JJ309 Fluid Mechanics Unit 6

7/30/2019 JJ309 Fluid Mechanics Unit 6

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BELTING J 3010/6/1

UNIT 6

BELTING

OBJECTIVES

General Objective : To understand and apply the concept of belting

Specific Objectives : At the end of this unit you will be able to:

> state the difference between open and close belt.

> explain that power transmitted by the flat belt and V belt.

> explain that ratio tension for flat and V belt.

> calculate power transmitted by the belt with consider

centrifugal force.


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BELTING J 3010/6/2

INPUT

6.0 INTRODUCTION

In factories, the power has to betransmitted from one shaft to another,

then belt driving between pulleys on

the shafts may be used.

The pulley rotating shaft is called driver. The pulley intended to rotate is

known, as follower or driven. When the driver rotates, it carries the belt thatgrip between its surface and the belt. The belt, in turn, carries the driven

pulley which starts rotating. The grip between the pulley and the belt is

obtained by friction, which arises from pressure between the belt and the

pulley.

(a) Types of belts

The flat belt is mostly used in the factories and work shops,

where a moderate amount of power is to be transmitted, from one

pulley to another, when the two pulleys are not more than 10 m a

part.

The V-belt is mostly used in the factories and work shops

where a great amount of power is to be transmitted, from one pulley

to another, when the two pulleys are very near to each other.


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C

BELTING J 3010/6/3

The types of belts:-

a. flat belt

b. V belt

6.1 LENGTH OF AN OPEN BELT DRIVE

A

B

K

1 2F

01 02

r2r1

D

E

d

Fig.6.1 0pen belt drive

01 and 02 = Centers of two pulleys

r1 and r2 = radius of the larger and smaller pulleys

d = Distance between 01 and 02L = Total length of the belt.Angle A0102 = 1Angle B02C = 2Angle AFE = 1 (radian)

Angle BCD = 2 (radian)


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BELTING J 3010/6/4

We know that the length of the belt,

L = Arc AFE + ED + Arc DCB + BA

r1

r2Cos 1

= dr1

r2

1 = cos-1

d

1 = 2 - 21= 2 ( - 1) (radian)

2 = 1 = 2 2

(radian) Arc AFE = r1

1

Arc DCB = r 2 2

And ED = BA = K02KO

2Sin 1 =d

K02 = d Sin 1

Finally the total of length of belt,

L = Arc AFE + ED + Arc DCB + BA= r11+ d Sin 1+ r2 2 + d Sin 1= r11 + r2 2 +2d Sin 1


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Example 6.1

Find the length of belt necessary to drive a pulley of 480 cm diameter running

parallel at a distance of12 meter from the driving pulley of diameter 80 cm.

This system is an open belt drive.

Solution 6.1

A

KB

F 1

01

r1

02 2 C

r2

D

E

d

Fig.6.2 0pen belt drive

01 and 02 = Centers of two pulleysr1 and r2 = radius of the larger and smaller pulleys

d = Distance between 01 and 02L = Total length of the belt.

Angle A0102 = 1Angle B02C = 2Angle AFE = 1 (radian)Angle BCD = 2 (radian)

Radius of smaller pulley = 80/2 = 40 cm.Radius of larger pulley = 480/2 = 240 cm.

Distance between the pulleys, d = 12m = 1 200cm

We know that the length of belt is,


Cos 1 =r1 r

2

d


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=240 40

1200


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1= cos-

1

240 40

1200

= cos-1

0.16667= 800

1= 1.396radian

1 = 2 - 21= 2 ( - 1.396 ) (radian)

= 3.491 radian

2 = 21= 2 2 (radian)

