Pump Power Calculator
Calculate the shaft power required to pump a fluid against a given head, accounting for flow rate, fluid density, and pump efficiency. Use it when sizing motors and selecting pumps for water, oil, or chemical service.
About this calculator
The power delivered by a pump to a fluid (hydraulic power) is P_hydraulic = ρ × g × Q × H, where ρ is fluid density (kg/m³), g = 9.81 m/s², Q is volumetric flow rate (m³/s), and H is total head (m). Because no pump is perfectly efficient, the shaft power (brake power) drawn from the motor is higher: P_shaft (kW) = (Q × ρ × 9.81 × H) / (η × 1000), where η is the dimensionless pump efficiency (0 to 1). The factor 1000 converts watts to kilowatts. Total head H includes static elevation difference, friction losses in pipes and fittings, and pressure head differences between inlet and outlet. Efficiency η accounts for hydraulic, volumetric, and mechanical losses inside the pump. Accurate power calculations prevent motor undersizing (causing overload trips) and oversizing (wasting capital and energy).
How to use
A pump circulates cooling water (density ρ = 1000 kg/m³) at Q = 0.02 m³/s against a total head H = 30 m, with pump efficiency η = 0.75. Apply the formula: P = (0.02 × 1000 × 9.81 × 30) / (0.75 × 1000) = (5,886) / (750) = 7.848 kW. So the pump requires approximately 7.85 kW of shaft power from the motor. Select a motor rated at least 8–9 kW to allow for a service factor and any future head increases in the system.
Frequently asked questions
What is the difference between hydraulic power and shaft power in a pump?
Hydraulic power (or water power) is the useful power actually transferred to the fluid, calculated as P_hydraulic = ρgQH. Shaft power (brake power) is the total mechanical power delivered to the pump shaft by the motor, which is always greater because of internal losses. These losses include hydraulic losses (turbulence and recirculation inside the impeller), volumetric losses (leakage past seals and wear rings), and mechanical losses (bearing and seal friction). Pump efficiency η is defined as P_hydraulic / P_shaft, and typical values range from 0.60 for small pumps to 0.90 for large, well-designed centrifugal pumps.
How does total head differ from pressure difference in pump calculations?
Total head H (meters) is a more complete measure of the energy added per unit weight of fluid. It includes the static pressure head (ΔP/ρg), the elevation head (Δz), and the velocity head (Δv²/2g) between suction and discharge flanges. Expressing pump duty in meters of head makes results independent of fluid density, which is why pump curves are always plotted in meters rather than pascals. When you convert head to pressure for a specific fluid, you multiply by ρg: ΔP = ρgH. Using pressure difference alone would require re-rating the pump for every different fluid.
Why does pump efficiency matter so much for operating costs?
Because pumps often run continuously, even small differences in efficiency translate into large energy cost differences over time. For example, a 10 kW pump running 8,000 hours per year at $0.10/kWh costs $8,000/year in electricity. Improving efficiency from 70% to 80% reduces shaft power by 12.5%, saving $1,000/year. Over a 10-year plant life that is $10,000 in savings from a single pump. In large facilities with hundreds of pumps, optimizing pump selection and maintaining impeller clearances to preserve efficiency delivers substantial reductions in both operating costs and carbon footprint.