	9	10
0.1	0.43616	0.43756	0.4386	0.43897	0.43915	0.43925	0.4393	0.43934	0.43937	0.43939
0.15	0.41391	0.41599	0.41755	0.41812	0.41838	0.41853	0.41861	0.41867	0.41871	0.41874
0.2	0.39435	0.39711	0.39919	0.39993	0.40029	0.40048	0.40059	0.40067	0.40072	0.40076
0.25	0.37678	0.3802	0.38279	0.38372	0.38416	0.3844	0.38455	0.38464	0.38471	0.38475
0.3	0.36079	0.36486	0.36795	0.36907	0.3696	0.36988	0.37006	0.37017	0.37025	0.3703
0.4	0.33253	0.33784	0.34191	0.34339	0.34409	0.34447	0.3447	0.34486	0.34496	0.34503
0.5	0.30814	0.3146	0.31961	0.32145	0.32232	0.3228	0.32309	0.32327	0.3234	0.3235
0.6	0.28678	0.29429	0.30019	0.30238	0.30342	0.30399	0.30433	0.30456	0.30471	0.30482
0.7	0.26789	0.27632	0.28306	0.28559	0.28679	0.28745	0.28786	0.28812	0.2883	0.28843
0.8	0.25104	0.26029	0.26781	0.27066	0.27203	0.27278	0.27324	0.27354	0.27375	0.27389
0.9	0.23594	0.24588	0.25412	0.25729	0.25881	0.25966	0.26017	0.2605	0.26074	0.2609
1	0.22232	0.23285	0.24176	0.24522	0.2469	0.24783	0.2484	0.24877	0.24903	0.24921
1.5	0.17077	0.18286	0.19415	0.19889	0.20126	0.2026	0.20343	0.2097	0.20435	0.20462
2	0.13715	0.1493	0.16169	0.16731	0.17024	0.17193	0.17299	0.1737	0.17419	0.17455
3	0.09708	0.1078	0.12019	0.1266	0.13023	0.13245	0.13388	0.13486	0.13555	0.13606
4	0.07458	0.08365	0.09495	0.10136	0.10526	0.10776	0.10944	0.11062	0.11147	0.1121
5	0.06038	0.06809	0.07814	0.0842	0.0881	0.09072	0.09254	0.09385	0.09481	0.09554
6	0.05065	0.05731	0.06622	0.07184	0.0756	0.07821	0.08009	0.08147	0.08251	0.08331
7	0.04359	0.04943	0.05738	0.06254	0.0661	0.06865	0.07053	0.07194	0.07302	0.07386
8	0.03825	0.04343	0.05057	0.05531	0.05865	0.0611	0.06294	0.06435	0.06545	0.06632
10	0.03069	0.03491	0.04082	0.04484	0.04777	0.04998	0.0517	0.05306	0.05414	0.05502

Surface 1:
Since it is flat, the self-angle factor [latex]F_{11} = 0[/latex]. The angle factor [latex]F_{12}[/latex] has already been obtained. By the use of the reciprocity rule, the angle factor between 1 and 3 is obtained as

[latex]F_{13}=\frac{A_{3}}{A_{1}}F_{31}=\frac{0.5 imes 0.5}{1 imes 0.5} imes 0.233=0.116[/latex]

Hence, the sum rule will give the remaining view factor as

[latex]F_{14}=1-F_{11}-F_{12}-F_{13}-1-0-0.286-0.116=0.598[/latex]

Surface 2:
Again [latex]F_{22} = 0[/latex] because the surface is flat. By symmetry [latex]F_{21} = F_{12} = 0.286[/latex]. Also, by symmetry [latex]F_{23} = F_{13} = 0.116[/latex]. The sum rule then gives

[latex]F_{24} = 1 − F_{22}− F_{21} − F_{23}= 1 − 0 − 0.286 − 0.116 = 0.598[/latex]

Surface 3:
[latex]F_{33}= 0[/latex] since it is flat. By symmetry [latex]F_{32} = F_{31} = 0.233[/latex]. Lastly, by the sum rule we get

[latex]F_{34} = 1 − F_{33} − F_{32} − F_{31} = 1 − 0 − 0.233 − 0.233 = 0.534[/latex]

