Here are the steps to solve this problem: a) Power of the engine - Power stroke area = 6000 mm2 = 6000 x 10 Nm = 60,000 Nm - Resisting torque is uniform, so work done = Area x Resisting torque - Assuming resisting torque is 10 Nm, work done per cycle = 60,000 Nm - Work done per cycle x rpm = Power - Given: Work done per cycle is 60,000 Nm - Assuming rpm is N - Then, Power = 60,000 x N W b) Energy to be stored by flywheel - Total area under the diagram = Power stroke + Exhaust + Suction + Compression

1. Turning Moment (Crank Effort) Diagram
for a 4-stroke I C engine
Crank Angle 
TorqueN-m
0   
Suction Compression
Expansion
Exhaust

T
T
max
mean
Excess Energy
(Shaded area)

2. Turning Moment (Or Crank Effort) Diagram (TMD)
Turning moment diagram is a graphical
representation of turning moment or torque
(along Y-axis) versus crank angle (X-axis) for
various positions of crank.
Uses of TMD
1. The area under the TMD gives the work done
per cycle.
2. The work done per cycle when divided by the
crank angle per cycle gives the mean torque Tm.

3. Uses of TMD
3. The mean torque Tm multiplied by the angular
velocity of the crank gives the power consumed by
the machine or developed by an engine.
4. The area of the TMD above the mean torque
line represents the excess energy that may be
stored by the flywheel, which helps to design the
dimensions & mass of the flywheel.

4. FLYWHEEL
Flywheel is a device used to store energy when
available in excess & release the same when
there is a shortage.
Flywheels are used in IC engines, Pumps,
Compressors & in machines performing
intermittent operations such as punching,
shearing, riveting, etc.
A Flywheel may be of Disk type or Rim Type
Flywheels help in smoothening out the
fluctuations of the torque on the crankshaft &
maintain the speed within the prescribed limits.

5. DISK TYPE FLYWHEEL
DISK TYPE FLYWHEEL
D

6. RIM TYPE FLYWHEEL
Section X-X
X X
b
t
D

7. 2
, where m=Mass of the flywheel.
k=Radius of gyratio
Flywheels posess inertia due to its heavy mass.
Mass moment of inertia of a flywheel is given by
I = mk
Comparision between Disk Type & Rim Type Flywheel :
2 2
n of the flywheel.
For rim type, k= where D=Mean diameter of the flyheel
2
For Disk type, k= where D=Outer diameter of the flywheel
2 2
Hence I=m and I=m
4 8
Rim Disk
D
D
D D   
   
   
Hence for a given diameter & inertia,the mass of the
rim type flywheel is half the mass of a disk type flywheel

8. 1 2
It is the difference between the maximum & minimum speeds
in a cycle. (= )n n
Important Definitions
(a) Maximum fluctuation of speed :
(b)Coefficient of fluctuation of
1 2 1 2
)
It is the ratio of maximum fluctuation of speed to the mean speed.
It is often expressed as a % of mean speed.
(or K )
2
where =Angular velocity=
60
s s
s s
or K
n n
C
n
n
 



    
    
   
 

 
speed :(C


9. 1 2
)
It is the ratio of maximum fluctuation of energy to the
mean kinetc energy.
(or K )
e e
e e
or K
E E
C


Important Definitions
(c)Coefficient of fluctuation of energy :(C
It is the reciprocal of coeffi
E e
E E E
     
     
    
 
 
 
(d)Coefficient of steadiness :
e e
** It is often expressed as the ratio of excess energy
e
to the work done per cycle.C ( or K ) =
W.D / cycle
1 2
cent of fluctuation of speed.
Coefficient of steadiness=

 
 
  
 

10. 1 2
Let
be the mass moment of inertia of the flywheel
& be the max & min speeds of the flywheel
Mean speed of the flywheel
m=Mass of the flywheel, k=Rad
I
 
 
EXPRESSION FOR ENERGY STORED BY A FLYWHEEL
 
 
s
2 2 2 2
1 2 1 2 1 2
1 2 1 2
ius of gyration of the flywheel
C =Coefficient of fluctuation of speed
The max fluctuation of energy (to be stored by the flywheel)
1 1 1
2 2 2
1
( )
2
e E E I I I
e I
   
   
     
   

11.  1 2
2
1
Putting the mean agular speed = ,
2
We get Multiplying & dividing by ,
Also ,the coefficient of fluctuation osC
  





EXPRESSION FOR ENERGY STORED BY A FLYWHEEL
1 2
1 2
1 2
e = Iω(ω - ω )
(ω - ω )
e = Iω
(ω - ω )
2
2
2
2
2 2
f speed
Hence
Putting I=mk ,we get
1
Alternatively, if Mean kinetic energy E= ,
2
2 , e=2EC
Note: 1
2 But O
.
R
s
s e
I
I E
e e
C C
E E

