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## Q. 5.2

Use the simple model for solar radiation under cloudless skies (p. 64) to estimate the direct, diffuse, and total solar irradiance on a horizontal surface at 45°N at local solar noon on 21 June for three values of aerosol optical thickness, 0.05, 0.20, 0.40. Assume that the optical thickness for molecular attenuation is 0.30.

## Verified Solution

At solar noon $\psi =\phi -\delta =45.0-23.5=21.5$. Hence m = sec 21.5 = 1.075. Then $S_{p} = 1366 \exp (-0.30\times 1.075) (\exp -(1.075\times \tau_{a} )) =990\exp (-1.075\times \tau_{a} )$ and $S_{s} = S_{p} \cos \psi$.

Diffuse radiation is calculated as $S_{d} = 0.3 \times 1366 \times \cos 21.5(1-\exp (-(\tau_{m} + \tau_{a} )\times 1.075))$, and $S_{t} =S_{s} +S_{d}$. The table shows results of the calculations in W $m^{-2}$ for the three values of $τ_{a}$.

 $τ_{a}$ $S_{p}$ $S_{s}$ $S_{d}$ $S_{t}$ 0.05 941 875 118 993 0.20 802 746 160 906 0.4 644 599 202 801