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Conveyors and Continuous Transport

CHAPTER 5

Automation of continuous transport mechanisms: conveyors, escalators, ropeways, excavators

Electrical Control of Industrial Mechanisms

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

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

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What you will be able to do

CHAPTER 5 · LEARNING OUTCOMES

1

Size a conveyor drive from its belt pull, and its take-up tension

2

Design a start that respects the elastic belt

3

Design a master–slave or droop load-sharing scheme

4

Analyse escalators and ropeways: safety chain, limits, power

5

Match a bucket-wheel excavator to its conveyors

Lectures, Fig. 5.1

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

Chapter 5 · 2

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Meet the reference conveyor of the chapter

5.1 · REFERENCE CONVEYOR

Quantity	Value	Remember it as
Length, speed	400 m horizontal, 2.5 m/s	a plant or quarry conveyor
Mass per metre	material 90, belt 25, idlers 20 kg/m	64 t moving in total
Resistance (DIN 22101)	17.3 kN, 48 kW	twice the lumped estimate
Belt stiffness	EA = 15 MN → c = 306 m/s	natural period 5.2 s
Drives	2 × 30 kW, head + intermediate	master–slave or droop

!

The same conveyor is simulated in Simscape and sized in Tutorial Exercise 5.1

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

Chapter 5 · 3

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A constant-torque load sized from the belt pull

5.1 · CONVEYOR DRIVES

Belt, roller and chain conveyors: torque nearly independent of speed
Effective pull F_e = T₁ − T₂ at the drive pulley
No slip while T₁/T₂ ≤ e^μθ: take-up must supply T₂ ≥ F_e/(e^μθ − 1)
Torque fixed by F_e; power proportional to belt speed — ideal for a VFD

Lectures, Fig. 5.1

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

Chapter 5 · 4

The effective pull combines friction of all moving parts, with an equivalent friction coefficient f over the length L, and the lift of the material through the height H. Point at the figure: T1 on the tight side and T2 on the slack side both pull the pulley towards the tail; their difference is F_e. The ratio is limited by the capstan relation T1/T2 <= e^(mu theta); if the drive asks for more, the belt slips on the pulley. Eliminate T1 and you get the take-up tension: for the reference conveyor on one head pulley, lagged (mu 0.35) over 200 degrees, e^(mu theta) = 3.39 and T2 >= 17.3/2.39 = 7.2 kN; wet, 12.4 kN. A gravity take-up pulls two strands, so it needs about 1.5 t of weight. Splitting the pull between two drives roughly halves it.

Then torque F_e times pulley radius, independent of speed, and power F_e v / eta, proportional to speed: a constant-torque load, the perfect match for a VFD.

Ask self-check 5.1: 17.3 kN over a pulley with e^(mu theta) = 3.39? (7.2 kN.)

Timing: 5 minutes. Reference: Lectures sections 5.1.1 and 5.1.2, eqs 5.1 to 5.3, Fig. 5.1.

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Sizing an inclined conveyor drive

5.1 · WORKED EXAMPLE

GIVEN

L = 120 m, lift H = 14 m, v = 1.6 m/s; belt 22 kg/m, material 55 kg/m; f = 0.022; η = 0.9

FIND

The effective pull and the standard motor rating

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

SOLUTION

Friction: 0.022 × 9.81 × 120 × 77 ≈ 1.99 kN
Lift: 55 × 9.81 × 14 ≈ 7.55 kN
F_e ≈ 1.99 + 7.55 = 9.54 kN
P = 9540 × 1.6 / 0.9 ≈ 17.0 kW → 18.5 kW standard motor

✓

Check. Lift = potential-energy rate (55 × 1.6) × 9.81 × 14 ≈ 12.1 kW, nearly four times the friction term.

