DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF.

/EEE

UNIT IV

INDUCTION MOTORS

Output equation of Induction motor – Main dimensions – Length of air gap- Rules for selecting rotor
slots of squirrel cage machines – Design of rotor bars & slots – Design of end rings – Design of
wound rotor -– Magnetic leakage calculations – Leakage reactance of polyphase machines-
Magnetizing current - Short circuit current – Circle diagram - Operating characteristics.

- 1-
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

OUTPUT EQUATION

------------------------------------------------- (1)

Equ. (1) is known as the output equation of an a.c. machine. Quantity C0 is called the output co-efficient.

MAIN DIMENSIONS

- 2-
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

CHOICE OF SPECIFIC LOADINGS

1. Choice of specific magneticloadings

- 3-
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

2. Choice of specific electricloadings

DESIGN OF STATOR

1. Statorturns/phase

ES B  DL
Turns/phase TS = and  m = av
4.44mfK WS P

2. Area of statorconductors

IS
Area of each stator conductor a =
S
S
Input kVA
Stator current per phase I S =
3ES
Current density in stator S = 3 to 5 A/mm 2

3. Shape of Stator slots

Slots may be completely open or semi closed. Semi closed slots are preferred for induction motors
because with their use the gap contraction factor is small giving a small value of magnetizing current.
The use of semi enclosed slots results in low tooth pulsation loss and a much quieter operation as
compared with that with open slots.

4. Stator slots

Number of stator slots SS = 3 pqS

Where p = Number of poles
qS = Number of stator slots/pole/phase  2

- 4-
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

=
D
Number of stator slots S S
ySS

Where ySS = stator slot pitch

= 10 to 15 mm for single layer winding
= 15 to 25 mm for double layer winding

5. Statorconductors

Number of stator conductors

ZS = 2 TS for single phase
ZS = 6TS for three phase

Number of stator conductor per slot

ZS
Z SS =
SS
= Integer for single layer winding
= Even integer for double layer winding

6. Slotloading

Slot loading = IZ ZSS and IZ = IS

NOTE: 1. The stator is provided with radial ventilating ducts if the core length exceeds 100 to 125 mm.

2. The width of each duct is about 8 to 10 mm.

kW
3. Input kVA Q =
 cos

4. Input kVA Q = hp  0.746

 cos

5. Kws = 0.955

- 5-
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 01

- 6-
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 02

- 7-
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

- 8-
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 03

- 9-
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

- 10 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 04

GIVEN DATA
P=3.7 kW V=400 V 3phase p=4 f=50 Hz Squirrel cage IM Bav=0.45 Wb/m2
Ac=23000 η=0.84 design machine for minimum cost Kw=0.955 Ki=0.9
Started by star-delta starter η = 0.85

- 11 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 05

- 12 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

GIVEN DATA

P=15 kW 3 phase V=400 V f=50 Hz N=2180 rpm η=0.88 cosΦ=0.9

Bav=0.5 Wb/m2 ac=25000 Vcr=20 m/s Squirrel cage IM

EXAMPLE: 06

GIVEN DATA

H.P=250 3 phase V=400 V N=1410 rpm Slip ring IM

Bav=0.5 Wb/m2 ac=30000 A/m η=0.9 cosΦ=0.9 Kw=0.955
δ=3.5 A/mm2 Sf=0.4 L/τ=1.2 f=50 Hz
Delta connected machine

- 13 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 07

GIVEN DATA
(i) P=15 kW V=440 V p=4 f=50 Hz 3 phase D=0.25 m L=0.16m
ac=23000 a/m
(ii) P=11 kW V=460 V p=6 f=50 Hz η=0.84 cosΦ=0.82 Kw=0.955

- 14 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

- 15 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

LENGTH OF AIR GAP

RELATIONS FOR CALCULATION OF LENGTH OF AIR GAP

- 16 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

CHOICE OF ROTOR SLOTS FOR SQUIRREL CAGE MACHINES

NOTE:

1. CRAWING

2. COGGING

- 17 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

RULES FOR SELECTING ROTOR SLOTS OF SQUIRREL CAGE MACHINES

DESIGN OF ROTOR BARS AND SLOTS

1. Design of rotor bars

- 18 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

Length of bar Lb = L + 0.045 m

Lb
Total Copper loss in bar = S Ir b 2
ab

In case of squirrel cage motor the cross-section of bars will take the shape of the slot and
insulation is not used between bars and rotor core.

2. Design ofslots

The semi closed slots provides better overload capacity.

DESIGN OF END RINGS

SI
End ring current I e= r b
p
Area of cross section of end ring a =e Ie
e

Also

Area of cross section of end ring = Depth of end ring x Thickness of end ring
ae = d e t e

- 19 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

Fig. shows the dimensions of end ring.

