UNIT IV design of Electrical Apparatus

Output equation of Induction motor; Main dimensions; Separation of D and L; Choice of Average flux density; length of air gap; Design of Stator core; Rules for selecting rotor slots of squirrel cage machines; Design of rotor bars and slots; Design of end rings; Design of wound rotor; Magnetic leakage calculations; Leakage reactance of polyphase machines; Magnetizing current; Short circuit current; Operating characteristics; Losses and Efficiency.

1. DESIGN OF INDUCTION MOTORS

2. Introduction
Inductionmotorsaretheprimemoversinmostofthe
Industries(WorkHorseofMotionIndustries)
simpledesign,rugged,low-price,easymaintenance
widerangeofpowerratings:fractionalhorsepowerto
10MW
runessentiallyasconstantspeedfromno-loadtofull
load
Itsspeeddependsonthefrequencyofthepower
source
applicationssuchascentrifugalpumps,conveyers,
compressorscrushers,anddrillingmachinesetc
2

3. Construction Details
ACinductionmotorcomprisestwoelectromagnetic
parts
Stationarypartcalledthestator
Rotatingpartcalledtherotor
Differsfromadcmachineinthefollowingaspects.
Laminatedstator
Absenceofcommutator&Uniformandsmallairgap
Practicallyalmostconstantspeed
Thestatorandtherotorareeachmadeupof
Anelectriccircuit-usuallymadeofinsulated
copperoraluminumwinding,tocarrycurrent
Amagneticcircuit-usuallymadefromlaminated
siliconsteel,tocarrymagneticflux
3

4. Stator -Construction
Thestatoristheouterstationarypart
ofthemotor
outercylindricalframe-yokewhichis
madeeitherofweldedsheetsteel,
castironorcastaluminumalloy
Themagneticpath-comprisesaset
ofslottedsteellaminationscalled
statorcorepressedintothecylindrical
spaceinsidetheouterframe.
Itislaminatedtoreduceeddy
currents,reducinglossesand
heating
Smooth Yoke
Ribbed Yoke
4

5. Stator -Construction
Asetofinsulatedelectricalwindings,whichareplacedinsidetheslots
ofthelaminatedStator.
Fora3-phasemotor,3setsofwindingsarerequired,oneforeachphase
connectedineitherstarordelta..
Stator Laminations
cross sectional view of an induction motor
5

6. Rotor -Construction
Rotoristherotatingpartof
theinductionmotor
of a set of slotted silicon
steel laminations pressed
together to form of a
cylindrical magnetic circuit
and the electrical circuit
The electrical circuit of the
rotor is of
Squirrel cage rotor
Wound rotor (Slip Ring
Rotor)
6

7. Squirrel Cage Rotor
setofcopperoraluminumbarsinstalled
intotheslots,whichareconnectedtoan
end-ringateachendoftherotor
windingsresemblesa‘squirrelcage
barsring
ring
bar
s
ring
ring
Even though the
aluminumrotorbarsare
indirectcontactwiththe
steel laminations,
practicallyalltherotor
currentflowsthroughthe
aluminumbarsandnotin
thelamination
7

8. Wound Rotor
consistsofthreesetsofinsulatedwindings
withconnectionsbroughtouttothreeslip
ringsmountedononeendoftheshaft
The external connections to the rotor are
made through brushes onto the slip rings
Brushe
s
Slip rings
8

9. Some more parts
Two end-flanges to support
the two bearings, one at the
driving-end and the other at
the non driving-end.
Two sets of bearings to
support the rotating shaft
Steel shaft for transmitting
the mechanical power to the
load
Cooling fan located at the
non driving end
Terminal box on top of the
yoke or on side to receive the
external electrical connections
9

10. 10

11. Main purpose of designing
istoobtainthecompletephysicaldimensionsofallthe
parts
tosatisfythecustomerNeeds
PhysicalDimensions
Themaindimensionsofthestator.
Detailsofstatorwindings.
Designdetailsofrotoranditswindings
Performancecharacteristics(ironandcopperlosses,no
loadcurrent,powerfactor,temperatureriseand
efficiency)
CustomerNeeds
outputpower,voltage,numberofphases,speed,
frequency,connectionofstatorwinding,typeofrotor
winding,workingconditions,shaftextensiondetailsetc
11