 
  
s
se = mk ω C
e = Iω C
e
s
C
= 2
C

12. 2 21
e= I , Putting mean Kinetic energy E=
2
and expressing C as a percentage,
2EC
100
1
Alternatively, if Mean kinetic
0.02
energy ENote: =.
2
2
s
s
s
se E
e
k
C
C I
m
 



EXPRESSION FOR ENERGY STORED BY A FLYWHEEL
2
2
2
2 2 2
,
1
( ) , E=
2
s
k v mv
e mv c

 
 

13. 2
We know that
mass m=Density Volume
For Disk type flywheel, Volume = t
4
For Rim type flywheel, Volume= D( )
where A= Cross section of t
D
A





MASS OF FLYWHEEL IN TERMS OF
DENSITY & CROSSECTION AREA
2
he rim =b t
b= width of rim & t= thickness of the rim
(i)Velocity of the flywheel
(ii) Hoop Stress (Centrifugal stress) in the flywheel
where = density of flywheel mate
Note:
v=
ri
/ sec
60
= v al
Dn
m

 


14. Problem 1
A single cylinder 4 stroke gas engine develops
18.4 KW at 300 rpm with work done by the
gases during the expansion being 3 times the
work done on the gases during compression.
The work done during the suction & exhaust
strokes is negligible. The total fluctuation of
speed is 2% of the mean. The TMD may be
assumed to be triangular in shape. Find the
mass moment of inertia of the flywheel.

15. Crank Angle 
TorqueN-m
0   
Suction
Compression
Expansion
Exhaust

T
T
max
meanx
Excess Energy
TURNING MOMENT DIAGRAM

16. 3
Power P=18.4 KW=18.4 10 W, Mean speed n=300 rpm
Work done during expansion W 3 Work done during compression
2% 0.02
Given 4-stroke cycle engine
Crank angle per cycle=
E
sC

 
 

Data :
4π radians( 2 rev of cra
3
3
31.416 rad
2
Angular Velocity of flywheel =
60
2 300
. .
60
Also power P=T 18.4 10 T 31.416
/sec
Mean tor
18.4 10
31.41
que T 585. N-m
6
7
m
m
m
n
i e






 
    

  
Solution :
nk shaft)

17. W.D/C
Work done per cycle=T Crank angle per cycle
i.e. W.D/Cycle =T 4 585.7 31.416
W.D/Cycle W.D during expansio
y
n
cl
W.D during compression
(As the W.D during suction &
e 7360 N-m
co
m
m 

  

 

Work done per cycle
max
mpression are neglected)
7360=(W W )
Given W 3 Or W , we can write
3
2
7360= W
3 3
1
. . 11040
2
11040 N-m
Max
E C
E
E C C
E
E E E
W
W
W
W
W
i e T

 
 
 
   
 
  

This work represents the area under triangle for expansion stroke
max 7tor 028.qu 3 -e mNT 

18. max
max
The shaded area represents the excess energy.
1
. .excess energy stored by flywheel e= ( )
2
where is the base of shaded triangle, given by
( )
mean
mean
i e x T T
x
T Tx

  


Excess energy stored by the flywheel
max
max
max
( ) (7028.3 585.7)
2.88
7028.3
1
Hence e= 2.88 (7 9276.67 N-028.3 585.7)
2
m
mean
T
T T
x rad
T
 
 
     
   

19. 2
2
2
We know that excess energy is given by
e=I 9276.64 (31.416) 0.02
Hence mass moment of inertia of flywheel
I=470 Kg-m
sC I    

20. Problem 2
A single cylinder internal combustion engine
working on 4-stroke cycle develops 75 KW at
360 rpm. The fluctuation of energy can be
assumed to be 0.9 times the energy
developed per cycle. If the fluctuation of
speed is not to exceed 1% and the maximum
centrifugal stress in the flywheel is to be 5.5
MN/m2, estimate the diameter and the cross
sectional area of the rim. The material of the
rim has a density 7.2 Mg/m3.

