Two very large parallel planar surfaces are maintained at 400 K and 800 K, respectively. The surfaces are non-gray and have the emissivity variation given below:

Question

Surface 1: [latex]\varepsilon_{1,1}=0.6[/latex] for [latex]0<\lambda<3 \mu m[/latex] ; [latex]\varepsilon_{1,2}=0.2[/latex] for [latex]\lambda>3 \mu m[/latex]

Surface 2: [latex]\varepsilon_{2,1}=0.2[/latex] for [latex]0<\lambda<3 \mu m[/latex] ; [latex]\varepsilon_{2,2}=0.6[/latex] for [latex]\lambda>3 \mu m[/latex]

Determine the heat transfer between the two planes per unit area using a non-gray analysis. Compare this with a gray analysis based on the equivalent gray emissivities of the two surfaces at their respective temperatures.

Accepted Answer

Step 1 There are two common bands in this problem. The first band spans the wavelength region 0 to [latex]3 \mu m[/latex] and the second band spans the wavelength region from [latex]3 \mu m[/latex] to ∞. Let [latex]f_{j, i}[/latex] represent the fraction of black body power for the jth surface and the ith band.

Step 2

Surface 1:

With [latex]T_{1}=800 K[/latex] and [latex]\lambda_{1}=3 \mu m[/latex], we have [latex]\lambda_{1} T_{1}=2400 \mu m K[/latex] From Table 8.1, we then have [latex]f_{0-2400}=0.150256[/latex] and hence [latex]f_{1,1}=0.150256[/latex]. With [latex]\lambda_{2}=\infty[/latex] we then have [latex]f_{1,2}=1-f_{1,1}=1-0.150256=0.859744[/latex] .

Surface 2:

With [latex]T_{2}=400 K[/latex] and [latex]\lambda_{1}=3 \mu m[/latex], we have [latex]\lambda_{1} T_{2}=1200 \mu m K[/latex] From Table 8.1, we then have [latex]f_{0-2400}=0.150256[/latex] and hence [latex]f_{2,1}=0.002134[/latex]. With [latex]\lambda_{2}=\infty[/latex] we then have [latex]f_{2,2}=1-f_{2,1}=1-0.002134=0.997866[/latex].

Table 8.1 Black body fraction function

[latex]\lambda T[/latex]	[latex]f_{0-\lambda T}[/latex]	[latex]\lambda T[/latex]	[latex]f_{0-\lambda T}[/latex]	[latex]\lambda T[/latex]	[latex]f_{0-\lambda T}[/latex]	[latex]\lambda T[/latex]	[latex]f_{0-\lambda T}[/latex]
1000	0.000321	10000	0.914155	19000	0.983405	56000	0.999210
1200	0.002134	10200	0.918135	19200	0.983867	57000	0.999250
1400	0.007790	10400	0.921877	19400	0.984312	58000	0.999287
1600	0.019718	10600	0.925400	19600	0.984741	59000	0.999321
1800	0.03934	10800	0.928717	19800	0.985155	60000	0.999354
2000	0.066727	11000	0.931845	20000	0.985553	61000	0.999384
2200	0.100886	11200	0.934795	20200	0.985938	62000	0.999413
2400	0.140253	11400	0.937580	20400	0.986309	63000	0.999439
2600	0.183115	11600	0.940211	20600	0.986668	64000	0.999464
2800	0.227884	11800	0.942699	20800	0.987014	65000	0.999488
3000	0.273223	12000	0.945052	21000	0.987348	66000	0.999510
3200	0.318091	12200	0.947280	22000	0.988858	67000	0.999531
3400	0.361722	12400	0.949391	23000	0.990137	68000	0.999551
3600	0.403592	12600	0.951392	24000	0.991229	69000	0.99957
3800	0.443366	12800	0.953291	25000	0.992166	70000	0.999587
4000	0.480858	13000	0.955092	26000	0.992974	71000	0.999604
4200	0.515993	13200	0.956804	27000	0.993675	72000	0.999620
4400	0.548774	13400	0.958431	28000	0.994286	73000	0.999635
4600	0.579256	13600	0.959977	29000	0.994821	74000	0.999649
4800	0.607534	13800	0.961449	30000	0.995291	75000	0.999663
5000	0.63372	14000	0.962850	31000	0.995706	76000	0.999676
5200	0.657942	14200	0.964185	32000	0.996074	77000	0.999688
5400	0.68033	14400	0.965458	33000	0.996401	78000	0.999699
5600	0.701016	14600	0.966671	34000	0.996693	79000	0.999710
5800	0.720126	14800	0.967828	35000	0.996954	80000	0.999721
6000	0.737785	15000	0.968933	36000	0.997188	81000	0.999731
6200	0.754106	15200	0.969989	37000	0.997399	82000	0.999740
6400	0.769199	15400	0.970998	38000	0.997589	83000	0.999749
6600	0.783163	15600	0.971962	39000	0.997762	84000	0.999758
6800	0.796092	15800	0.972884	40000	0.997918	85000	0.999766
7000	0.808071	16000	0.973767	41000	0.998060	86000	0.999774
7200	0.819179	16200	0.974612	42000	0.998189	87000	0.999781
7400	0.829488	16400	0.975421	43000	0.998308	88000	0.999789
7600	0.839063	16600	0.976196	44000	0.998416	89000	0.999795
7800	0.847965	16800	0.976940	45000	0.998515	90000	0.999802
8000	0.856248	17000	0.977653	46000	0.998606	91000	0.999808
8200	0.863962	17200	0.978337	47000	0.998690	92000	0.999814
8400	0.871153	17400	0.978994	48000	0.998767	93000	0.999820
8600	0.877863	17600	0.979624	49000	0.998838	94000	0.999826
8800	0.88413	17800	0.980230	50000	0.998904	95000	0.999831
9000	0.889987	18000	0.980812	51000	0.998965	96000	0.999836
9200	0.895467	18200	0.981371	52000	0.999021	97000	0.999841
9400	0.900599	18400	0.981910	53000	0.999074	98000	0.999845
9600	0.905408	18600	0.982427	54000	0.999123	99000	0.999850
9800	0.90992	18800	0.982926	55000	0.999168	100000	0.999854

