I have 4 x Vision AGM 200AH batteries. 

My usage is approximately 500 watts consistently. 

For how many hours would I expect my batteries to run before I reach my cutoff point of 46 volts which I have set on my inverter?

Many Thanks 

Subject to correction please, just trying to figure this out for myself also.

P = V x I

500W = 48 x I

500/48 = 62.5 amps draw from a 200Ah battery bank.

So 200Ah / 62.5 amps draw = 3.2 hours

But then then it is completely discharged. So for 46V (and here is where I fall out of the bus lol) one must figure out what the state of charge is of the battery.

According to Uncle Google, at 46.4 Volt your battery is 90% discharged. Only 10% remaining thus.

And you do not want to be doing that to AGM's or Gel batteries.

Hi Shiraz

If you set your voltage to go to a 50% DOD, with the available 4.8 KW and a load of 500W, your batteries should last between 9 and 10 hours

8 minutes ago, Swartkat said:

Subject to correction please, just trying to figure this out for myself also.

P = V x I

500W = 48 x I

500/48 = 62.5 amps draw from a 200Ah battery bank.

So 200Ah / 62.5 amps draw = 3.2 hours

But then then it is completely discharged. So for 46V (and here is where I fall out of the bus lol) one must figure out what the state of charge is of the battery.

Hi Swartkat

Watts = V x A

Thus, he has a potential of 48V x 200ah = 9600Wah, but at 50% DOD that just gives him usable 4800Wah. If we devide this by 500W, it will theoretically give him the 9 hours I stated. If I am not maybe missing something here

Hi PhilFM,

 

Thanks for this. I see where I went wrong. The available power to him is not 500W. This is what he wants to draw consistently. 

Happiness, it makes sense to me.

16 minutes ago, Swartkat said:

According to Uncle Google, at 46.4 Volt your battery is 90% discharged. Only 10% remaining thus.

And you do not want to be doing that to AGM's or Gel batteries.

Here I need to be corrected, hopefully someone here with more knowledge will help. But under load, 50% SOC will be close to 46V. If you take the load off, the voltage will probably shoot up to 48V again, which is basically 50% SOC without load.

This is what I got from google.

50% SOC will be 48.8V

 

State of Charge (%)	12V	24V	48V
100	12.83	25.66	51.32
90	12.72	25.44	50.88
80	12.60	25.20	50.4
70	12.47	24.94	49.88
60	12.34	24.68	49.36
50	12.20	24.40	48.80
40	12.06	24.12	48.24
30	11.91	23.82	47.64
20	11.76	23.52	47.04
10	11.61	23.22	46.44

Swartkat, yes I agree with those figures, but Voltage under load will read lower if I am right, and that would be the voltage the inverter uses to determine when to switch back to utility

So what I am basically saying, voltage on a battery not connected to anything will be higher than if it was under load. Same thing as you will see with your panels, no load, high voltage, the higher the load gets, the lower the voltage gets

Yes I absolutely agree with you.

I would hope that his inverter has the capability to use SOC instead of voltage across the battery terminals.