Hydraulic Pump Power Calculator

An introduction to hydraulic power, shaft power and motor input power

A pump does not have one power figure; it has at least three, and confusing them is the most common error in pump and motor sizing. Hydraulic power — also called fluid power, or water power when the liquid is water — is the rate at which the pump adds energy to the liquid. Shaft power, normally called brake power or bhp, is the mechanical power the coupling must deliver to the pump shaft. It is larger than the hydraulic power because the pump recirculates liquid internally past its wear rings, drags on bearings and seals, and churns fluid inside the casing. Motor input power is the electrical power drawn from the supply, larger again because the motor has its own copper, iron, friction and stray losses. This calculator reports all three, states which boundary each one sits at, and never collapses them into a single unnamed "efficiency" figure.

That distinction matters because it fixes the direction of the arithmetic. The US Department of Energy sourcebook written jointly with the Hydraulic Institute defines pump efficiency as fluid power divided by pump shaft power. Efficiency is therefore a ratio whose numerator is hydraulic power and whose denominator is shaft power, which means shaft power is hydraulic power divided by efficiency. Some calculators multiply instead, and a shaft power smaller than the fluid power the pump is already delivering is thermodynamically impossible — it would mean the pump created energy. If a tool ever shows you a shaft power below its own hydraulic power figure, stop using it.

In SI terms, the hydraulic power delivered to a liquid of density that is raised through a total head at volumetric flow is

Formula: P_h = ρ g Q H

The differential pressure a pump generates across its flanges is related to that head by the hydrostatic relation

Formula: Δ p = ρ g H

so the same hydraulic power can be written with no density term in it at all:

Formula: P_h = Q Δ p

Both forms are exact and both are used on this page. Which one you should feed the calculator depends on how your duty point is written down. Rotodynamic pump curves for water service are published in head, because head is very nearly independent of the liquid pumped, so a curve drawn for water still applies to a denser liquid. Hydraulic power-unit and fluid-power circuits are specified in pressure, because the relief-valve setting is the number the designer controls. Enter whichever you actually have, and the calculator converts to the other and shows you both.

The two efficiency steps then complete the chain. With pump efficiency and motor efficiency expressed as fractions,

Formula: P_shaft = P_h / η_p, P_elec = P_shaft / η_m = P_h / (η_p η_m)

The product is the wire-to-water efficiency of the pump set, and it is the number that governs the electricity bill. A 78 per cent pump on a 94 per cent motor is a 73 per cent pump set; more than a quarter of the energy you buy never reaches the liquid.

How to use this pump power calculator with a pump curve or a commissioning test sheet

1. Enter the duty flow and pick its unit. Litres per minute, litres per second, cubic metres per hour and US gallons per minute are all accepted, and the calculator normalises internally to cubic metres per second, so you never have to convert by hand. Use the flow the system actually needs at the operating point you are analysing, not the pump maximum.
2. Enter the duty as head or as differential pressure. Choosing metres or feet tells the calculator you are giving total dynamic head; choosing bar, kilopascals or psi tells it you are giving the pressure rise across the pump. There is no separate mode switch to get wrong — the unit you pick decides the interpretation.
3. Set the fluid density. The default is 998.2 kg/m³, the NIST reference density of liquid water at 20 °C and one atmosphere. Density only affects the answer when you have entered head; if you entered pressure it cancels out, and the calculator says so in the result. For a mineral hydraulic oil, read the density off the product data sheet rather than guessing — most ISO 3448 viscosity-graded mineral oils sit somewhere near 860 to 890 kg/m³ at 15 °C, but that is a range, not a constant.
4. Enter pump efficiency in per cent, taken from the manufacturer curve at your flow, not the peak value at the best efficiency point. A pump running well to the right or left of its best efficiency point can easily be ten to twenty points below its catalogue peak.
5. Enter motor (or drive) efficiency in per cent. Leave it at 100 if you only want shaft power and do not want a motor-input figure; the calculator will then report the two figures as equal and label the motor row accordingly. If the pump is driven through a variable frequency drive, use the combined motor-plus-drive efficiency at the operating speed.
6. Set the annual running hours if you want an energy figure. Continuous duty is 8,760 hours a year; a single-shift process pump is often nearer 2,000.
7. Move the speed slider after calculating to see the affinity-law scenario for a trimmed or variable-speed duty, and read the scenario table underneath for a set of fixed speed steps.

Every field is validated in JavaScript rather than by the browser, so an out-of-range entry produces a written explanation in the result panel instead of a silent refusal to submit. Zero and negative flows, zero and negative heads, non-positive densities, efficiencies above 100 per cent and efficiencies typed as a fraction such as 0.85 are each rejected with a message naming the field and the reason.

