G.H. RAISONI COLLEGE OF ENGINEERINGG.H. RAISONI COLLEGE OF ENGINEERING

STUDY CIRCLE GROUP TWO

GUIDED BY

Prof.P.S.KADU

DEPARTMENT OF MECHANICAL ENGINEERING

SEMINAR ON

TURNING MOMENT DIAGRAM AND FLYWHEEL

INTRODUCTIONINTRODUCTION

TURNING MOMENT DIAGRAM:-TURNING MOMENT DIAGRAM:-

Turning moment diagram is Turning moment diagram is the graphical representation the graphical representation of turning moment or crank of turning moment or crank – effort for various positions – effort for various positions of crank,it is plotted on of crank,it is plotted on cartesian co-ordinatescartesian co-ordinates

Fig of an engine having flywheel

TURNING MOMENT DIAGRAM FOR SINGLE TURNING MOMENT DIAGRAM FOR SINGLE CYLINDER DOUBLE ACTING STEAM ENGINCYLINDER DOUBLE ACTING STEAM ENGIN

A B C D E

C

db

ae

FTmean

Tmax

P 90360

270

MEAN TORQUE

TURNING MOMENT DIAGRAM FOR A FOUR TURNING MOMENT DIAGRAM FOR A FOUR SROKE CYLINDER INTERNAL CUMBUSTION SROKE CYLINDER INTERNAL CUMBUSTION

ENGINEENGINE

In four stoke IC engine ,there is one working In four stoke IC engine ,there is one working stroke after the crank has turned through two stroke after the crank has turned through two revolutions i.e 720%revolutions i.e 720%Whenever pressure inside cylinder is less than Whenever pressure inside cylinder is less than atmospheric pressure ,then negative loop is atmospheric pressure ,then negative loop is formedformedTherefore during suction ,compression and Therefore during suction ,compression and exhaust negative loop is formed and during power exhaust negative loop is formed and during power stroke positive loop is formedstroke positive loop is formed

MULTICYLINDERMULTICYLINDER ENGINE ENGINE

The resultant turning The resultant turning moment diagram is sum of moment diagram is sum of the turning moment diagram the turning moment diagram for three cylindersfor three cylinders. .

First cylinder is high First cylinder is high pressure, second is pressure, second is intermediate and third is low intermediate and third is low pressure cylinder.pressure cylinder.

The cranks are usually The cranks are usually placed at an angle of 120placed at an angle of 120

Fig of multi cylinder engine

Example of multicylinder engine

Fluctuation of EnergyFluctuation of Energy

Definition Definition

It may be determined by the turning moment diagram It may be determined by the turning moment diagram for one complete cycle of operation for one complete cycle of operation

The difference between the maximum and the The difference between the maximum and the minimum energies is known as minimum energies is known as maximum maximum fluctuation of energyfluctuation of energy

Determination of max. Fluctuation of energyDetermination of max. Fluctuation of energy

A B C D E FG

a1 a3 a4

a2 a5 a6Mean torque line

Crank Angle

Turning

moment

ΔE = (E+a1) – (E+a1-a2+a3-a4) = a2-a3+a4

Coefficient of Fluctuation of EnergyCoefficient of Fluctuation of Energy

Definition Definition

It may be defined as the ratio of the maximum fluctuation of energy to the work It may be defined as the ratio of the maximum fluctuation of energy to the work

done per cycledone per cycle..

CCEE = = Maximum fluctuation of energy Maximum fluctuation of energy

work done per cyclework done per cycle The work done per cycle may be obtained by using two relationsThe work done per cycle may be obtained by using two relations

WorkWork done per cycle = T done per cycle = Tmean * mean * θ

Tmean = P*60 = P

2πN ω

Work done per cycle = Work done per cycle = P*60P*60

nn

Flywheel in Punching PressFlywheel in Punching Press

What is Flywheel ?

• Stores energy

• Absorbs excess energy during power stroke

• Controls speed variations

• Used in punching machines, shearing machines,rivetting machines, crushers etc.

• Reduces fluctuation of speed when load on crankshaft is constant and i/p torque varies during cycle.

