Pump Head and Power Calculator
Calculate the shaft power required to drive a centrifugal pump given flow rate, static head, friction losses, and pump efficiency. Use this when sizing pumps for piping systems.
About this calculator
The power delivered by a pump to a fluid — called hydraulic power — is given by P_hydraulic = ρ × g × Q × H, where ρ is the fluid density (kg/m³), g is gravitational acceleration (9.81 m/s²), Q is the volumetric flow rate (m³/s), and H is the total head (m). The total head H is the sum of static head (elevation difference) and friction losses in the pipe network. Because real pumps are not perfectly efficient, the shaft power drawn from the motor is higher: P_shaft = P_hydraulic / η, where η is the pump efficiency as a decimal. This calculator converts flow rate from L/min to m³/s (dividing by 60,000) and computes shaft power in watts. Selecting the correct pump power prevents motor overloading and ensures energy-efficient operation across the desired flow range.
How to use
Example: Flow rate Q = 200 L/min, static head = 15 m, friction loss = 5 m, fluid density ρ = 1000 kg/m³ (water), pump efficiency = 75%. Step 1: Convert flow rate: 200 / 60,000 = 0.003333 m³/s. Step 2: Total head H = 15 + 5 = 20 m. Step 3: Hydraulic power = 0.003333 × 1000 × 9.81 × 20 = 654 W. Step 4: Shaft power = 654 / 0.75 ≈ 872 W (0.87 kW). You would therefore select a motor rated at least 1 kW to drive this pump safely.
Frequently asked questions
What is pump head and how is it different from pump pressure?
Pump head is expressed in meters of fluid column and represents the energy added per unit weight of fluid passing through the pump. Pressure, in contrast, is expressed in Pascals (Pa) or bar and depends on the fluid density. The relationship is ΔP = ρ × g × H, so a pump delivering 20 m of head for water (ρ = 1000 kg/m³) produces ΔP = 1000 × 9.81 × 20 ≈ 196,200 Pa (1.96 bar). Head is a useful concept because a given pump delivers the same head regardless of the fluid density, while the pressure it produces changes with density.
How does pump efficiency affect the motor power required for a centrifugal pump?
Pump efficiency accounts for hydraulic losses inside the pump (flow separation, turbulence) and mechanical losses (bearing friction, seal drag). A pump with 60% efficiency requires 67% more shaft power than a theoretically perfect pump to deliver the same hydraulic output. For example, if the required hydraulic power is 1 kW, a 60%-efficient pump needs 1.67 kW from the motor, while an 80%-efficient pump needs only 1.25 kW. Choosing a pump operating near its best efficiency point (BEP) reduces energy consumption and extends equipment life significantly.
What causes friction loss in a pump and piping system and how is it estimated?
Friction loss arises from viscous drag between the fluid and the pipe wall, as well as turbulence at fittings, valves, elbows, and other obstructions. In straight pipes it is calculated using the Darcy-Weisbach equation, which accounts for pipe length, diameter, flow velocity, and a friction factor that depends on the Reynolds number and surface roughness. Minor losses from fittings are usually expressed as equivalent pipe lengths or loss coefficients (K-values). The total friction loss added to the static head gives the system curve, which when intersected with the pump curve identifies the actual operating flow rate and head.