9. In case of internal combustion
engines, 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 four stroke it takes two revolution
of crank shaft.
9
10. 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.
10
11. 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.
11
12. 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. 12
13. In machines where the operation is
intermittent like *punching machines,
shearing machines, riveting 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.
13
14. The function of a **governor in an
engine is entirely different from that of
a flywheel. It regulates the mean
speed of an engine when there are
variations in the load, e.g., when the
load on the engine increases, it
becomes necessary to increase the
supply of working fluid. On the other
hand, when the load decreases, less
working fluid is required..
14
15. The governor automatically controls
the supply of working fluid to the
engine with the varying load condition
and keeps the mean speed of the
engine within certain limits.
The flywheel does not maintain a
constant speed, it simply reduces the
fluctuation of speed.
It does not control the speed
variations caused by the varying load
15
16. 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.
16
17. Let N1 and N2 = Maximum and
minimum speeds in r.p.m. during the
cycle
N = Mean speed in r.p.m.
∴ Coefficient of fluctuation of speed,
17
18. Where, v is the linear speed
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. 18
19. The reciprocal of the coefficient of
fluctuation of speed is known as
coefficient of steadiness and is
denoted by m.
19
20. Energy Stored in a Flywheel
A flywheel is shown in Fig.
When a flywheel absorbs energy, its
speed increases and when it gives up
energy, its speed decreases.
20
22. 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.k2,
N1 and N2 = Maximum and minimum
speeds during the cycle in r.p.m.,
ω1 and ω2 = Maximum and minimum
angular speeds during the cycle in rad/s
22
23. N = Mean speed during the cycle in r.p.m.
N=(N1+N2)/2
ω = Mean angular speed during the cycle in
rad/s
ω = (ω1+ω2)/2
CS = Coefficient of fluctuation of speed,
CS=(N1- N2)/N or (ω1−ω2)/ω
23
24. Mean kinetic energy of the flywheel
As the speed of the flywheel changes
from ω1 to ω2, the maximum fluctuation
of energy,
∆E = Maximum K.E. – Minimum K.E.
24
26. 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,
where v = Mean linear velocity (i.e. at
the mean radius) in m/s = ω.R
26
28. In the above expressions, only the
mass moment of inertia of the flywheel
rim (I) is considered and the mass
moment of inertia of the hub and arms
is neglected. This is due to the fact that
the major portion of the mass of the
flywheel is in the rim and a small
portion is in the hub and arms. Also the
hub and arms are nearer to the axis of
rotation, therefore the mass moment of
inertia of the hub and arms is small.
28
29. 29
Coefficient of Fluctuation of energy CE
The ratio of maximum fluctuation of
energy to the work done per cycle is
known as coefficient of fluctuation of
energy.
CE= maximum fluctuation of energy
work done per cycle
30. Problem 1.The mass of flywheel of an
engine is 6.5 tonnes and the radius of
gyration is 1.8 metres. It is found from
the turning moment diagram that the
fluctuation of energy is 56 kN-m. If
the mean speed of the engine is 120
r.p.m.,
find the maximum and minimum
speeds.
30
31. Given : m = 6.5 t = 6500 kg
k = 1.8 m
∆ E = 56 kN-m = 56 × 103 N-m
N = 120 r.p.m.
Let N1 and N2 = Maximum and
minimum speeds respectively.
Fluctuation of energy (∆ E)
31
34. Problem 2. The flywheel of a steam
engine has a radius of gyration of 1 m
and mass 2500 kg. The starting
torque of the steam engine is 1500 N-
m and may be assumed constant.
Determine:
1. Angular acceleration of the
flywheel, and
2. Kinetic energy of the flywheel after
10 seconds from the start.
34
35. Solution. Given : k = 1 m
m = 2500 kg, T = 1500 N-m
Angular acceleration of the flywheel
Let α = Angular acceleration of the
flywheel.
