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Transcript of ME 361 Unit 6
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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
= + ; = +
= ;