12. Output Equation
mathematicalexpressionwhichgivestherelationbetween
thevariousphysicalandelectricalparametersofthe
electricalmachine
12

13. Output Equation
Vph= phase voltage
Iph= phase current
Iz= Current in each
Conductor
Z = Total no of conductors
Tph= no of turns/phase
Ns = Synchronous speed in
rpm
ns = synchronous speed in
rps
p = no of poles,
ac = Specific electric
loading
Ø= air gap flux/pole
Bav= Average flux density
kw= winding factor
eff= efficiency
cosØ = power factor
D = Diameter of the stator,
L = Gross core length
Co = Output coefficient
13

14. Output Equation
For a 3 Ø machine,
kVAratingQ = 3 VphIph10
-3
kW
Assuming, Vph= Eph
Eph = 4.44 f Ø TphkW
f = PN
S/120 = P ns/2
Output = 3 x 4.44 x Pns/2 x Ø TphKwIphx 10
-3
kW
Output = 6.66 x PØ x IphTphx ns x Kwx 10
-3
kW
Iz= Iph/ a
Let Iz= Iph(1 Parallel Path)
Z = 3 x 2 Tph( Tph= Z/6)
14

15. Output Equation
Total Magnetic Loading PØ = BavπDL
Total Electric Loading ac = IzZ/π. D
Output = 1.11 x PØ x IzZ x ns x Kwx 10
-3
kW
Output = 1.11 x Bavπ. DLx ac x π. D x ns x Kwx 10
-3
kW
Output = 11 Bavac Kw 10
-3
x D
2
Lns kW
Output(Q) = Co D
2
L ns kW
Where, Co = 11 Bavac Kw 10
-3
kVAinput Q = H.P x 0.746 (kW)/ effcosØ
15

16. Choice of Specific loadings
SpecificMagneticloadingorAirgapfluxdensity(Bav)
Ironlosseslargelydependuponairgapfluxdensity
Limitations :
Magnetisingcurrent high –Poor power factor
Flux density in teeth < 1.8 Tesla
Flux density in core 1.3 –1.5 Tesla
Advantages of Higher value of Bav
Large Flux/Pole –Tphless –Leakage reactance less -Overload
capacity increases
Size of the machine reduced
Cost of the machine decreases
For50Hzmachine,ThesuitablevaluesofBavis0.35–0.6Tesla.
16

17. Choice of Specific loadings
Specific Electric Loading (ac)
Advantages of Higher value
Reduced size
Reduced cost
Disadvantages of Higher value
Higher amount of copper
More copper losses
Increased temperature rise
Lower overload capacity
Normal range 10000 ac/m –450000 ac/m.
For Machines high voltage rating –ac value Small
17

18. Choice of power factor and
efficiency
powerfactorandefficiencyunderfullloadconditionswill
increasewithincreaseinratingofthemachine
Percentagemagnetizingcurrentandlosseswillbe
lowerforalargermachinethanthatofasmaller
machine
thepowerfactorandefficiencywillbehigherforahigh
speedmachinethanthesameratedlowspeedmachine
becauseofbettercoolingconditions
Squirrelcage–Efficiency–0.72to0.91&P.F–0.66to
0.9
Slipring-Efficiency–0.84to0.91&P.F–0.7to0.92
18

19. Separation of D and L
TheoutputequationgivestherelationbetweenD
2
Lproduct
andoutputofthemachine
The separation of D and L for this product depends on a
suitable ratio between gross length and pole pitch ( L / τ)
to obtain the best power factor the following relation will be
usually assumed for separation of D and L.
Pole pitch/ Core length = 0.18/pole pitch





















DesignEconomicalOverall
higherfor
PFGoodfor
DesignOverallGood
L
:0.25.1
:5.1
:25.11
:1


D = 0.135 P Sqrt
(L)
19

20. Peripheral Speed
DandLhavetosatisfytheconditionimposedonthevalue
ofperipheralspeed
For the normal design of induction motors the calculated
diameter of the motor should be such that the peripheral
speed must be below 30 m/s.
In case of specially designed rotor the peripheral speed can
be 60 m/s.
Ventilating Ducts : Provided when core length exceeds
100 –125 mm. The width of Duct –8 to 10 mm
20

