Group D Section 1

8/7/2019 Group D Section 1

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BELT CONVEYOR

Prepared by:Dawit GudetaGedion TadelleBisrat YhannesDawit NegussieAnteneh Tesfaye

G/selase


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INTRODUCTION

A conveyor system is a common piece of mechanical handling equipmentthat moves materials from one location to another.

among the most commonly known conveyor systems1. Belt conveyors2. Line shaft roller conveyor3. Chain conveyor

What are belt Conveyors?

Belt conveyors are the most commonly used bulk hand conveyors inhistory due to their reliability, versatility, range of capacities.Materials from fine powders to large, lumpy stones can be handled on abelt conveyor.


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according to their path of motion belt conveyors are classified as:horizontal, inclined and combined(inclined- horizontal, horizontal-inclined andwith two or more bends )

BeltsDifferent types of textiles belts are employed in belt conveyors: camel

hair; cotton(woven or sewed), duck cotton and also rubberized textile beltsof various types

Recommended belt plies

Belt width in mm 300 400 500 650 800 1000 1200

9--14minimum and maximum numberof plies 3--4 3--5 3--6 3--7 4--8 5--10 6--12 7--12


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A belt conveyor consists of two or more pulleys, with a continuous loop of materials-the conveyor belt that rotates about them.

One or both of the pulleys are powered, moving the belt and the materialon the belt foreword.

The powered pulley is called the drive pulley while the unpowered pulleyis called the idler.

There are two main industrial classes of belt conveyors, those in generalmaterial handling such as those moving boxes along inside a factory and

bulk material handling such as those used to transport industrial andagricultural materials, such as grain, coal, ores, etc generally in outdoorlocations.


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The belt consists of one or more layers of materials.many belts in generalmaterial handling have two layers.

1. Carcass :- is under layer of material to provide linear strength and shape.

2. The cover :- is the over layer which oftenly is rubber or plasticcompounds specified by use of the belt.

Belts are joined directly on the conveyor by rawhide splicing, coldcementing with special glues, hot cementing(vulcanizing), rigid hingedmetal belt fastener of different design and type.In selection of the belt width should be analyzed two different case:

When we transport boxes or packagesWhen we transport bulky materials


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Essential parts of a belt conveyorFrameDriving pulleyTake up pulleyUpper idler roller or pulleysLower idler roller or pulleys

Drive unitBelt cleaners


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Belt conveyor system

Belt conveyors are similar in construction. It consists of a metal framewith rollers at either end of a flat metal bed and the belt is looped aroundeach of the rollers and when one of the rollers is powered by an electricmotor as shown in figure 1&2 below. The belt slides across the solid metalframe bed, moving products.

In heavy use application, the bed which the belting is pulled over arereplaced with rollers. The rollers allow weight to be conveyed as theyreduce the amount of friction generated from the heavier loading on thebelt.But during a long distance conveying operation, the belting is providedwith idle pulley or pulleys to increase tension to loaded side of belt.


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The intermediate roller supports should also have a characteristics of

- minimum resistance to motion of belt- low weight and simple design- high strength and resistance toward friction and- should have reliable service, life and easy maintenance.

Belt conveyors system are selected because of there advantage totransport (carry load) and transmit pull simultaneously, belt weight islight, permit high conveying speed and do not contain quickly wearingjoint.They also have the advantage in lower initial investment and runningcost and also can travel long distance. and are commonly used to

conveyor items with irregular bottom surface, items that would fall inbetween rollers or bags that sag between rollers.


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Figure 1 Figure 2

Figure 3


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Belt conveyors are now also manufactured with curved sections which usetapered rollers and curved belting to convey products around a corner, Asshown in figure 3 above. These conveyor systems are commonly used inpostal sorting offices and airport baggage handling systems.

Belt conveyor systems are commonly used in many industries includingautomotive, agricultural, food processing, electronics, pharmaceutical,bottling and canning, etc

Belt conveyor systems are used across a range of industries due to

there ability to safely transport materials from one level toanother with less expense.

can be installed almost anywhere can move loads of all shape, size and weight

they have advanced safety features that help to preventaccidents.


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Design of belt conveyor

This calculation is used to determineDimension of the belt

Motor power required

So to determine the above we must know the following data Characteristics of the conveyor load Average and pick calculated capacity per hour Geometry of the conveyor and its main dimension

Operation condition (humidity, location out door or indoor,method of feed and discharge, etc .


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Belt widthWhen conveying bulk load, the belt width is determined by the conveyor

capacity and size of the material conveyed. i.e. in case of pieces goods bynumber of pieces and their overall dimension.

The belt supported by flat idler a free flowing material will assume theshape of an isosceles triangle .