Mass moment of inertia of the flywheel,
I = m.k2 = 2500 × 12 = 2500 kg-m2
∴ Starting torque of the engine (T),
T=1500 = I.α = 2500 × α or
α = 1500 / 2500
α = 0.6 rad /s2
35
36. 2. Kinetic energy of the flywheel
The angular speed of the flywheel after
10 seconds from the start (i.e.from rest),
Assuming uniform acceleration.
Let ω1 = Angular speed at rest = 0
ω2 = Angular speed after 10 seconds,
t = Time in seconds.
We know that ω2 = ω1 + α t
= 0 + 0.6 × 10
= 6 rad /s
∴ Kinetic energy of the flywheel
= 45 kN-m
36
37. Problem 3. A horizontal cross compound
steam engine develops 300 kW at 90
r.p.m. The coefficient of fluctuation of
energy as found from the turning
moment diagram is to be 0.1 and the
fluctuation of speed is to be kept within
± 0.5% of the mean speed. Find the
weight of the flywheel required, if the
radius of gyration is 2 metres
37
38. Solution. Given :
P = 300 kW = 300 × 103 W
N = 90 r.p.m.
CE = 0.1
k = 2 m
We know that the mean angular
speed, ω = 2 π N/60
= 2 π × 90/60
= 9.426 rad/s
38
39. Let ω1 and ω2 = Maximum and minimum
speeds respectively.
Since the fluctuation of speed is ± 0.5%
of mean speed, therefore
total fluctuation of speed,
ω1 – ω2 = 1% ω = 0.01 ω
coefficient of fluctuation of speed,
39
40. Work done per cycle = P × 60 / N
= 300 × 103 × 60 / 90
= 200 × 103 N-m
∴ Maximum fluctuation of energy,
∆E = Work done per cycle × CE
= 200 × 1000× 0.1 = 20×103 N-m
Let m = Mass of the flywheel.
Maximum fluctuation of energy ( ∆E ),
= m × 22 × (9.426)2 × 0.01
= 3.554 m
∴ m=20 × 1000/3.554 = 5630 kg
40
41. 2. The work done per cycle may also be
obtained by using the following relation :
Work done per cycle
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.
41
42. 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 the turning moment is taken as the
ordinate and crank angle as abscissa.
42
46. Turning moment diagram for Single
Cylinder Double Acting Steam Engine
Torque acting on crank shaft bearing of steam
engine or IC engine is expressed by following
equation.
where FP = Piston effort,
r = Radius of crank,
n = Ratio of the connecting rod length &
radius of crank
θ = Angle turned by the crank from inner dead
centre 46
48. 48
Turning moment diagram
Turning moment diagram is a diagram which
shows the variation of turning moment (or
torque)on a crank on for various positions of
crank.
Turning moment diagram is a graph
between T and θ as shown
50. From the above expression, we see that the
turning moment (T ) is zero, when the crank
angle (θ) is zero.
It is maximum when the crank angle is 90°
and it is again zero when crank angle is
180°.
50
52. The curve abc in
Figure
represents the
turning moment
for outstroke.
The curve cde is the
turning moment diagram for in-stroke.
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.
52
53. 53
Work done = force x displacement
Work done per revolution = (F x r/r) x 2πr
=T x 2π
But T x θ = Area of turning moment diagram
Work done per revolution = Area of turning
moment diagram for one revolution
= Area abcde
In actual practice, the engine is assumed to
work against the mean resisting torque
Tmean= (Area abcde / 2π)
54. The mean resisting torque is shown by a
horizontal line AF.
The height of the ordinate aA represents the
mean height of the turning moment diagram.
Area of rectangle aAFe= aA x ae
= Tmean x 2π
=Area of turning moment dia abcde
= Work done per revolution
54
55. 55
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.
56. 1. When the turning moment is positive (i.e. when
the engine torque is more than the mean resisting
torque) as shown between points B and C (or D
and E) the crankshaft accelerates and the work is
done by the steam.
