A pump motor is sized from two numbers, the flow and the head, through two efficiencies and one margin. The arithmetic takes a minute. The judgement is in the numbers you feed it: where the duty point really is, what the pump's efficiency really is there, and what happens when the system does not behave as the drawing said. Get those right and the motor is neither the 30 hp that a supplier rounds up to "to be safe" nor the one that trips every summer afternoon.
Flow, head and the duty point
Flow is what the process needs: cubic metres per hour into a tank, litres per second through a chiller, an irrigation demand. Head is the energy per unit weight the pump must add to the water to get it there, expressed as a height in metres of the fluid. It has two parts. The static head is fixed by geometry: the vertical distance from the water level at the source to the discharge level, plus any pressure the outlet must be delivered at, converted to metres (10.2 m of water per bar). The friction head is lost in the pipe, the valves, the bends and the meters, and it rises with roughly the square of the flow; the Darcy–Weisbach equation gives it from the pipe size, length, roughness and velocity, and the Hazen–Williams formula is a quicker empirical route for water.
Plot the total head against flow and you have the system curve: it starts at the static head and bends upward as friction grows. The pump has its own curve, from the manufacturer's test to ISO 9906, and it falls as flow rises. The pump runs where the two curves cross, and only there. That crossing is the duty point, and it is the point the motor must be sized for. A pump does not deliver its "rated" flow because the brochure says so; it delivers whatever the system curve lets it.
Hydraulic power: the physics
Lifting a mass flow of water through a height takes power, and the expression is the same one as for a lift or a conveyor: weight per second times height.
with ρ the density in kg/m³ (1 000 for water), g = 9.81 m/s², Q the flow in m³/s and H the head in metres; the result is in watts. Head, not pressure, is the natural variable because a centrifugal impeller of a given diameter at a given speed imparts a fixed head to whatever liquid passes through it. Pump seawater (density 1 025 kg/m³) instead of fresh and the head is unchanged, the pressure at the outlet is 2.5 % higher, and the shaft power is 2.5 % higher. Pump diesel at 850 kg/m³ and the same pump draws 15 % less. A pump curve drawn in bar would have to be redrawn for every liquid; drawn in metres it is universal, and the density is accounted for in the power.
Pump efficiency: the number the brochure buries
Hydraulic power is what the water receives; shaft power is what the motor delivers, and the ratio is the pump efficiency. It is not one number. Every pump has a best efficiency point (BEP) at one flow, and the efficiency falls away on either side, slowly at first and then steeply. Large, well-matched centrifugal pumps reach 80–90 % at BEP; a 50 m³/h water pump does well to reach 70 %; small pumps under a kilowatt are often below 50 %; and any pump run far from its BEP is worse than all of these. Wear on the impeller and wear rings takes more off with the years.
The published curve has efficiency contours drawn across it. Read the value where your duty point actually sits, not the headline figure in the catalogue, which is the BEP of the largest impeller. Hydraulic Institute guidance (ANSI/HI 9.6.3) describes a preferred operating region of roughly 70 to 120 % of the BEP flow; outside it the penalties are not only efficiency but radial thrust on the bearings, recirculation, vibration and a rising NPSH requirement at the high end. If your duty point falls outside that band, you have the wrong pump, and no motor will fix it.
From shaft power to a motor rating
Divide the hydraulic power by the pump efficiency for the shaft power at the duty point, then add a margin. The margin exists because the shaft power is a single point with tolerances all around it: ISO 9906 allows the tested pump to differ from the published curve, the water may be colder and denser than assumed, the supply voltage in a Pakistani industrial estate sags in the afternoon, and above all the system head may turn out lower than designed. A widely used rule of thumb, and it is only that, is 25 % margin below 7.5 kW, 15 % from 7.5 to 40 kW and 10 % above. Round the result up to the next standard IEC rating: 0.37, 0.55, 0.75, 1.1, 1.5, 2.2, 3, 4, 5.5, 7.5, 11, 15, 18.5, 22, 30, 37, 45, 55, 75, 90 kW and so on. IEC motors carry no service factor; the nameplate kW is the continuous rating at 40 °C and up to 1 000 m (IEC 60034-1), and above either the motor must be derated.
The run-out case deserves its own thought. If the head is lower than designed, because the tank is fuller, the pipe is shorter, or a valve is left open, a radial-flow pump slides down its curve to more flow and, for most such pumps, more power. A motor with a 15 % margin covers a modest excursion; a pump that can run against no head at all, filling an empty reservoir through a wide-open line, may need a motor rated for the power at the end of its curve, which is called a non-overloading selection. Axial-flow pumps behave the other way, drawing most power at low flow, which is why they are started with the discharge valve open.
