Question 12.118: KNOWN: Dimensions, spectral absorptivity, and temperature of......
KNOWN: Dimensions, spectral absorptivity, and temperature of solar receiver. Solar irradiation and ambient temperature.
FIND: (a) Rate of energy collection q and collector efficiency η, (b) Effect of receiver temperature on q and η.
ASSUMPTIONS: (1) Steady-state, (2) Uniform irradiaton, (3) Opaque, diffuse surface.
PROPERTIES: Table A.4, air \left(T_{ f }=550 K \right): \nu =45.6 \times 10^{-6} m ^2 / s , k =0.0439 W / m \cdot K, \alpha=66.7 \times 10^{-6} m ^2 / s, \operatorname{Pr}=0.683.
SCHEMATIC:
ANALYSIS: (a) The rate of heat transfer to the receiver is q = A _{ s }\left(\alpha_{ S } G _{ S }- E – q _{\text {conv}}^{\prime \prime}\right), or
q =\pi DL \left[\alpha_{ S } G _{ S }-\varepsilon \sigma T _{ s }^4-\overline{ h }\left( T _{ s }- T _{\infty}\right)\right]
For λT = 3 µm × 5800 K = 17,400, F_{(0→λ)} = 0.979. Hence,
\alpha_{ S }=\alpha_1 F_{(0 \rightarrow \lambda)}+\alpha_2\left(1- F _{(0 \rightarrow \lambda)}\right)=0.9 \times 0.979+0.2(0.021)=0.885
For λT = 3 µm × 800 K = 2400 µm·K, F_{(0→λ)} = 0.140. Hence,
\varepsilon=\varepsilon_1 F _{(0 \rightarrow \lambda)}+\varepsilon_2\left(1- F _{(0 \rightarrow \lambda)}\right)=0.9 \times 0.140+0.2(0.860)=0.298.
With Ra _{ L }= g \beta\left( T _{ s }- T _{\infty}\right) L ^3 / \alpha \nu=9.8 m / s ^2(1 / 550 K )(500 K )(12 m )^3 / 66.7 \times 10^{-6} m ^2 / s \times 45.6 \times 10^{-6} m ^2 / s = 5.06 \times 10^{12}, Eq. 9.26 yields
\overline{ Nu }_{ L }=\left\{0.825+\frac{0.387 Ra _{ L }^{1 / 6}}{\left[1+(0.492 / Pr )^{9 / 16}\right]^{8 / 27}}\right\}^2=1867
\overline{ h }=\overline{ Nu }_{ L } \frac{ k }{ L }=1867 \frac{0.0439 W / m \cdot K }{12 m }=6.83 W / m ^2 \cdot K
Hence,
q =\pi(7 m \times 12 m )\left[0.885 \times 80,000 W / m ^2-0.298 \times 5.67 \times 10^{-8} W / m ^2 \cdot K ^4(800 K )^4-6.83 W / m ^2 \cdot K (500 K )\right]
q =263.9 m ^2 (70,800-6,920-3415) W / m ^2=1.60 \times 10^7 W
The collector efficiency is \eta=q /A_s G_s. Hence
\eta=\frac{1.60 \times 10^7 W }{263.9 m ^2\left(80,000 W / m ^2\right)}=0.758
(b) The IHT Correlations, Properties and Radiation Toolpads were used to obtain the following results.
Losses due to emission and convection increase with increasing T_s, thereby reducing q and η.
COMMENTS: The increase in radiation emission is due to the increase in T_s, as well as to the effect of T_s on ε, which increases from 0.228 to 0.391 as T_s increases from 600 to 1000 K.
