Pump Power

Hydraulic Power · Mechanical Power · Efficiency η · Dewatering · Pressure Supply

Pump Calculator

Select Calculation:

Hydraulic Power P_hydr

Mechanical Power P_mech and Efficiency η

Calculate Efficiency η

Back-Calculate Flow Rate Q

Back-Calculate Pump Head H

Fluid / Density ρ [kg/m³]

Volume Flow Q

10 l/s ≈ 36 m³/h ≈ 0.01 m³/s

Pump Head H [m] Difference of pressure head + geodetic height + velocity head

Overall Efficiency η [%]

Centrifugal pump: 60–85 % | Positive displacement pump: 80–95 %

Formulas & Fundamentals

Hydraulic Power (Delivery Power):
P_hydr = ρ · g · Q · H [W]
ρ = Density [kg/m³] | g = 9.81 m/s²
Q = Volume flow [m³/s] | H = Pump head [m]

Mechanical Power (Drive Power):
P_mech = P_hydr / η [W]
η = Overall efficiency (pump efficiency + motor efficiency)
Typical: Centrifugal pump 60–85 %, Positive displacement pump 80–95 %

Specific Energy (Pump Head):
H = P_hydr / (ρ · g · Q) [m]
Alternatively: H = (p₂ – p₁) / (ρ·g) + (z₂ – z₁) + (v₂² – v₁²) / (2g)
(pressure increase + geodetic + velocity increase)

Efficiency:
η = P_hydr / P_mech · 100 %
Composition: η = η_vol · η_hydr · η_mech
(Volumetric, hydraulic, mechanical efficiency)

Typical Efficiency Values

Pump Type	η	Range
Centrifugal pump	70–82 %	low to medium
Positive displacement (axial)	85–95 %	high
Gear / Screw pump	80–92 %	medium to high
Centrifugal + Motor 90%	77 · 0.9 = 69 %	overall

Pump Power – Fundamentals of Pump Selection

What Does Pump Power Describe?

Pump power describes the energy transfer from a pump to the fluid. The hydraulic (or delivery) power P_hydr is the actual power absorbed by the water; the mechanical drive power P_mech is the power supplied by the motor. The efficiency η describes how much of the drive power is actually converted into hydraulic power.

P_hydr = ρ · g · Q · H
Hydraulic power is proportional to the fluid density, the volume flow, and the pump head.

Hydraulic Energy Form

Pressure & Potential Energy: P_hydr is equivalent to
Weight × Height = ρ·g·Q·H

Example: lift 10 l/s by 15 m
= 998 · 9.81 · 0.01 · 15 = 1.47 kW

Losses in Pump & Drive

Overall Efficiency:
η = η_vol × η_hydr × η_motor

E.g.: 82 % (pump) × 90 % (motor)
= 74 % overall efficiency

Application Examples: Dewatering, Water Supply, Pressure Boost

Dewatering & Construction

Groundwater control in excavations: pump water masses from construction pits.
Pump head: geodetic height + pressure in discharge line (5–10 bar = 50–100 m).

Drainage & Discharge

Land drainage, stormwater, wastewater protection against backflow.
Typical: small pumps (1–10 l/s), moderate head (2–6 m).

Pressure Supply & Long-Distance Water

Drinking water to high-rise buildings, irrigation systems, pressure pipelines.
High pressures = large pump head = high power.

Worked Example: Construction Pit Dewatering

Task:

Water is to be pumped from an 8 m deep excavation at 15 l/s. Discharge line operates at 3 bar. Pump efficiency η = 75 %. Water 20°C (ρ = 998 kg/m³). What drive power P_mech?

Solution:

Geodetic height: 8 m
Pressure head: 3 bar = 30 m water column (p/(ρ·g) = 300,000/(998·9.81))
Pump head: H = 8 + 30 = 38 m
P_hydr = ρ · g · Q · H = 998 · 9.81 · 0.015 · 38 = 5.56 kW
P_mech = P_hydr / η = 5.56 / 0.75 = 7.41 kW ≈ 10 HP

Frequently Asked Questions

Hydraulic power P_hydr is the energy the pump transfers to the fluid (pressure + height + velocity). Drive power P_mech is the power supplied by the motor. The difference is energy loss due to friction, turbulence, and mechanical losses in the pump and drive. The efficiency η describes this difference.

Pump head is composed of several components:
H = (p₂ – p₁)/(ρ·g) + (z₂ – z₁) + (v₂² – v₁²)/(2g)

• Pressure increase: (p₂ – p₁) in Pascal, then divide by ρ·g (yields meters)
• Geodetic height: physical height difference in meters
• Velocity increase: usually negligible

Example: Water from 2 m depth to pressure line at 5 bar = (50 + 2) m = 52 m H.

Centrifugal pumps work with centrifugal force and create high velocities, which are then converted to pressure. This causes turbulence losses and leakage. Positive displacement pumps (axial or radial piston) displace fluid directly with fewer turbulence losses. Thus: Centrifugal pump 70–82 %, Positive displacement 85–95 %. However, centrifugal pumps are simpler, more robust, and cheaper.

Pipe friction losses are calculated using the Darcy-Weisbach formula Δp = λ·(L/D)·(ρ·v²/2). The pressure losses appear as additional pump head: H_total = H_static + Δp/(ρ·g). This calculator shows only static + dynamic head; for accurate sizing, piping losses must be added separately.

Yes, but with energy losses. Throttling (outlet valve) dissipates excess pressure but still consumes drive power. Better: Speed control (frequency converter, hydromat) or bypass regulation (excess flow back to suction tank). Modern direct-drive or high-efficiency motors save significant energy.

Summary

Hydraulic Power

P_hydr = ρ·g·Q·H
[Watt / Kilowatt]

Drive Power

P_mech = P_hydr / η
[kW / HP]

Efficiency

η = P_hydr/P_mech
60–95 %

Typical Applications

Dewatering & Construction: Groundwater lowering in excavations, water control pumps
Drainage: Land drainage, wastewater pumping stations, stormwater management
Pressure Supply Pipelines: High-rise water supply, distant water systems, irrigation
Industrial Use: Cooling water, process pressure buildup, hydraulic systems
Emergency Response: Flood protection, emergency pump deployment, fire protection

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