First determine what the static water level is in your well, which is the height of the water surface above the ground when the water pump is operating normally. Since we have not installed a water pump, we cannot measure this distance temporarily. so measure the distance between the water surface and the ground when there is no pumping. Subtract 10 meters from the value measured as net height. At this time, calculate the shaft power P = flow × head ÷ 102 ÷ 3.6 ÷ efficiency. The general flow rate of a 2-inch agricultural pump is between 20 and 30 cubic meters, and the efficiency is generally between 55% and 55%. About 60%, calculated based on the static water level of 30 meters, the flow rate takes the larger value, and the efficiency takes the smaller value P = 4.45 kW, which gives a margin of 20% = 5.3 kW, so the water pump power is 7.5 kW, calculated based on the die motor powersalt: 1 horsepower = 0.73549875 kilowatts, so the diesel engine chooses a minimum of 11 horsepower.
This method of watering the soil is quite expensive!
Are you sure you have a 350 KW generator? It should be a 35 kW generator. . An 11 kilowatt water pump is already extremely large. . It would be a bit difficult if all the numbers were removed. It is difficult to start at the same time and the load is very heavy in normal operation. Typically 35 kilowatts, with a water pump of around 15 kilowatts, so that the water pump can operate at full capacity