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Lab Manual

Theory of Machines (2151902)

Darshan Institute of Engineering & Technology, Rajkot

DEPARTMENT OF MECHANICAL ENGINEERING

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Mechanical Engineering has completed the term work

satisfactorily in Theory of Machines (2151902) for the

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DARSHAN INSTITUTE OF ENGG. & TECH.

Department of Mechanical Engineering

B.E. Semester – V

Theory of Machines (2151902)

List of Experiments

Sr.

No. Title

Date of

Performance Sign Remark

1. Performance on gravity controlled

governors.

2. Performance on spring controlled

governors.

3. Analysis of gyroscopic effect.

4. Measurement of mass moment of

inertia.

5. Measurement of radius of gyration of

various components.

6. Power measurement using

dynamometers.

7. Analysis of clutch.

8. Analysis of brakes.

9.

Dynamic force analysis of 4-bar

mechanism and slider crank mechanism

(Analytical Methods)

10. Design of flywheel for IC engine and

Punch press.

Gravity Controlled Governor

Department of Mechanical Engineering Theory of Machines (2151902)

Darshan Institute of Engineering and Technology, Rajkot Page. 1.1

Experiment

Governor Apparatus

1. INTRODUCTION:

The function of a governor is to regulate the mean speed of an engine, when there are variations in

loads e.g. when load on an engine increase or decrease, obviously its speed will, respectively

decrease or increase to the extent of variation of load. This variation of speed has to be controlled by

the governor, within small limits of mean speed. This necessitates that when the load increase and

consequently the speed decreases, the supply of fuel to the engine has to be increased accordingly to

compensate for the loss of the speed, so as to bring back the speed to the mean speed. Conversely,

when the load decreases and speed increases, the supply of fuel has to be reduced.

2. THEORY:

The function of the governor is to maintain the speed of an engine within specific limit whenever

there is a variation of load. The governor should have its mechanism working in such a way, that the

supply of fuel is automatically regulated according to the load requirement for maintaining

approximately a constant speed. This is achieved by the principle of centrifugal force. The

centrifugal type governors are based on the balancing of centrifugal force on the rotating balls by an

equal and opposite radial force, known as the controlling force.

1. Centrifugal Governors.

2. Inertia Governors.

The centrifugal governors are based on the balancing of the centrifugal force on the rotating balls by

an equal and opposite radial force, known as controlling force.

In Inertia governors the position of the balls are affected by the forces set by an angular acceleration

or deceleration of the given spindle in addition to centrifugal forces on the balls.

The apparatus is designed to exhibit the characteristics of the spring-loaded governor and centrifugal

governor. The experiments shall be performed on followings centrifugal type governors:

1. Watt governor

2. Porter governor

3. Proell governor

4. Hartnell governor

The drive unit consists of a DC motor connected to the shaft through V belt. Motor and shaft are

mounted on a rigid MS base frame in vertical position. The spindle is supported in ball bearing.

The optional governor mechanism can be mounted on spindle. The speed control unit controls the

precise speed and speed of the shaft is measured with the help of hand tachometer. A counter sunk

has been provided at the topmost bolt of the spindle. A graduated scale is fixed to measure the sleeve

lift.

The center sleeve of the Porter and Proell governors incorporates a weight sleeve to which weights

can be added. The Hartnell governor consists of a frame, spring and bell crank lever. The spring

tension can be increased or decreased to study the governor.




3. EXPERIMENTAL PROCEDURE:

3.1 STARTING PROCEDURE:

1. Assemble the governor to be tested.

2. Complete the electrical connections.

3. Switch ON the main power.

4. Note down the initial reading of pointer on the scale.

5. Switch ON the rotary switch.

6. Slowly increase speed of governor until sleeve is lifted from its initial position by variac.

7. Let the governor be stabilized.

8. Note down the sleeve’s height and relative RPM (with help of hand tachometer).

9. Increase speed of governor in steps to get different positions of sleeve lift at different RPM.

3.2 CLOSING PROCEDURE:

1. Decrease the speed of governor gradually by bringing the variac to zero position and then

switch off the motor.

2. Switch OFF all switches.

3. Disconnect all the connections.

4. NOMENCLATURE:

a Distance of pivot to center of spindle mm

Ftheo Theoretical centrifugal force kg

Fact Actual centrifugal force Kg

g Acceleration due to gravity m/sec2

H Height of governor mm

h Final height mm

h' Initial height mm

L length of link mm

Ntheo theoretical speed of governor RPM

Nasc Actual speed of governor while ascending RPM

Ndes Actual speed of governor while descending RPM

R radius of rotation mm

w Weight of balls on one side kg

X Final height of Sleeve mm

X' height of sleeve at N rpm mm

X'' initial reading of pointer on sleeve mm

ω Angular velocity rad/sec

W Total weight on sleeve kg

W1 Weight of cast iron sleeve kg

W2 Dead weight applied on sleeve kg

W3 Weight of arms on one side kg

α Initial Angle

γ Initial Angle




Experiment No. 1

Performance on Gravity Controlled Governors.

