Find the insolation in Wm^-2 at summer solstice for a low obliquity of 22 degrees?

Question

I used the following:

h_0=cos^-1(-tan(phi)* tan(delta))

cos(theta)=[sin(phi)*sin(delta)+cos(phi)*cos(delta)*cos(h_0)]

Result was 397 Wm^-2

This looked like an easy problem, but I am having difficulty with it.

Thanks for your help.

Answer 1

I don't follow your work, however, it is laid out plainly here:

http://education.gsfc.nasa.gov/experimental/July61999siteupdate/inv99Project.Site/Pages/solar.insolation.html

Answer 2

Thanks Bob. I neglected to give the latitude of 65 degrees North.
I also looked at the reference material. From what I can see, you need an hour variable. The question does not indicate a time. This is why I find the question difficult.

Answer 3

put in 12 for noon.

Answer 4

Hi,

Hour angle= 15 degrees*(time-12)
15 degrees*(12-12)=0
H=cos(0)=1

Z=cos^-1(sin(65 degrees)*sin(22 degrees)+cos(65 degrees)*cos(22 degrees)*cos(0)
Z=0.7504915

I=ScosZ

S=1368 Wm^-2
I=1368*cos(0.7504915)
I=1000.492 Wm^-2

Is this correct in your opinion?