= 2 (1.396 )= 2.792 radian

Arc AFE = r11

= 240 x 3.491= 837.84 cm

Arc DCB = r 2 2= 40 x 2.792

= 111.68

cm

And ED = BA = K02KO

2Sin 800 =

d

K02 = d Sin 1=1 200 x 0.98481= 1181.77

m

Finally the total of belt length is


= r1 1+ d Sin 1+ r2 2+ d Sin 1

= r1 1 + r2 2 +2d Sin 1= 240 x 3.491 + 40 x 2.792 +2 x 1181.77

= 837.84 + 111.68 + 2363.54

= 3313.06 cm= 33.13 m


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6.2 LENGTH OF CLOSE BELT DRIVE

K

A

D

F01 1 2 02

C

r2

r1B

E

d

Fig.6.3 Cross belt drive


d = Distance between 01 and 02

L = Total length of the belt.Angle A0102 = 1Angle D0201 = 2

Angle AFE = 1 (radian)


We know that the length of the belt is


Cos 1=

r1+

r2

d

r1+

r21= cos-1

d

1 = 2 - 21= 2 ( - 1) (radian)

1 = 2


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Arc AFE = r11

Arc DCB = r 2 2

And ED = BA = K02KO

2Sin 1 =d

K02 = d Sin 1

Finally the total of length belt,


= r1 1+ d Sin 1+ r2 2 + d Sin 1= r1 1 + r2 2 +2d Sin 1

Example 6.2

Find the length of belt for a cross belt drive system. The diameter of

the drive pulley is 480 cm which running parallel at a distance of12 meter from the driving pulley which has a diameter of 80 cm.

Solution 6.2

K

A

D

1 2 CF

01 02

r2r1

B

E

d

Fig.6.4 Cross belt drive


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d = distance between 01 and 02L = Total length of the belt.

Angle A0102 = 1Angle D0201 = 2Angle AFE = 1 (radian)


Radius of smaller pulley = 80/2 = 40 cm.Radius of larger pulley = 480/2 = 240 cm.

Distance between the pulleys, d = 12m = 1 200cm

We know that the length of the belt,


Cos 1 =r1+

r2

d

=240 +40

1200

1= cos-1 240 +40

1200

= cos-1

0.23333

= 76.501= 1.335

radian

1 = 2 -

21= 2 ( - 1) (radian)

= 2 ( - 1.335) (radian)= 3.613 radian

1 = 2

Arc AFE = r11

= 240 x 3.613

= 867.12 cm

Arc DCB = r2 2= 40 x 3.613= 144.52 cm

And

ED = BA = K02


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KO2

Sin 1 =d

K02 = d Sin 76.50 =1 200 x 0.9724

ED = BA = K02 =1166.88 cm


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Finally the total of length belt,


= r1 1 + d Sin1+ r2 2 + d Sin 1= r1 1 + r2 2 +2d Sin 1= 240 x 3.613 + 240 x 3.613 + 2 x 1 200 Sin 76.50

= 867.12 + 144.52 + 2333.76= 3345.4 cm

L = 33.45 m


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Activity 6A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE

NEXT INPUT!

6.1 Two pulleys, one with a 450 mm diameter and the other with a 200 mmdiameter are on parallel shaft of 1.95 m apart and are connected by a

cross belt. Find the length of the belt required and the angle of contact

between the belt and each pulley.

6.2 It is required to drive a shaft B at 620 rpm by means of a belt from a parallel

shaft A. A pulley of 30 cm diameter on shaft A runs at 240 rpm. Determine

the size of pulley on the shaft B.

6.3 Find the length of belt necessary to drive a pulley of 1.4 m diameter

running parallel at a distance of 1.7 meter from the driving pulley ofdiameter 0.5 m. It is connected by an open belt.

N1

=d

2

N2

d1

N1 = speed diver in rpm

N2 = speed follower in rpm

d1 ,d2 = diameter pulley driver and

follower.


14/30

Feedback to Activity 6A

Have you tried the questions????? If YES, check your answers now

6.1 4.975 m; 199 or 3.473 radian.

