DR flywheel - Copy.pptx

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Department of Automobile EngineeringPSG College of Technology

9/11/20131

Flywheel




9/11/20132

• Necessity

In a combustion engine, & especially in one

with one or two cylinders, energy is imparted to

the crankshaft intermittently, & in order to keep

it rotating at a fairly uniform speed under a

substantially constant load, it is necessary to

provide it with a flywheel.

Flywheel




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• In a single cylinder engine(4 Stroke), in which

there is only one power stroke in two

revolutions of the crankshaft, a considerable

fraction of energy generated per cycle is stored

in the flywheel, & the proportion thus storeddecreases with an increase in the No. of

cylinders

• In a 4 cylinder engine about 40% of the energyof the cycle is temporarily stored.

Flywheel




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However,not all of this energy goes into flywheelDuring the 1st half of the power stroke, when

energy is being supplied in excess by the burninggases, all of the reciprocating parts of the engine

are being accelerated & absorb energy; besides, therotating parts other than the flywheel also havesome flywheel capacity, & this reduces the

proportion of the energy of the cycle which must bestored in the flywheel.

Flywheel




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• In a 6 cylinder engine the proportion of the energy

which must be absorbed & returned by the moving

parts amounts to about 20%.

• The greater the No. of cylinders the smaller the

flywheel capacity required per unit of piston

displacement, because the overlap of power

strokes is greater & besides other rotating parts of the engine have greater inertia.

Flywheel




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• However, the flywheel has by far the greatest

inertia even in a multi cylinder engine.

• Aside from its principle function, the fly wheel

serves as a member of the friction clutch, & it

usually carries also the ring gear of the electric

starter.

Flywheel




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Flywheel




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Flywheel




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Flywheel




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Flywheel

•Energy accumulator•Energy re-distributor

•A flywheel serves as a reservoir which stores

energy during the period when the supply of energy is more than the requirement & releases, it

during the period when the requirement of energy

is more than supply.•A flywheel helps to keep the crankshaft rotating at

a uniform speed




9/11/201311

In internal combustion engines,• the energy is developed during one stroke and the

engine is to run for the whole cycle on the energyproduced during this one stroke.

• the energy is developed, only during power stroke which is much more than the engine load; and noenergy is being developed during suction,

compression and exhaust strokes in case of 4 strokeengines & during compression in case of 2 strokeengines.

Flywheel




9/11/201312

• The excess energy developed during power stroke is

absorbed by the flywheel and releases it to thecrankshaft during other strokes in which no energyis developed, thus rotating the crankshaft at a

uniform speed.• When the flywheel absorbs energy, its speed

increases and when it releases, the speeddecreases. Hence a flywheel does not maintain aconstant speed, it simply reduces the fluctuation of speed.

Flywheel




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• The function of a governor in engine is entirely different

from that of a flywheel

Governor regulates the mean speed of an enginewhen there are variations in the load,e.g., when the loadon the engine increases it becomes necessary to increasethe supply of Working fluid. On the other hand, when theload decreases, less working fluid is required.

The governor automatically; controls the supply, of working fluid to the engine with the varying load conditionand keeps the mean speed within certain limits.

Flywheel




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• The flywheel does not maintain constant speed

• It simply reduces the fluctuation of speed.

• A flywheel controls the speed variations caused bythe fluctuation of the engine turning moment

during each cycle of operation.

It does not control the speed variations causedby the varying load.

Flywheel




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Capacity & Diameter

• The flywheel capacity of a given mass increases with itsdistance from the axis of rotation ; consequently, if the

flywheel is made large in diameter it need not be so heavy.

•

On the other hand there are two reasons for limiting thediameter.

– At high rpm, flywheel is subjected to disruptive or bursting force,

& by keeping down the diameter, the F O S can be kept high.

– As it is cast or cast & pressed steel housing, this need not weigh somuch if the diameter is smaller

Flywheel




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Flywheel

A Flywheel is given a high rotational inertia; i.e.,

most of its weight is well out from the axis




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Construction of Flywheels




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The flywheels of smaller size

(up to 600 mm diameter) are

casted in one piece.

The rim & hub are joined

together by means of web.

The holes in the web may

be made for handling purposes.




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In case the flywheel isof larger size (up to 2.5m

diameter), the arms are

made instead of web.The number of arms

depends upon the size of

flywheel & its speed of rotation.





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The split flywheels are above 2.5m diameter & are

usually casted in two piece.

It has advantage of relieving shrinkage stresses inarms due to unequal rate of cooling of casting.





