Turning moment-diagram-flywheel

flywheel

1. Turning Moment (Crank Effort) Diagram
for a 4-stroke I C engineCrank Angle
T
o
r
q
u
e
N
-
m
0 


SuctionCompression
Expansion
Exhaust

T
T
max
mean
Excess Energy
(Shaded area)

2. Turning Moment (Or Crank Effort) Diagram (TMD)
Turningmomentdiagramisagraphical
representationofturningmomentortorque
(alongY-axis)versuscrankangle(X-axis)for
variouspositionsofcrank.
Uses of TMD
1.TheareaundertheTMDgivestheworkdone
percycle.
2.Theworkdonepercyclewhendividedbythe
crankanglepercyclegivesthemeantorqueT
m.

3. Uses of TMD
3.ThemeantorqueT
mmultipliedbytheangular
velocityofthecrankgivesthepowerconsumedby
themachineordevelopedbyanengine.
4.TheareaoftheTMDabovethemeantorque
linerepresentstheexcessenergythatmaybe
storedbytheflywheel,whichhelpstodesignthe
dimensions&massoftheflywheel.

4. FLYWHEEL
Flywheelisadeviceusedtostoreenergywhen
availableinexcess&releasethesamewhen
thereisashortage.
FlywheelsareusedinICengines,Pumps,
Compressors&inmachinesperforming
intermittentoperationssuchaspunching,
shearing,riveting,etc.
AFlywheelmaybeofDisktypeorRimType
Flywheelshelpinsmootheningoutthe
fluctuationsofthetorqueonthecrankshaft&
maintainthespeedwithintheprescribedlimits.

5. DISK TYPE FLYWHEELDISK 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 :
22
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
22
Hence I=m and I=m
48
Rim Disk
D
D
DD   
   
   
Hence for a given diameter & inertia, the mass of the
rim type flywheel is half the mass of a disk type flywheel

8. 12
It is the difference between the maximum & minimum speeds
in a cycle. (= )nn
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
ss
ss
or K
nn
C
n
n




   

   
   



speed : (C


9. 12
)
It is the ratio of maximum fluctuation of energy to the
mean kinetc energy.
(or K )
ee
ee
or K
EE
C


Important Definitions
(c) Coefficient of fluctuation of energy : (C
It is the reciprocal of coeffi
Ee
E E E
    

   
   



(d) Coefficient of steadiness :
ee
** It is often expressed as the ratio of excess energy
e
to the work done per cycle. C ( or K ) =
W.D / cycle
12
cent of fluctuation of speed.
Coefficient of steadiness=



 



10. 12
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 FLYWHE EL
 
 
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
eI
   
   
     
   

11.  
12
2
1
Putting the mean agular speed = ,
2
We get Multiplying & dividing by ,

Also ,the coefficient of fluctuation o
s
C
  





EXPRESSION FOR ENERGY STORED BY A FLYWHE EL
12
12
12
e = Iω(ω - ω )
(ω - ω )
e = Iω
(ω - ω )
2
2
2
2
22
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
se
I
IE
ee
CC
EE


  
s
s
e = mkωC
e = IωC
e
s
C
=2
C

12. 22 1
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
s
eE
e
k
C
CI
m




EXPRESSION FOR ENERGY STORED BY A FLYWHE EL
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
Asinglecylinder4strokegasenginedevelops
18.4KWat300rpmwithworkdonebythe
gasesduringtheexpansionbeing3timesthe
workdoneonthegasesduringcompression.
Theworkdoneduringthesuction&exhaust
strokesisnegligible.Thetotalfluctuationof
speedis2%ofthemean.TheTMDmaybe
assumedtobetriangularinshape.Findthe
massmomentofinertiaoftheflywheel.

15. Crank Angle
T
o
r
q
u
e
N
-
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
s
C




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
ie







    

  
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
33
1
. . 11040
2
11040 N-m
Max
EC
E
E C C
E
EE E
W
W
W
W
W
i e T 




  


  

This work represents the area under triangle for expansion stroke
max
7 tor 028.qu 3-e m NT

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
TTx

  


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
TT
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

s
CI   

20. Problem 2
Asinglecylinderinternalcombustionengine
workingon4-strokecycledevelops75KWat
360rpm.Thefluctuationofenergycanbe
assumedtobe0.9timestheenergy
developedpercycle.Ifthefluctuationof
speedisnottoexceed1%andthemaximum
centrifugalstressintheflywheelistobe5.5
MN/m
2
,estimatethediameterandthecross
sectionalareaoftherim.Thematerialofthe
rimhasadensity7.2Mg/m
3
.

