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UNIT 6FLYWHEELS AND GOVERNORS

ME 361DYNAMICS OF MACHINERY

BSC MECHANICAL ENGINEERING, 3RD YEAR

INSTITUTE OF DISTANCE LEARNING/DEPARTMENT OF MECHANICAL ENGINEERING

KWAME NKRUMAH UNIVERSITY OF SCIENCE AND TECHNOLOGY, KUMASI

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Introduction

Large variation in accelerations within mechanisms can causesignificant variations in inertia and pin forces, supports reactions andtorque required to drive it at a constant or near constant speed.

To stabilize the back-and-forth flow of kinetic energy of rotatingequipments, flywheels are often attached to shafts.

Flywheels are used in machines as an energy reservoir. They storeenergy during periods when the energy is more than required, andrelease energy to the system when the energy requirement is morethan supplied.

Governors are also used to control the mean speed of machines byregulating the flow of working fluid/fuel, which power the machines.They rely on centrifugal force to function and control variation ofspeeds caused by load variation.

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FUNCTIONS OF FLYWHEELS

Reduce amplitude of speed fluctuation.

Reduce maximum torque required.

Store and release energy when needed during

cycling.

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Speed Fluctuation

2

21

If 1and 2 are the maximum and minimum speeds, respectively, then the

average speed is defined as:

21

S

C

Coefficient of speed fluctuation, Cs, is defined as the change in speed per

the average speed. Thus,

Another terminology which is sometimes used to describe speed

fluctuation is coefficient of steadiness, S, defined as the inverse of the

coefficient of speed fluctuation.

SC

S1

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Energy Fluctuation

Coefficient of Energy fluctuation CEis defined as the change in energy per work done

in one cycle.

TCycleperdoneWork

EEC

E

Where

= the angle of turn per cycle. The angles of turn for two-stroke and four-strokeengines are 2and 4, respectively.

The average torque, T, may be defined in terms of the power transmitted, P, and

angular speed, , as

PT

n

P60CycleperdoneWork where n = number strokes per minute

n = N in case of steam engines and two-stroke engines

n = N/2 in case of four-stroke engines

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Maximum Energy Fluctuation

S

S

CIE

Therefore

CIIE

IIIEEE

2

2121

2

2

2

1

2

2

2

121

)())((2

1

)(

2

1

2

1

2

1

isenergyofnfluctuatiomaximumThe

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Turning Moment Diagram of One-

Cylinder Four-Stroke IC Engine

Turning moment diagram (also known as crank effort diagram) is aplot of turning moment, T, on the vertical axis against the crank

angle on the horizontal axis.

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Example 6-1

The turning moment diagram for a two-stroke one-cylinderengine is shown below. If the mean torque is 77 N-m, find (a)

the maximum energy fluctuation (b) coefficient of energy

fluctuation.

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Solution

Let the energy at A be E. Then, the energy at various points are:

Energy at B = E +52

Energy at C = E + 52-70 = E-18

Energy at D = E + 52-70+140 =E +122

Energy at E = E + 52-70+140-45 =E + 77

The maximum energy fluctuation is, E = maximum of energy from Ato E - minimum of energy from A to E

E = (E +122) - (E-18) = 140.

The coefficient of Energy fluctuation is given by

289.0

277

140

TCycleperdoneWork

E

E

C

EEC

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Example 6-2A machine running at an average speed of 300 rev/min isdriven through a single reduction gear from an enginerunning at an average speed of 60 rev/min. The moment ofinertia of the rotating parts on the machine shaft isequivalent to 110 kg at a radius of 0.3 m and that of the

rotating parts of the engine shaft 18 kg at a radius of 0.3 m.

The torque transmitted to the machine from the engine is2500 + 675 sin 2N m, where is the angle of rotation of

the machine from some datum. The torque required to drivethe machine is 2500 + 270 sin N m. Find the coefficient offluctuation of speed.

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SolutionThe torque/crank angle curves for the engine and

machine are shown below

The engine and resisting torques are-equal when

2500 + 675 sin 2= 2500 + 270 sin

i.e. when 5 sin cos = sin , either sin = 0 or cos =0.5

i.e. = n, n=0,1,2etc; and 7828', 28123' etc.