To prevent spill of the belt edges the base of the triangle is taken asb =0 .8B

the angle at the triangle base 1= 0 .35where

B is belt width and is static angle of repose of the load

To define possible spillage of the load on inclined belt, correction factorC1( depending of the conveyor slope)Cross sectional area of the load on a flat belt is( F1)

F1=(bh/2)C1


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Substituting the above valuesF1= 0 .8B*0 .4BC1 tan 1

2then finally we get the simplified equation-

F1= 0 .16B^ 2 C1 tan( 0 .35 )

On a belt supported by troughed idler the cross section area of the load Fis equal to the sum of the area of trapezoid F 2 and triangle F 1.If the inclined side of the roller is 2 0 and the length of the central roller

lr 0 .4BThe total cross sectional area will be :

F= F1+F2Substituting F1 and F2 then,

F = 0 .16 B^2 C1tan 1+ 0 .0 435B^2F = B^2[0 .16C1tan( 0 .35 ) + 0 .0 435]


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Substituting these value of cross sectional area in the equation of conveyance of material in a continuous stream.

Qf = 3600 FV Where Qf- capacity of the belt

F cross sectional area of the loadV speed of the belt

- load characteristicsBut F= F1 for a belt supported by flat idlerQf = 3600 * F1 * V *Qf = 3600 * 0 .16 Bf^2 * C1 * tan( 0 .35 ) * V * Bf = Qf m

576C1 V tan( 0 .35 )


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When the conveyor with intermediate discharge by means of plough beltspeed should not be exceed 1.25-1.6

For rough estimation it can be taken = 45For this value of C1 taken from table C1 = 0 .85Substituting the angle value in to the flat belt equation

Bf = Qf m576C1 V tan( 0 .35 )

Bf = Qf m160 V C1

Applying the same procedure for the troughed idler equationBtr = Qtr m

324 V C1


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For a belt supported by troughed idlerQtr = 3600 FV

Where F = F1 + F2, butF1 = 0 .16B tr^2C1tan( 0 .35 )F2 = 0 .0 435B tr^2 thenF = 0 .16B tr^2C1tan( 0 .35 ) + 0 .0 435B tr^2F = Btr^2[0 .16C1tan( 0 .35 ) + 0 .0 435]

Qtr = 3600 * F * V * Qtr = 3600 Btr^2[ 0 .16C1tan( 0 .35 ) + 0 .0 435] * V *

Btr = Qtr m160 V [3.6C1tan( 0 .35 ) +1]

To get the value of C1(inclination) based on the following table

C1 1.0 0.95 0.9 0.85

Inclination in degrees 0 -10 10 -15 15-2 0 > 20


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Speed of the beltDepend of the nature of the material to be transported, the width of thebelt , the conveyor inclination and if there are or not intermediate point of unloading the recommended speed of the belt is given in the table below.

Note that V = m/s

300-400 500 650 800 1000-12001400-1800

mean abrasive 1.5 2 2.25 2.5 3 3.3high abrasive 1.5 2 2.25 2.5 3 3

mean abrasive 1.5 2 2.25 2.5 3.5 3.5high abrasive 1.5 2 2.25 2.5 3.3 3.3

2 2.5 3 3.5 4 4lightheavy

Width of the belt in mmMaterial Charactersticsbig and medium size lumped,non abrasive 2 2.5 3 3.5 4 4

Fragile loads 0.75-1.25

big sizelumped

mean sizelumpedGrain,Rye wheatthinlumped

1.251.5


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idlerBelt is supported by idler, in rare case by solid wood or steel run way.Idler is used in conveyors handling bulk loads less frequently unit loadAccording to the location of the conveyor idler can be classified as

UpperLower

Upper idler support loaded strand of the belt

Lower idler support return strand .

400 500 650 800 1000 1200 1400 1600-20002 1300 1300 1200 1200 1100 1100 1000 1000

bulk weight of load(ton/m)

spacing l for width b in mm


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Drive units

The pulley motor and transmission gear between the motor and pulleyFriction

the motive power in belt conveyor is transmitted to the belt byfriction as the belt wraps around the pulley which is rotated by themotor.

Friction drive theory (Euler s law) There is no belt slip on the pulley when St < Ssl

Where, St and Ssl tight and slack side tensions of the belt atthe driving pulley

-coefficient of friction-wrap angle of the belt in radians-Naperian system of logarithm


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Neglecting losses on the driving pulley due to belt stiffness, the peripheralpull

w =St-Ssl< Ssl -Sl=Ssl( -1)orw ( -1)St/

When there is pressure of rollers (p) on the pulley (snubbed)

The Concentrated friction forced of the belt against the pulley acting inthe spot where the roller pushes will be:

p The Euler equation thus assumes the form of :

St


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Neglecting belt stiffness W


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Driving pulley :made of cast iron or welded of sheet metal

