Pump Power and Efficiency Calculator
Calculates the shaft power a pump must deliver to move fluid against a given head, accounting for pump efficiency. Used when selecting pumps for water supply, irrigation, HVAC, and industrial process systems.
About this calculator
The power required to drive a pump is derived from the hydraulic power needed to raise fluid pressure, divided by the pump's efficiency. The formula used here is: P (kW) = (Q × H × ρ × g) / (η / 100) / 1000, where Q is volumetric flow rate (m³/s), H is total head (m), ρ is fluid density (kg/m³), g is gravitational acceleration (9.81 m/s²), and η is pump efficiency (%). The numerator Q × H × ρ × g gives the useful hydraulic power in watts — the rate at which energy is transferred to the fluid. Dividing by efficiency accounts for mechanical and hydraulic losses inside the pump. Dividing by 1000 converts the result to kilowatts. A lower efficiency means the motor must supply more electrical energy to achieve the same fluid power, directly increasing operating costs. Pump efficiency typically ranges from 50% for small centrifugal pumps to over 90% for large axial-flow machines.
How to use
A centrifugal pump moves water (density 1000 kg/m³) at a flow rate of 0.05 m³/s against a total head of 30 m, and has an efficiency of 75%. Apply the formula: P = (0.05 × 30 × 1000 × 9.81) / (75 / 100) / 1000. Step 1: hydraulic power = 0.05 × 30 × 1000 × 9.81 = 14,715 W. Step 2: divide by efficiency = 14,715 / 0.75 = 19,620 W. Step 3: convert to kW = 19,620 / 1000 = 19.62 kW. Enter Q = 0.05 m³/s, H = 30 m, density = 1000 kg/m³, and efficiency = 75% into the calculator to get the 19.62 kW shaft power requirement instantly.
Frequently asked questions
What is total head in a pump calculation and how do I determine it?
Total head (H) represents the total energy per unit weight that the pump must add to the fluid, expressed in metres. It combines static head (the vertical height difference between suction and discharge), pressure head (any difference in system pressure at inlet and outlet), and friction head (pressure losses in pipes, fittings, and valves expressed as equivalent metres of fluid). To determine total head, sum the static elevation difference, convert any pressure differences using H = ΔP/(ρg), and add the friction losses calculated from the Darcy-Weisbach equation. Accurate total head estimation is essential for selecting a pump whose curve intersects the system curve at the desired operating point.
How does pump efficiency affect electricity costs over time?
Pump efficiency directly multiplies the electrical energy consumed for a given duty. A pump running at 60% efficiency instead of 80% consumes 33% more electrical power for identical flow and head conditions. For a 20 kW hydraulic duty running continuously, the difference in shaft power is (20/0.60) − (20/0.80) = 33.3 − 25.0 = 8.3 kW. Over one year at $0.15/kWh that gap costs approximately $10,900 in extra electricity. This is why lifecycle cost analysis — not just purchase price — drives pump selection in industrial and municipal applications, and why high-efficiency motors and variable speed drives are rapidly adopted.
What is the difference between hydraulic power and shaft power for a pump?
Hydraulic power (also called water power) is the useful energy delivered to the fluid per second: P_hydraulic = Q × H × ρ × g. Shaft power (brake power) is the mechanical power that the motor or engine must supply to the pump shaft. The difference between them is due to internal losses — impeller friction, recirculation, leakage, and bearing losses — all captured by the pump efficiency η: P_shaft = P_hydraulic / η. The motor must then supply even more power to overcome its own electrical and mechanical losses, so motor efficiency is a separate factor. Always confirm you are comparing the correct power stage when reading manufacturer pump curves.