Question 12.9: The cover glass on a flat-plate solar collector has a low ir...

Known: Spectral transmissivity of solar collector cover glass.

Find: Total transmissivity of cover glass to solar radiation.

Assumptions: Spectral distribution of solar irradiation is proportional to that of blackbody emission at 5800 K.

Analysis: From Equation 12.61 the total transmissivity of the cover is

τ = \frac{\int_{0}^{∞}G_{λ,tr}(λ)  dλ}{\int_{0}^{∞}G_{λ}(λ)  dλ} = \frac{\int_{0}^{∞}τ_{λ}(λ) G_{λ}(λ)  dλ}{\int_{0}^{∞}G_{λ}(λ)  dλ}              (12.61)

τ = \frac{\int_{0}^{∞}τ_{λ}G_{λ}  dλ}{\int_{0}^{∞}G_{λ}  dλ}

where the irradiation G_{λ} is due to solar emission. Having assumed that the sun emits as a blackbody at 5800 K, it follows that

G_{λ}(λ)  ∝  E_{λ,b}(5800  K)

With the proportionality constant canceling from the numerator and denominator of the expression for τ, we obtain

τ = \frac{\int_{0}^{∞}τ_{λ}E_{λ,b}(5800  K)  dλ}{\int_{0}^{∞}E_{λ,b}(5800  K)  dλ}

or, for the prescribed spectral distribution of τ_{λ}(λ),

τ = 0.90\frac{\int_{0.3}^{2.5}E_{λ,b}(5800  K)  dλ}{E_{b}(5800  K)}

From Table 12.2

λ_{1} = 0.3  μm, T = 5800 K:\qquad λ_{1}T = 1740  μm · K, F_{(0→λ_{1})} = 0.0335

λ_{2} = 2.5  μm, T = 5800 K:\qquad λ_{2}T = 14,500  μm · K, F_{(0→λ_{2})} = 0.9664

Hence from Equation 12.35

F_{(λ_{1}→λ_{2})} = \frac{\int_{0}^{λ_{2}}E_{λ,b}  dλ  –  \int_{0}^{λ_{1}}E_{λ,b}  dλ}{σT^{4}} = F_{(0→λ_{2})}  –  F_{(0→λ_{1})}              (12.35)

τ = 0.90[F_{(0→λ_{2})}  –  F_{(0→λ_{1})}] = 0.90(0.9664  –  0.0335) = 0.84

Comments: It is important to recognize that the irradiation at the cover plate is not equal to the emissive power of a blackbody at 5800 K, G_{λ} ≠ E_{λ,b} (5800  K). It is simply assumed to be proportional to this emissive power, in which case it is assumed to have a spectral distribution of the same form. With G_{λ} appearing in both the numerator and denominator of the expression for τ, it is then possible to replace G_{λ} by E_{λ,b}.