Question 12.56: KNOWN: Spectral transmissivity of a plain and tinted glass. ......

ANALYSIS: To compare the energy transmitted by the glasses, it is sufficient to calculate the transmissivity of each glass for the prescribed spectral range when the irradiation distribution is that of the solar spectrum. From Eq. 12.55,

\tau_{ S }=\int_0^{\infty} \tau_\lambda \cdot G _{\lambda, S } d \lambda / \int_0^{\infty} G { }_{\lambda, S } d \lambda=\int_0^{\infty} \tau_\lambda \cdot E _{\lambda, b }(\lambda, 5800  K )  d \lambda / E _{ b } (5800  K ).

Recognizing that \tau_\lambda will be constant for the range \lambda_1 \rightarrow \lambda_2, using Eq. 12.31, find

\tau_{ S }=\tau_\lambda \cdot F _{\left(\lambda_1 \rightarrow \lambda_2\right)}=\tau_\lambda\left[ F _{\left(0 \rightarrow \lambda_2\right)}- F _{\left(0 \rightarrow \lambda_1\right)}\right].

(a) For the two glasses, the solar transmissivity, using Table 12.1 for F, is then

Plain glass:       \lambda_2=2.5  \mu m \quad \lambda_2 T =2.5  \mu m \times 5800  K =14,500  \mu m \cdot K \quad F _{(0 \rightarrow \lambda 2)}=0.966

                            \lambda_1=0.3  \mu m \quad \lambda_1 T =0.3  \mu m \times 5800  K =1,740  \mu m \cdot K ~~\quad~~ F_{(0 \rightarrow \lambda_1)}=0.033

\tau_{ S }=0.9[0.966  –  0.033]=0.839.

Tinted glass:    \lambda_2=1.5  \mu m \quad \lambda_2 T =1.5  \mu m \times 5800  K =8,700  \mu m \cdot K \quad F _{\left(0 \rightarrow \lambda_2\right)}=0.881

                            \lambda_1=0.5  \mu m \quad \lambda_1 T =0.5  \mu m \times 5800  K =2,900  \mu m \cdot K \quad F _{\left(0 \rightarrow \lambda_1\right)}=0.033

\tau_{ S }=0.9[0.886-0.250]=0.568.

(b) The limits of the visible spectrum are \lambda_1=0.4 \text { and } \lambda_2=0.7  \mu m \text {. } For the tinted glass, \lambda_1=0.5  \mu m rather than 0.4 μm. From Table 12.1,

\lambda_2=0.7  \mu m \quad \lambda_2 T =0.7  \mu m \times 5800  K =4,060  \mu m \cdot K \quad F _{\left(0 \rightarrow \lambda_2\right)}=0.491

\lambda_1=0.5  \mu m \quad \lambda_1 T =0.5  \mu m \times 5800  K =2,900  \mu m \cdot K \quad F _{\left(0 \rightarrow \lambda_1\right)}=0.250

\lambda_1=0.4  \mu m \quad \lambda_1 T =0.4  \mu m \times 5800  K =2,320  \mu m \cdot K \quad F _{\left(0 \rightarrow \lambda_1\right)}=0.125

Plain glass:      \tau_{\text{vis}}=0.9[0.491-0.125]=0.329

Tinted glass:    \tau_{\text{vis}}=0.9[0.491-0.250]=0.217

COMMENTS: For solar energy, the transmissivities are 0.839 for the plain glass vs. 0.568 for the plain and tinted glasses. Within the visible region, \tau_{\text{vis}} is 0.329 vs. 0.217. Tinting reduces solar flux by 32% and visible solar flux by 34%.