# Solar Energy Calculation

1. Sep 12, 2011

### athar.walile

Can we calculate the daily energy generation of 20KW power pack if have the following information:

Date Time GHI DNI DHI
01-01-2008 08:00 0 0 0
01-01-2008 09:00 149 398 75
01-01-2008 10:00 348 686 98
01-01-2008 11:00 512 792 112
01-01-2008 12:00 626 845 119
01-01-2008 13:00 674 859 123
01-01-2008 14:00 659 860 119
01-01-2008 15:00 574 827 112
01-01-2008 16:00 431 753 100
01-01-2008 17:00 245 497 106
01-01-2008 18:00 36 63 30
01-01-2008 19:00 0 0 0

Thank you.

2. Sep 12, 2011

### Staff: Mentor

What is the context of the question? Can you please provide more details? Is this question for schoolwork?

3. Sep 12, 2011

### athar.walile

Based on in information i have,

Location of Project : Latitude- 26.9° N, Longitude- 72.0° E
Size of Project : 20 KW

I want to find out the possible generation of energy for that plant in KW. I have provided the hourly based meteo data.

4. Sep 13, 2011

### athar.walile

Here is the solution

Total generation of units (kWh) = Insolation factor x Yield Factor x Total installed capacity

5. Sep 13, 2011

### Zryn

Are your panels fixed? What is their orientation?

How are you using the GHI, DNI and DHI to arrive at a value of Insolation, and is this data your own?

Can you expand upon what Yield Factor incorporates?

6. Sep 14, 2011

### athar.walile

Yield factor is 0.89

Little confused about using GHI, DNI and DHI values in my insulation calc. but i m using GHI in my calculation. I am not aware of how to use meteorological data here.

I got this data from NREL website: http://rredc.nrel.gov/solar/new_data/India/nearestcell.cgi?

7. Sep 14, 2011

### Zryn

Find out your Latitude & Longitude:

http://www.whatsmygps.com/"

http://eosweb.larc.nasa.gov/cgi-bin/sse/grid.cgi?email"

2. Enter your Latitude & Longitude
3. Select the data you want ('Average Insolation' only will suffice)

You now have your average monthly Insolation.

Alternatively you can calculate the contribution from GHI, DNI and DHI to your Hourly Insolation relative to the orientation of your solar panel(s) and get much more accurate specific site data, but if you don't need it, the NASA information is a good enough approximation.

So if you have a 2kW system, and have 5 Hours Insolation, you can expect to get (2kW * 5H =) 10kWH from your system that day.

I assume the Yield Factor is an expression of efficiency, so at 89% you would end up with (10kWH * 0.89 =) 8.9kWH instead.

At $0.20 per kWH you would have made/saved (8.9kWH *$0.20 =) \$1.78 that day!

Last edited by a moderator: Apr 26, 2017