2.Inertia Forces-1ppt.pdf .

2.Inertia Forces-1ppt.pdf

1. Inertia Forces
DYNAMICS OF MACHINERY
Chapter 2
1

2. Introduction
2
•Dynamicforcesareassociatedwithacceleratingmasses.Asallmachines
haveacceleratingparts,dynamicforcesarealwayspresentwhenthe
machinesoperate.
•Insituationswheredynamicforcesaredominantorcomparabletothe
magnitudesofexternalforces,andoperatingspeedsarehigh,dynamic
analysishastobecarriedout.
•Forexample,incaseofrotorswhichrotateatspeedsmorethan80,000rpm,
eventheslightesteccentricityofthecenterofmassfromtheaxisofrotation
producesveryhighdynamicforces.
•Thismayleadtovibrations,wear,noiseorevenmachinefailure.

3. D’ Alembert’s Principle
3

4. D’ Alembert’s Principle
4

5. Dynamic Analysis of Slider-Crank Mechanism
-Velocity and Acceleration of Piston
5
r

7. 7
Velocity of Piston

8. 8
Acceleration of Piston
•Note that this expression of acceleration has
been obtained by differentiating the
approximate expression for velocity.
•It is usually very cumbersome to differentiate
the exact expression for velocity. However,
this gives satisfactory results.

9. Thedrivingforceactingonthepistonistermedaspistoneffort.InaverticalcylinderIC
engine,followingthreetypesofforcesact:
GasForce
Theforceduetovariationofworkingfluidpressureisknownasgasforce,orGasforce
F
g=
πD
2
4
×p ……..(1)
where
D = diameter of the piston and
p = gas pressure
Engine Force Analysis
Piston Effort

10. InertiaForce
•InanICengine,duringthefirsthalfofthestroke,thereciprocatingmassaccelerates
andtheinertiaforcetendstoresistthemotion.Thus,thenetforceonthepistonis
decreased.
•However,duringthesecondhalfofthestroke,thereciprocatingmassdecelerate,
andinertiaforceopposesthisdeceleration.Thus,itincreasestheeffectiveforceon
thepiston.
•The inertia force of the piston is given as
�
??????=��=��??????
2
cos�+
cos2�
�
…(2)

11. WeightoftheReciprocatingMass
•Theweightofreciprocatingmassassiststhepistonduringitsmovementtowards
bottomdeadcentre(BDC).Therefore,pistoneffortisincreasedbyanamountequal
totheweightofthepiston.
•However,whenthepistonmovestowardstopdeadcentre(TDC),thepistoneffortis
decreasedbythesameamount.
Net piston effort:
??????=�
�+�
??????±�…(3)

12. 12
�
??????cos??????=??????���
??????=
??????
cos??????
�
�=�
??????sin??????=??????tan??????
Where, ��??????�??????=��??????��⇒??????=�??????�
−1
�??????�??????
�

13. 13
(iv)CrankPinEffort
Forceisexertedonthecrankpinasaresultoftheforceonthepiston.Thecomponentofforceactingalong
connectingrodperpendiculartothecrankisknownascrank-pineffort.
�
??????�∗�=�
??????��??????��+??????
�
??????�=�
??????�??????��+??????
�
??????�=
??????
���??????
�??????��+??????
(v)ThrustontheBearings
Thecomponentof�
??????�alongthecrankintheradialdirectionproducesathrustonthecrankshaftbearings.
�
??????�=�
??????����+??????
�
??????�=
??????
���??????
����+??????
�
??????�
D
E
90−�+φ
�
�

14. Crankeffortistheneteffortappliedat
thecrankpinperpendiculartothecrank
(�
??????�)whichgivestherequiredturning
momentonthecrankshaft.

15. 15

16. Graphical Method to Determine Crank Effort or Torque
T = P x distance OY
WhereOYisthedistancemeasuredbetweencentreofcrankandapointof
intersectionofYaxisandextensionofconnectingrodP
2B

17. Note:
•ThecrankeffortisafunctionofpistoneffortPandcrank
rotationangle.Further,thepistoneffortisalsoafunctionof
crankangle.
•Thediagramshowingthecrankeffortortorqueasafunctionof
crankrotationangleforanyreciprocatingengineiscalled
crank-effortdiagramorturningmomentdiagram.