56
57. 2. When the turning moment is negative (i.e.
when the engine torque is less than the
mean resisting torque) as shown between
points C and D in Figure, the crankshaft
retards and the work is done on the steam.
57
58. 58
3. If T = Torque on the crankshaft at any
instant
Tmean = Mean resisting torque.
Then accelerating torque on the rotating
parts of the engine = T – Tmean
59. 59
4. If (T –Tmean) is positive, the flywheel
accelerates and if (T – Tmean) is
negative, then the flywheel retards.
60. Turning Moment Diagram for a Four
Stroke Cycle Internal Combustion Engine
A turning moment diagram for a four stroke
cycle internal combustion engine is shown in
figure. 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 4 π radians). 60
62. 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
62
63. 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.
63
64. 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 Figure.
64
66. 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.
66
67. 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.
We see that the mean resisting torque line
AF cuts the turning moment diagram at
points B, C, D and E.
67
69. 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.
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 the flywheel
and hence the speed of the flywheel decreases.
69
70. 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
70
71. 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 CcD.
71
72. 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 DdE and the speed again increases.
72
73. 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.
73
74. 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.
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.
74
75. Determination of Maximum Fluctuation
of Energy
A turning moment diagram for a multi-
cylinder engine is shown by a wavy curve in
Figure.
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.
75
76. 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 Figure, 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)
76
77. 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
77
79. 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,
79
80. 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,
θ = Angle turned (in radians), in one
revolution.
=2 π, in case of steam engine and two
stroke internal combustion engines
=4 π, in case of four stroke internal
combustion engines.
80
81. The mean torque (Tmean) in N-m may be
obtained by using the following relation :
where
P = Power transmitted in watts
N = Speed in r.p.m.
ω=Angular speed in radians/s
= 2 πN/60
81
82. 2. The work done per cycle may also be
obtained by using the following relation :
Work done per cycle
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.
82
83. The following table shows the values of
coefficient of fluctuation of energy for steam
engines and internal combustion engines.
83
84. 84
Dimensions of the Flywheel Rim
Consider a rim of the flywheel as shown in Fig.
85. 85
Let D = Mean diameter of rim in metres,
R = Mean radius of rim in metres,
A = Cross-sectional area of rim in m2
,
ρ=Density of rim material in kg/m3
,
N = Speed of the flywheel in r.p.m.,
ω = Angular velocity of the flywheel in rad/s,
v = Linear velocity at the mean radius in m/s
= ω .R = π D.N/60, and
σ = Tensile stress or hoop stress in N/m2
due to the
centrifugal force.
86. 86
Consider a small element of the rim as shown
shaded in figure.
Let it subtends an angle δθ at the centre of the
flywheel.
Volume of the small element
= A × R.δθ ` A= bxt
∴ Mass of the small element
dm= Density × volume = ρ.A.R.δθ
Centrifugal force on the element,
acting radially outwards,
dF = dm.ω2.R = ρ.A.R.δθ.ω2R
dF = ρ.A.R2.ω2.δθ
89. 89
Problem 4. The turning moment diagram for a multi-
cylinder engine has been drawn to a scale of 1 mm to
500 N-m torque and 1 mm to 6° of crank displacement.
The intercepted areas between output torque curve and
mean resistance line taken in order from one end, in sq.
mm are – 30, + 410, – 280, + 320, – 330, + 250, – 360, +
280, – 260 sq. mm, when the engine is running at 800
r.p.m. The engine has a stroke of 300 mm and the
fluctuation of speed is not to exceed ± 2% of the mean
speed. Determine a suitable diameter and cross-section
of the flywheel rim for a limiting value of the safe
centrifugal stress of 7 MPa. The material density may be
assumed as 7200 kg/m3. The width of the rim is to be 5
times the thickness
90. 90
Given : N = 800 r.p.m. or
ω = 2π × 800 / 60 = 83.8 rad/s
Stroke = 300 mm
σ = 7 MPa = 7 × 106 N/m2
ρ = 7200 kg/m3
Fluctuation of speed is ± 2% of mean speed,
Total fluctuation of speed,
ω1 – ω2 = 4% ω = 0.04 ω
91. 91
Coefficient of fluctuation of speed,
Cs= (ω1- ω2)/ω = 0.04
Diameter of the flywheel rim
D = Diameter of the flywheel rim in metres, and
v = Peripheral velocity of the flywheel rim in m/s.
