#### Question

On an overcast day the directional distribution of the solar radiation incident on the earth’s surface may be approximated by an expression of the form $I_{i}=I_{n} \cos \theta$, where $I_{n}=80 \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{sr}$ is the total intensity of radiation directed normal to the surface and $\theta$ is the zenith angle. What is the solar irradiation at the earth’s surface?

#### Solution

Verified#### Step 1

1 of 2Given: $I_i = I_n \cos \theta$, $I_n = 80$ W/m$^2$.sr

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The total intensity is given by,
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$\begin{align*} G & = \int_{0}^{ 2 \pi} \int_{0}^{\pi /2 } I_i(\theta) \cos \theta \sin \theta d \theta d \phi \\ & = 2 \pi I_n \int_{0}^{\pi /2 } \cos ^2 \theta \sin \theta d \theta\\ & = 2 \pi I_n \bigg[ - \frac13 \cos ^3 \theta \bigg] \bigg|_0^{\pi /2}\\ & = 2 \pi I_n \cdot - \frac13 \bigg[ \cos ^3 (\pi /2) - \cos ^3 0 \bigg]\\ & = 2 \cdot \pi \cdot 80 \frac{W}{m^2 \cdot sr} \cdot \frac13 sr\\ & = 167.55 \text{ W/m}^2 \end{align*}$

$\boxed{ G = 167.55 \text{ W/m}^2}$