Selecting a motor · DriveLab

Everything in this chapter now comes together. Selecting a motor means checking one candidate against four constraints: power, torque (running, starting and peak), duty and environment. Then you pick the smallest motor that passes all four.

The procedure

Describe the load. Its torque-speed curve (lesson 1), breakaway torque, inertia and duty cycle.
Carry it to the motor shaft. Speed, torque and inertia through the transmission, with its efficiency.
Choose the type. A cage induction motor unless there is a reason not to: exact speed or plant power-factor correction (synchronous), highest efficiency on a VFD (permanent magnet), an existing DC installation.
Size it thermally. Rated power or torque ≥ the RMS value of the duty cycle, at the right duty type (lesson 4).
Check torque margins. Starting torque > breakaway torque; peak torque < 80 % of breakdown; both with the supply voltage at its lowest.
Check starting. Can the supply take the starting current? If not, choose a starting method (Chapter 2).
Match the environment. IP code for dust and water; derate for ambient temperature above 40 °C or altitude above 1000 m; hazardous-area certification if needed.
Finish the specification. Voltage and connection, frequency, efficiency class, mounting (IM B3 foot, IM B5 flange, IM V1 vertical…) and frame size.

FoundationStart here if this is new to you

It is like buying shoes. First you know what they are for: running, hiking or an office (the load). Then the size must fit, not too small (it hurts, overheats), not two sizes too big (you trip, it is inefficient). Then you check the conditions: rain, mud, snow (the IP rating). A good choice passes every check, and among those that pass, you take the smallest.

Worked example: power sizing (course §1.4.2)

A belt conveyor carries 50 kg/s of gravel up a 10 m incline at 15°, belt speed 1.2 m/s, gearbox efficiency 0.95, friction coefficient 0.05.

Material on the belt: each kilogram stays $10 / 1.2 = 8.33$ s, so $m = 50 \times 8.33 = 416.5$ kg.
Forces: $F_\text{lift} = m g \sin 15° = 1056$ N; $F_\text{friction} = \mu m g \cos 15° = 197$ N; total 1253 N.
Power: $P_\text{belt} = F v = 1253 \times 1.2 = 1.504$ kW; $P_\text{motor} = 1.504 / 0.95 = 1.58$ kW.
Next standard size: 2.2 kW. A 1.5 kW motor would run about 5 % overloaded for ever.

The course's integrated example puts all four constraints together. An outdoor wastewater aeration blower is a variable-torque S1 load needing 35 kW: next size 37 kW; rain and dust outdoors call for at least IP55, so a TEFC motor; the blower's specification fixes a 200L frame. Result: 37 kW, 4 poles, IP55, TEFC, frame 200L.

Reading the nameplate

The nameplate is the motor's contract. Every number on it has a meaning, and several useful numbers follow from it: pole count, slip, rated torque, input power. Click through the fields:

Try it: read a motor nameplate

3~ Mot. · 160L

Rated speed: 1,465 min⁻¹

Speed at full load, a little below synchronous speed. The nearest 120 f / p above it is 1,500 rpm, so the motor has 4 poles and a full-load slip of 2.33 %. Rated torque = P / ω = 97.8 N·m.

Poles: 4
Synchronous speed: 1,500 rpm
Full-load slip: 2.33 %
Rated torque: 97.8 N·m
Input power: 16.39 kW
Current from P, η, cos φ: 28.2 A

Worked example: Tutorial 1.1

A pump motor's nameplate reads 25 HP, 460 V, 32 A, 3 550 rpm, 60 Hz, frame 284T, NEMA nominal efficiency 93.6 %.

Poles. At 60 Hz the candidate synchronous speeds are 3 600 rpm (2 poles) and 1 800 rpm (4 poles). 3 550 rpm sits just below 3 600: a 2-pole motor.
Slip. $s = (3600 - 3550) / 3600 = 1.39$ %.
Torque. $P = 25 \times 746 = 18\,650$ W; $T = 18\,650 / (2\pi \times 3550 / 60) = 50.2$ N·m.
Input power and power factor. $P_\text{elec} = 18\,650 / 0.936 = 19\,925$ W; $\cos\varphi = 19\,925 / (\sqrt{3} \times 460 \times 32) = 0.78$ .
On 50 Hz at constant V/Hz the synchronous speed falls to 3 000 rpm, and the torque capability stays about the same: power and speed both scale by 5/6.
The low slip means low rotor losses ( $P_\text{rotor} = s\,P_\text{ag}$ ): a premium-efficiency design. For a NEMA Design B motor, expect a starting torque of about 150–200 % and a breakdown torque of 200–300 % of this 50 N·m.

Practise it with your own numbers in the exercise Analysis of motor nameplate data.

