Pump Hydraulic and Shaft Power Calculator

Calculate the hydraulic power transferred to a liquid and the mechanical power required at the pump shaft. Enter actual volumetric flow, total pump head, fluid density, gravity, and pump efficiency, with SI and engineering unit choices.

Hydraulic power and shaft power

P_hydraulic = ρgQH and P_shaft = P_hydraulic / η_pump. The pump power difference is P_loss = P_shaft − P_hydraulic.

Hydraulic power is the useful mechanical energy transferred to the liquid per unit time. Shaft power is the mechanical input at the pump shaft or coupling. The difference represents the pump’s modeled internal losses; it is not an independently calculated fluid heating duty. The calculator normalizes Q to m³/s, H to m, ρ to kg/m³, g to m/s², and efficiency to a fraction, giving power in W before result conversion.

The intermediates are mass flow ṁ = ρQ in kg/s and specific energy gH in J/kg. Their product gives W. Imperial mass inputs are normalized to SI, so no separate numerical gravitational conversion constant is required.

Enter total pump head and pump efficiency

• Total pump head: supply the total head added by the pump at the operating point. Across suction and discharge measurement sections for a constant-density liquid, H = (p_d − p_s)/(ρg) + (V_d² − V_s²)/(2g) + (z_d − z_s). Use consistent pressure references. Static lift alone or pressure rise alone may omit required contributions.
• Flow and density: use actual delivered volumetric flow through the pump and liquid mass density at the operating conditions. No water density, specific gravity, fluid-property lookup, or standard-gas conversion is assumed.
• Pump efficiency: enter overall pump efficiency, the ratio of hydraulic output to mechanical shaft input, at the same duty point. Use a fraction greater than 0 and at most 1, or a percentage greater than 0 and at most 100. The 100% endpoint is an ideal limit.
• Electrical input is a separate quantity. Motor, drive, gearbox, and coupling losses outside the selected pump boundary require their own efficiencies. Do not substitute a combined wire-to-water efficiency when seeking pump shaft power.
• Gravity: the editable initial value 9.80665 m/s² is standard gravity. Use a known local value where appropriate. Head is a length of the supplied liquid, not a fixed water-column pressure unit.

The Pressure–Head Converter can convert the pressure-head component using explicit density and gravity. Add the applicable velocity and elevation contributions before using that component as total pump head.

Assumptions and limits

• This is a steady, single-phase, approximately incompressible liquid duty-point calculation with constant representative density, gravity, and supplied efficiency. It does not derive a pump curve, system curve, or operating point.
• Flow, head, density, and gravity must be positive. Negative flow/head belongs to a different operating regime. At zero flow or zero head, hydraulic output is zero but actual shaft input may remain positive; use a manufacturer’s performance curve rather than extrapolating a fixed efficiency. Such inputs are rejected here.
• No viscosity correction, cavitation/NPSH check, minimum-flow assessment, transient load, service factor, motor sizing margin, or equipment selection is performed. Shaft demand at this one point does not establish a motor rating.
• Results display up to 8 significant digits and steps up to 10, using full stored precision in arithmetic. All normalized inputs, mass flow, specific energy, powers, and output conversions must be finite and representable. Nonzero underflow is rejected.

Worked example

A liquid has supplied density 1000 kg/m³ and actual flow 50 m³/h. Its required total pump head is 30 m, gravity is 9.80665 m/s², and pump efficiency at that duty is 75% (η = 0.75). This density and efficiency are illustrative inputs, not a property prediction or pump recommendation.

Convert Q = 50/3600 = 0.01388888889 m³/s. Then ṁ = 1000 × Q = 13.88888889 kg/s and gH = 9.80665 × 30 = 294.1995 J/kg. Thus P_hydraulic ≈ 4086.104167 W = 4.086104167 kW and P_shaft = 4086.104167/0.75 ≈ 5448.138889 W = 5.448138889 kW.

The modeled pump loss is approximately 1.362034722 kW. At the same flow, head, density, and gravity, changing pump efficiency to 60% leaves hydraulic power unchanged and raises shaft demand to approximately 6.810173611 kW. Load the example to reproduce these results.

Sources and calculation version

• KSB Centrifugal Pump Lexicon: pump power output, pump efficiency, and pump affinity laws — hydraulic output ρgQH and the relationship between output, efficiency, and shaft input.
• KSB Centrifugal Pump Lexicon: Power input — the shaft/coupling boundary and zero-flow input power.
• NIST SP 811, Appendix B.9 — unit conversions. US liquid and Imperial gallons are distinct; hp means mechanical horsepower, 745.6998715822702 W. International feet/inches and avoirdupois mass units are used.

Changelog: engine 1.0.0 introduces separate hydraulic and shaft power, pump-loss accounting, explicit density/gravity/efficiency, unit conversion, SI intermediates, examples, and range checks. Factor registry 1.0.0 is shared with the Engineering Unit Converter. No fluid-property or pump-performance dataset is used. Engineering review status: independent review pending.