Previous in Forum: minimum input to an inverter Next in Forum: bus bar
Member

Join Date: Feb 2009
Posts: 5

# home inverter

06/13/2009 7:38 AM

I am having 800VA INVERTER , WITH 150 AH BATTERY AND SUPPLY VOLTAGE IS 230 VOLT, WHAT WILL SIMPLE FORMULA FOR LOAD TIME CALCULATION WITH FULL LOAD ????

ABHISHEK

Interested in this topic? By joining CR4 you can "subscribe" to
Guru

Join Date: Oct 2007
Location: Sour Lake, TX 30°08'59.68"N 94°19'42.81"W
Posts: 671
#1

### Re: home inverter

06/13/2009 10:57 AM

Here is some very approximate calculation:

If your inverter has an efficiency of 90%, than the load that your battery side sees is about 890W. You haven't mention the voltage of the battery. If it is 24 V, than a 150 Ah battery will provide (remember that you cannot suck dry the current from the battery - you need to leave some there) a couple of hours of working output.

The 890 Watt will mean some 37 amp load. This is a bit too a heavy load for your battery.If you had a perfect battery, 2/3 of its capacity would mean some 99 Ah therefore some two hours of full load working. But you should use a much larger capacity battery, your load is heavy.

P.S. Don't use capital letters, it is a bit aggressive

__________________
Bridge rule #1: Nobody is as good as he thinks about himself nor as dumb, as his partner thinks...
Associate

Join Date: Nov 2007
Location: Denmark, Aalborg
Posts: 53
#2

### Re: home inverter

06/15/2009 4:26 PM

Hi, Indels post is accurate, but lacks the formula. You have the 890 VA which we for the simplification (and powerfactor of 1) assumes is equal to watts, so here goes:

890/battery voltage = xx Amps of continous draw. Indel is correct with the 24 V example. Then you take the Amp-hours of your batterys and divide with your Amp-maxdraw, and you have the time:150/37= 4,05 hours. Be careful to not suck the batterys dry, as they will be destroyed. As a thumb rule for lead-acid cells you can draw power until 10 % under rated voltage, e.g. 12 V - 1,2 V is 10,8 V.

It is not a linear draw, because to start with there is a little higher voltage, so that the Amperage draw is less. But when the voltage drops, the Amperage rises to compensate for the Voltage drop and the time shortens. And it is faster and faster dropping, when voltage comes under e.g 11 V.

Good luck, moe

__________________
Don't think, play!