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Solution contd

The greatest fluctuation of energy occurs between points B andC, or between C and D, and is represented by the areas a or b,

i.e. fluctuation of energy =

= (270 sin 675 sin 2)d = 972 N m281 23

78 28

Equivalent moment of inertia of machine shaft

= (110 x 0.32+ 18 x 0.32)x = 16.38 kg

= Therefore

=

=

. =0.06 or 6%

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Acceptable Values Coefficient of Speed

Fluctuation for Various Machines

Type of Equipment Coefficient of Fluctuation

Crushing Machinery 0.0200

Electrical Machinery 0.003

Direct Driven Electrical Machinery 0.002

Engine with belt transmission 0.030

Flour Milling Machinery 0.020

Gear Wheel Transmission 0.02

Hammering machine 0.200

Machine Tools 0.030Paper-making Machinery 0.025

Pumping Machinery 0.03-0.050

Shearing Machinery 0.03-0.050

Spinning Machinery 0.010-0.020

Textile Machinery 0.025

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Example 6-3

A four-stroke engine is to deliver 450 kW at a

mean speed of 100 rpm. The coefficient of

energy fluctuation is 0.2 and operational speeds

must be kept within 4% of the mean speed.Determine the radius of gyration of the flywheel

if the mass of the flywheel must not exceed

5000 kg.

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Example 6-3 conti;

ave

10041

maxave

04.1max

ave

100

41

min ave 96.0

min

ave

aveave

ave

sC

96.004.1minmax

08.0s

C

2/100

10x45060

2/

6060CycleperdoneWork

3

N

P

n

P

kJ/cycle900CycleperdoneWork

2.010x900cycleperdonework 3

EE

CE

180000E

2

2

602x10008.0180000

IICE aves

2m-kg20517.53I

2mkI 5000

5.20517mIk

m03.2k

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GOVERNORSGovernors are broadly classified into centrifugal and

inertia governors. Centrifugal governors are based onbalancing of centrifugal force and moments of a

rotating mass by an equal force and moments

provided by either by action of a weight or springs.

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GOVERNORS

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Centrifugal Governors

The centrifugal governors may further be classified as follows

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Centrifugal Governors

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Watt Governor

hg

mgrrhm

rmFF

rwhM

CC

O

/

force,lcentrifugatheis

0F0

2

2

2

C

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Example 6-4Find the vertical height of a Watt governor rotating at 65 rpm.

Also determine the change in height corresponding to 1 rpmincrease in the speed.

Solution:

mm-6.515

60

2x1

60

2x65

81.922

givesequationabovetheintorpm1andrpm65ngSubstituti

2yieldsatingdifferentiPartially

mm212

60

2x65

81.9

33

32

22

dg

dh

d

dg

dhg

h

gh

P G

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Porter Governor

From (a) = 0 ; = +1 = 0 ; + = 2From (b)

= 0 ; ( ) = n 3(3) into (1)

= = + /

Substituting these into (1) = : / +

= +

+

= +

( 1 + ) =

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Porter Governor

n= number of balls

m= mass of each ball

g= acceleration due to gravityr= ball radius

W= weight of sleeve

F= frictional force

= + ( 1 + )

where =

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Example 6-5

The upper and lower arms of a three-ball Porter governor are 300 mm

and 250 mm long, respectively. The upper arms are pivoted on the axis

of rotation, and the lower arms pivoted on the sleeve at a distance 40

mm from the axis of rotation. The total mass of the sleeve and the load

are 40 kg, and mass of each fly ball is 2 kg. Neglecting friction,

determine the speed of rotation of the governor if the radius of each

ball is 200 mm.

Solution

= s i n = 4 1 . 8; = s i n

= 3 9 . 8

= =0.9318; =0.894

= 0; = 3; = = 392.4 N; w = mg = 19.62 N = +

( 1 + )2 0.2

0.894 =19.62+

392.40

3 1+0.9318

=24.67rad/s 235.6 rpm

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Proell Governor

G

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Proell Governor

21

43

21

13

431321

0

(a),FigureofCpointatmomentTaking

0

hh

xxA

hh

xxwF

xxAxxwhhF

M

FWnA

F

yC

yC

C

y

y

21

43

21

132

21

43

21

132

1

1

1

hh

xxFW

nmrhh

xx

r

g

FWn

A

hh

xxFW

nhh

xxwrmF

y

C

P ll G

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Proell Governor

Let the lengths of DE, AD and DI be

denoted by L1, L2and L3, respectively.

