No subject


Thu May 8 09:17:00 PDT 2014


Wiley & Sons Publisher, Chap 2.10 - Beam and Diffuse Components of Hourly
Radiation (taken from Erbs et al):

I_d / I = 1.0 - 0.09 * k_t   for k_t <= 0.22
          = 0.9511 - 0.1604 * k_t + 4.388*k_t^2 - 16.638*k_t^3 +12.336*k_t^4
for 0.22 < = k_t <= 0.8
          = 0.165 for k_t > 0.80

where:
   I_d / I = diffuse fraction of hourly total horiz radiation
   k_t = hourly clearness index

You should read the section before you apply this. Your engineering library
should have a copy. If not, let me know and I'll fax you a copy.

Good luck.

- Jim

________________________
Jim Kelsey P.E.
KW Energy Engineering
175 Filbert Street, Suite 205
Oakland, CA 94607-2541
510.834.6420
http://www.kw-energy.com



Vikas Shrestha wrote:

> Hello!
>
> I have hourly data on total horizontal radiation for Kathmandu, Nepal. I
> need to derive direct beam radiation and diffused radiation. I also do
> have hourly data on total sunshine hours and cloud cover.
>
> Does anybody have the algorithm handy to calculate direct beam and
> diffused components? I would really appreciate your help.
>
> Vikas Shrestha
> Graduate student
> Department of Architecture and Urban Design
> UCLA
>
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