Help with calculations: length of time running on battery

[qwerty11]

I need help with this as I'm getting confused. Could somebody break down the calculations to figure:

How long will a 12V battery with an amp/hr rating of Y (assuming its based on 20hrs) maintain a steady watt load of X before the battery reaches 50% discharge?

Need to build a spreadsheet I can play around with where I can change the wattage and see how long I can go.

Thanks for the help!

[pianotuna]

Hi,

Here is a link to the rather special spreadsheet that N8GS has created to help size solar battery charging systems! It will answer your question about "how long" and help you with an energy audit too!

Solar spreadsheet by N8GS

[HiTech]

qwerty11 wrote:

BFL13 wrote:


Now it gets sort of ugly. The voltage of the battery keeps changing when it is being used , so the amps change if the watts stay the same.

Does the voltage drop follow any standard curve (i.e. logarithmic, exponential)?

It has to have some type of x/y value relationship. I'm guessing x would be voltage and y would be % discharge perhaps?

I know this is getting very academic here and giving me a level of fidelity not necessary, but I live in an engineering world (obviously not electrical) so I get excited when dealing with calculations.

Yes the curve is determined mostly by the Peukert curve as amps increase. But that constant varies by battery and battery type. At low to medium draws golf cart and true deep cycle batteries rule. At medium draws hybrid and marine batteries do best. At high draws starting and AGMs generally have the advantage. But that is very generalized.

Jim

[BFL13]

You might need a valium then, before going to this link. It has more in other articles in that group linked near the top under "Battery Applicaction..."

Note the discharge side voltage is fairly linear at a steady draw.

http://www.engineersedge.com/battery/specific_gravity_battery.htm

[qwerty11]

BFL13 wrote:


Now it gets sort of ugly. The voltage of the battery keeps changing when it is being used , so the amps change if the watts stay the same.

Does the voltage drop follow any standard curve (i.e. logarithmic, exponential)?

It has to have some type of x/y value relationship. I'm guessing x would be voltage and y would be % discharge perhaps?

I know this is getting very academic here and giving me a level of fidelity not necessary, but I live in an engineering world (obviously not electrical) so I get excited when dealing with calculations.

[BFL13]

You need to use 10 hours of the 20hr rate to get to 50%. the 20hr rate is just the battery AH rating divided by 20. EG 232/20 =11.6a or 100/20 =5a

the 20hr rate is different for each amount of AH capacity rating.

Now you want to know the time for different amps instead of the 20hr rate, starting with knowing the watts.

Now it gets sort of ugly. The voltage of the battery keeps changing when it is being used , so the amps change if the watts stay the same.

If you use light bulbs though, they get dimmer when voltage drops, so their amps go down! To keep the same amps you have to keep turning on more lights as the battery goes down. 🙂

OTOH an inverter load will have amps creeping up as the inverter maintains the watts while voltage goes down.

So ignoring that ugly stuff, pretend you have steady amps and that battery voltage is your pick- a- number like 12.4v for your spreadsheet info. Now you have the amps to use for the various watt loads.

With those amps, you compare with the 20hr rate amps using Peukert (ok it just got ugly again) But--to make it easy to do Peukert, now you can go back to that US battery table linked above to see how that would work--not forgetting about you going to 50% so 10 hrs worth instead of going 20hrs to flat.

[hmknightnc]

Your probably making this more complicated than necessary for estimating battery usage. Instead of trying to deal with watts convert everything to amps and then add up the various uses of amps for your total draw. For 12v devices just divide your watts by 12 to get amps and for 110 devices running off an inverter just divide watts by 10 (accounts for inverter efficiency)

You can use half of the 20 AH rating to get to 50%. So as an example using a 100 AH battery to power a 60 Watt LCD TV with an inverter. TV uses 6 amps/hour at 12volt DC (60/10) so you can run that TV approximately 8 hours (48 AHs)

[HiTech]

That is the normal but not universal measure.

Take a look at my second calculator above.

[qwerty11]

BFL13 wrote:
This would be part of the puzzle, but you need to know the AH capacity of each battery type there at its 20hr rate.
http://www.usbattery.com/usb_images/USB%20Ampere%20Chart.pdf

The Ah in my data set is based on 20hrs.

[HiTech]

Here's an extremely crude one. It will be wrong for big loads on small batteries but it is a starting point.

Calc

This one is a little better. Divide your watts by 12 and enter that in then hit calculate.

Calc2

[HiTech]

An interesting question given non linear discharge curves that vary by battery, and that constant watt loads are not constant amp. A battery that drops volts faster on discharge has to sacrifice more amps to keep constant watts. And then there is of course the non linearity of peukert losses as amp draws increase.

But I bet there is a calculator somewhere.

Jim

[BFL13]

This would be part of the puzzle, but you need to know the AH capacity of each battery type there at its 20hr rate. eg a U-2200 (see 6v table) is 232AH at its 20hr rate (which is 11.6a)

Also note they go to flat instead of just down to 50%, so halve the times if they do that

http://www.usbattery.com/usb_images/USB%20Ampere%20Chart.pdf