# Determining Battery Bank Size

#### chris1971

##### Senior Member
Just curious how would you determine the size battery back up bank needed for a 240 volt single phase 40 amp resistive load? Pardon my lack of understanding on this subject.

#### drcampbell

##### Senior Member
Multiply the 40 amps that your load consumes by the number of hours you'd like it to run on a charge and install a battery bank with that many amp-hours capacity.
(assuming that it's a 240-volt battery bank)

For example, If it needs to play for two days: 40 amps * 48 hours = 1920 amp*hours.

Caution: Switches, fuses, relay contacts and the like rated for 250 volts AC will be woefully inadequate for 250 volts DC. Be sure that every component has an adequate DC rating.
Caution: Batteries cannot be switched off. Hot-work procedures will be required to safely assemble this battery bank.
Suggestion: An engine-driven generator will almost certainly be a more economical solution.

#### LarryFine

##### Master Electrician Electric Contractor Richmond VA
Wouldn't the capacity need to be higher than that to make sure the voltage doesn't sag near the end of the desired run time?

#### drcampbell

##### Senior Member
Maybe.

A battery's voltage will start dropping the moment you connect a load to it and will continue dropping for as long as it's powering a load.

Is it necessary to prevent (excessive) voltage sag? I don't know. That depends on the nature of the load. If it's a resistance heater, probably not.
There are a gazillion variables to consider when designing a battery bank. I didn't go into any detail because I suspect that this project is a non-starter as soon as someone investigates the cost of batteries.

#### winnie

##### Senior Member
drcampbell's post gives you a bare minimum which might run your load correctly.

The key requirement is that the battery store sufficient energy to serve the required load for the required time.

It would be convenient if batteries were rated in watt hours, but instead they are rated in amp-hours. This is simply the current delivered by the battery times the duration of the current flow, as the battery discharges from 'full' to 'empty'. The devil is in the details of this definition; how do you define 'full' and 'empty', and what current level do you use....

For many storage batteries,'empty' is defined either by a particular open circuit voltage, or by a particular voltage under load. So the battery is 'empty' when the voltage reaches a threshold. You could keep discharging the battery (there is more energy in there) but this may damage the battery.

Batteries have internal resistance, so they deliver less total energy at higher currents. So the amp-hour rating will be at a particular current level. For lead-acid storage batteries, this is usually at the 'C/20' rate, meaning a current that is 1/20 of the amp-hour capacity. If you need to run for only 1 hour than your battery will have a lower effective amp-hour capacity.

In general, deep discharge reduces the life of a battery. So you probably want to have larger than minimum capacity for your load requirements.

With more details, we could probably work through a battery sizing problem.

-Jon

#### SceneryDriver

##### Senior Member
Wouldn't the capacity need to be higher than that to make sure the voltage doesn't sag near the end of the desired run time?
Assuming lead-acid batteries, roughly half of the Ah capacity is usually usable before the voltage really begins to drop. DC-DC converters will extend that quite a bit, as they output a steady voltage while their input voltage begins to drop. It's not a linear relationship - discharge rate VS usable capacity - thanks to the dark magic that is battery chemistry. Also remember that deeply discharging lead-acid batteries will drastically shorten their lifespan, "deep cycle" or not.

At that size and voltage however, have your checkbook ready. Better yet, have someone else's checkbook ready. Why do you need so large a battery bank? How will the battery bank be recharged? A generator will likely be much more cost effective. If you're looking at generating heat only (OP said resistive load), there are much more efficient ways than making electricity first.

SceneryDriver