21. Design of Stator
TheDesignconsiderationofStatorInvolvesinestimationof
Stator Winding
Stator Turns per Phase
Length of Mean Turn
Stator Conductors
Shape & No of Stator Slots
Area of Stator Slot
Stator Teeth
Depth of Stator Core
21

22. Stator Winding
ForSmallMotorsupto5HPSinglelayerWindinglike
MushWinding
WholecoilConcentricWinding
Bifurcatedconcentricwindingisused.
GenerallyDoublelayerWinding(LaporWave)with
diamondshapedcoilsisused.
Thethreephasesofwindingcanbeconnectedineitherstar
orDeltadependingontheStartingMethodsEmployed.
Squirrel cage –Star Delta Starter –Stators designed -Delta
Slip ring –Rotor resistance –Either star or Delta
22

23. 23

24. 24

25. 25

26. Stator Turns per phase
StatorPhaseVoltageEs=4.44fØTsKws
StatorTurnsperphaseTs=Es/4.44fØKws
Where,Kws=0.955Windingfactor
SpecificMagneticLoading,Bav=Fluxperpole/Areaundera
pole
=pØ/pi.DL
Ø=Bavxpi.DL/p
26

27. Length of Mean Turn of winding
ForStatorsthatuseupto650V
LengthofMeanturn,Lmts=2L+2.3τ+0.24
Where,L–LengthofStatorCore
τ-PolePitch
Resistance of the stator winding per phase is calculated
using the formula = (0.021 x lmt x Tph ) / a
swhere lmt is in
meter and a
sis in mm2
27

28. Stator Conductors
kVAratingQ =3EsIs10
-3
kW
StatorCurrent/Phase,Is=Q/3Esx10
-3
AreaofCrossSection,a
s=Is/g
s
Where,g
s–CurrentDensity–3to5A/mm
2
AreaofCrossSection,a
s=pid
s
2
/4
Where,d
s–DiameterofStatorConductor
Round conductors are generally used
For diameter more than 2 or 3 mm –Bar or Strip
conductors are used
28

29. Stator Slots
Ingeneraltwotypesofstator
slots-openslotsand
semiclosedslots
OpenSlots:
slotopeningwillbeequalto
thatofthewidthoftheslots.
assemblyandrepairofwinding
areeasy.
slotswillleadtohigherairgap
contractionfactorandhence
poorpowerfactor
open slots
29

30. Stator Slots
SemienclosedSlots:
slotopeningismuchsmaller
thanthewidthoftheslot.
assemblyofwindingsismore
difficultandtakesmoretime
comparedtoopenslots
costlier
Airgapcharacteristicsare
bettercomparedtoopentype
slots
30

31. Choice of Stator Slots
numberofslots/pole/phasemaybeselectedasthreeor
moreforintegralslotwinding
fractionalslotwindingsnumberofslots/pole/phasemaybe
selectedas3.5
SlotPitchforopentypeofSlotsshouldbe15to25mm.
SlotPitchforSemienclosedtypeofSlotsshouldbe<15
mm.
Statorslotpitch,Yss=GapSurface/TotalNoofSlots
=π.D/S
s
So,S
s=π.D/Yss
Stator Slots Ss = Number of phases x poles x slots/pole/phase
31

32. Conductors per Slot
TotalNoofStatorConductorsZs=Phasex
Conductors/Phase
=3x2Ts=6Ts
ConductorsperSlot,Zss=TotalNoofZs/TotalNoofS
s
=6Ts/S
s
Where,Ts–StatorTurnsperPhase
Ss–TotalStatorSlots
Zss–MustbeEvenfordoublelayer
winding
32

33. Area of Stator Slot
Areaofeachslot=CopperAreaperslot/SpaceFactor
=Zssxa
s/Spacefactor
Where,Zss–NoofConductorsperslot
a
s–AreaofeachConductor
SpaceFactor–0.25to0.4
33

34. Stator Teeth
TheDimensionsofslotdeterminethefluxdensityinthe
teeth.
HigherFluxDensity–ironloss–GreaterMagnetisingmmf.
MeanFluxdensityintooth<1.7Wb/m
2
MinimumteethAreaperpole=Øm/1.7
Teethareaperpole=Ss/pxLixWts(Widthofstator
Tooth)
So,Øm/1.7=(Ss/p)xLixWtsmin
Wtsmin = Øm/ 1.7 (Ss / p ) x Li
34

