Question 12.8: The spectral, hemispherical absorptivity of an opaque surfac...

Known: Spectral, hemispherical absorptivity and irradiation of a surface. Surface temperature (500 K) and total, hemispherical emissivity (0.8).

Find:
1. Spectral distribution of reflectivity.
2. Total, hemispherical absorptivity.
3. Nature of surface temperature change.

Schematic:

Assumptions:
1. Surface is opaque.
2. Surface convection effects are negligible.
3. Back surface is insulated.

Analysis:
1. From Equation 12.63, ρ_{λ} = 1  –  α_{λ}. Hence from knowledge of α_{λ}(λ), the spectral distribution of ρ_{λ} is as shown.

α_{λ} + ρ_{λ} = 1              (12.63)

2. From Equations 12.51 and 12.52,

α \equiv \frac{G_{abs}}{G}              (12.51)

α = \frac{\int_{0}^{∞}{α_{λ}(λ)G_{λ}(λ)}  dλ}{\int_{0}^{∞}{G_{λ}(λ)}  dλ}              (12.52)

α = \frac{G_{abs}}{G} = \frac{\int_{0}^{∞}{α_{λ}G_{λ}}  dλ}{\int_{0}^{∞}{G_{λ}}  dλ}

or, subdividing the integral into parts,

α = \frac{0.2\int_{2}^{6}{G_{λ}}  dλ + 500 \int_{6}^{8}{α_{λ}}  dλ + 1.0 \int_{8}^{16}{G_{λ}}  dλ}{\int_{2}^{6}{G_{λ}}  dλ + \int_{6}^{12}{G_{λ}}  dλ + \int_{12}^{16}{G_{λ}}  dλ}

α = \left\{0.2(\frac{1}{2})500  W/m^{2} · μm  (6  –  2)  μm + 500  W/m^{2} · μm  [0.2(8  –  6)  μm + (1  –  0.2)(\frac{1}{2})(8  –  6)  μm] + [1 × 500  W/m^{2} · μm  (12  –  8)  μm + 1(\frac{1}{2})500  W/m^{2} · μm  (16  –  12)  μm]\right\} ÷ [(\frac{1}{2})500  W/m^{2} · μm  (6  –  2)  μm + 500  W/m^{2} · μm  (12  –  6)  μm + (\frac{1}{2})500  W/m^{2} · μm  (16  –  12)  μm]

Hence

α = \frac{G_{abs}}{G} = \frac{(200 + 600 + 3000)  W/m^{2}}{(1000 + 3000 + 1000)  W/m^{2}} = \frac{3800  W/m^{2}}{5000  W/m^{2}} = 0.76

3. Neglecting convection effects, the net heat flux to the surface is

q''_{net} = αG  –  E = αG  –  εσT^{4}

Hence

q''_{net} = 0.76(5000  W/m^{2})  –  0.8 × 5.67 × 10^{-8}  W/m^{2} · K^{4} (500  K)^{4}\\ q''_{net} = 3800  –  2835 = 965  W/m^{2}

Since q''_{net} > 0, the surface temperature will increase with time.