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Selecting an Electric Motor for Industry

CHAPTER 1

How to choose the right electric motor for a machine

Electrical Control of Industrial Mechanisms

Master 1, Semester 3 · 2025–2026 · Prof. Z. TIR — University of El Oued

Photo: Tmw11w1, Public domain, Wikimedia Commons

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Motors are everywhere around you

GETTING STARTED

At home: fan, washing machine, refrigerator, air conditioner
In the region: borehole pumps for the palm groves, water towers, elevators
In industry: conveyors, cranes, crushers, cement kilns
Each machine needs its own motor — choosing it is this chapter

Photo: SAF, CC BY-SA 4.0, Wikimedia Commons

!

A motor turns electrical energy into rotation. A wrong choice fails, however good the control.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 2

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What you will be able to do

CHAPTER 1 · LEARNING OUTCOMES

1

Say what a machine needs: how much torque, at which speed

2

Read a motor nameplate — the motor's identity card

3

Compute synchronous speed, slip, torque and power

4

Choose the motor size for continuous or on–off work

5

Choose the protection (IP) and cooling for the place it works

Lectures, Fig. 1.9

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 3

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Five quantities you already know

REFRESHER

Quantity	Symbol	Unit	Formula	Think of…
Force	F	newton (N)	—	pushing on a spanner
Torque	T	N·m	T = F × r	the spanner turning the nut
Speed	n or ω	rpm or rad/s	ω = 2π n / 60	turns of the shaft
Power	P	watt (W), kW	P = T × ω	how fast work is done
Efficiency	η	% (no unit)	η = P_out / P_in	the part not lost as heat

!

Before using P = T×ω, always convert rpm into rad/s: ω = 2πn/60

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 4

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Warm-up: your first three numbers

YOUR TURN · PEN AND PAPER

a

A mechanic pushes with 50 N at the end of a 0.30 m spanner. What torque is applied?

b

A motor turns at 1440 rpm. What is its speed in rad/s?

c

The same motor gives 63.6 N·m at 1440 rpm. What power does it deliver?

YOU WILL NEED

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 5

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Key words: English – Français

VOCABULARY · PART 1

English	Français	In simple words
motor	moteur	turns electrical energy into rotation
stator / rotor	stator / rotor	the fixed part / the turning part
winding	enroulement (bobinage)	coils of copper wire
torque	couple	the turning effort, in N·m
speed	vitesse de rotation	turns per minute (rpm) or rad/s
power	puissance	torque × speed, in W or kW
synchronous speed	vitesse de synchronisme	speed of the rotating magnetic field
slip	glissement	how far the rotor is behind the field

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 6

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The motor is an energy converter

1.1 · ELECTRO-MECHANICAL SYSTEMS

Electrical power goes in, mechanical power comes out (P = T·ω)
Part of the power is lost: in the copper, the iron and the bearings
The lost power becomes heat — heat is what limits a motor
A control system acts on two things: torque and speed

Lectures, Fig. 1.1

!

Power in = power out + losses. The losses are heat.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 7

Follow the diagram from left to right: the supply, the drive or starter, the motor, the machine. At each stage a little power is lost as heat (orange boxes). Everyday picture: a phone charger or a laptop gets warm — that warmth is lost power. A motor is the same: its efficiency may be 90 %, so 10 % of the input becomes heat inside the motor.

Ask: where exactly is the heat produced? Expected: in the copper windings (current), in the iron core (changing magnetic field), and in the bearings (friction).

Misconception: students see efficiency as just a number. Say: losses are heat, and too much heat burns the insulation of the windings. That is why choosing the size of a motor is mainly a question of temperature — we come back to this at the end of the chapter.

Timing: 5 minutes. Reference: Lectures section 1.1.1, Fig. 1.1.

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Why does a motor turn?

1.1 · TORQUE PRODUCTION

A wire carrying current in a magnetic field is pushed sideways: F = I·L·B
Bend the wire into a loop: one side is pushed up, the other down
Two opposite forces around the shaft make it turn
More current → more force → more torque

Lectures, Fig. 1.2

!

Current makes torque. A drive controls the torque by controlling the current.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 8

Left panel: one wire between the N and S poles, current coming out of the page. It is pushed upwards. Use the right-hand rule with the class; let them try it with their own hand. Right panel: the same wire bent into a loop. The current goes out of the page on one side and into the page on the other, so one side is pushed up and the other down. Two equal and opposite forces at distance 2r from each other make the loop turn: T = 2rF. Everyday picture: turning a car steering wheel with both hands — one hand pushes up, the other pulls down. The formula T = N I A B sin phi just says: more turns, more current, bigger loop, stronger field give more torque. Do not ask students to memorise it; ask them to remember "current makes torque".

