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Case Studies and Mini-Projects

CHAPTER 6

Applied case studies across industry, system integration, and the capstone mini-project

Electrical Control of Industrial Mechanisms

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

Photo: Dylan Moore, CC BY-SA 2.0, Wikimedia Commons

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What you will be able to do

CHAPTER 6 · LEARNING OUTCOMES

1

Recognise the principles of Chapters 1–5 in other industries

2

Size a regenerative supply, an inching drive and a winder drive

3

Check an installation against IEEE 519, cable overvoltage and availability

4

Write a testable specification and a control philosophy

5

Size, simulate and defend a complete design

Photo: FFC1966, CC BY-SA 4.0, Wikimedia Commons

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

Chapter 6 · 2

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Cross-sector industrial case studies

6.1

Five questions read any machine: load, quadrants, loop, sizing quantity, failure.

Photo: Unknown author, Public domain, Wikimedia Commons

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

Chapter 6 · 3

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Five questions for every machine

6.1 · METHOD

✓

Load character: constant torque, quadratic, constant power or active? (Ch. 1)

✓

Quadrants, and where does the braking energy go? (Ch. 3, 4)

✓

Which quantity is controlled, and by which loop? (Ch. 2, 5)

✓

Which single number sizes the equipment?

✓

What happens when the drive fails — is that acceptable?

Photo: Dylan Moore, CC BY-SA 2.0, Wikimedia Commons

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

Chapter 6 · 4

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The reversing rolling mill

6.1 · METALLURGY

The slab passes back and forth; the drive reverses at full torque many times a minute
Four quadrants at tens of MW, energy recovered in every deceleration
Constant-torque braking: power falls linearly to zero
So the peak regenerated power is twice the mean

Lectures, Fig. 6.1

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

Chapter 6 · 5

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Regenerated energy and power per mill reversal

6.1 · WORKED EXAMPLE

GIVEN

J_eq = 4200 kg·m² referred to the motor; constant-torque braking from +60 rad/s to rest in 2.5 s

FIND

The energy returned, and the mean and peak power the supply converter must accept

Photo: Unknown author, Public domain, Wikimedia Commons

SOLUTION

E_rev = ½ × 4200 × 60² = 7.56 MJ
Mean power 7.56 / 2.5 ≈ 3.0 MW
Braking torque 4200 × 60 / 2.5 = 100.8 kN·m
Peak 2 × 7.56 / 2.5 ≈ 6.0 MW, at the start of braking

✓

Check. 100.8 kN·m × 60 rad/s = 6.05 MW: the same peak. The regenerative front end is rated on 6 MW, not 3.

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

Chapter 6 · 6

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The cement kiln must never stop while hot

6.1 · PROCESS INDUSTRIES

Photo: Dylan Moore, CC BY-SA 2.0, Wikimedia Commons

Lectures, Fig. 6.2

→

An independent inching drive, through an overrunning clutch and on its own supply, keeps the shell turning when the main drive fails

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

Chapter 6 · 7

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Sizing a kiln inching drive

6.1 · WORKED EXAMPLE

GIVEN

Shell at 2.5 rpm; main motor 450 kW at 1490 rpm; inching at 0.1 rpm with a 1470 rpm motor on the main gearbox input

FIND

Shell torque, inching power, ratio of the auxiliary gearbox

SOLUTION

ω_s = 2.5 × 2π/60 = 0.262 rad/s → T = 450 000/0.262 ≈ 1.72 MN·m
Constant torque at 0.1 rpm (0.0105 rad/s): P ≈ 18 kW
Main train 1490/2.5 = 596 → its input turns at 59.6 rpm
Auxiliary gearbox 1470/59.6 ≈ 25

✓

Check. At constant torque power scales with speed: 450 × 0.1/2.5 = 18 kW. Full torque, 4 % of the power.

