What is the formula for hydraulic pump electric motor power?

In metric units: P_motor [kW] = (Pressure [bar] * Flow [L/min]) / (600 * eta_total), where eta_total is overall pump efficiency (typically 0.82 to 0.92). In imperial units: P_motor [HP] = (Pressure [psi] * Flow [GPM]) / (1714 * eta_total).

How is hydraulic pump displacement (cc/rev) calculated from flow rate and RPM?

Pump geometric displacement per revolution (V_g) is: V_g [cm^3/rev] = (Flow [L/min] * 1000) / (N [RPM] * eta_volumetric). For example, a pump delivering 60 L/min at 1,500 RPM with 95% volumetric efficiency requires a displacement of 42.1 cc/rev.

What is the difference between volumetric efficiency and hydromechanical efficiency?

Volumetric efficiency (eta_v) accounts for internal fluid slippage/leakage past rotating gear teeth, vanes, or piston clearances. Hydromechanical efficiency (eta_hm) accounts for mechanical friction in bearings, shaft seals, and fluid viscous shear drag. Overall efficiency is the product: eta_total = eta_v * eta_hm.

How is input drive shaft torque calculated for a hydraulic pump?

Theoretical pump drive shaft torque is: T_theo [N*m] = (V_g [cm^3/rev] * Pressure [bar]) / (20 * pi). The actual required drive torque accounting for friction is: T_actual = T_theo / eta_hm.

Hydraulic Pump Power Calculator

Applied Fluid Power & Hydraulic Prime Mover Sizing

A hydraulic pump is the heart of every fluid power system, converting mechanical rotary power from an electric motor or internal combustion engine into high-pressure fluid energy. Sizing the drive motor is one of the most critical engineering steps in power unit (HPU) design. Over-sizing the motor multiplies capital cost and causes poor electrical power factor ($\cos\varphi < 0.70$), while undersizing causes the motor to stall, trip thermal overloads, or burn windings when the hydraulic system reaches full relief valve pressure.

Rigorous sizing per ISO 4409 and NFPA Fluid Power Standards accounts for the thermodynamic transition from mechanical shaft torque to fluid hydrostatic work, including both volumetric slip leakage and mechanical bearing friction losses.

Core Mathematical Equations for Hydraulic Power

The useful fluid output power ($P_{\text{hyd}}$) delivered into the hydraulic circuit is the product of operating pressure and volumetric flow rate:

Metric System Formulations (bar and L/min)

Because $1\text{ bar} = 10^5\text{ N/m}^2$ and $1\text{ L/min} = 10^{-3} / 60\text{ m}^3/\text{s}$:

$$P_{\text{hyd}} = \frac{P \cdot Q}{600} \quad [\text{kW}]$$

Imperial System Formulations (psi and GPM)

Because 1 horsepower equals 33,000 ft-lbf/min and 1 gallon equals 231 cubic inches:

$$P_{\text{hyd}} = \frac{P \cdot Q}{1714} \quad [\text{HP}]$$

Required Motor Shaft Drive Power & Efficiency Derating

No hydraulic pump operates without losses. The input mechanical power ($P_{\text{shaft}}$) demanded from the electric motor or diesel engine must supply both fluid work and internal losses:

$$P_{\text{shaft}} = \frac{P_{\text{hyd}}}{\eta_{\text{total}}} = \frac{P \cdot Q}{600 \cdot \eta_{\text{total}}} \quad [\text{kW}] \quad \Longleftrightarrow \quad P_{\text{shaft}} = \frac{P \cdot Q}{1714 \cdot \eta_{\text{total}}} \quad [\text{HP}]$$

Overall pump efficiency ($\eta_{\text{total}}$) is the product of two distinct physical loss mechanisms:

$$\eta_{\text{total}} = \eta_{\text{volumetric}} \cdot \eta_{\text{hydromechanical}}$$

Volumetric Efficiency ($\eta_v$): Quantifies fluid slippage escaping through micro-clearances between rotating pistons, valve plates, or gear teeth from the high-pressure discharge port back to suction or case drain (typically 0.90 to 0.98).
Hydromechanical Efficiency ($\eta_{hm}$): Quantifies mechanical friction inside shaft bearings, swashplate pivots, and oil viscous shear resistance (typically 0.85 to 0.95).

Drive Shaft Input Torque Formulation

The mechanical torque ($T$) demanded by the pump shaft determines coupling selection and shaft shear safety:

$$T = \frac{V_g \cdot P}{20 \cdot \pi \cdot \eta_{hm}} \quad [\text{N}\cdot\text{m}] \quad \Longleftrightarrow \quad T = \frac{V_g \cdot P}{24 \cdot \pi \cdot \eta_{hm}} \quad [\text{lb}\cdot\text{ft}]$$

Where $V_g$ is pump geometric displacement per revolution ($\text{cm}^3/\text{rev}$ or $\text{in}^3/\text{rev}$).

