Belt Conveyor Systems

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• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering

Sixth Edition

CHAPTER

H

Construction Planning, Equipment,and Methods

BELT-CONVEYOR SYSTEMS

By

Dr. Ibrahim Assakkaf

ENCE 420 – Construction Equipment and Methods

Spring 2003Department of Civil and Environmental Engineering

University of Maryland, College Park

CHAPTER H (handout). EQUIPMENT FOR PUMPING WATERENCE 420 ©Assakkaf

Slide No. 2

Belt-Conveyor Systems

• Belt-conveyor systems are used extensively in the field of

construction.

• Belt-conveyor systems frequently provide the most

satisfactory and economical method of handling and

transporting materials, such as earth, sand, gravel, crushed

stone, mine ores, cement, concrete, etc.

• Because of the continuous flow of materials at relatively highspeeds, belt conveyors have high capacities.

Note: During the construction of the Channel Tunnel (between

England and France) conveyors were used to move up

to 2,400 tons of spoil per hour from the tunnel headings.




Slide No. 3

Belt-Conveyor System

The essential parts of a belt-conveyor system

include:

– A continuous belt, idlers

– A driving unit, driving and tail pulleys

– Take-up equipment

– A supporting structure.


Slide No. 4

• A conveyor for transporting materials a short distance may

be a portable unit or a fixed installation. This machine is

available in lengths of 33-60 ft, with belt widths of 18, 24, and

30 in. It is self-powered with a gasoline-engine drive through

a shaft and gearbox to the driving pulley. The operating

features include swivel wheels, a V-type truck, a hydraulic

hoist, a low mast height, and anti-friction bearings

throughout.

Portable Belt-Conveyor System




Slide No. 5

El Abra, Chile

Conveyor Belt


Slide No. 6

El Abra, Chile

Conveyor Belt




Slide No. 7

Ruhrkohle,

Germany

Newman,

Western

Australia

Conveyor Belt


Slide No. 8

THE ECONOMY OF TRANSPORTING

MATERIALS WITH A BELT CONVEYOR

Is the use of a belt conveyor

the most dependable and

economical when compared

with other methods?•The proper way to answer this

question is to estimate the cost of

transporting the material by each

method under consideration




Slide No. 9

• The net total cost of the conveyor systemwill include: – The installed cost of the system

– An access road for installing and servicing the system

– Maintenance

– Replacements and repairs

– Fuel, or electrical energy

– Labor

– Interest on the investment, plus taxes and insurance,

– Cost of obtaining a right-of-way for the system.

Less

– The net salvage value of the system upon completion of

the project.




Slide No. 10

•The unit cost of moving the material,

per ton or cubic yard, may be

obtained by dividing the net total cost

of the system by the quantity of

material to be transported.






Slide No. 11

• In the construction of the Bull Shoals Dam

in Texas, more than 4.5 million tons of

aggregate was transported on belt

conveyers at a reported cost of $0.045 per

ton-mile

• The contractor saved $560,00 on purchase

and installation of the conveyer system as

compared with a fleet of trucks, plus a haul

road• Furthermore, there was a saving of

$375,000 on labor




Slide No. 12

• The "belt" is the moving and supporting

surface on which the material is

transported.

• Many types, sizes, and grades are

available from which the most suitable beltfor a given service may be selected.

• Belt-conveyor systems are based on the

location of the drive pulley, the number of

drive pulleys, and the take-up method of

maintaining the necessary tension in the

belt.

Components of Conveyor Belt




Slide No. 13

Length

Tail drive Snub pulley

Screw takeup

Head pulley

Belt

Feeder

H e i g h t

Components of Conveyor Belt


Slide No. 14

• Belts are manufactured by joining several

layers or plies of woven cotton duck into a

carcass which provides the necessary

strength to resist the tension in the belt.

•The layers are covered with an adhesive,which combines them into a unified

structure.