18. P-V diagram of petrol engine
Theturningmomentdiagramofanyenginecanbeplottedifthegas
pressurepisknownforallpositionsofthecrank.Thevalueofgaspressure
canbefoundfromagivenpressure-volume(P-V)diagram

19. Variation of Gas force and Inertia force
Usingthesepressurevalues,gasforcescanbecomputedandplottedas
shownhere:

20. Variation of Piston Effort
Further,thevariationofinertiaforceduetomassofreciprocatingpartscan
beplottedasshownhere:

21. T= piston effort x OY
WhereOYisthecrankeffortarmlength.Thevariationincrankeffortarmlength
fordifferentcrankpositionisshownbelow.
Variation of Crank Effort Arm Length

22. Turning Moment Diagram for a Single Cylinder Double Acting Steam Engine
Double Acting Steam Engine:
•When steam is coming in both side
of the piston and it produces double
working stroke in each revolution,
is called double acting steam
engine.
•It produces double power than
single acting steam engine.

23. Turning Moment Diagram for a Single Cylinder Double Acting Steam Engine
Double Acting Steam Engine:
•Since the work done is the product
of the turning moment and the angle
turned, therefore the area of the
turning moment diagram represents
the work done per revolution.

24. Turning Moment Diagram for a Single Cylinder
Four Stroke Engine

25. Turning Moment Diagram for a Single Cylinder
Four Stroke Engine
•Turningmomentdiagram
showsthattorqueTisentirely
positiveinexpansionstrokeof
enginewhereasinsuction,
compressionandexhaust
strokes,itisnegative.
•Thisindicatesthatinthese
strokes,powerisconsumed.
Thusthereislargevariationof
torquewhichmaycause
fluctuationofspeed.

26. Turning Moment Diagram for a Multi Cylinder
Four Stroke Engine
Inmulti-cylinderengine,theturning
momentdiagramofeachcylinderis
obtainedseparatelyandtheyare
superimposedovereachotherwithstarting
pointshiftedtophasedifferenceofangle
betweenrespectivecrankpositions.

27. Flywheel
•Aflywheelisaninertialenergystoragedevice.Itabsorbsmechanical
energyandservesasareservoir,storingenergyduringtheperiod
whenthesupplyofenergyismorethantherequirementandreleasesit
duringtheperiodwhentherequirementofenergyismorethanthe
supply.
•Internalcombustionengineswithoneortwocylindersareatypical
example.Pistoncompressors,punchpresses,rockcrushersetc.arethe
othersystemsthathaveflywheel.

28. Flywheel
Dependinguponthesourceofpowerandthetypeofdrivenmachinery,
therearethreedistinctsituationswhereaflywheelisnecessitated.
1.Whentheavailabilityofenergyisatafluctuatingratebutthe
requirementofitfordrivenmachineryisatuniformrate.For
example,ICEnginedrivenwaterpumps,generators,compressors,
fans,etc.Insuchasituationflywheelisneededtostoresurplus
energy.

29. Flywheel
2.Inapplications,namely,electricmotordriven
punching,shearingandrivetingmachines,
rollingmills,etc.,thoughtheenergyisavailable
atauniformrate,thedemandforitisvariable.
Figureonrightshowsatypicalenergy
requirementandavailabilitycurveforan
electricmotordrivenrollingmillwhichshows
thatforasmallfractionofthecycleperiod,
thereishugerequirementofenergy.Thusagain
aflywheelisneeded.
3.Inthistypeofsituation,boththerequirement
andavailabilityofenergyrepresentavariable
rate,e.g.ICenginedrivenreciprocatingair
compressororpump.
Turning moment diagram
of electric motor driven
rolling mill

30. Types of Flywheel
Threetypesofflywheel–disctype,webtypeandarmtypearemost
commonlyused.

31. Types of Flywheel

32. Fluctuation of energy (??????
�):
•Aflywheelisusedtocontrolthevariationsin
speedduringeachcycleofanengine.
•Aflywheelofsuitabledimensionsattachedto
thecrankshaft,makesthemomentofinertiaof
rotatingpartsquitelargeandtherebyitactsasa
reservoirofenergy.
•Duringtheperiodswhenthesupplyofenergy
ismorethanrequired,itstoresenergyand
duringtheperiodwhenthesupplyislessthan
required,itreleasestheenergy.
•Thedifferencebetweenmaximumand
minimumkineticenergiesofflywheelis
knownasmaximumfluctuationofenergy,??????
�.

33. Coefficientof Fluctuation of energy (??????
�):
Ratioofmaximumfluctuationofenergyandworkdonepercycleisknownas
coefficientoffluctuationofenergy.
�
�= Max. fluctuation of energy / work done per cycle.
Therefore, �
�=
??????
??????
??????
Work done per cycle = �
��??????�×
where �
��??????�= Mean torque and = angle turned by the crank in radians.
Also, work done per cycle is = ??????×60,000/??????
where P = Power in KW and N is in rpm.