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

Chapter 5 · 5

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Where the resistance really comes from

5.1 · DIN 22101

Primary F_H: idler friction, belt indentation and flexing, along both strands
Secondary F_N = (C − 1)F_H: loading point, skirts, cleaners, pulleys
Slope F_St = m'_LgH: the only recoverable term
Special F_S: tilted idlers, ploughs, extra cleaners
Worked example: 1.99 kN simplified, 3.06 kN by DIN

Lectures, Fig. 5.2

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

Chapter 5 · 6

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Short conveyors are power-hungry per tonne

5.1 · DIN 22101

Lectures, Fig. 5.2

→

Secondary resistance: about one third of the total at 20 m, barely 1 % at 2 km (C falls from 3.0 to 1.05)

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

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The belt is an elastic body

5.1 · STARTING AND STOPPING

Tension changes travel at c = √(EA/m'): 306 m/s; natural period T_n ≈ 4L/c = 5.2 s
An abrupt start launches a tension wave: the peak damages splices
Huge translational inertia: ramps of tens of seconds are normal
Fluid coupling, wound rotor, now VFD with S-curve
Incline: backstop. Decline: regenerates for its whole life

Photo: byrev, CC0, Wikimedia Commons

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

Chapter 5 · 8

A belt is not a rigid coupling but a long spring. When the head drive applies torque, the change travels down the belt at c = root of EA over m', about 300 m/s for the loaded belt simulated in the booklet — on a 400 m conveyor the tail learns about the start more than a second later. Two consequences: a tension wave that reflects and superimposes, whose peak governs splice life; and an enormous inertia, dominated by the belt and material mass, so direct-on-line starting is out of the question. Historical solutions: the fluid coupling, robust and still common; the wound-rotor motor. Today the VFD with a programmable, jerk-limited profile. Stopping: a loaded incline runs backwards without power, so a mechanical backstop is fitted; a loaded decline overhauls and regenerates continuously — an active front end pays for itself.

Timing: 5 minutes. Reference: Lectures section 5.1.4.

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How long must the start ramp be?

5.1 · STARTING

Lectures, Fig. 5.3

→

Ramp several belt periods long. Below t_a/T_n ≈ 2 the S-curve is worse than a linear ramp: its peak acceleration is 1.5× the mean

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

Chapter 5 · 9

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The tension wave, simulated along the belt

5.1 · STARTING

Lectures, Fig. 5.4

→

2 s ramp: the tail moves 1.31 s after the head (c ≈ 306 m/s) and the head peaks near 7× running. 16 s S-curve: below 2×

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

Chapter 5 · 10

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Starting the reference conveyor

5.1 · WORKED EXAMPLE

GIVEN

Reference conveyor: 64 t moving, 17.3 kN running pull, two 30 kW drives (10.8 kN each at the belt, 150 % short-time); 16 s S-curve to 2.5 m/s

FIND

The ramp in natural periods, the peak starting force, the drive overload

SOLUTION

16 / 5.2 ≈ 3 natural periods: safely right of the crossover
Mean a = 2.5/16 = 0.156 m/s²; S-curve peak 1.5× = 0.234 m/s²
Inertial force 64 000 × 0.234 = 15.0 kN; total 32.3 kN = 1.87 × running
Per drive 16.1 kN = 1.5 × rated for a few seconds

✓

Check. An 8 s ramp would double the inertial force (47 kN in total, 2.2× rated per drive): too short for the drives and for the belt.

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

Chapter 5 · 11

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Control of multi-motor systems

5.2

Several motors on one belt will fight each other unless the controller tells them how to share.

Lectures, Fig. 5.5

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

Chapter 5 · 12

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Master–slave or droop

5.2 · LOAD SHARING

Master–slave: one drive in speed control, the others follow its torque over the fieldbus
Droop: every drive in speed control, reference lowered in proportion to its own load
Master–slave: stiff, exact speed, but needs the bus and a master
Droop: no communication, degrades gracefully, small speed error