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

Chapter 6 · 8

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The winder: constant power with a tension loop

6.1 · PAPER AND FILM

Photo: FFC1966, CC BY-SA 4.0, Wikimedia Commons

Lectures, Fig. 6.3

→

Constant tension F at constant surface speed v: T = F·r, ω = v/r, so T·ω = F·v stays constant

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

Chapter 6 · 9

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Torque and speed range of a winder

6.1 · WORKED EXAMPLE

GIVEN

Rewind roll grows from a 0.15 m core to 0.75 m full diameter; web tension F = 600 N at v = 8 m/s

FIND

The torque, speed and power range of the winder drive

SOLUTION

T_core = 600 × 0.075 = 45 N·m; T_full = 600 × 0.375 = 225 N·m
ω_core = 8/0.075 ≈ 107 rad/s; ω_full = 8/0.375 ≈ 21 rad/s
Torque ×5 while speed ÷5: power F·v = 4.8 kW throughout
The drive needs a 5:1 constant-power range

✓

Check. 45 × 107 ≈ 225 × 21 ≈ 4.8 kW = 600 × 8: torque times speed equals tension times surface speed.

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

Chapter 6 · 10

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What the build-up ratio and the roll inertia cost

6.1 · PAPER AND FILM

Field weakening covers only 2:1 to 3:1: a 2:1 motor for the 5:1 roll is rated 12 kW for 4.8 kW
Roll inertia grows as r⁴: every line ramp needs extra torque
Full roll, 0.5 m/s²: +33 N·m, 15 % — else tension sags 88 N
Diameter computed as r = v/ω; held at standstill

Lectures, Fig. 6.3

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

Chapter 6 · 11

Two consequences of the winder law that a design must handle. First, the range. An induction motor holds constant power by field weakening over only about 2:1 to 3:1. With 2:1 the base speed is half of 107 rad/s, 53 rad/s, where the motor must already give 225 N·m: rated 12 kW for a process that uses 4.8 kW. Without field weakening, 24 kW. The build-up ratio, not the power, sizes a winder motor. Second, inertia, eq. 6.4. The full roll of the example, 1 m wide at 800 kg per cubic metre, weighs 339 kg and has J = 24.8 kg m squared. A 0.5 m/s squared ramp needs 33 N·m more, 15 % of the tension torque; without it the tension would sag 88 N on every acceleration. The drive computes the radius as v over omega and must hold it at standstill.

Timing: 5 minutes. Reference: Lectures section 6.1.4, eq. 6.4, Fig. 6.3.

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Six machines, one method

6.1 · SYNTHESIS

Machine	Load character	Quadrants	Sizing quantity
Rolling mill	high inertia, fast reversal	I–IV	peak regenerated power
Arc furnace	fast electrode position	I, III	loop bandwidth; flicker
Cement kiln	constant torque, uninterruptible	I	shell torque at inching speed
Draw-works	active (hoist)	I–IV	hook load × speed; braking
Pipeline pump	quadratic (centrifugal)	I	duty point, static head
Winder	constant power, tension loop	I, II	full-roll torque; build-up ratio

!

Every industrial drive is a combination of the loads of Chapters 1–5

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

Chapter 6 · 12

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System integration: what surrounds the drive

6.2

A correctly chosen drive still fails in service if the system around it is inconsistent with it.

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

Chapter 6 · 13

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Six-pulse rectifiers inject harmonics

6.2 · POWER QUALITY

The DC-link capacitor charges only near the voltage peaks: pulsed line current
Orders 6k ± 1: 5th, 7th, 11th, 13th… amplitudes falling about 1/h
Current THD 80–120 %; simulated 81 %, 37 % with a 3 % line reactor
Extra heating in transformers and cables: K-factor derating

Lectures, Fig. 6.4

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

Chapter 6 · 14

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IEEE 519 judges the TDD at the PCC

6.2 · POWER QUALITY

Limit on total demand distortion at the point of common coupling
TDD = drive THD × the drives' share of the plant demand
I_sc/I_L 20–50: TDD 8 %, each harmonic below the 11th 7 %
Harmonic voltage: v_h ≈ h·x_s·i_h

Lectures, Fig. 6.5

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

Chapter 6 · 15

IEEE 519 sets the limit at the point of common coupling, the PCC, not at the drive — a distinction students routinely miss. And it limits the total demand distortion, eq.

