Prof' B'Dinesh Prabhu PESCEMandya

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CHAPTER-7Flywheel Camshaft
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Flywheel
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Flywheel

• 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.

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Flywheel

• 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 stored decreases
  with an increase in the No. of cylinders
• In a 4 cylinder engine about 40 of the energy of
  the cycle is temporarily stored.

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Flywheel

• However,
• not all of this energy goes into flywheel
• During the 1st half of the power stroke, when
  energy is being supplied in excess by the burning
  gases, all of the reciprocating parts of the
  engine are being accelerated absorb energy
  besides, the rotating parts other than the
  flywheel also have some flywheel capacity, this
  reduces the proportion of the energy of the cycle
  which must be stored in the flywheel.

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Flywheel

• 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.

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Flywheel

• 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.

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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

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Flywheel

• In internal combustion engines,
• the energy is developed during one stroke and the
  engine is to run for the whole cycle on the
  energy produced during this one stroke.
• the energy is developed, only during power stroke
  which is much more than the engine load and no
  energy is being developed during suction,
  compression and exhaust strokes in case of 4
  stroke engines during compression in case of 2
  stroke engines.

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Flywheel

• The excess energy developed during
• power stroke is absorbed by the flywheel and
  releases it to the crankshaft during other
  strokes in which no energy is developed, thus
  rotating the crankshaft at a uniform speed.
• When the flywheel absorbs energy, its speed
  increases and when it releases, the speed
  decreases. Hence a flywheel does not maintain a
  constant speed, it simply reduces the fluctuation
  of speed.

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Flywheel

• The function of a governor in engine is
• entirely different from that of a flywheel
• Governor regulates the mean speed of an engine
  when there are variations in the load,
• e.g., when the load on the engine increases it
  becomes necessary to increase the supply of
  Working fluid. On the other hand, when the load
  decreases, less working fluid is required.
• The governor automatically controls the
  supply, of working fluid to the engine with the
  varying load condition and keeps the mean speed
  within certain limits.

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Flywheel

• The flywheel does not maintain constant speed
• It simply reduces the fluctuation of speed.
• A flywheel controls the speed variations caused
  by the fluctuation of the engine turning moment
  during each cycle of operation.
• It does not control the speed variations caused
  by the varying load.

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Flywheel

• Capacity Diameter
• The flywheel capacity of a given mass increases
  with its distance 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 the diameter.
• 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 so much if the diameter is
  smaller

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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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Construction of Flywheels

• 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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Construction of Flywheels

• In case the flywheel is of 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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Construction of Flywheels

• The split flywheels are above 2.5m diameter
  are usually casted in two piece.
• It has advantage of relieving shrinkage stresses
  in arms 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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Flywheel

• The Maximum Fluctuation of Speed
• the difference between the maximum
• minimum speeds during a cycle. i.e., (N1-N2)
• Where, N1Maximum speed in r.p.m. during the
  cycle,
• N2Minimum 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.

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Flywheel
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• Fluctuation of Energy

The Fluctuation of Energy may be determined by
the turning moment diagram for one complete cycle
of operation
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Flywheel

• turning moment is zero when the crank angle is
  zero
• It rises to a max. value when crank angle reaches
  90 and it is again zero when crank angle is 180.

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Flywheel

• work doneturning moment x angle turned
• The area of the turning moment diagram represents
  the work 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
  done against the mean resisting torque.

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Flywheel

• 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.

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Flywheel

• 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
  hence the speed of the flywheel increases while
  the crank moves from p to q.

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Flywheel

• Similarly when the crank moves from q to r,
• more work is taken from engine than is
  developed area CcD.
• To supply this loss, the flywheel gives up some
  of its energy and thus the speed decreases while
  the crank moves 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.

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Flywheel

• The variations of energy above and below the mean
  resisting torque line are called fluctuation of
  energy.
• The areas BbC, CcD, DdE etc. represent
  fluctuations of energy.

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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.

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Flywheel

• The Fluctuation of Energy the variation of
• Energy above below mean resisting torque line.

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Flywheel

• 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 is formed
• During compression stroke, the work is done on
  the gases, 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

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Maximum Fluctuation of Energy
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• Maximum
• Fluctuation of Energy

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• Coefficient of
• Fluctuation of Energy

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• Energy stored in a flywheel

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Example-1
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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
  every half revolution of the engine and the areas
  above and below mean turning moment line, taken
  in order are 295, 685, 40, 340, 960, 270 mm2.
• Determine the mass of 300 mm diameter flywheel
  rim when the coefficient of fluctuation of speed
  is 0.3 and the engine runs at 1800 r.p.m..
• Also determine the cross-section of the rim when
  the width of the rim is twice of thickness.
  Assume density of rim material as 7250 kg/m3.