35. Depth of Stator Core
fluxdensityinStatorCore<1.5
Wb/m
2
.
So,Fluxincore=halfofflux/
pole
=Øm/2
Area = Flux / Flux Density
= (Øm/ 2 ) / Bcs
Also,Area=Lixdepth(d
cs)
(Øm/2)/Bcs=Lixd
cs
Depthofthecore,d
cs=Øm/2
BcsLi
OuterDiameter,Do=D+2(dss+
dsc)
Do
dcs
dss
D
35

36. 36

37. 37

38. 38

39. 39

40. Length of Air Gap
Advantages larger air gap length :
Increased overload capacity
Increased cooling
Reduced unbalanced magnetic pull
Reduced in tooth pulsation
Reduced noise
Disadvantages oflarger air gap length
Increased Magnetising current
Reduced power factor
ForSmallInductionMotor–lg=0.2+2Sqrt(DL)mm
Lg=0.125+0.35D+L+0.015Vamm
ForGeneralUse -lg=0.2+Dmm
Forjournalbearings-lg=1.6sqrt(D)–0.25mm
40

41. 41

42. 42

43. 43

44. Design of Rotor
squirrelcagetypeareruggedandsimpleinconstruction
andcomparativelycheaper&haslowerstartingtorque.
Inthistype,therotorconsistsofbarsofcopperor
aluminumaccommodatedinrotorslots.
Slipringinductionmotorsarecomplexinconstructionand
costlierwiththeadvantagethattheyhavethebetter
startingtorque.
Thistypeofrotorconsistsofstarconnecteddistributed
threephasewindings.
44

45. Design of Rotor
CoggingandCrawlingarethetwophenomenawhichare
observedduetowrongcombinationofnumberofrotorand
statorslots.
Inaddition,inductionmotormaydevelopunpredictable
hooksandcuspsintorquespeedcharacteristicsorthe
motormayrunwithlotofnoise
45

46. Crawling
Therotatingmagneticfield
producedintheairgapofthewill
beusuallynonsinusoidaland
generallycontainsoddharmonics
oftheorder3rd,5thand7
th
Thethirdharmonicfluxwill
producethethreetimesthe
magneticpolescomparedtothatof
thefundamental.Similarlythe5th
and7thharmonics
Themotorwithpresenceof7th
harmonicsistohaveatendencyto
runthemotoratoneseventhofits
normalspeed
46

47. Cogging
Whenthenumberofrotorslots
arenotproperinrelationto
numberofstatorslotsthe
machinerefusestorunand
remainsstationary.
Undersuchconditionstherewill
bealockingtendencybetween
therotorandstator–Cogging
rotorslotswillbeskewedby
oneslotpitchtominimizethe
tendencyofcogging,torque
defectslikesynchronoushooks
andcuspsandnoisyoperation
whilerunning.
47

48. Squirrel Vz Wound
NoSliprings
HigherEfficiency
Star–Deltastartersufficient
Cheaper
Smallcopperloss
BetterP.F,greaterOverload
Capacity
48
Possibletoinsertresistance
inrotor–Increasesstarting
Torque
Lowstartingcurrent
RotorResistancestarter

49. Design of Squirrel Cage Rotor
TheDesigninvolves
DiameteroftheRotor
ChoiceandDesignofRotorbars&slots
DesignofEndRings
DiameterofRotor
ShouldbeSlightlylessthanthatofStatortoavoidMechanicalFriction
DiameterofRotor,Dr=D–2lg
Where, D–DiameterofStatorBore
lg–lengthofairgap
49
bar
s
ring
ring

50. Choice of Rotor Slots
Toavoidcoggingandcrawling:(a)SsSr(b)Ss-Sr±3P
Toavoidsynchronoushooksandcuspsintorquespeedcharacteristics
±P,±2P,±5P.
TonoisyoperationSs-Sr±1,±2,(±P±1),(±P±2)
DesignofRotorBars
Rotorbarcurrent,I
b=(6IsTsKwscosØ)/Sr=0.85x(6IsTs)/Sr
(approx)
Where, Is–StatorCurrentperPhase
Ts–StatorTurnsperphase
Sr–Numberofrotorslots
Kws–Windingfactorofstator
50