6.6: the same harmonic current as the THD, but divided by the maximum demand current of the whole installation. A drive with 80 % THD is acceptable if it is a small part of a large demand, and unacceptable if it is most of it. The allowance rises with supply stiffness: 5 % below 20, 8 % between 20 and 50, up to 20 % above 1000. The voltage distortion follows from the source reactance, eq. 6.7, which grows with harmonic order. The figure shows the four front ends when the drive is the dominant load, where TDD equals THD: only a 12-pulse on a stiff supply, or an active front end, complies.

Timing: 5 minutes. Reference: Lectures section 6.2.1, eq. 6.6, eq. 6.7, Fig. 6.5.

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IEEE 519 compliance of a drive installation

6.2 · WORKED EXAMPLE

GIVEN

1000 kVA, 6 %, 400 V transformer; LV bus is the PCC; I_L = 1000 A; 250 kW of six-pulse drives, PF 0.95

FIND

The TDD for each front end, and the drive power a line reactor can serve

SOLUTION

I_sc = 1443/0.06 = 24.1 kA → I_sc/I_L = 24: limit 8 %
I₁ = 250 000/(√3 × 400 × 0.95) = 380 A
TDD: bridge 30.9 %, reactor 14.2 %, 12-pulse 3.8 %, AFE 1.5 %
Reactor complies only up to I₁ = 214 A: about 140 kW

✓

Check. Bare bridge: I₅ = 256 A is 26 % of I_L (limit 7 %), and v₅ = 5 × 0.06 × 256/1443 = 5.3 % (limit 5 %).

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

Chapter 6 · 16

Work it on the board. The transformer rated current is 1443 A, so the short-circuit current is 24.1 kA and the stiffness ratio 24: the 8 % band. The drives draw 380 A of fundamental. Multiply by each THD and divide by 1000 A: 30.9, 14.2, 3.8 and 1.5 %. Only the 12-pulse and the active front end comply. With a reactor, 0.373 I1 must stay below 80 A: about 140 kW of drives. The check uses both criteria: the bare bridge's 5th harmonic exceeds its individual current limit and, through eq. 6.7, distorts the bus voltage by 5.3 %. The spectrum is that of the stiff-supply simulation — conservative, since the transformer itself smooths the current a little.

Self-check 6.3 is the same calculation with other numbers. Tutorial Exercise 6.4 adds the 5th-harmonic limit, which turns out to be the tighter one.

Timing: 6 minutes. Reference: Lectures section 6.2.1, worked example, eq. 6.6.

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Long motor cables: reflections and bearing currents

6.2 · MOTOR CABLES

PWM edges on a cable reflect: up to 2·V_DC, about 1130 V on a 400 V supply
400 ns edges tolerate 30 m; 50 ns edges reach full reflection under 4 m
Remedies: dv/dt or sine filter, inverter-duty motors
Common-mode voltage drives bearing currents: insulated bearing, shaft earthing

Lectures, Fig. 6.6

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

Chapter 6 · 17

The inverter output is a train of steep pulses and the motor cable is a transmission line. When the rise time is short compared with the cable delay, the reflection at the motor adds to the incident pulse and the terminal voltage approaches twice the DC link. Full reflection occurs above the critical length, eq. 6.8, with v_p about 1.5 × 10⁸ m/s. Below it, eq. 6.9 takes the overshoot as growing in proportion to length — that is how Fig. 6.6 is drawn. Faster switching is more efficient and quieter, and makes this problem far worse, which is why it appears when an old drive is replaced by an apparently identical new one. Second effect of the same edges: a common-mode voltage drives current through the bearing film, pitting the races — failures in one to two years, usually misdiagnosed as mechanical.

