Turning moment-diagram-flywheel

Turning Moment (Crank Effort) Diagram

for a 4-stroke I C engine

Crank Angle

Torq

ue

N-m

0

Suction Compression

Expansion

Exhaust

T

Tmax

mean

Excess Energy

(Shaded area)

Turning Moment (Or Crank Effort) Diagram (TMD)

Turning moment diagram is a graphicalrepresentation of turning moment or torque(along Y-axis) versus crank angle (X-axis) forvarious positions of crank.

Uses of TMD

1. The area under the TMD gives the work doneper cycle.

2. The work done per cycle when divided by thecrank angle per cycle gives the mean torque Tm.

Uses of TMD

3. The mean torque Tm multiplied by the angularvelocity of the crank gives the power consumed bythe machine or developed by an engine.

4. The area of the TMD above the mean torqueline represents the excess energy that may bestored by the flywheel, which helps to design thedimensions & mass of the flywheel.

FLYWHEEL

Flywheel is a device used to store energy whenavailable in excess & release the same whenthere is a shortage.

Flywheels are used in IC engines, Pumps,Compressors & in machines performingintermittent operations such as punching,shearing, riveting, etc.

A Flywheel may be of Disk type or Rim Type

Flywheels help in smoothening out thefluctuations of the torque on the crankshaft &maintain the speed within the prescribed limits.

DISK TYPE FLYWHEEL

DISK TYPE FLYWHEEL

D

RIM TYPE FLYWHEELSection X-X

X X

b

t

D

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 flyheel2

For Disk type, k= where D=Outer diameter of the flywheel2 2

Hence I=m and I=m4 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

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 )

2where =Angular velocity=

60

s s

s s

or K

n nC

n

n

speed : (C

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 EC

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

eto the work done per cycle. C ( or K ) =

W.D / cycle

1 2

cent of fluctuation of speed.

Coefficient of steadiness=

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

1 2

2

1Putting the mean agular speed = ,

2

We get Multiplying & dividing by ,

Also , the coefficient of fluctuation osC


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

1Alternatively, if Mean kinetic energy E= ,

2

2 , e=2EC

Note: 1

2 But O

.

R

s

s e

I

I E

e eC C

E E

s

se = mk ω C

e = Iω C

e

s

C= 2

C

2 21e= I , Putting mean Kinetic energy E=

2

and expressing C as a percentage,

2EC

100

1Alternatively, if Mean kinetic

0.02

energy ENote: =.2

2

s

s

s

se E

e

k

C

C I

m


2

2

2

2 2 2

,

1( ) , E=

2

s

k v mv

e mv c

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

/ sec60

= v al

Dnm

Problem 1A single cylinder 4 stroke gas engine develops18.4 KW at 300 rpm with work done by thegases during the expansion being 3 times thework done on the gases during compression.The work done during the suction & exhauststrokes is negligible. The total fluctuation ofspeed is 2% of the mean. The TMD may beassumed to be triangular in shape. Find themass moment of inertia of the flywheel.

Crank Angle

Torq

ue

N-m

0

Suction

Compression

Expansion

Exhaust

T

Tmax

meanx

Excess Energy

TURNING MOMENT DIAGRAM

3Power 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

2Angular Velocity of flywheel =

60

2 300. .

60

Also power P=T 18.4 10 T 31.416

/sec

Mean tor18.4 10

31.41

que T 585. N-m6

7

m

m

m

n

i e

Solution :

nk shaft)

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 write3

27360= W

3 3

1. . 11040

2

11040 N-m

Max

E C

EE C C

EE E E

W

W

W

WW

i e T

This work represents the area under triangle for expansion stroke

max 7 tor 028.qu 3 -e m NT

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

1Hence e= 2.88 (7 9276.67 N-028.3 585.7)

2m

mean

T

T Tx rad

T

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

Problem 2A single cylinder internal combustion engineworking on 4-stroke cycle develops 75 KW at360 rpm. The fluctuation of energy can beassumed to be 0.9 times the energydeveloped per cycle. If the fluctuation ofspeed is not to exceed 1% and the maximumcentrifugal stress in the flywheel is to be 5.5MN/m2, estimate the diameter and the crosssectional area of the rim. The material of therim has a density 7.2 Mg/m3.