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Maximum Fluctuation of Speed

&

Coefficient of Fluctuation of Speed




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•The Maximum Fluctuation of Speed

the difference between the maximum &

minimum speeds during a cycle. i.e., =(N1-N2 )

Where, N1=Maximum speed in r.p.m. during the cycle,N2=Minimum speed in r.p.m. during the cycle

•The Coefficient of Fluctuation of Speed is the ratio

of the maximum fluctuation of speed to the mean

speed.

Flywheel




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Flywheel




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Fluctuation of Energy

The Fluctuation of Energy may bedetermined by the turning moment

diagram for one complete cycle of

operation




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• turning moment iszero when thecrank angle is zero

• It rises to a max.value when crankangle reaches 90°and it is again zero

when crank angle is180°.

Flywheel




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• work done=turning moment x angle turned

• The area of the turning moment diagram represents thework done per revolution

• Engine is assumed to work against the mean resisting

torque,• Since it is assumed that

work done by the turning moment / revolution =work done against the mean resisting torque

Therefore,area of rectangle (aAFe) is proportional to work doneagainst the mean resisting torque.

Flywheel




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• When crank moves from 'a' to 'p'

work done by the engine is = area aBp,

whereas the energy required = area aABp.• In other words, the engine has done less work than

the requirement. This amount of energy is taken

from the flywheel and hence the speed of the

flywheel decreases.

Flywheel



Department of Automobile EngineeringPSG College of Technology 9/11/201328

Now the

• crank moves from p to q,

• work done by the engine = area pBbCq

• requirement of energy = area pBCq• Therefore the engine has done more work than the

requirement.

This excess work is stored in the flywheel and hencethe speed of the flywheel increases while the crankmoves from p to q.

Flywheel




• Similarly when the crank moves from q to r,

more work is taken from engine than is developed= areaCcD.

To supply this loss, the flywheel gives up some of its

energy and thus the speed decreases while the crankmoves from q to r.

• As the crank moves from r to s,

excess energy is again developed = area DdE

& the speed again increases.• As the piston moves from s to e,

again there is a loss of work & the speed decreases.

Flywheel




•The variations of

energy above andbelow the meanresisting torque lineare called fluctuation

of energy.

The areas BbC,CcD, DdE etc.

represent fluctuationsof energy.

Flywheel




• Engine has a max. speed either at q or at s.

This is due to the fact that flywheel absorbs energy

while the crank moves from p to q and from r to s.

• Engine has a minimum speed either at p or at r.The reason is that flywheel gives out some of its

energy when the crank moves from a to p and q to r.

• The difference between the maximum and the minimum

energies is known as maximum fluctuation of energy.




Flywheel

MeanSpeed

Maximum Speed

S p e e d

Crankangle Min. Speed




Flywheel

•The Fluctuation of Energy the variation of

Energy above & below mean resisting torque line.




For a 4 Stroke IC Engine,

•During suction stroke, since the pr. inside the engine

cylinder is less than the atmospheric pr., a negative loop isformed

•During compression stroke, the work is done on thegases, there a higher negative loop is obtained

•In the working stroke, the fuel burns & the gases expand,therefore a large positive loop is formed

•During exhaust stroke, the work is done on the gases,therefore a negative loop is obtained

Flywheel




Maximum Fluctuation of Energy




Maximum

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Department of Automobile Engineering

PSG College of Technology 9/11/201339

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Example-1

E l 1





Example-1

The turning moment diagram for a

Petrol engine is drawn to following scales:Turning moment, 1 mm =5 Nm; Crank angle, I mm = 1°;

The turning moment diagram repeats itself at everyhalf revolution of the engine and the areas above and belowmean turning moment line, taken in order are 295, 685, 40,340, 960, 270 mm2.

Determine the mass of 300 mm diameter flywheel rimwhen the coefficient of fluctuation of speed is 0.3% and theengine runs at 1800 r.p.m..

Also determine the cross-section of the rim when thewidth of the rim is twice of thickness. Assume density of rimmaterial as 7250 kg/m3.





Example-1… •Solution;

Given:

ρ = 7250 kg/m3;

Cs = 0.3% = 0.003;

D = 300 mm or

R = 150 mm= 0.15 m;

N =1800r.p.m. or

ω = 2 π x 1800 /60= 188.5rad/s

m, b, t = ?

