Hydraulic Horsepower Calculator
Calculate hydraulic horsepower from flow rate (GPM) and pressure (PSI) instantly.
Hydraulic horsepower measures the power a hydraulic system delivers through fluid flow and pressure. It's the key figure for sizing pumps, motors, and power units in fluid-power and industrial systems.
Hydraulic Horsepower Formula
The constant 1714 converts gallons-per-minute and pounds-per-square-inch into horsepower. Real systems aren't perfectly efficient, so the motor driving the pump must supply more than the theoretical hydraulic horsepower — typically 80–90% efficiency for a good pump.
How to Use This Calculator
- Enter flow rate in gallons per minute (GPM).
- Enter pressure in PSI.
- Set efficiency (default 85%) to see the input horsepower the motor needs.
Worked Example
To convert the result to kilowatts use the HP to kW calculator, or for pump sizing from head instead of pressure see the pump horsepower calculator.
Flow and Pressure: The Two Halves of Hydraulic Power
Hydraulic power is the product of flow (how fast fluid moves, in GPM) and pressure (the force behind it, in PSI). High pressure with low flow suits slow, forceful tasks like a press; high flow with lower pressure suits fast actuators. The 1714 constant converts these into horsepower. Because both terms multiply, doubling either one doubles the power — which is why system designers balance pump displacement against relief-valve pressure.
Efficiency and Input Power
The formula gives ideal hydraulic horsepower delivered to the fluid. Real pumps lose energy to internal leakage and friction, so the electric motor or engine driving the pump must supply more — divide by pump efficiency (typically 0.80–0.90) to get the true input horsepower. Always size the prime mover on that larger input figure, with margin for pressure spikes.
Hydraulic HP by Flow and Pressure
| Flow (GPM) | Pressure (PSI) | Hydraulic HP |
|---|---|---|
| 5 | 1,500 | 4.4 |
| 10 | 2,000 | 11.7 |
| 20 | 3,000 | 35.0 |
| 40 | 2,500 | 58.3 |
Where the 1714 Constant Comes From
The 1714 isn't an empirical fudge factor. It is four unit conversions collapsed into a single number, and following the chain makes the formula far easier to trust and to adapt.
Hydraulic power is pressure multiplied by flow. Working in the units on the gauge: a US gallon is 231 cubic inches, so GPM × 231 gives cubic inches per minute. Multiply by PSI and you get inch-pounds of work per minute. Divide by 12 to reach foot-pounds per minute, then divide by 33,000 — the definition of one horsepower — to finish:
Knowing the derivation tells you immediately when the constant doesn't apply. Imperial gallons are larger than US gallons, so a UK flow figure needs a different number entirely. The metric version is cleaner: kilowatts equal liters per minute multiplied by bar, divided by 600. If your gauges read in bar and L/min, use that directly rather than converting to imperial and back — every conversion is a chance to introduce an error. Convert the result with the HP to kW calculator if you need it the other way.
Choosing a Realistic Efficiency Figure
The calculator asks for an efficiency percentage, and the default will rarely match your actual system. Overall efficiency combines volumetric losses — fluid slipping back past clearances instead of leaving the port — with mechanical losses from friction in bearings and seals. Pump type is the largest single determinant:
| Pump type | Typical overall efficiency | Usual pressure ceiling | Character |
|---|---|---|---|
| External gear | 80–87% | 3,000 PSI | Inexpensive and tolerant of contamination |
| Internal gear | 85–90% | 3,000 PSI | Quieter, smoother flow delivery |
| Balanced vane | 82–88% | 2,500–3,000 PSI | Long service life at fixed displacement |
| Axial piston | 88–93% | 5,000–6,000 PSI | Variable displacement, high efficiency |
| Radial piston | 90–95% | 10,000 PSI and above | Very high pressure, specialised applications |
Two cautions apply to every figure in that column. Efficiency falls as a pump wears, so a unit with years of service will sit below its catalogue rating — often by several points. And efficiency is quoted at the pump's design point; running well below rated pressure or speed drops it further. If you're sizing a motor from this calculation, we'd use the low end of the band rather than the optimistic end.
