Hydraulic Pump Flow Calculator

Flow from displacement and speed, or the displacement a given flow needs.

Calculator

cm³/rev
RPM
%

From the datasheet at your pressure. It falls as pressure rises.

mm

Gives the cylinder speed at this flow.

mm
L/min

Gives the real volumetric efficiency from a flow test.

Example Result

Your result will appear here

The calculated result will appear here after you press Calculate.

About the Hydraulic Pump Flow Calculator

A hydraulic pump moves a fixed volume on every turn of its shaft, so its flow is that volume multiplied by the speed. A 28 cm³/rev pump at 1,450 rpm moves 40.6 L/min before losses. Some oil slips back through internal clearances, more at high pressure, and volumetric efficiency accounts for it. Enter the displacement, the speed and the efficiency to find the flow, or switch modes and enter the flow you need to find the displacement to buy. Add a cylinder bore to see how fast that flow moves it. Flow sets how fast the machine moves; pressure, a separate calculation, sets how hard it pushes.

How to Use the Hydraulic Pump Flow Calculator

  1. 1

    Choose the flow mode, or the required displacement mode for sizing.

  2. 2

    Enter the displacement or the flow you need, and the shaft speed.

  3. 3

    Enter the volumetric efficiency at your working pressure.

  4. 4

    Add a cylinder bore and rod for the cylinder speed.

  5. 5

    Press Calculate.

Hydraulic Pump Flow Calculator Examples

Frequently Asked Questions

How do I calculate hydraulic pump flow?

Flow (L/min) = displacement (cm³/rev) × rpm ÷ 1,000 × volumetric efficiency.

How do I convert cc/rev to gpm?

gpm = cc/rev × rpm ÷ 3,785.

What displacement do I need for a given flow?

D = 1,000 × Q ÷ (rpm × ηv), then round up to a catalogue size.

Does pressure change pump flow?

Not the theoretical flow, but internal leakage rises with pressure, so actual flow falls.

What is volumetric efficiency?

Actual flow divided by theoretical flow. The rest leaks internally.

How is it different from overall efficiency?

Overall efficiency also includes friction: it is volumetric × mechanical efficiency, and it decides the motor size.

Why does the same pump flow less on 50 Hz power?

A 4-pole motor runs near 1,450 rpm on 50 Hz and 1,750 rpm on 60 Hz, and flow follows speed.

By O.E. — Partner, chairman of the board and CEO of a metal products manufacturer for 21 years; the company exports to 45 countries.

Drafted with AI assistance; calculations checked against primary sources and tested with the site's own code. About FlowUtil

Last reviewed

Flow and displacement

D is in cm³/rev or in³/rev, n in rpm, ηv as a decimal. One US gallon is 231 in³.

SymbolMeaning
QFlow: D × n × ηv ÷ 1,000 (L/min), or ÷ 231 (gpm)
DDisplacement needed: 1,000 Q ÷ (n × ηv), or 231 Q ÷ (n × ηv)
vCylinder speed: Q ÷ piston area
ηvVolumetric efficiency, actual ÷ theoretical flow

A 28 cm³ pump, and sizing in reverse

  1. 128 cm³/rev on a 4-pole motor at 1,450 rpm, 95% volumetric efficiency.

    28 × 1,450 ÷ 1,000 = 40.6 L/min; × 0.95

    38.6 L/min (10.2 US gpm)

    On a 60 Hz supply at 1,750 rpm the same pump delivers 20.7% more. Driving a Ø63 mm cylinder (area 31.17 cm²), that flow extends the piston at 0.21 m/s.

  2. 2You need 40 L/min at 1,450 rpm and expect 92% efficiency at working pressure.

    40,000 ÷ (1,450 × 0.92)

    30.0 cm³/rev

    Choose the next catalogue size up, not the nearest: 32 cm³/rev gives 42.7 L/min. In US units, 15 gpm at 1,750 rpm at 93% needs 2.13 in³/rev.

What volumetric efficiency really tells you

Textbooks give typical ranges, and every pump has its own curve. The figure falls as pressure rises and oil thins, and it can be far lower at part displacement on a variable pump. Take it from the datasheet at your operating point. To find it on a working machine, measure flow at pressure and divide by the theoretical flow: 35 L/min from a pump with 40.6 L/min theoretical is 86%. A value falling over time points to wear.

Pump typeTypical volumetric efficiency
Gear80–90%
Vane82–92%
Piston90–98%

From flow to power

Flow and pressure together give hydraulic power: 38.6 L/min at 200 bar is 38.6 × 200 ÷ 600 = 12.9 kW. A pump with 88% overall efficiency needs 14.6 kW at its shaft.

Formula basis

  1. Positive-displacement pump relationship Q = D n ηv. The efficiency ranges are typical textbook values, not limits; 1 US gallon = 231 in³.

Explore Knowledge

Deepen your understanding with in-depth engineering guides, formulas, and worked examples.

Browse Engineering Guides