Timing: 5 minutes. Reference: Lectures section 1.1.2, Fig. 1.2.

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The main families of motors

1.2

Induction, synchronous, DC — and the servo and stepper motors.

Photo: Zureks, CC BY-SA 3.0, Wikimedia Commons

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 9

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Three fixed windings make a turning field

1.2 · INDUCTION MOTORS

Lectures, Fig. 1.3

→

Like a stadium wave: each winding only pulses in place, yet the field goes round. n_s = 120 f / p

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 10

Start with the analogy before the figure. In a football stadium, the wave goes round the stadium, but each spectator only stands up and sits down in his or her own seat. Nobody moves around the stadium, yet the wave turns. The three windings of the stator are like three groups of spectators. Each one only pulses on its own axis (coloured arrows), but their sum, the black arrow, turns round at constant length. Top of the figure: the three currents. Bottom: the field at four moments t1 to t4. Let students see the black arrow turn. The speed of this field is the synchronous speed n_s = 120 f / p. On our 50 Hz grid: 2 poles give 3000 rpm, 4 poles give 1500 rpm, 6 poles give 1000 rpm. Write this little table on the board.

Timing: 6 minutes. Reference: Lectures section 1.2.1, Fig. 1.3, eq. for n_s.

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Slip: the rotor is always a little slower

1.2 · INDUCTION MOTORS

The rotating field pulls the rotor round
If the rotor turned as fast as the field, nothing would push it
So it always lags a little: this is the slip s
Normal motors: 2 to 5 % slip at full load

Photo: Zureks, CC BY-SA 3.0, Wikimedia Commons

!

Example: n_s = 1500 rpm, n = 1440 rpm → s = 60 / 1500 = 0.04 = 4 %

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 11

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Meet the reference motor of the book

1.2 · REFERENCE MOTOR

Quantity	Value	Remember it as
Rated output	9.6 kW, S1	a medium pump or fan motor
Supply	400 V, 50 Hz, delta, 4 poles	n_s = 1500 rpm
Full load	1440 rpm, s = 0.04	63.6 N·m, 17.6 A, cos φ = 0.86
Starting	106 N·m, 104 A	1.67 × torque, 5.9 × current
Breakdown	191 N·m at 1094 rpm	3 × full-load torque

!

Every induction-motor number in this chapter comes from this one motor

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 12

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Where the power goes: the equivalent circuit

1.2 · EQUIVALENT CIRCUIT

Lectures, Fig. 1.4

→

Of the 10.0 kW crossing the air gap, the slip (4 %) heats the rotor: 0.40 kW lost, 9.60 kW to the shaft

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 13

One phase of the motor can be drawn as a simple circuit of resistances and coils. Read panel (a) from left to right: the stator winding (R1, X1), the coil that makes the magnetic field (Xm), then the rotor seen from the stator (X2', R2'). The last resistance is the clever part: the moving rotor behaves like a resistance R2'(1 - s)/s, and the power it 'burns' is in fact the mechanical power on the shaft. At full load, s = 0.04, it is 13.2 ohm. Panel (b) follows the power: 10.46 kW in, 0.46 kW lost in the stator copper, 10.0 kW crosses the air gap. Of that, the fraction s = 4 % heats the rotor bars (0.40 kW) and the rest, 9.60 kW, turns the shaft. Rule to remember: rotor loss = s x air-gap power. A motor with small slip is an efficient motor. The same rule explains the next slide: a wound rotor slowed to 0.76 n_s burns 24 % of the air-gap power in its resistors. Every torque curve of the booklet is computed from this circuit; its maximum, 191 N·m (3 times full load), is reached at s = 0.27, that is 1094 rpm.

Timing: 6 minutes. Reference: Lectures section 1.2.1, Fig. 1.4, eqs. 1.9 and 1.10.

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Your turn: the borehole pump motor

YOUR TURN · PEN AND PAPER

a

A 2-pole pump motor works on the 50 Hz grid. What is its synchronous speed?

b

It turns at 2900 rpm. What is its slip, in %?

c

It delivers 7.5 kW at 2900 rpm. What torque does it give?