cos1

Lh

sin12

Lx

uxxrx

21

Using Pythagoras theorem, the vertical

extension of the lower arm is

2

1

2

31 xLh

llu xLxxxx sin

242

2

2sin

L

xxxlu

cos22

Lh

tan23 hx

tan24

hx

Example 6 6

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Example 6-6

A Proell governor has four flyballs, each of mass 3.5 kg. The

lengths of all the arms are 250mm and distance of each pivot ofarm from the axis of rotation is 30mm. The length of extension oflower arms on which each fly ballis attached is 80 mm, and inclined

at 30o

to the lower arm. The massof the sleeve is 60 kg. Knowingthat the arms are inclined at 50o tothe axis of rotation, determine theequilibrium speed of the governor.

Solution; W=60x9.81=588.6 N; m=3.5 kg; F=0; n=4;

=80cos 20 = 75.2 , =250 cos 50 = 160.7 , r=+ +30=248.9 ===250 sin 50 = 191.5 , = 80sin20 = 27.4 mmTherefore=17.7 rad/s

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Hartnell Governor

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Hartnell Governor

sin

geometrylevertheFrom

tan1

tan1

cos1

sincos

0

2

2

yRr

r

g

y

xSW

nmr

wyxSW

nrmF

xSWn

wyyF

M

C

C

pivot

From the lever geometry,

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Example 6-7

The extreme radii of rotations of fly balls of Hartnell type of governor are 80

mm and 120 mm. The governor has a central sleeve spring and three right-angled bell crank levers, which vertical and horizontal arms 120 mm and 80mm, respectively. The levers are pivoted 120 mm from the axis of rotation.Each fly ball has a mass of 3 kg. Knowing that the two extreme speeds of thegovern are 400 rpm and 450 rpm for a sleeve lift of 18 mm, and that thehorizontal arm is perpendicular to the axis of rotation at the minimum speed,determine (a) total load on the sleeve at the lowest and highest speeds, (b)

stiffness of the spring, and (c) deflection of the spring at the minimum speed ifthe mass of the sleeve is 50 kg. Neglect friction at all contacts.

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Solution

0tan81.9312.060

2x40033

80

120tan

2

11

2

11

mgrmn

xySW

N4.28421 SW

The inclination of the horizontal arm to the horizontal axis is given by

80

18sin

x

h

o

0.13

13sin120120sin2

yRr mm99.1462 r

o

2

222

22 0.13tan81.9314699.060

2x45033

80

120tan

mgrmn

xySW

N3.44982 SW

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Solution contd

m0196.0

10x2.09

4.1165

N4.1165

9.81x509.16559.1655

kN/m0.92

10x18

9.1655

9.1655

9.1655

N9.1655

4.28423.4498

0

3

1

0

01

1

1

3-

12

12

1212

h

k

Sh

khS

S

WS

k

hk

khSS

SS

SSSWSW

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HARTUNG GOVERNOR

tan1

tan1

coscos1

sincos

0

2

2

r

g

y

y

mr

S

y

xW

nmr

wy

yS

y

xW

nrmF

SyWxn

wyyF

M

s

sC

sC

pivot

E l 6 8

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Example 6-8

The extreme radii of rotations of the three fly balls of Hartung type of governor are 80mm and 120 mm. The governor has a central sleeve spring and three right-angled bell

crank levers, which vertical and horizontal arms are 120 mm and 80 mm, respectively.The levers are pivoted at 120 mm from the axis of rotation. Each fly ball has a mass of3 kg. A retaining spring is attached to each the vertical arm of each crank bell lever at100 mm from the pivot of the crank and to the frame of the governor. The axis of eachspring is perpendicular to the axis of rotation. Knowing that the two extreme speeds ofthe govern are 400 rpm and 450 rpm for a sleeve lift of 18 mm, and that the horizontalarm is perpendicular to the axis of rotation at the minimum speed, determine (a)stiffness of the spring, and (b) deflection of the spring at the minimum speed if themass of the sleeve is 50 kg. Neglect friction at all contacts.

