A positive-displacement pump's flow is set by how much it sweeps each turn and how fast it turns: displacement × RPM. To read that as gallons per minute in US units, you divide by 231 — the number of cubic inches in a gallon — giving GPM = displacement × RPM / 231. Both bigger displacement and higher speed raise the flow directly. The figure is the theoretical maximum; real output sits a few percent below it because some fluid slips back past internal clearances.
Reviewed: June 20, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: the positive-displacement flow relation Q = displacement × speed, 231 in³/gallon.
The pump flow equations
Displacement times speed is the swept volume per minute; dividing by 231 converts cubic inches to US gallons. To find the displacement a pump needs for a target flow, rearrange to disp = GPM × 231 / RPM; to find the drive speed, RPM = GPM × 231 / disp. In SI, use cc/rev and divide by 1000 for L/min. Multiply the theoretical flow by the volumetric efficiency for the real delivered flow.
Worked example — sizing a power pack
Scenario: A pump with a 2 in³/rev displacement is driven at 1800 RPM. What flow does it deliver?
The pump delivers about 15.58 GPM, or 59 L/min, theoretically. Run it at 3600 RPM and the flow doubles to 31.17 GPM; throttle to 1200 RPM and it falls to 10.39 GPM — flow tracks speed directly. At 92% volumetric efficiency the real output at 1800 RPM is closer to 14.3 GPM. Pair this flow with a cylinder's bore area and you get the actuator's extend speed from v = Q/A.
Frequently Asked Questions
GPM = displacement(in³/rev) × RPM / 231. 2 in³ at 1800 RPM → 15.58 GPM.
231 in³ = 1 US gallon. It converts in³/min of swept volume to gallons/min.
Volume pumped per revolution (in³/rev or cc/rev), fixed by the pump's geometry.
× 3.785 (1 US gal = 3.785 L). 15.58 GPM ≈ 59 L/min. The calc shows both.
Actual ÷ theoretical flow, ~90–95%. Multiply the theoretical GPM by it for real output.