We know that centrifugal stress (σ)
7 × 106
= ρ.v2
= 7200 v2
Or v2
= 7 × 106
/7200
= 972.2
∴ v = 31.2 m/s
We know that
v = π D.N/60
∴ D = v × 60 / π N = 31.2 × 60/π × 800 = 0.745 m
92. 92
Cross-section of the flywheel rim
Let t = Thickness of the flywheel rim in metres
b = Width of the flywheel rim in metres = 5 t
∴ Cross-sectional area of flywheel rim,
A = b.t = 5 t × t = 5 t2
First of all, let us find the mass (m) of the
flywheel rim. The turning moment diagram is
shown in figure
93. 93
Since the turning moment scale is 1 mm = 500 N-m and
crank angle scale is 1 mm= 6°= 6π /180 rad =π /30 rad
1 mm2 on the turning moment diagram = 500 × π / 30
= 52.37 N-m
Let the energy at A = E, then referring to Fig.
Energy at B = E – 30 . . . . . . . . . . . . . (Minimum energy)
Energy at C = E – 30 + 410 = E + 380
Energy at D = E + 380 – 280 = E + 100
Energy at E = E + 100 + 320 = E + 420 . . . (Max energy)
Energy at F = E + 420 – 330 = E + 90
Energy at G = E + 90 + 250 = E + 340
Energy at H = E + 340 – 360 = E – 20
94. 94
Energy at K = E – 20 + 280 = E + 260
Energy at L = E + 260 – 260 = E = Energy at A
We know that maximum fluctuation of energy,
∆E = Maximum energy – Minimum energy
=( E + 420) – (E – 30) = 450 mm2
= 450 × 52.37 = 23 566 N-m
We also know that maximum fluctuation of
energy (∆E),
23 566 = m.v2.CS
= m × (31.2)2 × 0.04
= 39 m
∴ m = 23566 / 39 = 604 kg
95. 95
We know that mass of the flywheel rim (m),
604 = Volume × density
= π D.A.ρ
= π × 0.745 × 5t2 × 7200
= 84 268 t2
∴ t2 = 604 / 84 268
= 0.007 17 m2 or
t = 0.085 m = 85 mm
b =5 t = 5 × 85
= 425 mm
96. 96
Problem 5. A single cylinder double acting
steam engine develops 150 kW at a mean
speed of 80 r.p.m. The coefficient of fluctuation
of energy is 0.1 and the fluctuation of speed is
± 2% of mean speed. If the mean diameter of
the flywheel rim is 2 metre and the hub and
spokes provide 5% of the rotational inertia of
the flywheel, find the mass and cross-sectional
area of the flywheel rim. Assume the density of
the flywheel material (which is cast iron) as
7200 kg/m3.
97. 97
Given : P = 150 kW = 150 × 103 W
N = 80 r.p.m. or ω = 2 π × 80 /60 = 8.4 rad/s
CE = 0.1
D = 2 m or R = 1 m
ρ = 7200 kg/m3
Since the fluctuation of speed is ± 2% of mean
speed, therefore total fluctuation of speed,
ω1 – ω2 = 4% ω = 0.04 ω
coefficient of fluctuation of speed,
Cs= (ω1 – ω2)/ ω = 0.04
98. 98
Mass of the flywheel rim
Let m = Mass of the flywheel rim in kg, and I =
Mass moment of inertia of the flywheel in kg-m2.
