What the calculator does
A pump's duty is two numbers: the flow it must deliver and the total head it must deliver it against. Those two, with the density of the fluid, fix the hydraulic power, which is the useful work done on the water. The pump's efficiency at that point turns it into the shaft power the motor must supply, and a margin and the list of standard IEC ratings turn that into a motor you can order. The calculator then goes on to the electrical side: the input power at the duty point, the line current at 400 V three-phase or 230 V single-phase, how loaded the chosen motor will be, and the specific energy in kilowatt-hours per cubic metre, which is the figure that decides what the pump costs to run.
It is the arithmetic for a water-supply scheme, an irrigation tube well, a chiller circulation pump or a process transfer pump, and for checking whether an existing motor is the right size for the duty its pump actually runs at.
Formula
Hydraulic power Ph = ρ × g × Q × H ÷ 1 000 (kW)
Shaft power Ps = Ph ÷ ηpump
Motor rating ≥ Ps × (1 + margin), rounded up to the next IEC rating
Electrical input Pe = Ps ÷ ηmotor
Three-phase I = Pe ÷ (√3 × V × PF) Single-phase I = Pe ÷ (V × PF)
Loading = Ps ÷ rating × 100 % Specific energy = Pe ÷ Q (kWh/m³, with Q in m³/h)
where SG is the specific gravity of the fluid relative to water, ηpump the pump efficiency read from its curve at the duty point, ηmotor the motor efficiency at its load, and PF the motor power factor. Head is in metres of the fluid being pumped, not in bar: a pump adds the same head to any liquid, and the density enters through ρ. One horsepower is taken as 0.7457 kW.
Worked example
A water-supply pump for a housing scheme near Lahore: 50 m³/h against 40 m of total head, pump efficiency 65 % at that point, a 90 % efficient motor with 15 % margin, on 400 V three-phase at 0.85 power factor.
- Q = 50 ÷ 3 600 = 0.013889 m³/s.
- Hydraulic power = 1 000 × 9.81 × 0.013889 × 40 ÷ 1 000 = 5.45 kW.
- Shaft power = 5.45 ÷ 0.65 = 8.38 kW (8.385 before rounding).
- Motor: 8.385 × 1.15 = 9.64 kW, so the next IEC rating is 11 kW (14.8 hp). At the duty point the motor is 8.385 ÷ 11 = 76.2 % loaded, in the healthy range.
- Electrical input = 8.385 ÷ 0.90 = 9.32 kW; line current = 9 316 ÷ (√3 × 400 × 0.85) = 9 316 ÷ 588.9 = 15.8 A.
- Specific energy = 9.32 ÷ 50 = 0.186 kWh per cubic metre.
Twelve hours a day of this duty moves 600 m³ and uses about 112 kWh. A pump chosen closer to its best efficiency point, at 75 % instead of 65 %, would need 7.27 kW at the shaft, still an 11 kW motor, and use about 97 kWh for the same water: roughly 15 kWh a day, every day, for the same motor size and a different pump selection.
Conversions used
| From | To | Factor |
|---|---|---|
| m³/h | m³/s | ÷ 3 600 |
| L/s | m³/s | ÷ 1 000 |
| L/min | m³/s | ÷ 60 000 |
| US gal/min | m³/s | × 0.0000630902 (1 US gallon = 3.785 L) |
| feet | metres | × 0.3048 |
| kW | hp | ÷ 0.7457 |
| metres of water | bar | × 0.0981 (10.2 m of water ≈ 1 bar) |
Standard IEC motor ratings
| kW | hp | kW | hp |
|---|---|---|---|
| 0.37 | 0.5 | 18.5 | 24.8 |
| 0.55 | 0.7 | 22 | 29.5 |
| 0.75 | 1.0 | 30 | 40.2 |
| 1.1 | 1.5 | 37 | 49.6 |
| 1.5 | 2.0 | 45 | 60.3 |
| 2.2 | 3.0 | 55 | 73.8 |
| 3 | 4.0 | 75 | 100.6 |
| 4 | 5.4 | 90 | 120.7 |
| 5.5 | 7.4 | 110 | 147.5 |
| 7.5 | 10.1 | 132 | 177.0 |
| 11 | 14.8 | 160 | 214.6 |
| 15 | 20.1 | 200 | 268.2 |
| 250 | 335.3 | ||
| 315 | 422.4 |
Assumptions and limitations
- Head must be the total dynamic head at the duty flow. That is the static lift from the source level, the delivery height or the pressure required at the outlet, and the friction loss in the pipe, valves and fittings at that flow. The calculator does not compute the friction part; it is a separate step with the Darcy–Weisbach or Hazen–Williams equation, and it rises roughly with the square of the flow.