Diameter is dependent on the number of plies Dp>ki in mmWhere dp-diameter of the pulley

I-number of pliesK-factor of proportionality, usually b/n values of 125to15 0 for 2-6 plies k =125for 8-12 plies k =150


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Resistance on separate section of conveyor

When the belt runs on rectilinear sections over idlers resistance losseswill be caused by the friction;In the roller bearing.The belt rolling on the idler rollers and by

It s bending over the roller.The forces of resistance on the rectilinear section of the inclinedconveyer can be determined by the equation;for loaded strand: W 1=(q+qb+q p)*Lh*w (q+qb)*H

W1=(q+qb+qp)*Lw cos (q+qh)*Lsinfor idler strand: W i=(qb+qp)*Lh*w qb*H

W1=(qb+qp )Lw cos qbLsin


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where q ,qb ,qp are weight of the load ,belt and rotatingparts of the idler respectively, q p loaded and q idlers strand.

Lh is length of horizontal projection of rectilinearsection, m.

H is difference elevation at the beginning and end of the sections , m.

w is coefficient of resistance of the belt on the roller.

is the angle of inclination of the conveyor from horizontal.In the equation the plus sign is taken for up ward motion of the belt andthe minus sign for down ward.

The weight of the idler rotating parts depends on their design andspecification. For rollers of 159mm diameter it can roughly obtained as afunction of belt width .

For troughed idler; Gp =10 B+7 k.gFor flat idler ; Gp =10 B+ 3 k.g


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The weight per meter of the idler rotating part is determined fromq p=GP/l and q p=Gp/l2 k.g/m.

where l and l2 are the idlers spacing on the loaded and idle runs in m.

Table showing resistance factor for idlers on roller bearing.

O perationcondition

C haracteristics of the operating condition Factor wof theidler

flat troughed

favorable Operation in clean dry premise ,non abrasive

dust

0 .0 18 0 .0 2

medium Operation heated premise in the presence of alimited amount of abrasive dust

0 .0 22 0 .0 25

adverse Unheated premise, large amount of abrasivemoisture or other factors

0 .0 35 0 .0 4


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If the loaded strand of the belt slides on a stationary runway witha friction factor resistance on a rectilinear inclined section is

W1=(q+qb)*(Lr*+H) and on a horizontal

W1=(q+qb)*Lr* )belt friction factor on steel runway is from 0 .35 to 0 .6; on planed

wooden runway from 0.4 to 0.7

T he resistance on the deflecting pulleys is determined fromSsl=St+Wc but Wc=kSt

substituting and rearranging the aboveSsl=(1+k)*St withk=1.05 to 1.07 for 180 wrap angle and

k=1.03 to 1.05 90 wrap angle

Resistance on driving pulleys with the losses in the bearing can be

determined byWdr =k(St+Ssl) Where St istention in spot where the belt runs on the roller bank


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T he resistance on the loading feeders is taken asW=C1*Q*N*(v-v )/3.6 g

Where C1 =factor that takes in to account the frictionbetween the material and the sides.Q=equipment capacity in KN/h. V=belt speed (m/s).

V =initial speed of the material.


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DETERMINATION OF BELT TENTION,PULL & MOTOR POWERThe detailed calculation of belt tension started at the point where beltruns off the driving pulley & completed where it enters the pulleySt=tension the tight side of the diver drumSsl=tension on the slack side of the driver drumSt=k Ssl + A

where k =Numerical value determined by the magnitudeof load resistanceA =numerical quantity in Kg , sum of linear resistances many rectilinearsectionsCHECKING THE NUMBER OF PLIES

i=> KSmax/BKtWhere K =safety factor

Smax=maximum calculated tensionKt =braking limit of belt[n/cm]I =no of plies


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F=((q+qb)/8*Stmin)*l A

F= sag L=length between the idlers q= weight of the load qb = weight of the beltPractically for bulk loads maximum sag is obtained

Fmax (0 .0 25 0 .0 3)When piece goods are transported the weight of the belt is uniformly

distributed load while that of material ( load ) is concentrated load. If the section between the idlers holds only one unit of load weight Gload, the total sag will be

F=qb*l/8Slmin +Gl/4Slmin B

1) If the section between the idlers holds several units, the belt sag isdetermined from equation .A

2) If the unit loads are available use .B


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MOTER POWER

N=

W v*hp/75ng=

W v/10

2ng kwThe rating of the motor installed should exceed by 15-2 0%

DRUM DIAMETEROnce the diameter of the drum is calculated

D t ikDt 2Wo/uaBPadm

Where i =number of pliesK=coefficient of pliesWo =coefficient of pulloutu= friction coefficient between the belt and druma=angle of embranceB=wedith of the belt

Padm =allowable pressure over the drum in Pascal ( 0 .1- 0 .11mpa)

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Group D Section 1