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    Calculates locally in your browser · duty point to running cost
    Rotating equipment / pump sizing

    Pump power

    Go from duty point to electricity bill: hydraulic power from flow and head, shaft power through pump efficiency, motor input through motor efficiency, then annual running cost.

    Wire-to-water64.8%Specific speed21.6Losses1.70 kW
    Hydraulic power
    3.132 kW
    ρgQH — the useful work
    Shaft power
    4.350 kW
    At 72.0% pump efficiency
    Motor input
    4.833 kW
    At 90.0% motor efficiency
    Annual energy cost
    $2,900
    1.933e+4 kWh at $0.15/kWh
    Power chain
    How electrical input converts into hydraulic output through each efficiency stage.
    32.0 m head
    Power chain
    StagePowerEfficiencyLoss
    Hydraulic output3.132 kW——
    Shaft input4.350 kW72.0%1.218 kW
    Motor input4.833 kW90.0%0.483 kW
    Overall4.833 kW64.8%1.701 kW
    Annual energy1.933e+4 kWh—$2,900

    Overall wire-to-water efficiency is the product of the two stage efficiencies, not their average.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Power chain
    Hydraulic, shaft and motor power side by side.
    Power chain0.0001.142.283.424.565.703.13HydraulicHydraulic4.35ShaftShaft4.83MotorMotorkW
    Power vs pump efficiency
    At fixed duty, shaft power falls hyperbolically as efficiency improves.
    Power vs pump efficiency0.0001.753.515.267.028.7740%40%50%50%60%60%70%70%80%80%90%90%Shaft kW
    Canonical equations
    The exact relationships implemented by this engine.

    P_h = ρgQH · the physics of lifting fluid

    P_s = P_h/η_p · pump efficiency penalty

    P_m = P_s/η_m · electrical input

    n_s = n√Q/H^0.75 · selects the impeller family

    engineering methodology

    A clear path from inputs to results.

    A pump converts electrical power into fluid power through two efficiency stages. Hydraulic power is the physics — ρgQH. Shaft power is what the pump needs at its coupling. Motor input is what the meter records. The gaps between them are where running costs live.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Enter the design flow rate.
    2. Enter the total head, including static lift, friction and discharge pressure.
    3. Set the pump efficiency at the duty point.
    4. Set the motor efficiency.
    5. Enter the operating speed for specific speed, and the annual hours and energy price for running cost.
    6. Check that the duty point is near the best efficiency point of the selected pump.

    The formula

    Ph=ρgQH,Ps=Phηp,Pm=Psηm,ns=nQH0.75P_h = \rho g Q H, \qquad P_s = \frac{P_h}{\eta_p}, \qquad P_m = \frac{P_s}{\eta_m}, \qquad n_s = \frac{n\sqrt{Q}}{H^{0.75}}

    Reference: Hydraulic Institute HI 14.6 / ISO 9906 pump performance standards.

    Worked example

    36 m³/h against 32 m of head (default inputs)

    Inputs
    • Flow 36 m³/h.
    • Head 32 m.
    • Pump efficiency 72%; motor efficiency 90%.
    • Water at 998 kg/m³, 4,000 hours a year at $0.15/kWh.
    Calculation path
    1. 1

      Convert flow: 36/3600 = 0.01 m³/s.

    2. 2

      Hydraulic power: 998 × 9.807 × 0.01 × 32 = 3.132 kW.

    3. 3

      Shaft power: 3.132 / 0.72 = 4.350 kW.

    4. 4

      Motor input: 4.350 / 0.90 = 4.833 kW.

    5. 5

      Annual cost: 4.833 × 4,000 × 0.15 = $2,900.

    Result

    3.13 kW hydraulic · 4.35 kW shaft · 4.83 kW motor · 64.8% wire-to-water · $2,900 a year.

    Interpreting the result

    Overall efficiency is 64.8%, the product of 72% and 90%. The 1.7 kW lost between the terminals and the water is the pump’s inefficiency, and at 4,000 hours a year it is 6,800 kWh — usually more than the pump costs to buy. A 10-point efficiency gain would save about $400 a year on this duty.

    Assumptions & limitations

    • The duty point is on the pump curve at its best efficiency point. Off-BEP operation raises losses and vibration.
    • Head is total head including friction, static lift and discharge pressure.
    • Efficiency figures are at the operating point, not the nameplate maximum.
    • No variable-speed drive losses are included; add 2–4% if one is fitted.
    • Fluid density is constant and the fluid is Newtonian.

    Frequently asked questions

    Why is motor input higher than shaft power?

    Because the motor itself has losses — copper, iron and mechanical. A 90% efficient motor turns 90% of electrical input into shaft power and dissipates the rest as heat.

    How much does efficiency really matter?

    Enormously on continuous duty. A pump running 8,000 hours a year spends roughly its purchase price in electricity every few months. Efficiency dominates lifecycle cost.

    How do I account for a variable-speed drive?

    Add the VSD’s own losses (typically 2–4%) to the motor input, and use the affinity laws: flow scales with speed, head with speed squared, power with speed cubed.

    What is specific speed used for?

    It classifies impeller geometry. Low values (under 1,000) indicate radial-flow impellers for high head and low flow; high values indicate mixed or axial flow for high flow and low head.

    What is wire-to-water efficiency?

    The product of pump and motor efficiency — the fraction of electrical input that becomes fluid power. It is the number to compare between pumping solutions.

    Can I use this for a fan or compressor?

    The structure is similar but the constants differ. Fans use pressure rather than head; compressors need an isentropic efficiency and a compression ratio, not a static head.

    Methodology last reviewed 2026-10-18. Results are informational and not financial or engineering advice — see our disclaimer.