YOU WILL NEED

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 14

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Wound rotor: resistance moves the peak, not its height

1.2 · WOUND-ROTOR MOTORS

Lectures, Fig. 1.5

→

More resistance in the rotor moves the peak towards zero speed: the start gets 3× full-load torque instead of 1.67×

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 15

In a wound-rotor motor the rotor has copper windings brought out through slip rings, so we can add resistors outside the motor. Read the figure with the class. Each curve is the same motor with more rotor resistance. Point out: the top of every curve is at the same height, 3.0 — the maximum torque does not change. Only its position moves to the left, towards zero speed. With 3.7 times the normal resistance, the peak reaches zero speed, so the motor starts with 3 times its full-load torque instead of 1.67. With even more resistance (6 times), the peak goes too far and the starting torque falls again. The dots show the speeds with a constant load: 0.96, 0.92, 0.85, 0.76 of n_s — this is how old cranes changed speed. The price: the energy is burned in the resistors — at 0.76 n_s, 24 % of the air-gap power (rotor loss = s x air-gap power, previous slide).

Timing: 5 minutes. Reference: Lectures section 1.2.1, Fig. 1.5, self-check 1.3.

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Synchronous motors: no slip at all

1.2 · SYNCHRONOUS MOTORS

The rotor is a magnet: it turns exactly at n_s
More load → the rotor lags the field by a bigger angle δ
Beyond δ = 90° it loses step and stops
It cannot start alone: a small cage starts it
Used for large constant-speed machines

Lectures, Fig. 1.6

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 16

The rotor of a synchronous motor is a magnet (permanent magnets or a DC-fed winding). A magnet follows the rotating field exactly, like a compass needle that follows a turning magnet, so the speed is exactly n_s — no slip. With more load, the rotor stays at the same speed but lags behind the field by an angle delta. In the figure the rated load is carried at 30 degrees. The torque is largest at 90 degrees; above that, more lag gives less torque and the rotor drops out of step. It cannot start by itself, because the field turns too fast for the rotor to catch it from rest; a small cage in the rotor starts it as an induction motor first. Mention only briefly: by over-exciting the field it can also improve the plant power factor.

Timing: 5 minutes. Reference: Lectures section 1.2.2, Figs. 1.6 and 1.7.

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DC motors: the easiest to control

1.2 · DC MOTORS

Torque follows the current, speed follows the voltage
Shunt: almost constant speed (machine tools)
Series: very strong start, dangerous without load (trains, trams)
Compound: between the two (presses)
Today replaced by AC motors with drives — but the idea remains

Lectures, Fig. 1.8

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 17

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Servo and stepper motors

1.2 · SPECIAL MOTORS

Servo = motor + position sensor (encoder) + drive: very precise
Stepper: one step for each pulse, no sensor needed
A stepper can miss steps if it is overloaded
You will drive a real stepper in the lab (PLC-200)

Photo: Dolly1010, CC BY 3.0, Wikimedia Commons

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 18

A servomotor is a motor with a position sensor and a drive that corrects every error: robots and CNC machines use them. A stepper motor moves by fixed steps; send it 200 pulses and it turns 200 steps. The photo shows the toothed rotor of a stepper. Everyday example: the printer at the office and 3D printers. The step angle is one turn divided by the number of steps per turn.

Example from our lab: the stepper of the PLC-200 trainer makes 48 steps per turn (96 steps for two turns, in TP 3), so each step is 360 / 48 = 7.5 degrees.

Lab link: on the PLC-200 trainer, one pulse on Q0.1 is one step and Q0.0 sets the direction; an encoder checks that no step was lost.

Ask: why can a stepper lose its position? (If the load is too heavy, it misses steps without knowing it.)

Timing: 4 minutes. Reference: Lectures section 1.2.4, Table 1.2.

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Which motor for which job?

1.2 · OVERVIEW

Motor	How it works, simply	Typical uses
Squirrel-cage induction	field turns, rotor follows with slip	pumps, fans, conveyors
Wound-rotor induction	same, with resistors in the rotor	old cranes, crushers
Synchronous	rotor magnet locked to the field	large compressors and mills
DC	torque ∝ current, speed ∝ voltage	old drives, traction
Servo	motor + encoder + drive (closed loop)	robots, CNC machines
Stepper	one step per pulse (open loop)	printers, indexing tables

!

