BIZOMAT MAKEOS · Hydraulics
Pump Shaft Power
What shaft power does a pump need for this flow and head?
Calculate the hydraulic power delivered to a liquid and the shaft power a pump needs at a given flow, head and efficiency.
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How this tool works
Inputs, units and where to find them
| Input | Unit | Typical source |
|---|---|---|
| Pump flow | L/min | Flow at the duty point. |
| Total dynamic head | m | Total dynamic head the pump must develop at that flow, including static lift and losses. |
| Liquid density | kg/m³ | Liquid density at the pumping temperature. |
| Overall pump efficiency (as a fraction) | fraction 0–1 | Pump efficiency as a fraction from 0.01 to 1, read from the pump curve at the duty point. Enter 0.7 for 70%. |
What each input means
Open an input to see what it is, its units, the worked example and the common mistakes, with a small diagram.
What this means: Pump flow
- What it is
- The volume of liquid the pump delivers each minute at the operating (duty) point.
- Units and format
- Enter a number in L/min: litres per minute. The form accepts 0 L/min to 1000000 L/min. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 120 L/min as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Hydraulic and shaft power are directly proportional to flow.
- Get it right
- Use the flow at the duty point on the system curve, not the pump’s maximum catalogue flow.
What this means: Total dynamic head
- What it is
- The total energy per unit weight the pump must add, expressed as a height of liquid: static lift between the liquid levels plus friction and fitting losses and any pressure difference.
- Units and format
- Enter a number in m: metres. The form accepts 0 m to 10000 m. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 30 m as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Hydraulic and shaft power are directly proportional to head.
- Get it right
- Include system losses as well as the vertical lift; static height alone understates the head.
What this means: Liquid density
- What it is
- The mass per unit volume of the liquid being pumped at its pumping temperature.
- Units and format
- Enter a number in kg/m³: kilograms per cubic metre. The form accepts 1 kg/m³ to 25000 kg/m³. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 1000 kg/m³ as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Power is directly proportional to density for the same flow and head.
- Get it right
- Use the actual liquid, not water, when pumping oils, brines or slurries.
What this means: Overall pump efficiency (as a fraction)
- What it is
- The share of shaft power the pump turns into hydraulic power at this duty point, written as a fraction and read from the manufacturer’s pump curve.
- Units and format
- Enter a number in fraction 0–1: fraction 0–1. The form accepts 0.01 fraction 0–1 to 1 fraction 0–1. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 0.7 fraction 0–1 as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Shaft power is inversely proportional to efficiency.
- Get it right
- Type a fraction, not a percent: 0.7 means 70 %. Read it at the actual duty point, not the best efficiency point.
Method and formulas
- Hydraulic power (kW) = ρ × g × Q × H / 1,000, with Q in m³/s (L/min ÷ 60,000) and g = 9.80665 m/s².
- Shaft power = hydraulic power / efficiency.
- Example: 120 L/min, 30 m, 1,000 kg/m³ and efficiency 0.7 give a hydraulic power of 0.588399 kW and a shaft power of 0.840570 kW.
Limitations
- Efficiency varies along the pump curve; the result applies only at the entered duty point.
- Motor losses, drive losses and the motor sizing margin are not included. Viscous liquids need viscosity corrections to the pump curve.
- No certification or pump selection is implied.
- Example values are illustrations. Replace them with values for your actual task.
Typical uses
- Estimate pump input power at a duty point
- Compare duty points or pump efficiencies
- Check a motor size before supplier selection
Who it is for
- Design engineers
- Maintenance engineers
What you get
- Pump shaft power (kW)
- Hydraulic power (kW)
Interpreting the result
- Size the motor for the highest shaft power over the operating range, not only the normal duty point.
Before relying on the result
- Read efficiency and power from the manufacturer’s pump curve for the actual impeller and speed.
- Apply the pump or motor manufacturer’s sizing allowance before choosing a motor.
References
- US Department of Energy, DOE-HDBK-1012/3-92, Fluid Flow — https://www.energy.gov/sites/default/files/2026-04/DOE-HDBK-1012-92_VOL3.pdf
- Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 365–368. Topic reference; no source tables or recommended settings are reproduced.
Learn the Concepts Behind This Tool
These BIZOMAT Library readings explain the method this tool uses, with worked examples.
Method explained here · Library collection
Pipe flow, pressure drop and pump selection: getting fluid where it needs to go efficiently
In Fluids, HVAC and Building Services
What it covers: How head, velocity, friction and fittings set pressure drop, how system and pump curves fix the operating point, why efficiency and NPSH matter, and how pipe size affects energy cost.
Method explained here · BIZOMAT original book
Pumps, energy and the least work of separation
In Reverse Osmosis by the Numbers
What it covers: High-pressure and booster pump power, specific energy per cubic metre, the thermodynamic least work of separation from the entropy of mixing, where the energy goes and what saving it is worth.
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