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

2lar 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

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 20 =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

2 2

2 2

The energy stored by the flywheel is given by

.

1.466For rim type, radius of gyration k= 0.733

2 2

37.7) 0.01

But , for rim type, m

s

Dm

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

2A=0.09m

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 :

Problem 3The crank effort diagram for a 4-stroke cycle gasengine may be assumed to for simplicity of fourrectangles, areas of which from line of zeropressure are power stroke =6000 mm2, exhauststroke =500 mm2, Suction stroke=300 mm2,compression stroke = 1500 mm2. Each Sq mmrepresents 10 Nm. Assuming the resisting torqueto 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 thetotal fluctuation of speed to ±2% of the meanspeed of 150 rpm.

Crank Angle

Torq

ue

N-m

0

Suction

Compression

Expansion

Exhaust

Tmean

Tmax

Excess energy

(Shaded area)

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

2ywheel =

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-mCrank angle/cycle 4

m

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

Torq

ue

N-m

0

Suction

Compression

Expansion

Exhaust

Tmean

Tmax

Excess energy

(Shaded area)

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

Problem 4

A multi cylinder engine is to run at a speed of600 rpm. On drawing the TMD to a scale of1mm=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 meanspeed. Density of Cast iron flywheel=7250kg/mm3 and hoop stress is 6 MPa. Assumingthat the rim contributes to 92% of theflywheel effect, determine the dimensions ofthe rectangular cross section of the rimassuming width to be twice the thickness.

1 2 3 4 5 6

160

172 191

197

162

168T

urn

ing M

om

ent

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

2

2 2 600Angular velocity = 62.84 / sec,

60 60

1% 2% 0.02

31mm 250 13.1

180

Max Fluctuation of

s

Nrad

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

2 6 2Using = v ; 6 10 7250

Velocity

600Also 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

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.

Problem 5

Torque –output diagram shown in fig is asingle cylinder engine at 3000 rpm. Determinethe weight of a steel disk type flywheelrequired to limit the crank speed to 10 rpmabove and 10 rpm below the average speed of3000 rpm. The outside diameter of theflywheel is 250 mm. Determine also theweight of a rim type flywheel of 250 mmmean diameter for the same allowablefluctuation of speed.

0 90 180 360 450 540 630 720

25

50

75

100

-25

-50

-75

-100

T

N-m

(Degrees)

0Crank angle per cycle=720

Mean speed n=3000 rpm,

250Radius of gyration k= =125 mm =0.125 m

2

0.25Radius of gyration k= =0.0884 m

2 2

10 20C 0.00

3000 3000s

(For rim type)

(For disk type)

Data :

= 4π radians

667


60

2 3000. . 314.16 rad/ e

0c

6s

n

i e

Solution :

=75 50 100 75 50 100 752 2 2 2 2 2

W.D per c

WD/c

ycle 87.5Mean 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

090 180 360 450 540 630 720

25

50

75

100

-25

-50

-75

-100

T

N-m

(Degrees)

Tmean

1 2 3 4

5

6 7 8

Let the energy at 1=E

Energy at 2=E+(75-21.875) 26.56252

Energy at 3=( 26.5625 ) - (71.875) 9.7352

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) 20.31252

Energy at 6=( 20.3125 ) (50-21.875) 34.3752

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

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

Problem 6

The torque required for a machine is shown infig. The motor driving the machine has a meanspeed of 1500 rpm and develop constanttorque. The flywheel on the motor shaft is ofrim type with mean diameter of 40 cm andmass 25 kg. Determine;

(i) Power of motor

(ii) % variation in motor speed per cycle.

