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
 Induction motors are the prime movers in most of the
Industries (Work Horse of Motion Industries)
 simple design, rugged, low-price, easy maintenance
 wide range of power ratings: fractional horsepower to
10 MW
 run essentially as constant speed from no-load to full
load
 Its speed depends on the frequency of the power
source
 applications such as centrifugal pumps, conveyers,
compressors crushers, and drilling machines etc
2

3. Construction Details
 AC induction motor comprises two electromagnetic
parts
 Stationary part called the stator
 Rotating part called the rotor
 Differs from a dc machine in the following aspects.
 Laminated stator
 Absence of commutator & Uniform and small air gap
 Practically almost constant speed
 The stator and the rotor are each made up of
 An electric circuit - usually made of insulated
copper or aluminum winding, to carry current
 A magnetic circuit - usually made from laminated
3

4. Stator - Construction
 The stator is the outer stationary part
of the motor
 outer cylindrical frame - yoke which is
made either of welded sheet steel,
cast iron or cast aluminum alloy
 The magnetic path - comprises a set
of slotted steel laminations called
stator core pressed into the cylindrical
space inside the outer frame.
 It is laminated to reduce eddy
currents, reducing losses and
heating
Smooth Yoke
Ribbed Yoke
4

5. Stator - Construction
 A set of insulated electrical windings, which are placed inside the slots
of the laminated Stator.
 For a 3-phase motor, 3 sets of windings are required, one for each phase
connected in either star or delta..
Stator Laminations
cross sectional view of an induction motor
5

6. Rotor - Construction
 Rotor is the rotating part of
the induction motor
 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
 set of copper or aluminum bars installed
into the slots, which are connected to an
end-ring at each end of the rotor
 windings resembles a ‘squirrel cage
bars ring
ring
bar
s
ring
ring
 Even though the
aluminum rotor bars are
in direct contact with the
steel laminations,
practically all the rotor
current flows through the
aluminum bars and not in
the lamination
7

8. Wound Rotor
 consists of three sets of insulated windings
with connections brought out to three slip
rings mounted on one end of the shaft
 The external connections to the rotor are
made through brushes onto the slip rings
Brushe
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
9

10. 10

11. Main purpose of designing
 is to obtain the complete physical dimensions of all the
parts
 to satisfy the customer Needs
 Physical Dimensions
 The main dimensions of the stator.
 Details of stator windings.
 Design details of rotor and its windings
 Performance characteristics (iron and copper losses, no
load current, power factor, temperature rise and
efficiency)
 Customer Needs
 out put power, voltage, number of phases, speed,
frequency, connection of stator winding, type of rotor
11

12. Output Equation
 mathematical expression which gives the relation between
the various physical and electrical parameters of the
electrical machine
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
cos Ø = power factor
D = Diameter of the stator,
L = Gross core length
Co = Output coefficient
13

14. Output Equation
For a 3 Ø machine,
kVA rating Q = 3 Vph Iph 10-3 kW
Assuming, Vph = Eph
Eph = 4.44 f Ø Tph kW
f = PNS/120 = P ns/2
Output = 3 x 4.44 x Pns/2 x Ø Tph Kw Iph x 10-3 kW
Output = 6.66 x PØ x Iph Tph x ns x Kw x 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 = Bav π DL
Total Electric Loading ac = Iz Z/ π. D
Output = 1.11 x PØ x Iz Z x ns x Kw x 10-3 kW
Output = 1.11 x Bav π. DL x ac x π. D x ns x Kw x 10-3 kW
Output = 11 Bav ac Kw 10-3 x D2L ns kW
Output (Q) = Co D2L ns kW
Where, Co = 11 Bav ac Kw 10-3
kVA input Q = H.P x 0.746 (kW) / eff cos Ø
15