34. Coefficient of Fluctuation of speed (??????
??????):
•Thedifferencebetweenthegreatestandtheleastangularspeedsofthe
flywheeliscalledthemaximumfluctuationofspeed.
•Theratioofgreaterfluctuationofspeedpercycletothemeanspeedis
calledcoefficientoffluctuationofspeed.

37. Example 1
Amulti-cylinderenginerunsataspeedof1500rpm.Theturningmomentdiagram
repeatsitselfforeveryrevolutionofthecrankshaft.Thescaleoftheturningmoment
is1cm=6,000N-mandthecrankangleisplottedtoascaleof1cm=60°.Theareas
belowandabovethemeanturningmomentline,takeninorderareasfollows:-0.3,
+4.1,-2.8,+3.2,-3.3,+2.5,-2.6,+2.8and-3.6cm
2
.Findoutthefluctuationof
energy.Alsofindoutthecoefficientoffluctuationofspeediftheweightofthe
rotatingpartsis5000N,andtheradiusofgyrationis0.2m.

38. Example 1
EnergyatA=�
EnergyatB=�−0.3
EnergyatC=�−0.3+4.1=�+3.8
EnergyatD=�+3.8−2.8=�+1
EnergyatE=�+1+3.2=�+4.2
EnergyatF=�+4.2−3.3=�+0.9
EnergyatG=�+0.9+2.5=�+3.4
EnergyatH=�+3.4−2.6=�+0.8
EnergyatI=�+0.8+2.8=�+3.6
EnergyatJ=�+3.6−3.6=�
Thus,
Greatestenergy(atE)=�+4.2
Leastenergy(atB)=�−0.3
Therefore,
�
�=�+4.2−�−0.3
⇒�
�=4.5��
2
⇒�
�=4.56000×60×
�
180
∴

39. Example 1
Further,
�
�=????????????
��??????�
2
�
�=��
2
??????
��??????�
2
�
�
28274=
5000
9.81
×0.2
2
2�×1500
60
2
�
�
Therefore,thecoefficientoffluctuationofspeed
Or

40. Example 2
Theturningmomentdiagramsfora4strokecyclegasenginemaybeassumedtobe
representedbyfourrectangles,theareasofwhicharemeasuredfromthelineofzero
pressureareasfollows.Expansionstroke=8.5cm
2
,Exhaust=0.8cm
2
,Suction=
0.7cm
2
,Compression=2.2cm
2
.Assumingtheresistingtorquetobeuniform,find
theweightoftherimofthewheelrequiredtokeepthespeedbetween116&124
rpm.Assumethatmeanradiusoftherim=1mandeachcm
2
ofareaofthediagram
represents150N-mofenergy.

41. Example 2
NetworkdoneduringthecycleinanICengineisthedifferenceofenergygenerated
duringpowerstrokeandtheenergyconsumedintheremainingstrokes.
Thatis,networkdoneduringthecycle=8.5−0.8−0.7−2.2��
2
Therefore,networkdoneduringthecycle=4.8��
2
=4.8×150??????�
=720??????�

42. Example 2
Therefore,maximumfluctuationofenergy,
�
�=8.5×150−
1
4
×720

43. Example 2
�
�=
1
2
????????????
�??????�
2
−
1
2
????????????
�??????�
2
1095=
1
2
??????
2�×124
60
2
−
2�×116
60
2

44. Example 2
Also,
??????=
�
�
�
2
104.01=
�
9.81
1
2

45. Example 3
Asinglecylinder,singleacting4strokecyclegasenginehasapistondiameterof33
cmandastrokeof60cm.Meanspeed=200rpm.Meanpressuresinthecylinder
aboveatmosphereareasfollows:
Suctionstroke=0.7N/cm2belowatmosphere,
Compressionsstroke=20N/cm
2
,
Expansionstroke=70N/cm
2
,
Exhauststroke=1.4N/cm
2
.
Assumingconstantresistanceandminimumandmaximumspeedtooccuratthe
beginningandendofexpansionstroke,determinethemomentofinertiaofflywheel
iftotalfluctuationofspeedisnottoexceed1%ofmeanspeed.Findalsothedropin
speedwhichwillthenoccurduringacycleinwhichthereisnoadmission.