21. 3
3 3
Power P=75 KW=75 10 W, Mean speed n=360 rpm
Fluctuation of energy =0.9 W.D/cycle
4 stroke cycle Crank angle per cycle=
Density =7.2 Mg/m 7200 Kg/m ,Hoop stress =5.5 MPa
Angu
 




Data :
Solution :
e
4π radians
3
3
2
lar Velocity of flywheel =
60
2 360
. .
60
Also
37.7 rad/sec
Mean torque T 1989.4
power P=T 75 10 T 37.7
75 10
37.7
N-m
m
m
m
n
i e 





 
    

  

22. 2
W.D/Cycle
Work done per cycle=T Crank angle per cycle
i.e. W.D/Cycle =T 4 1989.4 4
Also given
Hoop s
25000
tress
N-m
= v 5.5 1
m
m  
 


  

 
Work done per cycle :
Diameter of the flywheel :
e = 0.9 ×W.D / cycle = 22500 N - m
6 2
0 =7200 (v )
Hence,
360
60 60
Dn D 


 

velocityof flywheel v = 27.64m / sec
Also
Diameter of the flywheel =
v = 2
1.4
7.64 =
66 m

23. 2 2
2 2
The energy stored by the flywheel is given by
.
1.466
For rim type, radius of gyration k= 0.733
2 2
37.7) 0.01
But , for rim type, m
s
D
m

 
   
Hence, Mass of the flywheel m = 2946.4 Kg
e = mk C
22500 = m ( 0.733) (
ass m= DA
(where A=cross section area of the rim)
1.466 7200A
 


    

2946.4
2
A= 0.09m

24. 2
2
If it is given that the rectangular cross section of the
rim has width (b)=3 thickness ( t),
Then A=b t=3t t=3t
t=0.1732m 173 mm
b=3t=
0.09 3
520 mm
t  

 

Note :

25. Problem 3
The crank effort diagram for a 4-stroke cycle gas
engine may be assumed to for simplicity of four
rectangles, areas of which from line of zero
pressure are power stroke =6000 mm2, exhaust
stroke =500 mm2, Suction stroke=300 mm2,
compression stroke = 1500 mm2. Each Sq mm
represents 10 Nm. Assuming the resisting torque
to be uniform, find
a) Power of the engine
b) Energy to be stored by the flywheel
c) Mass of a flywheel rim of 1m radius to limit the
total fluctuation of speed to ±2% of the mean
speed of 150 rpm.

26. Crank Angle 
TorqueN-m
0   
Suction
Compression
Expansion
Exhaust

T
mean
T
max
Excess energy
(Shaded area)

27. 4 stroke cycle Crank angle per cycle=
Radius of gyration k 1 meter, Mean speed n=150 rpm
C 2% 4% 0.04 ( Total fluctuation=2 Fluctuation on either side)
Angular Velocity of fl
s


    
Data :
Solution :
4π radians
  2
15.71 rad/sec
WD/cycle=W.D during Expansion-(W.D during other strok
2
ywheel =
60
2 150
. .
60
W.D/cycle= 6000-(300 1500 500) 3700mm
W.D/cycle 3700 scale of diagram=3700 10=37000 N-m
Mean Torque
es)
n
i e 


 
 
   
  

W.D/cycle 37000
T 2944.4 N-m
Crank angle/cycle 4
m

  

28. max
max
max
2944.4 15.71
But W.D during expansion =T
6000 10 T
Substituting for T
46.256m WP T K


    

   

(i) Power developed by engine :
(ii) Energy stored by flywheel :
maxT = 19098.6N - m
max me = Shaded area = π( T -T )
max
,
( ) (19098.6 2944.6)me T T    
e = 50749.27 N - m
Crank Angle 
TorqueN-m
0   
Suction
Compression
Expansion
Exhaust

T
mean
T
max
Excess energy
(Shaded area)

29. 2 2
2 2
We know that energy stored by flywheel
50749.27 (1) (15.71) 0.04
se mk C
m

  

(iii) Mass of flywheel
Mass of flywheel m = 5140.64 Kg

30. Problem 4
A multi cylinder engine is to run at a speed of
600 rpm. On drawing the TMD to a scale of
1mm=250 Nm & 1mm=30, the areas above &
below the mean torque line are +160, -172,
+168, -191, +197, -162 mm2 respectively.
The speed is to be kept within ±1% of the mean
speed. Density of Cast iron flywheel=7250
kg/mm3 and hoop stress is 6 MPa. Assuming
that the rim contributes to 92% of the
flywheel effect, determine the dimensions of
the rectangular cross section of the rim
assuming width to be twice the thickness.

31. 1 2 3 4 5 6
160
172 191
197
162
168TurningMoment
Crank angle
Mean Torque
line
Let the energy at 1=E
Energy at 2=(E+160)
Energy at 3=(E+160)-172=(E-12)
Energy at 4=(E-12)+168=(E+156)
Energy at 5=(E+156)-191=(E-35)
7
Energy at 6=(E-35)+197=(E+162)
Energy at 7=(E+162)-162=E= Energy at 1
Hence, Maximum fluctuation of energy
(in terms of area) = (E+162)-(E-35)=197 Sq mm

32. 2
2 2 600
Angular velocity = 62.84 / sec,
60 60
1% 2% 0.02
3
1mm 250 13.1
180
Max Fluctuation of
s
N
rad
C
Nm
 



 
   
 
   