16. Choice of Specific loadings
 Specific Magnetic loading or Air gap flux density (Bav)
 Iron losses largely depend upon air gap flux density
 Limitations :
 Magnetising current 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 – Tph less – Leakage reactance less - Overload
capacity increases
 Size of the machine reduced
 Cost of the machine decreases
 For 50 Hz machine, The suitable values of Bav is 0.35 – 0.6 Tesla.
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
 power factor and efficiency under full load conditions will
increase with increase in rating of the machine
 Percentage magnetizing current and losses will be
lower for a larger machine than that of a smaller
machine
 the power factor and efficiency will be higher for a high
speed machine than the same rated low speed machine
because of better cooling conditions
 Squirrel cage – Efficiency – 0.72 to 0.91 & P.F – 0.66 to
0.9
 Slip ring - Efficiency – 0.84 to 0.91 & P.F – 0.7 to 0.92
18

19. Separation of D and L
 The output equation gives the relation between D2L product
and output of the machine
 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
 D and L have to satisfy the condition imposed on the value
of peripheral speed
 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
 The Design consideration of Stator Involves in estimation of
 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
 For Small Motors up to 5 HP Single layer Winding like
 Mush Winding
 Whole coil Concentric Winding
 Bifurcated concentric winding is used.
 Generally Double layer Winding ( Lap or Wave ) with
diamond shaped coils is used.
 The three phases of winding can be connected in either star
or Delta depending on the Starting Methods Employed.
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
 Stator Phase Voltage Es = 4.44 f Ø Ts Kws
Stator Turns per phase Ts = Es / 4.44 f Ø Kws
Where, Kws = 0.955 Winding factor
Specific Magnetic Loading, Bav = Flux per pole / Area under a
pole
= p Ø / pi . D L
Ø = Bav x pi . D L / p
26

27. Length of Mean Turn of winding
 For Stators that use up to 650 V
Length of Mean turn , Lmts = 2L + 2.3 τ + 0.24
Where, L – Length of Stator Core
τ - Pole Pitch
Resistance of the stator winding per phase is calculated
using the formula = (0.021 x lmt x Tph ) / as where lmt is in
meter and as is in mm2
27

28. Stator Conductors
 kVA rating Q = 3 Es Is 10-3 kW
Stator Current / Phase , Is = Q / 3 Es x 10-3
Area of Cross Section , as = Is / gs
Where, gs – Current Density – 3 to 5 A/mm2
Area of Cross Section , as = pi ds
2 / 4
Where, ds – Diameter of Stator Conductor
Round conductors are generally used
For diameter more than 2 or 3 mm – Bar or Strip
conductors are used
28

29. Stator Slots
 In general two types of stator
slots - open slots and
semiclosed slots
Open Slots :
 slot opening will be equal to
that of the width of the slots.
 assembly and repair of winding
are easy.
 slots will lead to higher air gap
contraction factor and hence
poor power factor
open slots
29

30. Stator Slots
Semi enclosed Slots :
 slot opening is much smaller
than the width of the slot.
 assembly of windings is more
difficult and takes more time
compared to open slots
 costlier
 Air gap characteristics are
better compared to open type
slots
30

31. Choice of Stator Slots
 number of slots/pole/phase may be selected as three or
more for integral slot winding
 fractional slot windings number of slots/pole/phase may be
selected as 3.5
 Slot Pitch for open type of Slots should be 15 to 25 mm.
 Slot Pitch for Semi enclosed type of Slots should be < 15
mm.
Stator slot pitch, Yss = Gap Surface / Total No of Slots
= π . D / Ss
So, Ss = π . D / Yss
Stator Slots Ss = Number of phases x poles x slots/pole/phase
31

32. Conductors per Slot
 Total No of Stator Conductors Zs = Phase x
Conductors/Phase
= 3 x 2 Ts = 6 Ts
 Conductors per Slot, Zss = Total No of Zs / Total No of Ss
= 6 Ts / Ss
Where, Ts – Stator Turns per Phase
Ss – Total Stator Slots
Zss – Must be Even for double layer
winding
32

33. Area of Stator Slot
 Area of each slot = Copper Area per slot / Space Factor
= Zss x as / Space factor
Where, Zss – No of Conductors per slot
as – Area of each Conductor
Space Factor – 0.25 to 0.4
33

