Consider a 1 m thick layer of a gas mixture at 1000 K and 5 bar that consists of 10% carbon dioxide, 20% water vapor, and 70% nitrogen. What is the total normal intensity escaping from this layer?

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Accepted Answer

From equations (10.33) [latex]I_{\eta }(X)=I_{\eta }(0)e^{-\kappa _{\eta }X}+ I_{b\eta }(1-e^{-\kappa _{\eta }X})[/latex] and (10.138) [latex]\epsilon \equiv \frac{\int_{0}^{\infty }{}I_{b\eta }\epsilon _{\eta }d\eta }{\int_{0}^{\infty }{}I_{b\eta }d\eta } =\frac{\int_{0}^{\infty }{}I_{b\eta (1-e^{-\kappa _{0}X})d\eta } }{\int_{0}^{\infty }{}I_{b\eta }d\eta } =\sum\limits_{i=1}^{N}{}\left(\frac{\pi I_{b\eta 0}}{\sigma T^{4}} ight) _{i}\int_{ riangle \eta b and }^{}{} (1e^{-\kappa _{0}X})d\eta =\sum\limits_{i=1}^{N}{} \left(\frac{\pi I_{b\eta 0}}{\sigma T^{4}} ight) _{i} A_{i}[/latex] we see that the exiting total intensity is

[latex]I=\int_{0}^{\infty }{}I_{b\eta }(1-e^{-\kappa _{\eta }X})d\eta =\int_{0}^{\infty }{} I_{b\eta }\epsilon _{\eta }d\eta =\frac{\epsilon \sigma T^{4}}{\pi }[/latex]

where ε is the total emissivity of the water vapor-carbon dioxide mixture. First we calculate the emissivity of [latex]CO_{2}[/latex] at a total pressure of 1 bar from Table 10.5:

With [latex]p_{CO_{2}}L=0.1 imes 5 m bar=50[/latex] bar cm and [latex]T_{g}=1000K[/latex] we find [latex]\epsilon _{CO_{2}.0}(1 bar)=0.157[/latex] (which may also be estimated from Fig. 10-25); for a total pressure of 5 bar we find from Table 10.4

the effective pressure is [latex]P_{E}[/latex]= 5.14, a = 1.1, b = 0.23, c = 1.47 and [latex](p_{a}L)_{m}[/latex]= 0.225 bar cm. Thus, from equation (10.145) [latex]\frac{\epsilon (p_{a}L,p,T_{g})}{\epsilon _{0}(p_{a}L,1bar,T_{g})}=1-\frac{(a-1)(1-P_{E})}{a+b-1+P_{E}}exp \left\lgroup -c \left[log_{10}\frac{(p_{a}L)_{m}}{p_{a}L} ight]^{2} ight group[/latex]

[latex]\left(\frac{\epsilon }{\epsilon _{0}} ight)_{CO_{2}}=1-\frac{0.1 imes (-4.14)}{0.33+5.14}exp \left[-1.47 imes \left(log_{10}\frac{0.225}{50} ight)^{2} ight]\approx 1.00,[/latex]

and

[latex]\epsilon _{CO_{2}}\approx 0.157[/latex] .

Similarly, for water vapor with [latex]p_{H_{2}O}L=0.2 imes 5[/latex] m bar = 100 bar cm we find [latex]\epsilon _{H_{2}O}(1 bar) \approx 0.359[/latex] and the pressure correction factor becomes, with [latex]P_{E} = 7.56, a = 1.88, b = 1.1, c = 0.5 and (p_{a}L)_{m} = 13.2 bar cm ,[/latex]

[latex]\left(\frac{\epsilon }{\epsilon _{0}} ight)_{H_{2}O}=1-\frac{0.888 imes (-6.56)}{1.988+7.56}exp \left[-0.5 imes \left(log_{10}\frac{13.2}{100} ight)^{2} ight] =1.414,[/latex]

and

[latex]\epsilon _{H_{2O}}\approx 0.359 imes 1.414=0.508[/latex] .

Finally, since we have a mixture of carbon dioxide and water vapor, we need to deduct for the band overlaps: From equation (10.146), with [latex]\zeta =\frac{2}{3},\Delta \epsilon =0.072[/latex] . Thus, the total emissivity is ε = 0.157 + 0.508 - 0.072 = 0.593. Alternatively, and more easily, using subroutine totemiss with ph2o = 1., pco2 = .5, ptot = 5, L = 100, and Tg = 1000 returns the same numbers.

The total normal intensity is then

[latex]I=0.593 imes 5.670 imes 10^{-8}W/(m^{2}K^{4}) imes (500K)^{4}/\pi sr = 669 W/m^{2 }sr[/latex]

Question 10.11: Consider a 1 m thick layer of a gas mixture at 1000 K and 5 ...

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