 
Scale of t
Energy stored by the fl
he dia
ywhee
gra
l :
m is
2
2 2
energy (Max.K.E-Min K.E)
e=(E+162)-(E-35)=197 mm
. .2581=I I (62.84) 0.02
Mass moment of inertia
s
e
i e C



  
 2
e =197 ×13.1 = 2581Nm
I = 32.7Kg - m

33. 2 6 2
Using = v ; 6 10 7250
Velocity
600
Also v= 28.8=
60 60
Mean dia of flywheel D=0.92 m
v=28.8 m ec/s
v
DN D
 
 
  

 


Dia
G
meter of the flywhee
iven 92%of the
l :
flywh
2 2 2 2
2
0.92 2581 2375
( )
2375 (28.8) 0.02
rim
rim s s s
Nm
mk c m k c mv c
m
 
  
  
   

eel effect is provi
Mass of rim m =
ded by
143 kg.
the rim,
e
e

34. 2
We know that mass of the flywheel rim
m=Volume of rim density=( DA)
143 ( 0.92 A) 7250
A=0.00682
As cross section of rim is rectangular with b=2t,
A (
4m
= b t
 

 
    


Dimensions of the crossection of the rim :
2 2
)=2t 0.006824 2t 
Hence t = 58.4 mm,b = 2t = 116.8 mm.

35. Problem 5
Torque –output diagram shown in fig is a
single cylinder engine at 3000 rpm. Determine
the weight of a steel disk type flywheel
required to limit the crank speed to 10 rpm
above and 10 rpm below the average speed of
3000 rpm. The outside diameter of the
flywheel is 250 mm. Determine also the
weight of a rim type flywheel of 250 mm
mean diameter for the same allowable
fluctuation of speed.

36. 0 90 180 360 450 540 630 720
25
50
75
100
-25
-50
-75
-100
T
N-m
 (Degrees)

37. 0
Crank angle per cycle=720
Mean speed n=3000 rpm,
250
Radius of gyration k= =125 mm =0.125 m
2
0.25
Radius of gyration k= =0.0884 m
2 2
10 20
C 0.00
3000 3000
s
(For rim type)
(For disk type)

  
Data :
= 4π radians
667
2
Angular Velocity of flywheel =
60
2 3000
. . 314.16 rad/ e
0
c
6
s
n
i e





 
Solution :

38. =75 50 100 75 50 100 75
2 2 2 2 2 2
W.D per c
WD/c
ycle 87.5
Mean torque T
Crank angl
ycle=Net are
e per c
a under TM
ycle 4
D
m
     



           
                
           
 
 m
W.D / cycle = 87.5π N - m
T = 21.875N - m

39. 0
90 180 360 450 540 630 720
25
50
75
100
-25
-50
-75
-100
T
N-m
 (Degrees)
Tmean
1 2 3 4
5
6 7 8

40. Let the energy at 1=E
Energy at 2=E+(75-21.875) 26.5625
2
Energy at 3=( 26.5625 )-(71.875) 9.735
2
Energy at 4=( 9.735 )+(1 68.7500-21.875)
Energy
E
E E
E E



 
  
 
  
 
 
   
 
  
Excess energy stored by flywheel
at 5=( 68.75 )-(96.875) 20.3125
2
Energy at 6=( 20.3125 ) (50-21.875) 34.375
2
Energy at 7=( 34.375 )-(121.875)
2
Energy at 8=( 26.5625 )+(75-21.875)
2
2
6.5625
E E
E
E
E
E
E E

 


 




 
   
 
 
    
 
 
  
 
 
  
 


41. 2 2
2 2
= Max Energy-Min energy
e=(E+68.75 ) (E-26.5625 )
We know that energy stored by flywheel
299.43 (0.125) (314.16) 0.00667
se mk C
m
 

 

  

Mass of flywheel
Excess energy e
Rim type
Mass of flywheel m
299.43Nm

= 29.11 Kg
Fordisk type,k = 0.0884 m
Mass of flywheel m = 58.221 Kg

42. Problem 6
The torque required for a machine is shown in
fig. The motor driving the machine has a mean
speed of 1500 rpm and develop constant
torque. The flywheel on the motor shaft is of
rim type with mean diameter of 40 cm and
mass 25 kg. Determine;
(i) Power of motor
(ii) % variation in motor speed per cycle.