46. Example 3
Part-I:Tofindthemomentofinertiaofflywheelfor1%fluctuationinmeanspeed.
Workdoneduringpowerstroke
=??????�������
���������??????�����������
=70×
�33
2
4
×0.6
Workexpendedduringsuction,compression&exhauststrokes
=??????
���+??????
??????��+??????
���
=0.7+20+1.4
�33
2
4
×0.6
Networkdonepercycle=35,922−11,341

47. Example 3
Fluctuationofenergy=workdoneduringpowerstroke-1/4
th
ofnetworkdoneper
cycle.
�
�=35,922−
1
4
×24,581
�
�=????????????
��??????�
2
�
�
Or
29,776=??????
2�×200
60
2
0.01
Therefore,

48. Example 3
Part-II:Tofindthedropinspeedwhenthereisnoadmissionduringacycle
Whenthereisnoadmissionduringacycle,theenginehastoperformfourmoreidle
strokesandtheenergystoredinflywheelisfurtherexpended.Thereforenetworkdone
inonecyclewillhavetobedistributedover8strokes.
Therefore,
�
�=35,922−
1
8
×24,581
Also,
�
�=????????????
��??????�
2
�
�
⇒32,849.375=6,798×
2�×200
60
2
�
�

49. Example 3
And,
�
�=
??????
�??????�−??????
�??????�
??????
��??????�
=
??????
�??????�−??????
�??????�
??????
��??????�
Since??????
��??????�=200���
0.011=
??????
�??????�−??????
�??????�
200

50. Flywheel of Punching Presses

51. Flywheel of Punching Presses
•Fromthepreviousdiscussions,itcanbeobserved
thatwhentheloadonthecrankshaftisconstantor
variesandtheinputtorquevariescontinuouslyduringa
cycle,aflywheelisusedtoreducethefluctuationsof
speed.
•Aflywheelcanperformthesamepurposeinapunching
pressorarivetingmachineinwhichthetorqueavailable
isconstantbuttheloadvariesduringthecycle.
•Sincepunchingoperationsoccursonlyduringthe
downwardstrokeandonlywhenthepunchtravels
throughthethicknessoftheplate,intheabsenceofa
flywheelthedecreaseinthespeedofthecrankshaftwill
beverylargeduringactualpunchingcomparedtono-
loadperiods.

52. Flywheel of Punching Presses
Moment

53. Formula’s Used:
Flywheel of Punching Presses
??????=���??????
�

54. Example 4
Amachineisrequiredtopunch4holesof4cmdiameterinaplateof3cmthickness,
perminute.Theworkrequiredis700N-m/cm
2
ofshearedarea.Thepunchhasa
strokeofl0cm.Maximumspeedofflywheelatitsradiusofgyrationis30m/s.Find
theweightofflywheelifspeedshouldnotfallbelow28m/sattheradiusofgyration.
Alsofindthepowerofthemotor.
Solution:
Workrequired/punching�
�=����,where�istheworkrequired/cm
2
ofshearedarea.
�
�=�×4×3×700
Therefore,
??????�??????��=
�
�×��.��ℎ����/�??????�
60×1000
=
26,390×4
60×1000

55. Example 4
�
�=�
�1−
�
2�
=26,3901−
3
2×10
Also,
�
�=
1
2
????????????
�??????�
2
−
1
2
????????????
�??????�
2
=
1
2
????????????
�??????�
2
−??????
�??????�
2
�
�=
�
2�
�
2
??????
�??????�
2
−??????
�??????�
2
�
�=
�
2�
??????
�??????�
2
�
2
−??????
�??????�
2
�
2
⇒22,431=
�
2×9.81
30
2
−28
2
Therefore,weightoftheflywheel,

56. Example 5
Findweightofflywheelneededbyamachinetopunch22mmholesin18mmsteel
plates.Themachineistomake30revperminuteandistobecapableofpunchinga
holeeveryrevolution.Theholeistobeformedduring30°ofrotationofcrankshaft
ofthepunch.Thecrankshaftistobeconnectedtotheflywheelshaftbyageartrain
of12:1ratiosothatthemaxrpmofflywheelwillbe12timesthatofthemachine.
Assumemechanical�tobe85%.Minimumspeedofflywheelistobe90%ofthe
maximum.Meandiameterofflywheelrimis1m.Ultimateshearstrengthofthe
plateis35,000N/cm
2
.
Solution:
Workrequired/punching�
�=Τ
1
2??????�=Τ
1
2���??????
��
�
�=ൗ
1
2
�2.2×1.8×35,000×1.8
Since�
��??????ℎis0.85,workrequired/punching�
�=
3918.82
0.85

57. Example 5
??????�??????��=
�
�×��.��ℎ����/�??????�
60×1000
=
4610.3×30
60×1000
⇒
�
�=�
�1−
�
2−�
1
2�
=4610.31−
30
360
Maximumspeedofflywheel=30×12=360rpm.
Therefore,
=??????�=
2�×360
60
×0.5=
=0.9×18.84=

