Hydraulic Pump Horsepower & Displacement Calculator

Determine pump output flow rate (GPM), displacement per revolution, required electric or gas drive engine horsepower, and heat generation.

Displacement per Rev ≈ 41.0 cc / revolution

Drive Shaft Speed (RPM)

Max System Pressure (PSI)

Pump Mechanical Type

Prime Mover Power Source

Required Drive Motor Power

31.2 HP

Standard 30 or 40 HP NEMA Electric Motor

Actual Delivery Flow

18.2 GPM

68.9 Liters / Minute

Fluid Output Power

26.5 HP

19.8 kW Hydraulic Power

Theoretical Flow (100% Volumetric): 18.94 GPM

Drive Shaft Input Torque: 93.6 ft-lbs (1,123 in-lbs)

Heat Generation in Fluid: 11,960 BTU/hr (4.7 HP loss)

Gasoline Engine Recommendation: ≈ 42 - 45 HP Gas Engine

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How Hydraulic Pump Flow & Drive Horsepower Work

Hydraulic pumps displace fluid mechanically against system resistance:

Flow (GPM) = (Displacement cu in/rev × Shaft RPM) / 231
Actual Flow = Theoretical Flow × Volumetric Efficiency (~0.92)
Drive Horsepower (HP) = (Flow GPM × Pressure PSI) / (1,714 × Overall Efficiency)
Input Torque (ft-lbs) = (Horsepower × 5,252) / Shaft RPM

Why Gas Engines Need More HP Than Electric Motors: Electric motors maintain full breakdown torque across their RPM curve and can temporarily overload by 150% to 200% when a hydraulic cylinder deadheads. Internal combustion gas engines lack overload reserve and stall easily under pressure spikes, requiring a 1.3× to 1.5× horsepower derate multiplier.

Frequently Asked Questions

Where does the constant 1714 in the hydraulic horsepower formula come from?

The constant 1714 is derived mathematically from mechanical work units: 1 Horsepower = 33,000 ft-lbs/min. Converting gallons to cubic inches (231 cu in/gal) and feet to inches yields: 33,000 × 12 / 231 = 1,714.28.

What is the difference between volumetric efficiency and overall efficiency?

Volumetric efficiency accounts for internal fluid slippage bypassing gears or pistons back to the case drain (~90-95%). Mechanical efficiency accounts for friction in bearings and shaft seals (~90%). Overall efficiency is the product of both: $\eta_o = \eta_{vol} \times \eta_{mech}$ (typically 80% to 88% for gear pumps, 90% to 94% for piston pumps).

How do you calculate heat generation in a hydraulic system?

All mechanical and volumetric pump inefficiencies convert directly into thermal heat within the hydraulic oil. 1 Horsepower of wasted energy produces 2,545 BTU/hr of heat that must be dissipated by the oil reservoir or an external air-to-oil heat exchanger.