43. 
400 N-m
2000 N-m
Torque
Crank angle
   

44. Crank angle per cycle=
Mean speed n=1500 rpm,
40
Radius of gyration k= =20 cm =0.2 m
2
m=25 kg
2
Angular Velocity of flywheel
1
=
60
2 1500
. .
6
57.08 e
0
rad/s
(For rim type)
n
i e



 
 
Data :
Solution :
2π radians
c

45. W.D/cycle area 1+area 2+area 3
1
=400 2 (2000 400) (2000 400)
4 2 2
W.D per cycle 1600
Mean torque T
Crank angle per cycle 2
m
 




     
           
     
 

(i) Power developed by the engine :
m
W.D / cycle =1600π N - m
T
P=T 800 157.08m 

   
Power developed by the en
125.664 KW
gine
= 800 N - m

46. 
400 N-m
800 N-m
2000 N-m
Crank angle
   
mean
Excess energy e
(shaded area)
1
2 3
Torque
x

47.  
   
1200
From the similar triangles, 1.178
1600
2
Energy stored by flywheel = Shaded area
1
= 2000 800 1.178 2000 800
4 2
We know that energy st
s
x
x rad
e


  
 

   
(ii) Coefficent of fluctuation of speed
e
C
e =1649.28Nm
2 2
2 2
ored by the flywheel
1649.28 25 (0.2) (157.08)
s
s
e mk C
C

  
Coefficient of fluctuation of speed = 0.0668 = 6.68%

48. Problem 7
A 3 cylinder single acting engine has cranks set
equally at 1200 and it runs at 600 rpm. The TMD
for each cylinder is a triangle, for the power
stroke with a maximum torque of 80 N-m at 600
after dead center of the corresponding crank.
The torque on the return stroke is zero. Sketch
the TMD & determine the following;
(i) Power developed
(ii) Coefficient of fluctuation of speed if mass of
flywheel is 10 kg and radius of gyration is 8 cm.
(iii)Maximum angular acceleration of flywheel.

49. T (N-m)
0
60 120 180 240 300 360
80N-m
 degrees

50. Crank angle per cycle=
Mean speed n=600 rpm,
Radius of gyration k=8 =0.08 m
m=10 kg
2
Angular Velocity of flywheel =
60
2 600
. .
6
62.83 r e
0
ad/s c
cm
n
i e 


 
 
Data :
Solution :
2π radians

51. W.D/cycle area of 3 triangles
1
=3 80
2
W.D per cycle 377
Mean torque T
Crank angle per cycle 2
m
m



 
    
 
 
 mT = 60 N - m
max
mean
377 N - m
As the maxim
(i) Mean to
um torque (T ) is 80 Nm,
and
rque
T
T :
= 60 minNm,the minimum torque (T )
will be = 40 N - m. Hence the modified TMD
may be drawn as shown in fig.

52. T (N-m)
0
60 120 180 240 300 360
80 N-m
 degrees
60 Nm
40 Nm
Modified TMD for 3 Cylinder engine

53.  
 
60 62.83
20
From the similar triangles,
32 40
3
Due to symmetry,the energy stored by flywheel
=Area of (Shaded portion)
1
= 80 60
32
meanP T
x
x rad
e




    
  
   
(i) Power developed :
any one traingle
3.77 Kw
e =10.47 N - m

54. 2 2
2 2
We know that energy stored by the flywheel
10.47 10 (0.08) (62.83)
s
s
e mk C
C

  
Coefficient of fluctuation of speed = 0.04
(ii) Coefficeint of fluctuation of speed :
(iii) Maximum angular
14 = 4.1
acce
4%
lera
 
max
2
2
2
We know that T=I ,where
T=Max fluctuation of torque=(T )
I=mk ,the mass moment of inertia of flywheel
= Max angular acceleration, rad/sec
20 10(0.08)
meanT




 
   ∴α = 312.5 2
tion of flywheel
rad / sec

55. Problem 8
A torque delivered by a two stroke is represented
by T=(1000+300 sin 25 cos 2 Nm where 
is the angle turned by crank from IDC. The engine
speed is 250 rpm. The mass of the flywheel is 400
kg and the radius of gyration is 400 mm. Determine
(i) The power developed
(ii) Total percentage fluctuation of speed
(iii) The angular acceleration and retardation of
flywheel when the crank has rotated through an
angle of 600 from the IDC
(iv) Max & Min angular acceleration & retardation of
flywheel.

56. 0 0
As the torque is a function of 2 ,
equate 2 360 180
Mean speed n=250 rpm,
Radius of gyration k=400 =0.4 m
m=400 kg
Angular Velocity of fl
mm

   
 
Data :
Solution :
0
The crank angle per cycle = 180 π radians
2
ywheel =
60
2 250
. .
6
26.18 rad/se
0
c
n
i e





 

57. 0 0
W.D per cycle
Mean torque T
Crank angle per cycle
1 1
(1000 300sin 2 500cos2 )
1000 26.18
m
Td d
P
 
   
 

 
   

 
  
 
Mean torque
Power developed by eng
(i) Power developed by
ine P
engine :
=
m
m
T =1000 N - m
T
26.18 KW

