Lecture 1

MATERIAL TRANSPORTION ENGINEERING

ASSOC. PROF. DR. HASHIM HUSSINASSOC. PROF. DR. HASHIM HUSSIN

Course Synopsis :•The course covers the material handling / transportation methods and equipments that are widely used in mining and mineral industries.

• Students are introduced to theories (where related), principles, mechanism and the performance of the equipments.

•The topics covered are belt conveyors, chain conveyors and bucket elevators, screw conveyors and elevators, shaking and vibratory conveyors, fluid transport, rope haul systems, monorails and aerial ropeways, locomotive haulage and hoist and mine winders.

•Students are also introduced to the basic of calculation of tonnage, speed, motor power, and the efficiency of the equipments.

•Designing of flow sheet using real plant examples with the awareness of its impact on environment at cost effective.

Course Outcomes (CO) :At the end of the course, the students should be able to:

distinguish and describe the materials handling methods that are widely used in mining and mineral industries.

explain the basic concept and principle in materials transportation

describe and apply the technology, types and characteristics of materials transportation equipments available in the market.

design a proper flow sheet to transport materials and ores from one place to another place or from one unit to another unit.

evaluate and calculate the equipment performance.

classify and apply knowledge of materials transportation to real situations with an awareness of its impact on environment at cost effective.

Asses a range of learning resources and to take responsibility for own learning with appropriate support.

explain and work effectively with others as a member of a group and meet obligations to others.

Teaching Plans / Syllabus :

No Topic ContentsTeaching Weeks

1 Introduction to materials handlingmaterials handling methods, basic concept applied mechanic

1

2 Belt conveyor introduction to the equipment, mechanism of the equipment, calculation of the power, tonnage, speed, wide, number ply. The efficiency of the belt conveyors. etc.

2,3

3 Chain conveyor and bucket elevatorintroduction to the equipmentmechanism of the equipmentcalculation of the power, tonnage, speed, the

efficiency of the chain conveyors. etc.

3,4

Teaching Plans / Syllabus :No Topic ContentsTeaching Weeks

4 Screw conveyors and elevators introduction to the equipment mechanism of the equipment calculation of the power, tonnage, speed, the efficiency of the chain conveyors. etc.

Shaking conveyor introduction to the equipment mechanism of the equipment calculation of the power, tonnage, speed, the efficiency of the shaking conveyors. etc.

Vibratory conveyor introduction to the equipment mechanism of the equipment calculation of the power, tonnage, speed, the efficiency of the vibratory conveyor. etc.

4 & 5

Teaching Plans / Syllabus :No Topic ContentsTeaching Weeks

5 Fluid transport introduction to the equipment mechanism of the equipment calculation of the tonnage, speed, pulp density, slope, efficiency etc.

5

6 Rope haulage systemo introduction to the equipmento mechanism of the equipmento calculation of the tonnage, speed, efficiency etc.

Monorails introduction to the equipment mechanism of the equipment

calculation of the tonnage, speed, efficiency etc.

Aerial ropeways introduction to the equipment mechanism of the equipment

calculation of the tonnage, speed, efficiency etc.

7 & 8

Teaching Plans / Syllabus :No Topic ContentsTeaching Weeks

7 Locomotive haulage introduction to the equipment mechanism of the equipment calculation of the ideal gradient, optimum gradient and work example.

8 & 9

8 Hoist and mine winders• introduction to the equipment• mechanism of the equipment• the mechanics of hoisting• calculation of the speed, and acceleration, drum torques and duty cycle

diagrams

10,11& 12

9 Designing of flow sheet, industrial examples, both local and international, with an awareness of health and safety, social, political, economic and impact on environment.

13 &14

**Week 6 – Mid-semester Break (1 week)

Assessment methods %

Test (2 tests)

Quizzes

20

Assignment2 assignments (individual and group)

20

Final Exam 60

Total 100

Contribution of assessment

• Exam 60 %

• Course work 40 %

References• Brook, N. 1971. Mechanics of Bulk Materials

Handling. London Butterworths.

• Handbook Society of Mining Engineers. 1979. New York.

• Hartman H.L., 1987. Introductory Mining Engineering. New York. John Wiley & Son.

• Ramlu M. A. 1996. Mine Hoisting. A.A Balkema/Rotterdam/Brookfield.

• Recent journals /publications related to this subject

Objective

• To introduce most of materials transportation equipments (mechanisms, principles and calculation) that normally used in mining, mineral processing plant, and quarrying.

• Give some idea to mineral resources engineering students to make a preliminary calculation to narrow the field before selecting the equipment.

