# Calculating Battery Run Time

KnowledgeSponge
Solar Expert Posts:

**173**✭✭✭
Hi,

I'm using this formula for battery run time calculations...but I don't think it takes into consideration Inverter loss.

Backup Time = Battery AH x 12V x N x Efficiency of Battery / Load in Watts

where,

Battery AH = Ampere Hour Capacity of Battery

N = Number of 12 V Batteries needed

Efficiency of Battery = Generally it is 0.8, which is the max. power factor of home standard

Is there a better formula somewhere here ?

Thx

I'm using this formula for battery run time calculations...but I don't think it takes into consideration Inverter loss.

Backup Time = Battery AH x 12V x N x Efficiency of Battery / Load in Watts

where,

Battery AH = Ampere Hour Capacity of Battery

N = Number of 12 V Batteries needed

Efficiency of Battery = Generally it is 0.8, which is the max. power factor of home standard

Is there a better formula somewhere here ?

Thx

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## Comments

173✭✭✭Ok,

Found out the ProWatt SW 2000 has an "Optimum Efficiency" of 90%

But I have no idea what that means in the real world

Does that mean I can just add 10% usage to all my calculations?

10,300✭✭✭✭you're on the right track with that 10%, but it is at least adding that 10% as inverter efficiency tends to change with varying load levels. the 10% represents the least amount of power the inverter will consume as it can be more and often is. to be on the safe side go with 20% and don't forget that an inverter will consume power even if your loads to it are not on. as long as the inverter is powered it will have a minimum power draw so keep that in mind.

17,615✭✭Oh if only it were that simple! But it's not.

Why not? Because batteries vary in their Peukart curves. Because temperatures affect things. Because efficiency isn't linear over Voltage/current/Watts/time ....

The closest you can come is an approximation based on nominals and averages.

Something like this:

Amp hours (usable) * nominal battery Voltage = approximate Watt hour capacity DC.

From that subtract the power required by the inverter.

What's left over times the inverter's conversion efficiency (that 90% figure) = approximate AC Watt hours available.

What skews it:

Heavy current draw = lower actual capacity of batteries. So if you draw 10 Amps from a 100 Amp hour battery its real capacity is less than it is if you draw 5 Amps, regardless of the total Watt hours consumed.

Temperature; low = less capacity, high = less capacity. The capacity is figured at a consistent 25C which doesn't actually happen in the real world.

Inverter efficiency; 90% at a load of so much percent. Becomes less efficient with more load or even less load.

How you deal with this: margins for error. Tolerances. Buffers. Rounding numbers in the right direction to be on the safe side.

1,925✭✭Battery effiiciency has nothing to do with that. It only matters while charging. However, inverter efficiency does. Also, very important factor is the depth of discharge because you won't discharge your battery 100%. So, it should be:

Backup Time = Battery AH x V x N x Depth of Discharge x Inverter Efficiency (including wiring)/ Load in Watts

For example, I have 8 6V batteries rated 673AH and I want to run a 3kW and discharge battery to 50%

Backup Time = 673 x 6 x 8 x 0.5 x 0.85 / 3000 = 4.6 hr

Once I calculated this, I notice that 673AH is a 20-hr rating, but I'm discharging to 50% in 5-hr. So, my usage more resembles 10-hr full discharge. So, I go back to the spec sheet and lookup up 10-hr rate, which is 604AH. I go back and re-calculate:

Backup Time = 604 x 6 x 8 x 0.5 x 0.85 / 3000 = 4.1 hr

Battery capacity depends on temperature and also falls as batteries age. So, I want to derate battery capacity to account for these factors. I do

604 x 0.8 = 483AH

Now I'm ready for the final calculation

Backup Time = 483 x 6 x 8 x 0.5 x 0.85 / 3000 = 3.3 hr

173✭✭✭Good stuff guys !

Very helpful forum.