• Special types of reinforcing, such as

rayon, nylon, and steel cables, are

sometimes employed to increase the

strength of a belt.

Conveyor Belt




Slide No. 15

Conveyor Belt


Slide No. 16

• A belt should be selected that is wide enough to

transport the material at the required rate. Most

belts used on construction projects travel over

troughing rollers to increase the carrying

capacities. The number of tons that can be

transported in an hour is determined by:

T = weight of material, tons per hour

A = cross-sectional area of material, sq ft.

S = speed of the belt, ft per min

W = weight of material, lb per cu ft

000,2

60 ASW T =

Conveyor Belt




Slide No. 17

3

w

3

w

3

w

200

α = 200 to 300

α = angle of slope

Cross-section Area of a Load on a

Conveyor Belt

Surface of Load

Edge of Load

0. 0 5 w

+ 1

Line B

Figure 1


Slide No. 18


Conveyor Belt

• The area of the cross section depends on:

—the width of the belt

—angle of repose for the material

—the extent to which the belt is loaded to capacity• To avoid side spillage, the material should

not be placed closer than 0.05w + 1 inches

from the sides of the belt

• The area of surcharge is the area above

the B line in Figure 1




Slide No. 19

Table 1. Loaded Belt Material Cross-sectional Area for troughing Idlersat an Angle of 200


Conveyor Belt


Slide No. 20

Example 1

Determine the carrying capacity of a 42-in

belt, moving 100 fpm, loaded with sand

weighing 100 lb per cu ft, with 200 angle of

repose.

Hour

Ton5.334

2000

)100)(100)(115.1(60

000,2

60===

ASW T




Slide No. 21

Belt Idlers


Slide No. 22

Belt Idlers

• Idlers provide the supports for a belt conveyor.

• For the load-carrying portion of a belt the idlers are

designed to provide the necessary troughing,

whereas for the return portion of a belt idlers provide

flat supports.

• The essential parts of a troughing idler include therolls, brackets, and base.

• Antifriction bearings are generally used in idlers, with

high pressure grease fittings to permit periodic

lubrication of the bearings.




Slide No. 23

Figure 2. Belt Idlers, (A) Heavy-duty Troughing, (B) Return

Belt Idlers


Slide No. 24

Belt Idlers

• The rolls may be made of steel tubing or cast iron,

either plain or covered with a composition, such as

rubber, where it is necessary to protect a belt against

damage due to impact.




Slide No. 25

• Spacing of Idlers:

Troughing idlers should be spaced close enoughto prevent excessive deflection of the loaded belt

between the idlers (Table 2)

• Training Idlers:

Sometimes a conveyor belt is operated under

conditions which make it difficult to keep the belt

centered on the troughing idlers. If the conditions

cannot be corrected sufficiently to keep the belt

centered, it may be necessary to install training

idlers, spaced 50-60 ft apart (Figure 3)

Belt Idlers


Slide No. 26

Table 2. Recommended Maximum Spacing of Troughing Idlers

Weight of Material (lb/ft3)Width of Belt

(in) 30 to 70 70 to 120 120 to 150

14 5.50 ft 5.00 ft 4.75 ft

16 5.50 ft 5.00 ft 4.75 ft

18 5.50 ft 5.00 ft 4.75 ft

20 5.50 ft 5.00 ft 4.75 ft

24 5.50 ft 5.00 ft 4.75 ft

30 5.00 ft 4.50 ft 4.25 ft

36 5.00 ft 4.50 ft 4.25 ft

42 4.50 ft 4.00 ft 3.75 ft

48 4.00 ft 3.25 ft 3.00 ft

54 4.00 ft 2.75 ft 2.50 ft

60 4.00 ft 2.25 ft 2.00 ft

Belt Idlers




Slide No. 27

Figure 3. Training Idlers, (A) Reversible Troughing, (B) Return

Belt Idlers


Slide No. 28

• Idler Friction:

In analyzing a belt conveyor to determine the

horsepower required, it is necessary to include the

power needed by the idlers. This power will

depend on the type and size of idler, the kind of

bearings, the weight of the revolving parts, the

weight of the belt, and the weight of the load.