Lectures, Fig. 5.5

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

Chapter 5 · 13

Master-slave: the PLC sends the speed reference to the master drive, which runs in speed control and sets the line speed; the slaves run in torque control and follow the master's torque reference broadcast over the fieldbus. Each takes its share without contesting the speed. Droop: every drive is in speed control, but its reference is reduced in proportion to its own torque. A drive taking more than its share slows slightly and sheds load to the others — exactly how parallel generators share load through frequency droop. Table 5.2 of the booklet, orally: master-slave is precise but needs fast communication and fails if the master is lost; droop is simpler and robust, at the price of a load-dependent speed error of typically 2 to 5 %. One more design rule: droop must be large compared with mechanical mismatch. A pulley 0.5 % larger shifts the split by about epsilon / k_d: with 3 % droop, from 80/80 to 88/72 %.

Timing: 5 minutes. Reference: Lectures section 5.2.2, Fig. 5.5, eq. 5.6, Table 5.2.

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Choosing a droop setting

5.2 · WORKED EXAMPLE

GIVEN

Two 30 kW drives on one belt, k_d = 0.03 (3 % droop). A mismatch loads drive 1 to 110 % and drive 2 to 90 %. Belt 2.5 m/s.

FIND

The corrective speed difference, and the price paid at full load

SOLUTION

Δω*/ω*₀ = k_d(1.10 − 0.90) = 0.03 × 0.20 = 0.6 %
Drive 2 now runs faster and picks up load until the imbalance closes
Price: at full load both run 3 % slow: 2.5 × 0.97 ≈ 2.43 m/s
A larger k_d stiffens the sharing but worsens the speed error

✓

Check. 2.5 × (1 − 0.03) = 2.425 m/s; the simulation shows 2.44 m/s because each drive is a little below rated torque.

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

Chapter 5 · 14

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Both schemes simulated, including a fieldbus failure

5.2 · LOAD SHARING

Lectures, Fig. 5.6

→

Healthy bus: both share 50/50. Bus lost at 35 s: master–slave drifts to 42/58, droop stays 50/50 — but runs at 2.44 m/s

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

Chapter 5 · 15

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Protection distributed along the belt

5.2 · PROTECTION

Lectures, Fig. 5.7

Pull cords both sides, full length; reset only at the switch
Belt-drift switch pairs: outer alarms, inner trips
Zero-speed sensor on the tail (non-driven) pulley: detects slip
Chute-blockage and belt-tear detectors; backstop on inclines

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

Chapter 5 · 16

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Start against the flow, stop with it

5.2 · SEQUENCE INTERLOCKING

Lectures, Fig. 5.8

→

Start from the discharge end, each belt proven at speed. Normal stop from the feed end with run-on. Emergency stop: all at once

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

Chapter 5 · 17

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Screws, bucket elevators, chains and air

5.2 · OTHER MECHANISMS

Lectures, Fig. 5.9

Screw conveyor: sized on breakaway torque; shear pin or torque limit
Bucket elevator: backstop and zero-speed monitor; dust ignition risk
Chain: polygon effect — the speed dips once per tooth
Pneumatic conveying is a blower problem (Roots blower: constant torque)

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

Chapter 5 · 18

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Passenger transport systems

5.3

Escalators and ropeways: the drive is subordinate to safety.

Photo: Stig Nygaard from Copenhagen, Denmark, CC BY 2.0, Wikimedia Commons

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

Chapter 5 · 19

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Escalator: seven devices, one series chain

5.3 · ESCALATORS

Photo: Stig Nygaard from Copenhagen, Denmark, CC BY 2.0, Wikimedia Commons

Lectures, Fig. 5.10

→

All contacts NC and in series: any one — or a broken wire — removes the enable and applies the brake

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

Chapter 5 · 20

EN 115-1 limits: at most 0.75 m/s up to 30 degrees, 0.5 m/s for the 35-degree units allowed up to a 6 m rise. The handrail runs at 0 to +2 % of the step speed; the monitor stops the escalator if it falls more than 15 % behind. The brake must stop within 0.20 to 1.00 m at 0.5 m/s, without exceeding about 1 m/s². The drive is a constant-torque motor, usually on a VFD with a slow standby speed. Over it sits the safety chain, the same philosophy as the elevator. Name the devices from the figure: 1 emergency stops at both landings, 2 combplate switches, 3 broken drive-chain switch, 4 handrail speed monitor compared with the step band, 5 step-upthrust switch, 6 skirt-panel switches, 7 handrail-entry switch. All normally closed, all in series to the safety relay; any open contact or broken wire stops the drive and applies the brake.