Pump Displacement & Delivery Flow Rate Relationships

Theoretical discharge flow rate ($Q_{\text{theo}}$) is purely geometric, whereas actual delivered flow ($Q_{\text{actual}}$) reflects volumetric slippage:

$$Q_{\text{theo}} = \frac{V_g \cdot N}{1000} \quad [\text{L/min}] \quad \implies \quad Q_{\text{actual}} = Q_{\text{theo}} \cdot \eta_v$$

Hydraulic Pump Architecture	Typical Pressure Rating	Volumetric Efficiency ($\eta_v$)	Hydromechanical ($\eta_{hm}$)	Overall Efficiency ($\eta_t$)
External Gear Pump	150 to 250 bar (2,200 to 3,600 psi)	0.88 to 0.93	0.85 to 0.90	0.75 to 0.84
Internal Gear / Gerotor	100 to 210 bar (1,500 to 3,000 psi)	0.90 to 0.95	0.88 to 0.92	0.80 to 0.87
Vane Pump (Balanced)	140 to 210 bar (2,000 to 3,000 psi)	0.90 to 0.96	0.88 to 0.93	0.80 to 0.88
Axial Piston (Swashplate)	250 to 420 bar (3,600 to 6,000 psi)	0.95 to 0.98	0.92 to 0.96	0.88 to 0.94
Radial Piston Pump	350 to 700 bar (5,000 to 10,000 psi)	0.96 to 0.99	0.93 to 0.97	0.90 to 0.95

Step-by-Step Worked Case Study: CNC Machine Tool HPU Sizing

Design Scenario: An engineer is sizing the electric motor and pump for a modern CNC machine tool clamping and pallet changer hydraulic power unit:

Required flow rate: $Q = 60\text{ L/min}$ ($15.85\text{ GPM}$).
Maximum operating pressure: $P = 210\text{ bar}$ ($3,045\text{ psi}$).
Drive motor speed: 4-pole 60 Hz electric motor running at $N = 1,750\text{ RPM}$.
Pump type: Variable-displacement axial piston pump.
Efficiencies: Volumetric $\eta_v = 0.95$ ($95\%$), Hydromechanical $\eta_{hm} = 0.92$ ($92\%$).

Step 1: Compute overall total pump efficiency ($\eta_{\text{total}}$)

$$\eta_{\text{total}} = \eta_v \cdot \eta_{hm} = 0.95 \times 0.92 = 0.874 \quad (87.4\%)$$

Step 2: Calculate hydraulic fluid output power ($P_{\text{hyd}}$)

$$P_{\text{hyd}} = \frac{P \cdot Q}{600} = \frac{210\text{ bar} \cdot 60\text{ L/min}}{600} = \frac{12,600}{600} = 21.00\text{ kW} \quad (28.16\text{ HP})$$

Step 3: Calculate required electric motor shaft power ($P_{\text{shaft}}$)

$$P_{\text{shaft}} = \frac{P_{\text{hyd}}}{\eta_{\text{total}}} = \frac{21.00\text{ kW}}{0.874} = 24.03\text{ kW} \quad (32.22\text{ HP})$$

The engineer selects the next standard commercial electric motor size: 30 kW (40 HP).

Step 4: Calculate required pump displacement ($V_g$)

$$V_g = \frac{Q \cdot 1000}{N \cdot \eta_v} = \frac{60 \cdot 1000}{1750 \cdot 0.95} = \frac{60000}{1662.5} = 36.09\text{ cc/rev}$$

The engineer selects a standard 45 cc/rev variable displacement pump derated to 36 cc/rev.

Step 5: Compute required drive shaft input torque ($T$)

$$T = \frac{V_g \cdot P}{20 \cdot \pi \cdot \eta_{hm}} = \frac{36.09 \cdot 210}{20 \cdot \pi \cdot 0.92} = \frac{7578.9}{57.805} = 131.1\text{ N}\cdot\text{m} \quad (96.7\text{ lb}\cdot\text{ft})$$

Frequently Asked Questions

Why does electric motor power depend on overall pump efficiency?

If overall efficiency ($\eta_{\text{total}} = 0.80$) is ignored, calculating power purely as $P \cdot Q / 600$ underestimates required shaft power by 25%. A 20 kW hydraulic load with 80% pump efficiency actually demands 25 kW from the electric motor. Sizing the motor without efficiency losses causes motor overload trips, blown fuses, and premature winding insulation failure.

What happens to the energy lost to pump inefficiency?

Every watt of energy lost due to volumetric leakage and hydromechanical friction transforms directly into heat within the hydraulic fluid. In a 30 kW system operating at 85% efficiency, 4.5 kW of continuous thermal energy enters the oil reservoir. If an air/oil heat exchanger is not sized to dissipate this parasitic heat, fluid temperatures will rapidly exceed 80°C, destroying seals and degrading oil viscosity.

How do pressure-compensated variable displacement pumps save power?

A fixed displacement pump constantly pumps its full rated flow ($Q_{\text{max}}$). When actuators are stationary, the entire flow dumps over the relief valve at full pressure, wasting 100% of motor power as boiling heat. In contrast, a pressure-compensated pump automatically destroke its swashplate to near-zero displacement ($Q \to 0$) when pressure reaches setpoint, holding full pressure while drawing only 5% to 10% standby motor power.

How does altitude affect hydraulic pump power and performance?

While hydraulic pressure is generated mechanically, atmospheric pressure pushes oil from the open reservoir into the pump suction port. At high altitudes (> 2,000 meters / 6,500 ft), lower atmospheric pressure reduces net positive suction head (NPSH), causing severe pump cavitation. To prevent pump destruction at high elevations, reservoirs must be pressurized to 0.2 to 0.5 bar, or pump suction lines must be enlarged to lower fluid velocity below 0.8 m/s.

Pump Power, Drive Motor Sizing & Torque Results