jj309 fluid mechanics unit 6
TRANSCRIPT
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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,
L = Arc AFE + ED + Arc DCB + BA
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
L = Arc AFE + ED + Arc DCB + BA
= 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
01 and 02 = Centers of two pulleysr1 and r2 = radius of the larger and smaller pulleys
d = Distance between 01 and 02
L = Total length of the belt.Angle A0102 = 1Angle D0201 = 2
Angle AFE = 1 (radian)
Angle BCD = 2 (radian)
We know that the length of the belt is
L = Arc AFE + ED + Arc DCB + BA
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,
L = Arc AFE + ED + Arc DCB + BA
= 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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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 D0201 = 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 the belt,
L = Arc AFE + ED + Arc DCB + BA
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,
L = Arc AFE + ED + Arc DCB + BA
= 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.
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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
Have you tried the questions????? If YES, check your answers now
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
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=
= e
ration of tension for flat belt
T2
Ti
= eisin
ration of tension for V beltT2
where = half angle of groove
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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
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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.
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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
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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.
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Feedback to Activity 6C
Have you tried the questions????? If YES, check your answers now
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.