# Assistance with Amp Hour calculations

Discussion in 'General Electronics Chat' started by ShockBoy, Jan 10, 2010.

1. ### ShockBoy Thread Starter Active Member

Oct 27, 2009
186
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A brief rundown on what I'd like to do. This may seem silly, but baby steps for me here. I want to take one of my 18.2cf refrigerators off the grid. I've calculated it pulling approximately 600W 5.22A 115V.
I need to figure what amp hour battery size to cover the refrigerator for approximately 15 hours. I have completed my two solar panels each at 24V with 3.5A. More than likely will parallel them to give me 7A (in full direct sun of course) 3-4A more like it I suspect.
Only running the battery down to 50% of it's full capacity doubles the amp hours needed. With a 100Ah rated battery pulling 5.22A you'd get 19.16Ah and then split that by 50% so as not to run the battery completely down gives me 9.58Ah with the 100Ah rated battery. Am I correct? Anything else I have overlooked? Thanks All!

Jun 1, 2009
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I think you should be close, if I'm not mistaken the AH capacity of a lead acid battery is rated at 1/20th of it's capacity, so it's 100AH capacity is based off a 5 amp discharge. Discharge currengts higher than this will reduce it's effective capacity.
However that 19.16 hours is to a completely discharged battery. If you run it for 15 hours you'll only have about 4 hours of time left, that's 80% discharged. I'm pretty sure your battery pack would be happier if it you didn't discharge it so much. Even deep cycle batteries will last dramatically longer if you use less of their capacity. I think about 20% discharge before topping back off is typically a good idea. Discharging further than that will increase sulfation over time, but it depeneds on how much overall life you expect to get out of a battery pack.

3. ### ShockBoy Thread Starter Active Member

Oct 27, 2009
186
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So at 20% as opposed to 50% instead of 9.58Ah I am down to 3.83Ah the refrigerator would run until the charge threshold is reached, assuming I am using a 100Ah battery?

Jun 1, 2009
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Yes, with your numbers so far I'd say that's a decent ballpark, so 4 hours of run time from an absolutely fully charged pack till 80% discharge. I'm a little fuzzy on the details but anything more than that is considered 'reserve charge' I believe. If you're using deep cycle batteries they're designed to handle deeper discharges better so you might be able to go to 50% and still maintain a decent battery life. But the rule is pretty standard, the higher the charge state you keep a lead acid battery in the longer it will last. Which is why car batteries can last so long, typically they're only drained for starting current and maybe radio or headlight usage with the alternator off most of the time, the rest of the time they're kept constantly full with a float charge from the alternator. I think ballpark for a battery used in that way is 5-10 years. You can discharge the battery more than to that 80% mark if you immediatly recharge it and still maintain decent battery life, basically the longer the battery is in a discharged state the more sulfation occurs, and all lead acid batteries eventually die of sulfation.

5. ### SgtWookie Expert

Jul 17, 2007
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Thing is; it doesn't run continuously. You'll need to figure out how long it actually runs over a period of time to get a good idea of actual demand. Since you haven't mentioned actual demand, the following assumes 100% demand.

Batteries are AH rated based on a 20-hour discharge time. Your mileage will be lower if you discharge them at a faster rate.

Always rate them conservatively.

If you run them down to 50%, you'll be replacing them after about 250 charge/discharge cycles. If you discharge them only 30%, you will get a lot more life out of them. See the battery manufacturer's datasheets for more details.

You're overlooking the fact that you have to run the battery power through an inverter to get the 120v required by the refridgerator; and that inverters are not 100% efficient.

Let's say you have an inverter that's really efficient; around 92%.
5.22A x 120VAC = 626.4 Watts. Divide that by 92%, and we get 681 Watts (rounded off).

Now let's translate that to your 12v current requirement.
681 Watts/12V = 57 Amperes (rounded off). Whoops! Looks like you were off by quite a bit.

You'll need a battery bank rated for at least 1140 AH just to have a 5-hour reserve, for 100% discharged.
But let's say you want that just 30% discharge for maximum battery lifespan.
1140 / 30% = 3,800 AH.

OK, now you've run the refrigerator all night and need to charge the batteries back up - PLUS keep the 'fridge running.

Note that lead-acid battery charging is at best around 75%-85% efficient - you get back around 80% of what you put in.

So, you've been drawing roughly 57A from the bank for 15 hours, or 855 AH.

Let's say you're charging at 80% efficiency, so 855/80%= 1069 AH you'll have to put back in - oh, and don't forget you'll need to run the fridge in the meantime, too.

So, 24-15= 9 hours of sunlight (pretty optimistic for this time of year, unless you're near the equator, but anyway...)
1069 AH to charge the batteries
57A * 9 = 513AH to keep the fridge running
1582AH required
1582/9 = 176 Amperes from a 14v+ charge source (your solar array) required.

Your 7A solar array is about 4% of the capacity that you need.

Taking a fridge off the grid is pretty tough, as you can see.

