A flat-plate solar collector with no cover plate has a selective absorber surface of emissivity 0.1 and solar absorptivity 0.95. At a given time of day the absorber surface temperature Ts is 120°C when the solar irradiation is 750 W/m², the effective sky temperature is -10°C, and the ambient air

Question

A flat-plate solar collector with no cover plate has a selective absorber surface of emissivity 0.1 and solar absorptivity 0.95. At a given time of day the absorber surface temperature $T_{s}$ is 120°C when the solar irradiation is 750 W/m², the effective sky temperature is -10°C, and the ambient air temperature $T_{∞}$ is 30°C. Assume that the heat transfer convection coefficient for the calm day conditions can be estimated from

$\overline{h} = 0.22(T_{s} – T_{∞})^{1/3} W/m^{2} · K$

Calculate the useful heat removal rate (W/m²) from the collector for these conditions. What is the corresponding efficiency of the collector?

[latex]\overline{h} = 0.22(T_{s} - T_{∞})^{1/3} W/m^{2} · K[/latex]

Calculate the useful heat removal rate (W/m²) from the collector for these conditions. What is the corresponding efficiency of the collector?

Accepted Answer

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

Find:
1. Useful heat removal rate per unit area, [latex]q''_{u} (W/m^{2})[/latex].
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,

[latex]\dot{E}_{in} - \dot{E}_{out} = 0[/latex]

or, per unit surface area,

[latex]α_{S}G_{S} + α_{sky}G_{atm} - q''_{conv} - E - q''_{u} = 0[/latex]

From Equation 12.73,

[latex]G_{atm} = σT_{sky}^{4}[/latex] (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

[latex]α_{sky} ≈ ε = 0.1[/latex]

With

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

it follows that

[latex]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}[/latex]

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

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

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

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

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