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 eta_vol and eta_mh the calculator returns every quantity live – enter either speed or flow rate.
Positive Displacement Pump Calculator
Model: hydrostatic positive displacement pump with constant displacement (gear, piston, vane), steady-state operation. Q = V·n·eta_vol, M = V·dp/(2·pi·eta_mh), P = Q·dp/eta_ges with eta_ges = eta_vol·eta_mh. Cavitation (NPSH), pulsation and thermal effects are not considered. Sizing tool for fluid power.
Results
Calculating …
Formulas and fundamentals
Hydrostatic positive displacement pumps are defined by a fixed displacement V per revolution. The delivered flow is proportional to speed: theoretical Q_th = V·n, effective Q = V·n·eta_vol. The volumetric efficiency eta_vol captures internal leakage (gap flow), which grows with pressure and falls with speed. Rearranged, the speed required for a given flow is n = Q/(V·eta_vol).
The required drive torque follows from the pressure build-up: M = V·dp/(2·pi·eta_mh). The factor V/(2·pi) is the displacement per radian, dp the pressure difference between outlet and inlet, and the mechanical-hydraulic efficiency eta_mh captures friction in bearings, seals and gearing. The theoretical torque M_th = V·dp/(2·pi) is increased by eta_mh in the denominator because the drive must additionally overcome friction.
The drive power is P = Q·dp/eta_ges with the overall efficiency eta_ges = eta_vol·eta_mh. It agrees with the mechanical route P = M·omega (omega = 2·pi·n): the hydraulic output P_hyd = Q·dp divided by the overall efficiency gives the same shaft power. Internally SI units are used: V in m³/rev, dp in Pa (1 bar = 10^5 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 dp = 150 bar at a mechanical-hydraulic efficiency eta_mh = 0.8. The required drive torque is M = V·dp/(2·pi·eta_mh) = 70·10^-6 m³ · 150·10^5 Pa / (2·pi·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 eta_vol = 0.8, the required speed is n = Q/(V·eta_vol) = 100,000 cm³/min / (70 cm³ · 0.8) = 1786 rpm.
At Q = 100 l/min (= 0.001667 m³/s), dp = 150 bar and an overall efficiency eta_ges = eta_vol·eta_mh = 0.64, the drive power is P = Q·dp/eta_ges = 0.001667 · 150·10^5 / 0.64 = 39,063 W, about 39 kW. The mechanical check P = M·omega 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·eta_vol is smaller by the internal leakage; the volumetric efficiency eta_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 eta_vol reduces the usable flow (leakage), hence Q = V·n·eta_vol. The mechanical-hydraulic efficiency eta_mh describes friction losses the drive must additionally supply, so the required torque rises: M = V·dp/(2·pi·eta_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·eta_vol, M = V·dp/(2·pi·eta_mh) and P = Q·dp/eta_ges. 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·omega with omega = 2·pi·n. This is identical to P = Q·dp/eta_ges: inserting Q = V·n·eta_vol and M = V·dp/(2·pi·eta_mh) the efficiencies combine to eta_ges = eta_vol·eta_mh. The calculator outputs both routes; they must agree.
Which pressure should be used – operating pressure or pressure difference?
The pressure difference dp between outlet and inlet is decisive. Since the inlet pressure is near ambient in standard applications, dp practically equals the operating pressure read at the outlet gauge. With a charged inlet, use the difference.
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