FLYWHEEL (Conclusion):-FLYWHEEL (Conclusion):-

Flywheel used in machines serves as a reservior,which Flywheel used in machines serves as a reservior,which stores energy during the period when the supply of stores energy during the period when the supply of energy is more than the requirement , and releases it energy is more than the requirement , and releases it during the period when the requirement of energy is during the period when the requirement of energy is more than the supplymore than the supply

Flywheel controls the speed Flywheel controls the speed variations caused by the fluctuation of the engine variations caused by the fluctuation of the engine turning moment during each cycle of operationturning moment during each cycle of operation

Flywheel attached with an engine

COEEFICIENT OF FLUCTUATION OF ENERGY:-COEEFICIENT OF FLUCTUATION OF ENERGY:-

TThe difference between the maximum and he difference between the maximum and minimum speeds during a cycle is called the minimum speeds during a cycle is called the maximum fluctuation of speed .the ratio of maximum fluctuation of speed .the ratio of maximum fluctuation of speed to mean speed is maximum fluctuation of speed to mean speed is called the called the coefficient of fluctuation of speed.coefficient of fluctuation of speed.

N1 and N2 is maximum and minimum speeds in N1 and N2 is maximum and minimum speeds in r.p.m during the cycles r.p.m during the cycles

N is mean speed in r.p.m=(N1+N2)/2N is mean speed in r.p.m=(N1+N2)/2Coefficient of fluctuation of speed ,Coefficient of fluctuation of speed ,

Cs=(N1-N2)/N=2(N1-N2)/(N1+N2)Cs=(N1-N2)/N=2(N1-N2)/(N1+N2) The reciprocal of coeeficient of fluctuation of The reciprocal of coeeficient of fluctuation of

speed is known asspeed is known as coefficient of stediness coefficient of stediness m=1/Cs=N/(N1-N2)m=1/Cs=N/(N1-N2)

Energy stored in a flywheelEnergy stored in a flywheel

M=mass of flywheel in kg,M=mass of flywheel in kg,

K=Radius of gyration of flywheel in metres,K=Radius of gyration of flywheel in metres,

i=mass moment of inertia of flywheel about its axis of rotation in i=mass moment of inertia of flywheel about its axis of rotation in kg-m^2kg-m^2

N1and N2 = maximum and minimum angular speeds during the N1and N2 = maximum and minimum angular speeds during the cycles in r.p.m,cycles in r.p.m,

W=mean angular speed during the cycle in rad/sec=(w1+w2)/2W=mean angular speed during the cycle in rad/sec=(w1+w2)/2

Delta E=MR^2W^2Cs = mv^2CsDelta E=MR^2W^2Cs = mv^2Cs

Dimension of the Flywheel RimDimension of the Flywheel Rim

D = Mean dia. Of rim in meter,D = Mean dia. Of rim in meter,

R = Mean radius of rim in meter,R = Mean radius of rim in meter,

A = Cross sectional area of rim A = Cross sectional area of rim ρ = density of rim material N = Speed of the flywheel in r.p.m ω = Angular velocity of the flywheelLinear velocity at mean radius = ע σ = Tensile stress or hoop stressσ = Tensile stress or hoop stress

A = mA = m

ρ σ σ π

Flywheel in Punching Press

Crank

Crank shaft

Motor

Flywheel

Plate

Punch

Die

Crank is driven by motor which supplies constant torque.Crank is driven by motor which supplies constant torque. Load acts only during rotation of crank from Load acts only during rotation of crank from θθ==θθ11 to to θθ==θθ22,, Unless a flywheel is used speed of crankshaft will increase Unless a flywheel is used speed of crankshaft will increase

during rotation of crank frm during rotation of crank frm θθ22 to 2∏. to 2∏. While drop in speed of crankshaft is very large during rot’n of While drop in speed of crankshaft is very large during rot’n of

crank frm crank frm θθ11 to to θθ22.. Thus flywheel absorbs energy available at one stage & makes Thus flywheel absorbs energy available at one stage & makes

up deficient energy at other stage to keep fluctuations of speed up deficient energy at other stage to keep fluctuations of speed within limits. within limits.

Relation for maximum fluctuation of energyRelation for maximum fluctuation of energy

Let E1 be the energy required for punching a hole punched,the Let E1 be the energy required for punching a hole punched,the thickness of the material and the physical properties ofthickness of the material and the physical properties of

the material.the material.

