# confused

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Solar Expert Posts: 41
4-each L-16h=360ah@24V
The 36 Amps comes from Trojan's recommended peak charge current of 10% of the batteries' total Amp hours. In this case 360 * 0.10 = 36 Amps. To get that at 24 Volts from solar: 36 * 24 = 864 at 77% typical efficiency = 1530 Watts of array in 1 hour. would 5 hours of sunlight 1530/5 = 306 watts of array. Whats wrong here?

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Re: confused

Welcome to the forum.

What's wrong here is you're confusing the peak potential charge current (36 Amps) with the daily Watt hour (or Amp hour) harvest.

The math as you did it comes out to the right size array for coming up with the potential for 36 Amps peak charge current. That doesn't mean it's going to charge at that rate all the time, or even ever.

The daily Watt hour harvest is that array size (1530 Watts derated to 864) times the hours of "equivalent good sun" which is usually at least 4: 3456 Watt hours (not including losses incurred when converting to AC).

In terms of "AC Watt hours daily" that's figured roughly like this: 1530 Watts (total array size) * 4 (hours of good sun) / 2 (50% over-all system efficiency) = 3060 Watt hours per day.

Is that any clearer?
• Solar Expert Posts: 41
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Re: confused

4-each L-16h=360ah@24V
The 36 Amps comes from Trojan's recommended peak charge current of 10% of the batteries' total Amp hours. In this case 360 * 0.10 = 36 Amps. To get that at 24 Volts from solar: 36 * 24 = 864 at 77% typical efficiency = 1530 Watts of array derated by 77% = 864 watts * 4 hours = 3456 Wh minus conversion factor making your 3060 / 3 days storage = rounded 1,000 watt hours (1KWh) available for usage for a healthy system.
My problem is I have 6-130 watt panels = 780 watts
8-75 watt panels = 600 watts (thru 2 MX 60’s) totaling 1380 watts * 4 hrs = 5520 Wh delivered into 12-L16’s (3 sets of 4) hence 1080 AH
1080 * 0.10 = 108 amps 108 * 24 = 2592 watts of panels needed for said batteries.
1350 * 4 = 5400 watt hours per day / 2 = 2700
I consistently produce the 5000 or better and am in float around 2 pm. I use on average in the 4000 watt hour range.
Don’t I have any storage?
• Banned Posts: 17,615 ✭✭✭
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Re: confused

Break it down in to two parts: the panels and the batteries.

The panels:
1380 Watts total. 1062 @ 77% efficiency. Divide by 24 = 44 Amps peak potential current.
That is enough to support 440 Amp hours of battery (especially with L16's).

The batteries:
1080 Amp hours total @ 24 Volts.
First of all, that would want 108 Amps of peak potential current, which would require two arrays & controllers because no controller can handle that much current on its own.
Second, the total amount of panels necessary to supply that much current would be:
108 Amps * 24 Volts = 2592, less 77% derating = 3366 Watt array.

Obviously these two sets of numbers don't work together. Your 44 Amp potential current is only 4% of the total battery Amp hour capacity, below the 5% minimum recommendation and far below the 10% Trojan recommends (especially for the L16's which are known to suffer from electrolyte stratification). This means your batteries will not recharge properly, resulting in a shortened lifespan.

The peak current rate is one thing. The Watt hour potential is another:

The 1380 Watts of panels have a potential of:
1380 @ 77% = 1062 * 4 hours minimum = 4250 Watt hours
Since you're managing 5 kW hours per day you're evidently getting better than 4 hours of equivalent sun and/or running a bit better than 77% efficiency.

The 1080 Amp hours of batteries have a potential of:
1080 @ 50% maximum DOD = 540 * 24 Volts = 12,960 Watt hours

In short you have far too much battery capacity for that amount of panel. As long as the batteries aren't discharged below about 32% you will be able to "replace the used Amp hours", but the low charge rate potential will shorten the battery life.

Your best bet is to recalculate the battery capacity according to your needs and see if you can't reduce it somewhat. Failing that, keep an eye on the specific gravity of the battery cells and find some way to do an equalization charge (probably once a month).
• Solar Expert Posts: 41
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Re: confused

I guess the thing I cant get my mind around is the three days storage. I use a max of 4000 watt hours per day. Usually in the 3200 to 3600 range. I am old and have converted from wood to pellets. My understanding is that I shouldn’t have the panel capacity to replace the added storage in a one day period. Is this correct? I can run the generator if need be. I have equalized said batteries on days when heat wasn’t necessary. I live above 8500 feet. If this is incorrect should I drop 4 batteries out of the system. If I do this what happens to the 3 day factor?
• Banned Posts: 17,615 ✭✭✭
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Re: confused

There was a time when panels were expensive and batteries were cheap. The formula then was to recharge slowly over time. But this shortens battery life. Now that batteries are expensive and panels are a lot cheaper than they were the formula is to go for quick recharge which extends battery life.

As for storage capacity, the recommendation of having "three days, no sun" storage is in my opinion seriously flawed. Without enough panel to quickly (in one day) fully recharge the whole bank you run the risk of the batteries being only partially charged for too long. That leads to accelerated sulphation and shortened battery life.

You can divide the battery bank in three and use them in rotation, having enough panel to do 1/3 at a time. But that requires some battery switches and the need to remember to change them 'round. Make a mistake and forget about a set for too long and you come up against the same problem; shortened battery life.

The least aggravating situation is to plan your bank so that 25% DOD will supply one day's worth of power. Have enough panel so that you can recharge that bank fully (10% peak current rate potential) daily. Know that you've got another 25% capacity in reserve to get you through another day without good sun. On the third day, start the generator and recharge the whole thing fully.

With a daily consumption of around 3600 Watt hours maximum, you could use just two parallel strings of those batteries: 720 Amp hours. That would give a maximum of 8.6 kW hours - more than two days worth.

At 8000 feet of elevation you have an advantage: your panels are probably working at around 84% efficiency instead of 77%. Ideally then you would only need around 2 kW of panel to have the full charge rate available. That should also suffice for supplying your daily Watt hour needs.

It really looks like you don't need that third string of batteries.
• Solar Expert Posts: 41
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Re: confused

I missed the change from 3 to 2 so I will change and go from there. I could not understand what the difference in calculations were
THANKS for the patience and the help..