Timing: 5 minutes. Reference: Lectures section 6.2.2, eq. 6.8, eq. 6.9, Fig. 6.6.

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Is a dv/dt filter needed?

6.2 · WORKED EXAMPLE

GIVEN

400 V supply, V_DC = 566 V; 50 ns IGBT edges; 25 m of motor cable; v_p = 1.5 × 10⁸ m/s; filter stretches edges to 1 µs

FIND

The peak motor terminal voltage without and with the filter

SOLUTION

Without filter: L_c = 50 ns × 1.5 × 10⁸/2 = 3.75 m
25 m is almost 7 × longer: V_m,pk ≈ 2 × 566 ≈ 1130 V
With filter: L_c = 75 m
V_m,pk ≈ 566 × (1 + 25/75) ≈ 755 V

✓

Check. The filter does not remove the reflection; it lengthens the edge until the cable is electrically short.

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

Chapter 6 · 18

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Protecting a variable-speed drive

6.2 · PROTECTION

Lectures, Fig. 6.7

Breaker Q1: short circuit and the supply cable
RCD type B only, with raised threshold: normal HF leakage is large
Semiconductor fuses (I²t) protect the rectifier
Drive's current limit and thermal model protect motor and cable — no overload relay
PTC / Pt100 in the winding: the only valid protection at low speed

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

Chapter 6 · 19

Read the chain left to right and ask, for each device, which fault it is fastest to clear. The breaker protects the supply cable; it must ride through the DC-link charging inrush. A type B RCD, where one is required, must tolerate the normal leakage at switching frequency, tens to hundreds of milliamperes. Semiconductor fuses protect the diodes and IGBTs — breakers are far too slow. The drive limits its output current within microseconds, so it is the best protection for the motor and cable; an overload relay after the drive would see variable-frequency current and is redundant. But at low speed a self-ventilated motor loses its cooling and the drive's current model cannot know it: only a PTC or Pt100 in the winding measures the real temperature — this is self-check 6.5.

Timing: 5 minutes. Reference: Lectures section 6.2.3, Fig. 6.7.

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Availability: units in series, units in redundancy

6.2 · AVAILABILITY

Shorter repairs raise availability as much as fewer failures — and cost less
Process stops if any unit fails: availabilities multiply
N+1: k of n units needed — losing it takes two failures
Only if failures are independent: beware the common switchboard

Photo: Suyash.dwivedi, CC BY-SA 4.0, Wikimedia Commons

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

Chapter 6 · 20

The availability formula, eq. 6.10, carries a lesson that is easy to miss: availability improves just as much by reducing the repair time as by extending the time between failures, and reducing repair time is almost always cheaper — a spare drive on the shelf, a documented procedure, backed-up parameters. A plant combines many units, eq. 6.11. If the process stops when any unit fails, the availabilities multiply. If it needs only k of n identical units — N+1 is k equal to n minus 1 — it is available whenever at least k are up, and losing it needs two failures at once. The proviso is the whole of the engineering: four pump sets on one switchboard fail together. That is a common-cause failure, and the redundancy calculation then becomes fiction.

Timing: 5 minutes. Reference: Lectures section 6.2.4, eq. 6.10, eq. 6.11.

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Value of an N+1 pump set

6.2 · WORKED EXAMPLE

GIVEN

A pumping station needs three pump sets; each set MTBF = 8000 h, MTTR = 8 h

FIND

The station availability with three sets, and with a fourth (N+1)

SOLUTION

One set: A = 8000/8008 = 0.999001
Three in series: A³ = 0.99701 → 26 h a year at reduced capacity
Three of four: A⁴ + 4A³(1 − A) = 0.999994
→ about 3 minutes a year

✓

Check. Two simultaneous failures needed: 1 − A_3/4 ≈ 6 q² = 6 × (0.001)² = 6 × 10⁻⁶ — the same result.