Let dLet d11=Diameter of the hole punched,=Diameter of the hole punched, tt11=Thickness of the plate,=Thickness of the plate, ζζuu=Ultimate shear stress for the =Ultimate shear stress for the plate plate

material.material. Maximum shear forceMaximum shear force,,

Fs=Area sheared x Ultimate shear stressFs=Area sheared x Ultimate shear stress =∏ d=∏ d11 t t1 1 ζζuu

Workdone or energy required for punching a hole,Workdone or energy required for punching a hole,

E1=1/2.Fs.tE1=1/2.Fs.t The energy supplied by the motor to the crankshaft during The energy supplied by the motor to the crankshaft during

actual punching operation,actual punching operation,

E2=E1(E2=E1(θθ2-2-θθ1)/2∏ 1)/2∏ Therefore Balance energy required for punching=E1{1-(Therefore Balance energy required for punching=E1{1-(θθ2-2-

θθ1)/21)/2∏∏}}

Thus maximum fluctuation of energy,Thus maximum fluctuation of energy, ΔΔE=EE=E11-E-E22=E{1-(=E{1-(θθ2-2-θθ1)/21)/2∏∏}} The values of The values of θθ1 and 1 and θθ2 may be determined only if the crank 2 may be determined only if the crank

radius (r) ,radius (r) , length of connecting rod (l) and the relative of job with respect length of connecting rod (l) and the relative of job with respect

to the crankshaft axis are known.to the crankshaft axis are known. In absence of relevant data ,we assume In absence of relevant data ,we assume ((θθ22--θθ11)/2∏=t/2s=t/4r)/2∏=t/2s=t/4r t=Thickness of the materialt=Thickness of the material s=Stroke of punch=2rs=Stroke of punch=2r

ProblemProblem

The equation of the turning moment curve of a three crank The equation of the turning moment curve of a three crank engine is (5000+1500 sin 3engine is (5000+1500 sin 3θθ)N-m, where )N-m, where θθ is the crank angle is the crank angle in radians. The moment of inertia of the flywheel is 1000 kg-in radians. The moment of inertia of the flywheel is 1000 kg-m² and the mean speed is 300 rpm. m² and the mean speed is 300 rpm.

Calculate: 1.Power of the engine,Calculate: 1.Power of the engine,

2. the maximum fluctuation of speed of 2. the maximum fluctuation of speed of flywheel in percentage when the flywheel in percentage when the resisting torque resisting torque is constant .is constant .

Given:Given:

T=(5000+1500sin3T=(5000+1500sin3θθ)N-m,)N-m,

I= 1000 kg-m²,I= 1000 kg-m²,

N=300 rpm,N=300 rpm,

ωω=31.42 rad/sec=31.42 rad/sec 1. Power of the engine1. Power of the engine

Workdone per revolutionWorkdone per revolution

=∫{(5000+1500sin3=∫{(5000+1500sin3θθ)d)dθθ,0,2∏},0,2∏}

={[5000={[5000θθ-1500cos3-1500cos3θθ/3],0,2∏}/3],0,2∏}

=10000 ∏ N-m=10000 ∏ N-m

ThereforeTherefore

mean resisting torque,mean resisting torque,

TTmeanmean= (workdone / rev)/2∏= (workdone / rev)/2∏

= 10000∏/2∏= 10000∏/2∏

= 5000 N-m= 5000 N-m

P=TP=Tmeanmean..ωω

=5000x31.42=5000x31.42

=157.1 kW=157.1 kW

2.Maximum fluctuation of the speed of flywheel2.Maximum fluctuation of the speed of flywheel Let CLet Css=Max or total fluctuation of speed=Max or total fluctuation of speed (i) When resisting torque is constant:(i) When resisting torque is constant: The turning moment dia. is shown . Since the The turning moment dia. is shown . Since the resisting resisting

torque is constant , therefore the torque is constant , therefore the torque exerted on shaft torque exerted on shaft is equal the mean is equal the mean resisting torque on the flywheel. resisting torque on the flywheel.

T=TT=Tmeanmean

5000+1500sin35000+1500sin3θθ=5000=5000 33θθ=0 or 180 =0 or 180 θθ= 0 or 60= 0 or 60

Maximum fluctuation of energy,Maximum fluctuation of energy,

ΔΔE=∫{(T-Tmean) dE=∫{(T-Tmean) dθθ,0,60},0,60}

=∫{(1500sin3=∫{(1500sin3θθ) d) dθθ,0,60},0,60}

=1000 N-m=1000 N-m Also , Also , ΔΔE=I.E=I.ωω.Cs.Cs

1000=1000(31.42) Cs1000=1000(31.42) Cs

Cs=0.001 or 0.1% ---------ANSCs=0.001 or 0.1% ---------ANS

ProblemProblem

A riveting machine is driven by a constant torque 3 kW motor. A riveting machine is driven by a constant torque 3 kW motor. The moving parts including the flywheel are equivalent to 150 The moving parts including the flywheel are equivalent to 150 kg at 0.6 m radius. One riveting operation takes 1 sec and kg at 0.6 m radius. One riveting operation takes 1 sec and absorbs 10000 N-m of energy.The speed of the flywheel is absorbs 10000 N-m of energy.The speed of the flywheel is 300rpm before riveting. Find the speed immediately after 300rpm before riveting. Find the speed immediately after riveting. How many rivets can be closed per minute.riveting. How many rivets can be closed per minute.