6.2. 11.6 cm

6.3 6.51 m


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INPUT

6.3 POWER TRANSMITTED BY A BELT

Power = (TiT2) v

Where, Ti =Tension in tight sidein Newton.T2 = Tension in slack side.

V = velocity of belt

Fig. 6.5, shows the driving pulley (i.e., driver) A and the follower B.

The driving pulley pulls the belt from one side, and delivers the same to the

other.The maximum tension in the tight side will be greater than that

slack side.

Fig. 6.5

Torque exerted on the driving pulley= ( Ti - T2) ri

Torque exerted on the driven or follower

= ( Ti - T2) r2

Power transmitted = Force x distance

= ( Ti - T2) v


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Example 6.3

The tension in the two sides of the belt are iOO N and 8O N respectively.

If the speed of the belt is 75 meters per second. Find the power transmitted

by the belt.

Solution 6.3

Given,

Tension in tight side,

Ti= iOO N

Tension in slack side,

T2 = 8O N

Velocity of belt, v = 75 mIs

Let P = Power transmitted by the belt

Using the relation,

Power = (TiT2) v

= ( iOO 8O) 75

= i5OO watt

= i.5 kw.


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Activity 6B

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE

NEXT INPUT!

6.4. Find the tension in the tight side, if the tension in slack side i50 N. The speed

of the belt is 58 meters per second and the power transmitted by this

belt is 2 kw.

6.5 Diameter of driver is 50 mm. If the driver transmits 8 kw when it is rotatingat 300 revl min. Calculate velocity of driver.

6.6 The tension in the two sides of the belt are 300 N and i80 N respectively. If

the speed of the belt is 50 meters per second, find the initial tension and

power transmitted by the belt.

6.7 A i00 mm diameter pulley is belt-driven from a 400 mm diameter pulley.

The 400 mm pulley rotates at 480 revlmin. The number of revlsec of the

i00 mm diameter pulley isA. 2 B. 32 C. i20 D. i920

T0 =Ti +T2

2

Where, T0 = Initial tension in the belt.


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Feedback to Activity 6B


6.4 ii5.5 N

6.5 0.785 mis

6.6 6 000 watt or 6 kw; 240 N

6.7 B. 32


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INPUT

6.4 RATIO OF TENSIONS.

Fig. 6.6

Consider a driven pulley rotating in the clockwise direction as in fig 6.6.

Let Ti = Tension in the belt on the tight side.T2 = Tension in the belt on the slack side.

= An angle of contact in radians (i.e , angle subtended by the arc

AB, along which the belt touches the pulley, at the centre)

Now consider a small position of the belt PQ, subtending an angle at thecentre of the pulley as shown fig 6.6. The belt PQ is in equilibrium under the

following force:

i. Tension T in the belt at P.

ii. Tension T +T in the belt at Q.

iii. Normal reaction R, and

iv. Frictional force F = x R

Where is the coefficient of friction between the belt and

pulley. Resolving all the forces horizontally and equating the

same,


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R = ( T + T) sin

+ T sin2

(6.i)

2

Since the is very small, therefore, substituting,sin =

2 in equation 6.i2

R = ( T + T)

+T

2 2

=T

2+

T

2+T.

2

= T. ( neglectingT

2) (6.2)

Now resolving the forces vertically,

x R = (T + T) cos

- T cos2

(6.3)

2

Since the angle .is very small, therefore substituting cos

equation 6.3 or, x R = T + T T = T

=i in

2

R =T

(6.4)

Equating the values of R from equation 6.2 and 6.4,

Or T. =T

OrT

T= .

Integrating both sides from A to B,

Ti

T

1=

1

T2

T

or loge

0

Ti

T2

Ti


21/30

=

= e

ration of tension for flat belt

T2

Ti

= eisin

ration of tension for V beltT2

where = half angle of groove


22/30

Example 6.4

Find the power transmitted by a belt running over a pulley of 60 cm diameter

at 200 rpm. The coefficient of friction between the pulley is 0.25, angle of lap

i600 and maximum tension in the belt is 250 KN.

