Question 12.59: KNOWN: Spectrally selective, diffuse surface exposed to sola......
KNOWN: Spectrally selective, diffuse surface exposed to solar irradiation.
FIND: (a) Spectral transmissivity, \tau_\lambda, (b) Transmissivity, \tau_S, reflectivity, \rho_S, and absorptivity, \alpha_S , for solar irradiation, (c) Emissivity, ε, when surface is at T _{ s } = 350 K, (d) Net heat flux by radiation to the surface.
ASSUMPTIONS: (1) Surface is diffuse, (2) Spectral distribution of solar irradiation is proportional to E _{\lambda, b }(\lambda, 5800 K).
SCHEMATIC:
ANALYSIS: (a) Conservation of radiant energy requires, according to Eq. 12.56, that \rho_\lambda+\alpha_\lambda+\tau_\lambda =1 \text { or } \tau_\lambda=1-\rho_\lambda-\alpha_\lambda. Hence, the spectral transmissivity appears as shown above (dashed line). Note that the surface is opaque for λ > 1.38 μm.
(b) The transmissivity to solar irradiation, G_S, follows from Eq. 12.55,
\tau_{ S }=\int_0^{\infty} \tau_\lambda G _{\lambda, S } d \lambda / G _{ S }=\int_0^{\infty} \tau_\lambda E _{\lambda, b }(\lambda, 5800 K ) d \lambda / E _{ b }(5800 K )
\tau_{ S }=\tau_{\lambda, b } \int_0^{1.38} E _{\lambda, b }(\lambda, 5800 K ) d \lambda / E _{ b }(5800 K )=\tau_{\lambda, 1} F _{\left(0 \rightarrow \lambda_1\right)}=0.7 \times 8.56=0.599
where \lambda_1 T _{ S }=1.38 \times 5800=8000 \mu m \cdot K and from Table 12.1, F _{\left(0 \rightarrow \lambda_1\right)}=0.856. From Eqs. 12.52 and 12.57,
\rho_{ S }=\int_0^{\infty} \rho_\lambda G _{\lambda, S } d \lambda / G _{ S }=\rho_{\lambda, 1} F _{\left(0 \rightarrow \lambda_1\right)}=0.1 \times 0.856=0.086
\alpha_{ S }=1-\rho_{ S }-\tau_{ S }=1-0.086-0.599=0.315.
(c) For the surface at T_s = 350 K, the emissivity can be determined from Eq. 12.38. Since the surface is diffuse, according to Eq. 12.65, \alpha_{\lambda} = \varepsilon_{\lambda}, the expression has the form
\varepsilon=\int_0^{\infty} \varepsilon_\lambda E _{\lambda, b }\left( T _{ s }\right) d \lambda / E _{ b }\left( T _{ s }\right)=\int_0^{\infty} \alpha_\lambda E _{\lambda, b }(350 K ) d \lambda / E _{ b }(350 K )
\varepsilon=\alpha_{\lambda, 1} F _{(0-1.38 \mu m )}+\alpha_{\lambda, 2}\left[1- F _{(0-1.38 \mu m )}\right]=\alpha_{\lambda, 2}=1
where from Table 12.1 with \lambda_1 T_S = 1.38 × 350 = 483 μm·K, F_{(0-\lambda T)} ≈ 0.
(d) The net heat flux by radiation to the surface is determined by a radiation balance
q _{ rad }^{\prime \prime}= G _{ S }-\rho_{ S } G _{ S }-\tau_{ S } G _{ S }- E
q _{ rad }^{\prime \prime}=\alpha_{ S } G _{ S }- E
q _{ rad }^{\prime \prime}=0.315 \times 750 W / m ^2 – 1.0 \times 5.67 \times 10^{-8} W / m ^2 \cdot K ^4(350 K )^4 = -615 W / m ^2.
