A circular disk of diameter 1 m and at 600 K is kept directly opposite to a disk of diameter 0.5 m and 400 K. The distance between the two disks is 0.3 m. Both the disks have an emissivity of 0.6. The room in which the two disks are placed is very large and is at a temperature of 300 K.

Question

A circular disk of diameter 1 m and at 600 K is kept directly opposite to a disk of diameter 0.5 m and 400 K. The distance between the two disks is 0.3 m. Both the disks have an emissivity of 0.6. The room in which the two disks are placed is very large and is at a temperature of 300 K. Determine the heat transfer from each surface. Use electrical analogy to solve the problem.

Accepted Answer

Step 1 The nomenclature used in the problem is introduced through Fig. 10.23a. The electrical analog is shown in Fig. 10.23b. The given data is written down as

[latex]r_{1}=\frac{1}{2}=0.5 m[/latex] ; [latex]T_{1}=600 K[/latex] ; [latex]\varepsilon_{1}=0.6[/latex] ; [latex]r_{2}=\frac{0.5}{2}=0.25 m[/latex] ; [latex]T_{2}=400 K[/latex] ; [latex]\varepsilon_{2}=0.6[/latex]

Node 3 may be visualized as a black surface at the temperature of the room, with zero surface resistance.

[latex]\varepsilon_{3}=1[/latex] ; [latex]T_{3}=300 K[/latex] ; and [latex]J_{3}=E_{b 3}=\sigma T_{3}^{4}=5.67 imes 10^{-8} imes 300^{4}=459.3 W/m^2[/latex]

Step 2 Calculation of geometric parameters:

Areas of surfaces 1 and 2 are, respectively, given by [latex]A_{1}=\pi r_{1}^{2}=\pi imes0.5^2 = 0.7854 m^2[/latex] and [latex]A_{2}=\pi r_{2}^{2}=\pi imes0.25^2 = 0.1964m^2[/latex] . The angle factor requires the following non-dimensional ratios:

[latex]\frac{L}{r_{1}}=\frac{0.3}{0.5}=0.6[/latex] ; [latex]\frac{r_{2}}{L}=\frac{0.25}{0.3}=0.8333[/latex]

where L is the distance between the disks of 0.3 m. From Table G.5, the required angle factor is obtained by interpolation as [latex]F_{12}=0.175[/latex] . Since surfaces 1 and 2 are flat, [latex]F_{11}=F_{22}=0[/latex]. The sum rule applied to surface 1 gives [latex]F_{13}= 1 - F_{12} = 1 - 0.175 = 0.825[/latex] Reciprocity is used to get

Table G.5 Angle factors for diffuse radiation interchange between parallel coaxial disks

[latex]\frac{1}{R_{1}}[/latex]	[latex]R_{2}[/latex]
[latex]\frac{1}{R_{1}}[/latex]	0.2	0.4	0.6	0.8	1	1.5	2	3	4	5
0.1	0.00040	0.00158	0.00356	0.00634	0.00990	0.02227	0.03959	0.08902	0.15812	0.24672
0.2	0.00154	0.00615	0.01384	0.02459	0.03840	0.08623	0.15279	0.33944	0.58388	0.81900
0.3	0.00330	0.01320	0.02965	0.05260	0.08196	0.18242	0.31803	0.64586	0.86472	0.93590
0.4	0.00551	0.02200	0.04930	0.08713	0.13502	0.29352	0.48769	0.80000	0.91153	0.95015
0.5	0.00799	0.03179	0.07092	0.12446	0.19098	0.39754	0.60961	0.84861	0.92482	0.95492
0.6	0.01056	0.04187	0.09278	0.16121	0.24388	0.47901	0.67889	0.86822	0.93063	0.95716
0.7	0.0131	0.05169	0.11360	0.19495	0.28993	0.53614	0.71722	0.87817	0.93375	0.95841
0.8	0.01551	0.06090	0.13258	0.22442	0.32784	0.57481	0.73985	0.88397	0.93564	0.95919
0.9	0.01776	0.06929	0.14937	0.24934	0.35811	0.60123	0.75423	0.88769	0.93688	0.95970
1	0.01980	0.07681	0.16393	0.27005	0.38197	0.61980	0.76393	0.89023	0.93774	0.96006
2	0.03120	0.11586	0.23196	0.35532	0.46887	0.67544	0.79176	0.89770	0.94035	0.96118
3	0.03487	0.12729	0.24941	0.37438	0.48612	0.68493	0.79640	0.89899	0.94081	0.96138
4	0.03636	0.13177	0.25595	0.38125	0.49219	0.68818	0.79798	0.89943	0.94097	0.96145
5	0.03709	0.13393	0.25906	0.38447	0.49500	0.68967	0.79871	0.89964	0.94105	0.96148
6	0.03750	0.13513	0.26077	0.38623	0.49653	0.69048	0.79911	0.89975	0.94109	0.96150
7	0.03775	0.13587	0.26181	0.38729	0.49745	0.69097	0.79935	0.89982	0.94111	0.96151
8	0.03791	0.13634	0.26248	0.38798	0.49805	0.69128	0.79950	0.89986	0.94113	0.96152
9	0.03803	0.13668	0.26295	0.38846	0.49846	0.69150	0.79960	0.89989	0.94114	0.96152
10	0.03811	0.13691	0.26328	0.38879	0.49875	0.69165	0.79968	0.89991	0.94114	0.96152