Introduction

• Material transportation/handling methods are normally classified into 3 main groups ( from the standpoint of mechanics involved);

Continuous methods

Semi-continuous, or small batches methods and

Batch methods

• Some example of material transportation/handling equipments and techniques:

– Belt conveyor (penghantar tali sawat) – Chain conveyor (penghantar rantai)– Screw conveyor and feeder (penghantar skru dan penyuap)– Suction pipe (paip sedutan)– pressure pipe (paip tekanan)– suction pneumatic – pressure pneumatic– Monorails– bucket elevator– aerial ropeway– Chute– shaker conveyor– Locomotive– Hoists and lift

Belt Conveyor (penghantar tali sawat)

Chain conveyor (penghantar rantai)

Bucket elevator (pengangkat timba)

Shaker Conveyor (penghantar bergetar)

Suction Pipe (paip sedutan)

Screw Conveyor (penghantar skru)

Locomotive Haulage

Hoist

Mine Winder (pengangkat lombong)

Chute

Lecture 2

Belt Conveyor (Talisawat Penghantar)

Belt conveyor is basically an endless strap stretched between two drums.

Suitable for very short distances and low outputs.

It is necessary to support the top strand of the conveyor at regular intervals to prevent undue sagging and to reduce the spillage of material which may occur if the belt does not run truly, while maintaining a high carrying capacity.

these requirements usually met by trough idlers.

Idler (Pemelahu)

Consist of three separate rollers to support the belt and also bend it into a trough shape.

The two outer rollers are tilted upwards at an angle of 25o to 30o.

For very wide belts a design using five separate rollers is sometimes used.

For narrow belts two angled idlers only may be used.

• impact idlers = Pemelahu hentaman• trough idlers = Pemelahu paluh• return idlers = Pemelahu kembali

Idlers for small conveyor belt

Top strand (Lembar atas )

Bottom strand (Lembar bawah)

Material Belt Conveyor

(Talisawat Penghantar)

Normal Trough Idler (three idlers) Pemelahu paluh yang biasa (tiga pemelahu)

Belt Conveyor

Trough idler

Idler for wide belt conveyor (five idlers) Pemelahu untuk talisawat yang lebar (lima pemelahu)

Return Idlers (Pemelahu Kembali )

Trough Idler (Pemelahu Paluh)

Feed Chute (Pelongsor suapan)

Impact Idler (Pemelahu Hentaman )

Drive Drum (Gelendung Pemacu)

Idler Drum (Gelendung Pemelahu )

Bottom Strand (Lembar bawah)

Top Strand (Lembar atas )

Exercise 1

Lecture 3

• The driving drum relies on the friction between drum and belt to provide the drive to the belt.

• If the two tensions in the belt at the driving drum are P1 and P2 with P1 the bigger tension in the top strand, the limiting ratio of tensions when about slip is about to occur is given by:

θ

P1

P2

P1 / P2 = eµθ

Log10 P1 / P2 = 0.434µθµ is coefficient of friction or coefficient of grip between the belt & drum

θ Is the angle of wrap

• For a long conveyor with a large hauling duty P1 requires to be large.

• Value of P1 possible is obtained by three method:

– Increasing µ by lagging the driving drum with a suitable rubber-like material.

– Increasing the value of θ

– Increasing the value of P2

No slip P1/ P2 ≤eµθ

The value of θ can be increased by:

1. Using a snub pulley

θ

θ = 250o

2. Using more than one driving drum.

θ = θ1 + θ2 ;

θ=400~450o

θ2

θ1

θ1

θ2

θ3

θ = θ1 + θ2 + θ3; θ= ~600o

The value of P2 can be increased by

3. pre-tensioning belt

•By a screw tightening device at the tail end of the belt conveyor

2. Take-up loop

Tension

Rail mounted carriage

Driving drums

• as part of a belt storage or take-up loop

Driving drums

Weighted roller

3. Gravity tensioning device

•For higher powered conveyors a gravity operated tension devise may be used, either by pulled via steel cables on the loop take-up carriage or by a heavy roller mounted in a frame

Pemelahu Kembali (Return Idlers)

Pemelahu Paluh (trough idler)

Gelendung Pemacu (Drive drum)

Gelendung Pemelahu (Idler drum)

Lembar bawah (Bottom strand)

Lembar atas (top strand)

Pemelahu Hentaman (impact idler)

Pelongsor suapan

(feed chute)

Berat Pengambang (Counter weight)

Kapi ambil (take-up Pulley)

Factors affecting the use of belt conveyors

1. A straight line plan is usually required (some small deviations of a few degrees are possible). If the line of the conveyor system must be angled it is often necessary to use separate conveyors but some systems have a complex belt lacing which permits a fix angle at some point in the conveyor, the material being discharged from one section of the belt to the other as if the two sections were separate conveyor.

