Irrigation system

Welcome to the forum Chibubi,

Your system, as proposed (3 hp @ relatively high pressure and volume) will consume a lot of energy. Solar power is not cheap--Upwards of $1-$2 per kWH or more (just a SWAG at this point). Powering the system for 10 minutes a day vs 5+ hours of irrigation per day is a big question too.

I guess you are in Zambia--So I don't know the cost or availability of power in your area, or the available solar + battery equipment you would like to use....

Just as a start--A generic 3 HP motor:

• 3 HP * 746 Watts per HP = 2,238 Watts (ideal 3 hp motor)
• 2,238 Watts * 1/0.50 PF (losses and inductive losses) = 4,476 VA inverter

Guess at motor actual power usage (losses):

• 2,238 Watt * 1/0.65 generic AC motor eff = 3,443 Watts motor running power (guess)

Standard larger AC inverters (home/small business) are 48 VDC systems. The minimum battery bank just to reliably run a 5 kwatt AC inverter would be:

• 5,000 Watt inverter * 100 AH / 1,000 Watt @ 48 volt battery bank =  500 AH @ 48 volt battery bank minimum (does not account for how many hours per day pumping--Just the minimum suggested battery bank capacity)

You do not say how many hours per day of pumping you plan... Just a starting guess at 1 hour per day pumping with two days of stored/no sun energy:

• 3,443 Watt motor * 1 hour per day = 3,443 WH per day
• 3,443 WH per day * 0.52 AC Solar off grid system efficency * 1/48 volt battery bank * 2 days of storage * 1/0.50 max planned discharge for longer battery life = 552 AH @ 48 volt lead acid battery bank 

There are two calculations for sizing solar array... One based on minimum battery charging (amps into bank) of 10% (13%-25% max):

• 552 AH * 58 volts charging * 1/0.77 solar panel+controller deratings * 0.10 rate of charge = 7,176 Watt array minimum (10% rate of charge)

• The second is based how many Watt*Hours per day you use, and what are the hours of sun per day for you location. Guessing Lusaka Zambia, fixed array, 15 degree tilt facing north.:
  https://pvwatts.nrel.gov/pvwatts.php
  
  Don't know the system/irrigation is year round, or just part of year... Pick 4.97 Hours of sun per day (December) for now. The solar array suggested for 1 hour per day pumping would be
• 3,443 WH per day pumping * 1/0.52 off grid system eff = 6,621 Watt array December "break even"

Depending on your energy needs... If you "need" 3,443 WH per day (sunny or cloudy weather), then a solar "fudge factor" of 65% to 50% of planned harvest for reliable solar harvest (i.e., minimum usage of generator--save fuel), then an array of:

• 6,621 Watt array (ideal) * 1/0.65 solar fudge factor = 10,186 Watt array (minimize genset usage in December)

With solar energy, you really need to work closely with your pump supplier... The above assumes "off the self" generic AC motor. There are more efficient motors (permanent magnet type) and using a VFD (variable frequency drive--A kind of AC inverter that varies the output frequency to give "soft start" and variable pump RPM depending on pumping needs and available energy). 

And there are integrated pumps+VFDs that can run directly from solar panels (no batteries needed, just pump when the sun is up and the sky is relatively clear.

For pumping, VFDs + appropriate 3 phase motor can be a cost effective solution. Note that VFDs can be "hard" on AC 3 phase motors, and you need to pick a motor that is rated to operate with a VFD.

The above is just an example of how the math works out based on your questions an my guesses on what such an irrigation system may look like with a battery bank.

One high quality well pump vendor that makes "solar friendly pumps" is Grondfos.

https://www.grundfos.com/

The usual answer, for off grid solar power systems--Design the most efficient and "solar friendly" system you can (low energy usage, low power surges, minimum pressure/runtime for pumps, try to avoid battery banks for irrigation systems, etc.)--It is almost always cheaper in the to conserve energy than to generate it.

-Bill