Last edited: Jan 10, 2010
6. ### SgtWookie Expert

Jul 17, 2007
22,182
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Here's a link to some info on various battery types and capacities:
http://www.windsun.com/Batteries/Battery_FAQ.htm

I'm not trying to discourage you. I'm trying to show you what the numbers are.

You can reduce the energy required very significantly - but you will need to make your residence more energy-friendly first. Like, you said the house has no insulation for one thing. I don't recall what you'd said about the roof, but simply using a white coating on shingles will drop the attic temperature around 30°F, which will help a great deal.

7. ### beenthere Retired Moderator

Apr 20, 2004
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Stapling aluminum foil to the underside of the roof sheathing will do about a 15% reduction in heat coming in.

8. ### SgtWookie Expert

Jul 17, 2007
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Now that you've seen what some still-possibly-optimistic numbers are for a fridge, you might consider using your array as a beginning to replacing some of your incandescent lighting.

You don't need a big expensive inverter, so you don't get that up-front cost.

You can make do with a much smaller battery.

Whatever LED lighting you put in will eliminate the expense of running the incandescent lighting.

You can start with just one room, maybe a couple of rooms.

LEDs are cheap, and you can get quite a bit of light out of them on not a lot of power.

Remember, seven 100W bulbs use more power than your fridge does.

Let's say you can get 7A out of your array for 8 hours a day. That's 56AH. Remember charging efficiency is about 80%.
So, you have roughly 45AH usable charge in an 8-hour day.
If you got a 150AH battery, and didn't discharge it more than 30%, the array should be able to charge it back up in a day's time. Fortunately, lights usually aren't much needed when the sun's up, and are normally off much of the night anyway.

But even if you did have lights on constantly, you could sustain a 3A constant discharge rate for 15 hours before the battery was 30% discharged.

You can run quite a few LEDs on 3A.

Last edited: Jan 10, 2010

Jun 1, 2009
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hehe, Wookies response is exactly the difference between theory and practicality =P Nothing wrong with my numbers, but I didn't take into account the reality of it or what you were actually trying to do.

10. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
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There is one other thing you should consider.

The (short term) startup current draw of the fridge motor may well be several times the running current. This is characteristic of electric motors.

Unfortunately it leads to blown breakers/fuses all too often with mains driven equipment. It will simply overload or stall an inadequate inverter/battery bank.

11. ### SgtWookie Expert

Jul 17, 2007
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1,728
Well, ShockBoy's trying hard to cut down his grid electric usage, which is a noble thing - and no laughing matter.

However, one can't expect to keep using the old technologies and expect to be able to power everything using a few solar panels and a couple of car batteries. It just doesn't work that way.

However, as I'd already mentioned, begin using auxiliary LED lighting. They're not quite up to incandescent light output standards yet, but they're far more efficient - and getting more so every year.

Even if he started off using some small 20mA LEDs, it would be a start.

Three white LEDs could be run in series from 12v, with a dropping resistor; not the most efficient way, but it's a start, and inexpensive. You could run 150 such strings (450 LEDs total) on 3A for 15 hours, discharging a 150AH battery less than 30%.

450 white LEDs would put out a pretty fair amount of light.

Jun 1, 2009
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Wouldn't it be more sensibly to simply try to produce as much extra power as possible and use grid tie inverters to supplement your current usage? The worst you can do in that case is slow your energy usage, in some cases you can cause your energy meter to actually spin backwards and the electric company will pay you =) Obviously yes the goal is to reduce energy usage as a whole, but you could do that by stapling on more insulation to the fridge, or the house itself. Being off the grid is not perhaps the only sollution.

13. ### SgtWookie Expert

Jul 17, 2007
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ShockBoy has a long, long, long way to go before he'll have "extra" power to sell back to the power company. He's probably spent weeks building his first solar array. I can only imagine how disappointed he was to find out that he needs to build 24 more like the 1st one just to power his 'fridge.

And he'd probably have to build another 80 or more to run his hot water heater.

We haven't gotten to his heating or A/C yet. That's another biggie.

Meanwhile, his home has no insulation, and when we talked about it a month ago, he didn't have a good idea of his peak energy demands - only the total KWH for the month; and they were quite high.

However, there ARE a lot of ways to simply reduce consumption.

He can put his solar array to work right away by using LEDs and a relatively small battery; like a deep-cycle marine battery. He could save quite a bit of money on up-front costs just by adding LED lighting to his home, instead of trying to power the 'fridge.

It's about the only inexpensive way I can think of offhand for him to put his efforts to work right away. It's not as easy as just plugging in a 'fridge, but... he'll wind up saving about the same amount of money if he replaces the incandescent lamps with LEDs, powered by his solar array.

14. ### thyristor Active Member

Dec 27, 2009
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Battery capacity is not always quoted at its 20 hour rate - just usually. Some vendors will play specsmanship to get a better apparent rating. For example a 110AH battery (at the 20 hour rate) could be advertised as a 130AH battery at the 40 hour rate.

