Inveter backup with 100ah battery

Registered Users Posts: 10 ✭✭

I want to know that what would be the backup of 700 watt inverter with 250 watt, 220 v load running on 12 volt with 100ah lead acid flooded battery.

Very roughly:

• 250 Watt load * 1/0.85 inverter eff * 1/12 volt nominal = 24.5 Amp load @ 12 volts

You have a 12 volt @ 100 AH battery, to take the battery dead:

• 100 AH / 24.5 amps = 4.1 hours to "dead"

The reality is that you should only take a lead acid battery to 50% state of charge if you want it to have a long cycle life. So, that would be ~2 hours of use.

And the 100 AH rating is typically at a 20 Hour discharge rate:

• 100 AH / 20 hour = 5 amp load

When you draw at a C/4 hour discharge rate, the battery has less "apparent" capacity. And it does differ based on brand/model/type of battery (Deep cycle Lead Acid, AGM, Li Ion, etc.). Your 12 volt (assuming it is a deep cycle Flooded Cell Lead Acid) battery will probably have less than 80% capacity at C/4 discharge rate:

• 100 AH * 80% = 80 AH
• 80 AH * 0.50 = 40 AH @ normal cycling
• 40 AH / 24.5 Amp load = 1.6 hours suggested usage

You can take a FLA battery (deep cycle type, not a Marine/Car type battery) to 20% SoC if it is recharge quickly after the "job" is done:

• 80 AH (C/4 discharge rate) * 0.80 discharge = 64 Amps maximum suggested draw
• 64 AH / 24.5 amp load = 2.6 hours maximum usage before ruining battery

-Bill

Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
• Registered Users Posts: 10 ✭✭

How can we know that that how much of power an inverter take amps/watt from dc12 volt battery to generate an ac watt 220. I mean 10 watt of ac 220 volt generated by modified sine wave inveter will take how much power from dc 12 volt battery source

The 85% (typically 80-90+%) inverter efficiency only applies to a "moderately to heavily" loaded AC inverter output.

When you very lightly load a large AC inverter the efficiency is much less. For example, say your inverter has a 15 Watt "tare" load (the amount of energy it take just the AC inverter being turned on, and then add the AC loads such as your 10 Watt 220 VAC load, roughly, the current draw on a 12 volt battery would be:

• 15 Watts tare
• 10 Watt * 1/0.85 AC inverter eff = 11.8 Watts of DC power to supply AC load
• 15 Watts Tare + 11.8 Watts DC load = 26.8 Watts total load on DC bus
• Assuming battery is running at ~12.3 volts from a partially discharged battery bank:
• 26.8 Watt DC load / 12.3 volts DC battery bus = 2.2 amp estimated DC bus draw
• Assuming battery is being charged at 14.5 volts:
• 26.8 Watt DC load / 14.5 volts DC battery bus = 1.8 amp estimated DC bus draw

You can also directly measure the battery (or AC inverter) current with a DC shunt (precision Power Resistor) and a voltmeter calibrated to read amps from the shunt.

Or you can get a DC Current Clamp Digital Multi-Meter and measure the DC current draw that way.

On the AC side, you can use a Kill-a-Watt type meter to measure the AC power draw too.

-Bill

Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
• Registered Users Posts: 390 ✭✭✭✭

BB,

You can't use the 1/.85 figure for calculating efficiency at small loads , you need to use the effeciency at that power level , I've seen Harbor Freight inverters to draw 220-250 watts at 12 volts to power a 13 watt CFL. Effeciency at 10% load is less than 40% on most inverters , worse on a Chinese inverters, add that to the tare load and things get ugly real fast.

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