1. Watt Governor

It is assumed that mass of the arms; links & sleeve are negligible in comparison with the mass of the

balls and are neglected in the analysis.

L

L

h'

H

R

Fc

A

h'

a a

w = mg

WATT GOVERNOR

Figure 1

GIVEN DATA:

1. Acceleration due to gravity g = 9.81m/sec2

2. Length of link, L = 105 mm

3. Initial height, h' = 100 mm

4. Distance of pivot to centre of spindle, a = 52.5 mm

5. Weight of balls on one side, w = 1.498 kg

OBSERVATION TABLE:

X’’ = mm

Sr. No X’, mm Nasc, rpm Ndes, rpm




CALCULATIONS:

' ''X (X X )

' Xh h

2

1 hcos

L

aH h

tan

theo

60 g 1000N

2 H

R a Lsin

act2 N

60

2

act

w R ωF

g 1000

theo

w R F

H

RESULT TABLE:

Ascending

X, mm h, mm α H, mm Ntheo,

rpm R, mm

ω,

rad/sec Fact, kg

Ftheo,

kg R/H




Descending


rpm R, mm

ω,

rad/sec Fact, kg

Ftheo,

kg R/H

Plot the graph for following curves:-

1. R/H vs Ntho

2. R/H vs Nact

3. X vs Ntho

4. X vs Nact

CONCLUSION:




2. Porter Governor

Porter Governor differs from Watt’s Governor only in extra sleeve weight, else is similar to Watt

Governor.

L

W1

PORTER GOVERNORFigure 2

W2

W

HL

h'

R

w = mg

Fc

h'

Sleeve Weight

C

W2

A

aaE

GIVEN DATA:

Acceleration due to gravity g = 9.81m/sec2

Length of link, L = 105 mm

Initial height, h' = 100 mm

Dead weight applied on sleeve, W2 = 1.011+0.542 = 1.553 kg

Distance of pivot to center of spindle, a = 52.5 mm

Weight of balls on one side, w = 0.749 +0.749 = 1.498 kg

Weight of cast iron sleeve, W1 = 2.042 kg

Weight of arms on one side W3 = 0.165 kg

OBSERVATION TABLE:

X’’ = mm





CALCULATIONS:

' ''X (X X )

' Xh h

2

1 hcos

L

aH h

tan

R a Lsin

2 N

60

2

act

w R ωF

g 1000

1 2 3W W W W

theo

60 w W g x1000N

2 w H

theo

WF w 1 k tan (k = 1) w W tan α

2

RESULT TABLE

Ascending


rpm R, mm W, kg

ω,

rad/sec

Fact,

kg

Ftheo,

kg R/H




Descending


rpm R, mm W, kg

ω,

rad/sec

Fact,

kg

Ftheo,

kg R/H

Plot the graph for following curves: -

1. R/H vs Ntho

2. R/H vs Nact

3. X vs Ntho

4. X vs Nact

CONCLUSION:




3. Proell Governor

L

PROELL GOVERNOR

Figure 3

WW2

W1

H

h'

R

w = mg

B

O

Fc

G

h'

Sleeve Weight

DW2

C

A

aaE

L

GIVEN DATA:



Initial height, h' = 100 mm

Distance of pivot to center of spindle, a = 52.5 mm

Weight of balls on one side, w = 0.506 +0.506 = 1.012 kg


Dead weight applied on sleeve, W2 = 0.542 + 1.011 = 1.553 kg

Weight of arms on one side W3 = 0.165 kg

Displacement between points G & C of lower link, GC = 155.33 mm

Initial Angle γ’ = 23.6110

Initial Angle, α’ = 17.7530

OBSERVATION TABLE:

X’’ = mm





CALCULATIONS:

' ''X (X X )

' Xh h

2

1 hcos

L

aH h

tan

'α α γ'

R a GCsin

2 N

60

2

act

w R F

g 1000

DG GC cos γ

' XBD h

2

1 2 3W W W W

tanw

cos

tan

GC

BDw2WFtheo

theotheo

F g 100060N

2 w R

CALCULATIONS TABLE

Ascending




X,

mm

h,

mm α

H,

mm ϒ

R,

mm

ω,

rad/s

Fact,

kg

DG,

mm

BD,

mm

W,

kg

Ftheo,

kg

Ntheo,

RPM R/H

Descending

X,

mm

h,

mm α

H,

mm ϒ

R,

mm

ω,

rad/s

Fact,

kg

DG,

mm

BD,

mm

W,

kg

Ftheo,

kg

Ntheo,

RPM R/H


1. R/H vs Ntho

2. R/H vs Nact

3. X vs Ntho

4. X vs Nact

CONCLUSION:

Spring Controlled Governor



Experiment No. 2

Performance on Spring Controlled Governors.