Ask: why compare handrail and step speeds? (A divergence means a mechanical failure; passengers would be pulled off balance.)

Timing: 5 minutes. Reference: Lectures section 5.3.1, Fig. 5.10; Tutorials Exercise 5.6.

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Power of a busy escalator

5.3 · WORKED EXAMPLE

GIVEN

Rise 6 m at 30°, 0.5 m/s; 4500 passengers an hour of 75 kg; no-load losses 3 kW

FIND

The power upwards, and what changes downwards

Photo: Stig Nygaard from Copenhagen, Denmark, CC BY 2.0, Wikimedia Commons

SOLUTION

Mass flow 4500 × 75 / 3600 = 93.8 kg/s
Lifting 93.8 × 9.81 × 6 = 5.5 kW; with losses about 8.5 kW upwards
Downwards the passengers deliver 5.5 kW > 3 kW losses
The motor generates about 2.5 kW

✓

Check. 12 m incline at 0.5 m/s = 24 s per passenger: about 30 people, 2.2 t, on the steps at once — the load the brake must stop.

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

Chapter 5 · 21

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Ropeways: redundancy because nobody can get off

5.3 · ROPEWAYS

Lectures, Fig. 5.11

Main motor turns the bullwheel; an independent diesel-hydraulic drive backs it up
Service brake on the motor shaft, emergency brake on the bullwheel rim
Tension carriage keeps the rope tension — and the friction grip — constant
Continuous supervision: rope speed, rope tension, carrier spacing

Photo: PetrS., CC BY-SA 3.0, Wikimedia Commons

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

Chapter 5 · 22

A ropeway stoppage can leave passengers suspended above a valley, as in the photo of an aerial tramway cabin. So redundancy is the design principle. Read the plan view: at the drive terminal the main motor turns the bullwheel through a gearbox; an independent diesel-hydraulic backup can bring the carriers to a station. Braking is duplicated: a service brake on the motor shaft and an emergency brake on the bullwheel rim. At the return terminal the bullwheel sits on a tension carriage pulled by a cylinder or counterweight, keeping the rope tension — and so the friction grip on the drive bullwheel — constant as the rope stretches. Supervision is continuous: rope speed, tension, carrier spacing.

Ask: what would happen to the drive if the rope tension were not maintained as the rope stretches? (The slack-side tension T2 would fall; by the capstan relation T1/T2 <= e^(mu theta) the bullwheel could then transmit less pull before the rope slips.)

Timing: 4 minutes. Reference: Lectures section 5.3.2, Fig. 5.11.

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Integrated systems: the bucket-wheel excavator

5.4

Every mechanism of the course in one machine, coordinated so the material won matches the material carried.

Photo: byrev, CC0, Wikimedia Commons

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

Chapter 5 · 23

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Coordinating many high-power drives

5.4 · BUCKET-WHEEL EXCAVATOR

Lectures, Fig. 5.12

Photo: byrev, CC0, Wikimedia Commons

→

Output 60·z·n·V_b·φ must fit the conveyors; the cut thins as cos φ, so the slew speeds up as 1/cos φ

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

Chapter 5 · 24

Map each mechanism to an earlier chapter. The bucket wheel is a high-inertia constant-torque drive riding through the impact of every bucket. The slewing drive is a large-inertia positioning drive, like the crane trolley but with a boom over a hundred metres long. The boom hoist adjusts the cutting height and is an active, counterweighted, four-quadrant load — Chapter 3 and 4. The crawlers advance the machine. The boom and discharge conveyors are multi-motor load-shared belts — this chapter. The supervisory control matches cutting rate, slew speed and conveyor speed. The wheel delivers at most 60 z n V_b phi cubic metres per hour — buckets times speed times volume times fill — and the conveyors must carry it. But the buckets fill only if the slew feeds them: the cut is crescent-shaped and thins as cos phi, so the slew must speed up as 1 / cos phi — twice as fast at 60 degrees. Tutorials Exercise 5.7 works the numbers.

