danishdeshmuk
Advanced Member level 1
Are these calculations right about solar panel ?
To generate regularly (24hrs) 24Amps load , how many solar panels & cells are required ? What capacity battery & battery charger would be required to charge the battery ............
P=V*I = 24*230 = 5520W
Conversion efficiency for DC to AC will limit out at 90%, so you need 5520/0.9 = 6133W into the converter.
Battery storage/recovery efficiency will be a max of 90% since this is a PV application. If we were looking at an electric vehicle, it is a max of 70%. So now, power into storage is 6133/0.9 = 6814W and a day's worth of storage is 24*6814= 164KW-hr
Since the sun only shines part of the day, we will need to generate all the electricity in that part of the day. Assuming no clouds, we can get about an average of 50% of peek power for no more than 8 hours. We need 164KW-hr into storage, so we need to generate 328KW-hr in 8 hours or 41KW per hour for our 8 hour period.
The typical solar cell efficiency is 12%, and the sun is 1.1KW/m^2 so that means we need:
(41/0.12)/1.1 m^2 of cells = 311square meters of cells.
To generate regularly (24hrs) 24Amps load , how many solar panels & cells are required ? What capacity battery & battery charger would be required to charge the battery ............
P=V*I = 24*230 = 5520W
Conversion efficiency for DC to AC will limit out at 90%, so you need 5520/0.9 = 6133W into the converter.
Battery storage/recovery efficiency will be a max of 90% since this is a PV application. If we were looking at an electric vehicle, it is a max of 70%. So now, power into storage is 6133/0.9 = 6814W and a day's worth of storage is 24*6814= 164KW-hr
Since the sun only shines part of the day, we will need to generate all the electricity in that part of the day. Assuming no clouds, we can get about an average of 50% of peek power for no more than 8 hours. We need 164KW-hr into storage, so we need to generate 328KW-hr in 8 hours or 41KW per hour for our 8 hour period.
The typical solar cell efficiency is 12%, and the sun is 1.1KW/m^2 so that means we need:
(41/0.12)/1.1 m^2 of cells = 311square meters of cells.