I want to use solar panels to power my existing borehole pumps (5 of them). Someone says I'll have to power each pump separately with several solar panels. I am however thinking; I could build a small solar farm and connect the pumps in parallel on the grid i would have created, what do you think. The pumps range from 1.5hp to 3 hp and i am considering a 100W or 200W panel.

My other consideration is replacing all the ac pumps with dc pumps.

  • 2
    \$\begingroup\$ What duty cycle do the pumps run? I think that is key. You may be able to store energy and power them, but that is a huge load. \$\endgroup\$ May 22, 2012 at 11:19
  • \$\begingroup\$ Do those pumps have to run continuously? What are you gonna do at night? \$\endgroup\$
    – stevenvh
    May 22, 2012 at 13:43

3 Answers 3


That's not a single 200 W panel, I hope. 1 hp = 746 W, so 5 \$\times\$ 3 hp is good for 11.2 kW, if that 3 hp is the consumed electric power (most likely). If it's delivered mechanical power you'll have to divide by the pump's efficiency, in practice that could be doubling.

If your solar farm can supply the power there's no reason why the pumps should not be connected together.

  • \$\begingroup\$ clabacchio used the metric definition of the hp, that's 736W instead of 746W. Not that it makes much difference in the answer. \$\endgroup\$
    – stevenvh
    May 22, 2012 at 11:28
  • 1
    \$\begingroup\$ You know that there is no other correct measurement system :) \$\endgroup\$
    – clabacchio
    May 22, 2012 at 13:32
  • \$\begingroup\$ Either way, it's a gross underestimation of what a horse can do. The value was chosen so low to have steam engines look better in comparison. \$\endgroup\$
    – stevenvh
    May 22, 2012 at 13:40
  • \$\begingroup\$ Take a look at wiki, it's very interesting :) The comparison was made with mills as benchmark \$\endgroup\$
    – clabacchio
    May 22, 2012 at 13:50

1 hp = 0.735 kW = 735 W


1.5 hp = 1.1 kW is the minimum power that you need from the panel to run one pump. It's already quite a number, and you need to drive 5 pumps, for a total of 5.5 to 11 kW.

It's a very big solar system, with an estimated cost of about (rounded to the bottom) 100k$.

Replacing the pumps won't help you that much in this sense.

Consider that if you have such a big system you will want to use the power as efficiently as possible, so it'll be better to drive the pumps independently, even if using only one solar farm.


Use pump load = 10 kW to make easy sums easier.

200 W panel storing energy at 100% efficiency and running pump at 100% efficiency will run pumps for Power_Panel/Power_pump
= 200/10,000 = 0.02 hours per hour of full sun.
ie a single panel will give a minute of operation per hour of full sun.

Available average sunlight can be found from the marvellous www.gaisma.com site. If you are lucky enough (sun wise) to be near Phoenix in Arizone then The Gaisma Phoenix page shows that sunshine-hours per day are: Insolation, kWh/m²/day Jan Feb Mar ... Dec:

2.98 Jan
3.78 Feb
6.99 Jul
2.74 Dec.

ie from about 2.75 hours in December up to about 7 hours average daily in June.
7 hours is exceptionally high. Close to the the best on earth (but, then uyou'd have to live there.)

So in December a 200W panel will run you pumps for about 3 minutes per day AT 100% efficiency for battery output and motors. And about 20 minutes/day in July.

If you want to run 8 hours/day in July you'll need 7 hrs/20 mins = 20+ 200W panels at 100% efficiency or 30+ in practice. And about 200+ panels in mid winter !

Running less than 8 hours/day (eg water trough filling as opposed to eg irrigation will take proportionately less.


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