Table 3. Friction Factors for Idlers Equipped with Antifriction

BearingsDiameter of Idler Pulley

4 in 5 in 6 in 7 in

FrictionFactors

0.0375 0.036 0.030 0.025

Belt Idlers




Slide No. 29

Example 2

Consider a conveyor 100 ft long, 30 in-wide

belt weighing 6.8 lb per ft. The load will

weigh 100 lb/ft3, or 54 lb per foot of

conveyor. The revolving parts will weigh 50

lb for a troughing idler and 31 lb for a return

idler. Both idlers are 6 in in diameter

From Table 3, for 6 in-diameter idler:

Idler Friction Factor = 0.030


Slide No. 30

From Table 2, for belt width = 30 in, and

weight of material = 100 lb/ft

No. of Idlers Required = 100 ÷ 4.5 = 22 Idlers

Add Extra Idlers at Loading Point = 3 IdlersTotal No. of Troughing Idlers = 25 Idlers

No. of Returning Idlers = 100 ÷ 10 = 10 Idlers

Example 2 (continued)




Slide No. 31

Total Weight of the Revolving Parts of Idlers:

Troughing: 25 X 50 = 2500 lb

Return: 10 X 31 = 310 lb

Weight of Belt: 200 X6.8 = 1360 lb

Weight of Load: 100 X 54 = 5400 lb

Total Weight = 8,320 lb

Force Required to overcome Idler Friction:

Force = 8,320 X 0.03 = 249.6 lb



Slide No. 32

For a belt speed of 100 fpm:

The horsepower required to overcome idler friction:

For other belt speeds, the required horsepower:

lb-ft24,960249.6100 per Minutedq'Energy ReThe =×=

hp76.0000,33

960,24 == P

=

100

fpmspeed,76.0 P





Slide No. 33

• The total external power required to drive a loadedbelt conveyor is the algebraic sum of the power

required by each of the following:

1. To move the empty belt over the idlers

2. To move the load horizontally

3. To lift or lower the load vertically

4. To turn all pulleys

5. To compensate for drive losses

6. To operate a tripper, if one is used

•The power required for each of these operationscan be determined with reasonable accuracy for

any given conveyor system.

Power Required to Drive a Belt

Conveyor


Slide No. 34

• The power required to move an empty conveyor belt over the

idlers will vary with the type of idler bearings; the diameter

and spacing of the idlers; and the length, weight, and speed

of the belt. The energy required to move an empty belt is:

E = energy, ft-lb per min

L = length of conveyor, ft

S = belt speed, fpm

C = idler-friction factor, from Table 11-6

Q = weight of moving parts per foot of conveyor

000,33

(hp)orlb/min)-(ft LSCQ P LSCQ E ==

Power Required to Move an Empty

Belt




Slide No. 35

• This is valid for systems which operate attemperatures above freezing (32ºF). A

correction factor must be added for cold

weather applications. The factor is about 1.2

at 0º'F and goes to 1.5 at -15ºF.

Note: Representative values of Q are given in Table 4


Belt


Slide No. 36

Table 4. Representative Values of Q


Belt




Slide No. 37

Determine the horsepower required to move a

30-in wide belt on a conveyor whose length is1,800 ft, equipped with 5-in diameter idler

pulleys with antifriction bearings. Assume a

belt speed of 100 fpm.

From Table 3, for 5 in-diameter idler:

Idler Friction Factor = 0.036

From Table 4, for 30-in wide belt, Q = 26 lb/ft

The power required to move the empty belt:

hp1.5000,33

)26)(036.0)(100(800,1

000,33===

LSCQ P

Example 3


Slide No. 38

The power required to move a load horizontally may be

expressed as:

where W = weight of the load in pounds per foot of belt Above equation may be expressed in terms of the load

moved in tons per hour.

