Question 12.12: A flat-plate solar collector with no cover plate has a selec...

Known: Operating conditions for a flat-plate solar collector.

Find:
1. Useful heat removal rate per unit area, q''_{u}  (W/m^{2}).
2. Efficiency η of the collector.

Schematic:

Assumptions:
1. Steady-state conditions.
2. Bottom of collector well insulated.
3. Absorber surface diffuse.

Analysis:
1. Performing an energy balance on the absorber,

\dot{E}_{in}  –  \dot{E}_{out} = 0

or, per unit surface area,

α_{S}G_{S} + α_{sky}G_{atm}  –  q''_{conv}  –  E  –  q''_{u} = 0

From Equation 12.73,

G_{atm} = σT_{sky}^{4}              (12.73)

Since the atmospheric irradiation is concentrated in approximately the same spectral region as that of surface emission, it is reasonable to assume that

α_{sky} ≈ ε = 0.1

With

q''_{conv} = \overline{h}(T_{s}  –  T_{\infty}) = 0.22(T_{s}  –  T_{\infty})^{4/3}     and     E = εσT^{4}_{s}

it follows that

q''_{u} = α_{S}G_{S} + εσT_{sky}^{4}  –  0.22(T_{s}  –  T_{\infty})^{4/3}  –  εσT_{s}^{4}\\ q''_{u} = α_{S}G_{S} + 0.22(T_{s}  –  T_{\infty})^{4/3}  –  εσ(T_{s}^{4}  –  T^{4}_{sky})\\ q''_{u} = 0.95 × 750  W/m^{2}  –  0.22(120  –  30)^{4/3}  W/m^{2}  –  0.1 × 5.67 × 10^{-8}  W/m^{2} · K^{4} (393^{4}  –  263^{4})  K^{4}\\ q''_{u} = (712.5  –  88.7  –  108.1)  W/m^{2} = 516  W/m^{2}

2. The collector efficiency, defined as the fraction of the solar irradiation extracted as useful energy, is then

η = \frac{q''_{u}}{G_{S}} = \frac{516  W/m^{2}}{750  W/m^{2}} = 0.69

Comments:
1. Since the spectral range of G_{atm} is entirely different from that of G_{S}, it would be incorrect to assume that α_{sky} = α_{S}.
2. The convection heat transfer coefficient is extremely low (\overline{h} ≈ 1  W/m^{2} · K). With a modest increase to \overline{h} = 5  W/m^{2} · K, the useful heat flux and the efficiency are reduced to q''_{u} = 154  W/m^{2} and η = 0.21. A cover plate can contribute significantly to reducing convection (and radiation) heat loss from the absorber plate.