Hartnell Governor

F

SLEEVEWEIGHT

b

a

SPRING

HARTNELL GOVERNOR

Figure 4

R

GIVEN DATA:



Length of vertical arm, a = 75 mm

Length of horizontal arm, b = 130 mm

Weight of balls on one side, w = 0.362 + 0.362 = 0.724 kg


Dead weight applied on sleeve, W2 = 0.542 + 0.441 = 0.983

Stiffness of spring, s = 0.27 kg/mm

Initial radius of rotation, R’ = 195 mm

OBSERVATION:

X’’ = mm

Sr. No X’, mm Nasc, rpm Ndsc, rpm




CALCULATIONS:

' ''X (X X )

' aR R X

b

act2 N

60

2

cw R

Fg 1000

c

aS 2 F W

b

'RR

F

b

a2s c

2

If = 0, then Fc’ = 0

RESULT TABLE

Ascending

X, mm R, mm ω Fc, kg S, kg s, kg/mm




Descending

X, mm R, mm ω Fc, kg S, kg s, kg/mm


1. X vs Nact

CONCLUSION:

Gyroscope



Experiment No. 3

Analysis of gyroscopic effect of a rotating disc

AIM: Experimental justification of the equation T = I..p for calculating the gyroscopic couple by

observation and measurement of results for independent variation in applied couple T and precession

p.

INTRODUCTION:

AXIS OF SPIN:

If a body is revolving about an axis, the latter is known as axis of spin (Refer Fig.1, where OX is the

axis of spin).

PRECESSION:

Precession means the rotation about the third axis OZ (Refer Fig. 1) that is perpendicular to both the

axis of spin OX and that of couple OY.

AXIS OF PRECESSION:

The third axis OZ is perpendicular to both the axis of spin OX and that of couple OY is known as

axis of precession.

GYROSCOPIC EFFECT:

To a body revolving (or spinning) about an axis say OX, (Refer Fig.1) if a couple represented by a

vector OY perpendicular to OX is applied, then the body tries to process about an axis OZ which is

perpendicular both to OX and OY. Thus, the couple is mutually perpendicular.

The above combined effect is known as processional or gyroscopic effect.

GYROSCOPE:

It is a body while spinning about an axis is free to rotate in other directions under the action of

external forces.

Fig. 1- OX – Axis of spin, OY – Axis of Couple, OZ – Axis of Precession

Gyroscope



NOMENCLATURE

dθ Angle of precession

dt Time required for this precessions sec

g Acceleration due to gravity m/sec2

I Moment of inertia of disc kg m /sec2

L Distance of weight for the center of disc m

N Revolution of Disc spin RPM

r Radius of disc m

Tthe Theoretical Gyroscopic couple kg-m

Tact Actual Gyroscopic couple kg-m

W Weight of rotor disc kg

w Weight on pan kg

Angular velocity of disc rad/sec

p Angular velocity of precession of yoke about vertical axis rad/sec

Fig. 2

Gyroscope



THEORY:

GYROSCOPIC COUPLE OF A PLANE DISC:

Let a disc of weight ‘W’ having a moment of inertia I be spinning at an angular velocity about axis

OX in anticlockwise direction viewing from front (Refer Fig.2). Therefore, the angular momentum

of disc is I. Applying right–hand screw rule the sense of vector representing the angular momentum

of disc which is also a vector quantity will be in the direction OX as shown.

Fig. 3

A couple whose axis is OY perpendicular to OX and is in the plane Z, is now applied to prices the

axis OX.

Let axis OX turn through a small angular displacement from OX to OX’ in time t. The couple

applied produces a change in the direction of angular velocity, the magnitude & the magnitude

remaining constant. This change is due to the velocity of precession.

Therefore, ‘OX’ represents the angular momentum after time dt.

Change of angular momentum = OX’ – OX = XX’

dt

dOX

dt

XXntdisplacemeAngular

'

As, dOXXX ' in direction of XX’

Now as rate of change of angular momentum

Couple applied = C = T

We get dt

dOXT

Gyroscope



But IOX

Where

I = Moment of Inertia of disc

= Angular Velocity of disc.

dt

d.IT

And in the limit dt is very small

We have P

dt

d

Where ωP = Angular velocity of precession of yoke about vertical axis.

Thus, we get –

The direction of the couple applied on the body is clockwise when looking in the direction XX’ and

in the limit this is perpendicular to the axis of and of p.

The reaction couple exerted by the body on its frame is equal in magnitude to that of C, but opposite

in direction.

DESCRIPTION:

The set up consists of heavy disc mounted on a horizontal shaft, rotated by a variable speed motor.

The rotor shaft is coupled to a motor mounted on a trunion frame having bearings in a yoke frame,

which is free to rotate about vertical axis. A weight pan on other side of disc balances the weight of

motor. Rotor disc can be move about three axis. Weight can be applied at a particular distance from

the center of rotor to calculate the applied torque. The gyroscopic couple can be determined with the

help of moment of inertia, angular speed of disc and angular speed of precession.

EXPERIMENTAL PROCEDURE:

Set the rotor at zero position.

Start the motor with the help of rotary switch.

Increase the speed of rotor with dimmer stat & stable it & measure the R.P.M. with the help

of tachometer.

Put the weight on weight pan then yoke rotate at anticlockwise direction.

Measure the rotating angle (30o, 40o) with the help of stopwatch.

Repeat the experiment for the various speeds and loads.