We know that work done per cycle = P × 60/N
= 150 × 103 × 60 / 80
= 112.5 × 103 N-m
Maximum fluctuation of energy
∆E = Work done /cycle × CE
= 112.5 × 103 × 0.1
= 11 250 N-m
Maximum fluctuation of energy (∆E)
11 250 = I.ω2.CS = I × (8.4)2 × 0.04 = 2.8224 I
∴ I = 11 250 / 2.8224 = 3986 kg-m2
99. 99
Since the hub and spokes provide 5% of the
rotational inertia of the flywheel, therefore, mass
moment of inertia of the flywheel rim (Irim) will be
95% of the flywheel,
Irim = 0.95 I = 0.95 × 3986
= 3787
Irim = m.k2
m= Irim / k2 = 3787/12 = 3787 kg
Cross-sectional area of flywheel rim in m2.
We know that the mass of the flywheel (m)
m = 3787 = π D× A × ρ = π × 2 × A × 7200
= 45 245 A
∴ A = 3787/45 245 = 0.084 m2
103. 103
The function of a flywheel in an engine is to reduce
the fluctuations of speed, when the load on the
crankshaft is constant and the input torque varies
during the cycle.
The flywheel can also be used to perform the same
function when the torque is constant and the load
varies during the cycle.
Such an application is found in punching press or
in a rivetting machine. A punching press is shown
diagrammatically in Fig. The crank is driven by a
motor which supplies constant torque and the
punch is at the position of the slider in a slider-crank
mechanism.
104. 104
We see that the load acts only during the rotation
of the crank from θ 1 to θ 2, when the actual
punching takes place and the load is zero for the
rest of the cycle.
Unless a flywheel is used, the speed of the
crankshaft will increase too much during the
rotation of crank from θ = 0 to θ=θ1 and again from
θ = θ2 to θ= 0 or 2pi, because there is no load
while input energy continues to be supplied.
On the other hand, the drop in speed of the
crankshaft is very large during the rotation of crank
from θ = θ1 to θ = θ2 due to much more load than
the energy supplied.
105. 105
Thus the flywheel has to absorb excess energy
available at one stage and has to make up the
deficient energy at the other stage to keep to
fluctuations of speed within permissible limits.
This is done by choosing the suitable moment of
inertia of the flywheel.
Let E1 be the energy required for punching one
hole. (For one punching operation)
This energy is determined by the size of the hole
punched, the thickness of the material and the
physical properties of the material.
Let d1= Diameter of the hole punched
t1 = Thickness of the plate
Ƭu = Ultimate shear stress for the plate material.
106. 106
∴ Maximum shear force required for punching,
FS = Area sheared × Ultimate shear stress .
=π d1 t1 Ƭu
It is assumed that as the hole is punched, the
shear force decreases uniformly from maximum
value to zero.
For one revolution (crank rotation of 2 π radians)
energy supplied by motor is E1
For crank rotation of (θ2 –θ1) radians
E1……2 π
θ2 –θ1
energy supplied by motor will be equal to is
= E1 (θ2 –θ1 )
2 π
107. 107
Assuming one punching operation per revolution,
the energy supplied to the shaft per revolution
should also be equal to E1 .
The energy supplied by the motor to the
crankshaft during actual punching operation,
E2= E1 (θ2 –θ1 )
2 π
∴ Balance energy
E1- E2 = E1- E1 (θ2 –θ1 )
2 π
This energy is to be supplied by the flywheel and
hence there is decrease in its kinetic energy of
flywheel and its speed falls from maximum to
minimum.
108. 108
Thus maximum fluctuation of energy,
The values of θ 1 and θ 2 may be determined only if the
crank radius (r), length of connecting rod (l) and the
relative position of the job with respect to the
crankshaft axis are known. In the absence of relevant
data, we assume that
(θ2 –θ1 ) = t = t
2 π 2s 4r
Where t = Thickness of the material to be punched
s = Stroke of the punch = 2 Crank radius = 2r
By using the suitable relation for the maximum fluctuation
of energy (∆E) as discussed in the previous articles, we
can find the mass and size of the flywheel.