But the 130AH will be in big letters whilst you can bet that the 40 hour rate will be in small letters!! So check the label.

Further, if you discharge at more than the hour-rate current figure, then you will get less AH's out of the battery than a straight linear calculation would suggest. The hour-rate current for this battery would be 110/20 = 5.5Amps

So a 110AH battery (at the 20 hour rate) will be half discharged after about 10 hours, ie: (110/2)/5.5

However, if one discharged it at say 28A, then one might think that it would be half discharged in (110/2)/28 = 2 hours. This will not be the case and the actual time to 50% discharge at 28A will be only around 1 hour.

The maths is a little complex but the difference in the two times (the apparent time and the actual time) is due to the "Peukert Factor".

All currents need to be converted to "Peukert currents" which are the actual current raised to the power of approximately 1.3 for a lead-acid battery. Further the Peukert capacity of a 110AH battery (at the 20 hour rate) is actually 183AH. One only gets the correct answer if one uses Peukert capacities and Peukert currents.

For currents close to the hour-rate the difference can be ignored but as the current increases or decreases then the Peukert factor can become significant.

Thus, if one discharges the battery at significantly less than the hour-rate current, then the battery will last much longer than a straight linear calculation would suggest. eg: a 1A discharge current, instead of discharging a 110AH to 50% in 55 hours, would actually require just over 90 hours.

The maths also reveals that paralleling batteries produces an overall capacity of more than just their arithmetic sum.

So two 110AH batteries in parallel will have not an overall capacity of 220AH (at the 20 hour rate) but a total capacity of around 25% greater than this simple linear calculation would suggest. That's why it is always better to parallel up batteries in, say, a boat system rather than have individual batteries running the inverter say or the fridge. You're getting an additional 25% free capacity.

PS: A good rule-of-thumb for real-life solar panels, which takes into account sun angle and direction, average cloud cover, hours of sun and the (in)efficiency of the physics of charging is to take the advertised watts and divide by a factor of 7 to get the AH's that will actually find their way into the battery. So a 100W panel will, on average, deliver only about 15AH per day into a battery. Nothing like the 40AH or so a tyro might expect from this panel per day.

Last edited: Jan 10, 2010
15. ### ShockBoy Thread Starter Active Member

Oct 27, 2009
186
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Sgt. Not discouraged one bit! I have a natural gas water heater, gas oven/stove, gas central heating, no central air conditioning and find it difficult to believe that I would need just under 30 solar panels to pull one fridge off the grid. Batteries, yes, probably quite a few of them for this goal, but the actual charging panels? Like I said, baby steps which is why I consulted with yu'all. I Really appreciate your input!
Just a thought, I understand the numbers and all but for me using a 10A/2A battery charger to charge a marine starter battery in 30min. does not make sense to me that my panels would not charge the same battery in 50-60min.
Do the amp hours rated to the battery relate to the time it takes to charge the battery, the same 12V battery?

16. ### ShockBoy Thread Starter Active Member

Oct 27, 2009
186
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It took me combined soldering the tabs, building the boxes probably 8 hours, One working day. Not disappointed, and not a child. Where the hell did you get the number 24? I have 84W panels each. At the most I'd need 6-8 more panels.

17. ### ShockBoy Thread Starter Active Member

Oct 27, 2009
186
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I agree, I have plenty of 4' double flourescent light fixtures I'd love to eliminate.

18. ### SgtWookie Expert

Jul 17, 2007
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That's good to hear. I didn't want to discourage you.

OK, keep in mind that you didn't give a duty cycle - so I based the numbers on a 100% duty cycle; fridge compressor running all the time. Without knowing how much of the time your fridge compressor actually runs, it's hard to make a good guess. It probably won't run much at all at night, as it's cooler, and nobody's opening the door. But during the day, it's warmer, and people keep going in there to put stuff in or pull stuff out.

Yes.
But, you also have to figure in how deeply the battery is discharged, and what kind of condition the battery is in.

If you have a brand-new deep-cycle marine battery that you've used to start your boat a couple of times, and it only takes you 30 minutes to charge it back up with a 10A/2A charger, you probably didn't discharge it much at all.

If the battery has several years on it, it might be heavily sulfated, and won't accept (or release) much of a charge anymore.

19. ### ShockBoy Thread Starter Active Member

Oct 27, 2009
186
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The boat battery is a starter battery for a 45hp outboard. I would not use a deep cycle battery to start a motor such as mine. I have not purchased a battery yet, but am looking to buy new. Actually 15 minutes to recharge an auto battery drained with lights left on over night. I agree not a full charge, but enough to get the car started on its own.

20. ### thyristor Active Member

Dec 27, 2009
94
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NO !!

Owing to battery physics, charging is only about 70% efficient. This means that if you remove say 100AH from your battery bank, then it will need around 140AH put back in to get it back to full charge.

If you take a rule-of-thumb that 50% more needs to be put back in than is taken out, you won't go far wrong. The additional AH, over the nominal capacity, are lost in heating up the battery.