After the test is over set dimmer stat to zero position and switch off main supply.

pIT

Gyroscope



OBSERVATION & CALCULATION:

GIVEN DATA:

Acceleration due to gravity g = 9.81 m/sec2

Radius of disc, r = 0.15 m

Weight of rotor disc, W = 5.42 kg

Distance of weight for the center of disc, L = 0.185 m

OBSERVATION TABLE:

CALCULATIONS:

2 N

60

p

d

dt 180

2W rI

g 2

the pT I

actT w L

RESULT TABLE

PRECAUTION & MAINTENANCE INSTRUCTIONS:

Never run the apparatus if power supply is less than 200Volts and above 230 Volts

Before start the motor set dimmer stat at zero position.

Increase the speed gradually.

CONCLUSION:

We concluded from experiment that when the right turns under CW condition the effect of reactive

gyroscopic couple will be to dip nose and rise the tail & anticlockwise condition effect of reactive

gyroscopic couple will be to raise the nose and dip the tail.

Sr.

No

N

(RPM)

w

(kg) d

(degree)

dt

(sec)

Sr. No. I, kg m/sec2 , rad/sec p, rad/sec Tthe, kg m Tact, kg m

Mass Moment of Inertia



Experiment No. 4

Measurement of mass moment of inertia of Various Components.

AIM: Measurement of mass moment of inertia of a steel bar.

NOMENCLATURE:

Kact Radius of gyration m

Ktheo Radius of gyration about the C.G. m

L Length of suspended pendulum m

N Nos. of oscillations

OG Distance of Centre of Gravity of the rod from support m

Tact Actual time period sec

Ttheo Theoretical time period sec

DESCRIPTION:

The compound pendulum consists of a steel bar. The bar is supported by knife -edge. Two

pendulums of different lengths are provided with the set-up.





Support the rod on knife -edge.

Note the length of suspended pendulum and determine OG.

Allow the bar to oscillate and determine T by knowing the time for say 10 Oscillations.

Repeat the experiment with different length of suspension.

Complete the observation table given below.


OBSERVATION TABLE:

Sr. No. L, m m, Kg OG, m n t, sec

1 0.64 1.240 0.315 10

2 0.84 1.520 0.415 10

CALCULATIONS:

actual

tT

n

2

2actact

TK g OG OG

2

Theo

Lk

2 3

2I mk

CALCULATION TABLE:

Sr. No Kact, m Ktheo, m




AIM: Measurement of mass moment of inertia of a connecting rod.

DESCRIPTION:

The connecting rod is suspended by end bearing.

NOMENCLATURE:

Kact Radius of gyration m

Ktheo Radius of gyration about the C.G. m

d1 Diameter of small end bearing m

d2 Diameter of big end bearing m

L1 Length of equivalent compound pendulum when suspended from the top of small

end bearing

m

L2 Length of equivalent compound pendulum when suspended from the top of big end

bearing

m

h1 Distance of centre of gravity, G, from the top of small end bearing m

h2 Distance of centre of gravity, G, from the top of big end bearing m

L Centre distance of connecting rod m


G Centre of Gravity of the connecting rod


Ttheo Theoretical time period sec





Suspend the connecting rod from small end bearing on apparatus.

Note the centre distance between small end and big end bearing of connecting rod.

Allow the connecting rod to oscillate and determine T by knowing the time for say 10

Oscillations.

Repeat the experiment with connecting rod suspended from big end bearing.



OBSERVATION TABLE:

Sr. No.

Time for 10 Oscillations, t (sec)

Suspended from small end bearing Suspended from big end bearing

CALCULATIONS:

For suspended from small end bearing,

Average time = sec

No.of oscillationFrequency, f

time

p1

1Time period t

f

For suspended from big end bearing,

Average time = sec

No.of oscillationFrequency, f

time

p2

1Time period t

f




p

LNow,t 2

g

1p1

Lt 2

g

L1 =

2p2

Lt 2

g

L2 =

2 2k hL

h

2 2

11

1

k hL

h

2 2

22

2

k hL

h

2 2

1 1k 0.2688h h

2 2

2 2k 0.2243h h

1 21 2

d dh h 304.5

2 2

1 2h 355.2 h

Put this value in equation (i)

2 2

2 2 2k 95.4778 0.2688h 126167.04 710.4h h

2 2

2 2k 126071.56 710.13h h

2 2

2 2 2 2126071.56 710.13h h 0.2243h h

2126071.56 709.9069h

2h 177.88 mm 0.177 m

1h 355.2 177.58 177.62 mm 0.177 m

2 2k 0.2688 0.1776 0.1776

2k 0.0477 0.0315 0.127 m

2I mk

Radius of Gyration



Experiment No. 5

Measurement of Radius of Gyration of Various Components.

Part 1: To determine the radius of gyration of given plate by using Bi-Filar suspension.

Fig. 1 Bi-Filar suspension system

DESCRIPTION:

A uniform rectangular section bar is suspended from the pendulum support frame by two parallel

cords. Top ends of the cords pass through the two small chucks fitted at the top. Other ends are

secured in the Bi-Filer bar. It is possible to adjust the length of the cord by loosening the chucks.