Solution 6.4

Given,

Diameter of pulley, d = 60 cm = 0.6 m

Speed of pulley N = 200 rpm

Speed of belt, v = dN

=60

Coefficient of friction,

= 0.25

x 0.6x 200

60= 2 = 6.28 mi sec

Angle of contact, 0= i600 = i60 x i80

= 2.7926 radian.

Maximum tension in the belt,

Ti = 250 kN

Let P = power transmitted by the belt.

Using the relation,

T2 = Ti

2.0i

Ti = e10

T2

= e0.25 x 2.7926

Ti

=2.0iT

2

=250

= i24.38 kN2.0i

Now using the relation,P = (TiT2 ) v

= ( 250 - i24.38 ) 6.28

= 788.89 watt

P = 0.79 kw


23/30

INPUT

6.5 CENTRIFUGAL TENSION.

Fig.6.7

The tension caused by centrifugal force is called centrifugal tension. At lower

speeds the centrifugal tension is very small, but at higher speeds its effect is

considerable, and thus should be taken into account.

Consider a small portion PQ of the belt subtending an angle dat the centre

of the pulley as show in fig 6.7.

Let M = mass of the belt per unit length,

V = linear velocity of the belt,

r = radius of the pulley over which the belt runs,

Tc = Centrifugal tension acting tangentially at P and Q.

Length of the belt PQ,

= rd

Mass of the belt PQ,

M = M rd

We know that centrifugal force,

= M v2ir


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Centrifugal force of the belt PQ

2

=M x r.dv

r

= M x dv2

The centrifugal tension (Tc) acting tangentially at P and Q keeps the belt in

equilibrium.

Now resolving the force (i.e, centrifugal force and centrifugal tension)

horizontally and equating the same,

d2 Tc sin2

= M x dv2

since angle dis very small, therefore, substituting

sind

2

d

=

d

2

2 Tc2

= M x dv2

Tc = M x dv2

i d

Tc = M v2

Ti T

c

=e

1

T2 T

c

Ti

T

c

= e1i

sin

T2 T

c

ratio tension for flat belt.

ratio tension for V belt

6.5.1 Condition for the transmission of maximum power

The maximum power,

When T c = !T i

It shows that the power transmitted is maximum ! of the maximum tension

is absorbed as centrifugal tension.

The velocity of belt for maximum transmission of power may be

obtained from equation Ti = 3Tc = M v2.


25/30

v2

=

v =

3Ti

M

Ti

3M

Example 6.5

An open belt drive connects two pulleys i.2 m and 0.5 m diameter, on

parallel shafts 3.6 m apart. The belt has a mass of 0.9 kgim length and the

maximum tension in it is not exceed 2 kN. The larger pulley runs at 200

revimin. Calculate the torque on each of the two shafts and the power

transmitted. Coefficient of friction is 0.3 and angle of lap on the smaller

pulley is i68 ( 2.947 radian ).

Solution 6.5

Given

Diameter of larger pulley,

di = i.2 m

Radius of the larger pulley,

ri = 0.6

Diameter of smaller pulley,

d2 =0.5 mRadius of the smaller pulley,

r2 = 0.25 m

Distance between two shaft,

D = 3.6 m

Mass of the belt per meter length,

M = 0.9 kgim

Maximum Tension, Ti = 2 kN = 2 000 N

Speed of the larger pulley

= 200 rpmVelocity of the belt,

V=

T c = Mv2

= 0.9 x i2.572

= i42.2 N

xi.2x 200

60= i2.57 mis


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2

ei

= e0.3 x 2.947

= 2.42i

Using the relation,Ti

T

C

= e1

T2 T

C

2000 i42 .2

T2

i42 .2

= 2.42i

(T2 i42.2) 2.42i = i857.8

(T i42.2) = i857 .82.42i = 767.37

T2 = 767.37 + i42.2

T2 = 909.57 N

We know that the torque on the larger pulley shaft (TL),

TL = ( Ti T2) ri

= ( 2000 909.57) 0.6

= 654.26 Nm.