58. Sl
No
Angle

Torque
T N-m
1 0 500
2 30 1010
3 60 1510
4 90 1500
5 120 990
6 150 490
7 180 500
0 180
Tmean =1000 N-m
500 Nm
T
(N-m)
Crank Angle
Excess
Energy 
mean
)

59. 1
2 1 2
The excess energy stored by the flywheel is
given by integrating between the limits
& where & correspond to points where
T=T Or (T-Tmean mean

  
(ii)Coefficient of fluctuation of speed :
ΔT
) = = 0
i.e. (300sin 2 500cos2 ) 0
500
Hence tan2 = 1.667
300
(As the torque curve intercepts the mean torque line at these points)
 

 
 
 
 
 0 0 0 0
1
0 0
1 2
22θ = 59
ΔT
Hence θ = 29.
&
5 &
2θ =( 180 +5
θ =119
9 ) = 2
.5
39

60. 2
1
119.5
29.5
.
(300sin 2 500cos2 )
T d
d



  

 


Excess energy e =
e =
(The above integration may be perf
(ii)Coefficient
ormed using
calcul
of fluctuation of
ator by keeping in
speed (contd..
radian mode
...)
and
2 2 2 2
Also e=mk 583.1 400 (0.4)
C 0.0
(
133 1.33
2 8)
%
6.1s s
s
C C


   
 

sub
e = 583.1 N - m
Coefficient of fluctua
sti
tio
tuting
the limits of integration in radia
n of speed
ns)

61. 0
0
Acceleration (or retardation) is caused by excess (or deficit)
torque measured from mean torque at any instant.
At =
i.e T
60 , 300sin(2 60) 50
.
  


(iii) Angular acceleration at 60 crank position
ΔT
 2
0
0cos(2 60)
Now,
509.8 400 (0.4)
Hence Angular acceleration at 60 crank positi
50
n
9 8
o
. N m
I




  




0
2
60
ΔT
ΔT
α =7.965 rad / sec

62. max
Maximum acceleration (or retardation) is caused by
maximum fluctuation of torque from mean, i.e
max(To find ΔT first find the crank positions at which
ΔT is ma
(iv) Maximum angular acceleration :
ΔT
0
0 0 0
For max value of T, ( T) 0
(300sin 2 500cos2 ) 0
. 600 co
tan2 =-0.6 2 =-31 &
2 (180 ( 31
s 2 +1000 sin2 0
H
) 149
ence
ximum &then substitute those values in the
equation of ΔT.)
d
d
d
d
i e

 

 

 
  


   
 


63. 0
0
max
max 2
At 2 =-31 , 583.1N-m (causes retardation)
At 2 =149 , 583.1 (Causes acceleration)
Max angular acceleration of flywheel
583.1
400(0.4)
T
T
T
I



  
  


  
(iv) Maximum angular acceleration (contd) :
9.11
min
min 2
Max angular retardation of flywheel
583.1
400(0.4)
(-ve sign indicates retardation)
T
I


 
  
2
2
rad / sec
-9.11 rad / sec

64. Problem 9
A machine is coupled to a two stroke engine which
produces a torque of (800+180 Sin 3 N-m where  is
the crank angle. The mean engine speed is 400 rpm.
The flywheel and the rotating parts attached to the
engine have a mass of 350 kg at a radius of gyration of
220 mm. Calculate;
(i) The power developed by the engine
(ii) Total percentage fluctuation of speed when,
(a) The resisting torque is constant
(b) The resisting torque is (800+80 Sin

65. Sl No Angle

Torque
T N-m
1 0 800
2 30 980
3 60 800
4 90 620
5 120 800
(a)When the resisting torque is Constant
0 120
T
(N-m)
Crank Angle
0 0
300
600
900
Excess Energy
T
E
T
E = Engine torque
=mean TorqueT
m
Tm Nm

66. 0 0
As the torque is a function of 3 ,
equate 3 360 120
Mean speed n=400 rpm, m=350 kg
Radius of gyration k=220 =0.22 m
Angular Velocity o
mm

   
 
Data :
Solution :
0 2
The crank angle per cycle = 120
3
π
radians
41.89 r
2
f flywheel =
60
2 400
. . ad/s
60
ec
n
i e 


 
 

67. 2 2
3 3
0 0
W.D per cycle
Mean torque T
Crank angle per cycle
1 1
(800 180sin3 )
2 2
3 3
800 41.89
m
Td d
P
 
  
 


  

 
  
 
Mean torque
Power developed by engine
(i) Power developed by engi
P
ne :
=
m
m
T = 800 N
33.
- m
T
51 KW

68. 1
2 1 2
The excess energy stored by the flywheel is
given by integrating between the limits
& ,where To find &

