Question 12.49: KNOWN: Spectral, hemispherical absorptivity of an opaque sur......
KNOWN: Spectral, hemispherical absorptivity of an opaque surface.
FIND: (a) Solar absorptivity, (b) Total, hemispherical emissivity for \mathrm T_{\mathrm s} = 340 K.
ASSUMPTIONS: (1) Surface is opaque, (2) ε_λ = α_λ, (3) Solar spectrum has \mathrm G_λ = \mathrm G_{λ,\mathrm S} proportional to \mathrm E_{λ,\mathrm b} (λ, 5800 K).
SCHEMATIC:
ANALYSIS: (a) The solar absorptivity follows from Eq. 12.47.
\alpha_{ S }=\int_0^{\infty} \alpha_\lambda(\lambda) E_{\lambda, b} (\lambda, 5800 K ) d \lambda / \int_0^{\infty} E _{\lambda, b }(\lambda, 5800 K ) d\lambda.
The integral can be written in three parts using \mathrm F_{(0 → \lambda)} terms.
\alpha_{ S }=\alpha_1 F _{(0 \rightarrow 0.3)}+\alpha_2\left[ F _{(0 \rightarrow 1.5)}- F _{(0 \rightarrow 0.3)}\right]+\alpha_3\left[1- F _{(0 \rightarrow 1.5)}\right].
From Table 12.1,
\lambda T = 0.3 \times 5800 = 1740 \mu m \cdot K ~~~~\quad~~~~ F _{(0 \rightarrow 0.3 \mu m )} = 0.0335
\lambda T = 1.5 \times 5800 = 8700 \mu m \cdot K ~~~~\quad~~~~ F _{(0 \rightarrow 1.5 \mu m )} = 0.8805.
Hence,
\alpha_{ S }=0 \times 0.0355+0.9[0.8805-0.0335]+0.1[1-0.8805]=0.774.
(b) The total, hemispherical emissivity for the surface at 340 K will be
\varepsilon=\int_0^{\infty} \varepsilon_\lambda(\lambda) E _{\lambda, b }(\lambda, 340 K ) d \lambda / E _{ b }(340 K ).
If \varepsilon_{\lambda} = \alpha_{\lambda}, then using the \alpha_{\lambda} distribution above, the integral can be written in terms of \mathrm F_{(0 → \lambda)} values. It is readily recognized that since
F_{(0 \rightarrow 1.5 \mu m, 340 K)} \approx 0.000 ~~~~\quad \text { at } \quad~~~~ \lambda T =1.5 \times 340=510 \mu m \cdot K
there is negligible spectral emissive power below 1.5 μm. It follows then that
\varepsilon=\varepsilon_\lambda=\alpha_\lambda=0.1
COMMENTS: The assumption \varepsilon_{\lambda} = \alpha_{\lambda} can be satisfied if this surface were irradiated diffusely or if the surface itself were diffuse. Note that for this surface under the specified conditions of solar irradiation and surface temperature \alpha_{\mathrm S} ≠ ε. Such a surface is referred to as a spectrally selective surface.