The pump power formula and where the constants 600, 1714, 3960, 367 and 102 come from

Trade practice is full of one-line shortcuts with a magic denominator in them. Every one of those denominators is just a unit conversion applied to , and it is worth deriving them rather than trusting them, because each one hides an assumption about the fluid.

The 600 for litres per minute and bar. One litre per minute is 10^-3/60 cubic metres per second and one bar is exactly 10⁵ pascals, so one L/min against one bar is 10⁵ × 10^-3/60 = 1.6667 W, which is one six-hundredth of a kilowatt. Hence

Formula: P_h[kW] = (Q[L /min] × Δ p[bar]) / 600

This constant is exact and carries no fluid assumption, because pressure and flow already contain everything the energy balance needs. Divide by pump efficiency to reach shaft power; this is the form used for hydraulic power units, where the relief setting is known in bar.

The 1714 for US gallons per minute and psi. A US gallon is exactly 231 cubic inches, so one gpm at one psi is 231 inch-pounds-force per minute, which is 19.25 ft·lbf/min. One mechanical horsepower is 33,000 ft·lbf/min, so the ratio is 33,000/19.25 = 1714.29:

Formula: P_h[hp] = (Q[gpm] × Δ p[psi]) / 1714

The 3960 for US gallons per minute and feet of head. This is the classic centrifugal-pump shortcut, quoted in the DOE and Hydraulic Institute sourcebook as fluid power = HQ(s.g.)/3,960 with head in feet, flow in gpm and s.g. the specific gravity of the liquid. Its derivation is a weight flow rate: gallons per minute times the weight of a gallon gives pounds-force per minute, times head in feet gives ft·lbf/min, divided by 33,000 gives horsepower. Working backwards, 33,000/3,960 = 8.3333, so the constant silently assumes a gallon of water weighs exactly 8⅓ pounds. Including pump efficiency, the shortcut for brake horsepower is

Formula: bhp = (Q[gpm] × H[ft] × SG) / (3960 η_p)

How good is 8⅓ lb/gal? The NIST Chemistry WebBook gives liquid water at 60 °F (15.56 °C) and one atmosphere a density of 999.017 kg/m³, which works out at 8.3365 pounds per US gallon. The exact denominator would therefore be 33,000/8.3365 = 3,958.6, so the traditional 3,960 runs about 0.035 per cent optimistic for cold water. That is far inside any realistic uncertainty on the efficiency you typed, which is why the rounded value has survived for a century — but it is an approximation, not an identity, and this calculator does not use it. It works from your density and the standard gravitational acceleration m/s² instead.

The 367 and 102 for metric head units. With water at 1,000 kg/m³, one cubic metre per hour lifted one metre is 1,000 × 9.80665 × (1/3600) × 1 = 2.724 W, giving Q·H/367.1 kilowatts; the same sum with flow in litres per second gives Q·H/101.97 kilowatts. Both constants embed a density of exactly 1,000 kg/m³, which no real water sample has, so for a dense brine or a light oil they are wrong by whatever the density error is.

All five rows describe the same physics. The calculator on this page evaluates in SI and converts on the way in and the way out, so switching units changes the displayed figure only by the rounding of the conversion factors themselves.

A worked example: 250 gpm of cooling water against 120 ft of total dynamic head

A plant cooling-water pump has to deliver 250 US gpm against 120 ft of total dynamic head. The water is at about 60 °F, so its density is the NIST value 999.017 kg/m³. The vendor curve shows 72 per cent pump efficiency at that flow, and the pump is close-coupled to a premium-efficiency motor rated at 94 per cent. The pump runs 4,000 hours a year.

Start by putting the duty into SI. A US gallon is exactly 3.785411784 litres, so 250 gpm is 0.0157726 m³/s, and 120 ft is 36.576 m. The differential pressure the pump must generate follows from the hydrostatic relation:

Formula: Δ p = 999.017 × 9.80665 × 36.576 = 358335 Pa = 3.583 bar

Hydraulic power is then simply the product of flow and pressure rise:

Formula: P_h = 0.0157726 × 358335 = 5652 W = 5.65 kW

which is 7.58 hp. The 3,960 shortcut gives 250 × 120 × 1.00 / 3,960 = 7.576 hp, agreeing to within 0.05 per cent, exactly the discrepancy predicted above from the 8⅓ lb/gal assumption. Dividing by the two efficiencies in turn:

Formula: P_shaft = 5.652 / 0.72 = 7.85 kW = 10.53 hp, P_elec = 7.85 / 0.94 = 8.35 kW

The practical conclusions follow immediately. A 10 hp motor would be undersized: the shaft demand is 10.53 hp, so the next standard frame, 15 hp (11 kW), is the correct choice, and it will run at roughly 70 per cent of nameplate where induction-motor efficiency is still close to peak. The wire-to-water efficiency is 0.72 × 0.94 = 67.7 per cent. Over 4,000 hours the pump set consumes about 33,400 kWh, and the specific energy is 8.35 kW ÷ 56.8 m³/h = 0.147 kWh per cubic metre pumped — a figure worth recording, because it is the single most useful benchmark when the same duty is re-tendered or when a second pump is compared against this one.