  
(ii)Coefficient of fluctuation of speed
(a)When the resisting torque is constant
:
ΔT
0 0
are crank positions
at which T=T (T-T
i.e. (800 180sin3 800) 0
180sin3 0 3 0 & 3 1 0
0
8
mean mean
(As the torque curve intercepts the mean torque line at these points)

  
  
    

Or
0
1 2
ΔT =
θ = 0 &
)
θ
=
= 600

69. 2
1
60
0
.
(180sin3 )
T d
d



 




Excess energy e =
e =
(The above integration may be performed using
calc
(ii)Co
ulator
efficient of fluctuation of sp
by keeping in radian mode and
eed (contd..
substitutin
...)
g
the
2 2 2 2
Also e=mk 120 350 (0.22) (41.89)
C 0.00404 0.404%s
s sC C     


 
e = 120 N - m
Coefficient of fluctuation of speed
limits of integration in radians)

70. (b)When the resisting torque is (800 + 80sin )
0 120
T
(N-m)
Crank Angle
0 0
300
600
900
Excess Energy
180
0
T
E
T
M
T
E
T
M
= Engine torque
=Machine Torque




71. 1
2 1
The excess energy stored by the flywheel is
given by integrating between the limits
& ,where To find &


 
(ii)Coefficient of fluctuatio
(b)When the resisting torque is (80
n of speed :
0 + 80sin )
ΔT
2
3
are crank positions
at which T =T (T -T
i.e. (800 180sin3 ) (800 80sin ) 0
180(3sin 4sin ) 0 i
0
8 s
E M E M
(As the engine torque curve intercepts the
machine torque curve at these points)

 
 
   
  
Or Δ) = T =
 2
n =0
(460-720sin 0 sin 0.799

   
 0 0
1 2θ = 53 &θ =( 180 -53) =127

72. 2
1
127
53
.
(180sin3 80sin )
T d
d



  

 


Excess energy e =
e =
(The above integration may be performed
(ii)Coeff
using
calcu
icient of f
lator by ke
luctuation of sp
eping in radian
e
m
ed (contd.....)
ode and substi
2 2 2 2
Also e=mk 208.3 350 (0.22) (41
C 0.007
.89)
0.7%
s
s
sC C     
 

tuting
the limits of integration
e = -208.3 N - m
Coefficient of fluctuation of sp
in radians)
(Take absolute valu
eed
e)

73. Problem 10
A certain machine requires a torque of
(500+50sin N-m to drive it, where  is the
angle of rotation of the shaft. The machine is
directly coupled to an engine which produces a
torque of (500+60 sin2 Nm. The flywheel and
the other rotating parts attached to the engine
have a mass of 500 kg at a radius of 400 mm. If
the mean speed is 150 rpm. Find;
(a) The maximum fluctuation of energy
(b)Total % fluctuation of speed
(c) Max & Min angular acceleration of the flywheel
& the corresponding shaft positions.

74. 0
T
(N-m)
Crank Angle
0
Excess Energy
180
0
TE
T
M
TE
T
M
= Engine torque
=Machine Torque




75. 0 0
As the torque is a function of 2 ,
equate 2 360 180
Mean speed n=150 rpm,
Radius of gyration k=400 =0.4 m
m=500 kg
Angular Velocity of fl
mm

   
 
Data :
Solution :
0
The crank angle per cycle = 180 π radians
2
ywheel =
60
2 150
. .
6
15.71 rad/se
0
c
n
i e





 

76. 1
2 1 2
The excess energy stored by the flywheel is
given by integrating between the limits
& ,where To find & are crank positions
at which T =T (T -TE M E M

  
(i) Excess energy stored by flywheel :
Or
ΔT
)
 
 
0 0
0
i.e. (500 60sin 2 ) (500 50sin ) 0
(60sin 2 50sin )=0
sin 12cos 5 0.
0
0 ,18Either sin 0
5or 12cos 5 0 cos
12
Considering max difference between consecutive
0
65.37
 
 
 
 


   
 
    
     
=
Put sin2θ = 2sinθ
Δ
cosθ
T =
crank positions,
0 0
1 2θ = 65.37 &θ = 180

77. 2
1
180
65.37
.
(60sin 2 50sin )
T d
d



  






Excess energy e =
e =
(The above integration may be performed using
calculator by keeping in radian mode and substituting
the limits of integration in radian
e = -
s)
120.
2 2 2 2
Also e=mk 120.42 500 (0.4) (15.71)
C 0.0061
s s
s
C C     
  
(Take absolute value)
Coefficient o
42 N - m
(i
f fluctuat
i)Coefficient of fluctuation of speed :
ion of speed 0.61%

78. max
Maximum acceleration (or retardation) is caused by
maximum fluctuation of torque from mean, i.e
max(To find ΔT first find the crank positions at wh
(iii) Maximum & Minimum angular acceleration :
ΔT
For max value of T, ( T) 0
(60sin 2 50sin ) 0
. .12 cos 2 - 5cos 0
ich
ΔT is maximum &then substitute those values in the
equation of ΔT.)
d
d
d
d
i e

