geewee
Electrical
- Aug 25, 2014
- 1
Hi,
I'm having questions about calculating the charge efficiency of electric batteries.
For example: A 100Ah battery in the 24V PV system setup is discharged with a current of 1A. So could last for 100 hours (if its new and allowed to discharge completely).
Now I thought it would be reasonable to use the avarage dischargevoltage of 23 volts. So thats 23Vx10Vx10h=2300Wh
Now I want to now the required amount of energy to charge the battery so a 100Ah will be available again.
So if I would want to charge it I will need more then the 2300Wh.
However the battery needs to be charged at about 28V. Assumed that the efficiency of the battery is 90% (I'm not taking the the effect of the state of charge on the efficiency in account for the sake of this example).
Now where I get stuck:
Do I need to correct the Energy in Wh or the Ah? In the case of Wh it would be 2300Wh/0.9 is about 2556Wh needs to be charged in to the battery. At 28V this is 2556/28 = 91Ah..
or...
Calculate the Ah going out first: 2300Wh/23V=100Ah. Increase the Ah number with the efficiency: 100Ah/0.9=111Ah. And calculate the Wh going in: 111Ah x 28V = 3111Wh.
This is quite a difference. In case 1 I need 2556Wh and in case 2 3111Wh. Because in case to the voltage difference between charging and discharging plays a role ass well.
What is the right way to do it.
Appreciate your comments.
regards Geert
I'm having questions about calculating the charge efficiency of electric batteries.
For example: A 100Ah battery in the 24V PV system setup is discharged with a current of 1A. So could last for 100 hours (if its new and allowed to discharge completely).
Now I thought it would be reasonable to use the avarage dischargevoltage of 23 volts. So thats 23Vx10Vx10h=2300Wh
Now I want to now the required amount of energy to charge the battery so a 100Ah will be available again.
So if I would want to charge it I will need more then the 2300Wh.
However the battery needs to be charged at about 28V. Assumed that the efficiency of the battery is 90% (I'm not taking the the effect of the state of charge on the efficiency in account for the sake of this example).
Now where I get stuck:
Do I need to correct the Energy in Wh or the Ah? In the case of Wh it would be 2300Wh/0.9 is about 2556Wh needs to be charged in to the battery. At 28V this is 2556/28 = 91Ah..
or...
Calculate the Ah going out first: 2300Wh/23V=100Ah. Increase the Ah number with the efficiency: 100Ah/0.9=111Ah. And calculate the Wh going in: 111Ah x 28V = 3111Wh.
This is quite a difference. In case 1 I need 2556Wh and in case 2 3111Wh. Because in case to the voltage difference between charging and discharging plays a role ass well.
What is the right way to do it.
Appreciate your comments.
regards Geert