Given: P=3 kW;Given: P=3 kW;

m=150kg;m=150kg;

k=0.6m;k=0.6m;

NN11=300 rpm;=300 rpm;

ωω11=31.42 rad/sec=31.42 rad/sec

Let Let ωω22== Angular speed of flywheel after rivetingAngular speed of flywheel after riveting

EE22= 3 kW=3000W=3000 N-m/s= 3 kW=3000W=3000 N-m/s

EE11=10000 N-m=10000 N-m

ΔΔE=EE=E11-E-E22=10000-3000=7000 N-m=10000-3000=7000 N-m

We know that max fluctuation of energy(We know that max fluctuation of energy(ΔΔE),E),

7000=1/2 .7000=1/2 . m.k[(m.k[(ωω11)-()-(ωω22)])]

ωω22=26.98 rad/sec =26.98 rad/sec

Corresponding speed in rpm;Corresponding speed in rpm;

N2=26.98x60/2∏N2=26.98x60/2∏

=257.6 rpm=257.6 rpm

Since the energy absorbed by each riveting operation which Since the energy absorbed by each riveting operation which takes 1 sec is 10000 Nm,therefore numbers of rivets that can takes 1 sec is 10000 Nm,therefore numbers of rivets that can be closed per minutebe closed per minute

= (E= (E22/E/E11)x60)x60

=(3000/10000)x60=(3000/10000)x60

=18 rivets --------------Ans=18 rivets --------------Ans

Given: N=800 rpm;Given: N=800 rpm;

ωω=83.77 rad/s;=83.77 rad/s;

Fluctuation of speed=2% Fluctuation of speed=2% ωω

Coeff.of fluctuation of speedCoeff.of fluctuation of speed

=(=(ωω11--ωω22)/)/ωω

=0.02=0.02

1mm=700 Nm vertically and1mm=700 Nm vertically and

3 degree=3x∏/180 rad3 degree=3x∏/180 rad

1mm=700x∏/601mm=700x∏/60

=36.65 Nm=36.65 Nm

ProblemProblem Turning moment dia for a six cylinder engine has been drawn Turning moment dia for a six cylinder engine has been drawn

to a scale 1mm=700Nm vertically and 1mm=3degree to a scale 1mm=700Nm vertically and 1mm=3degree horizontally.The intercepted area under the dia starting from A horizontally.The intercepted area under the dia starting from A w.r.t. mean resisting load line is -52,+120,-95,+145,-85,+71,-w.r.t. mean resisting load line is -52,+120,-95,+145,-85,+71,-106 mm.Engine speed is 800 rpm. Find moment of inerti of 106 mm.Engine speed is 800 rpm. Find moment of inerti of flywheel to present fluctuations of speed greater than 2% from flywheel to present fluctuations of speed greater than 2% from mean speedmean speed

Let Total energy at A=E then,Let Total energy at A=E then, total energy at B=E-52 min energytotal energy at B=E-52 min energy total energy at C=E-52+120total energy at C=E-52+120 total energy at D=E-52+120-95total energy at D=E-52+120-95 total energy at E=E-52+120-95+145total energy at E=E-52+120-95+145

total energy at F=E-52+120-95+145-85total energy at F=E-52+120-95+145-85 total energy at G=E-52+120-95+145-85+71total energy at G=E-52+120-95+145-85+71 total energy at H=E-52+120-95+145-85+71- total energy at H=E-52+120-95+145-85+71- 106 106

=E-2=E-2

Maximum Fluctuation of Energy,Maximum Fluctuation of Energy, ΔΔE=Max energy-Min energyE=Max energy-Min energy =( E+118)-(E-52)=( E+118)-(E-52) =170 mm=170 mm ΔΔE = 170 x 36.65E = 170 x 36.65 ΔΔE = 6230.5 Nm ------------AnsE = 6230.5 Nm ------------AnsAlso, Also, ΔΔE=IE=IωωCsCs I=6230.5/(83.77x0.02)I=6230.5/(83.77x0.02) I=44.4 kg m --------------------AnsI=44.4 kg m --------------------Ans