Chapter 4.
Three-phase Induction Machines
Introduction
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The induction machine is the most rugged and
the most widely used machine in industry.
Both stator and rotor winding carry alternating
currents.
The alternating current (ac) is supplied to the
stator winding directly and to the rotor winding
by induction – hence the name induction machine.
Application (1-phase): washing machines, ceiling
fans, refrigerators, blenders, juice mixers, stereo
turntables, etc.
2-phase induction motors are used primarily as
servomotors in a control system.
Application 3-phase: pumps, fans, compressors,
paper mills, textile mills, etc.
Induction Machine
Construction
Construction
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Unlike dc machines, induction machines have a uniform air gap.
The stator is composed of laminations of high-grade sheet steel.
A three-phase winding is put in slots cut on the inner surface of the
stator frame.
The rotor also consists of laminated ferromagnetic material, with
slots cut on the outer surface.
Squirrel-cage Rotor
Wound Rotor
Slip ring
Wound rotor
Construction
Cross-sectional view
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Y-connected stator
-connected stator
The three-phase winding are displaced from each other by 120
electrical degrees in space
Current flows in a phase coil produce a sinusoidally distributed mmf
wave centered on the axis of the coil.
Alternating current in each coil produces a pulsating mmf wave.
Mmf waves are displaced by 120 degrees in space from each other.
Resultant mmf wave is rotating along the air gap with constant peak.
Induction Motor Operation
RMF – rotating magnetic field
Rotating Magnetic Field – consider 2-pole
machine
a)
b)
Three phase stator winding, aa’, bb’ and cc’
displaced by 120o.
Mmf (pulsating) in space at various instants due
to a.c current in coil aa’
c)
Instantaneous 3 phase current
a. Graphical Method – Resultant mmf
(magnitude and direction)
Resultant mmf
Mmf phase a
at t = t0= t4
Graphical Method
Constant amplitude, move
around the air gap
n = synchronous speed
rpm
Synchronous speed in rad/s
1 
e
p

2f1
rads-1 ; p  pole pairs
p
f = f1= supply freq.,
p = no of poles
b. Analytical Method
Motion of the
resultant mmf
N = effective number of
turns
ia= current in phase ‘a’
Analytical Method
Induced Voltages
A
where r = radius of the stator;  = axial length of stator
Induced Voltages
V per phase
At Standstill operation
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E = 4.44f N  Kw
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E = 4.44f N  Kw ; f = f
1
1
2
1
2
p
2
2
1
2
2
p
E = 4.44f N  Kw
p
1
2
1
2
Running Operation
pn
f
120
at slip s
* E2 – induced rotor voltage at standstill
Can be reprensented in rads-1
e
s  m
2
; s 
; s
s
p
s
s
m  (1  s )s
where :
 s : synchronous frequency
e : supply frequency
m : rotorfrequency
p : pole pair
2 : slip frequency
Example 1
A three-phase, 100 hp, 460 V, four-pole, 60 Hz induction machine
delivers rated output power at a slip of 0.05. Determine the
(a) Synchronous speed and motor speed.
(b) Speed of the rotating air gap field.
(c) Frequency of the rotor circuit.
(d) Slip frequency (in rpm).
(e) Speed of the rotor field relative to the
(i) rotor structure
(ii) stator structure
(iii) stator rotating field
(f) Rotor induced voltage at the operating speed, if the stator-to-rotor
turns ratio is 1:0.5
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Pg 219: 1800 &1710rpm, 1800rpm, 3 Hz, 90 rpm, (90rpm, 1800rpm, 0rpm), 6.64 V/ph)
Sol_pg21
Equivalent Circuit Model
• To study and predict the performance of the induction machine
Equivalent Circuit Model
Equivalent Circuit Model
Equivalent Circuit Model
Stator voltage equation:
V1 = R1 I1 + j(2f)LlI1 + Eag;
Eag – air gap voltage or back e.m.f
Eag = E1 = k f1 ag
Rotor voltage equation:
E2 = R2 I2 + js(2f)Ll2
E2 = k f2 ag = k sf1 ag = sE1
E2 – induced emf in rotor circuit ; E1=R2/sI2+j2fLI2
Induction Motor Drives SEE4433 Dr Zainal / Dr
Awang
25
Equivalent Circuit Model
sE2 – rotor voltage at
standstill
Equivalent Circuit Model
This model is
not convenient
to use to predict
circuit
performance
Equivalent Circuit Model
Example 2
A three-phase, 15 hp, 460 V, four-pole, 60 Hz, 1728 rpm
induction motor delivers full output power to a load
connected to its shaft. The windage and friction loss of
the motor is 750 W. Determine the
(a) mechanical power
(b) air gap power
(c) rotor copper loss.
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Pg 226: 11940 W, 12437.5 W, 497.5 W
Sol_pg29
Equivalent Circuit Model
Assume
small volt
drop across R1
and X1 – ease
computation
of I and I2’,
V1 = E1
Due to machine air
gap, I is high- 3050% of full –load
current, X1 is high,
core loss (Rc) is
lumped into the
mechanical losses
Equivalent Circuit Model
For simplification, replace V1, R1,X1, Xm with Vth, Rth, Xth ( at
terminal Pag)
R12<<(X1+Xm)2
X1 << Xm
Equivalent Circuit Parameters
Rc, Xm, R1, X1, X2, R2
No-Load Test
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The parameters of the equivalent circuit, Rc, Xm, R1, X1,
X2, and R2 can be determined from the results of a noload test, a blocked-rotor test and from measurement of
the dc resistance of the stator winding.
The no-load test, like the open circuit test on a
transformer, gives information about exciting current and
rotational losses.
This test is performed by applying balanced polyphase
voltages (415V) to the stator windings at the rated
frequency(50Hz).
The rotor is kept uncoupled from any mechanical load.
I1
R1
X1
Xm
At no load, N r  N s
Ns  N r
s
 0;
Ns
R2 R2


