Uninterruptible Power Supply – Battery Duration Calculation

The setup in your answer is a bad one.

First you parallel two batteries, which you shouldn't, because their voltages are never exactly the same. Their low internal resistance will cause a current from one battery to the other. So feed three LEDs from 1 battery, and the other three from the other.

Then you place all LEDs parallel, which means that you lose (12 V - 3.2 V) x 20 mA = 176 mW per LED in its series resistor, while the LED itself uses only 64 mW. Total power loss is more than 1 W. That's because the large voltage difference between battery and LED. The best way to get a longer endurance is to keep losses in the series resistors as low as possible. So better place two times three LEDs in series, so that their total current is 40 mA instead of 120 mA. The power loss in the series resistors is then (12 V - 3 x 3.2 V) x 20 mA = 45 mW per 3 LEDs, or 96 mW in total. That's less than 10 % of the power loss for all LEDs in parallel.

Then your batteries will last 100 mAh / 40 mA = 2.5 hours or 150 minutes. This is pretty optimal. The batteries' capacity is 1200 mWh, and the LEDs consume 384 mW, so with an ideal conversion you can get a little over 3 hours out of them. But the most efficient conversion using a switching current regulator will get you maybe 85 % efficiency, and then you only gain 9 extra minutes.

edit re comments
An alkaline battery's voltage quickly drops by 10-15 %, and then remains more constant for a great part of the discharge cycle. So either you calculate the resistors for a larger current at the start, and 20 mA for the rest, or for 20 mA at the start, and a lower current later on. The latter solution will give you a longer battery life, but a bit less brightness.

jippie suggests to use a switcher anyway to get more out of the batteries, and it's a thought. You'll have to place the batteries in series to get 24 V to allow a voltage drop as high as possible. The larger Vin/Vout ratio of the switcher will make it less efficient, but overall you should get some extra time from the batteries.

I am no expert, but since nobody else has answered I thought I could help.

I understood that the C (coloumb) rating of a battery indicated the rate of discharge that was advised. For instance, a 5C battery can be discharged at up to 5 times the rated AH of the battery. So my understanding is that a 10C 12v battery with 150 AH could have an instantaneous draw of 1500 amps on the battery. I also understand that the actual capacity of a battery does depend on the draw. For instance, I understand that lead acid batteries tend to last longer AND exhibit greater total capacity if the discharge is small (e.g., 500 milliamps). Lead acid batteries that are discharged quickly (e.g., 500 amps) tend to discharge a lot faster AND exhibit much lower total capacity. The level to which this is true depends on the battery chemistry, from what I've read. For example, if I was drawing 1 amp from your 150AH battery, I may get 150 hours of usage. On the other hand, if I was drawing 150 amps from your battery, I may only get 30 minutes of usage. The numbers were randomly picked, but you get the idea. Hopefully that cleared up your confusion. Yes, your C5, C10, and C20 batteries all offer 150AH but only under specific load conditions as you said in your question.

Not only do higher C ratings on batteries result in a higher battery cost, but, in general, high amperage charge/discharge of lead acid batteries tends to crush a battery's life significantly. This could mean that with a high enough amperage on your charge/discharge, you may only see 10 cycles to the battery before its useful life is depleted.

[1]. No battery is really better, I think. This really depends on your use case. For example, you could do one of the two extremes:

[A]. Buy just a small number of C20 batteries for your system. They're designed with high charge and discharge rates in mind. A C20 rating would guarantee that you could get the 150AH out of the battery despite a high discharge. You'd get the same total AH rating but in a smaller package. You'll probably get fewer charge/discharge cycles before the batteries need to be replaced. If you were only planning on using this solar system as a backup when the power went out (you did say UPS), then this is the better option because you would buy fewer batteries (fewer dollars). But you may need to replace the whole thing sooner.

[B]. Buy a lot more C5 batteries. They're each designed to only handle a small charge/discharge rate to give you the same total AH rating. You'd need to have a lot more of them since you're maximum demand of amps probably couldn't be met with just a few C5 rated batteries. Also, the total number of amps flowing into the batteries from the charger would need to be lower for C5 rated batteries when compared with C20. You'll probably get a lot more charge/discharge cycles out of this setup before the batteries need to be replaced. You'd probably need to consider this option if you wanted to use the battery pack for more than just a UPS. Assuming you used these batteries daily, you'd definitely want them to have a much longer lifetime/cycles than, say, option A.

In either case, the maximum discharge and the charging rate would need to be considered when you decide how many batteries and which C rating you'll need.

[2]. Lower C ratings are probably better for solar power use because most people who install a solar system plan to use it on a daily basis (unlike a UPS). Even though the lower C rating translates to a need to have more total batteries, the cost is probably a lot lower and the total lifetime in cycles is probably a lot higher.

Here's a couple examples:

Assuming you need a maximum demand of 1000 amps at 12v (12000 watts), then here are some options:

With C20 batteries rated at 150AH, the maximum amp discharge is 3000 amps. So you could definitely run the load with a single battery since 1000 amps < 3000 amps. However, your useful running time with a single battery would only be, roughly, 150/1000 * 60 = 9 minutes.

With a C5 battery rated at 150AH, the maximum amp discharge is 750 amps. So even though you probably could run the load with a single battery, it wouldn't actually give you 150 AH (and wouldn't be advised). Instead, you'd want at least 2 batteries. By spreading the load across 2 batteries, the demand from a single battery (assuming constant voltage) is only 500 amps. In this scenario, you'd be pulling (150/500 * 60) * 2 batteries = 36 minutes. In addition, the C5 battery is probably less money.

When planning this system, remember to take into consideration heat losses in your wiring which could be high depending on the gauge and length. Also remember that your inverter is only 80-95% efficient depending on how much you spend on that component. Remember that the total efficiency of what you're powering may also be affected by the quality of the sine wave (pure sine wave or the cheaper jagged/stair step kind) being output by the inverter. In the case of a computer where the power supply will simply convert the AC back to DC again, my hunch is that the efficiency of the total load will not be improved by using a pure sine wave inverter.

If I'm horribly wrong on any of these points, please somebody more knowledgable speak up! I'd be happy to know I was wrong.

Hopefully this makes some sense and is on target.

In any case, have fun and good luck with your project!