Work done in water pumping?Water is to be brought from a well 60 meters bellow ground level to a tower 10 meters high, once there the water falls freely on a 5000 liter tank. A submerged pump 10 meters under water level and a two inch diameter pipe is to be used. What is the work done by the pump when 2000 liters have been served to the tank; what is the power of a pump selected to do this work in 20 minutes.
Power_W = (Mass_kg * Gravity_m/s/s * Height_m) / time_s Where: Mass kg is 1kg/l of 2000 liters Gravity is 9.81m/s/s Head is 60m + 10m (height). The suction head is ignored because the pump is submerged. In practice there is some suction head due to intake restrictions like pipes, check valves, strainers. The time is 20 minutes in seconds. The work is the energy used in the time allocated, which is: power * time in seconds, which is watt seconds = joules. Not needed here as it is in the formula above. Power is the rate of doing work. The depth below water is not really relevant except for pipe losses which increase the head slightly. The flow is 2000 liters/20 minutes = 100 l/min. This can be used with the head of 70m to determine the pipe restriction, which amounts to a pressure drop representing extra head added. It will indicate whether the pipe is too small (excessive head added) or whether the pipe is overkill (no head added). You can find on line calculators for this, e.g. search pipe resistance flow on line calculator. I am guessing a 2 inch pipe has little loss at this flow. This is the so called water power. It is the output power of the pump. The pump may only be 50% or so efficient, so the mechanical power delivered by the motor needs to be greater by this ratio. The electrical input power to the motor is greater again to allow for motor efficiency. This might be 60-90%. In practice the efficiency of pump and motor are determined from user manual or specification of the actual devices with the actual loads (head and flow).