∆ E = Iω2Cs

= m.R2.ω2Cs

m = ρV

= ρ.(πD.A)

= ρ.(2πR.A)

A = b x t,

b=2t





•Mass of the flywheel (m kg)

As per given scale,on the turning moment diagram,

1 mm2 = 5 Nm x 10 = 5 x(10 xπ /180) = 0.087 Nm

Max. fluctuation of energy = ∆ E = m.R2.ω2Cs

Let the total energy at A = E,

from fig., Energy at B = E + 295

Energy at C = E + 295 - 685 = E - 390

Energy at D = E - 390 + 40 = E - 350

Energy at E = E - 350 - 340 = E - 690

Energy at F = E - 690 + 960 = E + 270

Energy at G = E + 270 - 270 =E=Energy at A

Example-1…

E l 1





:. Maximum energy =E+295 at B&

Minimum energy =E-690 at EMaximum fluctuation of energy,

i.e. ∆ E =Max. energy – Min. energy

=(E + 295) - (E - 690) = 985 mm2

=985 x 0.087=86 Nm

Also, Maximum fluctuation of energy,

i.e. ∆ E = 86 =m.R2

.ω2

Cs

=m (0.15)2 (188.5)2 (0.003) = 2.4xm

:. m = 86/2.4 = 35.8 kg………Ans.

Example-1…

E l 1





Example-1…

•Cross-section of the flywheel rim

Mass of the flywheel rim, m =Vol. x density=(A x 2πR) xρ

Let, Width of rim, b=2 x t, thickness of rim

:. Cross-sectional area of rim, A=b x t = 2t x t = 2 t2

mass of the flywheel rim (m) = 35.8 = A x 2πRx ρ

=2 t2 X2 π x 0.15 x 7250

=13668 t2

:. t2 =35.8/13668 =0.0026or t =0.051 m = 51mm….Ans.

& b =2 t =2 x 51 = 102 mm….Ans.




PSG College of Technology9/11/2013

49

Example-2





50

Example-2A single cylinder, single acting,4

stroke oil engine develops 20 kW at300 r.p.m.

The work done by the gases during expan.

stroke is 2.3 times the work done on the gases duringthe compression and work done during the suction

and exhaust strokes is negligible.

The speed is to be maintained within ±1%.Determine the mass moment of inertia of the

flywheel.

l





51

Solution; Given:

P = 20kW=20x 103 W;

N = 300r.p.m. or

ω= 2πx300/60=31.42rad/s

Cs=±1% or

ω1

- ω2

= ±1%ω

Example-2

∆ E = Iω2Cs

Example 2





52

Example-2…

Nm /

/n P ycledone per calso, work

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60

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engine,bydtransmittetorqueMean

xCsIxE

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l 2





53

Example-2…

14160Nm

8000/0.565W

0.565W

/2.3WW8000Nmor WWcycleper workdonedoneworkNet

strokeexpn.duringdoneworkW&

strokecomprn.duringdoneworkW

let,

E

E

EE

CE

E

C

l





54

The work done during expan. stroke

is shown by triangle ABC in Fig., in whichbase AC =π radians & height BF =Tmax

:. Work done during expansion stroke,

WE =14160 =(1/2) X π X Tmax = 1.571 Tmax

Or Tmax= 14160/1.571 = 9013 Nm

Height above the mean torque line,

BG = BF - FG = Tmax- Tmean

= 9013 - 636.5 = 8376.5 Nm

Example-2…

Example 2





55

Example-2…

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rad .π .

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56

Example-2…

.Ans..........619.5kgm412230/19.7I 2

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57

Stresses in a Flywheel Rim





58


•A flywheel, consists of a rim at which major portion of massor weight is concentrated.

•Following types of stresses are induced in the rim

1. Tensile stress due to centrifugal force,

2. Tensile bending stress caused by the restraint

of the arms, and

3. The shrinkage stresses due to unequal rate of

cooling of casting. This stress is taken care of by a factorof safety.





59


b = Width of rim,

t =Thickness of rim,

A = Cross-sectional area

of rim = b x t.

D = Mean diameter of

flywheelR =Mean radius of

flywheel,

ρ =Density of flywheel

material,

ω =Angular speed of

flywheel,V =Linear velocity of

flywheel, and

ςt =Tensile/hoop stress

St i Ri





60

Stresses in Rim

1. Tensile stress due to centrifugal force

The tensile stress in the rim due

to the centrifugal force, assuming

that rim is unstrained by arms, is

determined in a similar way as a

thin cylinder subjected to internal pr..

