Centrifugal Pump Sizing Calculator
Calculate the shaft power required to drive a centrifugal pump given flow rate, total head, and combined pump and motor efficiency. Used when selecting a pump-motor set for fluid transfer systems.
About this calculator
The hydraulic power delivered to a fluid equals the product of flow rate, fluid weight per unit volume (ρg), and total head H, where H is the sum of static head and friction losses. Accounting for pump and motor efficiencies, the electrical power drawn from the supply is: P (kW) = (Q/3600 × ρ × g × H) / (η_pump × η_motor), where Q is in m³/h, ρ in kg/m³, g = 9.81 m/s², and efficiencies are expressed as decimals. Friction losses represent energy dissipated by pipe walls, fittings, and valves. Both pump hydraulic efficiency and motor electrical efficiency must be included because losses occur independently at each stage. The result guides motor selection and predicts operating energy costs.
How to use
Given: Q = 36 m³/h, static head = 20 m, friction losses = 5 m, ρ = 1000 kg/m³, pump efficiency = 75%, motor efficiency = 90%. Total head H = 20 + 5 = 25 m. Q in m³/s = 36/3600 = 0.01 m³/s. Numerator: 0.01 × 1000 × 9.81 × 25 = 2452.5 W. Denominator: 0.75 × 0.90 = 0.675. Power = 2452.5 / 0.675 / 1000 = 3.63 kW. Enter these values into the calculator to confirm — you would need at least a 3.63 kW motor for this application.
Frequently asked questions
How do pump efficiency and motor efficiency affect the power calculation?
Pump efficiency (η_pump) accounts for hydraulic losses inside the pump casing — turbulence, recirculation, and impeller slip — while motor efficiency (η_motor) accounts for electrical-to-mechanical conversion losses due to heat and friction in windings and bearings. They multiply together in the denominator, so even moderate losses in both stages compound significantly. For example, 75% pump and 90% motor efficiency yields a combined efficiency of just 67.5%, meaning you need 48% more electrical power than the ideal hydraulic power. Always use nameplate efficiencies at the operating duty point for accurate sizing.
What is the difference between static head and friction losses in pump design?
Static head is the elevation difference between the fluid source and delivery point — it is fixed by plant geometry and must be overcome regardless of flow rate. Friction losses arise from fluid viscosity and turbulence as the fluid moves through pipes, valves, and fittings; they increase roughly with the square of velocity (and therefore flow rate). Total head H = static head + friction losses is what the pump must supply. When sizing a pump, engineers plot the system curve (H vs. Q) and intersect it with the pump curve to find the operating point.
Why do centrifugal pumps lose efficiency at flows far from their design point?
Centrifugal pump impellers are designed for a specific best efficiency point (BEP) where the angle of fluid entering the impeller matches the blade geometry. Operating significantly above or below the BEP causes incidence losses — the fluid strikes the blades at an unfavorable angle — increasing turbulence and recirculation. Far below BEP, recirculation within the impeller can cause vibration and cavitation. Far above BEP, NPSH margin decreases and bearing loads increase. Selecting a pump whose BEP matches the actual operating flow rate maximizes efficiency and equipment life.