51. Design of Rotor Bars
Areaofeachrotorbar,a
b=I
b/g
binmm
2
Where,I
b–Rotorbarcurrent
g
b–Currentdensityofrotorbar,Normally4–7A/mm
2
Copperlossinrotorbars
Length of rotor bar L
b= L + allowance for skewing
Rotor bar resistance, r
b= 0.021 x L
b/ a
b
Copper loss in rotor bars = I
b
2
x r
bx number of rotor bars
51

52. Design of End Rings
Alltherotorbarsareshortcircuitedby
connectingthemtotheendrings
Therotatingmagneticfiledproducedwill
induceanemfintherotorbarswhichwillbe
sinusoidaloveronepolepitch.
Astherotorisashortcircuitedbody,there
willbecurrentflowbecauseofthisemf
induced.
Inonepolepitch,halfofthenumberofbars
andtheendringcarrythecurrentinone
directionandtheotherhalfintheopposite
direction.
Thusthemaximumendringcurrentmaybe
takenasthesumoftheaveragecurrentin
halfofthenumberofbarsunderonepole.
52

53. 53

54. Design of End Rings
Maximum value of End ring current,
I
e(max) = ½ x ( Number rotor bars / pole) x I
b(av)
= ½ x (Sr/p) x I
b(av)
Where I
b(av) = (2/π) x I
b(max)
I
b(max) = √2 I
b
Since Bar current is Sinusoidal
I
e(max) = √2 SrI
b/ πp
Rmsvalue of ring current = I
e= I
e(max) / √2
54

55. Design of End Rings
Area of end rings
Area of each end ring a
e= I
e/ g
emm
2
,
Where g
e= Current density in end ring –4 to 7 A/mm
2
Area of each end ring a
e= depth x Thickness of end
ring
Area of each end ring a
e= d
ex t
e
Copper loss in End Rings
Mean diameter of the end ring (D
me) -4 to 6 cmsless of the
rotor
Mean length of the current path in end ring l
me= πD
me
resistance of the end ring r
e= 0.021 x l
me/ a
e
Total copper loss in end rings = 2 x Ie
2
x r
e
55

56. Design of Wound Rotor
Rotorcarriesdistributedstarconnected3phasewinding
Threeendsofthewindingareconnectedtothesliprings
Externalresistancescanbeconnectedtothesesliprings
atstarting,whichwillbeinsertedinserieswiththewindings
whichwillhelpinincreasingthetorqueatstarting
TheDesigninvolvesin
Rotorwinding
NumberofRotorslots
Numberofrotorturns
RotorCurrent
Areaofrotorconductor
Dimensionsofrotorteeth
Rotorcore&Sliprings,brushes
56

57. Design of Wound Rotor
Rotorwinding:SmallMotors–MushType
:LargeMotors–DoublelayerBarType
:Motors>750kW–Barrelwinding
NoofRotorturns
TurnsratioEr/Es=KwrTr/KwsTs
Rotorturns/phase,Tr=KwsTsEr/KwrEs
rotorampereturn=0.85xstatorampereturn
IrTr=0.85xIsTs
RotorcurrentIr=0.85IsTs/Tr
57

58. Design of Wound Rotor
Areaofrotorconductor,a
r=I
r/g
r
Whereg
r–currentdensity–3to5A/mm
2
Choiceofrotorslots
Rotorslotsshouldnotbeequaltostatorslots
Generallyforwoundrotormotorsasuitablevalueis
assumedfornumberofrotorslotsperpoleperphase,
andthen
totalnumberofofrotorslotsarecalculated.
Semiclosedslotsareusedforrotorslots.
58

59. Design of Wound Rotor
Rotorteeth
fluxdensityinrotortooth<1.7Wb/m2
Minimumteetharea/pole=Fluxperpole/Maxfluxdensity
=Øm/1.7
Tootharea/pole =Noofrotorslots/polexNetiron
lengthxwidthofthetooth
=(Sr/p)xLixWtr
Equatingboth
(Sr/p)xLixWtr =Øm/1.7
Wtr(min) =Øm/(1.7x(Sr/p)xLi)
Wtr(min)actual=rootRotorslotpitch–rotorslotwidth
=π(Dr–2dsr)/Sr-Wsr
59