Every Lost Horsepower Becomes Heat in the Oil
This is the part of hydraulic sizing that gets missed most often, and it is the reason systems fail in service rather than on the drawing board. Energy that doesn't leave as useful work doesn't disappear — it goes into the fluid as heat.
Take the calculator's own case. Twenty GPM at 3,000 PSI is 35.0 hydraulic horsepower. At 85% efficiency the input requirement is 41.2 HP, so 6.2 HP is lost. Converted at 2,545 BTU per hour per horsepower, that's roughly 15,800 BTU/hr going straight into the oil, continuously, for as long as the system runs.
The heat load also drives reservoir sizing. A common starting rule is a reservoir of two to three times the pump's flow in GPM. A 20 GPM pump wants 40 to 60 gallons of oil. That gives the fluid time to shed heat before it recirculates. Where duty cycle is high or ambient temperature is elevated, a cooler becomes necessary regardless of tank size. Compare the numbers against the HP to BTU calculator to size that cooler.
Pressure Drop Is Not Useful Work
The formula uses system pressure, and it is easy to feed it the wrong pressure figure. What produces useful work is the pressure difference across the actuator — the cylinder or motor doing the job. Pressure consumed elsewhere still costs input power, but it accomplishes nothing except making heat.
| Where pressure is lost | Typical drop | Effect |
|---|---|---|
| Undersized hose or tubing | 50–300 PSI | Pure heat, rises steeply with flow rate |
| Directional control valve | 50–150 PSI | Unavoidable, but sizing the valve to flow helps |
| Return line filter | 10–50 PSI | Grows as the element loads with contamination |
| Fittings and elbows | 5–25 PSI each | Adds up quickly on a complex circuit |
| Relief valve dumping excess flow | Full system pressure | The single largest heat source in most systems |
That last row is worth dwelling on. A fixed displacement pump delivers its full flow whenever it turns, and any flow the actuator doesn't need goes over the relief valve at full system pressure, converting entirely to heat. A pump moving 20 GPM at 3,000 PSI while the actuator only needs 5 GPM is dumping 15 GPM across the relief — roughly 26 hydraulic horsepower turned into pure heat. This is the reason variable displacement pumps and load-sensing circuits exist, and why they pay for themselves on any system that spends time at partial flow.
Frequently Asked Questions
Hydraulic horsepower = (Flow in GPM × Pressure in PSI) ÷ 1714. This gives the theoretical power delivered to the fluid.
It's the unit-conversion factor that turns gallons per minute and pounds per square inch into horsepower. It comes from combining the definitions of those units.
Hydraulic HP is the useful power in the fluid; input HP is what the motor must supply, found by dividing hydraulic HP by the pump's efficiency.
Most hydraulic pumps run 80–90% efficient. We'd use 85% as a reasonable default if you don't have the manufacturer's figure.
Multiply hydraulic horsepower by 0.7457 to get kilowatts, or use our HP to kW calculator for an instant conversion. If your gauges already read in bar and liters per minute, skip imperial entirely: kilowatts equal L/min multiplied by bar, divided by 600.
Every horsepower that doesn't leave as useful work ends up as heat in the fluid, at roughly 2,545 BTU per hour per horsepower lost. A 20 GPM system at 3,000 PSI running 85% efficient loses about 6.2 HP, which is around 15,800 BTU/hr going continuously into the oil. The largest single source in most systems is a relief valve dumping unneeded flow at full pressure.
A common starting rule is two to three times the pump's flow rate in gallons, so a 20 GPM pump suits a 40 to 60 gallon reservoir. That gives the fluid time to shed heat before recirculating. High duty cycles or hot ambient conditions need a cooler regardless of tank size, since most systems aim to keep reservoir temperature below about 140 °F.
For useful work output, use the pressure difference across the actuator. Pressure lost in hoses, valves, filters and fittings still costs input power but accomplishes nothing except making heat. Using full system pressure will overstate how much work the cylinder or motor is actually doing.