400 N-m

2000 N-m

Torq

ue

Crank angle



40Radius of gyration k= =20 cm =0.2 m

2

m=25 kg

2Angular Velocity of flywheel

1

=60

2 1500. .

657.08 e

0rad/s

(For rim type)

n

i e

Data :

Solution :

2π radians

c

W.D/cycle area 1+area 2+area 3

1 =400 2 (2000 400) (2000 400)

4 2 2

W.D per cycle 1600Mean torque T

Crank angle per cycle 2m

(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

400 N-m

800 N-m

2000 N-m

Crank angle

mean

Excess energy e

(shaded area)

1

2 3

Torq

ue

x

1200From the similar triangles, 1.178

16002

Energy stored by flywheel = Shaded area

1= 2000 800 1.178 2000 800

4 2

We know that energy st

s

xx 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%

Problem 7

A 3 cylinder single acting engine has cranks setequally at 1200 and it runs at 600 rpm. The TMDfor each cylinder is a triangle, for the powerstroke 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. Sketchthe TMD & determine the following;

(i) Power developed

(ii) Coefficient of fluctuation of speed if mass offlywheel is 10 kg and radius of gyration is 8 cm.

(iii)Maximum angular acceleration of flywheel.

T (N-m)

060 120 180 240 300 360

80N-m

degrees



Radius of gyration k=8 =0.08 m

m=10 kg


60

2 600. .

662.83 r e

0ad/s c

cm

n

i e

Data :

Solution :

2π radians

W.D/cycle area of 3 triangles

1 =3 80

2

W.D per cycle 377Mean 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 min

Nm, the minimum torque (T )

will be = 40 N - m. Hence the modified TMD

may be drawn as shown in fig.

T (N-m)

060 120 180 240 300 360

80 N-m

degrees

60 Nm

40 Nm

Modified TMD for 3 Cylinder engine

60 62.83

20From the similar triangles,

32 403

Due to symmetry,the energy stored by flywheel

=Area of (Shaded portion)

1= 80 60

32

meanP T

xx rad

e

(i) Power developed :

any one traingle

3.77 Kw

e =10.47 N -m

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

Problem 8

A torque delivered by a two stroke is representedby T=(1000+300 sin 25 cos 2 Nm where is the angle turned by crank from IDC. The enginespeed is 250 rpm. The mass of the flywheel is 400kg 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 offlywheel when the crank has rotated through anangle of 600 from the IDC

(iv) Max & Min angular acceleration & retardation offlywheel.

0 0

As the torque is a function of 2 ,

equate 2 360 180



m=400 kg


mm

Data :

Solution :

0The crank angle per cycle = 180 π radians

2ywheel =

60

2 250. .

626.18 rad/se

0c

n

i e

0 0

W.D per cycleMean torque T

Crank angle per cycle

1 1 (1000 300sin 2 500cos 2 )

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

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

)

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 500cos 2 ) 0

500Hence tan2 = 1.667

300

(As the torque curve intercepts the mean torque line at these points)

0 0 0 01

0 01 2

2

2θ = 59

ΔT

Hence θ = 29.

&

5 &

2θ =( 180 +5

θ =119

9 ) = 2

.5

39

2

1

119.5

29.5

.

(300sin 2 500cos 2 )

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 2Also 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)

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

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 500cos 2 ) 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

0

0

maxmax 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

minmin 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

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

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 0300 600 90 0

Excess Energy

TE

TE

= Engine torque

=mean TorqueTm

Tm Nm

0 0


equate 3 360 120

Mean speed n=400 rpm, m=350 kg


Angular Velocity o

mm

Data :

Solution :

0 2The crank angle per cycle = 120

3

π radians

41.89 r

2f flywheel =

60

2 400. . ad/s

60ec

n

i e

2 23 3

0 0

W.D per cycleMean torque T

Crank angle per cycle

1 1 (800 180sin 3 )

2 23 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 K

W

1

2 1 2



& , where To find &

(ii) Coefficient of fluctuation of speed

(a) When the resisting torque is consta nt

:

ΔT

0 0

are crank positions

at which T=T (T-T

i.e. (800 180sin 3 800) 0

180sin 3 0 3 0 & 3 1 0

0

8

mean mean

(As the torque curve intercepts the mean torque line at these points)

Or

01 2

ΔT =

θ =0 &

)

θ

=

=60

0

2

1

60

0

.

