# PV system battery calculations question

Discussion in 'Power Electronics' started by irmanao, Jun 5, 2018.

1. ### irmanao Thread Starter Member

Apr 29, 2017
83
1
I'm designing a stand alone system and came across a problem: If all appliances working at the same time consume 1000Wh what capacity must the battery be in order to be able to feed the house for 10 hours?

Lets say the battery (pack) is 60V and the input AC voltage of the appliances 230V. What i thought is:1000*10=10000Wh, 10000Wh*(230/60)=38333Wh
and for the inverter efficiency: 38333Wh/0.95=40350Wh. So the battery pack's capacity must be 40.35kWh in order to feed the house for 10 hours. Am i right or am i missing something?

thanks

2. ### Hymie Active Member

Mar 30, 2018
619
160
Watt hours are independent of voltage (where you have factored Wh by the voltage difference 230V/60V).

With a power draw of 1kW/hour for 10 hours will be 10kWh.
Assuming an inverter efficiency of 95%, then the primary power source (battery) must have a capacity of 10/0.95 = 10.5kWh.

If you consider that you have a 60Vdc battery, then to deliver a constant 1.05kW the 60V battery will be discharging at a rate of 17.5A. To do this for 10 hours the battery must have a capacity of 175Ah (60V x 175Ah = 10.5kWh).

Apr 29, 2017
83
1
4. ### irmanao Thread Starter Member

Apr 29, 2017
83
1
You are actually right. The AC load is 6kWh and the DC load 6.66kWh (139*48), which is 6000/0.9. Correct?