The most common industrial motor: the squirrel-cage induction motor

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 19

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What we know so far

RECAP

✓

A motor converts electrical power into T·ω; its losses become heat

✓

Current in a magnetic field gives a force; a loop of wire gives a torque

✓

Three windings make a rotating field: n_s = 120 f / p

✓

An induction rotor always slips a little: s = (n_s − n) / n_s

✓

Synchronous: no slip · DC: torque ∝ current · Stepper: one step per pulse

Photo: Tmw11w1, Public domain, Wikimedia Commons

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 20

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Reading what a motor can do

1.3

Can this motor start the machine, keep it running, and survive the overloads?

Lectures, Fig. 1.9

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 21

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Key words: English – Français

VOCABULARY · PART 2

English	Français	In simple words
rated / full load	nominal / pleine charge	the normal working point
starting torque	couple de démarrage	torque at zero speed
breakdown torque	couple maximal (décrochage)	the highest torque; beyond it, stall
starting current	courant de démarrage	5 to 8 × the normal current
efficiency	rendement	output power ÷ input power
power factor	facteur de puissance	cos φ
nameplate	plaque signalétique	the motor's identity card
duty type	service type (S1…S10)	how the motor works over time
enclosure (IP)	indice de protection (IP)	protection against dust and water

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 22

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Reading the torque–speed curve

1.3 · TORQUE–SPEED CURVE

Lectures, Fig. 1.9

→

Four points: starting torque 1.67, peak (breakdown) 3.0, full load 63.6 N·m at 1440 rpm, starting current 5.9×

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 23

Read the curve like a map, from left (motor stopped) to right (full speed). Blue line, torque. At zero speed, the starting torque: 1.67 times the full-load torque.

Then the torque rises to a peak, the breakdown torque: 3.0 times full load — the reserve for short overloads.

Then it falls quickly. The normal working point is near the right end: full-load torque 63.6 N·m at 1440 rpm. Orange dashed line, current. At start the current is 5.9 times the normal current, and it stays high until the motor is almost at speed. That is why the lights may flicker when a big motor starts: Chapter 2 treats this problem.

Ask: compute the full-load torque from 9.6 kW and 1440 rpm. (omega = 150.8 rad/s, T = 63.7 N·m; the book gives 63.6 because the exact power is 9.596 kW.)

Timing: 6 minutes. Reference: Lectures section 1.3.1, Fig. 1.9, eq. T_FL.

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The gap between the two curves speeds the machine up

1.3 · ACCELERATING TORQUE

Lectures, Fig. 1.10

→

Motor torque minus load torque accelerates the shaft; where the curves meet (1440 rpm) the speed stays steady

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 24

Two curves: the motor (blue) and the fan it drives (orange). The motor pushes; the fan resists. Only the difference, the green area, makes the shaft go faster: T_acc = T_m - T_L = J d omega/dt. Analogy: pushing a car. If you push harder than the friction, the car speeds up. When your push equals the friction, the car moves at constant speed. Here the two curves meet at 1440 rpm and 63.6 N·m: the motor gives exactly what the fan needs, and the speed stays steady. With the inertia of this fan, J = 1.2 kg·m², the start lasts about 1.3 s.

Ask self-check 1.4: what if the two curves almost touch at some speed? (The gap is tiny there, the start becomes very slow and the motor overheats; if they cross, the motor stays stuck there.)

Timing: 5 minutes. Reference: Lectures section 1.3.1, Fig. 1.10, eq. 1.17.

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A real start is more violent than the curve

1.3 · REAL START

Lectures, Fig. 1.11

→

In the first moments of a direct start the torque jumps to 5.1× rated — shafts and couplings must survive it

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 25

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A motor works best near its rated load

1.3 · EFFICIENCY

Efficiency η = useful output ÷ input
Best between 75 % and 100 % of the rated load
A motor that is too big works half-loaded: lower η and poor cos φ
So never oversize “just to be safe”

Lectures, Fig. 1.12

!