The affinity laws and what a speed change really does to shaft power

If the pump is fitted with a variable frequency drive, or if you are comparing a 1,450 rpm against a 2,900 rpm selection, the affinity laws relate the two operating points. At constant impeller diameter and with efficiency assumed unchanged, flow follows speed, head follows the square of speed, and shaft power follows the cube:

Formula: Q_2 / Q_1 = N_2 / N_1, H_2 / H_1 = (N_2/N_1)^2, P_2 / P_1 = (N_2/N_1)^3

The DOE and Hydraulic Institute sourcebook publishes the same three relationships in their impeller-diameter form, where replaces , and notes that a 2 per cent diameter reduction gives about 2 per cent less flow, 4 per cent less head and 8 per cent less power. It also warns that these relationships are "not strictly accurate because of nonlinearities in flow" and should be treated as an approximation for small changes.

The cube law is the reason speed control beats throttling so decisively, but it comes with a condition that is easy to miss: it describes the pump, and it only translates into a cubic energy saving when the system curve is pure friction and passes through the origin. In a system with real static lift — a borehole, a tank farm, a high-rise riser — the pump must still produce the static head at any speed, so slowing it moves the duty point along a much flatter path, the saving is far less than cubic, and below some speed the pump simply stops delivering. The DOE sourcebook is explicit that variable speed drives "are not practical for all applications", naming high-static-head systems and prolonged low-flow operation. Treat the speed scenario table on this page as the best case.

Total dynamic head, static head, and NPSH available against NPSH required

The head you feed this calculator must be total dynamic head, the full energy per unit weight the pump has to add between its suction and discharge flanges. Static head is only part of it — the geometric lift plus any difference in the pressures acting on the two liquid surfaces. Total dynamic head adds the friction and fitting losses through the whole circuit at the duty flow, plus the small velocity-head difference between the suction and discharge nozzles. Because dynamic losses scale roughly with the square of velocity, the DOE sourcebook notes that doubling the flow rate quadruples them; sizing a pump on static head alone in a long or undersized pipe run therefore understates the power by a wide margin.

Suction-side conditions do not appear in the power equation at all, but they decide whether the power equation means anything, because a cavitating pump does not follow its published curve. Net positive suction head available is the absolute head at the pump inlet above the vapour pressure of the liquid:

Formula: NPSH A = (p_abs - p_vap) / (ρ g) + z_s - h_f

where is the absolute pressure on the suction liquid surface, the vapour pressure at the pumping temperature, the static suction lift or flooded head, and the suction-line friction loss. NPSH required is a property of the pump published on its curve; as the DOE and Hydraulic Institute sourcebook records, the Hydraulic Institute defines it as the suction condition at which total head has already fallen by 3 per cent because of cavitation, which is why it is also written NPSH3. A common system-design rule quoted in the same document is to keep NPSH available at least 25 per cent above NPSH required across the whole expected flow range, and it warns that oversized pumps operating far out to the right of their design point can erode that margin dangerously. Cavitation collapses head, wrecks efficiency and pits the impeller, so a power calculation for a cavitating duty point is fiction.

Reading the result, and how to turn it into a motor selection

The result panel gives four headline figures and a breakdown. Read them in this order. The hydraulic power is the physics floor: no pump can be built that draws less than this for your duty. The shaft power is the number to compare with a motor nameplate, because a nameplate states the mechanical output the motor can deliver continuously. The motor input power is the number to multiply by running hours and your tariff. The wire-to-water efficiency is the audit figure; if it falls below about 55 per cent for clean water at a reasonable size, something is wrong with the selection rather than with the pump.

For motor selection, do not simply round the shaft power up. Add margin for the fact that the pump may run out along its curve when a control valve opens, that its curve may be at the top of the acceptance tolerance band, and that the fluid may be colder and denser than design. Non-overloading selection means the motor covers the peak of the pump power curve across the whole operating range, not just the duty point. The calculator shows a 10 per cent and a 25 per cent margined shaft power to make that judgement easier, but a pump whose power curve rises steeply toward run-out may need more than either.

