## Chapter 5

## 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.

## Step-by-Step

## 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 |