109. 109
Problem 6. A punching press is driven by a
constant torque electric motor. The press is
provided with a flywheel that rotates at maximum
speed of 225 r.p.m. The radius of gyration of the
flywheel is 0.5 m. The press punches 720 holes
per hour; each punching operation takes 2
second and requires 15 kN-m of energy.
Find the power of the motor and the minimum
mass of the flywheel if speed of the same is not
to fall below 200 r. p. m.
Power = work done per second
=720 x 15 x 1000 / (60x60)
=3000 Nm/s
110. 110
Minimum mass of the flywheel
Let m = Minimum mass of the flywheel.
Since each punching operation takes 2 seconds,
therefore energy supplied by the motor in 2
seconds,
E2= 3000 x 2 = 6000 N-m
∴ Energy to be supplied by the flywheel during
punching or maximum fluctuation of energy,
∆ E=E1-E2= 15 x103 − 6000 = 9000N-m
Mean speed of the flywheel,
N=(N1+N2)/2 = (225+200)/2
= 212.5 r.p.m
∆ E=I W2Cs
112. 112
Problem 7. A machine punching 38 mm holes
in 32 mm thick plate requires 7 N-m of energy
per sq. mm of sheared area, and punches one
hole in every 10 seconds. Calculate the power
of the motor required. The mean speed of the
flywheel is 25 metres per second. The punch
has a stroke of 100 mm.
Find the mass of the flywheel required, if the
total fluctuation of speed is not to exceed 3%
of the mean speed. Assume that the motor
supplies energy to the machine at uniform rate.
113. 113
Given : d = 38 mm ; t = 32 mm ; E= 7 N-m/mm2 of
sheared area ; v = 25 m/s ; s = 100 mm ;
v1 – v2 = 3% v = 0.03 v
Power of the motor required
Sheared area,
A = π d t = π x 38 x32 = 3820 mm2
Since the energy required to punch a hole is 7 N-
m/mm2 of sheared area, therefore total energy
required per hole,
E1 = 7 N-m/mm2 x 3820 mm2 = 26 740 N-m
Also the time required to punch a hole is 10
second, therefore energy required for punching
work per second = 26 740/10 = 2674 N-m/s
∴ Power of the motor required
= 2674 W = 2.674 kW
114. 114
Mass of the flywheel required
Let m = Mass of the flywheel in kg.
The stroke of the punch is 100 mm, distance
travelled per stroke is 200 mm and it punches one
hole in every 10 seconds, therefore the time
required to punch a hole in a 32 mm thick plate
200…….10 sec
32………? 10 x 32 = 1.6 s
200
∴ Energy supplied by the motor in 1.6 seconds,
E2 = 2674 x 1.6 = 4278 N-m
Energy to be supplied by the flywheel during
punching or the maximum fluctuation of energy,
∆E = E1-E2 = 26 740 - 4278 = 22 462 N-m
115. 115
Coefficient of fluctuation of speed,
Cs = (v1 − v2)/25 =0.03
Maximum fluctuation of energy ∆E
22 462 = m v2 Cs = m 25 2 × 0.03 = 18.75
∴ m = 22 462 / 18.75 = 1198 kg
Maximum fluctuation of energy (∆E) can also be
found out by following method
Energy required for one punch,
E = 26 740 N-m
116. 116
Problem 8. A riveting machine is driven by a
constant torque 3 kW motor. The moving parts
including the flywheel are equivalent to 150 kg at
0.6 m radius. One riveting operation takes 1
second and absorbs 10 000 N-m of energy. The
speed of the flywheel is 300 r.p.m. before riveting.
Find the speed immediately after riveting. How
many rivets can be closed per minute?
117. 117
Given : P = 3 kW ; m = 150 kg ; k = 0.6 m ; 1
N = 300 r.p.m. or ω1 = 2π × 300/60 = 31.42 rad/s
Speed of the flywheel immediately after riveting
Let ω2 = Angular speed of the flywheel immediately
after riveting.