Surface 4:
By reciprocity, we have

[latex]F_{41}=\frac{A_{1} }{A_{4}} F_{14}=\frac{0.5 }{1.25} imes 0.598=0.239[/latex]

By symmetry [latex]F_{42} = F_{41} = 0.239[/latex]. By reciprocity

[latex]F_{43}=\frac{A_{3} }{A_{4}} F_{34}=\frac{0.25 }{1.25} imes 0.534=0.107[/latex]

Finally, by the sum rule we get

[latex]F_{44}= 1 − F_{41} − F_{42} − F_{43} = 1 − 0.239 − 0.239 − 0.107 = 0.415[/latex]

The angle factor data is conveniently given in the form of a matrix. The two entries shown in bold face are the only values obtained independently, i.e., without the use of angle factor algebra.

[latex]\left [ \begin{matrix} F_{11} & F_{12} & F_{13} & F_{14} \ F_{21} & F_{22} & F_{23} & F_{24} \ F_{31} & F_{32} & F_{33} & F_{34} \ F_{41} & F_{42} & F_{43} & F_{44} \end{matrix} ight ] =\left [ \begin{matrix} 0 & 0.286 & 0.116 & 0.598 \ 0.286 & 0 & 0.116 & 0.598 \ 0.233 & 0.233 & 0 & 0.534 \ 0.239 & 0.239 & 0.107 & 0.415 \end{matrix} ight ][/latex]

2. Radiosity formulation and solution:
Surface 1: Heat flux specified
With [latex]Q_{1} = 50 kW and A_{1}= 0.5 m^{2}[/latex], the heat flux at surface 1 is

[latex]q_{1}=\frac{Q_{1}}{A_{1}}=\frac{50000}{0.5}=100000 W/m^{2}[/latex]

But [latex]q_{1}= J_{1} − G_{1}[/latex] and hence

(a) [latex]q_{1} = J_{1} − G_{1} = J_{1} − 0.286J2 − 0.116J3 − 0.598J4 = 100000[/latex]

Surface 2: Temperature specified
Temperature has been specified on this surface as [latex]T_{2} = 500 K[/latex] whose emissivity is [latex]ε_{2} = 0.8[/latex]. Hence, the emissive power of this surface is

[latex]E_{2}=\varepsilon_{2}\sigma T^{4}_{2}=0.8 imes 5.67 imes 10^{-8} imes500^{4}=2835 W/m^{2}[/latex]

The irradiation on this surface is

[latex]G_{2}=F_{21}J_{1}+F_{23}J_{3}+F_{24}J_{4}=0.286J_{1}+0.116J_{3}+0.598J_{4}[/latex]

The radiosity equation for surface 2 is then written down as

[latex]J_{2} = E_{2} + (1 − ε_{2})G_{2} = 2835 + (1 − 0.8)(0.286J_{1} + 0.116J_{3}+ 0.598J_{4})[/latex]

This simplifies to

[latex](b) − 0.057J_{1} + J_{2} − 0.023J_{3} − 0.120J_{4} = 2835[/latex]

Surface 3: Temperature specified
Temperature has been specified on this surface as [latex]T_{3} = 600 K[/latex] whose emissivity is [latex]ε_{3} = 0.8[/latex]. Hence, the emissive power of this surface is

[latex]E_{3} = ε_{3}σ T^{4}_{2} = 0.8 × 5.67 × 10^{-8} × 600^{4} = 5878.7 W/m^{2}[/latex]

The irradiation on this surface is

[latex]G_{3} = F_{31} J_{1}+ F_{32} J_{2} + F_{34} J_{4} = 0.233J_{1} + 0.233J_{2}+ 0.534J_{4}[/latex]

The radiosity equation for surface 3 is then written down as

[latex]J_{3} = E_{3}+ (1 − ε_{2})G_{3} = 5878.7 + (1 − 0.8)(0.233J_{1} + 0.233J_{2} + 0.534J_{4})[/latex]

This simplifies to

[latex](c) − 0.047J_{1} − 0.047J_{2} + J_{3} − 0.107J4 = 5878.7[/latex]

Surface 4: Reradiating surface
Since this surface is reradiating, the heat flux [latex]q_{4}= 0 and hence J_{4} = G_{4}[/latex]. This may be written as

