# solar water pumping in Sierra Foothills California

Discussion in 'Photovoltaics' started by David Williams, Nov 26, 2007.

1. ### David WilliamsGuest

-> My friend lives on a ranch in the Sierra Foothills in California. He
-> is interested in using a PV powered water pumping system for his new
-> well. He wants to pump about 1500 gallons per day from a well depth
-> of

Well, let's look at how much power will be needed. He wants to raise
about 5000 litres, or 5 tons, of water about 300 feet or 100 metres. So
the amount of energy he'll need per day, assuming 100% efficiency, will
be about 5000 x 100 x 10 or 5 million joules. A kilowatt-hour is 3.6
million joules, so he'll need about 1.4 kWh per day. If the sun shines
for 5 hours per day, he'll need his PV system to produce about 300
watts, if the pump is 100% efficient. More realistically, he should aim

Suppose the PV system is 20% efficient in turning sunlight into
electricity. He'll need 2500 watts of sunlight to produce 500 watts of
electricity. If the sunlight falls perpendicularly onto the PV cells,
about 2.5 square metres of cells would be needed. But, unless he
installs a fancy tracking system, the sunlight usually won't be
striking the cells perpendicularly. I'd guess that he'd need something
like 4 square metres, or 40 square feet, of PV cells to do the job.

So the next question is, can he buy that area of PV cells for a price
that makes the whole project economically feasible?

dow

2. ### Guest

<snip>

This is a case where it isn't practical to calculate the energy
needed. There are too many variables and you don't have nearly enough
info from the OP. For example, you started by using what appears to be
the well depth instead of the static water level, and you don't know
if the 50' is to the top of the tank. The OP knows (or will find out)
the missing data, and from there it will be easier and more accurate
for him to peruse the measured performance curves for available pumps,
which take into account pump design, efficiency, MPPT, etc.

Wayne