34. Stator Teeth
 The Dimensions of slot determine the flux density in the
teeth.
 Higher Flux Density – iron loss – Greater Magnetising mmf.
 Mean Flux density in tooth < 1.7 Wb/m2
Minimum teeth Area per pole = Øm / 1.7
Teeth area per pole = Ss / p x Li x Wts (Width of stator
Tooth)
So, Øm / 1.7 = (Ss / p) x Li x Wts min
Wts min = Øm / 1.7 (Ss / p ) x Li
34

35. Depth of Stator Core
 flux density in Stator Core < 1.5
Wb/m2.
 So,Flux in core = half of flux /
pole
= Øm / 2
Area = Flux / Flux Density
= (Øm / 2 ) / Bcs
Also, Area = Li x depth (dcs)
(Øm / 2 ) / Bcs = Li x dcs
Depth of the core, dcs = Øm / 2
Bcs Li
Outer Diameter,Do = D + 2(dss +
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 of larger air gap length
 Increased Magnetising current
 Reduced power factor
For Small Induction Motor – lg = 0.2 + 2 Sqrt(DL) mm
Lg = 0.125 +0.35D+ L + 0.015 Va mm
For General Use - lg =0.2 + D mm
For journal bearings - lg = 1.6 sqrt (D) – 0.25 mm
40

41. 41

42. 42

43. 43

44. Design of Rotor
 squirrel cage type are rugged and simple in construction
and comparatively cheaper & has lower starting torque.
 In this type, the rotor consists of bars of copper or
aluminum accommodated in rotor slots.
 Slip ring induction motors are complex in construction and
costlier with the advantage that they have the better
starting torque.
 This type of rotor consists of star connected distributed
three phase windings.
44

45. Design of Rotor
 Cogging and Crawling are the two phenomena which are
observed due to wrong combination of number of rotor and
stator slots.
 In addition, induction motor may develop unpredictable
hooks and cusps in torque speed characteristics or the
motor may run with lot of noise
45

46. Crawling
 The rotating magnetic field
produced in the air gap of the will
be usually non sinusoidal and
generally contains odd harmonics
of the order 3rd, 5th and 7th
 The third harmonic flux will
produce the three times the
magnetic poles compared to that of
the fundamental. Similarly the 5th
and 7th harmonics
 The motor with presence of 7th
harmonics is to have a tendency to
run the motor at one seventh of its
normal speed
46

47. Cogging
 When the number of rotor slots
are not proper in relation to
number of stator slots the
machine refuses to run and
remains stationary.
 Under such conditions there will
be a locking tendency between
the rotor and stator – Cogging
 rotor slots will be skewed by
one slot pitch to minimize the
tendency of cogging, torque
defects like synchronous hooks
and cusps and noisy operation
while running.
47

48. Squirrel Vz Wound
 No Slip rings
 Higher Efficiency
 Star – Delta starter sufficient
 Cheaper
 Small copper loss
 Better P.F, greater Overload
Capacity
48
 Possible to insert resistance
in rotor – Increases starting
Torque
 Low starting current
 Rotor Resistance starter

49. Design of Squirrel Cage Rotor
 The Design involves
 Diameter of the Rotor
 Choice and Design of Rotor bars & slots
 Design of End Rings
Diameter of Rotor
 Should be Slightly less than that of Stator to avoid Mechanical Friction
Diameter of Rotor, Dr = D – 2 lg
Where, D – Diameter of Stator Bore
lg – length of air gap
49
bar
s
ring
ring

50. Choice of Rotor Slots
 To avoid cogging and crawling: (a)Ss Sr (b) Ss - Sr ±3P
 To avoid synchronous hooks and cusps in torque speed characteristics
±P, ±2P, ±5P.
 To noisy operation Ss - Sr ±1, ±2, (±P ±1), (±P ±2)
Design of Rotor Bars
 Rotor bar current, Ib= (6 Is Ts Kws cos Ø) / Sr = 0.85 x (6 Is Ts) / Sr
(approx)
Where , Is – Stator Current per Phase
Ts – Stator Turns per phase
Sr – Number of rotor slots
Kws – Winding factor of stator
50