Energy supplied by the motor,
E1= 3kW = 3000 W = 3000N-m/s (∵ 1 W = 1 N-m/s)
Energy absorbed during one riveting operation which
takes 1 second,
E2= 10 000 N-m E
∴ Energy to be supplied by the flywheel for each riveting
operation per second or the maximum fluctuation of
energy,
∆E = E1- E2 = 10 000 - 3000 = 7000 N-m
119. 119
Number of rivets that can be closed per minute
Since the energy absorbed by each riveting
operation which takes 1 second is 10000 N-m,
Number of rivets that can be closed per minute,
= Energy supplied by motor in one minute
Energy required for one rivetting
= (E2/E1) x 60
=(3000 /10000) x 60
= 18 rivets
120. 120
Problem 9. A punching press is required to
punch 40 mm diameter holes in a plate of 15
mm thickness at the rate of 30 holes per
minute. It requires 6 N-m of energy per mm2 of
sheared area. If the punching takes 1/10 of a
second and the r.p.m. of the flywheel varies
from 160 to 140, determine the mass of the
flywheel having radius of gyration of 1 metre
121. 121
Given: d = 40 mm; t = 15 mm
No. of holes = 30 per min.
Energy required = 6 N-m/mm2
Time = 1/10 s = 0.1 s
N1 = 160 r.p.m. N2 = 140 r.p.m. k = 1m
We know that sheared area per hole
= π dt = π × 40 ×15 = 1885 mm2
∴ Energy required to punch a hole,
E1 = 6 ×1885 = 11310 N-m
Energy required for punching per second
= Energy required per hole × No. of holes per
second
= 11310 × 30 = 5655 N-m/s
60
122. 122
Since the punching takes 1/10 of a second,
therefore, energy supplied by the motor in 1/10
second,
E2 = 5655 ×1/10 = 565.5 N-m
Or … Cutting in (1/10)sec.. Takes 2 seconds
? E1
E1x(1/10)/2 = 565.6 Nm
∴ Energy to be supplied by the flywheel during
punching a hole or maximum fluctuation of
energy of the flywheel,
E=(E1 - E2) = 11 310 - 565.5
= 10 744.5 N-m
Mean speed of the flywheel,
=(N1+N2)/2 = (160 + 140)/2
= 150 r.p.m.
124. 124
Problem 10. A punching machine makes 25 working strokes per
minute and is capable of punching 25 mm diameter holes in 18
mm thick steel plates having an ultimate shear strength 300 MPa.
The punching operation takes place during 1/10th of a revolution
of the crankshaft. Estimate the power needed for the driving
motor, assuming a mechanical efficiency of 95 percent. Determine
suitable dimensions for the rim cross-section of the flywheel,
having width equal to twice thickness. The flywheel is to revolve at
9 times the speed of the crankshaft. The permissible coefficient of
fluctuation of speed is 0.1. The flywheel is to be made of cast iron
having a working stress (tensile) of 6 MPa and density of 7250
kg/m3. The diameter of the flywheel must not exceed 1.4 m owing
to space restrictions. The hub and the spokes may be assumed to
provide 5% of the rotational inertia of the wheel.
125. 125
n = 25, d1 = 25 mm = 0.025 m, t1 = 18 mm = 0.018 m,
Ƭu = 300 MPa = 300 N/mm2, ηm = 95% = 0.95,
Cs = 0.1, σ = 6 MPa = 6 × 106 N/m2, ρ = 7250 kg/m3,
D = 1.4 m or R = 0.7 m
Power needed for the driving motor
The area of plate sheared
A = π d1×t1 = π × 25 × 18 = 1413.71 mm2
∴ Maximum shearing force required for punching,
Fs = As × Tu = 1414× 300 = 424 200N
Energy required per stroke = Average shear force ×
Thickness of plate
= (Fs × t1 )/2 = (½) × 424200 × 0.018 = 3817.8 N-m
∴ Energy required per min = Energy/stroke × No. of
working strokes/min
= 3817.8 × 25
= 95 450 N-m
126. 126
We know that the power needed for the driving motor
= Energy required per min = 95 450
60 ηm (60 × 0.95)
= 1675 W = 1.675 kW
Dimensions for the rim cross-section
Let t = Thickness of rim in metres
b = Width of rim in metres = 2t ... …..(Given)
∴ Cross-sectional area of rim
A= b x t = 2 t x t = 2 t2
127. 127
Since the punching operation takes place (i.e. energy is
consumed) during 1/10th of a revolution of the
crankshaft, therefore during 9/10th of the revolution of a
crankshaft, the energy is stored in the flywheel.
