Question

Show that, for places of latitude angle of l, the solar attitude angle at solar noon (when maximum solar radiation of the day is achieved) can be calculated by ? = 90 - ?l-?? where ? is the deflection angle of the day of consideration

Answer 1

Since Zenith angle= latitude of interest- solar declination

Latitude of interest=40 degree N

Solar declination=-231/2 degree S

Therefore zenith angle=40- 23/12

=16.5

Since the zenith angle is equal to the latitude separating the person at 40 degree latitude and the place recieving direct sun light. Therefore the value of this angle is 16.5 degree.

Now, declination angle= sin-1 sin(23.5 degree) sin 360/365(d-81)

where d= no of days=??

-231/2=sin-1sin(23.5 degree)sin 360/365( d-81)

-23/12=sin-1(0.398)sin0.98(d-81)

-231/2=sin-1(0.398)(0.017)(d-81)

-231/2=sin-1(0.00676)(d-81)

-231/2=0.387(d-81)

(231/2)/(0.387)=d-81

60.7=d-81

d=60.7+81

=141.7

=142

Therefore 142 will be either 22th or 23 th of May starting d=1 as January 1st.

3.Solar elevation angle= 90-zenith angle

=90-19.5

=73.5 degree