Calculate hydraulic horsepower from flow (GPM or L/min) and pressure (PSI or bar), plus pump shaft power in hp and kW at a typical 85% efficiency.

Shaft power assumes a typical 85% pump efficiency, which you can change under Optional settings. Electric-motor losses are excluded.

Optional settings
85% is a typical value; enter your pumpโ€™s rating if you know it.

Blank means no added margin.


Related Calculators

Calculator Functions

FunctionWhat it doesWhat it requires
Hydraulic powerFinds the hydraulic horsepower a pump delivers to the fluid, and the shaft power needed to drive itRequires the delivered flow rate and the pump pressure rise
Available flowFinds the flow a pump can deliver from a known shaft power at a required pressureRequires the available pump shaft power and the pump pressure rise
Available pressureFinds the pressure a pump can reach from a known shaft power at a required flowRequires the available pump shaft power and the delivered flow rate

Flow goes in as US gpm, L/min or mยณ/s, pressure as psi, bar, kPa or MPa, and shaft power as hp or kW. Optional settings holds the overall pump efficiency, which is 85% unless changed, and Hydraulic power also takes a sizing margin. Every result lists hydraulic power in kW, pump shaft power and the estimated pump loss as heat.

Hydraulic Horsepower Formulas

HHP = P ร— Q รท 1714 [1]
HHPHydraulic horsepower delivered to the fluid (hp)
PPressure rise across the pump (psi)
QDelivered flow rate (US gpm)

The 1714 is a unit conversion, not a property of the pump. One US gallon is 231 cubic inches and one horsepower is 33,000 ftยทlbf per minute [2], which makes the exact divisor 12,000 รท 7, or 1,714.2857. The calculator uses the exact value. In metric units the same relationship is kW = bar ร— L/min รท 600.

Hydraulic power

A pump delivering 10 US gpm at a 2,000 psi rise puts 2,000 ร— 10 รท 1,714.2857 = 11.67 hp into the fluid, and 32 US gpm at 71 psi is 1.33 hp. The shaft has to supply more than that, because the pump is not perfectly efficient:

Shaft hp = HHP รท ฮท [1]

ฮท is the overall pump efficiency as a decimal. At the 85% default, 11.67 hp of hydraulic power needs 13.73 hp at the shaft, and the 2.06 hp difference is the estimated pump loss. A sizing margin multiplies the shaft power, so 20% raises 13.73 hp to 16.47 hp, and the result names the next size up in a list of common motor ratings.

Available flow

Q = 1714 ร— hp ร— ฮท รท P

This is the main formula solved for flow, with the shaft power first reduced by the pump efficiency. A 10 hp shaft at 85% and 2,000 psi supports 1,714.2857 ร— 10 ร— 0.85 รท 2,000 = 7.29 US gpm.

Available pressure

P = 1714 ร— hp ร— ฮท รท Q

The same rearrangement gives pressure. A 10 hp shaft at 85% moving 5 US gpm can reach 1,714.2857 ร— 10 ร— 0.85 รท 5 = 2,914 psi.

Limits of This Estimate

  • The 85% efficiency is a typical figure, not a rating. Real pumps vary with type, speed, pressure, fluid viscosity and wear, so use the makerโ€™s figure when it is known.
  • Shaft power is what the pump needs from its driver. Losses inside an electric motor or engine are not included, and neither are losses in valves, hoses and actuators.
  • Pressure means the rise across the pump and flow means the delivered flow. Both must stay inside the pumpโ€™s ratings.
  • The motor size shown is the next value in an example list. It is not a motor selection, which also depends on starting torque, duty and service conditions.

Sources

  1. Muncie Power Products, Useful Formulas: pump input horsepower from gpm, psi, 1714 and pump efficiency
  2. NIST, Guide for the Use of the International System of Units (SP 811), Appendix B.8: the horsepower, US gallon, gallon per minute and psi factors behind the 1714 divisor