Torque on the smaller pulley shaft (Ts),

Ts = ( Ti T2) r2

= (2000 909.57) 0.25

= 272.6i Nm.

Power transmitted, P = ( TiT2)v

= ( 2000 909.57) i2.57

= i3706.7i watt

= i3.7i kw


27/30

Activity 6C

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE TO THE NEXT

INPUT!

6.8 Two pulleys, one with a 450 mm diameter and the other with a 200 mmdiameter are on parallel of shafts i.95 m apart. What power can be

transmitted by the belt when the larger pulley rotates at 200 revimin, if the

maximum permissible tension in the belt is i kN and the coefficient of

friction between the belt and pulley is 0.25. Angle of contact between the belt

and larger pulley is 3.477 radian.

6.9 Find the power transmitted by a V drive from the following data:

Angle of contact = 840

Pulley groove angle = 450

Coefficient of friction = 0.25

Mass of belt per meter length = 0.472 kgim

Permissible tension = i39 N

Velocity of V belt = i2.57 mis.

6.i0 An open belt drive connects two pulleys i.2 m and 0.6 m, on parallel shafts 3

m apart. The belt has a mass 0f 0.56 kgim and maximum tension is i.5 kN. The

driver pulley runs at 300 revimin. Calculate the inertial tension, power

transmitted and maximum power. The coefficient of friction between the belt

and the pulley surface is 0.3.


28/30

Feedback to Activity 6C


6.8 2 740 watt or 2.74 kw

6.9 59i watt.

6.i0 i074.5 N ; 8.0 kw ; i7.54 kw.


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TT

SELF-ASSESSMENT 6

You are approaching success. Try all the questions in this self-assessment section

and check your answers with those given in the Feedback on Self-Assessment 6

given on the next page. If you face any problems, discuss it with your lecturer.

Good luck.

i. A pulley is driven by a flat belt running at speed of 600 mimin. The

coefficient of friction between the pulley and the belt is 0.3 and the angle lap

is i60. If the maximum tension in the belt is 700 N, find the power

transmitted by a belt.

2. A leather belt, i25 mm wide and 6 mm thick, transmits power from a pulley

of 750 mm diameter which run at 500 rpm. The angle of lap is i50 and

i = 0.3. If the mass ofi m3 of leather is i Mg and the stress in the belt

does not exceed 2.75 MNim2, find the maximum power that can be

transmitted.

3. A flat belt, 8 mm thick and i00 mm wide transmits power between two

pulleys, running at i 600 mimin. The mass of the belt is 0.9 kgim length. The

angle of lap in the smaller pulley is i65 and the coefficient of friction

between the belt and pulleys is 0.3. If the maximum permissible stress in the

belt is 2 MNim2, find

(i) Maximum power transmitted; and

(ii) Initial tension in the belt.

4. The moment on a pulley, which produces the rotation of the pulley is called:

A. Momentum B. Torque C. Work D. Energy

5. If Tiand T2 are the tension in the tight and slack sides of a belt and is the

angle of contact between the belt and pulley. Coefficient of friction is , then

the ratio of driving tension will be:

A.T

2

Ti = e

1

B.Ti

T2= e1

C.Ti

T2

=

1

TiD. log

i0

2

= 1


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Feedback to Self-Assessment 6

Have you tried the questions????? If YES, check your answers now.

i. 3.974 kw ( see example 6.4)

2. i8.97 kw

3. (i) i4.28i kw

(ii) i.322i kN

4. B. Torque

5. B. Ti = e1

T2

CONGRATULATIONS!!!!..

May success be with you

always.

Documents

JJ309 Fluid Mechanics Unit 6