Hydraulic Pump Horsepower Calculator
Size pump shaft power and PTO horsepower from flow, pressure, efficiency, and reserve margin.
Presets load flow, pressure, pump efficiency, PTO efficiency, reserve margin, shaft RPM, duty factor, and pump type.
Simple PTO drives and compact utility circuits.
Smoother output for moderate pressure systems.
Best for high-duty or high-pressure applications.
Good for lower-pressure auxiliary hydraulic drives.
| Flow | 1500 psi | 2500 psi | 3500 psi |
|---|---|---|---|
| 5 GPM | 4.38 hp | 7.30 hp | 10.21 hp |
| 10 GPM | 8.75 hp | 14.58 hp | 20.42 hp |
| 15 GPM | 13.13 hp | 21.88 hp | 30.63 hp |
| 20 GPM | 17.51 hp | 29.10 hp | 40.84 hp |
| Duty | Margin | Engine sizing | Use case |
|---|---|---|---|
| Light | 10% | Short burst | Cylinder jogs |
| Normal | 15% | Mixed load | Field work |
| Continuous | 20% | Long runtime | Power units |
| Severe | 25% | Hot reserve | Heavy cycles |
| Class | PSI | Bar | Typical system |
|---|---|---|---|
| Low | 1,500 | 103 | Light utility |
| Mid | 2,500 | 172 | Tractor remotes |
| High | 3,500 | 241 | Log splitters |
| Very high | 5,000 | 345 | Piston circuits |
| RPM | 5 HP torque | 10 HP torque | 20 HP torque |
|---|---|---|---|
| 1800 | 14.6 lb-ft | 29.2 lb-ft | 58.4 lb-ft |
| 2200 | 11.9 lb-ft | 23.9 lb-ft | 47.8 lb-ft |
| 3000 | 8.8 lb-ft | 17.5 lb-ft | 35.0 lb-ft |
| 3600 | 7.3 lb-ft | 14.6 lb-ft | 29.2 lb-ft |
Use the highest real working pressure, not just the relief setting, so the horsepower result reflects the actual hydraulic load.
Apply PTO losses and reserve margin after efficiency losses, then round up to the next standard engine size for safer field use.
To calculate the engine powers that is required for a hydraulic system, you need to understand the relationship between hydraulic horsepower, efficiency of the hydraulic pump and drivetrains, and a safety margin. The horsepower that is required to move the oils in the systems cylinders and motors is referred to as the hydraulic horsepower. You can calculate hydraulic horsepower by taking the flow of the hydraulic system in gallons per minute and the systems pressure in pound per square inch and dividing that value by 1714.
The hydraulic horsepower that this equation calculates, however, isnt the same as the engine horsepower that is required to supply that hydraulic horsepower to the system. The pumps and drivetrains that provides the hydraulic fluid to the system are not 100% efficient in there operations. The efficiency of the hydraulic pump is one of the main factors in the power losses of the system.
How to Calculate Engine Power for a Hydraulic System
For instance, tractor PTO drive systems often utilize gear pumps due to the relatively low cost of gear pumps and their durability under heavy use. However, gear pumps are only approximately 85% efficient. Vane pumps are more efficient at approximately 88% efficiency, but cannot handle high levels of pressure.
Piston pumps is the most efficient at 90-94% efficiency and can handle high levels of pressure (5000 psi or higher). As a result, using piston pumps will result in less heat and fuel burn than either gear or vane pumps. Gerotors are a fourth type of pump that is typically used in applications with low required hydraulic pressures.
Beyond the efficiency of the pump is the efficiency of the drivetrain that delivers power from the PTO (Power Take Off) shaft to the components of the drive system. Drivetrains (like PTO drives and belt drives) typically lose 5-8% of their power due to friction and misalignment between their components. You must account for the efficiency of the pump and the efficiency of the drivetrain simultaneously in creating the power calculations for the drive system.
In addition to accounting for these losses, you must also incorporate a reserve margin for unexpected loads or demands upon the drive system into the calculations. This reserve margin is needed to provide for the demands of the drive system during cold mornings when the hydraulic fluid is thick, and for any unexpected increases in the load upon the drive system. In addition to calculating the horsepower that is required to drive the drive system, it is also necessary to calculate the torque that is required for the drives shaft.
You can calculate the torque of the drive system by taking the horsepower that the drive system is to provide and multiplying by 5252 to obtain the torque in lb-ft. For instance, at 10 horsepower, the torque is 5252 lb-ft. At 2200 RPM, the torque is 24 lb-ft.
Decreasing the RPM to 1800 will increase the torque to 29 lb-ft, but decreasing the RPM to 3600 will decrease the torque to 15 lb-ft. The torque that is calculated must be within the rating of the coupling or keyway that connects the drive system to the tractors transmission. It is common for drive system calculations to be based off the relief valve setting of the hydraulic system rather than the working pressure of the system.
The relief valve determines the maximum pressure of the system, but the working pressure is the actual pressure of the fluid within the system under load. For instance, if the component that is being driven by the hydraulic system stalls, the working pressure will be 2500 psi rather than the relief valve setting at 3500 psi. Using the wrong pressure will result in either insufficiently sized hydraulic system components (which will result in inefficient operation of the drive system when it encounters high resistances), or the components will be oversized for the systems requirements.
As with any system that loses efficiency, heat is created within the drive system. Each percent of efficiency that is lost as the fluid moves through the drive system is released in the form of heat. The more efficiency is lost, the more quickly heat is added to the drive system.
One way to prevent excessive heat within the drive system is to add a 15% reserve to the system for normal duty. Additionally, since the viscosity of the oil increases with decreasing temperatures, another consideration to the engine power requirement is to provide enough power to overcome the drag that will result from the thick oil when the engine is started in the early morning. Thus, the drive system will not bog down the engine when it is first started.
In order to determine the power requirement of the engine for the drive system, calculate the hydraulic horsepower that is required for the drive system. Account for the losses that occur in the hydraulic pump and the drivetrain. Account for any reserve power that may be required.
For example, if 10 GPM of flow at 2500 psi is required for the drive system, 14.6 horsepower is calculated. However, if 10% losses are made in the pump, drivetrain, and reserve, 20 or more horsepower is required of the engine. You should of round the calculated horsepower requirement up to the next available size for the engine.
Following these steps will ensure that the hydraulic system will provide reliable power to its components while the engine operates within its performance range (and does not lug the engine under heavy loads).
