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Solar powered water pumping system


A question from Girish Upadhyaya: I need information for a solar powered water pumping system. Say a power requirement of about 300 watts to pump 1 metre cube of water per hour to a height of about 20 metres. I would like to use a 12 v DC motor as the pumping unit drive. Can you help me with this?

2 Answers


Dear Girish Upadhyaya,

Please find below a link from one of our experts which we hope will be of use to you. Please get back in touch if you have any further queries.

> The website below might be a good starting point for the inquirer


>Please let me know if any help is required in understanding or accessing the material.

> mike

Posted on behalf of Mike by Pauline REDR

Neil Noble

There is some useful information in the Practical Action Technical Brief Solar Voltaic Waterpumping which also links to a good manual by Green Empowerment especially useful is the section on sizing your solar pump….

Sizing solar pumps

The hydraulic energy required (kWh/day)

= volume required (m³/day) x head (m) x water density x gravity / (3.6 x 106) = 0.002725 x volume (m³/day) x head (m)

The solar array power required (kWp) =

Hydraulic energy required (kWh/day)/ Av. daily solar irradiation (kWh/m²/day x F x E)

where F = array mismatch factor = 0.80 on average (a safety factor for real panel performance in hot sun and after 10-20 years) and E = daily subsystem efficiency = 0.25 - 0.40 typically


The volume required needs to be in cubic metres per day rather than per hour as the rate of pumping will vary over the day. Let’s say 5m3 just as an example.

0.002725 x 5 (m³/day) x 20 (m) = 0.2725 kWh/day

Av. daily solar irradiation (kWh/m²/day x F x E)

The total solar irradiation received in a day can vary from 0.5kWh/m2/day in the UK winter to 5kWh/m2 in the UK summer and can be as high as 7kWh/m2/day in desert regions of the world, such as regions of Nigeria (Solar Water Heating in Nigeria, 2006) and the Sahara in Algeria. (Survey of Energy Resources, 2010) Many tropical regions do not have large seasonal variations and receive an average 6kWh/m2/day throughout the year.

Therefore we could use 6kWh/m2/day as an example

     6x0.80 x 0.4

The solar array power required (kWp) = 0.2725 / 1.92 = 0.142 kWp which then needs to be match to the solar panels available.