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Example-1

• Solution
• Given
• ? 7250 kg/m3
• Cs 0.3 0.003
• D 300 mm or
• R 150 mm
• 0.15 m
• N 1800r.p.m. or
• ? 2 p x 1800 /60 188.5rad/s
• m, b, t ?

? E I?2Cs m.R2.?2Cs m ?V ?.(pD.A)
?.(2pR.A) A b x t, b2t
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Example-1

• Mass of the flywheel (m kg)
• As per given scale,
• on the turning moment diagram,
• 1 mm2 5 Nm x 10 5 x(10 xp /180) 0.087 Nm
• Max. fluctuation of energy ? E m.R2.?2Cs
• 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 EEnergy at A

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Example-1

• . Maximum energy E295 at B
• Minimum energy E-690 at E
• Maximum fluctuation of energy,
• i.e. ? E Max. energy Min. energy
• (E 295) - (E - 690) 985 mm2
• 985 x 0.08786 Nm
• Also, Maximum fluctuation of energy,
• i.e. ? E 86 m.R2.?2Cs
• m (0.15)2 (188.5)2 (0.003) 2.4xm
• . m 86/2.4 35.8 kgAns.

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Example-1

• Cross-section of the flywheel rim
• Mass of the flywheel rim, m Vol. x density
• (A x 2pR) x?
• Let, Width of rim, b2 x t, thickness of rim
• . Cross-sectional area of rim, Ab x t 2t x t
  2 t2
• mass of the flywheel rim (m) 35.8 A x 2pRx ?
• 2 t2 X2 p x 0.15 x 7250
• 13668 t2
• . t2 35.8/13668 0.0026
• or t 0.051 m 51mm.Ans.
• b 2 t 2 x 51 102 mm.Ans.

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• Example-2

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Example-2

• A single cylinder, single acting,4
• stroke oil engine develops 20 kW at
• 300 r.p.m.
• The work done by the gases during expan. stroke
  is 2.3 times the work done on the gases during
  the 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.

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Example-2

• Solution Given
• P 20kW
• 20x 103 W
• N 300r.p.m. or
• ? 2px300/60
• 31.42rad/s
• Cs1 or
• ?1- ?2 1?

? E I?2Cs
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Example-2
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Example-2
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Example-2

• The work done during expan. stroke
• is shown by triangle ABC in Fig., in which
• base AC p radians height BF Tmax
• . Work done during expansion stroke,
• WE 14160 (1/2) X p 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

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Example-2
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Example-2
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Stresses in a Flywheel Rim
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Stresses in a Flywheel Rim

• A flywheel, consists of a rim at which major
• portion of mass or 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 factor of safety.

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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,
• st ?.R2.?2 ?.v2 ...(v ?.R)
• (when ? is in kg/m3 and v is in m/s, then st will
  be in N/m2 or Pa)
• Note From the above expression the mean diameter
  (D) of the flywheel may be obtained by using the
  relation vpDN / 60

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Stresses in Rim
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Stresses in Rim
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• Example-3

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Example-3

• A 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 iron
  flywheel with a rim whose breadth is twice its
  radial thickness. The density of cast iron is
  7250 kg/m3, and its working stress in tension is
  6MPa.
• Assume that rim contributes 92 of flywheel
  effect.

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Example-3

• Solution. Given
• N 600 r.p.m. or
• ? 2p x 600/60 62.84 rad/s
• ? 7250 kg/m3 st 6 MPa 6 x 106 N/m2

? E I?2Cs st?.v2 v (pDN)/60 ? E
m.R2.?2Cs m ?V ?.(pD.A) ?.(2pR.A) A b x
t, b2t
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Example-3

• Moment of inertia of the flywheel
• We know that,
• Maximum fluctuation of energy, ?E Ix?2xCs
• Scale for the turning moment is
• 1 mm 250 Nm scale for the crank angle is
• 1 mm 3 3xp/180 p/60 rad, therefore
• 1 mm2 on the turning moment diagram
• i.e., 1 mm2 250 x p/60 13.1 Nm

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Example-3

• Let total energy at A E. . from Fig.,
• Energy at B E 160
• Energy at C E 160 - 172 E - 12 Energy at D
  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 A
• Max. fluctuation of energy,
• ? E Max. energy Min. energy
• (E 162) - (E - 35) 197 mm2
• ? E 197 x 13.12581 N-m

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Example-3

• Since the fluctuation of speed is1
• of the mean speed (?),
• therefore total fluctuation of speed,
• ?1-?2 2 ? 0.02?
• coefficient of fluctuation of speed,
• Cs (?1-?2)/? 0.02
• maximum fluctuation of energy, ?E,
• ?E, 2581 I.?2.Cs I (62.84)2 0.02 79xI
• . I 2581/79 32.7 kgm2..Ans

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Example-3

• Dimensions of a flywheel rim
• We know that, Mass of the flywheel rim,
• m Vol. x density 2pR x A x ? pD x (b x t )x
  ? b2t (given)
• Peripheral velocity (v) mean diameter (D)
• We know that, tensile stress, st?.v2
• i.e., st6 x 106?.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
  (pDN)/60
• i.e., v28.76 (pDN)/60 (pDx600)/60 31.42 D
• So D 28.76/31.42 0.915 m 915 mm Ans.