   open cct
s
0
Blocked-Rotor Test
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The blocked-rotor test, like the short-circuit test on a
transformer, gives information about leakage impedances.
In this test the rotor is blocked so that the motor cannot rotate,
and balanced polyphase voltages are applied to the stator
terminal ( increases voltage until stator current reaches its
rated value).
The blocked-rotor test should be performed under the same
conditions of rotor current and frequency that will prevail in the
normal operating conditions.
The IEEE recommends a frequency of 25% of the rated
frequency for the blocked-rotor test. However, for normal
motors of less than 20 hp rating, the effects of frequency are
negligible and the blocked-rotor test can be performed directly
at the rated frequency
R1
X1
X2
R2
At full load (block rotor test), N r  0;
Ns  N r
s
 1;
Ns
R2 R2


 R 2  R2   X m , X m open cct
s
1
Equivalent Circuit Parameters
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Measurement of average dc resistance per stator
phase : R1
No load test :
VNL
INL
PNL
Blocked-rotor test:
VBL
INL
PNL
Example 3
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The following test results are obtained from a three-phase, 60 hp, 2200 V,
six-pole, 60 Hz squirrel-cage induction motor.
No-load test:
supply frequency = 60 HZ
line voltage = 2200 V
line current = 4.5 A
input power = 1600 W
Blocked-rotor test:
frequency = 15 Hz
line voltage = 270 V
line current = 25 A
input power = 9000 W
Average DC resistance per stator phase:
R1 = 2.8 ohm
(a) Determine the no-load rotational loss.
(b) Determine the parameters of the IEEE-recommended equivalent circuit.
(c) Determine the parameters (Vth, Rth, Xth) for the thevenin equivalent circuit.
Pg: 230: 1429.9 W
Sol_pg38(IM)
Performance Characteristics
Performance calculation using SPEC
I1
R1
L1
I2
Rm
V1
Input Power :
L2
R2
S
Lm
Pin  3V1I1 cos 
Note : V1 and I1 must be phase voltage and current
Stator copper loss :
Core loss :
Power across the air - gap :
Rotor copper loss :
Pls  3I12 R1
3V12
Plc 
Rm
3I 2 2 R2
Pg 
s
 Pin  Pls  Plc
Plr  3I 2 2 R2
Induction Motor Drives SEE4433 Dr Zainal / Dr
Awang
40
Gross out put power :
Po  Pg  Plr
3I 2 R2
3I 2 R2 (1  s )
2

 3I 2 R 2 
 Pg (1  s )
s
s
2
2
P ower at t heshaft :
Psh  Po  PFW ; PFW : frictionand windage loss.
Developed(elect romagnetic) t orque:
3I 2 (1  s ) R2
Te 

m
m s
Po
2
Since
s
1  m
  m  (1  s )1 ,
1
2
3I 2 R2
 Te 
1 s
But 1 
e
p
; e is t hesupply frequency.
T hen,

Te 
3 pI2
e
2
R2
s
Example
A single phase equivalent circuit of a 6-pole SCIM that
operates from a 220 V line voltage at 60 Hz is given below.
Calculate the stator current, input power factor, output
power, torque and efficiency at a slip of 2.5%. The fixed
winding and friction losses is 350 W. Neglect the core loss.
Also calculate the starting current.
I1
R1
X1
0.2
0.5
V1
Solution
V1  220V line-to line  3
220V
 127V
3
 2.5%  0.025

I2
X2
0.2
Xm
R2
20
0.1
X 1  0.5, X 2  0.2, X m  20
R

Z in  ( R1  jX 1 )  jX m //  2  jX 2 
 s

0.1


 j 0. 2 

  4.220o 
 0.2  j 0.5  j 20 0.025
0
.
1

 j 0.2  j 20 

 0.025

I1 
V1
220 3

 30.0  20o A
o
Z in 4.220
Pin  3V1 I1 cos  3(220 3 )(30)(cos20o )
 10,758W
Pls  3I1 R1  3(302 )(0.2)  540W
2
P ower transferred to rotor(neglecting core loss)
Pg  Pin  Pls
 10,758 540  10,216W
Gross power
Po  Pg (1  s )  10,216(1  0.025)  9,961W
P ower at theshaft
Psh  Po  PFW  9,961 350  9,611W
Output power 9611

 89.3%
Input power 10758
Electromagnet icT orque
P
Po
9611
Te  o 

m (e1 p ) (1  s ) 2 (60 / 3)(1  0.025)
Efficiency
 78.4 N .m
Input power
factor
Solution :
At standstill , s  1
 0.1  j 0.2 
Z in   0.2  j 0.5  j 20

0
.
1

j
0
.
2

j
20


 0.76
V
220 3
I1  1 
 167 A
Z in
0.76
Note that the starting current is about 5 times than
full load current.
This is common for induction motors.Care should be
taken when starting induction motors.
Example
The following results were obtained on a 3 phase, star
connected stator, 75 kW, 3.3 kV, 6-pole, 50 Hz squirrel-cage
induction motor.
No-load (NL) test:
Blocked-rotor (BR) test:
Rated frequency, 50 Hz
VNL = 3.3 kV (line), INL = 5A, PNL = 2500 W
Frequency 50 Hz
VBR = 400 V (line), IBR = 27 A, PBR = 15000 W
DC test on stator resistance per phase = 3.75 .
i) Determine the parameters of the IEEE recommended equivalent circuit.
ii) Find the parameters of the Thevenin equivalent circuit as seen from the
rotor circuit.
iii) For a slip of 4%, calculate the stator current, power factor and
efficiency of the motor.
Sol_pg46