51. Design of Rotor Bars
 Area of each rotor bar, ab = Ib / gb in mm2
Where, Ib – Rotor bar current
gb – Current density of rotor bar , Normally 4 – 7 A/mm2
Copper loss in rotor bars
Length of rotor bar Lb = L + allowance for skewing
Rotor bar resistance, rb = 0.021 x Lb / ab
Copper loss in rotor bars = Ib
2 x rb x number of rotor bars
51

52. Design of End Rings
 All the rotor bars are short circuited by
connecting them to the end rings
 The rotating magnetic filed produced will
induce an emf in the rotor bars which will be
sinusoidal over one pole pitch.
 As the rotor is a short circuited body, there
will be current flow because of this emf
induced.
 In one pole pitch, half of the number of bars
and the end ring carry the current in one
direction and the other half in the opposite
direction.
 Thus the maximum end ring current may be
taken as the sum of the average current in
half of the number of bars under one pole.
52

53. 53

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

55. Design of End Rings
Area of end rings
Area of each end ring ae = Ie / ge mm2,
Where ge = Current density in end ring – 4 to 7 A/mm2
Area of each end ring ae = depth x Thickness of end
ring
Area of each end ring ae = de x te
Copper loss in End Rings
Mean diameter of the end ring (Dme) - 4 to 6 cms less of the
rotor
Mean length of the current path in end ring lme = π Dme
resistance of the end ring re = 0.021 x lme / ae
Total copper loss in end rings = 2 x Ie2 x re
55

56. Design of Wound Rotor
 Rotor carries distributed star connected 3 phase winding
 Three ends of the winding are connected to the slip rings
 External resistances can be connected to these slip rings
at starting, which will be inserted in series with the windings
which will help in increasing the torque at starting
 The Design involves in
 Rotor winding
 Number of Rotor slots
 Number of rotor turns
 Rotor Current
 Area of rotor conductor
 Dimensions of rotor teeth
 Rotor core & Slip rings , brushes
56

57. Design of Wound Rotor
 Rotor winding : Small Motors – Mush Type
: Large Motors – Double layer Bar Type
: Motors > 750 kW – Barrel winding
 No of Rotor turns
Turns ratio Er/Es = Kwr Tr / Kws Ts
Rotor turns/ phase, Tr = Kws Ts Er / Kwr Es
rotor ampere turn = 0.85 x stator ampere turn
Ir Tr = 0.85 x Is Ts
Rotor current Ir = 0.85 Is Ts / Tr
57

58. Design of Wound Rotor
 Area of rotor conductor, ar = Ir / gr
Where gr – current density – 3 to 5 A/mm2
 Choice of rotor slots
 Rotor slots should not be equal to stator slots
 Generally for wound rotor motors a suitable value is
assumed for number of rotor slots per pole per phase,
and then
 total number of of rotor slots are calculated.
 Semi closed slots are used for rotor slots.
58

59. Design of Wound Rotor
 Rotor teeth
flux density in rotor tooth < 1.7 Wb/m2
Minimum teeth area / pole = Flux per pole / Max flux density
= Øm / 1.7
Tooth area / pole = No of rotor slots/pole x Net iron
length x width of the tooth
= (Sr / p) x Li x Wtr
Equating both
(Sr / p) x Li x Wtr = Øm / 1.7
Wtr(min) = Øm / (1.7 x (Sr / p) x Li )
Wtr(min) actual = root Rotor slot pitch – rotor slot width
= π (Dr – 2 dsr) / Sr - Wsr
59

60. Design of Wound Rotor
 Rotor Core
The flux density in the rotor core = Stator core density
Depth of rotor core, dcr = Øm / (2 x Bcr x Li )
Where, Bcr – Flux density in rotor core
Inner Diameter of rotor lamination, Di = Dr – 2(dsr + dcr)
Where, dcr – depth of rotor core
dsr – depth of rotor slot
Slip ring & brushes:
Area of Slip ring = rotor current / Current density ( 4 to 7 A/mm2)
Dimension of Brushes - Current density ( 0.1 to 0.2 A/mm2)
60

