New to design with questions

System · July 2007

I'm trying to design a solar-only system that will handle a 1Kw A/C unit, a couple of lights (approx 80 watts) and a small electronics console (approx 20 watts). All are alternating current appliances.

This is in California's high desert, so solar insolation is NOT an issue. I was looking at using a 2500W (5000W peak) inverter with a battery system. I don't anticipate that the air conditioning will be used in the winter months - Nov-Feb, but about 10 hours a day, 6 days a week, for half of every month outside of that.

According to the calculator off solar4power.com, if I was using 12 volt/225a batteries and 75W 4.35a panels, I'd need something like 60 panels for power.

Is this estimate right? Since I'd only be using power half-time, isn't there a way to expand my batter pack and use fewer panels?

Generator is not really an option, as fuel would have to be backpacked to the location.

Thanks in advance for any input...

BB. · July 2007

First, work on conservation (insulation, new ground sourced heat pumps, etc.)... It is almost always cheaper to save energy than to collect/store/generate it.

Next, you really need to know how much you will be running the loads. If this is an existing home (or you have a place with similar loads), take a look at the utility bills and/or just go out and read your power meter daily/weekly to get some trends/ideas.

You will find that loads like A/C cannot be predicted just from the nameplate ratings... If your system runs only 10 minutes out the hour for 10 hours--that is ~1kW*1/6 on time per hour *10 hours per day=1.66 kWhrs per day...

But, if the 1kW system is running 10 hours per day that is 10kWhrs per day (or 6x as large).

You need to characterize each of your loads this way... Once you do that, you can then start sizing the battery system and the array(s).

Heavy, continuous loads (like A/C) is always a very difficult (read expensive) proposition for solar (or even generators). You are probably talking about a $20-$40,000 system--just based on what you have said so far...

Is this where you want to go?

-Bill

niel · July 2007

i noticed that the forum indicated not all was read here and i went digging to find your post i think on the 7th page. it should not have been shoved way back, but things are just getting going on this new bbs system and some blips and blurps i guess will be here.
now i'm not sure i read your post all correctly, but you list 1,100w of items running for 10 hours of the day 6 days a week. we'll leave that 7th day alone as that could account for a cloudy day. now there will be losses for storage, resistance, efficieny, etc. so this could tack on another 20-30% or much more if the inverter is a modsine(not recommended with any electrical motors, pumps, compressors, fans, etc.) that brings the hourly power needed to about 1,400w. now for 10hrs that becomes 14,000wh per day needed to be produced by the sun. if you are getting 6 full sun hours in the summer that means you need about 2,350w of output from pvs. note that this is not the stc ratings of pvs and would be closer to the cec ratings california rates pvs at.
now in the winter there's less sun so if that 10hr period was for the winter(that's what i'm unclear about) then it would be the total watthours of 14,000 divided by the number of full sun hours you would typically get in the winter. crewzer has listed the site many times that lists the number of full sun hours for a given solar angle at various locations, but i don't remember it. again if i was wrong about the number of summer hours and if you meant that for 24hrs then your requirement would be 2.4 times as great as i had indicated with 10hrs or 5,640w of pv output. it all does boil down to how many full sun hours you will typically get where you are and your losses. don't confuse this with daylight as there are many hours of less intense light that when added together could be equal to a full sun hour. a full sun hour is equal to 1000w/m^2 and is the stc rating on most pvs. nearly nobody sees that max rating much even in areas of the california desert or florida so that is why i stated use the cec ratings of the pvs.
depending on the pv you have chosen in the example you stated you had tried with a calculator, it may have been correct depending on if i read you right.

niel · July 2007

it was the 10th page and i was not able to modify my post this time.

mike95490 · July 2007

Ironmtneer wrote: »

Is this estimate right? Since I'd only be using power half-time, isn't there a way to expand my batter pack and use fewer panels?

Sure, cut your power demand !

Seriously, if you expand your battery pack, you need even MORE solar to recharge all those expensive batteries, so they don't die. You say you will not have a backup generator, so preserving the expensive batteries becomes even more important.
What town is this near? That will help us figure more, how much sun is available.

System2 · July 2007

More info

First this is near Barstow, CA, but it's WAY out in the sticks away from power.

Insulation is not really an option, since this is for a shipping container-like box.

We typically get only 1-2 cloudy days a month, and we get at LEAST 6 hours of sun in the Winter. Summer it's more like 14 or so- and that's when we "need" the a/c.

Honestly, my choice would be no A/C unit, but it's not my decision.

BB. · July 2007

Well, really you don't get 14 hours of "full sun" per day... Check this PDF File of solar data for California.

You get about 3-7 hours (flat plate fixed collector at latitude, winter-summer) worth of "full" sun (using Bakersfield as the base). If you are more like Dagget--then winter sun will be closer to 5 hours per day (forgot about that valley fog).

But, in any case, if this is a steel shipping container building--some sort of building insulation would be required to make the place livable--In my humble opinion. Should make heating and cooling much better--as well as avoiding hot/cold walls, condensation, etc. But, if you have already lived in it and are happy--...