 Dr
Total Copper loss in end ring 2I e
ae

Note:

s Rotor cu loss
1. =
1-s output

120 f
2. N s=
p

3. N r = (1− s)N s

REDUCTION OF HARMONICS TORQUE

- 20 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 01

- 21 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

- 22 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 02

- 23 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 03

GIVEN DATA

P=11 kW 3 phase p=6 f=50 Hz V=220 V stat connected Ss=54

Conductors/slot=9 Sr=64 η=0.86 cosΦ=0.85 δ=5 A/mm2
Rotor mmf=0.85 Stator mmf

- 24 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 03

GIVEN DATA

3 phase p=2 f=50 Hz Dr=0.20 m L=0.12m Bm=0.55 Wb/m2 Sr=33

R/bar=125μΩ L/bar=2 μH S=0.06
Solution

- 25 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 04

GIVEN DATA

P=15 kW 3 phase p=6 f=50 Hz D=0.32 m L=0.125 m Ss=54

Conductors/stator slot=24 Is=17.5 A cosΦ=0.85 N=950 rpm ρ=0.02Ω/m and mm2
Solution

- 26 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

- 27 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

- 28 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

DESIGN OF SLIP RING ROTOR

1. Rotor Windings

2. Number of Rotor Turns

3. Rotor Current and Area of Rotor Conductor

- 29 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

4. Number of Rotor Slots

NOTE:

1. DISPERSION COEFFICIENT

2. FULL LOAD SLIP

Rotor slots
3. Coil Span =
Number of poles

- 30 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 01

EXAMPLE: 02

- 31 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

- 32 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

- 33 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

LEAKAGE REACTANCE CALCULATION FOR POLYPHASE MACHINES (INDUCTION

MACHINES)

- 34 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 01

- 35 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 02

Solution

- 36 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

OPERATING CHARACTERISTICS

1. MAGNETIZINGCURRENT

AT60 = mmf for air gap+mmf for stator teeth+mmf for rotor teeth+ mf for stator core+mmf for rotor core

- 37 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

Note:

m 
sin 
 2 
1. Distribution factor K = d
 
msin  
2 
Where

slots
m=
poles  phase
o

 = 180
n
slots
n=
pole

2. Pitch factor K =P cos 

2
3. Stator winding factor KWS = Kd K P
4. Area per pole A =  DL
p

=
m
5. Average flux density B av
A

- 38 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 01

Given data

P = 75 kW EL = 3300V f = 50 Hz p =8 Im = 0.35 of full load current AT60 = 500 A

Kws = 0.95 η = 0.94 cosΦ = 0.86

EXAMPLE: 02

Given data

P = 15 kW EL = 400 V f = 50 Hz p = 6 D = 0.3 m L = 0.12 m SS =72 ZSS = 20 lg = 0.55mm

Kg = 1.2 mmf required for iron path = 0.35 x air gap mmf coil span = 11 slots

Solution

- 39 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

SHORT CIRCUIT CURRENT

- 40 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

- 41 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

FORMULAE USED FOR CALCULATION

2 Total rotor cu loss

rr ' =
mI'
I'r = I cos
s r

3 s

4 2msTs Kws
I = I'
b r
Sr
5 Ie =
Sr Ib
p

EXAMPLE: 01

- 42 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 02

- 43 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

EXAMPLE: 03

- 44 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

CIRCLE DIAGRAM

- 45 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

- 46 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

- 47 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

TWO MARKS QUESTION AND ANSWERS

01

02

03

04

05

06

07

08

- 48 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

09

10

11

12

13

14

- 49 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

15

16

17

18

19

20

21

- 50 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

22

23 State the effect of change of air-gap length in a three phase induction motor

If the air-gap of an induction motor is changed then the mmf and magnetizing current
also changes. i.e. Increase in air-gap length increases the overload capacity, offers better
cooling, reduces noise and reduces unbalanced magnetic pull.

24 Define unbalanced magnetic pull.

The unbalanced magnetic pull is the radial force acting on the rotor due to non
uniform air-gap around armature periphery.

25 How does the external resistance of slip-ring induction motor influence the
motor performance.

External resistance connected to slip-ring

1. Increases the starting torque

2. Decreases the starting current and
3. Used to control the speed of rotor.

26 State the main constructional differences between cage induction motor and slip-ring
induction motor.

Squirrel Cage Induction Motor Slip-ring Induction Motor

Slip rings and brushes are absent. Slip-rings and brushes are present to add
external resistances.
Rotor consists of bars which are shorted at Rotors consists of a three phase winding
the ends with the help of end rings. similar to the stator winding.
The rotor automatically adjusts itself for Rotor must be wound for the same
the same number of poles as that of stator. number of poles as that of stator.

27 What are the different losses in an induction motor.

a) Rotational or constant losses b) I2R losses or variable losses

Rotational losses are made up of: i) Friction and windage losses ii) Iron losses
I2R losses are made up of : i) Stator cu loss ii) Rotor cu loss

28 List the main parts of a slip-ring Induction motor.

1. Stator 2. Rotor 3. Slip rings 4. Metal collar

5. Brushes 6. Bearings 7. Fan

- 51 -
DESIGN OF ELECTRICAL MACHINES UNIT - IV D. RAJASEKARAN ASSOC. PROF./EEE

29 Write an empirical formula for finding the length of the air gap of an induction motor.
lg = 0.2 + 2 DL
Where D = Diameter of bore
L = Length of stator D & L are expressed in metre.

30 How do you select L/τ ratio for design of induction motor.

L/τ ratio is selected based on design feature.

31 Why the length of air gap in an induction motor is kept minimum possible range.

The mmf and the magnetizing current are primarily decided by length of air-gap.
If air-gap is small then mmf and magnetizing current will be low, which in turn increase
the value of power factor. Hence by keeping small air-gap, higher power factor is
achieved.

32 What factors govern the choice of air gap in induction motor.

The factors govern the choice of air gap in induction motor.

1. Power factor 2. Unbalanced magnetic pull 3. Overhead capacity
4. Pulsation loss 5. Noice 6. Cooling

33 Define dispersion coefficient.

- 52 -