PSI to HP Calculator

Convert hydraulic pressure (PSI) and flow (GPM) into horsepower — and see why PSI alone can't become HP.

PSI to HP Calculator
RESULT

PSI is pressure, not power — a sealed cylinder can hold 3,000 PSI forever while doing zero work. Power appears only when pressurized fluid flows. That's why converting PSI to horsepower always needs the flow rate in gallons per minute.

Quick answer: HP = (PSI × GPM) ÷ 1,714. At 2,500 PSI and 10 GPM, that's 14.6 hydraulic horsepower.

PSI to HP Formula

Hydraulic horsepower
HP = (PSI × GPM) ÷ 1714
1,714 converts gallon·PSI per minute into 33,000 ft-lb/min. Divide by pump efficiency (~0.85) for the input power a motor must supply.

Why You Can't Convert PSI Alone

Pressure is force per area; power is work per time. The same 5 HP can appear as 3,430 PSI at 2.5 GPM (a log splitter) or 429 PSI at 20 GPM (a high-flow circuit). If someone quotes "PSI to HP" without flow, they're implicitly assuming a flow rate — always find out which one. The reverse question ("1 HP = how many PSI?") has no fixed answer for the same reason.

Hydraulic HP at Common Pressure/Flow Combinations

PSI \ GPM5 GPM10 GPM15 GPM20 GPM
1,0002.95.88.811.7
1,5004.48.813.117.5
2,0005.811.717.523.3
2,5007.314.621.929.2
3,0008.817.526.335.0

Worked Example

Log splitter pump sizing
1. Splitter runs 2,750 PSI at 11 GPM
2. Hydraulic HP = (2750 × 11) ÷ 1714 = 17.6 HP
3. At 85% pump efficiency: 17.6 ÷ 0.85 ≈ 21 HP engine required

Pressure Buys Force, Flow Buys Speed

Understanding which knob does what is the difference between specifying a system that works and one that disappoints.

Increase thisYou getYou do not get
Pressure (PSI)More force at the cylinder — a heavier load lifted, a tougher log splitNo extra speed. The ram moves at the same rate.
Flow (GPM)Faster cylinder movement, shorter cycle timeNo extra force. It will stall on the same load.

Both cost horsepower, and they cost it equally — the formula multiplies them. Doubling either doubles the power the engine must supply. This is why a machine that "needs more grunt" and one that "needs to cycle faster" can require the same size engine upgrade for entirely different reasons.

Turning PSI Into Cylinder Force

Pressure alone does tell you something useful — just not power. It tells you force, once you know the piston area:

Cylinder force
Force (lb) = PSI × piston area (in²)
Piston area = π × (bore ÷ 2)². A 4-inch bore has an area of 12.57 in².

So a 4-inch cylinder at 2,500 PSI pushes 31,400 lb — about 15.7 short tons. That is the number that decides whether the machine can do the job. The horsepower figure decides how fast it does it, and how big an engine has to drive the pump.

BoreArea (in²)Force @ 1,500 PSIForce @ 2,500 PSIForce @ 3,000 PSI
2"3.144,712 lb7,854 lb9,425 lb
3"7.0710,603 lb17,671 lb21,206 lb
4"12.5718,850 lb31,416 lb37,699 lb
5"19.6329,452 lb49,087 lb58,905 lb
6"28.2742,412 lb70,686 lb84,823 lb

Note this is the extend stroke. On retract, the piston rod occupies part of the area, so force is lower for the same pressure — typically 20–30% less depending on rod diameter.

Sizing the Engine or Motor Properly

Hydraulic horsepower is what the fluid delivers. The prime mover has to supply more, because losses stack up on the way:

  1. Hydraulic HP — the useful output, from PSI × GPM ÷ 1,714.
  2. ÷ pump efficiency — typically 0.85 for a gear pump, 0.90–0.92 for a good piston pump. This gives the shaft power the pump demands.
  3. Add margin — an engine running continuously at 100% of rating will not last. Sizing at roughly 80% of the engine's rated output is normal practice.
Relief valve setting is not system pressure. A circuit set to relieve at 3,000 PSI only reaches 3,000 PSI when the actuator stalls against its load. Sizing the engine for full relief pressure at full flow specifies for a condition the machine rarely sees — but it is the safe way to do it, because that condition is exactly when the engine would otherwise stall.

The Metric Version

Outside North America, hydraulics work in bar and liters per minute. The relationship is the same, with a different constant:

Metric fluid power
kW = (bar × L/min) ÷ 600
600 = 60,000 ÷ 100. Force in kN = bar × area in cm² ÷ 10.
ImperialMetricConversion
PSIbar1 bar = 14.504 PSI
GPM (US)L/min1 US GPM = 3.785 L/min
HPkW1 HP = 0.7457 kW

Worth watching: a UK or imperial gallon is 4.546 liters, not 3.785. Mixing gallon definitions introduces a 20% error, which on a pump specification is the difference between adequate and stalling.

This Does Not Apply to Compressed Air

Pneumatic systems look similar — pressure and flow — but air is compressible, so the simple product of pressure and flow overstates the useful work available. Air power is normally specified in SCFM at a stated pressure, and compressor sizing follows the compression work required rather than PSI × flow ÷ 1,714. Do not use this page to size an air compressor.

How this calculator is checked

Uses the standard fluid-power relation HP = PSI × GPM ÷ 1,714, where 1,714 = 33,000 ÷ (231 in³/gal ÷ 12). Verified against fluid-power handbook examples.

Frequently Asked Questions

No — pressure alone isn't power. You need the flow rate too: HP = PSI × GPM ÷ 1,714.

There's no fixed answer — it depends on flow. 1 HP supports 1,714 PSI at 1 GPM, 343 PSI at 5 GPM, or 171 PSI at 10 GPM.

No — it gives fluid (output) power. Divide by pump efficiency (typically 0.80–0.90) to size the motor or engine driving the pump.

One gallon is 231 in³ (19.25 board-feet of work per PSI per minute). Converting gallon-PSI/min into 33,000 ft-lb/min (1 HP) yields the constant 1,714.

31,400 lb — about 15.7 short tons. Force = PSI × piston area, and a 4-inch bore has an area of 12.57 in². Retract force is 20–30% lower because the rod occupies part of the area.

Pressure buys force; flow buys speed. Raising either costs the same horsepower, because the formula multiplies them. Pick based on whether the machine stalls on load or simply cycles too slowly.

kW = (bar × L/min) ÷ 600. Watch the gallon definition when converting — a UK gallon is 4.546 L against the US 3.785 L, a 20% difference.

No. Air is compressible, so PSI × flow ÷ 1,714 overstates the useful work. Pneumatic systems are specified in SCFM at a stated pressure, and compressor sizing follows compression work instead.