Question 10.7: What fraction of incident solar radiation is absorbed by a t...
What fraction of incident solar radiation is absorbed by a thick layer of CO_2 at 10 atm and 830 K? Use the approximate absorption bands of Example 10.5.
The effective radiating temperature of the sun is T_s = 5780 K. The desired result is the fraction of the solar spectrum that lies within the four CO_2 bands, as this is the only portion of the incident radiation that will be absorbed. Using the F_{0→λTs}, factors obtained using T_s gives
| λ (μm) | λT_s (μm · K) | F_{0→λTs} | λ (μm) | λT_s (μm · K) | F_{0→λT_s} |
| 1.8 | 10,400 | 0.92195 | 4.0 | 23,120 | 0.99028 |
| 2.2 | 12,720 | 0.95251 | 4.6 | 26,590 | 0.99340 |
| 2.6 | 15,030 | 0.96909 | 9 | 52,020 | 0.99902 |
| 2.8 | 16,180 | 0.97455 | 19 | 109,800 | 0.99989 |
The fraction absorbed is then
\alpha =\sum\limits_{absorbing \ bands}{[F_{0\longrightarrow (\lambda T_s)_{upper}}-F_{0\longrightarrow (\lambda T_s)_{lower}}]_{band}}
=(0.95251− 0.92195)+ (0.97455− 0.96909)+ (0.99340 -0.99028 )+(0.99989-0.99902)=0.040
Even though the gas layer is thick, only 4.0% of the incident energy is absorbed since the gas transmits well in the λ regions between the absorption bands. For radiative exchange calculations in furnaces, an approximate procedure for determining α_g is in Hottel and Sarofim (1967). The α_g is obtained from the gas total emittance values by using
α_g= \alpha _{CO_2}+\alpha _{H_2O}-\Delta \alpha (10.117)
where
\alpha _{CO_2}=C _{CO_2}\epsilon _{CO_2} ^+\left\lgroup\frac{T_g}{T_w} \right\rgroup^{0.5} (10.118)
\alpha _{H_2O}=C _{H_2O}\epsilon _{H_2O} ^+\left\lgroup\frac{T_g}{T_w} \right\rgroup^{0.5} (10.119)
Δα = (Δ\epsilon )_{at \ T_w } (10.120)
The \epsilon _{CO_2} ^+ and \epsilon _{H_2O} ^+ are, respectively, \epsilon _{CO_2} and \epsilon _{H_2O} obtained from Equation 9.62
\epsilon (T,pL_e)=\exp \left\{a_0+\sum\limits_{j=1}^{M}{a_j[\log (pL_e)]^j} \right\} (9.62)
evaluated at T_w and at the respective parameters P_{CO_2}L_e^′ and P_{H_2O}L_e^′ where L^′_e = L_e T_w / T_g . It is pointed out in Edwards and Matavosian (1984) that the exponent 0.5 is now becoming more accepted to replace the values 0.65 and 0.45 originally used in Equations 10.118 and 10.119. At high temperatures and pressures when there is overlapping of absorption lines in the infrared spectrum, L^{′}_{e} = L_e (T_w / T_g)^{3/2} , which is discussed in Edwards and Matavosian,
Section 10.9.1 considered a uniform isothermal gas in a black enclosure. If the bounding walls are not black and hence are reflecting, radiation can pass through the gas by means of multiple reflections from the boundary. For an enclosure with a single wall, this can be included by integrating Equation 10.74
Q_g=-Q_1=-A_1\int_{\lambda =0}^{\infty }{q_{\lambda ,1}d\lambda } =A_1\int_{0}^{\infty }{\frac{E_{\lambda b,g}-E_{\lambda b,1}}{\frac{1}{\epsilon _{\lambda ,1}}+\frac{1}{\overline{\alpha }_{\lambda,1-1 } } -1 } }d\lambda (10.74)
over all wavelengths. In Edwards and Matavosian, a procedure is proposed to use total emittance values for situations with multiple reflections. The solution of enclosure heat transfer problems with reflecting walls was treated in Section 10.6.4 by integrating spectral relations over the wavelength absorption bands.
An example that considers a black enclosure with walls at differing temperatures is in Example A.2 in the Appendix A at the publishers web site at www.crcpress.com.