How to calculate hydraulic pump power
This is the rate at which a pump adds useful hydraulic energy to the water. It is not the electrical input power printed on a motor.
What the formula is saying
Each kilogram gains gH joules of hydraulic energy. The mass flow rate is ρQ kilograms per second. Their product gives Ph = ρgQH.
Read the symbols in plain language
- ρ
- Fluid densitykg/m³
- g
- Gravitym/s²
- Q
- Flow ratem³/s
- H
- Headm
Sort out the units first
Q must be m³/s, not litres per second: 20 L/s = 0.02 m³/s. Use total added head H in m; the output is W. Divide by 1,000 for kW.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
- ρ · Fluid density
- 1000 kg/m³
- g · Gravity
- 9.81 m/s²
- Q · Flow rate
- 0.02 m³/s
- H · Head
- 10 m
Convert volume flow to mass flow
Density tells us how many kilograms are carried each second.
(1000) × (0.02) = 20 kg/sFind energy added per kilogram
Head is energy per unit weight, so multiplying by g gives energy per mass.
(9.81) × (10) = 98.1 J/kgMultiply mass flow by energy per mass
Joules per second are watts.
(20) × (98.1) = 1962 W
Does this worked answer make sense?
The example output is 1.962 kW. At an assumed overall efficiency of 0.70, the corresponding input would be about 2.803 kW, not 1.373 kW.
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
Use these new values. Work it out first, then check your answer.
- ρ · Fluid density
- 1000 kg/m³
- g · Gravity
- 9.81 m/s²
- Q · Flow rate
- 0.03 m³/s
- H · Head
- 15 m
Find: hydraulic pump power
A hint, not the answer
Each kilogram gains gH joules of hydraulic energy. The mass flow rate is ρQ kilograms per second. Their product gives Ph = ρgQH.
Q must be m³/s, not litres per second: 20 L/s = 0.02 m³/s. Use total added head H in m; the output is W. Divide by 1,000 for kW.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Convert volume flow to mass flow
Density tells us how many kilograms are carried each second.
(1000) × (0.03) = 30 kg/sFind energy added per kilogram
Head is energy per unit weight, so multiplying by g gives energy per mass.
(9.81) × (15) = 147.15 J/kgMultiply mass flow by energy per mass
Joules per second are watts.
(30) × (147.15) = 4414.5 W
Avoid the common trap
Do not multiply by efficiency when converting hydraulic output to required input. Input is hydraulic power divided by the relevant efficiency.
When this method applies — and when it does not
Hydraulic power for steady flow with specified density and total added head. Pump selection also requires losses, efficiencies and operating-point analysis.
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.
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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.
Lesson updated: · Worked examples checked against the implemented formula; not an independent engineering certification.