Step 3 Heat transfer from surface 1 to surface 2 per unit area:

It is calculated using expression 10.42 for each band. For band 1, we have

[latex]q_{1-2}=\frac{\varepsilon_{1} \varepsilon_{2} \sigma\left(T_{1}^{4}-T_{2}^{4} ight)}{1-\left(1-\varepsilon_{1} ight)\left(1-\varepsilon_{2} ight)}=\frac{\sigma\left(T_{1}^{4}-T_{2}^{4} ight)}{\left(\frac{1}{\varepsilon_{1}}+\frac{1}{\varepsilon_{2}}-1 ight)}[/latex] (10.42)

[latex]q_{1}=\frac{\sigma\left(f_{1,1} T_{1}^{4}-f_{2,1} T_{2}^{4} ight)}{\frac{1}{\varepsilon_{1,1}}+\frac{1}{\varepsilon_{2,1}}-1}[/latex] [latex]=\frac{5.67 imes 10^{-8}\left(0.150256 imes 800^{4}-0.002134 imes 400^{4} ight)}{\frac{1}{0.6}+\frac{1}{0.2}-1}=581.7 W / m ^{2}[/latex]

Similarly, for band 2 we have

[latex]q_{2}=\frac{\sigma\left(f_{1,2} T_{1}^{4}-f_{2,2} T_{2}^{4} ight)}{\frac{1}{\varepsilon_{1,2}}+\frac{1}{\varepsilon_{2,2}}-1}[/latex] [latex]=\frac{5.67 imes 10^{-8}\left(0.859744 imes 800^{4}-0.997866 imes 400^{4} ight)}{\frac{1}{0.2}+\frac{1}{0.6}-1}=3260.6 W / m ^{2}[/latex]

Hence, the net heat transfer from surface 1 to surface 2 is

[latex]q=581.7+3260.6=3842.3 W / m ^{2}[/latex]

Heat transfer under the gray model:

The gray model calculation is based on the equivalent gray body emissivities of the two surfaces at the respective surface temperatures.

[latex]\varepsilon_{1}=f_{1,1} \varepsilon_{1,1}+f_{1,2} \varepsilon_{1,2}=0.150256 imes 0.6+0.002134 imes 0.2=0.257[/latex]

[latex]\varepsilon_{2}=f_{2,1} \varepsilon_{2,1}+f_{2,2} \varepsilon_{2,2}=0.859744 imes 0.2+0.997866 imes 0.6=0.599[/latex]

The gray calculation thus yields

[latex]q=\frac{\sigma\left(T_{1}^{4}-T_{2}^{4} ight)}{\frac{1}{\varepsilon_{1}}+\frac{1}{\varepsilon_{2}}-1}=\frac{5.67 imes 10^{-8}\left(800^{4}-400^{4} ight)}{\frac{1}{0.257}+\frac{1}{0.599}-1}=4771.8 W / m ^{2}[/latex]

The example shows that a gray analysis is grossly in error.

Question 10.11: Two very large parallel planar surfaces are maintained at 40...

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