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Solution

m0113.0

80

10010x9.1849.267

N49.267

99.75726667.0

kN/m9.18

1000

18

80

10026.425

9.7572.118326667.026667.0

2.118326667.0

0.13tan9.81x314699.060

2x4503

120

80

3

1

120

100

1

1

3

11

1

1

12

1212

12

2

2

2

h

hhx

ykS

S

WS

k

kSS

hxykhh

xykSS

WSWS

WS

WS

s

ss

o

mm99.146

13sin120120sin

0.13

80

18sin

positionmaximumthe@

99.75726667.0

0tan9.81x312.060

2x4003

120

80

3

1

120

100

tan1

positionminimumthe@

2

2

1

2

1

11

2

11

r

yRr

x

h

WS

WS

wrmy

xW

ny

yS

o

s

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Pickering Governors

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Pickering Governors

The maximum deflection of a leaf spring with both ends fixed

and carrying load P at the centre is given by

Neglecting the mass of the left spring, the total load on the spring

is due to centrifugal force given by

The lift is given in terms of the deflection and the length lby

the empirical formula

EI

Pl

192

3

3

12

1btI

amP

2

EI

lam

192

32

l

h

24.2

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Example 6-9Each left spring of Pickering governor which drives gramophone

is 5 mm wide and 0.12 mm thick. When the governor is at rest,the effective length of each leaf spring is 45 mm and the distance

from the axis of rotation to the centre of gravity of each 22-g

governor mass attached to each leaf spring is 10 mm. Find the

speed of the governor when the lift of the sleeve is 1.0 mm. Take

E = 200 GPa.

rad/s103.2

10x2210x4410x282.410x01

10x.282410x.2710x200192192

mm282.4;145

4.21

4.2

m10x2.71000

12.0

1000

5

12

1

12

1

3-33-3-3-

3-16-9

3

2

2

2

416-

3

3

mla

EI

l

h

btI

Wil H t ll d

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Wilson-Hartnell governors and

Auxiliary Control Force Elements

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Example 6-10

The figure shows a throttle valvegovernor. Each ball has a mass of

6.5 kg and the arms are of equal

length. The speed range is 425 to

440 rev/min with a sleeve movementof 6 mm. The minimum ball path

radius is 122 mm. If the combined

strength of the ba11 springs is 15

kN/m of extension, find the rate of

the auxiliary spring.

S l i

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SolutionFBD

(a) (b)

From (a), = 0125P-2R(250)=0; R=0.25P

From (b), = 0

= = 0 . 2 5

F is the initial tension in the ball s rin

Therefore@top speed

6.5 425 260

0.122=0.25

1570.75=0.25 1

@Bottom speed

( + 180) 6.5 440 2

60

(0.122

+0.006)=0.251586.385=0.25 2

=

= 62.549

3 1 0 = 20.85 /

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Properties of Governors

The Effortof a governor is the force exerted at the sleeve for a given fractional change in speed.

The Powerof a governor is the work done at the sleeve for a given fractional change in speed, i.e. power=mean effort x sleeve movement

Sensitiveness(coefficient of fluctuation of speed)

The sensitiveness of a governor is defined as the ratio of the difference between the maximum and minimumequilibrium speeds to the mean speed.

Hunting

When a governor is too sensitive, the governor causes the speed of the engine or the system is controlling tofluctuate continuously above and below the equilibrium speed. Such a governor is said to be hunting.

Stability

A governor is said to be stable there is only one radius of rotation of the fly balls for a particular speed withthe operating range at which the governor is in equilibrium. For stable governor, the radius of rotation

increases with the speed increases.

Isochronous

A governor is said to be isochronous when the equilibrium speed is constant for all radii of rotation of flyballs with the working range. Gravity controlled governors such as Porter governor are not isochronous, butspring loaded governors are capable. The isochronous governors have no practical use because the sleevemove to its extreme position immediately the speed deviates from the isochronous speed. It is possible to

use this governor as angular speed-dependent toggle position switch.

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Properties of GovernorsFor stable governor, all radius of rotation

increases with the speed increase.

= ; =

= ; >

Isochronous

A governor is said to be isochronous when theequilibrium speed is constant for all radii of

rotation.

= ; =

= ; =

Unstable

= + ; = +

= ;