Question 12.117: KNOWN: Thermal conductivity, spectral absorptivity and inner......

ANALYSIS: From an energy balance at the outer surface, \dot{ E }_{\text{in}}=\dot{ E }_{\text {out}},

\alpha_{ S } q _{ S }^{\prime \prime}+\alpha_{\text{sur}} G _{\text{sur}}=\varepsilon \sigma T _{ s , o }^4+ h _{ o }\left( T _{ s , o }- T _{\infty, o }\right)+\frac{ T _{ s , o }- T _{\infty, i }}{( L / k )+\left(1 / h _{ i }\right)}

Since radiation from the surroundings is in the far infrared, α_{\text{sur}} = 0.2. From Table 12.1, λT = (3 μm × 5800 K) = 17,400 μm·K, find F_{(0→3  μm)} = 0.979. Hence,

\alpha_{ s }=\frac{\int_0^{\infty} \alpha_\lambda E _{\lambda, b }(5800  K ) d \lambda}{ E _{ b }}=\alpha_1 F _{(0 \rightarrow 3  \mu m )}+\alpha_2 F _{(3 \rightarrow \infty)}=0.9(0.979)+0.2(0.021)=0.885.

From Table 12.1, λT = (3 μm × 1000 K) = 3000 μm·K, find F_{(0→3  μm)} = 0.273. Hence,

\varepsilon_{ s }=\frac{\int_{ o }^{\infty} \varepsilon_\lambda E _{\lambda, b }(1000  K ) d \lambda}{ E _{ b }}=\varepsilon_1 F _{(0 \rightarrow 3)}+\varepsilon_2 F _{(3 \rightarrow \infty)}=0.9(0.273)+0.2(0.727)=0.391.

Substituting numerical values in the energy balance, find

L = 0.129 m.

The corresponding collector efficiency is

\eta=\frac{ q _{\text {use}}^{\prime \prime}}{ q _{ S }^{\prime \prime}}=\left[\frac{ T _{s , o }- T _{\infty, i }}{( L / k )+\left(1 / h _{ i }\right)}\right] / q _{ S }^{\prime \prime}

COMMENTS: The collector efficiency could be increased and the outer surface temperature reduced by decreasing the value of L.