Average Monthly Insolation on a Tilted Collector. Average horizontal insolation in Oakland, California (latitude 37.73ºN) in July is 7.32 kWh/m²-day. Estimate the insolation on a south-facing collector at a tilt angle of 30º with respect to the horizontal. Assume ground reflectivity of 0.2.

Question

Average Monthly Insolation on a Tilted Collector. Average horizontal insolation in Oakland, California (latitude 37.73ºN) in July is 7.32 kWh/m²-day. Estimate the insolation on a south-facing collector at a tilt angle of 30º with respect to the horizontal. Assume ground reflectivity of 0.2.

Accepted Answer

Begin by finding mid-month declination and sunrise hour angle for July 16 (n = 197):

[latex]\begin{matrix} \delta & = \ 23.45 \ \sin \ \left[\frac{360}{365} \ \left(n \ − \ 81 ight) ight] \ = \ 23.45 \ \sin \ \left[\frac{360}{365} \ \left(197 \ − \ 81 ight) ight] & \ & = \ 21.35º \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ & \left(7.6 ight) \end{matrix}[/latex]

[latex]\begin{matrix} H_{SR} & = \ \cos^{−1} \left(− an \ L \ an \ \delta ight) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \left(7.17 ight) \ & = \ \cos^{−1} \left(− an \ 37.73º \ an \ 21.35º ight) \ = \ 107.6º \ = \ 1.878 \ ext{radians} & \end{matrix}[/latex]

Using a solar constant of 1.37 kW/m², the E.T. horizontal insolation from (7.43) is

[latex]\begin{matrix} \bar{I}_{0} & = \ \left(\frac{24}{\pi} ight) \ SC \ \left[1 \ + \ 0.034 \ \cos \ \left(\frac{360n}{365} ight) ight] \ \left(\cos \ L \ \cos \ \delta \ \sin \ H_{SR} \ + \ H_{SR} \ \sin \ L \ \sin \ \delta ight) \ & = \ \left(\frac{24}{\pi} ight) 1.37 \ \left[1 \ + \ 0.034 \ \cos \ \left(\frac{360 \ \cdot \ 197}{365} ight)º ight] \ ( \cos \ 37.73 \ \cos \ 21.35º \ \sin \ 107.6º \quad \quad \quad \ \ & \ + \ 1.878 \ \sin \ 37.73º \ \sin \ 21.35º ) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ & = \ 11.34 \ {kWh}/{m^{2}} ext{-day} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \end{matrix}[/latex]

From (7.42), the clearness index is

[latex]K_{T} \ = \ \frac{\bar{I}_{H}}{\bar{I}_{0}} \ = \ \frac{7.32 \ {kWh}/{m^{2}} \ \cdot \ ext{day}}{11.34 \ {kWh}/{m^{2}} \ \cdot \ ext{day}} \ = \ 0.645[/latex]

From (7.44) the fraction diffuse is

[latex]\begin{matrix} \frac{\bar{I}_{DH}}{\bar{I}_{H}} & = \ 1.390 \ − \ 4.027K_{T} \ + \ 5.531 {K_{T}}^{2} \ − \ 3.108 {K_{T}}^{3} \ \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ & = \ 1.390 \ − \ 4.027 \ \left(0.645 ight) \ + \ 5.531 \ \left(0.645 ight)^{2} \ − \ 3.108\ \left(0.645 ight)^{3} \ = \ 0.258 \end{matrix}[/latex]

So, the diffuse horizontal radiation is

[latex]\bar{I}_{DH} \ = \ 0.258 \ \cdot \ 7.32 \ = \ 1.89 \ {kWh}/{m^{2}} ext{-day}[/latex]

The diffuse radiation on the collector is given by (7.45)

[latex]\bar{I}_{DC} \ = \ \bar{I}_{DH} \ \left(\frac{1 \ + \ \cos \ \Sigma}{2} ight) \ = \ 1.89 \ \left(\frac{1 \ + \ \cos \ 30º}{2} ight) \ = \ 1.76 \ {kWh}/{m^{2}} ext{-day}[/latex]

The reflected radiation on the collector is given by (7.46)

[latex]\bar{I}_{RC} \ = \ ho \ \bar{I}_{H} \ \left(\frac{1 \ - \ \cos \ \Sigma}{2} ight) \ = \ 0.2 \ \cdot \ 7.32 \ \left(\frac{1 \ - \ \cos \ 30º}{2} ight) \ = \ 0.10 \ {kWh}/{m^{2}} ext{-day}[/latex]

From (7.41), the beam radiation on the horizontal surface is

[latex]\bar{I}_{BH} \ = \ \bar{I}_{H} \ - \ \bar{I}_{DH} \ = \ 7.32 \ − \ 1.89 \ = \ 5.43 \ {kWh}/{m^{2}} ext{-day}[/latex]

To adjust this for the collector tilt, first find the sunrise hour angle on the collector from (7.49)

[latex]\begin{matrix} H_{SRC} & = \ ext{min} \left\{\cos^{-1} \left(- \ an \ L \ an \ \delta ight), \ \cos^{-1} \left[- \ an \left(L \ - \ \Sigma ight) \ an \ \delta ight] ight\} \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ & = \ ext{min} \left\{\cos^{-1} \left(- \ an \ 37.73º \ an \ 21.35º ight), \ \cos^{-1} \left[- \ an \left(37.73 \ - \ 30 ight)º \ an \ 21.35º ight] ight\} \ & = \ ext{min}\left\{107.6º, \ 93.0º ight\} \ = \ 93.0º \ = \ 1.624 \ ext{radians} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \end{matrix}[/latex]

The beam tilt factor (7.48) is thus

[latex]\begin{matrix} \bar{R}_{B} & = \ \frac{\cos\left(L \ - \ \Sigma ight) \ \cos \ \delta \ \sin \ H_{SRC} \ + \ H_{SRC} \ \sin\left(L \ - \ \Sigma ight) \ \sin \ \delta}{\cos L \ \cos \ \delta \ \sin \ H_{SR} \ + \ H_{SR} \ \sin L \ \sin \ \delta} \quad \quad \quad \quad \quad \ & = \ \frac{\cos\left(37.73 \ - \ 30 ight)º \ \cos \ 21.35º \ \sin \ 93º \ + \ 1.624 \ \sin\left(37.73 \ - \ 30 ight)º \ \sin \ 21.35º}{\cos \ 37.73º \ \cos \ 21.35º \ \sin \ 107.6º \ + \ 1.878 \ \sin \ 37.73º \ \sin \ 21.35º} \ & = \ 0.893 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \end{matrix}[/latex]

So the beam insolation on the collector is

[latex]\bar{I}_{BC} \ = \ \bar{I}_{BH} \bar{R}_{B} \ = \ 5.43 \ \cdot \ 0.893 \ = \ 4.85 \ {kWh}/{m^{2}} ext{-day}[/latex]

Total insolation on the collector is thus

[latex]\bar{I}_{C} \ = \ \bar{I}_{BC} \ + \ \bar{I}_{DC} \ + \ \bar{I}_{RC} \ = \ 4.85 \ + \ 1.76 \ + \ 0.10 \ = \ 6.7 \ {kWh}/{m^{2}} ext{-day}[/latex]

Question 7.13: Average Monthly Insolation on a Tilted Collector. Average ho......

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