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

Chapter 6 · 21

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Commissioning has a fixed order

6.2 · COMMISSIONING

Lectures, Fig. 6.8

→

FAT → install → loop checks → drive (nameplate, auto-tune, direction, encoder, speed loop, outer loop) → SAT → safety validation

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

Chapter 6 · 22

A design is not complete when the drawings are issued. Factory acceptance proves the panel against the specification with the plant simulated; site acceptance repeats the critical tests on the real machine. In between, loop checks verify every field device end to end — tedious, and where most installation errors are actually found. Drive commissioning has its own fixed order: nameplate data; auto-tune, so the vector model knows stator resistance, leakage and magnetising curve; direction checked with the motor uncoupled; encoder direction and scale; the speed loop; only then any outer process loop. Tuning an outer loop over an untuned inner one wastes a day and must be redone. Finally, safety validation: every safety function proof-tested by creating the demand, and recorded.

Timing: 5 minutes. Reference: Lectures section 6.2.5, Fig. 6.8.

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The mini-project: a capstone design

6.3

From analysing a given mechanism to conceiving, specifying and defending your own.

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

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

Chapter 6 · 23

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From specification to defended design

6.3 · METHODOLOGY

Lectures, Fig. 6.9

Specification: motions, loads, duty cycle, performance targets, environment, safety
Control philosophy: axes, drive types, sensors, interlocks, operator interface
Sizing: every component justified by calculation and manufacturer data
Simulation tied to requirements; then the defence

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

Chapter 6 · 24

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Each stage ends with a deliverable and a test

6.3 · DELIVERABLES

Stage	Deliverable	Accepted when
Specification	numbered requirements; loads, duty, environment	every requirement is measurable and testable
Philosophy	axis table; safety functions; GRAFCET	every hazard has a safety function
Sizing	calculation sheets; assumption register	every choice traced to a calculation and a datasheet
Simulation	model and results	each result names the requirement it verifies
Defence	report and presentation	the design answers the specification point by point

!

A stage is finished when its test is passed, not when its time has run out

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

Chapter 6 · 25

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Testable requirements, fixed sizing order

6.3 · METHOD

Lectures, Fig. 6.9

Not “the hoist shall be fast” but “1000 kg at 0.5 m/s, ±5 mm at each level”
State the environment: a summer ambient above 40 °C changes the frame size
Size in order: load → motor → drive → braking → supply → protection
Assumption register: every number the specification does not give

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

Chapter 6 · 26

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How the mini-project is assessed

6.3 · ASSESSMENT

Specification (4 marks) and control philosophy (4)
Component sizing (5): correctness and justification
Safety and reliability (3)
Presentation and defence (4)
The reasoning is assessed, not only the result

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

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

Chapter 6 · 27

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Common mistakes in system design

WATCH OUT

✗

MYTH

“The safety function can be implemented in the PLC program.”

✓

WHAT IS TRUE

A standard PLC can fail silently. Safety functions need certified safety hardware at the required PL (ISO 13849-1) or SIL (IEC 62061), independent of the control program.

✗

MYTH

“Selecting the drive is the design; the rest is wiring.”

✓

WHAT IS TRUE

Transformer, cables, protection, harmonics, cable length and dv/dt, earthing and EMC, braking path: any inconsistency makes a correct drive fail in service.

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

Chapter 6 · 28

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Two traps in the numbers

WATCH OUT

✗

MYTH

“The datasheet quotes 37 % THD, so the plant meets IEEE 519.”

✓

WHAT IS TRUE

The limit is on the TDD at the PCC, referred to the whole plant's demand. The same drive complies in a large plant and violates in a small one.

✗

MYTH

“With N+1 drives the process cannot stop.”

✓

WHAT IS TRUE

The calculation assumes independent failures. Units on one switchboard, one control supply or one parameter set fail together.