60. Design of Wound Rotor
RotorCore
Thefluxdensityintherotorcore=Statorcoredensity
Depthofrotorcore,dcr=Øm/(2xBcrxLi)
Where,Bcr–Fluxdensityinrotorcore
InnerDiameterofrotorlamination,Di=Dr–2(dsr+dcr)
Where,dcr–depthofrotorcore
dsr–depthofrotorslot
Slipring&brushes:
AreaofSlipring=rotorcurrent/Currentdensity(4to7A/mm
2
)
DimensionofBrushes-Currentdensity(0.1to0.2A/mm
2
)
60

61. Performance Evaluation
Theparametersforperformanceevaluationareironlosses,
noloadcurrent,noloadpowerfactor,leakagereactance
etc
Ironlosses:Ironlosshastwocomponents,hysteresisand
eddycurrentlossesoccurringintheironpartsdepend
uponthefrequencyoftheappliedvoltage
Thefrequencyoftheinducedvoltageinrotorisequalto
theslipfrequencywhichisverylowandhencetheiron
lossesoccurringintherotorisnegligiblysmall.
Hence the iron losses occurring in the induction motor is
mainly due to the losses in the stator alone
Total iron losses in induction motor = Iron loss in stator core + iron losses in stator
teeth.
61

62. Operating Characteristics

62

63. 63

64. 64

65. 65

66. 66

67. Leakage Calculations –Poly
Phase

67

68. Short Circuit (Blocked Rotor)
Current
ResistanceandLeakageReactance-Needstobeevaluated
FindtheStatorandRotorResistance
Findtotalresistanceofmotorasviewedfromstator
FinallyFindRotorcurrent
Ifrotorleakagereactance&LosscomponentofNoloadcurrent–
Neglected
Statorcurrentequivalenttorotorcurrent=IscosØ
WhereIs–Statorcurrent
cosØ-PowerFactor
68

69. Circle Diagram
69

70. Circle Diagram
We should know following for drawing the circle diagram
No load current and no load power factor
Short circuit current and short circuit power factor
Draw I
0at an angle from vertical line assuming some scale for current.
Draw I
scat an angle from vertical line.
Join AB, which represents the o/p line of the motor to power scale.
Draw a horizontal line AF, and erect a perpendicular bisector on the o/p
line AB so as to meet the line AF at the point O’. Then O’ as center and
AO’ as radius, draw a semi circle ABF.
Draw vertical line BD; divide line BD in the ratio of rotor copper loss to
stator copper loss at the point E.
Join AE, which represent the torque line
70

71. Circle Diagram
Full load current & power factor
Draw a vertical line BC representing the rated o/p of the
motor s per the power scale. From point C, draw a line
parallel to o/p line, so as to cut the circle at pint P. Join OP
which represents the full load current of the motor to
current scale. Operating power factor can also be found
out.
Full load efficiency
Draw a vertical line from P as shown in above figure.
PL = O/p Power
PX = I/p Power
71

72. Circle DiagramPowerInputRotor
lossCuRotor
Slip
SlipPS
LS

Starting torque
Line BE represents the starting torque of the motor in
synchronous watts to power scale
72

73. Problems
Solution:
kVAinput, Q = output / effx P.F
Co = 11 Bavac Kw 10
-3
Ns = 2f / p rps
Sub the value of Co & ns in o/p equation
Q = Co D
2
L ns Find D
2
L
Estimatethestatorcoredimensions,numberofstatorslots,noof
statorConductorsperslotfora100kW,3300V,50Hz,12pole,Star
connectedslipringinductionmotorBav=0.4Wb/m2,ac=25000
amp.cond/m,Eff=0.9,P.F=0.9,Choosethemaindimensionstogive
bestpowerfactor.Theslotloadingshouldnotexceed500amp.cond
73

74. Problems
For best power factor
τ= √(0.18L)we get a equation in terms of D & L
Sub and solve for D & L.
Stator star connected Es = El/√3
Flux per pole Ø = Bav πDL/ p
Find Ts using Flux Ts = Es / 4.44 f Ø Kws
Slot pitch should be 15 to 25 mm
Find Ss when Yss = 15 & 25 mm S
s= π. D /
Yss
74