The suspension may be used to determine the radius of gyration of any body. In this case, the body

under investigation is bolted to the center. Radius of gyration of the combined bar and body is then

determined.


Suspend the bar from chuck, and adjust the length of the cord ‘L’ conveniently. Note that the

suspension length of each cord must be same.

Allow the bar to oscillate about the vertical axis passing through center and measure the

periodic time T by knowing the time for say 10 oscillations.

Repeat the experiment by mounting the weights at equal distance from center.


Radius of Gyration



NOMENCLATURE:

a Half of the distance between the two strings. m

G Acceleration due to gravity m/sec2

Kact Actual radius of gyration of Bi – Filar suspension m

Ktheo Theoretical radius of gyration of Bi – Filar suspension m

L Length of the suspended string m



T time taken for 10 oscillations sec


DATA :

a = m

OBSERVATION TABLE:

Sr. No L, m n t, sec

CALCULATIONS:

act

tT

n

actact

T aK

L2

g

theo

Lk

2 3

CALCULATION TABLE:

Sr. No Kact, m Ktheo, m

Radius of Gyration



Part 2: To determine the Radius of Gyration of Trifilar Suspension.

DESCRIPTION:

A uniform circular disc is suspended from the pendulum support frame by three parallel cords. Top

ends of the cords pass through the three small chucks fitted at the top. Other ends are secured in the

Tri-Filer disc. It is possible to adjust the length of the cord by loosening the chucks.


Suspend the disc from chucks, and adjust the length of the cord ‘L’ conveniently.

Note that the suspension length of each cord must be same.

Allow the disc to oscillate about the vertical axis.

Measure the oscillation with time.

Repeat the experiment for different lengths & different radius.

NOMENCLATURE:

fa Actual frequency sec-1

L Length of the cord m

Kact Actual radius of gyration m

Ktheo Theoretical radius of gyration m

R Radius of disc m



T Time taken for n oscillations sec

OBSERVATION TABLE

Sr. No R, m n t, sec L, m

CALCULATIONS:

act

tT

n

a

1f

T

Radius of Gyration



2

act

a

1 gRk

2 f L

theo

Lk .

2 3

CALCULATION TABLE:

Sr. No Tact, sec fa, sec-1

Kact, m Ktheo, m

CONCLUSION

Dynamometer



Experiment No. 6

Dynamometer

INTRODUCTION

A Dynamometer is a brake but in addition it has a device to measure the frictional resistance.

Knowing the frictional resistance, may obtain the torque transmitted and hence the power of the

engine.

TYPES OF DYNAMOMETERS:

There are mainly two types of dynamometers:

1. Absorption Dynamometers: In this type, the work done is converted into heat by friction

while being measured. They can be used for the measurement of moderate powers only.

Examples are prony brake dynamometer and rope brake dynamometer.

2. Transmission Dynamometers: In this type, the work is not absorbed in the process, but is

utilised after the measurement. Examples are the belt – transmission dynamometer and the

trosion dynamometer.

PRONY BRAKE DYNAMOMETER:

Fig. 6.1

A prony brake dynamometer consists of two wooden blocks clamped together on a revolving pulley

carrying a lever (Fig. 6.1). The friction between the blocks and the pulley tends to rotate the blocks

in the direction of rotation of the shaft. However, the weight due to suspended mass at the end of the

lever prevents this tendency. The grip of the blocks over the pulley is adjusted using the bolts of the

clamp until the engine runs at the required speed. The mass added to the scale pan is such that the

arm remains horizontal in the equilibrium position; the power of the engine is thus absorbed by the

friction.

Frictional torque W l M g l

2 NPower of the machine under test T M g l

60M N k

where k is a constant for a particular brake.

Note that the expression fro power is independent of the size of the pulley and the coefficient of

friction.

Dynamometer



ROPE BRAKE DYNAMOMETER:

In a rope brake dynamometer (Fig. 6.2), a rope is wrapped over the rim of a pulley keyed to the shaft

of the engine. The diameter of the rope depends upon the power of the machine. The spacing of the

ropes on the pulley is done by 3 to 4 U-shaped wooden blocks which also prevent the rope from

slipping off the pulley. The upper end of the rope is attached to a spring balance where as the lower

end supports the weight of suspended mass.

t

Power of the machine TF r

2 NM g s r

60

Fig. 6.2

If the power produced is high, so will be the heat produced due to friction between the rope and the

wheel, and a cooling arrangement is necessary. For this, the channel of the flywheeel usually has

flanges turned inside in which water from a ripe is supplied. An outlet pipe with a flattenend end

takes the water out.

A rope brake dynamometer is frequently used to test the power of engines. It is easy to manufacture,

inexpensive, and requires no lubrication.

If the rope is wrapped several times over the wheel, the tension on the slack side of the rope, i.e., the

spring balance reading can be reduced to a negligible value as compared to the tension of the tight

side (as T1 / T2 = eµθ

and θ is increaased). Thus, one can even do away with the spring balance.