- Efficiency comes from the curve. The default 65 % is a placeholder for a mid-sized centrifugal pump near its best efficiency point. Read the real figure off the manufacturer's curve at the duty flow and head; a pump run well away from its best point can be 20 points worse.
- No NPSH check. The net positive suction head available at the pump inlet must exceed what the pump requires, with a margin, or it cavitates. That is a separate calculation from the suction side of the installation.
- No starting-current check. A direct-on-line start draws about six times the full-load current for a second or two. That matters for the cable, the protection and above all for a generator; the generator sizing calculator handles the motor-starting step.
- The current shown is at the duty point, not the nameplate. Size the cable, the overload relay and the contactor on the motor's rated full-load current from its nameplate, which for the 11 kW motor in the example is around 20 A, not the 15.8 A it draws at this duty.
- Motor ratings assume IEC reference conditions. The nameplate kW applies up to 1 000 m altitude and 40 °C ambient (IEC 60034-1); above either, the motor must be derated or the next size chosen. Quetta and the northern districts are above 1 000 m; a pump house in Sindh in June is above 40 °C.
- Variable-speed drives change the arithmetic. A pump on a drive follows the affinity laws: flow with speed, head with speed squared, power with speed cubed. The motor is still rated for full speed; check that the pump makes the static head at the minimum speed you intend to use.
Frequently asked questions
Why is the motor bigger than the shaft power?
Because the shaft power is a single point on a curve that has tolerances, and the pump will not stay on that point. The ISO 9906 acceptance grades allow the delivered power to differ from the curve by several percent; the fluid may be denser or colder than assumed; the supply voltage sags; and the system head may end up lower than designed, which pushes the pump to more flow and more power. The margin covers those, and the rounding up to a standard rating usually adds a little more.
What happens if the pump runs at a lower head than designed?
It moves more water and, for a normal radial-flow centrifugal pump, draws more power: the pump "runs out on its curve". A pump designed for 50 m³/h at 40 m may deliver 65 m³/h at 32 m with the shaft power up by a tenth or more, depending on where the efficiency goes, and a motor with no margin will trip on overload or run hot. If the system head is uncertain, size the motor for the power at the end of the curve (a "non-overloading" selection), or fit a drive or a throttling valve to hold the duty.
Horsepower or kilowatts?
Both describe the shaft output. One mechanical horsepower is 745.7 W, so 10 hp is 7.46 kW and the IEC 7.5 kW rating is what a "10 hp" motor is. Motors sold in Pakistan are often labelled in hp; convert once and work in kW.
Can I use a 230 V single-phase motor above 2.2 kW?
In practice, no. Single-phase induction motors are commonly made up to about 2.2–3 kW; beyond that the starting current on one phase, the capacitor and the cost make three-phase the only sensible choice. If only single-phase supply is available, a small drive with single-phase input and three-phase output serves motors up to about 2.2 kW; above that, get a three-phase connection.
Why does specific energy matter more than the motor size?
The motor is bought once; the energy is bought every hour the pump runs, and over a pump's life the electricity is usually the largest item in its cost, larger than the pump, the motor and the installation together. Two pumps that both need an 11 kW motor can differ by 15 % in kWh per cubic metre if one sits at its best efficiency point and the other does not. Compare pump offers on kWh/m³ at the actual duty, not on price.
References
- ISO 9906:2012, Rotodynamic pumps — Hydraulic performance acceptance tests — Grades 1, 2 and 3 — tolerances on flow, head, power and efficiency
- IEC 60034-1, Rotating electrical machines — Part 1: Rating and performance — reference conditions of 1 000 m and 40 °C
- IEC 60034-30-1:2014, Rotating electrical machines — Part 30-1: Efficiency classes of line operated AC motors (IE code) — IE1 to IE4 efficiency levels
- Europump and Hydraulic Institute, Pump Life Cycle Costs: A Guide to LCC Analysis for Pumping Systems, 2001
- Karassik, Messina, Cooper and Heald (eds.), Pump Handbook, 4th edition, McGraw-Hill, 2008 — centrifugal pump performance, efficiency and selection