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Hydraulics, open channels and fluid power
Hydraulic motor output torque
Shaft torque of a hydraulic motor, from the pressure drop Δp across it, the displacement per revolution Vd and the mechanical efficiency ηm, in steady motoring.
- Ref
- HYD-43
- Rev
- v1
- Field
- Hydraulics, open channels and fluid power
- Result
- T [N·m]
A hydraulic motor with displacement Vd per revolution turns the pressure drop Δp into torque. The ideal torque is Δp Vd/(2π), since one revolution takes energy Δp Vd, and the mechanical efficiency ηm accounts for friction, so T = ηm Δp Vd/(2π). Use the pressure drop between the motor’s inlet and outlet ports, and the output during steady motoring.
Shaft torque of a hydraulic motor, from the pressure drop Δp across it, the displacement per revolution Vd and the mechanical efficiency ηm, in steady motoring.
Calculator
- Ideal torque, Δp Vd/(2π)79.58 N·m
- Shaft power at 1000 rev/min7,500 W
- Flow needed at 1000 rev/min with no leakage8.333×10−4 m³/s
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The motor delivers 71.62 N·m, 90 % of its ideal 79.58 N·m.
- At 1000 rev/min that is 7.5 kW, needing 50 L/min of flow before leakage.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N·m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Vd Displacement per revolution | 1e-6 to 0.01 m³/rev | 1.432 | 14,320 | 14,320 | +10% | +10 % |
| Δp Pressure drop across the motor | 100,000 to 4e7 Pa | 0.7162 | 286.5 | 285.8 | +10% | +10 % |
| ηm Mechanical efficiency | 0.3 to 1 | 23.87 | 79.58 | 55.7 | +10% | +10 % |
Sweep of Pressure drop across the motor
| Pressure drop across the motor (Pa) | Output torque (N·m) |
|---|---|
| 100,000 | 0.7162 |
| 5,088,000 | 36.44 |
| 1.008e7 | 72.16 |
| 1.506e7 | 107.9 |
| 2.005e7 | 143.6 |
| 2.504e7 | 179.3 |
| 3.003e7 | 215 |
| 3.501e7 | 250.8 |
| 4e7 | 286.5 |
Sweep one variable
Variables and how each one contributes
Output torque
What the equation gives you. Everything else feeds it.
Mechanical efficiency
The mechanical efficiency, between 0 and 1. Raises the torque in direct proportion. Raising it by 10 % (to 0.99) moves the result by +10 %, from 71.62 to 78.78 N·m.
Pressure drop across the motor
The pressure drop across the motor, in Pa. Raises the torque in direct proportion. Raising it by 10 % (to 1.1×107) moves the result by +10 %, from 71.62 to 78.78 N·m.
Displacement per revolution
The geometric displacement per revolution, in m³/rev. Raises the torque in direct proportion. Raising it by 10 % (to 5.5×10−5) moves the result by +10 %, from 71.62 to 78.78 N·m.
Worked example with the values above
- Ideal torque: Δp Vd/(2π) = 1×107 × 5×10−5 / (2π) = 79.577 N·m.
- Apply the efficiency 0.9.
- T = 71.62 N·m.
Assumptions and limits
- An incompressible fluid with no leakage, except where an efficiency is given.
- Gauge pressures and steady operation, with friction and inertia not included unless stated.
Common mistakes
- Using the piston area where the annular area is needed, or the reverse.
- Mixing rev/min and rad/s.
- Forgetting back pressure on the other side of the piston.
Statistics and uncertainty
Real actuators lose force to seal friction and flow to leakage, and oil bulk modulus drops sharply with entrained air, so the stiffness and speed here are best-case figures.
Where it is used
- Sizing hydraulic cylinders, pumps and motors.
- Estimating drive speed, stiffness and natural frequency of a hydraulic actuator.
Maths you need
- Pascal’s law: pressure acts equally in all directions in a closed fluid
- Displacement: volume moved per revolution or stroke
- Bulk modulus: B, the pressure per unit volume change of the fluid