75. Problems
We get a range of Ss from the above step
Now Assume q = 2, 3, 4..etc and find Ss
Stator Slots Ss = Number of phases x poles x slots/pole/phase
Select Ss that is within the range
Check for Slot loading = Is .Zss
Since Star connected I
l= I ph = Is
Is = kVAx 10
3
/ 3 x V
L& Zss = 6 Ts / Ss
Finalisethe no of slots Ss
Now find the new Total stator conductors = Ss .Zss
& Turns / Phase, Ts = Zss . Ss / 6
75

76. Problems
76
Estimatethemaindimensions,airgaplength,numberofstatorslots,
statorturns/phase&crosssectionalareaofstatorandrotorconductors
fora3phase,15HP,400V,50Hz,6,975rpminductionmotor,Themotor
issuitableforstardeltastarting.Bav=0.45Wb/m2,ac=20000
amp.cond/m,Eff=0.9,P.F=0.85,L/τ=0.85.
Solution:
kVAinput, Q = HP X 0.746 / effx P.F
Co = 11 Bavac Kw 10
-3
Ns = 2f / p rps
Sub the value of Co & ns in o/p equation
Q = Co D
2
L ns Find D
2
L

77. Problems
77
Given L/τ= 0.85, Where τ= πd/p we get a equation
in terms of D & L
Substitute and solve for D & L.
Delta connected Es = El = Eph
Flux per pole Ø = Bav πDL/ p
Find Ts using Flux Ts = Es / 4.44 f Ø Kws
Slot pitch should be 15 to 25 mm
Find Ss when Yss = 15 & 25 mm S
s= π. D /
Yss

78. Problems
78
We get a range of Ss from the above step
Now Assume q = 2, 3, 4..etc and find Ss
Stator Slots Ss = Number of phases x poles x slots/pole/phase
Select Ss that is within the range
Finalisethe no of slots Ss
Find Zss = 6 Ts / Ss
Now find the new Total stator conductors = Ss .Zss
& Turns / Phase, Ts = Zss . Ss / 6

79. Problems
79
Assume g
s= 3 A/mm2 find Area Stator conductor a
s=
I
s/ g
s
Where Is = Iph= Q x 10
3
/ 3 Eph
Length of Airgap, lg = 0.2 + 2 Sqrt(DL) mm
Choose the No of Rotor slots Such that Ss –Srare
not equal
Find Rotor bar current, I
b= 0.85 x (6 Is Ts) / Sr
Assume g
r= 4 A/mm
2
find Area of rotor a
r= I
r/ g
r
Find End Ring Current, I
e = SrI
b / πp
Assume g
e= 4 A/mm
2
find Area of End rings a
e= I
e/ g
e

80. Problems
80
Designacagerotorfora40HP,3Phase,400V,50Hz,6pole,delta
connectedinductionmotorhavingafullloadefficiency87%andafull
loadP.Fof0.85.TakeD=33cmandL=17cm.Statorslots=54,
conductorsperslot=14.Assumesuitablythemissingdataifany.
Given: 3 phase, p=6, Ss=54, Zss= 14, Q=40
HP, Delta connected, V=400 V, Eff. = 0.87,
P.f= 0.85, D=0.33m, L= 17m
Solution:
kVA input, Q = HP X 0.746 / effx P.F

81. Problems
81
Choose the No of Rotor slots Such that Ss –Srare not
equal to 0,±p, ,±2p, ,±3p, ,±5p, ±1, ±2, ±(p ±1), ±(p ±2)
0,±6, ,±12, ,±18,±30, ±1, ±2, ±7,±5, ±8, ±4
Ss–Sr= ±3, ±9
Choose the minimum and Sr= 54-3 = 51
Find Rotor bar current, I
b= 0.85 x (6 Is Ts) / Sr
Find Is and Ts
Zss= 6 Ts/ Ss, Ts= Zssx Ss/6
find Is from input KVA = 3 Ephx Iphx 10
-3
Since delta connected, Eph=V = 400, Iph= Q / 3 Ephx
10
-3

82. Problems
82
Find End Ring Current, I
e= SrI
b/ πp
Assume g
r = 4 A/mm
2
find Area of rotor Bar a
r= I
r/ g
r
Assume g
e= 4 A/mm
2
find Area of End rings a
e= I
e/ g
e
Find length of rotor core = length of stator core
Find Diameter of rotor = Dr= D –2lg
Find Length of Airgap, lg= 0.2 + 2 Sqrt(DL) mm