BELT TRANSMISSION DYNAMOMETER:

The belt transmission dynamometer occupies a prominent position among transmission

dynamometers. When a belt transmits power from one pulley to another, there exists a difference in

tensions between the tight and slack sides. A dynamometer measures directly the difference in

tensions (T1 – T2) while the belt is running.

Dynamometer



Fig. 6.3

Fig. 6.3 shows a Tatham dynamometer. A continuous belt runs over the driving and the driven

pulleys through two intermediate pulleys. The intermediate pulleyshave their pins fixed to a lever

with its fulcrum at the midpoint of the two pulley centres. As the lever is not pivoted at its midpoint,

a mass at the left end is used for its initial equilibrium. When the belt transmits power, the lever

tends to rotate in the counter – clockwise direction due to the difference of tensions on the tight and

slack sides. To maintain its horizontal position, a weight of the required amount is provided at the

right end of the lever. Two stops, one on each side of the lever arm, are used to limit the motion of

the lever.

Taking moments about the fulcrum,

1 2

1 2

1 2

M g l 2 T a 2 T a 0

M g l 2a T T 0

M g lT T

2a

1 2Power, P T T v

where v =belt speed in metres per second.

EPICYCLIC - TRAIN DYNAMOMETER:

An epicyclic – train dynamometer is another transmission type of dynamometer. As shown in Fig.

6.4, it consists of a simple epicyclic train of gears. A spur gear A is the driving wheel which drives

an annular driven wheel B through an intermediate pinion C. The intermediate gear C is mounted on

a horizontal lever, the weight of which is balanced by a counterweight at the left end when the

system is at rest. When the wheel A rotates counter – clockwise, the wheel B as well as the wheel C

rotates clockwise. Two tangential forces, each equal to F, act at the ends of the pinion C, one due to

the driving force by the wheel A and the other due to reactive force of the driven wheel B. Both

forces are equal if friction is ignored. This tends to rotate the lever in the counter – clockwise

direction and it no longer remains horizontal. To maintain it in the same position as earlier, a

balancing weight W is provided at the right end of the lever. Two stops, one on each side of the lever

arm, are used to limit the motion of the lever.

Dynamometer



Fig. 6.4

For the equilibrium of the lever,

W l2 F a W l or F

2 a

and torque transmitted = F . r where r is the radius of the driving wheel

Thus power,

2 NP T F r

60

BEVIS – GIBSON TORSION DYNAMOMETER:

Fig. 6.5

A Bevis – Gibson torsion dynamometer consists of two discs A and B, a lamp and a movable torque

finder arranged as shown in Fig. 6.5(a). The two discs are similar and are fixed to the shft at a fixed

distance from each other. Thus, the two discs revolve with the shaft. The lamp is masked and fixed

on the bearing of the shaft. The torque finder has an eyepiece capable of moving circumferentially.

Each disc has a small radial slot near its periphery. Similar slots are also made at the same radius on

the mask of the lamp and on the torque finder.

Dynamometer



When the shaft rotates freely and does not transmit any torque, all the four slots are in a line and a

ray of light from the lamp can be seen through the eyepiece after every revolution. When a torque is

transmitted, the shaft twists and the slot in the disc B shifts its position. The ray of light can no

longer pass through the four slots. However, if the eyepiece is moved circumferentially by an amount

equal to the displacement, the flash will again be visible once in each revolution of the shaft. The

eyepeice is moved by a micrometer spindle. The angle of twist may be measured up to one

hundredth of the degree.

In case the torque is varied during each revolution of the shaft as in reciprocating engines and it is

required to measure the angle of twist at different angular position, then each disc can be perforated

with several slots arranged in the form of a spiral at varying radii [Fig. 6.5(b)]. The lamp and the

torque finder have to be moved radiallly to and from the shaft so that they come opposite each pair of

slots in the discs.

Flywheel



Experiment No. 10

Design of flywheel for IC engine and Punch press

INTRODUCTION

A flywheel used in machines serves as a reservoir, which stores energy during the period

when the supply of energy is more than the requirement, and releases it during the period

when the requirement of energy is more than the supply.

In case of steam engines, internal combustion engines, reciprocating compressors and pumps,

the energy is developed during one stroke and the engine is to run for the whole cycle on the

energy produced during this one stroke.

For example, in internal combustion engines, the energy is developed only during expansion

or power stroke which is much more than the engine load and no energy is being developed

during suction, compression and exhaust strokes in case of four stroke engines and during

compression in case of two stroke engines.

The excess energy developed during power stroke is absorbed by the flywheel and releases it

to the crankshaft during other strokes in which no energy is developed, thus rotating the

crankshaft at a uniform speed.

A little consideration will show that when the flywheel absorbs energy, its speed increases

and when it releases energy, the speed decreases. Hence a flywheel does not maintain a

constant speed, it simply reduces the fluctuation of speed.

In other words, a flywheel controls the speed variations caused by the fluctuation of the

engine turning moment during each cycle of operation.

In machines where the operation is intermittent like punching machines, shearing machines,

rivetting machines, crushers, etc., the flywheel stores energy from the power source during

the greater portion of the operating cycle and gives it up during a small period of the cycle.