∴ Maximum fluctuation of energy,
∆E = (9/10) x Energy/stroke = (9/10)× 3817.8
= 3436N-m
Let m = Mass of the flywheel in kg.
Since the hub and the spokes provide 5% of the
rotational inertia of the wheel, therefore the maximum
fluctuation of energy provided by the flywheel by the rim
will be 95%.
∴ Maximum fluctuation of energy provided by the rim,
∆Erim = 0.95 x ∆E = 0.95 x 3436 = 3264 N-m
128. 128
Since the flywheel is to revolve at 9 times the speed of
the crankshaft and there are 25 working strokes per
minute, therefore, mean speed of the flywheel,
N = 9 × 25 = 225 rpm.
mean angular speed ω =2π × 225/60 = 23.56rad/s
Maximum fluctuation of energy ∆Erim
3264 = m R2 ω2 Cs = m x 0.7 x 23.562 x 0.1
=27.2 m
∴ m = 3264/27.2 = 120 kg
mass of the flywheel (m),
120 = π D × A x ρ = π × 1.4 × 2 t2 × 7250
= 63782 t2
t2 = 120 / 63782 = 0.00188 or
t = 0.044 m = 44 mm
b=2t = 2 × 44 = 88 mm
129. 129
Formulas used
Thus maximum fluctuation of energy,
The values of θ 1 and θ 2 may be determined only if the
crank radius (r), length of connecting rod (l) and the
relative position of the job with respect to the
crankshaft axis are known. In the absence of relevant
data, we assume that
(θ2 –θ1 ) = t = t
2 π 2s 4r
Where t = Thickness of the material to be punched
s = Stroke of the punch = 2 Crank radius = 2r
By using the suitable relation for the maximum fluctuation
of energy (∆E) as discussed in the previous articles, we
can find the mass and size of the flywheel.
131. 131
Let E1 be the energy required for punching one
hole. (For one punching operation)
This energy is determined by the size of the hole
punched, the thickness of the material and the
physical properties of the material.
Let d1= Diameter of the hole punched
t1 = Thickness of the plate
Ƭu = Ultimate shear stress for the plate
material.
∴ Maximum shear force required for punching,
FS = Area sheared × Ultimate shear stress .
=π d1 t1 Ƭu
134. 134
we know that
θ2 –θ1 ) = t = t
2 π 2s 4r
Where t = Thickness of the material to be punched
s = Stroke of the punch = 2 Crank radius = 2r
135. 135
Problem 1.
P= 4 kW
M 140 kg
k =0.5
Rivetting operation takes 1 s
E1 9000 Nm
N =420 rpm Before rivetting
N2= ?
Number of rivets closed per hour
Problem 2.
A punching press is required to punch 40 mm diameter holes in a
plate of 30 mm thickness at the rate of 4 holes per minutes. It
requires 6 Nm of energy per square mm of sheared area. The punch
has a stroke of 100 mm. The rpm of the flywheel varies from 320 to
280. If the radius of gyration of flywheel is 1 m find 1. mass of the
flywheel
2. Power of the motor for punching
136. 136
Flywheel
Vlab address
Disc type and rim type
https://dom-nitk.vlabs.ac.in/exp/rim-type-flywheel/simulation.html
https://dom-nitk.vlabs.ac.in/exp/disc-type-flywheel/simulation.html