Choose a motor that works near 75–100 % of its rating most of the time

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 26

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The nameplate: the motor's identity card

1.3 · NAMEPLATE

Lectures, Fig. 1.13

VOLTS, AMPS: choose cables, contactor and overload relay
HP or kW, RPM: rated power and speed
S.F. 1.15: a small safety margin, not for every day
INS. CL. F: windings up to 155 °C · DUTY CONT = S1
This plate is American (60 Hz); our grid is 50 Hz, 400 V

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 27

Compare the nameplate with an identity card: it gives the motor's name (model, serial number) and its main data. Go through it line by line with the class: 25 HP, 460 V, 28 A, 3550 rpm, 60 Hz, service factor 1.15, insulation class F, TEFC IP55 enclosure, efficiency 93.6 %, power factor 0.89. Explain each in one sentence. HP is horsepower: 1 HP is about 746 W. The service factor allows 15 % more load for short periods, not every day — it shortens the life of the insulation. Point out the frequency: this is an American plate, 60 Hz. In Algeria the grid is 50 Hz and 400 V between phases; motors bought here carry 50 Hz plates. Three more fields: CODE G means 5.6 to 6.3 kVA per HP at standstill, so this motor draws 176 to 198 A when it starts, 6 to 7 times its 28 A. Bearings DE/ODE: drive end and opposite end, for ordering spares. A European plate reads 400 V delta / 690 V star: on our 400 V grid it runs in delta and can be started star-delta; a 230 V delta / 400 V star motor must stay in star. The next slide asks students to compute from this plate.

Timing: 5 minutes. Reference: Lectures section 1.3.3, Fig. 1.13; Tutorials Exercise 1.1.

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Your turn: read this nameplate

YOUR TURN · PEN AND PAPER

a

Rated power in kW? (1 HP ≈ 746 W)

b

Full-load torque at 3550 rpm?

c

Input power, with η = 93.6 %? Check it with √3·V·I·cos φ

d

It is a 2-pole motor on 60 Hz: synchronous speed and slip?

Lectures, Fig. 1.13

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 28

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How to choose a motor

1.4

Start from the machine, not from the motor catalogue.

Lectures, Fig. 1.14

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 29

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Choosing a motor: four simple questions

1.4 · METHOD

Lectures, Fig. 1.18

1. What does the machine need? (torque at each speed)
2. How much power? (load power ÷ efficiency)
3. How does it work over time? (always on, or on–off)
4. Where does it work? (dust, water, heat)

!

Answer the four questions, then check — and go back if a check fails

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 30

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Question 1: what does the machine need?

1.4 · LOAD PROFILES

Lectures, Fig. 1.14

→

Conveyor: same torque at every speed · Fan or pump: torque grows as speed² · Winder (3:1 roll): 3× torque at ⅓ speed, same power

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 31

Three families of machines. Solid line torque, dashed line power. (A) Constant torque: a conveyor carrying gravel needs the same push at low and at high speed. (B) Variable torque: a fan or a pump needs little torque when slow and much more when fast — torque grows with the square of speed, and power with the cube. Everyday proof: a fan at low speed is quiet and easy; at high speed it pushes much more air and uses much more electricity. (C) Constant power: a machine that winds paper on a roll needs more torque when the roll is big and turns slowly. Panel (C) has its own scale: for a roll that grows from 1 to 3 times its diameter, the speed falls to one third and the torque rises to 3 times rated, while the power stays the same. The grey zone is outside the winding range. The motor must be sized for that 3 times torque at low speed.

Ask the class to classify: a crane hoist, a ceiling fan, the borehole pump, a paper winder. (Constant torque, variable torque, variable torque, constant power.)

Timing: 5 minutes. Reference: Lectures section 1.4.1, Fig. 1.14.

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The conveyor of the example, drawn to scale

1.4 · WORKED EXAMPLE

Lectures, Fig. 1.15

→

Rise = 10 × sin 15° = 2.59 m · the weight splits into 3949 N on the idlers and 1058 N down the slope

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 32

Show the picture before any calculation: students must see the machine before they compute it. Panel (a) is drawn to scale: the belt is 10 m long, at a real 15 degrees, so it rises 10 x sin 15 = 2.59 m. Point out where the gravel enters (the tail) and where the motor and gearbox are (the drive pulley at the top). Panel (b): the weight of the gravel on the belt, mg = 4088 N, points straight down. We split it into two parts that follow the slope: one presses on the idlers, mg cos 15 = 3949 N; the other pulls the gravel back down the slope, mg sin 15 = 1058 N. The dashed triangle closes exactly on the weight: that is what "splitting a force" means.

Ask: which part must the motor fight? (The one along the slope.) Panel (c): at constant speed the belt pull equals the lift plus the friction, 1058 + 197 = 1255 N. The friction is small: 5 % of the force pressing on the idlers.