61. Performance Evaluation
 The parameters for performance evaluation are iron losses,
no load current, no load power factor, leakage reactance
etc
 Iron losses: Iron loss has two components, hysteresis and
eddy current losses occurring in the iron parts depend
upon the frequency of the applied voltage
 The frequency of the induced voltage in rotor is equal to
the slip frequency which is very low and hence the iron
losses occurring in the rotor is negligibly small.
 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
 Resistance and Leakage Reactance - Needs to be evaluated
 Find the Stator and Rotor Resistance
 Find total resistance of motor as viewed from stator
 Finally Find Rotor current
If rotor leakage reactance & Loss component of No load current –
Neglected
Stator current equivalent to rotor current = Is cos Ø
Where Is – Stator current
cos Ø - Power Factor
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 I0 at an angle from vertical line assuming some scale for current.
 Draw Isc at 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 Diagram
PowerInputRotor
lossCuRotor
Slip 
 Slip
PS
LS

Starting torque
Line BE represents the starting torque of the motor in
synchronous watts to power scale
72

73. Problems
 Solution:
 kVA input, Q = output / eff x P.F
 Co = 11 Bav ac Kw 10-3
 Ns = 2f / p rps
 Sub the value of Co & ns in o/p equation
Q = Co D2L ns Find D2L
Estimate the stator core dimensions, number of stator slots, no of
stator Conductors per slot for a 100kW, 3300V, 50Hz, 12 pole, Star
connected slip ring induction motor Bav = 0.4 Wb/m2, ac = 25000
amp.cond/m, Eff = 0.9, P.F = 0.9, Choose the main dimensions to give
best power factor. The slot loading should not exceed 500 amp.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 π DL / 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 Ss = 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 Il = I ph = Is
 Is = kVA x 103 / 3 x VL & Zss = 6 Ts / Ss
 Finalise the no of slots Ss
 Now find the new Total stator conductors = Ss .Zss
& Turns / Phase, Ts = Zss . Ss / 6
75

76. Problems
76
 Estimate the main dimensions, air gap length, number of stator slots,
stator turns / phase & cross sectional area of stator and rotor conductors
for a 3 phase, 15HP, 400V, 50Hz, 6, 975rpm induction motor, The motor
is suitable for star delta starting. Bav = 0.45 Wb/m2, ac = 20000
amp.cond/m, Eff = 0.9, P.F = 0.85, L/τ = 0.85.
 Solution:
 kVA input, Q = HP X 0.746 / eff x P.F
 Co = 11 Bav ac Kw 10-3
 Ns = 2f / p rps
 Sub the value of Co & ns in o/p equation
Q = Co D2L ns Find D2L

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 π DL / 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 Ss = 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
 Finalise the 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 gs = 3 A/mm2 find Area Stator conductor as =
Is / gs
 Where Is = Iph = Q x 103 / 3 Eph
 Length of Airgap, lg = 0.2 + 2 Sqrt(DL) mm
 Choose the No of Rotor slots Such that Ss – Sr are
not equal
 Find Rotor bar current, Ib = 0.85 x (6 Is Ts) / Sr
 Assume gr = 4 A/mm2 find Area of rotor ar = Ir / gr
 Find End Ring Current, Ie = Sr Ib / πp
 Assume ge = 4 A/mm2 find Area of End rings ae = Ie / ge

80. Problems
80
 Design a cage rotor for a 40 HP, 3 Phase , 400 V, 50 Hz, 6 pole, delta
connected induction motor having a full load efficiency 87% and a full
load P.F of 0.85. Take D =33 cm and L = 17 cm. Stator slots = 54,
conductors per slot = 14. Assume suitably the missing data if any.
 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 / eff x P.F

81. Problems
81
 Choose the No of Rotor slots Such that Ss – Sr are 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, Ib = 0.85 x (6 Is Ts) / Sr
 Find Is and Ts
 Zss = 6 Ts / Ss, Ts = Zss x Ss /6
 find Is from input KVA = 3 Eph x Iph x 10-3
 Since delta connected, Eph =V = 400, Iph = Q / 3 Eph x
10-3

82. Problems
82
 Find End Ring Current, Ie = Sr Ib / πp
 Assume gr = 4 A/mm2 find Area of rotor Bar ar = Ir / gr
 Assume ge = 4 A/mm2 find Area of End rings ae = Ie / ge
 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