Check the loading afterwards. A motor at 50–90 % of its rating at the duty point is well chosen. Below 50 % it runs at a poorer power factor and efficiency and cost more than it needed to; above 90 % there is nothing left for the run-out case or a hot day.
Motor efficiency, current and the cable
The motor has its own losses. IEC 60034-30-1 grades them into efficiency classes IE1 to IE4; an IE3 motor of 5–15 kW is about 88–92 % efficient at full load, a little less at three-quarters load, and noticeably less below half. Electrical input is the shaft power divided by that efficiency, and the current follows from the supply: for three-phase, I = P ÷ (√3 × V × PF), with a power factor of 0.8–0.9 at full load and lower at part load. The power and current calculator does the conversion for any supply.
Two currents matter and they are not the same. The current at the duty point is what the meter will show and what the energy bill is built on. The nameplate full-load current is larger, because it belongs to the rated output of the motor you rounded up to, and it is the figure the cable, the contactor and the overload relay are sized on. A direct-on-line start draws roughly six times the full-load current for a second or two, which the cable shrugs off but a generator does not; if the pump will ever run on standby power, the generator sizing guide covers the motor-starting step.
Specific energy: the bill per cubic metre
Divide the electrical input in kW by the flow in m³/h and you have the specific energy in kWh per cubic metre. It is the most useful single number about a pumping system, because it is what the electricity bill is proportional to and it exposes every inefficiency at once: an oversized pump throttled back, a worn impeller, a motor two classes below IE3, a pipe a size too small. Two things drive it. The head: at 65 % pump and 90 % motor efficiency, every metre of unnecessary head costs about 0.0047 kWh on every cubic metre, so a throttling valve holding back 10 m is burning a quarter of the energy on this system. And the efficiencies: the same water through a pump at 75 % instead of 65 % is 13 % less energy, forever.
Over a pump's working life the electricity is usually the largest item in its cost, larger than the pump, the motor and the installation together, which is the point of the Europump and Hydraulic Institute life-cycle cost guide. Ask for kWh/m³ at the real duty point when comparing offers, and be suspicious of a quotation that leads with the motor size.
The example in numbers
A water-supply pump near Lahore: 50 m³/h at 40 m total head, 65 % pump efficiency at the duty point, a 90 % efficient motor, 15 % margin, 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 = 5 450 W = 5.45 kW.
- Shaft power = 5.45 ÷ 0.65 = 8.38 kW. With 15 % margin, 9.64 kW, so an 11 kW (14.8 hp) motor. Loading at the duty point: 76.2 %.
- Electrical input = 8.385 ÷ 0.90 = 9.32 kW; current at the duty point = 9 316 ÷ (√3 × 400 × 0.85) = 15.8 A. The nameplate current of the 11 kW motor is about 20 A, and the cable is sized for that.
- Specific energy = 9.32 ÷ 50 = 0.186 kWh/m³. Twelve hours a day is 600 m³ and about 112 kWh.
The pump power calculator reproduces these figures and lets you change the pump efficiency, the fluid or the supply and watch the motor size and the kWh per cubic metre move.
Variable-speed drives and the affinity laws
Slow a centrifugal pump down and its flow falls in proportion to speed, its head with the square of speed, and its power with the cube. At 80 % speed the pump moves 80 % of the flow, makes 64 % of the head and draws about 51 % of the power. That is why a variable-frequency drive on a pump that is throttled for much of its life pays for itself: the throttling valve was turning the surplus head into heat and noise, and the drive simply does not generate the surplus. The saving is largest where the system is mostly friction and the flow varies; a pump lifting water up a fixed 40 m with little friction gains far less, because below the speed at which the pump can make the static head it delivers nothing at all, and that minimum speed may be 80 % or more.
Three caveats. The motor is still sized for full speed and full duty; the drive does not reduce the rating it needs. A standard motor cooled by its own shaft fan loses cooling at low speed, which is rarely a problem for a pump (the torque falls faster than the cooling does) but worth checking for continuous running below about half speed. And the drive itself loses about 2–3 %, adds harmonics to the supply, and needs its motor cable and earthing done properly. Ahmedonics designs pumping control with and without drives; the right answer depends on the shape of the system curve and the hours at each flow, not on the hardware.
References
- ISO 9906:2012, Rotodynamic pumps — Hydraulic performance acceptance tests — Grades 1, 2 and 3
- ANSI/HI 9.6.3, Rotodynamic Pumps — Guideline for Operating Regions — preferred and allowable operating regions around the best efficiency point
- IEC 60034-1, Rotating electrical machines — Part 1: Rating and performance — reference conditions and derating for altitude and ambient
- IEC 60034-30-1:2014, Rotating electrical machines — Part 30-1: Efficiency classes of line operated AC motors (IE code)
- 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