Learn: Pump input power
A pump must supply energy to raise liquid through a specified total head. Hydraulic output power is less than required input power because the pump or combined drive has losses.
What the formula is saying
Find liquid weight flow ρgQ, multiply by required head H to obtain hydraulic power, then divide by efficiency η. Division increases the required input when efficiency is below one.
Read the symbols in plain language
- ρ
- Fluid density
Mass per unit volume for the stated material and condition. This is density, not weight per volume.
kg/m³Use kg/m³ as the base unit shown here. Use density in kg/m³, Q in m³/s, H in m and g in m/s². The result is W; divide by 1000 for kW. Efficiency is a decimal fraction: 75% is 0.75.
- g
- Gravity
Gravity. Weight flow multiplied by delivered head gives energy transferred per second.
m/s²Use m/s² as the base unit shown here. Use density in kg/m³, Q in m³/s, H in m and g in m/s². The result is W; divide by 1000 for kW. Efficiency is a decimal fraction: 75% is 0.75.
- Q
- Flow rate
Flow rate. Weight flow multiplied by delivered head gives energy transferred per second.
m³/sUse m³/s as the base unit shown here. Use density in kg/m³, Q in m³/s, H in m and g in m/s². The result is W; divide by 1000 for kW. Efficiency is a decimal fraction: 75% is 0.75.
- H
- Pump head
Pump head. Weight flow multiplied by delivered head gives energy transferred per second.
mMetres measure length; 1 m = 1000 mm.
- η
- Efficiency as decimal
Efficiency as decimal. Divide useful power by the applicable efficiency to find required input power.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- Pin
- Result to find
Pump input power. Divide useful power by the applicable efficiency to find required input power.
W
Sort out the units first
Use density in kg/m³, Q in m³/s, H in m and g in m/s². The result is W; divide by 1000 for kW. Efficiency is a decimal fraction: 75% is 0.75.
Assumptions before calculating
Assume steady incompressible flow and a known required total dynamic head. η must match the desired input: pump efficiency for shaft input, or combined pump-and-motor efficiency for electrical input.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find Pin and explain the result in the stated output unit.
- ρ · Fluid density
- 1000 kg/m³
- g · Gravity
- 9.81 m/s²
- Q · Flow rate
- 0.05 m³/s
- H · Pump head
- 30 m
- η · Efficiency as decimal
- 0.75
Calculate useful hydraulic power
Weight flow multiplied by delivered head gives energy transferred per second.
(1000) × (9.81) × (0.05) × (30) = 14715 WAllow for efficiency losses
Divide useful power by the applicable efficiency to find required input power.
(14715) ÷ (0.75) = 19620 W
Does this worked answer make sense?
Input power must be at least hydraulic power for 0 < η ≤ 1. Halving efficiency doubles required input for the same flow and head.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- ρ · Fluid density
- 1000 kg/m³
- g · Gravity
- 9.81 m/s²
- Q · Flow rate
- 0.03 m³/s
- H · Pump head
- 25 m
- η · Efficiency as decimal
- 0.8
Calculate useful hydraulic power
Weight flow multiplied by delivered head gives energy transferred per second.
(1000) × (9.81) × (0.03) × (25) = 7357.5 WAllow for efficiency losses
Divide useful power by the applicable efficiency to find required input power.
(7357.5) ÷ (0.8) = 9196.875 W
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- ρ · Fluid density
- 1000 kg/m³
- g · Gravity
- 9.81 m/s²
- Q · Flow rate
- 0.04 m³/s
- H · Pump head
- 20 m
- η · Efficiency as decimal
- 0.7
Find: Learn: Pump input power
A hint, not the answer
Find liquid weight flow ρgQ, multiply by required head H to obtain hydraulic power, then divide by efficiency η. Division increases the required input when efficiency is below one.
Use density in kg/m³, Q in m³/s, H in m and g in m/s². The result is W; divide by 1000 for kW. Efficiency is a decimal fraction: 75% is 0.75.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Calculate useful hydraulic power
Weight flow multiplied by delivered head gives energy transferred per second.
(1000) × (9.81) × (0.04) × (20) = 7848 WAllow for efficiency losses
Divide useful power by the applicable efficiency to find required input power.
(7848) ÷ (0.7) ≈ 11211.42857 W
Avoid the common trap
Do not multiply hydraulic power by efficiency to estimate input power. Do not enter litres per second as m³/s, or confuse total dynamic head with elevation lift alone.
When this method applies — and when it does not
This does not choose a pump, operating point, motor rating or allowance for starting. Cavitation, suction conditions, efficiency variation, duty cycle and reliability margins remain separate checks.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
One idea understood. Keep going.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Pump input power. Water-measurement principles; for discharge devices, read the orifice/weir chapters and the installation and head-measurement conditions, not only the coefficient formula.
- U.S. Bureau of Reclamation — Water Measurement Manual, 3rd edition (1997; revised reprint 2001)
- Dawei Han, University of Bristol — Concise Hydraulics (2008, Ventus Publishing)
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.