A little math help, please

Steven Lake · April 2015

Hi all. Need to double check my math on something. I have an AC powered maintainer I use that averages about 120w of running power and runs about 20 minutes every hour on average in cycles of about 5min on, 10 off. I did the math and for 24hrs of runtime I need about 960wh, or about 40ah@24v of daily amperage. Accounting for system losses and Peukert numbers, I'm guessing to be safe I'd have to figure 50ah of total capacity for each day. If I want it to run for 3 total days without sun (as seems to be the general rule when designing a system) I'd need 150ah of capacity. Using the 60% capacity rule (ie, never let the battery get below 40%, ever, period) that'd mean about 250ah of total capacity. As for the panels I'm expecting to generate about 4-5 hours of power a day. Since winter isn't an issue (the unit doesn't run in winter, only late spring/summer/early fall) I'm guessing I need to figure for about 3 hours of average light. With that number, considering losses, 300w of panels seems a little on the thin side, so I'm figuring that 400w would be better.

Anyhow, that's my rough numbers. I'm pretty sure I've got this nipped, but that little voice in the back of my head keeps saying, "You don't have enough capacity. Go bigger." What do you guys think? Have I sized this out big enough, or am I short somewhere in the system? At this point it's only numbers on paper. I haven't actually gotten anything, save the maintainer, at this point, and that's being used via mains power right now Anyhow, can you guys double check my math? Thanks.

JoshK · April 2015

I'm a bit concerned with the 960wh per day number. That's less than 1kw, doesn't the average home use like 20kw per day? You could of course be right if you are running the basics like lights, fridge, computer... But throw in an A/C unit, a stove, oven, clothes dryer, or electric water heater and you completely blow through the 960wh. Are you just running the basics?

Steven Lake · April 2015

No, no, no. It's not for a house. This is for a farm. To save money we run a LOT of remote, shop and field equipment (especially monitors and maintainers) on solar power. We save so much money doing it this way that I can't even count the savings. So this proposed system is just another extension to what we're already doing. I haven't built it yet because I wasn't sure my math was right, so I wanted the guys here to double check my math to be sure I had this right. I'm wanting a system that has the right amount of capacity without being too much or too little. I can easily go a little over if I need to, but certainly not under. I'm already kicking myself for doing that on one of my prior builds.

tampasolar · April 2015

Since you are on the farm ?! Have you found a reasonable plan for remote pasture water pump off solar? A pump dropped about 100 feet into hole, pushing maybe 20 galloons a day, when the sun is up? Any ideas?

BB. · April 2015

OK... I would calculate as:

960 WH per day * 1/0.85 inverter eff * 1/24 volt battery bank * 3 days storage * 1/0.6 battery discharge = 235 AH @ 24 volt battery bank

Break even power usage on 3.0 hour of sun per day:

960 WH * 1/0.52 end to end system eff * 1/3.0 hours of sun = 615 Watt solar array minimum

Suggest a range of 5% to 13% rate of charge, with 10%+ rate of charge for daily used systems (happier battery bank, larger batteries should have 10% rate of charge minimum per some vendor's documentation):

235 AH * 29.0 volts charging * 1/0.77 panel+controller derating * 0.05 rate of charge = 442 Watt array minimum
235 AH * 29.0 volts charging * 1/0.77 panel+controller derating * 0.10 rate of charge = 885 Watt array nominal
235 AH * 29.0 volts charging * 1/0.77 panel+controller derating * 0.13 rate of charge = 1,151 Watt array "cost effective" maximum

Notice, as the battery bank gets larger, the solar array should get larger too--3x larger storage is a bit large for a typical lead acid battery bank. And discharging to 40% state of charge is a fairly deep discharge. But numbers wise, not really that far from "nominal" in my (fevered mind).

A 2 days / 50% discharge model would look like:

960 WH per day * 1/0.85 inverter eff * 1/24 volt battery bank * 2 days storage * 1/0.5 battery discharge = 188 AH @ 24 volt battery bank

Probably not worth "changing" your original system design by much--4x 6 volt @ 220 AH "golf cart" batteries would seem to be a good fit.

If your loads from the solar system are during the day time, that does draw from the array and reduces the current available for charging the battery (sizing up array to handle daytime loads + charging current).

Of course if much of your loads are during the day, that limits drawing power from the battery except during bad weather.

-Bill

Steven Lake · April 2015

BB: Hmm, then I'm not all that off on my math. Just a little short on overall power. So, based on your numbers, I should probably go for the happy median of 600w. That's above the array minimum, but not quite nominal. It's probably a little lean and may need to be expanded in the future. But I'll try it with this first and keep an eye on it, and if it needs a little bit of expansion I'll just engineer some breathing room in the design with an oversized controller (probably a 40a or 50a controller) to allow for some growth should I need it. And if not, hey, I've got plenty of controller to go around.

tampasolar: Actually, we cheat a bit on that. We have a couple of high volume ram pumps on the river that flows through our property that help feed a cistern from which all the water stations draw from. You'd be surprised how well that works, even in our frosty climate up here.

During the winter we only water from the south barn as it's too hard to keep the lines going out to the remote water stations from freezing. So come October they're blown clean and winterized and then put back into service around March or so. The ram pumps though run all year long. One of the ram pumps is older than I am. I think it was put into service in the 1920's and is still running today without a lick of maintenance that I'm aware of, save for a few repairs on the drive pipe, one of which was my fault when I accidentally drove over it with the tractor. ^_^;;

A little math help, please

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