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A-1

s4ForZnd YearMechonicol Section

ffi,ffi@

W,W'',,:*$, -ffi@,@

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Theory of mochines

m-k-I_

Mass of the flywheel in kg ,Radius of gyration of the flywheel in meters,Mass moment of inertia of the flywheel about

I-m* l*/V*u* and Nmin = Maximum and minimum speeds during the cycle in r.p,m.,/V - Mean speed during the cycle in r.p.m.

N *non = N*u* * N*tn2o)max ;tnd ttrmin= Maximum and minimum angular speeds during the cycle in radr's,co = Mean speed during the cycle in radlsec.

@*u* * O*in@ *"on = 2

Nr* - N*in 0*u* - 0*inN ^"o, @*"on

Energy Stored in a FlywheelA flywheel is shown in the Figure. We havediscussed that when a flywheel absorbs energy,its .speed increases dnd when it gives u penergy, its speed decreases.

Let:

its axis of rotation in kg-m2

: m"t"Fa

Ks - Coefficient of fluctuation of speed,

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Theory of mqchinesWe know that the mean kinetic energy of the flywheel,

-t*fltliEEl

[, -l* I*af2 in N.m or JouleAs the speed of the flywheel changes from

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Theory of rnqchines Ffuuhee.lWork done

The work done per cycle (in N-m or joules) may be obtained by using the followingtwo relations:- L.'Work done per cycle = Tmean X eWhere:

And

T^"rn = Mean torque,T^""n= Pmear/ O)^"rn

e - Angle turned (in radians), in one revolution.,@ 0 = 2o, in case of steam engine and two stroke internal combustion enginesj0 - 4n, in case of four strbke internal combubtion engines.

2. The work done per cycle may also be obtained by using the following relatlor1 :

work done per cycle - P * 60nWhere:

rr - Number of working strokes per minute,= N, in case of steam engines and two stroke internal combustion engdnes,- N 12, in case of four stroke internal combustion engines.

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Theory of mochines FlywheelExamplel. The turning moment diagram for a petror engine is drawn to thefotlowing scales: iurning moment, r mm = s N-m; crank angle, 7 mm = ro. Theturning moment diagram repeats itsetf at every half revolution of the engine andthe.areas above and below the mean turning moment line taken in order are 295,685, 40, 340, 960, 270 mm2. The rotating parts are equivarent to a mass of 36kg at a radius of gYration of 150 mm. Determine the coefficient of fluctuation ofspeed when the engine runs at 1800 r.p.m.

Crank aftgle

The turnirig firrlrneirt diagfaurLet the total energy at A = E, then referring to the shown diagram:EnergyEnergyEnergyEnergyEnergyEnergy

at8:E+atC=E+atD-E-atE- E-atF- E-atG=E+

29s295 -390 +350 -690 +270 -

... (Maximum energy)6Bs - E - 39040 = E- 350340 - E - 690

-s-...(Minimum energy)960-E+270

270-E=EnergyatAe ion of

AE = Maximum energy - Minimum energyAE = (E + 295) - (E - 690) = 985 mm2- The given scale is lmm2 - 5* l* Qf llg0; - 0.0g73 N.mSo, AE = 985 * 0.0873 = 86 JouleSince, AE:[*r'*.**Ks

86 : 36 * (0.150)' * (2+ r *1g00/60)' * K,Therefore, Ks : 0.003 : 0.3 o/o

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Theory of mqchines Fb'wheelExample 2- The mass of flywheet of an engine is 6.s tones and the radius ofgyration is 7-8 meters. It is found from the turning moment diagram that thefluctuation of energy is s6 kN.m. If the mean speed of the engine is 720 r.p.m.,-find the maximum and minimum speeds.

Givens:m = 6.5t = 6500k9 ; R = 1.8m ; AE= 56kN.m = 56 x 103N.m ; /V= 120 r.p.m.

Let /Vmax and /V-in = Maximum and minimum speeds respectively.We know that fluctuation of energy (AE),

LE - I * o' *"or* K, - I * @*ron* (@*u*- @*in)56 x 103 = m x R2 x N^"u, (ff.u* - /Vmin).x $Tt/$01 6Q))

IUrnu* : Iy'min - zr.p.m.' -......... (r)We also know that rnean spee d (N*uun),

L2O-(N*u**Nrnin)/2 orN.ur*Nm.n= L20xZ=240 r.p.m .(ii)

From equations (r) and (ii),/Umax = LZL I.P.ffi.,

andffnrin = 119 r.p.m.

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Theory of machines Flythu)Example 3. A shaft fitted with a ftywheel rotates at 250 r.p.m. and drives amachine. The torque of machine varies in a cyctic manner over a period of 3revolutions. The torque rises from 7so N-m to 3000 N-m uniformty during 1/2revolution and remains constant for the foilowing revolution. It then faltsuniformly to 750 N-m during the next 1/2 revotution and remains constant forone revolution, the cycle being repeated thereafter.Determine the power required to drive the machine and percentage fluctuation inspeed, if the driving torque applied to the shaft is constant and the mass of theflywheel is 500 kg with radius of gyration of 600 mm.

Given:N-250 r.p.m. or -z x250lG0:26.2radls ;nx:500kg; k:600mm:0.6m

;;The turning moment $iagrary tbr the comSilete cycle is shown'in the Figure:

30001125

1 875

750

2x 3n 4n 5n 6n-*lCrank angle

-+

tEiCoEoECt.gsFI

Inru. Re-v. ---*l |n*o k-r FIev.it[*rThe turninq nto{nent rliagra,m

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Theory of rnochines

we know that the torque required for one complete cycleFtyrlrce.l

Ans,

= Area of figure OA BCDEF= ArEA OAEF + ArCA ABG + ArCA BCHG + ArCA CDH=oFxoA+ x(0.5) xAGxBG+GHxcH* x(0.5) xHDxCH= 6n x 750 + (0.5) x zc (3000 - 750) + 2n( 3000 - 750 )

= tt 250 r N-m ... (r)If Tmean is the mean torque in N-m, then torque required for one complete cycle= Tmean x 6 zu N-mFrom equations (f) and (ii),

Tmean=11 250nfGn-1875...(ii)

'Power required'to drive the machine:N-m ;

Coefficient of fluctuation of soeed:Let K5 = Coefficient of fluctuation of speed.First of all, let us find the varues of LM and /vp. From similar triangles ABG and BLM,

we know that power required to drive the machine,P - T*"un X (Dmean = 1875 x 26.2 = 49 IZS W - 4g.L25 kW

LM BM=_AG BG OT

NP CtrHI} CH

or LH =0,$ A-OV LI'7=--":TC)Now, from similar triangles CHD and CNp,of NP 3m0 - 187s = 0.5 or AIP - 0,.5 lrTt, 30m - 75{)

From the turning moment diagram, we find that:BM = CN = 3000 - 1975 = 1125 N_m

since the area above the mean torque line represents the maximum fluctuation ofenergy, therefore, maximum fluctuation of energy,

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Theory of mochinesAE = Area LBCP = Area LBM + Area MBCN + ArCA PNC

= (0.5) x LM x BM + MN x BM + (0.5) x NP x CN= (0.50) x (0.5) zu x LI25* 2n x LL25 + (0.5) x (0.5) n x1t25

= BB37 N-mWe know that maximum fluctuation of energy (AE),

BB37 - rn.l?.a'.K, = 500 x (0.6)2 x Q6.2)'X K5 = 123 559 KsKs - 0.071 Ans,

FFirneel