Tensile stress,

ςt =ρ.R2.ω2 =ρ.v2 ...(v = ω.R)

(when ρ is in kg/m3 and v is in m/s, then σ t will be in N/m2 or Pa)

Note: From the above expression the mean diameter (D) of theflywheel may be obtained by using the relation v =πDN / 60

Stresses in Rim





61

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Stresses in Rim

Stresses in Rim





62

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Stresses in Rim





63

Example-3

Example 3





64

Example-3A multi-cylinder engine is to run at a constant load at a

speed of 600 r.p.m.On drawing the crank effort diagram

to scale of 1 m = 250 Nm and 1 mm = 3°,The areas in mm2 above & below mean torque line are: +160,- 172, + 168,- 191, + 197,- 162 mm2

The speed is to be kept within ±1% of the mean

speed of the engine. Calculate the necessary moment of inertia of the flywheel.

Determine suitable dimensions for cast ironflywheel with a rim whose breadth is twice its radial

thickness. The density of cast iron is 7250 kg/m

3

, and itsworking stress in tension is 6MPa.

Assume that rim contributes 92% of flywheel effect.

Example 3





65

Example-3

•Solution. Given:

N =600 r.p.m. orω =2π x 600/60 =62.84 rad/s;

ρ =7250 kg/m3; ςt =6 MPa =6 x 106 N/m2

∆ E = I

ςt=ρ.vv = (π

∆ E =

m = ρ

=

Example 3




PSG College of Technology

9/11/2013

66

•Moment of inertia of the flywheel

We know that,

Maximum fluctuation of energy, ∆E = Ixω2xCs

Scale for the turning moment is

1 mm = 250 Nm & scale for the crank angle is

1 mm = 3°= 3°xπ/180= π/60 rad, therefore

1 mm2 on the turning moment diagram

i.e., 1 mm2 =250 x π/60 = 13.1 Nm

Example-3

Example-3





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67

Let total energy at A = E. :. from Fig.,

Energy at B =E + 160

Energy at C =E + 160 - 172 = E - 12 Energy atD =E - 12 + 168 =E + 156

Energy at E =E + 156 - 191 =E – 35 (min. energy)

Energy at F = E - 35 + 197 =E + 162 (max. energy)

Energy at G =E + 162 - 162 = E =Energy at AMax. fluctuation of energy,

∆ E =Max. energy – Min. energy

= (E + 162) - (E - 35)= 197 mm2

∆ E =197 x 13.1=2581 N-m

Example-3





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Example-3





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69

Since the fluctuation of speed is±1%

of the mean speed (ω),therefore total fluctuation of speed,

ω1-ω2= 2% ω = 0.02ω

& coefficient of fluctuation of speed,

Cs =[(ω1-ω2)/ω] = 0.02

maximum fluctuation of energy, ∆E,

∆E, = 2581 = I.ω2.Cs = I (62.84)2 0.02 = 79xI

:. I = 2581/79 = 32.7 kgm2…..Ans

Example 3

Example-3





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•Dimensions of a flywheel rim

We know that, Mass of the flywheel rim,m = Vol. x density =2πR x A x ρ = πD x (b x t )x ρ b=2t (given)

•Peripheral velocity (v) & mean diameter (D)

We know that, tensile stress, ςt=ρ.v2

i.e., ςt=6 x 106=ρ.v2 =7250 X v2

So, v2 =(6 x 106)/7250 =827.6 or v =28.76 m/s

We also know that, peripheral velocity,v = (πDN)/60i.e., v=28.76 = (πDN)/60 = (πDx600)/60 = 31.42 D

So D = 28.76/31.42 = 0.915 m = 915 mm …Ans.

Example 3

Example-3





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•Mass of flywheel rim

Since rim contributes 92% of flywheel effect,

:. Energy of flywheel rim, Erim = 0.92 x total Energy of the flywheel, E

Max. fluctuation of energy, ∆E= E x 2 Cs

i.e., ∆E =2581 = E x 2 Cs =E x 2 x 0.02 = 0.04 E

:. E = 2581/ 0.04 = 64525 N-m& Energy of flywheel rim, E rim = 0.92 E

(v=ωR) = 0.92 x 64525 = 59363 Nm

Also, Erim= (1/2)Iω2 = (1/2)mk2ω2 = (1/2)mR2ω2 = (1/2)xmxv2

i.e., 59363 = 1/2 x m x v2 = 1/2 x m (28.76)2 = 413.6 m

:. m = 59363/413.6 = 143.5 kg

Example 3

Example-3





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72

The mass of the flywheel rim may also

be obtained by using following relations.