T = tons of material moved per hour

SW = pounds of material moved per minute

60SW = pounds of material moved per hour

000,33

LSCW P =

100

3

000,2

60 SW SW T ==

Power Required to Move a Load

Horizontally




Slide No. 39

Also

The horsepower required to move a load

horizontally:

where

L = length of conveyor

C = idler friction factor

T = tons of material moved per minute

NOTE: Table 11-9 gives values for the horsepower required to move loads

horizontally on conveyor belts

3

100min permovedMaterialof Pounds

T SW ==

990)000,33(3

100 LCT LCT P ==

Power Required to Move a Load

Horizontally


Slide No. 40

•When a load is moved up an inclined

belt conveyor, the power required may

be divided into two components:

1. The power required to move theload horizontally and

2. The power required to lift the load

through the net change in elevation.

Power Required to Move a Load Up

an Inclined Belt Conveyor




Slide No. 41

Horizontal Power

Vertical

Power

Figure 4. Two Components of Power Required




Slide No. 42

• The power required to move the load

horizontally may be determined from previous

equation.

• The power required to lift the load through the

net change in elevation may be determined by:

where H = net change in elevation, in ft.

990)000,33(3

100

:horsepower theand

3

100lb/min)-(ft

TH TH P

TH H SW E

==

=×=






Slide No. 43



Note: If the load is moved up an inclined

conveyor, the power must be supplied

from an outside source. If the load is

moved down an inclined conveyor, the

power will be supplied to the belt by the

load.


Slide No. 44

• If the load is moved up an inclined

conveyor, the power must be

supplied from an outside source• If the load is moved down an

inclined conveyor, the power will

be supplied to the belt by the load.






Slide No. 45

• A belt conveyor may be driven through the

head or tail pulley or through an intermediatepulley.

• In the event that high driving forces are

required, it may be necessary to use more

than one pulley, with the pulleys arranged in

tandem to increase the area of contact with the

belt.

• Smooth-faced or lagged pulleys may be used,

depending on the desired coefficient of frictionbetween the belt and the pulley surface.

Driving Equipment


Slide No. 46

• The pulley may be driven by an electric motor,

or a gasoline or diesel engine.

• It is usually necessary to install a suitable

speed reducer, such as gears, chain drives, or

belt drives, between the power unit and thedriving pulley.

• The power loss in the speed reducer should be

included in determining the total power

required to drive a belt conveyor. This power

loss may be 5-10% or more, depending on the

type of speed reducer.

Driving Equipment




Slide No. 47

• The coefficient of friction between the steel

shaft and bearings is approximately 0.10• Effective Driving Force

– When power is transmitted from a driving pulley to a

belt, the effective driving force, which is transmitted

to the belt, is equal to the tension in the tight side

less the tension in the slack side of the belt. The

effective driving force is given by

where T e = effective tension or driving force between pulley & beltT 1 = tension in the tight side of the belt

T 2 = tension in the slack side of the belt

21 T T T e −=

Driving Equipment


Slide No. 48

• The coefficient of friction between the rubber

belt and a bare steel or cast iron pulley is

approximately 0.25 to 0.35

• For a driving pulley with a given diameter and

speed, the effective tension T e required to

transmit a given horsepower to the belt can bedetermined from

where P = horsepower transmitted to belt

D = diameter of pulley, ft

T e = effective force between pulley and belt, lb

N = rpm (revolutions per minute)

000,33

N DT P eπ

=

Driving Equipment




Slide No. 49

•Tension Factor, F : – The ratio T 1/T e is defined as pulley tension factor.

