Medium solar water pumping system
Calculation for solar electrification
1
Pump (1 HP)
Example F:  Mediumsized solar waterpumping for 20 families
100 persons with 100 l/d up to 20 m hight
To calculate the energy consumption of the pump, you have to use the electrical power of the equipment and not the horsepower HP translation. Here it is important that there are two different power values: Nominal power (without load) and maximum power (with mecanical load).
In our example we work with an 1320 W (maximum power) pump.

There are two selection criterias in a solar system:
The water FLOW in the hight of the water tank or consumption. You have to measure the hight between the water entrance and the water outlet. Like you can see, the relation of flow and power consumption of this pump is better than of the pump of the last example.
The real electrical power CONSUMPTION for this flow.
En sistemas solares, que solamente sirven a bombear agua, la reserva diaria puede ser prevista en el tamaño de los baterías, también sirve instalar un tanque de agua más grande. Esto depende de los costos.
In the following page we are calculating a medium solar watersystem, and afterwards an example of waterpumping without batteries.
Calculation of consumption of the water pump
Generally you work with an average of 90 to 150 litres of water per person and day in rural instalations. This varies if you got water toilets or lavatories installed. You also have to calculate future water consumption and therefor to increase the solar system.
You multiply the daily water needs with the number of persons. In this case 100 persons multiplied by the daily average water needs 100 l/p/d equals 10000 l/d.
The pump got a flow of 5400 l/h in 20 meters of total hight. You divide the daily average need 10000 l/d by the flow of the pump 5400 l/h and get the needed working hours of the pump.

In this example the pump works more or less 1,9 hours a day.

If your solar system is working only for water pumping, you have got the option to buy a special direct current DC pump, and you could save the costs of the inverter, which produces the AC for your pump.
*** The inverter needs a bigger power output than the pump, because electrical motors and pumps consume about four times more current in the first seconds of the startup.
Quantity

A
Equipment

B
Power
W
C
Power
W Subtotal
D = (A x C)
Hours / day
de usage
E
Energy
Wh
F= (D x E)
Calculation of panels y bateries
The daily average consumtion is:

1 panel of 55 W produces with 3,5 hours sun/day:

For the energy generation you need:

You have to store (12 V system) la quantity of:

You need the quantity of stationary bateries  12 V 100 Ah (without reserve):

You need the quantity of stationary bateries  12 V 100 Ah (with a 1 day* reserve):

You need the quantity of stationary bateries  12 V 100 Ah (with a 3 days* reserve):
Wh/d

Wh/d

Paneles

Ah

Bateries

Bateries

Bateries
System costs
13 fotovoltaic solar panels of 55 W costs aproximately between:

8 stationary batteries** of 12 V 100 Ah costs more or less:

2 regulator (protector of the battery) of 12 V 30 A costs more or less:

1 inverter APS of 2000 W*** 110 V AC costs between:
(The APS transforms the direct current DC in AC,so you can connect any (small) normal equipment for 110 V AC)

Instalation costs, depending on the site:

Total of investment for this system is between:
1320
1320

Total =
1,9

Wh / day
2.444

2.444
2444

193

12,7

204

2,0

4,1

8,1
4290 - 6435 US\$

587 - 2347 US\$

160 - 300 US\$

1200 - 3000 US\$

200 - 1000 US\$

6437 - 13082 US\$
Like you can see, the relation of flow with 5400 l/h and power consumption of 1320 W of this pump is better than of the pump of the last example, which results in an more economic fotovoltaic instalation of 6500 to 13000 US\$.
This same system with the pump of the last example, with an flow of 900 l/h and power consumption of 500 W, results in costs for the solar system of 8500 to 16500 US\$.
In instalations with mayor hight or head, with an uneficient pump, the economic losses are even higher.
In all cases of waterpumping, it is useful installing a bigger tube, to get a minimum of internal friction losses in the tubes. The internal fricton menorizes the waterpressure and results in a long term in higher energetic expenses.
System CostsCalculo0
Sistemas solares para escuelas y colegios comunales