Since the rim contributes 92% of the flywheel effect,

•Irim = 0.92xIflywheel or m.k2= 0.92 x 32.7 =30 kgm2

Since radius of gyration, k =R =D/2 = 0.915/2 =0.4575m,

i.e., m=(30/k2) =(30/0.45752) =(30/0.2092)=143.5kg

•(∆ E) rim = 0.92 (∆ E )flywheel

m. V2.CS = 0.92 (∆ E )flywheel

m (28.76)2x0.02= 0.92x 2581

16,55xm = 2374.5

or m = 2374.5/16.55= 143.5kg

Example 3

Example-3





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73

m = 59363/413.6 = 143.5 kg

Also, mass of the flywheel rim,

m= (b x t )x πD x ρ & b=2t (given)

143.5 = b x t x πD x ρ

=2 t X t X π x 0.915 x 7250 = 41686 t2

t2 = 143.5/41686 =0.00344

t =0.0587 say 0.06 m =60 mm Ans.b =2 t =2 x 60 = 120 mm Ans.

Example 3





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Example-4





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75

Example-4

An otto cycle engine develops 50 kW at

150 r.p.m. with 75 explosions per minute.

The change of speed from the commencement to theend of power stroke must not exceed 0.5% of mean on eitherside.

Design a suitable rim section having width four timesthe depth so that the hoop stress does not exceed 4 MPa.

Assume that the flywheel stores 16/15 times the

energy stored by the rim and that the workdone during

power stroke is 1.40 times the workdone during the cycle.Density of rim material is 7200 kg/m3

Example-4





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76

•Solution.

Given:P =50 kW

=50 X 103W;

N = 150r.p.m.;n =75 ;

ςt =4 MPa

=4 x 106 N/m2;

ρ =7200 kg/m3

Example-4

Example-4





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77

We know that, Mass of the flywheel rim,m =Vol. x density=2πR x A x ρ

i.e., m=(b x t )x πD x ρ & [b=4t (given)]

Further, Energy of the flywheel rim,Erim= (1/2)Iω2 = (1/2)mk2ω2 = (1/2)mR2ω2 (v=ωR)

= (1/2)xmxv2 [Erim=(15/16)E …given]

And Max. fluctuation of energy, ∆E= E x 2 Cs

& hoop Stress=ςt=ρv2 & v=πDN/60

Example-4

Example-4





9/11/2013

78

Tmean transmitted by the engine or flywheel.

Power transmitted, P= (2xπxNxTmean)/60

i.e., 50x103= (2xπx150xTmean)/60 =15.71 Tmean

Tmean= 50 x 103/15.71= 3182.7 N-m

Workdone/cycle = Tmeanx θ = 3182.7 x 4 π= 40000 Nm{Or The workdone per cycle for a 4 stroke engine,

Workdone l cycle =[(Px60)/(No. of explosions/min)]

=(Px60)/n = (50000x60)/75=40000 Nm }

:. Workdone during power stroke =1.4xWorkdone/cycle=1.4x40000 = 56000 N-m

Example 4

Example-4





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79

The workdone during power stroke is shown

by ∆

le

ABC in Figin which base AC =π radians and height BF = Tmax

:. Workdone during working stroke= 1/2xπXTmax

= 1.571 Tmax

:. Also Workdone during working stroke=56000 Nm

:. Tmax=(56000/1.571) =35646Nm

Height above the mean torque line,BG = BF-FG = Tmax-Tmean

= 35646-3182.7= 32463.3Nm

Example 4

Example-4





9/11/2013

80

a p e

Nm46480E 0.83x00056

3564632463.3x56000

)(BF

(BG)x ABCtriangleof AreaEBDEof Area

BDE),triangleof area(i.e.,energyof nfluctuatioMaximum

)(BF

(BG)

ABCof Area

BDEof Arearelationlgeometrica

fromE),(energyof nfluctuatiomax.

representslinetorquemeantheabove

Fig)inshaded(shownBDEareatheSince

2

2

2le

2

2

le

le

Example-4





9/11/2013

81

Mean diameter of the flywheel (D)

Hoop stress,ςt=ρv

2

i.e., 4 x 106= ρ.v 2 = 7200 X v 2

:. v 2 =4 X 106/7200 = 556

or v = 23.58 m/s

Peripheral velocity, v = πDN/60

i.e., 23.58 = πDN/60 = πDx150/60 =7.855 D

D =23.58/7.855

D=3 m …Ans.

p

Example-4





9/11/2013

82

Cross-sectional dimensions of the rim

Cross-sectional area of the rim,

A=b x t= 4t x t =4t 2 (b=4t..given)

Since N1 - N2 = 0.5% N either side, (..given)

:. total fluctuation of speed,

N1 - N2 = 1% of mean speed =0.01 N& coefficient of fluctuation of speed,

Cs. = (N1 - N2 )/N = 0.01

E = Total energy of the flywheel.