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Example-3

• 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, Erim 0.92 E
• (v?R) 0.92 x 64525 59363 Nm
• Also, Erim (1/2)I?2 (1/2)mk2?2 (1/2)mR2?2
  (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

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Example-3

• 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)143.5k
  g
• (? 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

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Example-3

• m 59363/413.6 143.5 kg
• Also, mass of the flywheel rim,
• m (b x t )x pD x ? b2t (given)
• 143.5 b x t x pD x ?
• 2 t X t X p 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.

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• Example-4

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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 the
  end of power stroke must not exceed 0.5 of mean
  on either side.
• Design a suitable rim section having width four
  times the 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

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Example-4

• Solution.
• Given
• P 50 kW
• 50 X 103W
• N 150r.p.m.
• n 75
• st 4 MPa
• 4 x 106 N/m2
• ? 7200 kg/m3

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Example-4

• We know that, Mass of the flywheel rim,
• m Vol. x density2pR x A x ?
• i.e., m(b x t )x pD x ? b4t (given)
• Further, Energy of the flywheel rim,
• Erim (1/2)I?2 (1/2)mk2?2 (1/2)mR2?2
  (v?R)
• (1/2)xmxv2 Erim(15/16)E given
• And Max. fluctuation of energy, ?E E x 2 Cs
• hoop Stressst?v2 vpDN/60

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Example-4

• Tmean transmitted by the engine or flywheel.
• Power transmitted, P (2xpxNxTmean)/60
• i.e., 50x103 (2xpx150xTmean)/60 15.71 Tmean
• Tmean 50 x 103/15.71 3182.7 N-m
• Workdone/cycle Tmeanx ? 3182.7 x 4 p 40000
  Nm
• Or The workdone per cycle for a 4 stroke
  engine,
• Workdone l cycle (Px60)/(No. of
  explosions/min)
• (Px60)/n (50000x60)/7540000 Nm
• . Workdone during power stroke
  1.4xWorkdone/cycle 1.4x40000 56000 N-m

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Example-4

• The workdone during power stroke is shown
• by ?le ABC in Fig
• in which base AC p radians and height BF Tmax
• . Workdone during working stroke 1/2xpXTmax
• 1.571 Tmax
• . Also Workdone during working stroke56000 Nm
• . Tmax(56000/1.571) 35646Nm
• Height above the mean torque line,
• BG BF-FG Tmax-Tmean
• 35646-3182.7 32463.3Nm

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Example-4
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Example-4

• Mean diameter of the flywheel (D)
• Hoop stress, st?v2
• i.e., 4 x 106 ?.v2 7200 X v2
• . v2 4 X 106/7200 556
• or v 23.58 m/s
• Peripheral velocity, v pDN/60
• i.e., 23.58 pDN/60 pDx150/60 7.855 D
• D 23.58/7.855
• D3 m Ans.

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Example-4

• Cross-sectional dimensions of the rim
• Cross-sectional area of the rim,
• Ab x t 4t x t 4t2 (b4t..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

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Example-4

• 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 pD x ?
• i.e., 7837 A x pD x ? 4 t 2 X p X 3 x 7200
  271469 t2
• or t2 7837 / 271469 0.0288
• or t 0.17 m 170 mm Ans.
• b 4 t 4 x 170 680 mm Ans.

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Example-5
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Example-5
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Example-5
? E I?2Cs st?.v2 v (pDN)/60 ? E
m.R2.?2Cs m ?V ?.(pD.A) ?.(2pR.A) A b x
t, b2t
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Example-5
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Example-5
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Example-5
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Example-5
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Example-5
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Example-5
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Example-5
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• Stresses in Flywheel Arms

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Stresses in Flywheel Arms

• Following stresses are induced
• in the arms of a flywheel
• 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.

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Stresses in Flywheel Arms

• 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,
• st1 (3/4)st (3/4)?xv2

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Stresses in Flywheel Arms

• Bending stress due to torque transmitted
• T 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

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Stresses in Arms
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• Design of Flywheel Arms

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D/n of Arms
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• Design of Shaft, Hub and Key

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Design of Shaft, Hub and Key
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Shaft, Hub and Key
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Design of Shaft, Hub and Key
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Example-6
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Example-6

• 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 the
  flywheel is5.2 MPa, the total fluctuation of
  speed is to be limited to 3 of the mean speed.
  The work done during the power stroke is 1/3
  more than the average work done during the whole
  cycle.
• The max. torque on shaft is twice the mean
  torque. The density of cast iron is 7220 kg/m3.

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Example-6
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Example-6