How many 100amp hour batteries do you need to connect to a 96VDC inverter to produce the same output as a 5kva generator for a period of 8 hours?

Answer 1

We can only do this in a sketchy way, because some important numbers are missing.

-- You didn't mention the voltage of the 100-AH batteries.

-- We don't know whether the inverter can deliver 5,000 watts, like the generator can.

-- We also don't know the efficiency of the inverter.

So we'll go at it in terms of the total energy, we'll assume that the inverter can do

whatever we want it to do, and at the end, we can toss in a term for its efficiency.

-- At the back end, you specified 5 KVA for 8 hours. If the generator is 100% loaded,

with a power factor of 100%, then that's (5 x 8) = 40 KWH of energy in 8 hours.

-- If the inverter is 100% efficient, then 40 KWH is what the batteries need to supply, in 8 hours.

-- I'm going to assume that your batteries are the nominal-12-VDC "car battery" type.

So you need 8 of them in series to power the 96-VDC inverter.

-- The batteries keel over and die after delivering 100 Amp-Hours. At 96 volts,

that's 9,600 watt-hours of energy.

-- Throughout the 8 hours, you need 40,000 watt-hours. One string of batteries

delivers 9,600. That's 24% of the total energy required. Instead of 24%, I want to

write it as 6/25 of the total energy. You'll see why in the next line . . .

-- Since the one 96-volt string of 8 batteries produced 6/25 of the required total energy,

you need 25/6 = 4.1666 strings, in parallel.

You have to supply the batteries in sets of 96 volts, in order to run the inverter.

So far, the answer is 41/6 96-volt strings.

We assumed that a 100-AH battery could actually deliver 100-AH, and the voltage

would never sag. Realistically, you shouldn't count on more than maybe 90% of that.

Call that number the "battery margin".

We assumed that the inverter is 100% efficient ... all of the energy that comes

out of the batteries shows up at the inverter's output. That never happens. There's

always some energy lost in the inverter. I'm going to call the efficiency of the

inverter ' E '. It's a percent, it's less than 100, and I'll leave it up to you to find out.

So, here's the best I can do: You asked for the number of batteries.

-- Number of batteries in each series string = 96/battery voltage .

For 12-V car batteries, it's 8 batteries in each string.

Each battery assumed to be rated 100-AH.

-- Number of strings:

(41,666) divided by [ (battery margin percent) times ( ' E ' percent) ]

========================

Example:

Battery margin = 90%

Inverter efficiency = 80%

Number of strings = (41,666)/[90 x 80] = 41,666/7,200 = 5.79 . . . Use 6 strings.

RED FLAG . . .This whole discussion assumes that the inverter is rated for 5,000 watts output or more.

If it's not, then this has been an interesting exercise, but please forget the whole idea.

Your inverter CANNOT match the output of a 5 KVA generator for more than about 10 seconds.

After that, it'll be the clickety-clack of tripped breakers, echoing through the smoke.