Maximum fluctuation of energy (∆ E),46480 = Ex 2 Cs = E x 2 x 0.01 = 0.02 E

:. E = 46480/0.02 =2324 x 103 Nm

p

Example-4





9/11/2013

83

Since the energy stored by the flywheel is

16/15 times the energy stored by rim,

Therefore the energy of the rim,Erim = (16/15)E= (16/15)x 2324 x 103 = 2178.8 x 103Nm

Also Erim= (1/2)xmxv2

i.e., 2178.8 X 103 = (1/2)xmx(23.58)2 =278xm

:. m = 2178.8 x103 / 278 =7837 kg

Also, mass of the flywheel rim, m= A x πD x ρ

i.e., 7837 = A x πD x ρ =4 t 2 X π X 3 x 7200 =271469 t2

or t2 = 7837 / 271469 = 0.0288or t = 0.17 m = 170 mm …Ans.

b =4 t =4 x 170 = 680 mm …Ans.

p





9/11/2013

84

Example-5





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85

Example-5

.7200kg/mastaken bemayironcastof Density.39.23N/mm

astaken bemaykeyandshaftfor stressshear SafeThetorque.meanthetwiceisshafttheontorquemaximumThecycle.

wholetheduringdonework averagethetimes3

11isstroke power

theduringdonework thethatassumed bemayItspeed.meanof

4%tolimited betoisspeedof nfluctuatiototalThe.3.92N/mm

exceedtonotismaterialtheinstresshoopThe800rpm.at36.8kW

DevelopingEngineDieselstroke4afor FlywheelironcastaDesign

32

2

E l 5





9/11/2013

86

Example-5…

3

2

s

meanmax

s

2

7200kg/mironcastof Density

39.23N/mmf stressshear Safe

T2T

cyclewholetheduring

donework ave.3

11stroke power duringdoneWork

0.04K speedof nFluctuatio

3.92N/mmmaterialinstressHoop

[email protected]

FlywheelironCastGiven,

:Solution





9/11/2013

87

∆ E = Iω2

Csςt=ρ.v2

v = (πDN)/60

∆ E = m.R2.ω2Cs

m = ρV = ρ.(πD.A)= ρ.(2πR.A)

A = b x t,

b=2t

Example-5…





9/11/2013

88

p

Example-5…





9/11/2013

89

p

50mmisdiameter shaft,di.e.,

50mm49mm0.049md1039.23π

16878.54d

f 16

dπ

878.54Nm439.272T2TBut

439.27Nm800π2

6000036.8

Nπ2

60000Power T

kW60000

TNπ2Power that,knowWe

s

s

6

3

s

s

3s

meanmax

mean

mean

bookhanddataineqn.3.1ref.,

Example-5…





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90

Example-5…





9/11/2013

91

5980Nm4

55207360

EnergyMin.EnergyMax.energyof nFluctuatio

7360Nm3

115520

done/cycleWork311strokepower duringdoneWork

5520Nm4π439.27720439.27

turnedθ AnglemomentTTurningdone/cycleWork

100mm502d2diameter hubor diameter Boss

0

mean

s

bookhanddataineqn.17.8ref.,

E

p

Example-5





9/11/2013

92

Example-5…

546mmDrimof dia.Meani.e.,

0.5460mDor 1.02NDπ23.33V

,NDπ

V Also

23.33m/sVor V7200103.92

,VρσstressHoop As

ec21.30N.m.sINπ20.04I5980E

ωCIE Also,

1.02N)maxNspeed,meanof 4%tolimitedspeedof nfluctuatiototal(

rimof velocityV&densityρ

speedof nfluctuatioof tcoefficiens

C&speedangular Meanω

max

26

2

max

22

,2

s

60

60

60







9/11/2013

94

Example-5…

469.9mm

622.1mm

304.4mm

76.144tb

76.1mm,t

0.4699m

0.07610.546tDflywheeltheof diameter Inside

0.6221m

0.07610.546tDflywheeltheof diameter Outside

or

mean

mean





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95

Stresses in Flywheel Arms





9/11/2013

96


Following stresses are induced in the arms of aflywheel

•Tensile stress due to centrifugal force acting on the

rim.•Bending stress due to torque transmitted from the

rim to the shaft or from the shaft to the rim.