The specific energy figure, in kWh per cubic metre, is the most transferable number on the page. It normalises out the size of the machine and lets you rank a proposed replacement, a trimmed impeller, or a variable-speed retrofit against the incumbent on a single axis.

Limitations and assumptions built into this pump power estimate

• Single steady duty point. The calculation describes one flow and one head. Real systems move along a curve, and a pump that spends half its life at run-out has a quite different average power to one that sits at its design point.
• Efficiency is an input, not a prediction. Nothing here estimates how efficient your pump is. Efficiency varies strongly with flow, with size, with specific speed and with wear. The DOE sourcebook puts the practical range at 35 per cent to more than 90 per cent — a spread far too wide for any default to be meaningful, which is why this page does not claim typical efficiencies by pump type.
• Newtonian, gas-free, non-viscous-corrected. The equations assume a single-phase Newtonian liquid. For viscous liquids the pump's head, flow and efficiency all have to be corrected before the power calculation, using a published viscosity-correction method; entrained air, slurries and non-Newtonian fluids need dedicated treatment.
• Motor efficiency is treated as constant. Real induction-motor efficiency falls away below roughly half load, and a variable frequency drive adds its own 2 to 4 per cent loss plus harmonic heating. Fold both into the motor efficiency figure you enter if you are working at part load.
• Affinity scaling assumes constant efficiency and no static head. The speed scenarios are an upper bound on the saving, as described above.
• Acceptance tolerances are not modelled. Measured pump performance is only guaranteed within the tolerance band of the acceptance grade the pump was bought to, under ISO 9906 or ANSI/HI 14.6. A pump can be fully compliant and still absorb noticeably more power than its published curve suggests.
• Nothing here is a safety calculation. The result does not verify that hoses, pipework, seals or relief settings can contain the pressure, that the motor has starting torque for the load, that minimum continuous stable flow is respected, or that the fluid will stay below its temperature limit. Pressure containment, over-pressure protection and thermal management are separate design and code checks.

Frequently asked questions about pump, shaft and motor power

Sources for the equations, constants and fluid properties

Primary and institutional sources relied on for the formulas, constants and fluid properties used on this page:

• U.S. Department of Energy, Industrial Technologies Program, and the Hydraulic Institute, Improving Pumping System Performance: A Sourcebook for Industry, Second Edition, May 2006 (prepared by Lawrence Berkeley National Laboratory, Resource Dynamics Corporation and the Alliance to Save Energy). Section 1 gives fluid power = HQ(s.g.)/3,960 and defines pump efficiency as fluid power divided by pump shaft power; the impeller-trimming fact sheet gives the affinity laws in diameter form; the centrifugal-pump fact sheet gives the NPSH definitions, the Hydraulic Institute 3 per cent head-drop basis for NPSHR, and the 25 per cent NPSH margin rule.
• Hydraulic Institute, ANSI/HI 14.6, Rotodynamic Pumps for Hydraulic Performance Acceptance Tests — the US acceptance standard defining measured rate of flow, differential head, input power, speed and the acceptance tolerance bands, and the Hydraulic Institute Pump FAQs.
• International Organization for Standardization, ISO 9906:2012, Rotodynamic pumps — Hydraulic performance acceptance tests — Grades 1, 2 and 3, the international counterpart standard, whose grades 1B, 1E, 1U, 2B, 2U and 3B set the tolerance within which a measured duty point is deemed to meet its guarantee.
• National Institute of Standards and Technology, NIST Chemistry WebBook, Thermophysical Properties of Fluid Systems — liquid water at 1 atm: 999.975 kg/m³ at 4 °C, 999.017 kg/m³ at 15.56 °C (60 °F) and 998.207 kg/m³ at 20 °C. These are the densities used to check the 3,960 and 367 shortcut constants.
• International Organization for Standardization, ISO 3448, Industrial liquid lubricants — ISO viscosity classification, which defines each ISO VG number as the mid-point kinematic viscosity in mm²/s at 40 °C with a ±10 per cent band, so ISO VG 46 spans 41.4 to 50.6 mm²/s. Viscosity grade governs the friction and volumetric losses in an oil hydraulic circuit; take the density of your specific oil from its data sheet rather than from a grade number.
• Standard gravitational acceleration g = 9.80665 m/s² as fixed by the 3rd General Conference on Weights and Measures (CGPM, 1901) and listed in the BIPM SI Brochure; mechanical horsepower taken as 550 ft·lbf/s = 745.6998716 W, and the US gallon as exactly 231 in³ = 3.785411784 L.