Question 12.115: KNOWN: Glass sheet, used on greenhouse roof, is subjected to......

ANALYSIS: (a) Performing an energy balance on the glass sheet with \dot{ E }_{\text {in}}-\dot{ E }_{\text {out}}=0 and considering two convection processes, emission and three absorbed irradiation terms, find

\alpha_{ S } G _{ S }+\alpha_{ atm } G _{ atm }+ h _{ o }\left( T _{\infty, o }- T _{ g }\right)+\alpha_{ i } G _{ i }+ h _{ i }\left( T _{\infty, i }- T _{ g }\right)-2 \varepsilon \sigma T _{ g }^4=0                (1)

where \alpha_S = solar absorptivity for absorption of G _{\lambda, S } \sim E _{\lambda, b }(\lambda, 5800  K )

\alpha_{ atm }=\alpha_{ i }= absorptivity of long wavelength irradiation (λ >> 1 μm) ≈ 1

\varepsilon=\alpha_\lambda for λ >> 1 μm, emissivity for long wavelength emission ≈ 1

(b) For the prescribed conditions, T_{∞,i} can be evaluated from Eq. (1). As noted above, \alpha_{\text {atm }}=\alpha_{ i }=1 and ε = 1. The solar absorptivity of the glass follows from Eq. 12.47 where G _{\lambda, S } \sim E _{\lambda, b } (λ, 5800 K),

\alpha_{ S }=\int_0^{\infty} \alpha_\lambda G_{\lambda, S} d \lambda / G _{ s }=\int_0^{\infty} \alpha_\lambda E _{\lambda, b }(\lambda, 5800  K ) d \lambda / E _{ b }(5800  K )

\alpha_{ S }=\alpha_1 F _{(0 \rightarrow 1  \mu m )}+\alpha_2\left[1- F _{(0 \rightarrow 1  \mu m )}\right]=0 \times 0.720+1.0[1-0.720]=0.28.

Note that from Table 12.1 for λT = 1 μm × 5800 K = 5800 μm·K, F_{(0 – \lambda)} = 0.720. Substituting numerical values into Eq. (1),

0.28 \times 1100  W / m ^2+1 \times 250  W / m ^2+55  W / m ^2 \cdot K (24-27) K +1 \times 440  W / m ^2 + 10  W / m ^2 \cdot K \left( T _{\infty, i }-27\right) K -2 \times 1 \times 5.67 \times 10^{-8}  W / m ^2 \cdot K (27+273)^4 K ^4=0

find that

T _{\infty, i}=35.5^{\circ}C.