Positive Displacement Pump
Calculate flow rate, required speed, drive torque and drive power of hydrostatic positive displacement pumps (gear, piston and vane pumps). From displacement, operating pressure and the efficiencies η_vol and η_mh the calculator returns every quantity live - enter either speed or flow rate, live with every input.
Calculation
Model: hydrostatic positive displacement pump with constant displacement (gear, piston, vane), steady-state operation. Q = V·n·η_vol, M = V·Δp/(2·π·η_mh), P = Q·Δp/η_total with η_total = η_vol·η_mh. Cavitation (NPSH), pulsation and thermal effects are not considered. Sizing tool for fluid power.
Results
Calculating …
Calculation in your browser, inputs go to our server only when you export or save.
Formulas and fundamentals
Flow rate and speed
Hydrostatic positive displacement pumps are defined by a fixed displacement V per revolution. The delivered flow is proportional to speed:
The volumetric efficiency η_vol captures internal leakage (gap flow), which grows with pressure and falls with speed. Rearranged, the speed required for a given flow is:
Drive torque
The required drive torque follows from the pressure build-up:
The factor V/(2·π) is the displacement per radian, Δp the pressure difference between outlet and inlet, and the mechanical-hydraulic efficiency η_mh captures friction in bearings, seals and gearing. The theoretical torque is:
it is increased by η_mh in the denominator because the drive must additionally overcome friction.
Drive power
The drive power is:
It agrees with the mechanical route:
the hydraulic output P_hyd = Q·Δp divided by the overall efficiency gives the same shaft power. Internally SI units are used: V in m³/rev, Δp in Pa (1 bar = 10⁵ Pa), Q in m³/s - the interface converts cm³/rev, bar, l/min and m³/h automatically.
Worked example
An external gear pump with displacement V = 70 cm³/rev works against a pressure difference Δp = 150 bar at a mechanical-hydraulic efficiency η_mh = 0.8. The required drive torque is M = V·Δp/(2·π·η_mh) = 70·10⁻⁶ m³ · 150·10⁵ Pa / (2·π·0.8) = 208.9 Nm.
If the same size (V = 70 cm³/rev) is to deliver an effective flow of Q = 100 l/min and η_vol = 0.8, the required speed is n = Q/(V·η_vol) = 100,000 cm³/min / (70 cm³ · 0.8) = 1786 rpm.
At Q = 100 l/min (= 0.001667 m³/s), Δp = 150 bar and an overall efficiency η_total = η_vol·η_mh = 0.64, the drive power is P = Q·Δp/η_total = 0.001667 · 150·10⁵ / 0.64 = 39,063 W, about 39 kW. The mechanical check P = M·ω gives the same value.
Frequently asked questions
What is the difference between theoretical and effective flow?
The theoretical flow Q_th = V·n follows purely geometrically from displacement and speed. The effective flow Q = V·n·η_vol is smaller by the internal leakage; the volumetric efficiency η_vol is between 0.90 and 0.98 for good pumps and drops with pressure. For sizing the speed, the effective flow is always decisive.
Why does efficiency appear in the denominator for torque but as a factor for flow?
Because they describe different losses. The volumetric efficiency η_vol reduces the usable flow (leakage), hence Q = V·n·η_vol. The mechanical-hydraulic efficiency η_mh describes friction losses the drive must additionally supply, so the required torque rises: M = V·Δp/(2·π·η_mh).
Does the calculation apply equally to gear, piston and vane pumps?
Yes. All hydrostatic displacement pumps with constant displacement follow the same basic equations Q = V·n·η_vol, M = V·Δp/(2·π·η_mh) and P = Q·Δp/η_total. Only the typical efficiencies and pressure ranges differ. For variable pumps, V is the momentary swash-angle-dependent value.
How are drive power, torque and speed related?
Via P = M·ω with ω = 2·π·n. This is identical to P = Q·Δp/η_total: inserting Q = V·n·η_vol and M = V·Δp/(2·π·η_mh) the efficiencies combine to η_total = η_vol·η_mh. The calculator outputs both routes; they must agree.
Which pressure should be used - operating pressure or pressure difference?
The pressure difference Δp between outlet and inlet is decisive. Since the inlet pressure is near ambient in standard applications, Δp practically equals the operating pressure read at the outlet gauge. With a charged inlet, use the difference.
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