[latex]J_{4} = G_{4} = F_{41} J_{1} + F_{42} J_{2} + F_{43} J_{3} + F_{44} J_{4}= 0.239J_{1} + 0.239J_{2} + 0.107J_{3} + 0.415J_{4}[/latex]

This may be rearranged as

[latex](d) − 0.239J_{1} − 0.239J_{2} − 0.107J_{3}+ 0.585J_{4} = 0[/latex]

The four equations (a) − (d) are solved simultaneously to obtain the four radiosities. One possible way of obtaining the solution is to eliminate [latex]J_{4}[/latex] and reduce the number of equations to three for the radiosities of the other three surfaces. The three equations may be solved by Kramer’s rule. Another method of solution would be to use the Gauss method of iteration. This has been discussed earlier while dealing with numerical methods for solving heat conduction problems. The iteration would start with assumed radiosities for surfaces 1 to 3. For surfaces 2 and 3, one starts with [latex]J_{2} = E_{2} and J_{3} = E_{3}[/latex]. For surface 1 the starting value would be [latex]J_{1} = q_{1}[/latex]. Equation (d) would then be used to get the first guess for [latex]J_{4}[/latex]. The iteration would stop when the radiosities have converged to the desired level. This procedure was indeed used to arrive at the following radiosity values:

[latex]J_{1} = 153765.6; J_{2} = 21169.4; J_{3} = 22096 and J_{4} = 75526 all in W/m^{2}[/latex]

3. Required answers:
Now, we are in a position to evaluate the desired answers set out in the problem. Applying Eq. 10.50 to surface 1, the unknown temperature [latex]T_{1}[/latex] is obtained as

[latex]T_{i}={ \left[\frac{J_{i}}{\sigma }+\frac{1-\varepsilon_{i} }{\varepsilon_{i}}.\frac{q_{i}}{\sigma} ight] }^{\frac{1}{4} }[/latex] (10.50)

[latex]T_{i}={ \left[\frac{\frac{(1-0.8 imes 100000)}{0.8}+153765.6 }{5.67 imes 10^{-8} } ight]^{\frac{1}{4} }=1332.5 K }[/latex]

We apply Eq. 10.51 to surface 4 to obtain the unknown temperature [latex]T_{4}[/latex] as

[latex]T_{i}={ \left[ \frac{J_{i}}{\sigma } ight] }^{\frac{1}{4} }[/latex] (10.51)

[latex]T_{4}={ \left[\frac{75526}{5.67 imes 10^{-8} } ight]^{\frac{1}{4} }=1074.3 K }[/latex]

Applying Eq. 10.49 to surface 2, we get the unknown heat flux [latex]q_{2}[/latex] as

[latex]q_{i}=J_{i}-\frac{J_{i}-\varepsilon _{i}\sigma T^{4}_{i} }{1-\varepsilon _{i}}=\frac{\varepsilon _{i}}{1-\varepsilon _{i}}.\left[\sigma T^{4}_{i}-J_{i} ight][/latex] (10.49)

[latex]q_{2}=\frac{0.8}{1-0.8} imes (5.67 imes 10^{-8} imes 500^{4}-21169.4 )=-70502.7 W/m^{2}[/latex]

Hence, the heat transfer from surface 2 is given by

[latex]Q_{2} = q_{2} A_{2} = −70502.7 × (1 × 0.5) = −35251.4 W[/latex]

Similarly, we get [latex]q_{3}[/latex] as

[latex]q_{3}=\frac{0.8}{1-0.8} imes (5.67 imes 10^{-8} imes 600^{4}-22096)=-58990.9 W/m^{2}[/latex]

Hence, the heat transfer from surface 3 is given by

[latex]Q_{3} = q_{3}A_{3} = −58990.9 × (0.5 × 0.5) = −14747.7 W[/latex]

We may check the calculations by performing power balance for the enclosure. Thus

[latex]\sum\limits_{i=0}^{4}{q_{i}A_{i} }=50000-35251.4-14747.7=0.9 W[/latex]

Indeed the calculations have been performed with sufficient precision!

Question 10.9: A typical example of an enclosure with gray and diffuse wall...

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