Pump Power and Head Calculator
Determine the drive power required for a centrifugal or positive-displacement pump from flow rate and head. Enter the system head directly or build it from static lift, pressure difference and friction loss. The calculator returns hydraulic power, shaft power via the pump efficiency and optionally the electrical motor power – live with every input and as a PDF design report.
Pump Power Calculator
Model: incompressible medium, simplified system head H_A = H_geo + Δp/(ρ·g) + H_v without velocity head, g = 9.81 m/s². A key-value calculator for sizing and quotation; no cavitation check (NPSH) and no strength verification. The friction loss H_v must be determined beforehand via a pipe pressure-loss calculation.
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Calculating …
Formulas and fundamentals
The system head H_A is the specific work the system demands, expressed as a column of liquid. In simplified form it consists of the static lift H_geo (elevation difference between outlet and inlet levels), the pressure head Δp/(ρ·g) from the pressure difference between outlet and inlet vessels, and the friction loss H_v from pipe friction and local resistances: H_A = H_geo + Δp/(ρ·g) + H_v. The velocity head (v_a²−v_e²)/(2g) is small for near-equal cross sections and is neglected here.
The hydraulic or useful power is the power actually delivered to the fluid: P_hyd = ρ·g·Q·H, with density ρ, gravitational acceleration g = 9.81 m/s², volume flow Q and head H. It rises linearly with flow rate. For water (ρ = 1000 kg/m³) the shortcut P_hyd[kW] ≈ 2.725·Q[m³/h]·H[m]/1000 can be used.
The shaft power required at the coupling follows from the pump efficiency η_P, which lumps together internal hydraulic, volumetric and mechanical losses: P_shaft = P_hyd/η_P. If an electric motor drives the pump, the required motor input power follows from the motor efficiency η_motor as P_motor = P_shaft/η_motor; the overall efficiency is then η = η_P·η_motor.
Worked example
A centrifugal pump delivers Q = 50 m³/h of water (ρ = 1000 kg/m³) against a system head of H = 30 m. In SI units the flow rate is Q = 50/3600 = 0.01389 m³/s.
The hydraulic power is P_hyd = ρ·g·Q·H = 1000·9.81·0.01389·30 = 4087.5 W ≈ 4.09 kW. With a pump efficiency of η_P = 0.75 the shaft power is P_shaft = 4087.5/0.75 = 5450 W ≈ 5.45 kW.
Adding a motor efficiency of η_motor = 0.90 gives the electrical motor power P_motor = 5450/0.90 = 6056 W ≈ 6.06 kW, corresponding to an overall efficiency of 0.675. For motor selection the next larger standard motor is chosen and an operating margin is applied.
Frequently asked questions
What is the difference between hydraulic power and shaft power?
Hydraulic power P_hyd = ρ·g·Q·H is the pure useful power actually transferred to the fluid. Shaft power P_shaft = P_hyd/η_P is the mechanical power required at the coupling; it is larger by the internal pump losses. The pump efficiency η_P combines hydraulic, volumetric and mechanical losses and is typically between 0.4 and 0.85 for centrifugal pumps depending on size and operating point.
How is the system head composed?
From the static lift H_geo (pure elevation difference), the pressure head Δp/(ρ·g) from differing vessel pressures, and the friction loss H_v from pipe friction and local resistances: H_A = H_geo + Δp/(ρ·g) + H_v. The friction loss is not fixed but depends on the flow rate (approximately quadratically) and is determined from a pipe pressure-loss calculation.
Why g = 9.81 and how do I handle other media?
P_hyd = ρ·g·Q·H holds for any incompressible medium; only density ρ and head H change. For a denser medium the power rises in proportion to density. Note: a head given in metres is medium-independent, a pressure given in bar is not – Δp = ρ·g·H. The calculator consistently uses g = 9.81 m/s².
Does the calculator account for cavitation (NPSH)?
No. The calculator determines head and drive power, not the NPSH value. Cavitation safety requires comparing the available NPSH_A of the system with the required NPSH_R from the pump curve. This check must be carried out separately, especially for suction-lift operation, hot media or long suction lines.
How do I select the motor from this?
The governing value is the shaft power at the worst-case operating point. The required input power follows from the motor efficiency; a safety or operating margin is added and the next larger standard motor is chosen. For centrifugal pumps whose power curve rises beyond the design point, the whole possible operating range must be considered, not just the rated point.
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