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Pump Hydraulic Power

Calculates the electrical power a pump must draw to move fluid at a given flow rate against a given head, accounting for pump efficiency.

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Pump Hydraulic Power CalculatorP = ρ·g·Q·H / η

P = ρ · g · Q · H / η
P = power (W)  ·  ρ = density (kg/m³)  ·  g = gravity (m/s²)  ·  Q = flow rate (m³/s)  ·  H = head (m)  ·  η = efficiency (0–1)
⟹ SolveP, ρ, g, Q, H, η
W
kg/m³
m/s²
m³/s
m
Please fix the errors above.
Solve for:
Presets:
Hydraulic Power
P: ρ: g: Q: H: η:
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P = ρ·g·Q·H / η  ·  Hydraulic power delivered to the fluid; efficiency η accounts for losses.

Variables

SymbolQuantityUnit
PPump power requiredW
rhoFluid densitykg/m3
gGravitational acceleration9.81 m/s2
QVolumetric flow ratem3/s
HTotal dynamic headm
etaPump efficiency (dimensionless)

What it means

The hydraulic power required to drive a pump is the energy per unit time imparted to the fluid. It is given by P_hydraulic = ρ g Q H, where ρ is fluid density, g is gravity, Q is volumetric flow rate, and H is the total head (pressure head + elevation head + velocity head). However, pumps are not 100% efficient, so the actual input power (shaft power) is P = P_hydraulic / η, where η is the pump efficiency. This formula is essential for selecting pump motors, calculating energy consumption, and designing pumping stations. The head H is a measure of the energy added to the fluid per unit weight. In practice, the pump efficiency accounts for hydraulic, mechanical, and volumetric losses. The equation is also used for turbines, where efficiency relates output power to available hydraulic power. Properly sizing pumps reduces operational costs and ensures reliable performance. This formula is widely used in water supply, wastewater treatment, and chemical process industries.

Worked example

Pump Hydraulic Power – Two Examples

Real‑World
Scenario 1 – Water Pump: Flow 0.05 m³/s, head 20 m, efficiency 70%. Find shaft power.
ParameterValue
Q0.05 m³/s
H20 m
η0.7
1P = ρ·g·Q·H / η = 1000×9.81×0.05×20 / 0.7 = 9810×0.05×20/0.7 = 9810×1/0.7 ≈ 14,014 W ≈ 14.0 kW
ResultP ≈ 14 kW
Scenario 2 – Industrial Pump: Q=0.1 m³/s, H=30 m, η=0.8. Find power.
ParameterValue
Q0.1 m³/s
H30 m
η0.8
1P = 1000×9.81×0.1×30 / 0.8 = 9810×3 / 0.8 ≈ 36,787 W ≈ 36.8 kW
ResultP ≈ 36.8 kW
Key insight: Pump power depends on flow, head, and efficiency – higher efficiency reduces power input.

Common mistakes

  • Efficiency η: This is the overall efficiency (including hydraulic and mechanical losses).
  • Head H: The total dynamic head (including elevation, pressure, and velocity heads).
  • Units: ρ in kg/m³, g in m/s², Q in m³/s, H in m → power in Watts.
  • Input vs. output: This formula gives the hydraulic power (output of the pump). Input power = output / η.
  • Flow rate: Q is the volume flow rate, not mass flow.

Applications

Pump hydraulic power is the power delivered to the fluid, calculated from density, gravity, flow rate, head, and efficiency. This equation is used to size pumps and motors in water supply, wastewater treatment, and industrial fluid transport. It enables engineers to select appropriate pump types (centrifugal, positive displacement) and to determine the required driver power. The head term accounts for elevation differences, pressure losses, and friction in pipes. By optimising pump efficiency, significant energy savings can be achieved. The formula is also applied in the design of hydroelectric plants, where the head and flow rate determine the potential power generation. Accurate power calculations ensure reliable and cost‑effective fluid transport.

  • Sizing of pumps for water distribution systems
  • Design of wastewater and sewage pumping stations
  • Hydroelectric power plant feasibility studies
  • Industrial process fluid transport
  • Energy efficiency optimisation in pumping systems