Timing: 4 minutes. Reference: Lectures section 1.4.2, Fig. 1.15.

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Question 2: sizing a conveyor motor

1.4 · WORKED EXAMPLE

GIVEN

A conveyor lifts 50 kg/s of gravel up a 10 m slope at 15°; belt speed 1.2 m/s; gearbox efficiency 0.95; friction coefficient 0.05

FIND

The standard motor rating

Photo: Leonhard Lenz, CC BY-SA 4.0, Wikimedia Commons

SOLUTION

Mass on the belt: it stays 10 / 1.2 = 8.33 s, so m = 50 × 8.33 = 416.7 kg
Lift force m g sin 15° = 1058 N; friction μ m g cos 15° = 197 N; total 1255 N
Belt power = force × speed = 1255 × 1.2 = 1506 W
Motor power = 1.506 / 0.95 = 1.59 kW
Choose the next standard size: 2.2 kW (1.5 kW would be 6 % overloaded)

✓

Check. Lifting alone: 50 × 9.81 × 10 sin 15° = 1.27 kW; friction (237 W) adds less than a fifth.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 33

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Question 3: a motor heats slowly

1.4 · DUTY TYPES

Lectures, Fig. 1.16

→

Heating takes about 30 min: a motor that rests between short runs can carry more than its continuous rating

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 34

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Question 3: the equivalent (RMS) torque

1.4 · WORKED EXAMPLE

GIVEN

A machine needs 80 N·m for 10 s, then almost nothing for 15 s, again and again

FIND

The torque rating of a continuous-duty (S1) motor that will not overheat

SOLUTION

Heating grows as T², so we average T² over the cycle
T_rms = √(80² × 10 / 25) ≈ 50.6 N·m
Choose an S1 motor with full-load torque of at least 50.6 N·m
It must still give the 80 N·m peak: 1.6× its rating, below a 2–3× breakdown torque

✓

Check. T_rms lies between the average (32 N·m) and the peak (80 N·m), as it must.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 35

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Question 4: reading the IP and IC codes

1.4 · ENVIRONMENT

Lectures, Fig. 1.17

→

IP55 = dust-protected + water jets · IC411 = closed motor cooled by its own shaft fan (TEFC)

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 36

Each digit of these codes has a meaning; read them like a number plate. IP (International Protection): first digit for dust and solid objects, second for water. IP55: some dust may enter but never enough to harm, and water jets do no harm. In the sand of the Souf an outdoor motor needs at least IP55; in a cement works with fine dust, IP6x, dust-tight. A food factory washed with hot high-pressure water needs IP66 or IPX9 (the IP69K of catalogues is a road-vehicle code). IC (International Cooling): IC411 is the usual industrial motor, totally enclosed, fan cooled. If a variable-speed drive runs it slowly for hours, its own fan turns slowly too and cools badly: then choose IC416, with a separate fan.

Ask: what does IP23 allow? (Fingers kept out, spraying water; indoor, clean rooms only.)

Timing: 5 minutes. Reference: Lectures section 1.4.4, Fig. 1.17, Table 1.5.

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Question 4: heat and our hot climate

1.4 · ENVIRONMENT

Item	Value	What it means
Insulation class B / F / H	130 / 155 / 180 °C	the hottest the windings may get
Allowed rise, class F	105 K	above the 40 °C reference air
Reference air	40 °C, below 1000 m	the nameplate power holds up to here
Air at 45 °C	× 0.96	the motor may give 96 % of its rating
Air at 50 °C	× 0.92	El Oued in summer: only 92 %
Rule of thumb	+10 K	halves the life of the insulation

!

El Oued in summer, 50 °C: a 37 kW motor may give only 37 × 0.92 = 34.0 kW

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

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All four questions: an aeration blower

1.4 · WORKED EXAMPLE

GIVEN

An outdoor blower near El Oued needs 35 kW, runs all day, in sand and rain, summer air at 50 °C (derating 0.92); 4-pole motor

FIND

The complete motor specification

Photo: SAF, CC BY-SA 4.0, Wikimedia Commons

SOLUTION

Q1 — the machine: a blower, variable torque
Q2 — the power: 35 kW → next standard size 37 kW
Q3 — the duty: runs all day → S1
Q4 — the heat: 37 × 0.92 = 34.0 kW < 35 kW → too small; 45 kW × 0.92 = 41.4 kW ✓
Q4 — the place: sand and rain → IP55, TEFC (IC411); frame 225M