-Bill

mike95490 · July 2007

OK, Barstow is a good starting point, we can use standard charts to find out how much sun downtown Barstow gets.
National Solar Radiation Data Base:
http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/

Las Vegas is likely an OK match for comparison
http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/dsf/data/23169.txt

You can look up winter sun hours, and summer sun, and est your cooling loads for each season.
users manual:http://rredc.nrel.gov/solar/pubs/NSRDB/NSRDB_index.html

Insulation would help, or even shade cloth, the summer sun will make that container a nice toasty oven.

crewzer · July 2007

AC: 1 kW (~10,000 BTU?) x 10 hrs/day x 6 days/week = 60 kWh/week
Lights: 80 W total x 3 hours/day (summer) x 6 days/week = 1,440 Wh/week
Console: 20 W x 6 hours/day x 6 days/week = 720 Wh/week
Total energy requirement = 62.2 kWh/week = 8.9 kWh/day

Assuming 55% overall system efficiency, you’ll need to generate 8.9 kWh/day / 55% = ~16.2 kWh/day
Insolation on a south-facing array tilted up at latitude in nearby Daggett averages 6.8 hrs/day or better of “full Sun” from March through October.

http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/redbook/sum2/23161.txt

So, you’ll need an array rated at 16.2 kWh/day / 6.8 hrs/day = ~2,400 W STC. This array probably won’t keep up with the A/C load while the cabin is occupied, but it’ll do pretty well, and the system will catch up quickly during the 50% of the time the cabin is unoccupied.

As you no doubt aware, it can get both quite hot as well as rather cold in Daggett. Average highs in July are 105 F, and the record high is 118 F. Average lows in December are 34 F, and the record low is 5 F. Very hot ambient temperature will suppress mid-day array voltage, and cold temperatures will enhance early morning open circuit voltage. Accordingly, you should target an array spec of ~88 Vmp STC and ~110 Voc STC.

Configured for a 48 V nominal battery system, a single OutBack MX60 charge controller can handle an array of this size. Don’t forget the optional remote battery temperature sensor. The Outback FX3048T sine wave inverter/charger would be a good choice for the listed loads. You should probably also consider the Outback Mate to customize the inverter settings.

Assuming inverter efficiency of 90%, a daily load of 8.9 kWh/day means the battery bank will need to supply ~9.9 kWh/day. Allowing for 2 days of autonomy due to your location's regular abundant sunshine, and limiting battery discharge to 50%, you’ll need a 48 V battery bank rated at (9.9 kWh/day x 2 days / 50% ) / 48 V = 825 Ah.

HTH,
Jim / crewzer

crewzer · July 2007

Frankly, I doubt a 10K BTU A/C unit will keep up with the heat build up in a steel container. And, neither the batteries nor the electronics are going to be happy in that heat while you're away and the A/C is off.

niel · July 2007

crewzer wrote: »

Frankly, I doubt a 10K BTU A/C unit will keep up with the heat build up in a steel container. And, neither the batteries nor the electronics are going to be happy in that heat while you're away and the A/C is off.

i fully agree here and if it were me i'd be getting those insulating panels that have the insulation and the rigidity of a plywood together to attach to the outside walls and something for the roof too. even spray on foam insulation and painting that would work.(must be painted with a paint that won't negatively react with the foam and the paint protects the foam from deterioration) with temps like that on a metal box you would cook not only the electronics, but yourselves as well. it is easier to prevent the excess heat from getting into the metal housing than for the solar to power an ac unit to get rid of it that the ac unit would have trouble keeping up with.

Vic · July 2007

Hi Iron..

At our off-grid location we use a number of steel sea-going containers. The climate is more like Modesto, CA, and not really the desert. It is fairly easy to insulate these containers. The rigid polyisosanurate panels are the simplest to use as they are semistructural, but in a fire situation, may outgas cyanide and other toxics. Have also used fiberglass batts held in place with 2X3 or 2X4 stringers screwed to the steel box framing rails. Our power room has a sloped roof with solar panels on top, and an insulated floor deck above the container's roof. This roof is also part of the rain catchemnt system.

I also need some AC for the power room, as quite a lotta energy is dumped into that room.

In your case, Iron, if you have sufficient water you could consider Swamp cooling, altho I would not like to run an evaporative cooler in my absence. Here water is precious (rain catchment) and will probably use A/C on the power room, as the MX-60 has its tounge hanging out on hot days -- the MX fan runs constantly.

The containers are cool but they are too hot in warm sunny climates, without insulation, and COLD when they are not hot. But they are a great solution as instant, tight, semi-secure buildings.

Vic

Windsun · July 2007

Just a rough estimate, but assuming 700 watts average for 24 hours, that is abuot 17000 watt hours per day.

you will get around 6 hours of full sun per day, or 1/4 of the day.

so around 4200 watts of panel.

New to design with questions

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