Ask: which drive would you slow first if the discharge conveyor overloads? (The slew, to reduce the cutting rate.)

Timing: 5 minutes. Reference: Lectures section 5.4, Fig. 5.12.

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Common mistakes with conveyors

WATCH OUT

✗

MYTH

“Two identical motors on one belt share the load equally.”

✓

WHAT IS TRUE

Small differences in slip, gear ratio or pulley diameter make one take more — even motor while the other generates. Sharing needs master–slave or droop.

✗

MYTH

“Droop is inferior because it lets the speed sag.”

✓

WHAT IS TRUE

The sag is the sharing mechanism, and it needs no communication. Speed regulation against robustness: neither scheme is always better.

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

Chapter 5 · 25

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Two more traps

WATCH OUT

✗

MYTH

“More motor torque cures a slipping drive pulley.”

✓

WHAT IS TRUE

Slip is set by the tension ratio T₁/T₂ ≤ e^μθ. More torque makes it worse; the cure is take-up tension, wrap, lagging or a second drive.

✗

MYTH

“Starting a loaded conveyor is like starting it empty.”

✓

WHAT IS TRUE

A loaded belt has a large breakaway torque and huge inertia; an abrupt start sends a splice-damaging tension wave. Ramped, jerk-limited starts are a mechanical need.

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

Chapter 5 · 26

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Quick check — try before you look

SELF-CHECK 5.1–5.6

5.1

17.3 kN over a pulley with e^μθ = 3.39: slack-side tension the take-up must supply?

5.2

What keeps the belt from slipping on the drive pulley, and which sensor detects slip?

5.3

Why does a line of conveyors start from the discharge end and stop from the feed end?

5.4

Why must the start ramp of a long belt last several times its natural period?

5.5

Two drives with 3 % droop, momentarily at 105 % and 95 %: speed-reference difference?

5.6

Why is a pull-cord switch reset at the switch, not from the control room?

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

Chapter 5 · 27

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Summary

CHAPTER 5 · SUMMARY

Constant-torque load: size from the effective pull; power ∝ belt speed
DIN 22101 counts both strands, idlers and end losses; take-up T₂ ≥ F_e/(e^μθ − 1)
The belt is elastic: ramps of several belt periods; S-curve only beyond ~2
Protection along the belt; start against the flow, stop with it
Multi-motor: master–slave (stiff, needs the bus) or droop (robust, speed error)
Passenger systems: NC chains, EN 115-1 limits; excavator slews as 1/cos φ

KEY EQUATIONS

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

Chapter 5 · 28

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Practise this

CHAPTER 5 · NEXT STEPS

TD

Tutorials

Exercises 5.1–5.7: sizing and take-up, incline, start, load sharing, sequencing, escalator, excavator.

TP

Lab works

TP 5 — conveyor line and brick sorting on the PLC-200.

?

Before next week

Redo the self-check questions; then read Chapter 6 §6.1 on rolling mills.

Photo: byrev, CC0, Wikimedia Commons

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

Chapter 5 · 29

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Image credits

CHAPTER 5 · 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).
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
Bucket wheel excavator in winter — byrev, CC0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Bucket_wheel_excavator_in_winter.jpg
Copenhagen Metro escalators — Stig Nygaard from Copenhagen, Denmark, CC BY 2.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Copenhagen_Metro_escalators.jpg
Gondola of Skalnate pleso aerial tramway 1 — PetrS., CC BY-SA 3.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Gondola_of_Skalnate_pleso_aerial_tramway_1.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 5 · 30