21. 3
33
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
ie






    

  

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
62
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. 22
22
The energy stored by the flywheel is given by
.
1.466
For rim type, radius of gyration k= 0.733
22
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
Thecrankeffortdiagramfora4-strokecyclegas
enginemaybeassumedtoforsimplicityoffour
rectangles,areasofwhichfromlineofzero
pressurearepowerstroke=6000mm
2
,exhaust
stroke=500mm
2
,Suctionstroke=300mm
2
,
compressionstroke=1500mm
2
.EachSqmm
represents10Nm.Assumingtheresistingtorque
tobeuniform,find
a)Poweroftheengine
b)Energytobestoredbytheflywheel
c)Massofaflywheelrimof1mradiustolimitthe
totalfluctuationofspeedto±2%ofthemean
speedof150rpm.

26. Crank Angle
T
o
r
q
u
e
N
-
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
ie




   
  

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
46.256
m
WPT K


    

   

(i) Power developed by engine :
(ii) Energy stored by flywheel :
max
T = 19098.6N - m
max m
e =Shaded area =π( T -T )
max
,
( ) (19098.6 2944.6)
m
e T T   
e =50749.27 N -m Crank Angle
T
o
r
q
u
e
N
-
m
0   
Suction
Compression
Expansion
Exhaust

T
mean
T
max
Excess energy
(Shaded area)

29. 22
22
We know that energy stored by flywheel
50749.27 (1) (15.71) 0.04
s
e mk C
m

  

(iii) Mass of flywheel
Mass of flywheel m = 5140.64 Kg

30. Problem 4
Amulticylinderengineistorunataspeedof
600rpm.OndrawingtheTMDtoascaleof
1mm=250Nm&1mm=3
0
,theareasabove&
belowthemeantorquelineare+160,-172,
+168,-191,+197,-162mm
2
respectively.
Thespeedistobekeptwithin±1%ofthemean
speed.DensityofCastironflywheel=7250
kg/mm
3
andhoopstressis6MPa.Assuming
thattherimcontributesto92%ofthe
flywheeleffect,determinethedimensionsof
therectangularcrosssectionoftherim
assumingwidthtobetwicethethickness.

31. 1 23 4 5 6
160
172 191
197
162
168
T
u
r
n
i
n
g
M
o
m
e
n
t
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
22
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:
22
)=2t 0.006824 2t
Hence t = 58.4 mm, b = 2t = 116.8 mm.

35. Problem 5
Torque–outputdiagramshowninfigisa
singlecylinderengineat3000rpm.Determine
theweightofasteeldisktypeflywheel
requiredtolimitthecrankspeedto10rpm
aboveand10rpmbelowtheaveragespeedof
3000rpm.Theoutsidediameterofthe
flywheelis250mm.Determinealsothe
weightofarimtypeflywheelof250mm
meandiameterforthesameallowable
fluctuationofspeed.

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
22
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
ie






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
yc

le 4
D
m
     



           
     
           
           



m
W.D / cycle =87.5π N -m
T = 21.875N -m

39. 0
90 180 360 450540 630720
25
50
75
100
-25
-50
-75
-100
T
N-m
(Degrees)
T
mean
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
EE
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
EE
E
E
E
E
EE










  



   












41. 22
22
= 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
s
e 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
Thetorquerequiredforamachineisshownin
fig.Themotordrivingthemachinehasamean
speedof1500rpmanddevelopconstant
torque.Theflywheelonthemotorshaftisof
rimtypewithmeandiameterof40cmand
mass25kg.Determine;
(i)Powerofmotor
(ii)%variationinmotorspeedpercycle.

43. 
400 N-m
2000 N-m
T
o
r
q
u
e
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
ie





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 800 157.08
m


   
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 energye
(shaded area)
1
2 3
T
o
r
q
u
e
x

47. 
   
1200
From the similar triangles, 1.178
1600
2
Energy stored by flywheel = Shaded area
1
= 2000 800 1.178 2000 800
42
We know that energy st
s
x
x rad
e


  


   
(ii) Coefficent of fluctuation of speed
e
C
e =1649.28Nm
22
22
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
A3cylindersingleactingenginehascranksset
equallyat120
0
anditrunsat600rpm.TheTMD
foreachcylinderisatriangle,forthepower
strokewithamaximumtorqueof80N-mat60
0
afterdeadcenterofthecorrespondingcrank.
Thetorqueonthereturnstrokeiszero.Sketch
theTMD&determinethefollowing;
(i)Powerdeveloped
(ii)Coefficientoffluctuationofspeedifmassof
flywheelis10kgandradiusofgyrationis8cm.
(iii)Maximumangularaccelerationofflywheel.

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
ie




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




   





m
T =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
min
Nm, 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
3
2
mean
PT
x
x rad
e




    
  
   
(i) Power developed :
any one traingle
3.77 Kw
e =10.47 N -m

54. 22
22
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)
mean
T





   ∴α = 312.5
2
tion of flywheel
rad / sec

55. Problem 8
Atorquedeliveredbyatwostrokeisrepresented
byT=(1000+300sin25cos2Nmwhere
istheangleturnedbycrankfromIDC.Theengine
speedis250rpm.Themassoftheflywheelis400
kgandtheradiusofgyrationis400mm.Determine
(i)Thepowerdeveloped
(ii)Totalpercentagefluctuationofspeed
(iii)Theangularaccelerationandretardationof
flywheelwhenthecrankhasrotatedthroughan
angleof60
0
fromtheIDC
(iv)Max&Minangularacceleration&retardationof
flywheel.