2. The angle of inclination of the conveyor is limited by the friction of the material on the belt, to about 25o. The maximum gradient used for a conveyor must allow for restarting with the belt loaded on the incline and sufficient frictional grip must be provided to overcome the component of the weight tending to pull the material down the belt and also to accelerate the material.

3. The maximum lump size is limited to about half the belt width.

4. The carrying capacity of the belt depends on how the material can be piled up on the belt width. As the belt is continuously passing over sets of supporting idlers, the materials is slightly disturbed all the time and tends to spread out on the belt. The carrying capacity of the conveyor is given by the equation;

T= a (m2) b (t/m3) v (m/s) = abv (t/s)

where T is the carrying capacity, a is the average cross-sectional area of material, b is the bulk density and v is the speed of the conveyor belt.

For a belt of width W the value of the area a varies approximately between w2/10 (high loading) and w2 /12 depending on the nature of the material. A blocky type material such as coal, broken rock, or ore can be piled onto the belt as shown in Fig.(a) whereas a smooth material such as particle tends to run out over the belt as shown in Fig.(b).

The value of b the bulk density in t/m3 is numerically equal to the relative density (g/cm3) but it relates to the density of broken material including air spaces, and not to the solid relative density

(a)

(b)

Material Density (t/m3)

CoalSolid coalGravelDry ashesWet ashesBroken sandstoneSolid sandstoneBroken limestoneSlagDry sand

0.81.351.4 – 1.70.55 – 0.650.7 – 0.81.352.41.451.351.6

5. The belt strength affects the maximum force which can be taken by the belt, and the value of the maximum force depends on the power required and the drive head frictional grip. The power required by a belt conveyor can be divided into three components.

i. power for the empty belt, We

ii. Power to convey the material, Wm

iii. Power to raise the material, Wr

The total power required by a belt is then WT = We + Wm ± Wr

• The value of Wr is written as plus or minus as if the material is being lowered and help to run the belt thus the power requirement for this is negative.

• The power as calculated above is the power required at the driving drum of the conveyor, and so the motor power required will be greater because of power losses in the gearing at the drive head.

• Assuming an efficiency of 90% for this gearing the otor power is then given by

W = WT / 0.9

• The power required to drive the empty belt depends on the total force required to move the empty belt, and on the belt speed.

• The force required : Ne = total weight on idlers x friction coefficient

Ne = Mi g µ e

= mi ( l + lx ) g µ e

And then the power required We = Ne v

We = mi ( l + lx ) g µ ev

Note: l the length of the conveyor is increased by lx = 45 m to allow for end pulley friction

• The power required to convey the material; Wm =mml g µ mv

• The value of mm the mass of material per unit length is obtained from

mm = T tan/s

v m/s

Therefore,

Wm = T l g µm v

v

Or Wm = T l g µm

the value of µm the friction coefficient is again 0.03 for well maintained conveyors but sometimes raised to 0.04 if the conditions are unfavourble.

• The power required to raised the material at the rate T through height h is obtained directly as

Wr = T g h kW

Effective belt tension Pe = WT kN

v

The maximum tension P1 is obtained from the formula

Pe = P1 – P2

For no slip to occur P1 < eµ θ = n

P2

P2 = P1/n

P1–P2 = P1 – P1/n = P1 (n-1)/n

P1= (n/n-1) Pe

The calculated value of P1 is then used to find the belt stree f (kN/m per ply).

Fabric Density (kg/m2) Stress (kN/m plyU.S. cotton

Rayon and cotton

Rayon, cotton, nylon

Nylon and cotton

Steel reinforced(Steel cords in rubber and fabric belt)

0.8140.9301.0431.2201.3951.744

1.191.63

0.9301.02

3.0

16.4 – 42.3

4.255.255.757.08.7510.5

12.2515.8

7.09.65

35.0

80-450

• Exersice 2 A conveyor is 600 m long and conveys coal of bulk density

0.8 t/m3 up a gradient of 1 in 60 at the rate of 220t/h. Determine suitable speed and strength for the installation.

(Note: assume the area of the material is w2/11, width of the belt conveyor for transporting coal is 0.75 m, total mass acting on the idlers 60 w = 45kg/m, µm =0.04, Ø= 440o)