Motor power sizing for pumps and fans starts with the hydraulic power needed to move a fluid or air through a system. This hydraulic power depends on the flow rate, the head (or pressure rise) that must be achieved, and the physical properties of the fluid such as density and gravity.
Because real equipment is not perfectly efficient, the motor must supply more power than the pure hydraulic requirement. The pump (or fan) efficiency and the motor efficiency are applied as multiplicative factors that reduce the usable power, so the motor power is the hydraulic power divided by the product of these efficiencies.
When performing the calculation, keep all units consistent (e.g., meters, seconds, kilograms) and convert efficiencies from percent to decimal. The result is typically expressed in kilowatts (kW) and should be rounded up to the next standard motor size, adding a safety margin for overload and future capacity growth.
How do I calculate hydraulic power for a pump?
What factors affect motor power sizing?
How do I account for inefficiencies in the system?
Can this calculator be used for fans as well?
What units should I use for flow rate and head?
How do I convert hydraulic power to electrical power?
What is the significance of fluid density in this calculation?
Results are for informational purposes only and do not constitute professional advice.