58. Example 5
Now,usingtheequationoffluctuationofenergy,
�
�=
1
2
????????????
�??????�
2
−
1
2
????????????
�??????�
2
=
�
2�
??????
�??????�
2
�
2
−??????
�??????�
2
�
2
⇒4226=
�
2×9.81
18.24
2
−16.9
2
Theweightoftheflywheelisfoundas,

59. Example 6
AmachinePressisworkedbyanelectricmotordelivering2.25kWcontinuously.At
thecommencementofanoperation,aflywheelofmomentofinertia600Nm
2
onthe
machineisrotatingat250rpm.Thepressingoperationrequires5,300N-mofenergy
andoccupies0.75seconds.Findthemaximumnumberofpressingsthatcanbemade
in1hourandthereductioninspeedoftheflywheelaftereachpressing.Neglect
frictionlosses.
Solution:
Workrequired/punching�
�=Τ
1
2??????�=Τ
1
2���??????
��
�
�=ൗ
1
2
�2.2×1.8×35,000×1.8
Since�
��??????ℎis0.85,workrequired/punching�
�=
3918.82
0.85

60. Example 7
Inamachine,theintermittentoperationsdemandthetorquetobeappliedasfollows:
•Duringthefirsthalfrevolution,thetorqueincreasesuniformlyfrom800Nm
to3000Nm.
•Duringnextonerevolution,thetorqueremainsconstant.
•Duringnextonerevolution,thetorquedecreasesuniformlyfrom3000Nmto
800Nm.
•Duringlast1½revolutions,thetorqueremainsconstant.
Thusacycleiscompletedin4revolutions.Themotortowhichthemachineis
coupledexertsaconstanttorqueatameanspeedof300rpm.Aflywheelofmass
1500kgandradiusofgyrationof450mmisfittedtotheshaft.Determine:
i.Thepowerofthemotor
ii.Thetotalfluctuationofspeedofthemachineshaft.

61. Example 8
Asinglecylinderfourstrokepetrolenginedevelops20kWat600rpm.The
mechanicalefficiencyoftheengineis80%andtheworkdonbythegasesduring
expansionstrokeisthreetimestheworkconsumedincompressionstroke.Ifthetotal
fluctuationofspeedisnottoexceed±1.5percentofthemeanspeedandturning
momentdiagramduringexpansionstrokeisassumedtobetriangular.Determinethe
massoftheflywheelifthediameterofitis1m.Alsodeterminethecrosssectionof
flywheelrim.Thewidthtothicknessratiois1.5andmassdensityofflywheel
materialis7100kg/m
3
.

62. Example 8
Solution:
Permissiblespeedvariationof±1.5%impliesthatthecoefficientoffluctuationofspeed,
�
�=0.03
Angularvelocity,
??????=
2�×600
60
??????=62.83����
Indicatedpower,
????????????=
�??????
�
��??????ℎ
=
20
0.8
????????????=25��

63. Example 8
IP=Indicatedworkdone/cycle×Numberofexplosions/sec
25×10
3
=��������/�??????���×
600
2×60
Therefore,workdonepercycle
��=5000??????�
Sincetheworkduringsuctionandexhauststrokesisneglected,thenetworkdoneduring
thecycleisgivenby(Given:�
���=3�
??????���)
��=�
���−�
??????���
��=�
���−�
���/3
��=2/3�
���
Therefore,
�
���=3/2��=3/2×1500=7,500??????�

64. Example 8
�
���=7,500??????�=��������??????��������=
1
2
����
⇒��=�
�??????�=
2�
���
��
=
2×7500
�
⇒��=4,774.6??????�
Meantorque
�
��??????�=
��������/�??????���
4�
=
5000
4�
⇒�
��??????�=397.88??????�

65. Example 8
Excesstorque=��=��−��=4774.6−397.88=4376.72??????�
InsimilartrianglesABCandADF,
��
��
=
��
��
⇒��=��
��
��
=�×
4376.72
4774.6
⇒��=0.9167�
Fluctuationofenergy�
�=areaofABC
⇒�
�=
1
2
×0.9167�×4376.72=6302.2??????�

66. Example 8
Also,fluctuationofenergy
�
�=6302.2=��
2
??????
2
�
�=�×0.5
2
×62.83
2
×0.03
Thereforemassoftheflywheel
Assumingthat90percentofmassiscontributedbyrimandremaining10percentby
hub,arms,etc.,
0.9�=��×�×�×�⇒�×�=
0.9�
���
=
0.9×212.86
�×0.5×7100
⇒�×�=0.0086�
2
Since�=1.5�,1.5�
2
=0.0086.Therefore,

67. Thank you