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

Chapter 6 · 29

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Two more traps

WATCH OUT

✗

MYTH

“An energy saving at rated conditions is the annual saving.”

✓

WHAT IS TRUE

Annual energy depends on the load duration curve. Weight each operating point by its real hours; rated saving × 8760 h overstates it.

✗

MYTH

“A design is complete when it works.”

✓

WHAT IS TRUE

A defensible design states its assumptions, justifies components against standards, quantifies margins and identifies failure modes.

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

Chapter 6 · 30

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Quick check — try before you look

SELF-CHECK 6.1–6.6

6.1

J_eq = 4200 kg·m², constant-torque braking from 50 rad/s to rest in 2 s: energy and peak power?

6.2

Winder 10 m/s, 500 N, 0.2 → 1.0 m roll: torque range, power, motor rating with 2:1 field weakening?

6.3

I₁ = 200 A at 37 % THD; I_L = 800 A, I_sc = 20 kA. Does it meet the IEEE 519 TDD limit?

6.4

100 ns edges, 400 V supply, 20 m cable: critical length and peak terminal voltage?

6.5

Why no overload relay between drive and motor, and what protects the motor at low speed?

6.6

Two pump sets of A = 0.999, one needed: station availability? Both needed?

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

Chapter 6 · 31

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Summary

CHAPTER 6 · SUMMARY

Mill: rate the regenerative supply on the peak, twice the mean
Kiln: full torque at a few % power; winder: build-up ratio and inertia
Harmonics: TDD at the PCC, not the drive's THD
Long cables double the voltage beyond t_rv_p/2
Availability: short repairs; N+1 only with independent failures
Testable spec → axis table → sizing in order → simulation → defence

KEY EQUATIONS

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

Chapter 6 · 32

Close the course: a small set of principles spans the whole of manufacturing, read through five questions. The mill's supply is rated on the peak regenerated power; the kiln's inching drive needs full torque at a few per cent of the power; the winder motor is sized by its build-up ratio and must compensate the roll inertia. Around every drive sits a system: harmonics judged by the TDD at the PCC, cables that double the voltage beyond the critical length, protection, availability that depends on independent failures, and commissioning in a fixed order. The mini-project reverses the direction of study: the student conceives the mechanism, and the specification is the contract the design is judged against.

Ask each group to state, in one sentence, the mechanism they intend to choose for the project.

Timing: 5 minutes. Reference: Lectures, Chapter 6 summary box and key equations.

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Practise this

CHAPTER 6 · NEXT STEPS

TD

Exercises 6.1–6.7

Mill supply, kiln inching drive, winder, IEEE 519, cable filter, N+1 economics, and a model mini-project.

MP

Mini-project

Choose a mechanism, write the specification and control philosophy, size, simulate and defend.

TP

Lab works

TP 6 — PLC integration project: from a written specification to a commissioned, fault-tested program.

Photo: FFC1966, CC BY-SA 4.0, Wikimedia Commons

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

Chapter 6 · 33

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Image credits

CHAPTER 6 · 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).
Rotary kilns, Sundon cement plant - geograph.org.uk - 2421111 — Dylan Moore, CC BY-SA 2.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Rotary_kilns,_Sundon_cement_plant_-_geograph.org.uk_-_2421111.jpg
Burgos BOBINAS — FFC1966, CC BY-SA 4.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Burgos_BOBINAS.jpg
2015.028.040 Final rolling mill action forming finished steel rails — Unknown author, Public domain, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:2015.028.040_Final_rolling_mill_action_forming_finished_steel_rails.tif
Variable Frequency Drive (VFD) - 3 — Suyash.dwivedi, CC BY-SA 4.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Variable_Frequency_Drive_(VFD)_-_3.jpeg
MPI bus address setting — Mixabest, CC BY-SA 3.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:MPI_bus_address_setting.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 6 · 34