Load Calculation of Dual Batteries

[HungryDinosaur_imported]

Hi All,

We have a truck with 1 battery, and have installed a second battery for additional components in the truck (both are 24 V). How can I confirm that the auxiliary power system will not cause under voltage of the starter batteries, and how can I calculate based on the consumption, alternator specs etc.

Any good thumb rule of how much vehicle alternator capacity at idle or full load should be compared to the total load of the items (half, 1/3rd etc?).

Is there any calculations/literature on this load calculations anywhere? Thanks in advance.

 

[3DDave]

The alternator needs to supply the full electrical load unless you have added something like a 20kW amplifier for a sound system. Essentially if the alternator cannot keep up the battery will drain until there isn't enough current to operate something important. Mechanical injection diesel owners don't need to worry about that but owners of vehicles with electronic ignition certainly do.

It's like - "I have a tire with a leak. If I have an air pump does the air going in have to be 1/3 of the air going out?"

That's a tire that's going flat.

Whether it's necessary at idle depends on how long idling is to happen, but mainly, yes, the alternator needs to keep up with the electrical current and to recharge the battery for the next start.

 

[3DDave]

Oh, I forgot this - the bigger problem is to keep the battery charging balanced to prevent overcharging one of them or being unable to recharge both of them soon enough

There are devices called "isolators" that will direct current to the starting battery while it is below the full charge voltage. When the starting battery if fully charged then current is allowed to the secondary battery.

They also prevent a strong battery from being drained by a damaged weak battery.

 

[TugboatEng]

How old are the batteries?

 

[HungryDinosaur_imported]

Thanks Dave. Tugboat both are new batteries. It would be nice to have a literature related to this fundamentals.

 

[3DDave]

Add up the electrical load and then confirm the total is less than the amperage capacity of the alternator.

The only load that won't be directly added are intermittent loads, such as the starter motor or power windows, for which one needs a battery that can supply sufficient amperage and some amount of amperage to make up for the amp-hours drawn to start.

If the starter motor requires 500 amps for 15 seconds, is 500 Amps * (15/3600) hours = 5*15/36 Amp-hours, ~ 2 Amp-hours. So, if you want to recharge that in 20 minutes you need to provide 6 Amps, plus a little for inefficiency to do that. Not sure how conservative to make that for sub-zero weather.

If you want to see where you are now, use an Ammeter to measure the current out of the battery and turn the key to the run position. Remember to turn on all the accessories - headlights, inside fan on full blast, vehicle audio system; if you have electrically boosted power steering have someone crank that back and forth a bit. The only tough one is if the A/C compressor has a solenoid clutch to operate it. Some don't.

That reading on the Ammeter needs to be less than the rated output of the alternator, a value readily available for most alternators.

 

[GregLocock]

Peukert's law applies. So 500 A for 15 s from a 200 Ah battery will hit the battery for (500*15/3600)*(500/200)^1.4 =7.5 Ah





Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[3DDave]

Peukert's law may apply, but it is for the case of comparing constant discharge rates to get to a particular voltage.

Counting electrons in and electrons out should always apply unless water is split by hydrolysis.

Wattage is a different matter.

 

[HungryDinosaur_imported]

Thanks Dave, Greg.