Topic: Heat Transfer
A flat plate solar collector with no cover plate has a selective absorber of emissivity 0.2 and solar absorptivity 0.95. At a given time of day the absorber surface temperature is $125^{\circ}C$ when the solar irradiation is 1150 $W/m^{2}$, the effective sky temperature is $-10^{\circ}C$ and the ambient air temperature is $30^{\circ}C$. Assume the heat transfer convection coefficient to be $1~W/m^{2}K$. Calculate the useful heat removal rate from the collector. The sky radiation is in the same spectral region as that of surface emission.
- A. $561.3~W/m^{2}$
- B. $767.2~W/m^{2}$
- C. $913.8~W/m^{2}$
- D. None of the above
Answer: B. $767.2~W/m^{2}$
Explanation: Heat absorbed by the collector: $q_{abs} = \alpha G = 0.95 \times 1150 = 1092.5~W/m^{2}$.
Heat lost by convection: $q_{conv} = h(T_s - T_a) = 1 \times (125 - 30) = 95~W/m^{2}$.
Heat lost by radiation: The problem states sky radiation is in the same spectral region as surface emission, so absorptivity for sky radiation equals emissivity ($\epsilon = 0.2$).
$q_{rad} = \epsilon \sigma (T_s^4 - T_{sky}^4) = 0.2 \times 5.67 \times 10^{-8} \times ((125+273)^4 - (-10+273)^4) = 1.134 \times 10^{-8} \times (398^4 - 263^4) \approx 230.3~W/m^{2}$.
Useful heat removal rate: $q_{useful} = q_{abs} - q_{conv} - q_{rad} = 1092.5 - 95 - 230.3 = 767.2~W/m^{2}$.