Help for school project

Re: Help for school project

I've moved this to its own thread because it's starting off on a whole new adventure!

What you need is a Kill-A-Watt meter.
You need to measure the actual power used by the water pump, light, and fan during the course of a typical day. This will be a figure of Watt hours for each one. You add them up to get the total.

Then you have a rough idea of how big a battery and at what Voltage is needed to supply the power through what size inverter. The size of the battery determines what is needed for panels & charge controller.

If this is a standard water pump you will find it uses a lot of power and has a nasty high current demand on start-up. If it's a small one like a Shurflo http://www.solar-electric.com/2088-594-154.html its power demands will not be so great.

Let's look at the light for an example:
If it is a 100 Watt incandescent it will use 100 Watts * however many hours it is on. Let's say 4 hours at night. That's 400 Watt hours right there.
Then there is the conversion inefficiency of the inverter. This will add perhaps 10% to that figure. We're now at 440 Watt hours.
The inverter will consume power too. If it is only on for the light it may take as little as 6 Watts or as much as 40; this will depend on the actual inverter whose size is determined by the total load of everything running at once. If you are just running the light and can use a small inverter we'll add it in as 6 Watts for the 4 hours or 24 more Watt hours. The total is now 464 Watt hours.

Since that is a relatively small amount it could be handle on a 12 Volt system. So you divide by 12 and get 38.7 Amp hours. Since you don't want to discharge the battery more than 50% you need to multiply that by at least 2: 77.4 Amp hour 12 Volt battery. That will need to be rounded up to the nearest available size. Perhaps a 95 Amp hour Crown 12 Volt.

From there you target your peak charge current potential. If the loads are always off during charging you can use the minimum 5% rate. If they may be on you need to use a higher rate. 10% usually works in most cases. That means you want to provide the battery with up to 9.5 Amps of current. To get that you need enough panel to supply that rate. Since this is small, we'll expect to use a less expensive PWM type controller like this: http://www.solar-electric.com/ss-10.html You'll want "12 Volt" panels to supply it. They will have a Vmp around 17 to 18. The other important number is the current rating: "Imp". This should be near or even above the charge rate desired, total for all panels.

Here's where you can fudge things a bit. Kyocera makes some 140 Watt panels with an Imp of 7.9. That current is between the 5% minimum and 10% 'desired' peak rate of the 95 Amp hour battery, and so should work for this example.

But remember; that example is only for running one 100 Watt light for four hours. The other equipment and/or longer times will multiply out, requiring larger battery capacity and more solar panel.

Along about that time a long extension cord to the nearest outlet starts to look really good.

Re: Help for school project

Should be available at local hardware/electrical supply stores.
Or they certainly are available on-line. For instance from the people that host this forum: http://www.solar-electric.com/kiacpomome.html
About $30, and worth it as you can measure the electrical consumption of anything you plug in and get a good idea where the power is going. The "off" loads of many things are surprisingly high.