Lec 4 DOME II Flywheels

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Design of Machine Elements-II

IntroductionA flywheel used in machines serves as a reservoir which stores energy duringthe period when the supply of energy is more than the requirement and releases

it during the period when the requirement of energy is more than supply.

The flywheel absorb excess energy developed in power Stroke and itsspeed increases.

The flywheel release the energy to the crankshaft during the non-power

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.

Examples

Steam Engines

Internal Combustion Engines Reciprocating Compressor Pumps Machine Tools


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Coefficient of Fluctuation of SpeedThe ratio of the maximum fluctuation of speed to the mean speed is called

coefficient of fluctuation of speed.

Cs = N1 N2 / N

= 1 2 /

=

Where

N1 = Maximum speed in !p!m! duing the c"cle#

N2 = Minimum speed in !p!m! duing the c"cle# and

= =

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! ! !


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Fluctuation of Energy& tuning moment diagam fo single c"linde dou'le acting steam engine and a fousto(e intenal com'ustion engine is sho)

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Fluctuation of EnergyConsider diagram for four stroke engine, in this diagram, there are two revolutionsfor four strokes of the engine as explained following.

In the first stroke internal pressure of the cylinder is low than atmospheric

pressure so air fuel mixture enters. This work is done on the engine so in theturning moment diagram, it is represented by negative.

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In the next stroke this mixture compresses which means that large energy isrequired to compress it and is shown by large negative loop.

In the third stroke mixture burns and large amount of energy is produced whichis used to run our engine.

After that exhaust gases leaves cylinder which also demands some energy.

In the whole mechanism there are two positives loops exists one is made insuction stroke and the other one is made in exhaust stroke, both are produced

due to inertia.


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Mai!u! Fluctuation of EnergyTurning Moment diagram for a multi-cylinder engine

Let the energy in the flywheel at A = E,then from Fig.

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Energy at B = E + a1Energy at C = E + a1 a2Energy at D = E + a1 a2 + a3Energy at E = E + a1 a2 + a3 a4Energy at F = E + a1 a2 + a3 a4 + a5

Energy at G = E + a1 a2 + a3 a4 + a5 a6 = Energy at A

Maximum fluctuation of energy= E = Maximum energy Minimum energy


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Coefficient of Fluctuation of EnergyIt is defined as the ratio of the maximum fluctuation of energy to the work done

per cycle. It is usually denoted by CE

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Coefficient of Fluctuation of Energy

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Energy Stored in a Fly"#eel

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.

m = Mass of the flywheel in kg,k = Radius of gyration of the flywheel in meters,I = Mass moment of inertia of the flywheel about the axis of rotation in kg-m2

= m.k2,

N1 and N2 = Maximum and minimum speeds during the cycle in rpm.,


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.hee

he adius of g"ation k ma" 'e ta(en eual to the mean adius of the im R# 'ecause the

thic(ness of im is e" small as compaed to the diamete of im! heefoe su'stituting k = R

in euation# )e hae


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Stresses in Fly"#eel $i!Flywheel consist of :

Rim at which major weight of the flywheel is concentrated

Hub that connect the flywheel to the shaft Arms connect Rim to the Hub

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Stresses in Fly"#eel $i!Tensile/hoop Stress due to the centrifugal force:The tensile stress in the rim due to the centrifugal force, assuming that the rimis unstrained by the arms

b = Width of rim,

t = Thickness of rim,A = Cross-sectional area of rim = b t,D = Mean diameter of flywheel

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where is in kg / m3 and v is in m / s, then twill be in N / m2 or Pa.

R= Mean radius of flywheel,= Density of flywheel material,

= Angular speed of flywheel,

v = Linear velocity of flywheel, andt= Tensile or hoop stress.


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Stresses in Fly"#eel $i!Tensile Bending Stresses caused by restraint arms:he tensile 'ending stess in the im due to the estaint of the ams

b =

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Total Stress in the Rim

= t + b

n = $um'es of ams


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Design of Machine Elements-IIFly"#eel

b =

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Stresses in Fly"#eel %r!Tensile Stresses due to centrifugal force:Due to the centrifugal force acting on the rim, the arms will be subjected todirect tensile stress whose magnitude is experimentally found as of Tensile

stress in rim

= = v2

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Stresses in Fly"#eel %r!Bending Stresses due to the Torque transmitted:Due to the torque transmitted from the rim to the shaft or from the shaft to therim, the arms will be subjected to bending, because they are required to carry

the full Torque load

T = Maximum torque transmitted by the arm,R = Mean radius of the rim,

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Stresses in Fly"#eel %r!Total Stress in the Arm

= t1 + b1&esign of Fly"#eel %r!

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a1 = Major axis, andb1 = Minor axis.

-

minor axis


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&esign of s#aft' (u) and *eyThe diameter of shaft for flywheel is obtained from the maximum torquetransmitted. We know that the maximum torque transmitted

Shaft

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HubThe hub is designed as a hollow shaft, for the maximum torque transmitted. Weknow that the maximum torque transmitted

The diameter of hub is usually taken as twice the diameter of shaft and length from 2 to 2.5

times the shaft diameter. It is generally taken equal to width of the rim.


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&esign of s#aft' (u) and *ey

Key

A standard sunk key is used for the shaft and hub. The length of key is obtainedby considering the failure of key in shearing. We know that torque transmitted byshaft

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%ssign!ent

Example : 22.10 and 22.11Pr l m: n 12

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Lec 4 DOME II Flywheels