 

 
  
  


79. 2
0
0
2
0
&
. .12 cos 2 - 5cos 0
Put cos 2 (2cos 1), we get
12(2cos 1) - 5cos 0
At :
T=60 sin(2 35)-50sin(35)=
27.7
0.3
8
6
0
4max
i e  
 
 





 
 
 
 
27.7 N
Maximum
- m
raacce d /leration sec=
2
= 35 =12
24cos θ -5cosθ -12 = 0
5
7.6
= 3
0
At :
T=60 sin(2 127.6)-50sin(127.6)=-9
1
97.
.
0
6
22
2
8
max

  
 
2
2
7.62
Maximum retardation
N - m
rad / se= c
=127.6

80. Flywheel for Punch press
Crank
Plate
Die
Punching
tool
Flywheel
Flywheel for Punch Press
Crank shaft
connecting
rod
d
t

81. Flywheel for Punch press
• If ‘d’ is the diameter of the hole to be punched in
a metal plate of thickness ‘t’ , the shearing area
A=d t mm2
• If the energy or work done /sheared area is
given, the work done per hole =W.D/mm2 x
Sheared area per hole.
• As one hole is punched in every revolution,
WD/min=WD/hole x No of holes punched /min
• Power of motor required P=WD per min/60

82. 1 2
1
2
)
2(E = Energy supplied per sec× Actual time of punching)
e =( E E
where;
E = Energy required per hole
E = Energy
Excess energy Stored by Flywheel :
supplied during actual punching

83. Problem 11
A punching machine carries out 6 holes per min.
Each hole of 40 mm diameter in 35 mm thick
plate requires 8 N-m of energy/mm2 of the
sheared area. The punch has a stroke of 95 mm.
Find the power of the motor required if the
mean speed of the flywheel is 20 m/sec.
If the total fluctuation of speed is not to exceed
3% of the mean speed, determine the mass of
the flywheel.

84. 2
Mean speed of flywheel v=20m/sec
3% 0.03, Diameter of hole d=40 mm
Thickness of plate t=35 mm, Energy/mm =8 N-m
Stroke length =95 mm,
No of holes/min=6 Speed of crank=6rpm
Time required to pu
sC  


Data :
3
3
nch one hole=
W.D/hole=
W.D/hole holes/min
Power of motor= KW
60 10
35185.4 6
60 10
P
 



 

Solution :
2
10 secs
Sheared area per hole = πdt = π×40× 35 = 4398.23 mm
4398.23 8 = 35186 N
3.51
- m
86KW

85. Thickness of plate
Time taken per cycle
2×stroke length
35
10=
1 09
actual
actual

 1.842 Secs
Excess energy supplied by flyw
As the punch travels 95 ×2 =190 mm in10 secs
⇒actual time taken to punch one hole
T =
T =
2
e=Energy required/hole Energy supplied during actual punching
= (e


 
   
heel
2
s
35186 3518.6 1.842) = 28705 N - m
Also e = mv C 28705 m ( 20) 0.0
m = 239
3
2 Kg

86. Problem 12
A constant torque 2.5 KW motor drives a riveting
machine. The mass of the moving parts including
the flywheel is 125 kg at 700 mm radius. One
riveting operation absorbs 10000 J of energy and
takes one second. Speed of the flywheel is 240
rpm before riveting. Determine;
(i) The number of rivets closed per hour
(ii) The reduction in speed after riveting operation.

87. 1 1
2 240
Maximum speed of flywheel n =240 rpm
60
Energy required per rivet =10000 J
Time taken to close one rivet =1 sec
Energy supplied by motor=Power of motor =2.5KW=2500 J/sec
Ma



  
Data :
25.133 rad / sec
2 2
ss of flywheel m=125 kg, Rad. of gyration k=700 mm
Mass M.O.I of flywheel I=125 (0.7) 61.25
Energy supplied by motor =2.5KW=2500 J/sec
Energy supplied per ho
Kg m   

(i) :Number of rivets closed per hour
ur =2500 3600 J
Energy required per rivet =10000 J
2500 3600 J
Number of rivets closed per hour will be=
10000


 900 rivets / hr

88.  2 2
1 2
2 2
2
2 2
=Energy required/rivet-Energy supplied by motor
=(10000 2500)
1
2
1
7500 61.25 (25.13)
2
19.66 60
19.66 / sec 188
2
7500
e
e
I
rad n rpm
J
 



 

      
 
     
 
Excess energy supplied by flywheel :
Also e =
( ii)
1 2
.
Reduction in speed after riveting=( )n n
 (240 - 188) = 52 rpm