Calculate the average daily radiation on a horizontal surface in a location at $28^{\circ}$35N .If average sunshine hours is 9.5, a=0.35 b=0.4 on December 01.

Subject:- Renewable Energy Sources

Topic:- Solar Energy

Difficulty:- High

$∅=28^{\circ} 35^{1} = 28[\frac{35}{60}] = 28.583^{\circ}$

Average sun shine hours/day=9.5 hours on December 1

a=0.35,b=0.4

No. of days, n=365-30

=335 days on 1 December.

Solar declination, δ

δ=23.45 sin$([\frac{360}{365}](284+n))$

= 23.45 sin$([\frac{360}{365}](284+335))$

= $-22.06^{\circ}$

Sunrise hour angle is given by the equation

ωs= $cos^{-1}(-tan∅.tanδ )$

= $cos^{-1}[-tan 28.583 × tan(-22.06)⦌$

=77.244$^{\circ}$

=77.244×$[\frac{P}{180}]$ radians.

= 1.3482 radians .

Day length,

Smax= $\frac{2}{15}$ ωs

= $\frac{2}{15}$×77.244

= 10.299 Hrs.

$H_{0} =\frac{24}{P} Isc(1+0.033 cos[\frac{360}{365}])×(ωs Sin∅sinδ+cos∅cosδsinωs)$

$\frac{24}{P}$× 1.367 × 3600 $(1+0.033 cos [\frac{(360 \times 334)}{365}]$× $⦋(1.3482 sin(28.583)$

$sin(-22.06)+cos(28.583) \times cos(-22.06) sin(77.244)$

= 21328.6KJ/$m^{2}$ /day

=$[\frac{H_{g}}{H_{0}}](a+b s/S_{max})$

= 0.35+0.4(9.5/10.299)

= 0.718968

∴$H_{g} =H_{0}$ x 0.718968

= 21328.6x 0.718968

H = 15334.6KJ/$m^{2}$/day