This factor varies with:

• pulley surface

• bare or lagged

• arc of contact between belt and pulley

– Values for the factor F are given in Table 5

eT

T F 1=

Driving Equipment


Slide No. 50

Table 5. Tension Factors for Driving PulleysArc of Contact

(degrees)

Bare Pulley Lagged Pulley Arc of Contact

(degrees)

Bare Pulley Lagged Pulley

Single-pulley Drive Tandem Drive

200 1.71 1.42 360 1.26 1.13

210 1.70 1.40 380 1.23 1.11215 1.65 1.38 400 1.21 1.10

220 1.62 1.35 450 1.18 1.09

240 1.54 1.30 500 1.14 1.06

Driving Equipment




Slide No. 51

Determine the minimum tension in the tightand slack sides of the belt if the required

effective force T e is 3,000 lb and the arc of

contact is 2100. Assume a bare pulley.

From Table 5, for 2100 arc of contact:

F = 1.70

T 1 = FT e = 1.70 X 3,000 = 5,100 lb

T 2 = T 1 - T e = 5,100 - 3,000 =2,100 lb

Example 4


Slide No. 52

• A belt conveyor includes several pulleys, around

which the belt is bent.

• For the shaft of each pulley there is a bearing

friction that requires the consumption of power.

• The power required will vary with the tension in the

belt, the weight of the pulley and shaft, and thetype of bearing, babbitted or antifriction.

• For a given conveyor the friction factors for each

pulley may be determined reasonably accurately,

and from this information the additional power

required to compensate for the loss due to pulley

friction may be obtained.

Power Required to Turn Pulleys




Slide No. 53

• Because of the tendency of a conveyor belt to

elongate after it is put into operation, a method ofadjusting for the increase in length must be

provided.

• A screw take-up may be used to increase the

length of the conveyor by moving the head or tail

pulley. This adjustment may be sufficient for a

short belt but not for a long belt.

• Another take-up, which is more satisfactory,

depends on forcing the returning belt to travel

under a weighted pulley, which provides a uniformtension in the belt regardless of the variation in

length.

Conveyor Belt Take-Ups


Slide No. 54

• If a belt conveyor is operated on an incline, it is

advisable to install a holdback on the driving pulley

to prevent the load from causing the belt to run

backward in the event of a power failure.

• A holdback is a mechanical device which permits a

driving pulley to rotate in the normal direction butprevents it from rotating in the opposite direction.

• The operation of a holdback should be automatic.

At least three types are available. They are the

roller, ratchet, and differential band brake, all of

which operate automatically.

Holdbacks




Slide No. 55

• A holdback must be strong enough to resist the

force produced by the load less the sum of theforces required to move the empty belt, to move

the load horizontally, to turn the pulleys, to drive

the tripper, and to overcome drive losses.

• If a belt conveyor is operated on a decline, the

effect of the load is to move the belt forward. If this

effect exceeds the total forces of friction, it will be

necessary to install a suitable braking unit to

regulate the speed of the belt.

Holdbacks


Slide No. 56

• The purpose of a feeder is to deliver material to a belt

at a uniform rate.

• A feeder may discharge directly onto a belt, or it may

discharge the material through a chute in order to

reduce the impact of the falling material on the belt.

• Several types of feeders are available, each of which

has advantages and disadvantages when compared

with another type. Among the more popular types are

the following:

– Apron

– Reciprocating

– Rotary vane

– Rotary plow

Feeders




Slide No. 57

• When it is necessary to remove material

from a belt conveyor before the materialreaches the end of the belt, a tripper should

be installed on the conveyor.

• A tripper consists of a pair of pulleys which

are so located that the loaded belt must

pass over one pulley and under the other.

As the belt passes over the top pulley, the

load will be discharged from the belt into an

auxiliary hopper or chute.

Trippers


Slide No. 58

• A tripper may be stationary or a traveling

type. The latter type may be propelled by a

hand-operated crank, a separate motor, or

the conveyor belt. If a tripper is installed on

a conveyor, additional power should beprovided to operate it.

Trippers

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Belt Conveyor Systems