Thus, the energy from the power source to the machines is supplied practically at a constant

rate throughout the operation.

THE TURNING MOMENT DIAGRAM (also known as crank-effort diagram) is the graphical

representation of the turning moment or crank-effort for various positions of the crank. It is plotted

on cartesian co-ordinates, in which turning moment is taken as ordinate and crank angle as abscissa.

Turning Moment Diagram for a Single Cylinder Double Acting Steam Engine

A turning moment diagram for a single cylinder double acting steam engine is shown in Fig. 10.1.

The vertical ordinate represents the turning moment and the horizontal ordinate represents the crank

angle. The turning moment on the crankshaft,

2 2

sin 2sin

2 sinPT F r

n

where FP = Piston effort,

r = Radius of crank,

n = Ratio of the connecting rod length and radius of crank, and

= Angle turned by the crank from inner dead centre.

Flywheel



Fig. 10.1. Turning moment diagram for a single cylinder, double acting steam engine.

From the above expression, we see that the turning moment (T ) is zero, when the crank angle

() is zero. It is maximum when the crank angle is 90° and it is again zero when crank angle

is 180°. This is shown by the curve abc in Fig. 16.1 and it represents the turning moment

diagram for outstroke. The curve cde is the turning moment diagram for instroke and is

somewhat similar to the curve abc.

Since the work done is the product of the turning moment and the angle turned, therefore the

area of the turning moment diagram represents the work done per revolution. In actual

practice, the engine is assumed to work against the mean resisting torque, as shown by a

horizontal line AF.

The height of the ordinate aA represents the mean height of the turning moment diagram.

Since it is assumed that the work done by the turning moment per revolution is equal to the

work done against the mean resisting torque, therefore the area of the rectangle aAFe is

proportional to the work done against the mean resisting torque.

Turning Moment Diagram for a Four Stroke Cycle Internal Combustion Engine

Fig. 10.2. Turning moment diagram for a four stroke cycle internal combustion engine.

Flywheel



A turning moment diagram for a four stroke cycle internal combustion engine is shown in

Fig. 10.2. We know that in a four stroke cycle internal combustion engine, there is one

working stroke after the crank has turned through two revolutions, i.e. 720° (or 4π radians).

Since the pressure inside the engine cylinder is less than the atmospheric pressure during the

suction stroke, therefore a negative loop is formed as shown in Fig. 10.2. During the

compression stroke, the work is done on the gases, therefore a higher negative loop is

obtained.

During the expansion or working stroke, the fuel burns and the gases expand, therefore a

large positive loop is obtained. In this stroke, the work is done by the gases. During exhaust

stroke, the work is done on the gases, therefore a negative loop is formed. It may be noted

that the effect of the inertia forces on the piston is taken into account in Fig. 16.2.

Turning Moment Diagram for a Multi-cylinder Engine

A separate turning moment diagram for a compound steam engine having three cylinders and

the resultant turning moment diagram is shown in Fig. 10.3. The resultant turning moment

diagram is the sum of the turning moment diagrams for the three cylinders.

It may be noted that the first cylinder is the high pressure cylinder, second cylinder is the

intermediate cylinder and the third cylinder is the low pressure cylinder. The cranks, in case

of three cylinders, are usually placed at 120° to each other.

Fig. 10.3. Turning moment diagram for a multi-cylinder engine.

FLUCTUATION OF ENERGY

The fluctuation of energy may be determined by the turning moment diagram for one

complete cycle of operation. Consider the turning moment diagram for a single cylinder

double acting steam engine as shown in Fig. 10.1. We see that the mean resisting torque line

AF cuts the turning moment diagram at points B, C, D and E. When the crank moves from a

to p, the work done by the engine is equal to the area aBp, whereas the energy required is

represented by the area aABp.

Flywheel



In other words, the engine has done less work (equal to the area a AB) than the requirement.

This amount of energy is taken from flywheel and hence the speed of the flywheel decreases.

Now the crank moves from p to q, the work done by the engine is equal to the area pBbCq,

whereas the requirement of energy is represented by the area pBCq. Therefore, the engine has

done more work than the requirement.

This excess work (equal to the area BbC) is stored in the flywheel and hence the speed of the

flywheel increases while the crank moves from p to q. Similarly, when the crank moves from

q to r, more work is taken from the engine than is developed. This loss of work is represented

by the area C c D.

To supply this loss, the flywheel gives up some of its energy and thus the speed decreases

while the crank moves from q to r. As the crank moves from r to s, excess energy is again

developed given by the area D d E and the speed again increases. As the piston moves from s

to e, again there is a loss of work and the speed decreases. The variations of energy above and

below the mean resisting torque line are called fluctuations of energy.

The areas BbC, CcD, DdE, etc. represent fluctuations of energy. A little consideration will

show that the engine has a maximum speed either at q or at s. This is due to the fact that the

flywheel absorbs energy while the crank moves from p to q and from r to s.

On the other hand, the engine has a minimum speed either at p or at r. The reason is that the

flywheel gives out some of its energy when the crank moves from a to p and q to r. The

difference between the maximum and the minimum energies is known as maximum

fluctuation of energy.