Enclosure and efficiency

IP code (IEC 60529): two digits, solids then water. IP55 (dust-protected, water jets) is the usual industrial standard. IP65 (dust-tight, water jets) is for mines, quarries and cement works. Food plants washed down at high pressure need IP66 or IP69K. IP23 open motors are only for clean, dry rooms.

Efficiency class (IEC 60034-30-1): IE1 to IE4, IE5 for the best permanent-magnet motors. A motor running 8000 hours a year can spend its purchase price on electricity every few months, so the extra cost of a higher class is usually recovered quickly. The classes are IE1 (standard), IE2 (high), IE3 (premium), IE4 (super premium) and IE5 (ultra premium).

Cooling (IEC 60034-6, IC code): most industrial motors are totally enclosed, fan cooled (TEFC, IC411): a sealed frame with a shaft-mounted fan blowing over external fins. Open drip-proof (ODP) motors are for clean, dry rooms only; small motors may be totally enclosed and non-ventilated (TENV), cooled by natural convection.

Frame size fixes the shaft height, shaft diameter and bolt pattern (IEC 180L, NEMA 324T…), so motors from different makers are interchangeable. Service factor (SF, on NEMA plates) is a thermal margin: SF 1.15 on a 50 HP motor allows 57.5 HP, but it is not meant for continuous overload.

Predict first

A 15.3 kW load is given a 22 kW motor 'to be safe'. What gets worse?

Nothing, a bigger motor just runs coolerPower factor, as well as cost, size and starting currentOnly the price

Your turn: Tutorial 1.2

The booklet's three cases, one catalogue (Apex Industrial Motors, Global Series: 4-pole, 400 V, 50 Hz):

Case 1, process water circulation pump. Centrifugal pump, hydraulic power 13.5 kW at 88 % efficiency, runs 24/7 at constant speed in a clean dry indoor room. The motor needs $13.5 / 0.88 = 15.34$ kW: the next size is 18.5 kW. S1 duty; IP55 suffices indoors; IE3 is enough at fixed speed.
Case 2, intermittent parts hoist. Lifts 800 kg at 0.3 m/s through a 90 % gearbox and pulley; each cycle is 10 s lifting and 50 s at rest. Peak power $800 \times 9.81 \times 0.3 / 0.9 = 2.62$ kW; S3 duty with a cyclic duration factor of 17 %, so the RMS power is only $2.62\sqrt{10/60} = 1.07$ kW. Sizing a lifting motor on RMS alone is unsafe: it must deliver the peak and start the load. The booklet chooses the 11 kW motor with 220 % starting torque.
Case 3, quarry aggregate conveyor. 140 N·m at the motor shaft at about 1 470 rpm (gearbox losses already included), 8 to 10 h a day outdoors under a shelter, extremely dusty, with water-spray dust suppression. $P = 140 \times 153.9 = 21.55$ kW: a 22 kW motor. The environment decides: only the IP65 high-torque version fits.

Pick a motor for each. The checker applies the four constraints and tells you which one fails. The booklet marks the justification (duty, margin, protection), not the catalogue code alone.

Try it: choose the motor

Case 1, process water circulation pump: a centrifugal pump, hydraulic power 13.5 kW at 88 % mechanical efficiency, runs 24/7 at constant speed in a clean, dry indoor room.

Power needed: 15.3 kW
Starting torque: ≥ 100 %
Enclosure: ≥ IP54
Duty: S1

Model	kW	rpm	T rated (N·m)	T start	T breakdown	IP	IE
AG-110-IE3	11.0	1470	71.5	220 %	300 %	IP55	IE3
AG-110-IE4	11.0	1480	71.0	200 %	280 %	IP55	IE4
AG-150-IE3	15.0	1470	97.4	210 %	290 %	IP55	IE3
AG-150-IE4	15.0	1480	96.8	190 %	270 %	IP55	IE4
AG-185-IE3	18.5	1475	119.8	200 %	280 %	IP55	IE3
AG-220-IE3	22.0	1475	142.4	200 %	280 %	IP55	IE3
AG-220-IE3-HT	22.0	1470	142.9	280 %	350 %	IP65	IE3 (HT)

ExplorerGo deeper: derivations and open questions

Life-cycle cost. Compare an IE3 motor (η = 92.1 %) with an IE4 (η = 93.9 %) for a 15 kW load running 6000 h/year at 0.12 €/kWh. What is the yearly saving, and how many years pay back a 400 € price difference?

Ambient derating. Catalogues derate power by roughly 1 % per kelvin above 40 °C. At 50 °C in a Saharan pump station, which motor from the Tutorial 1.2 catalogue would you then choose for the pump? What about altitude?

VFD-fed motors. On a VFD, a self-cooled motor cannot deliver rated torque continuously at low speed, because its fan slows down. Look up a manufacturer's "loadability curve". How does it change step 4 of the procedure for a conveyor that runs at 30 % speed for hours?