56. 00
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 = 180π radians
2
ywheel =
60
2 250
. .
6
26.18 rad/se
0
c
n
ie







57. 00
W.D per cycle
Mean torque T
Crank angle per cycle
11
(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
5120 990
6150 490
7180 5000
180
T
mean=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-T
mean 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
00
12
2

2θ = 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 )
Td
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)
%
6.1
ss
s
CC


   
 

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
98
o
.Nm
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
ie






  


   



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
Amachineiscoupledtoatwostrokeenginewhich
producesatorqueof(800+180Sin3N-mwhereis
thecrankangle.Themeanenginespeedis400rpm.
Theflywheelandtherotatingpartsattachedtothe
enginehaveamassof350kgataradiusofgyrationof
220mm.Calculate;
(i)Thepowerdevelopedbytheengine
(ii)Totalpercentagefluctuationofspeedwhen,
(a)Theresistingtorqueisconstant
(b)Theresistingtorqueis(800+80Sin

65. SlNoAngle

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
30
0
60
0
90
0
Excess Energy
T
E
T
E
= Engine torque
=mean TorqueT
m
T
m
Nm

66. 00
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 :
02
The crank angle per cycle = 120
3
π
radians
41.89 r
2
f flywheel =
60
2 400
. . ad/s
60
ec
n
ie





67. 22
33
00
W.D per cycle
Mean torque T
Crank angle per cycle
11
(800 180sin3 )
22
33
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 K

W

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 consta nt
:
ΔT
00
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
12
ΔT =
θ =0 &
)
θ
=
=60

0

69. 2
1
60
0
.
(180sin3 )
Td
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
ss
CC    



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
30
0
60
0
90
0
Excess Energy
180
0
T
E
T
M
T
E
T
M
= Engine torque
=Machine Torque





71. 1
21
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
8s
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


  


00
12
θ =53 & θ =( 180 -53)=127

72. 2
1
127
53
.
(180sin3 80sin )
Td
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
s
CC    


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
Acertainmachinerequiresatorqueof
(500+50sinN-mtodriveit,whereisthe
angleofrotationoftheshaft.Themachineis
directlycoupledtoanenginewhichproducesa
torqueof(500+60sin2Nm.Theflywheeland
theotherrotatingpartsattachedtotheengine
haveamassof500kgataradiusof400mm.If
themeanspeedis150rpm.Find;
(a)Themaximumfluctuationofenergy
(b)Total%fluctuationofspeed
(c)Max&Minangularaccelerationoftheflywheel
&thecorrespondingshaftpositions.

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

75. 00
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 = 180π radians
2
ywheel =
60
2 150
. .
6
15.71 rad/se
0
c
n
ie







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 -T
E M E M

  
(i) Excess energy stored by flywheel :
Or

ΔT
)
 
 
00
0
i.e. (500 60sin 2 ) (500 50sin ) 0
(60sin 2 50sin )=0

sin 12cos 5 0.
0
0 ,18Either sin 0
5
or 12cos 5 0 cos
12
Considering max difference between consecutive
0
65.37



 


   

    
     
=
Put sin2θ = 2sinθ
Δ
cosθ
T=
crank positions,
00
12
θ = 65.37 &θ = 180

77. 2
1
180
65.37
.
(60sin 2 50sin )
Td
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
ss
s
CC    
  
(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
ie




  
  


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
4
max
ie 











27.7 N
Maximum
- m
ra acce d/leration sec =
2
=35 =12
24cosθ-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 pressCrank
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 mm
2
•If the energy or work done /sheared area is
given, the work done per hole =W.D/mm
2
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. 12
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
Apunchingmachinecarriesout6holespermin.
Eachholeof40mmdiameterin35mmthick
platerequires8N-mofenergy/mm
2
ofthe
shearedarea.Thepunchhasastrokeof95mm.
Findthepowerofthemotorrequiredifthe
meanspeedoftheflywheelis20m/sec.
Ifthetotalfluctuationofspeedisnottoexceed
3%ofthemeanspeed,determinethemassof
theflywheel.

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
s
C


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=
10

9
actual
actual

1.842 Secs
Excess energy supplied by flyw
As the punch travels 95 ×2 =190 mm in 10 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
Aconstanttorque2.5KWmotordrivesariveting
machine.Themassofthemovingpartsincluding
theflywheelis125kgat700mmradius.One
rivetingoperationabsorbs10000Jofenergyand
takesonesecond.Speedoftheflywheelis240
rpmbeforeriveting.Determine;
(i)Thenumberofrivetsclosedperhour
(ii)Thereductioninspeedafterrivetingoperation.

87. 11
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
22
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.  
22
12
22
2
22
=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)
12
.
Reduction in speed after riveting=( )

nn
(240 - 188) = 52 rpm