•Shrinkage stresses due to unequal rate of cooling of casting. These stresses are difficult to determine.





9/11/2013

97

•Tensile stress due to centrifugal force

Due to the centrifugal force acting on the

rim, the arms will be subjected to direct tensile stress

whose magnitude is,

:. Tensile stress in the arms,

ςt1= (3/4)ςt= (3/4)ρxv2






9/11/2013

98

•Bending stress due to torque transmittedT =Maximum torque transmitted by the shaft,

Z = Section modulus for the c/s of arms &

R =Mean radius of rim

r =Radius of the hub

n =Number of arms


Stresses in Arms





9/11/2013

99

11 bt

end,hubtheatarmstheinstresstensileTotal

r)-(R

ZnR

T

Z

Marms,instressBending

r)-(RnR

T Mi.e.,hub,theat

armtheonlieswhichmomentbendingmax.&

nR

T armeachonLoad

RTFrim,theof radiusmeantheatLoad

b1





9/11/2013

100

Design of Flywheel Arms

D/n of Arms





9/11/2013

101

D/n of Arms

(2)&(1)equationsbyobtaineddimensionsarms,b2aAssuming

2r).......-(RZnR

T

Z

Mσ

arms,instressdingMaximumBen

r)-(RnR

T M

moment,bendingmaximumthatknowWe

1.......abx32

π Zmodulus,Section

stress.bendingmax.for desgnednedisit&axis,minor thetwiceasaxismajor with

ellipticalusuallyisarmstheof c/sThe

11

b

2

11





9/11/2013

102

Design of Shaft, Hub and Key





9/11/2013

103


shaft.of materialthefor stressshear Allowable&shaft,theof Diameter d

where,

16 Td,transmittetorqueMaximum

d.transmittetorquemaximumthefromobtained

isflywheelfor shaftof diameter The

1

max

3

1d

Shaft, Hub and Key





9/11/2013

104

therim.of widthtoequaltakengenerallyisIt

diameter.shaftthetimes2.5to2fromlengthHub&

dia.shaftthetwiceastakenusuallyisdiameter Hub

shaft.of diameter or hubof diameter Inner d

&hub,of diameter Outer dwhere,

16 Td,transmittetorqueMax.

d.transmittetorquemaximumthe

for shaft,hollowaasdesignedishubThe

1

max

d

d d 4

1

4





9/11/2013

105

shaft.of Diameter d

alkey.materithefor stressShear

key,theof LengthLwhere,

2

dxLx w xTmaxshaft,bydtransmitteTorque

shearing.inkeyof failure

thegconsiderinbyobtainediskeyof lengthThe

hub.&shaftthefor usediskeysunkstandard A

1

1

&






9/11/2013

106

Example-6

Example-6





9/11/2013

107

Design and draw a cast iron flywheel used for a four

stroke I.C engine developing180 kW at 240 r.p.m.The hoop or centrifugal stress developed in theflywheel is 5.2 MPa, the total fluctuation of speed isto be limited to 3% of the mean speed. The work

done during the power stroke is 1/3 more than theaverage work done during the whole cycle.

The max. torque on shaft is twice the meantorque. The density of cast iron is 7220 kg/m3.

Example-6





9/11/2013

108

Solution.

Given:

P =180kW

=180X 103W;

N = 240r.p.m.;

ςt= 5.2MPa

= 5.2x 106N/m2 ;

Nl - N2 = 3% N;

ρ = 7220 kg/m3

.

Turning moment diagram of a 4

stroke engine is shown in Fig.

E)(ffl t tiM iExample-6





9/11/2013

109

Nm x xcycledoneWork :.

N ine, n stroke eng For g strokesof workin Nowhere n

,n

P cycleWorkdonetained asalso be obcycle mayworkdone

90000120

6010180/

2/4.min/.

60/,/

3

40

Nm900004x7161xTycleworkdone/c

Nm716122

60x10180x

N2

60PxT

flywheel,thebydtransmittetorqueMean

E)(energyof nfluctuatioMaximum

mean

3

mean

Example-6





9/11/2013

110

Nm763842120000xT

120000TXx

TXxstrokepower duringWorkdone

TBF height&radians ACbasewhich,inFig.in ABCabyshownisstrokepower theduringWorkdone

Nm12000090000x3

1 90000

strokeworking)(or power theduringWorkdone

cycle,wholeduringworkdoneaveragethanmore

times1/3isstrokepower duringworkdoneSince

max

max

max

max

le

21

21

Example-6litthbH i ht





9/11/2013

111

Nm98555(76384)

(69223)120000

(BF)

(BG)x ABC Δof Area

BDE Δof AreaEenergy,of nfluctuatioMax.

