A flat plate solar collector L = 2m long and W = 1m wide is inclined at θ = 13° to the horizontal. The cover plate is separated from the absorber plate by an air gap of s = 2.5 cm. The average temperature of the cover plate is Tc = 305 K, while the collector is at Th = 330 K. Estimate the

Question

A flat plate solar collector L = 2m long and W = 1m wide is inclined at [latex] heta=13^{\circ}[/latex] to the horizontal. The cover plate is separated from the absorber plate by an air gap of s = 2.5 cm. The average temperature of the cover plate is [latex]T_{c}=305 K[/latex], while the collector is at [latex]T_{h}=330 K[/latex]. Estimate the convective heat loss assuming air to be at 1 atm. Compare this with the solar energy input to the system at [latex]S=800 W / m ^{2}[/latex].

Accepted Answer

Air properties are taken at the mean air temperature of [latex]T_{m}=\frac{T_{h}+T_{c}}{2}=\frac{330+305}{2}= 317.5 K[/latex] and pressure of 1 atm.

Kinematic viscosity: [latex]v_{m}=17.46 imes 10^{-6} m ^{2} / s[/latex]

Thermal conductivity: [latex]k_{m}=0.0276 W / m ^{\circ} C[/latex]

Prandtl number: [latex]P r_{m}=0.704[/latex]

Isobaric expansion coefficient: [latex]\beta_{m}=\frac{1}{317.5}=0.0032 K ^{-1}[/latex]

The Rayleigh number based on air layer thickness is calculated.

[latex]R a_{s}=\frac{9.81 imes 0.0032(330-305) 0.025^{3}}{\left(17.46 imes 10^{-6} ight)^{2}} imes 0.704=27872[/latex]

The factor [latex]\psi[/latex] used in Hollands et al. correlation is then given by

[latex]\psi=\frac{1708}{17872 imes \cos 13}=0.0629[/latex]

Using the correlating equation n16.69, we have

[latex]\overline{N u}_{s}=1+\underbrace{1.44(1-\psi)\left[1-\psi \sin ^{1.6}(1.8 heta) ight]}_{0 ext { if negative }}+\underbrace{\left[\frac{0.664}{\psi^{\frac{1}{3}}}-1 ight]}_{0 ext { if negative }}[/latex] (16.69)

[latex]\overline{N u}_{s}=1+1.44(1-0.0629)\left[1-\psi \sin ^{1.6}(1.8 heta) ight]+\left[\frac{0.664}{0.0629^{\frac{1}{3}}}-1 ight]=3.00[/latex]

The convective heat transfer coefficient is then calculated as

[latex]\bar{h}=\frac{3 imes 0.0276}{0.025}=3.312 W / m ^{2 \circ} C[/latex]

The heat loss through the air layer is then given by

[latex]Q_{ ext {loss }}=\bar{h} L W\left(T_{h}-T_{c} ight)=3.312 imes 2 imes 1(330-305)=165.6 W[/latex]

Incident solar energy intercepted by the collector area is given by

[latex]Q_{ ext {Solar }}=L W S=2 imes 1 imes 800=1600 W[/latex]

The loss as a percentage of the incident solar flux is then given by

[latex]\frac{Q_{ ext {loss }} imes 100}{Q_{ ext {solar }}}=\frac{165.6 imes 100}{1600}=10.35 \%[/latex]

Question 16.10: A flat plate solar collector L = 2m long and W = 1m wide is ...

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