[latex]F_{21}=\frac{A_{1}}{A_{2}} F_{12}=\frac{0.7854}{0.1964} imes 0.175=0.6998 \approx 0.7[/latex]

Applying the sum rule to surface 2, we get [latex]F_{23}= 1 - F_{21} = 1 - 0.7 =0.3[/latex].

Step 3 The space resistances are calculated now.

[latex]R_{12}=\frac{1}{A_{1} F_{12}}=\frac{1}{0.7854 imes 0.175}=7.276 m ^{-2}[/latex]

[latex]R_{13}=\frac{1}{A_{1} F_{13}}=\frac{1}{0.7854 imes 0.825}= 1.543m ^{-2}[/latex]

[latex]R_{23}=\frac{1}{A_{2} F_{23}}=\frac{1}{ 0.1964 imes 0.3}= 16.977m ^{-2}[/latex]

Surface resistances are calculated next.

[latex]R_{s 1}=\frac{1-\varepsilon_{1}}{A_{1} \varepsilon_{1}}=\frac{1-0.6}{0.7854 imes 0.6}=0.849 m^{-2}[/latex]

[latex]R_{s 2}=\frac{1-\varepsilon_{2}}{A_{2} \varepsilon_{2}}=\frac{1-0.6}{ 0.0.1963 imes 0.6}= 3.395m^{-2}[/latex]

Step 4 Radiation network and radiosity equations:

All the resistances are now available for defining the radiation network shown in Fig. 10.23b. The black body emissive powers at temperatures of surface 1 and 2 are

[latex]E_{b 1}=\sigma T_{1}^{4}=5.67 imes 10^{-8} imes 600^{4}=7348.3 W / m ^{2}[/latex]

[latex]E_{b 2}=\sigma T_{2}^{4}=5.67 imes 10^{-8} imes 400^{4}= 1451.5W / m ^{2}[/latex]

Equations in the radiosity formulation are obtained from the electrical network by the application of Kirchhoff’s law to nodes 1 and 2. For node 1, we have

[latex]\frac{E_{b 1}-J_{1}}{R_{s 1}}+\frac{J_{2}-J_{1}}{R_{12}}+\frac{J_{3}-J_{1}}{R_{13}}=0[/latex]

This may be rearranged as

[latex]J_{1}\left(\frac{1}{R_{s 1}}+\frac{1}{R_{12}}+\frac{1}{R_{13}} ight)-\frac{J_{2}}{R_{12}}=\frac{E_{b 1}}{R_{s 1}}+\frac{J_{3}}{R_{13}}[/latex]

[latex]J_{2}[/latex] being known, has been taken to the right-hand side. Analogously, we may write the radiosity equation for surface 2 as

[latex]-\frac{J_{1}}{R_{12}}+J_{2}\left(\frac{1}{R_{s 2}}+\frac{1}{R_{12}}+\frac{1}{R_{23}} ight)=\frac{E_{b 2}}{R_{s 2}}+\frac{J_{3}}{R_{23}}[/latex]

Introducing the numerical values for the various quantities appearing in the above, we get the following two simultaneous equations for the radiosities of surfaces 1 and 2.

[latex]1.9635 J_{1}-0.1374 J_{2}=8954.6[/latex]

[latex]-0.1374 J_{1}+0.4909 J_{2}=454.6[/latex]

These equations are easily solved (use Kramer’s rule) to get

[latex]J_{1}=4717.8 W / m ^{2}[/latex] ; [latex]J_{21}= 2247.0W / m ^{2}[/latex]

Step 5 Surface heat transfer rates:

The heat transferred from the three surfaces, respectively, are

[latex]Q_{1}=\frac{E_{b 1}-J_{1}}{R_{s 1}}=\frac{7348.3-4717.8}{0.849}=3099.0 W[/latex]

[latex]Q_{2}=\frac{E_{b2}-J_{2}}{R_{s2}}=\frac{1451.5-2247.0}{3.395}=-234.3 W[/latex]

[latex]Q_{3}=\frac{J_{3}-J_{1}}{R_{13}}+\frac{J_{3}-J_{2}}{R_{23}}=\frac{459.3-7348.3}{1.553}+\frac{459.3-2247.0}{16.977}=-2864.7 W[/latex]

Overall energy balance for the enclosure gives

[latex]\sum_{i=1}^{3} Q_{i}=3099-234.3-2864.7=0 W[/latex]

The solution is indeed very satisfactory.

Question 10.10: A circular disk of diameter 1 m and at 600 K is kept directl...

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