Determining the Power Density and the Solar Efficiency The solar parameters at a certain time of a day at a particular location are the zenith angle θ = 35°, the transmittance of all gases αdt = 75%, the water vapor absorption βwa = 5%, the transmittance of aerosols αp = 85%, and the standard

Question

Determining the Power Density and the Solar Efficiency

The solar parameters at a certain time of a day at a particular location are the zenith angle [latex]θ= 35°[/latex], the transmittance of all gases [latex]α_{dt} = 75\%[/latex], the water vapor absorption [latex]β_{wa }= 5\%[/latex], the transmittance of aerosols [latex]α_p = 85\%[/latex], and the standard deviation of the solar distribution function [latex]σ= 3.5 h[/latex].

a. Compute the power density and the solar efficiency at that time.
b. Compute the solar power density at 2 pm if the standard deviation of the solar distribution function is σ= 3.5 h.

Accepted Answer

[latex] heta =35°, \alpha _{dt}=75\% , \beta _{wa}= 5\% ,\alpha _p= 85\% , \sigma =3.5h, ho _o= 1353 KW/m^2[/latex], and t =2 pm

a. Equation (16.15)

[latex] ho _{ir}= ho _o \cos( heta )(\alpha _{dt -} \beta _{\omega a})\alpha _p[/latex] (16.15)

gives the solar power density

[latex] ho_{ir} = 1353 imes \cos (35) imes (0.75 - 0.05) imes 0.85 = 659.45 W/m^2[/latex]

Equation (16.16)

[latex]\eta _s=\frac{ ho _{ir}}{ ho _0}=\cos( heta )(\alpha _{dt}-\beta _{\omega a})\alpha _p [/latex] (16.16)

gives the solar efficiency

[latex]\eta _s=\cos (35) imes (0.75 - 0.05) imes 0.85 = 48.74\%[/latex]

b. The maximum power occurs at [latex]θ= 0[/latex] and Eq. (16.15) gives the maximum power density

[latex] ho_{max} = 1353 imes \cos (0) imes (0.75 - 0.05) imes 0.85 = 805.04 W/m^2[/latex]

Equation (16.17)

[latex] ho _{ir} = ho _{max}e^{\frac{-(t - t_o)^2}{2\sigma ^2} }[/latex] (16.17)

gives the power density at time t = 2 pm

[latex] ho = 805.04e^{\frac{-(2+12 -12)}{2 imes 3.5 ^2} }= 743.93 W/m^2[/latex]

Question 16.1: Determining the Power Density and the Solar Efficiency The s...

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