,(BF)(BG) :

ABC Δof AreaBDE Δof Arearelation,lgeometricafrom

E),(energynof fluctuatiomax.therepresentslinetorque

meantheaboveFig.inshadedshownBDEareatheSince

Nm692237161-76384

T-TFG-BFBG

line,torquemeantheaboveHeight

2

2

2

2le

le

2

2

le

le

meanmax

Example-6ifl h lthfDi t1





9/11/2013

112

Ans.... m2.0426.8/13.1D

rimflywheeltheof Diameter 1.

D13.160

250D

ND

26.8v

velocity,Peripheral

or

720/722010X5.2V

VX7220.V10x5.2rim,flywheeltheindevelopedstressHoop

m/s26.8v

62

226

t

60

Example-6ifl h lthfM2





9/11/2013

113

.....Ans.kg4995555/19.7398m

rimflywheeltheof Mass2.

19.73m0.03(25.14)2

2.04 m98555E

C,.m.R98555E

energy,of nfluctuatioMax.

0.03N

N-N

Csspeed,of nfluctuatioof tCoefficien&

rad/s25.1460

x2502

60xN2 rim,flywheeltheof speed Angular

22

s

22

21

Example-6ifdi iti lC3





9/11/2013

114

Ansmm...4702352t2b......Ans.mm235tor

rimof dimensionssectional-Cross3.

&

m0.235say0.232tor

0.0544995/92556t

t925567220x2.04t2m

xD A x4995mrim,flywheeltheMassof

t2tt2tb Arim,theof areasectional-Cross

t(Assume)2metresinrimtheof Widthb&

metres,inrimof thicknessor DepthtLet

2

22

2

Example-6





9/11/2013

115

Ansmm........0.47mm470 bI&

m0.25mm250125x21

d2d,

........Ans125mmsay122

1

dor )2 N/mm40MPa40(Taking

)(d7.855)(d4016

)(d16

10x14322T

shaft,on theactingtorquemaximumAlso,

mm- N10x14322 Nm143227161x2Tx2T

shaft,on theactingtorqueMax.,Tx2TSince

hub, of Lengthl&dhub, of Diameter d shaft,of Diameter dLet

3

1

3

1

3

1

3

max

3

meanmax

meanmax

1

hubof lengthandDiameter 4.

that knowWe

Example-6di iti lC5





9/11/2013

116

Nmm10x2094.5Nm2094.5

Nm0.25)-(2.0462.04

14322d)-(D

nD

T r)-(R

nR

T M

by,giveniscantilever asassumediswhich

end,hubtheatarminmomentbendingmax.

)...(AssumeN/mm15MPa15

armsof materialfor stressBending

)...(Assumea0.5axisMinor baxis,Major a

)...(Assume6armsof Number n Let

3

2

b

111

armsellipticalof

dimensionssectionalCross5.

Example-6di iti lC





9/11/2013

117

Prof. B.Dinesh Prabhu

PESCE Mandya

....Ansmm70140x0.5 1

a0.51

b

....Ans.mm1401a

arms.....ellipticalof

dimensionssectionalCross

or

10x2793/1510x89041)(a

)(a

1089041

)(a0.05

102094.5x

Z

M 15

,stressbendingthethatknowWe

)(a0.05)(a0.5a)(abZ

arm.theof section-crossthefor modulussectionThe

333

1

3

1

3

3

1

3

b

3

1

2

11

2

11

&

3232

Example-6





9/11/2013

118

Prof. B.Dinesh Prabhu

PESCE Mandya

....Ans.mm160say159L

......Ansmm20keyof thickness&

s........Anmm36wkeyof Width

keyof Dimensions6.

33

33

max

1090/10x14322

L10902

125x40x36xL

d wL10x14322T

,shaftthebydtransmittetorqueMax.

.

:followsasaremm125diameter of shaftafor keysunkr rectangulaof dimensionsstandardThe

shearinginkeyof failuregconsiderinbyobtd.is(L)keyof Length

21

hi hiitt t lfCh k

Example-6



safeisdesignMPa15thanlessisSince

MPa.15thangreater benotshould

whichriminstresstotalfor Check

MPa6.97N/m10x6.970.595)(0.7510x5.18

N/m0.2356

22.044.935

0.7526.87220

tn

R4.9350.75ρvσ

rim,theinstresstotalthat,knowWe

266

2

2

2

2

2

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DR flywheel - Copy.pptx