(180sin 3 )

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 2Also 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)

(b) When the resisting torque is (800 + 80sin )

0 120

T

(N-m)

Crank Angle

0 0300 600 90 0

Excess Energy

1800

TE

TM

TE

TM

= Engine torque

=Machine Torque

1

2 1



& , 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 180sin 3 ) (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 01 2θ = 53 & θ =( 180 -53) =127

2

1

127

53

.

(180sin 3 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 2Also 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)

Problem 10

A certain machine requires a torque of(500+50sin N-m to drive it, where is theangle of rotation of the shaft. The machine isdirectly coupled to an engine which produces atorque of (500+60 sin2 Nm. The flywheel andthe other rotating parts attached to the enginehave a mass of 500 kg at a radius of 400 mm. Ifthe 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.

0

T(N-m)

Crank Angle

0

Excess Energy

1800

TE

TM

TE

TM

= Engine torque

=Machine Torque

0 0


equate 2 360 180



m=500 kg


mm

Data :

Solution :

0The crank angle per cycle = 180 π radians

2ywheel =

60

2 150. .

615.71 rad/se

0c

n

i e

1

2 1 2



& , 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

5 or 12cos 5 0 cos12

Considering max difference between consecutive

0

65.37

=

Put sin2θ = 2sinθ

Δ

cosθ

T =

crank positions,

0 0

1 2θ = 65.37 &θ = 180

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 2Also 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%

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

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)=

27.70.3

86

04max

i e

27.7 N

Maximum

- m

ra acce d /leration sec =

2

= 35 =12

24cos θ -5cosθ -12 =0

5

7.6

= 3

0At :

T=60 sin(2 127.6)-50sin(127.6)=-9

197.

.0

6

222

8max

2

2

7.62

Maximum retardation

N - m

rad / se= c

=127.6

Flywheel for Punch press

Crank

Plate

Die

Punching

tool

Flywheel

Flywheel for Punch Press

Crank shaft

connecting

rod

d

t

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

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

Problem 11

A punching machine carries out 6 holes per min.Each hole of 40 mm diameter in 35 mm thickplate requires 8 N-m of energy/mm2 of thesheared area. The punch has a stroke of 95 mm.Find the power of the motor required if themean speed of the flywheel is 20 m/sec.

If the total fluctuation of speed is not to exceed3% of the mean speed, determine the mass ofthe flywheel.

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/minPower of motor

= KW60 10

35185.4 6

60 10P

Solution :

2

10 secs

Sheared area per hole = πdt = π × 40 × 35 = 4398.23 mm

4398.23 8 = 35186 N

3.51

- m

86KW

Thickness of plateTime taken per cycle

2×stroke length

3510=

1 0

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

2s

35186 3518.6 1.842) = 28705 N - m

Also e = mv C 28705 m ( 20) 0.0

m= 239

3

2 Kg

Problem 12

A constant torque 2.5 KW motor drives a rivetingmachine. The mass of the moving parts includingthe flywheel is 125 kg at 700 mm radius. Oneriveting operation absorbs 10000 J of energy andtakes one second. Speed of the flywheel is 240rpm before riveting. Determine;

(i) The number of rivets closed per hour

(ii) The reduction in speed after riveting operation.

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 JNumber of rivets closed per hour will be=

10000

900 rivets / hr

2 2

1 2

2 2

2

2 2

=Energy required/rivet-Energy supplied by motor

=(10000 2500)

1

2

17500 61.25 (25.13)

2

19.66 6019.66 / 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

Date post:	12-Jan-2017
Category:	Engineering
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