Question 7.9: Beam Insolation on a Collector. In Example 7.8, at solar noo......

Renewable and Efficient Electric Power Systems [1073600]

Beam Insolation on a Collector. In Example 7.8, at solar noon in Atlanta (latitude 33.7º) on May 21 the altitude angle of the sun was found to be 76.4º and the clear-sky beam insolation was found to be 902 W/m². Find the beam insolation at that time on a collector that faces 20º toward the southeast if it is tipped up at a 52º angle.

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Using (7.26), the cosine of the incidence angle is

\begin{matrix} \cos \ \theta & = \ \cos \ \beta \ \cos\left(\phi_{S} \ – \ \phi_{C}\right) \ \sin \ \Sigma \ + \ \sin \ \beta \ \cos \ \Sigma \quad \quad \quad \quad \quad \quad \quad \quad \ \\ & = \ \cos \ 76.4º \ \cos\left(0 \ – \ 20º\right) \ \sin \ 52º \ + \ \sin \ 76.4º \ \cos \ 52º \ = \ 0.7725 \end{matrix}

From (7.24), the beam radiation on the collector is

I_{BC} \ = \ I_{B} \ \cos \ \theta \ = \ 902 \ {W}/{m^{2}} \ \cdot \ 0.7725 \ = \ 697 \ {W}/{m^{2}}

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