Determination of Maximum Fluctuation of Energy

A turning moment diagram for a multi-cylinder engine is shown by a wavy curve in Fig. 10.4. The

horizontal line AG represents the mean torque line. Let a1, a3, a5 be the areas above the mean torque

line and a2, a4 and a6 be the areas below the mean torque line. These areas represent some quantity

of energy which is either added or subtracted from the energy of the moving parts of the engine.

Let the energy in the flywheel at A = E, then from Fig. 10.4, we have

Energy at B = E + a1

Energy at C = E + a1– a2

Energy at D = E + a1 – a2 + a3

Energy at E = E + a1 – a2 + a3 – a4

Energy at F = E + a1 – a2 + a3 – a4 + a5

Energy at G = E + a1 – a2 + a3 – a4 + a5 – a6

= Energy at A (i.e. cycle repeats after G)

Let us now suppose that the greatest of these energies is at B and least at E. Therefore,

Maximum energy in flywheel = E + a1

Minimum energy in the flywheel = E + a1 – a2 + a3 – a4

Maximum fluctuation of energy,

E = Maximum energy – Minimum energy

= (E + a1) – (E + a1 – a2 + a3 – a4) = a2 – a3 + a4

Flywheel



Fig. 10.4. Determination of maximum fluctuation of energy.

COEFFICIENT OF FLUCTUATION OF ENERGY

It may be defined as the ratio of the maximum fluctuation of energy to the work done per cycle.

Mathematically, coefficient of fluctuation of energy,

Maximum fluctuation of energy

Workdone per cycleEC

The work done per cycle (in N-m or joules) may be obtained by using the following two relations:

1. Work done per cycle = Tmean × θ

where, Tmean = Mean torque, and

θ = Angle turned (in radians), in one revolution.

= 2π, in case of steam engine and two stroke internal combustion engines

= 4π, in case of four stroke internal combustion engines.

The mean torque (Tmean) in N-m may be obtained by using the following relation:

60

2mean

P PT

N

where, P = Power transmitted in watts,

N = Speed in r.p.m., and

ω = Angular speed in rad/s = 2πN/60

2. The work done per cycle may also be obtained by using the following relation :

60Workdone per cycle

P

n

where n = Number of working strokes per minute,

= N, in case of steam engines and two stroke internal combustion engines,

= N /2, in case of four stroke internal combustion engines.

The following table shows the values of coefficient of fluctuation of energy for steam engines and

internal combustion engines.

Flywheel



COEFFICIENT OF FLUCTUATION OF SPEED

The difference between the maximum and minimum speeds during a cycle is called the maximum

fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is called the

coefficient of fluctuation of speed.

Let N1 and N2 = Maximum and minimum speeds in r.p.m. during the cycle, and

1 2Mean speed in r.p.m.2

N NN

Coefficient of fluctuation of speed,

1 2 1 2

1 2

2( )

2S

N N N NC

N N

1 2 1 2

1 2

2( )

(In terms of angular speeds)

1 2 1 2

1 2

2( )v v v v

v v v

(In terms of linear speeds)

The coefficient of fluctuation of speed is a limiting factor in the design of flywheel. It varies

depending upon the nature of service to which the flywheel is employed.

The reciprocal of the coefficient of fluctuation of speed is known as coefficient of steadiness and is

denoted by m.

1 2

1

S

Nm

C N N

ENERGY STORED IN A FLYWHEEL

A flywheel is shown in Fig. 10.5. We have discussed in Art. 16.5 that when a flywheel absorbs

energy, its speed increases and when it gives up energy, its speed decreases.

Let m = Mass of the flywheel in kg,

k = Radius of gyration of the flywheel in metres,

I = Mass moment of inertia of the flywheel about its axis of rotation in kg-m2 = m.k

2,

N1 and N2 = Maximum and minimum speeds during the cycle in r.p.m.,

1 and 2 = Maximum and minimum angular speeds during the cycle in rad/s,

1 2Mean speed in r.p.m.2

N NN

1 2 Mean angular speed during the cycle in rad/s 2

1 2 1 2Coefficient of fluctuation of speed orS

N NC

N

We know that the mean kinetic energy of the flywheel,

2 2 21 1. . .

2 2E I m k

As the speed of the flywheel changes from 1 to 2, the maximum fluctuation of energy,

Flywheel



E = Maximum K.E. – Minimum K.E.

2 2 2 2

1 2 1 2

1 1 1.( ) .( ) ( ) ( )

2 2 2I I I

1 2 1 2 1 2

1.

2I I

……………………….(i)

2 1 2.I

2 2 2. . .S SI C m k C

…………………………..(ii)

2. . SE C

….……………………….(iii)

The radius of gyration (k) may be taken equal to the mean radius of the rim (R), because the

thickness of rim is very small as compared to the diameter of rim. Therefore, substituting k = R, in

equation (ii), we have

2 2

2

. . .

. .

S

S

E m R C

m v C

v = Mean linear velocity (i.e. at the mean radius) in m/s

lab manual theory of machines - darshan institute of ... · department of mechanical engineering...

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