✓

Check. 35 / 45 = 78 % load, still near the efficiency peak; at 40 °C the 37 kW motor would have done.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 38

This example uses the four questions in order — point at them on the board as you go. Q4 needs the IP code: the first digit is protection against dust (5 = dust-protected), the second against water (5 = water jets). IP55 is the usual outdoor motor. TEFC means totally enclosed, fan cooled: a fan on the shaft blows air over the fins of the closed body. The frame size is like a shoe size: it fixes the mounting dimensions, so the mechanical engineer can prepare the base. The hot-climate step is the new one: in a 40 degree plant room the 37 kW motor is right, but outdoors in an El Oued summer it can give only 34.0 kW. The climate alone moves us to 45 kW and frame 225M (shaft 225 mm above the feet).

Ask: is the 45 kW motor oversized? (No: it runs at 78 % of its rating, still on the flat top of the efficiency curve, and it is the smallest motor that survives the summer.)

Timing: 6 minutes. Reference: Lectures section 1.4.4, worked example; Tutorials Exercise 1.2.

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The full method: check, and go back if needed

1.4 · SELECTION PROCEDURE

Lectures, Fig. 1.18

→

If the heating or the torque check fails: a bigger motor, a duty-rated or wound-rotor motor, or a variable-speed drive

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

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Common mistakes when choosing a motor

WATCH OUT

✗

MYTH

“Choose the motor whose kW matches the load kW.”

✓

WHAT IS TRUE

kW alone is not enough. Check four things: heating over the duty cycle, starting torque, peak torque, and the environment.

✗

MYTH

“To be safe, take a much bigger motor.”

✓

WHAT IS TRUE

An oversized motor works at light load: lower efficiency, poor power factor, higher price and bill.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

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Quick check — try before you look

SELF-CHECK 1.1–1.6

1.1

A 4-pole motor on 50 Hz runs at 1440 rpm. Synchronous speed and slip?

1.2

Why can an induction motor never run at exactly synchronous speed?

1.3

More rotor resistance raises the starting torque, yet the maximum torque stays the same. Why?

1.4

Motor and load curves almost touch at one speed during a start. What happens?

1.5

100 N·m for 6 s, then 40 N·m for 14 s, again and again. Required S1 torque?

1.6

Why can a motor in S3 40 % duty carry more than its continuous rating?

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

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Summary

CHAPTER 1 · SUMMARY

A motor turns electrical power into T·ω; its losses become heat
Induction motors always slip a little; synchronous motors do not
The torque–speed curve tells if a motor can start and carry a machine
Choose in four steps: the machine, the power, the duty, the place
Too small burns; too big wastes energy and money

KEY EQUATIONS

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

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Practise this

CHAPTER 1 · NEXT STEPS

TD

Tutorials

Exercise 1.1 — analyse the nameplate of Fig. 1.13.

Exercise 1.2 — a case study in motor selection.

TP

Lab works

TP 0 — getting started with the PLC-200.

TP 1 — DC motor characterisation and the automatic barrier.

?

Before next week

Find a real motor nameplate (pump, fan, lift) and copy its data. Redo the self-check questions.

Photo: Tmw11w1, Public domain, Wikimedia Commons

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

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Image credits

CHAPTER 1 · SOURCES

Figures and diagrams: Lectures booklet, Electrical Control of Industrial Mechanisms (computed in Python/Simscape or drawn in TikZ from the chapter's own equations).
TMW 50906 Schnittmodell einer Drehstrommaschine (Asynchronmaschine) — Tmw11w1, Public domain, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:TMW_50906_Schnittmodell_einer_Drehstrommaschine_(Asynchronmaschine).jpg
High Pressure Centrifugal Fan — SAF, CC BY-SA 4.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:High_Pressure_Centrifugal_Fan.jpg
Wirnik by Zureks — Zureks, CC BY-SA 3.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Wirnik_by_Zureks.jpg
Stepper motor rotor — Dolly1010, CC BY 3.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Stepper_motor_rotor.jpg
Conveyor belt at surface mine Schleenhain 01 — Leonhard Lenz, CC BY-SA 4.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Conveyor_belt_at_surface_mine_Schleenhain_01.jpg
Licence texts: creativecommons.org/licenses (CC BY, CC BY-SA); creativecommons.org/publicdomain/zero/1.0 (CC0). Photos are shown unmodified except for cropping and scaling.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 1 · 44