Electrical Machines
Induction Motors_Note(1)
1
Induction Motor
• Comparing with synchronous motor No dc field
current is required to run the machine.
• Instead, amortisseur windings are installed in the
rotor.
• The machine is called “Induction” because the
rotor voltage is induced in the rotor windings
rather than physically connected by wires.
• It is possible to use an induction machine as motor
or generator, but there are many disadvantages to
use it as generator.
2
Induction Motor Components
1) Stator: Consisting of a steel frame
that supports a hollow, cylindrical
core of stacked laminations. Slots
on the internal circumference of the
stator house the stator winding
2) Rotor: There are two different types
of induction motor rotor:
Squirrel cage rotor (Cage rotor):
Consists of series of conducting
bars laid into slots carved in the
face of rotor and shorted at either
end by large shorting rings.
3
Wound rotor IM
•A wound rotor has a 3-phase
winding, similar to the stator
winding.
•The rotor winding terminals are
connected to three slip rings which
turn with the rotor. The slip
rings/brushes allow external
resistors to be connected in series
with the winding.
•The external resistors are mainly
used during start-up under normal
running conditions the windings
short circuited externally.
4
5
Induction Motors Operation
Principle
1) The three-phase stator is supplied by balanced
three-phase voltage that drives an AC
magnetizing current through each phase
winding.
2) The magnetizing current in each phase
generates a pulsating AC flux.
3) The total flux in the machine is the sum of the
three fluxes.
4) The summation of the three AC fluxes results
in a rotating flux, which turns with constant
speed and has constant amplitude.
6
5. The rotating flux induces a voltage in the
short-circuited bars of the rotor. This voltage
drives current through the bars.
6. The induced voltage is proportional with the
difference of motor and synchronous speed.
Consequently the motor speed is less than the
synchronous speed
7. The interaction of the rotating flux and the
rotor current generates a force that drives the
motor.
7
IM Rotating Field
Consider a simple stator with 6 salient poles - windings
AN, BN, CN. The windings are mechanically spaced at
120° from each other and connected to a 3-phase source.
AC currents Ia, Ib and Ic will flow in
the windings, but will be displaced in
time by 120°. Each winding
produces its own MMF, which
creates a flux across the hollow
interior of the stator. The 3 fluxes
combine to produce a magnetic field
that rotates at the same frequency
as the supply.
8
Rotating Field - Direction
of rotation
The phase current waveforms follow each other in
the sequence A-B-C.This produces a clockwise
rotating magnetic field.
If we interchange any two of the lines connected
to the stator, the new phase sequence will be AC-B.This will produce a counterclockwise
rotating field, reversing the motor direction.
9
Number of Poles – Synchronous Speed
1) The rotating speed of the
revolving stator flux can be
reduced by increasing the
number of poles (in
multiples of two). In a fourpole stator, the phase groups
span an angle of 90°. In a sixpole stator, the phase groups
span an angle of 60°.
2) This leads to the definition of synchronous speed (rotating
stator flux speed):
Ns = 120 f/p
where Ns = synchronous speed (rpm), f = frequency of the supply (Hz), p = number
of poles
10
Induction Motors Slip
The difference between the synchronous speed and
rotor speed can be expressed as a percentage of
synchronous speed, known as the slip:
X 100 %
where
s = slip, Ns = synchronous speed (rpm), N = rotor
speed (rpm)
At no-load, the slip is nearly zero (<0.1%). At full load, the slip for
large motors rarely exceeds 0.5%. For small motors at full load, it
rarely exceeds 5%. The slip is 100% for locked rotor.
11
Frequency Induced In the Rotor
The frequency induced in the rotor depends on the slip:
fR 
Ns  N
f
Ns
where fR = frequency of voltage and current in the
rotor, f = frequency of the supply and stator field, s =
slip
12
Examples
• Example 7-1
13
Equivalent circuit of induction motor
• Transformer Model of an Induction Motor
• Per-phase equivalent circuit of transformer:
Induction Motors Equivalent circuit
An induction motor can be described as rotating transformer, it
is input is three phase voltage and current, the output of IM is
shorted out so no electrical output exist, instead the output is
mechanical. The Per phase equivalent circuit of an induction
motor:
X sta
V sup
R sta
I sta
X rot_t
Rc Xm
V sta
Stator
R rot_t
I rot_t
R rot_t(1-s)/s
Rotor
Air gap
Stator: Resistance and self inductance , magnetization reactance XM and core resistance RC
Rotor: resistance and reactance
15
Transformer model of induction motor
Magnetization curve of induction motor
Transformer Model of Induction Motor
• Note: slope of induction motor’s magneto-motive
force-flux curve is much shallower than curve of a
good transformer
• because there is an air gap in an induction motor
which greatly increase reluctance of flux path &
therefore reduces coupling between primary &
secondary windings
• Higher reluctance caused by air gap means a higher
magnetizing reactance XM in equivalent circuit will
have a much smaller value (larger susceptance BM)
than its value in an ordinary transformer.
Power flow diagram
18
Power and torque in Induction Motors
• The supply power is:
Pin 
3VI cos 
• The power transferred through the air gap by the
magnetic coupling is the input power (Pin) minus the
stator copper loss and the magnetizing (stator iron) loss.
PAG  Pin  PSCL  Pcore
PAG  3 I
2
2
R2
s
19
• The stator copper and core losses are determined using the
following:
P SCL  3 I 1 R 1
2
2
P core 
3 E1
Rc
• The electrically developed power (Pdv) is the difference
between the air gap power (Pag) and rotor copper loss.
P dv ( conv )  PAG  PRCL   ind  m
PRCL  3 I 2 R 2  PRCL  sP AG
2
P dv ( conv )  PAG (1  s )  3 I
2
R2
s
(1  s )   ind  m
20
Induction Motors Output Power
• The subtraction of the mechanical ventilation and
friction losses (Pmloss) from the developed power gives
the mechanical output power (in Hp):
Pout  Pdv  Pmloss   load  m
21
Induction Motors Efficiency and Torque
• The motor efficiency:
 
Pout
Pin
• Motor torque:
T 
P out
m
22
23
Example(1)
A 460 V, 60 Hz, 25 hp, 4 pole, Y connected induction motor
has following impedances in Ω /phase referred to stator
circuit:
R1 = 0.641 Ω R2=0.332 Ω
X1 = 1.106 Ω X2 = 0.464 Ω XM=26.3 Ω
The total rotational losses are 1100 W, & assumed to be
constant core loss is lumped in with rotational losses. For
rotor slip of 2.2 % at rated voltage & rated frequency,
find:
(a) Speed (b) stator current (c) P.F. (d) Pconv & Pout
(e) Tind & Tload (f) Efficiency
24
IM Torque-Speed Characteristic
• How does the torque of IM change as the load
changes?
• At light loads: The rotor slip is very small and so
the relative motion between the rotor and magnetic
field is very small and the rotor frequency is also
very small. Current and ER is very small and in
phase so BR is relatively small, as the rotor
magnetic field is very small then the induced
torque is small:
 ind  k B R  B net sin 
   R  90 
Rotor power factor
25
IM Torque-Speed Characteristic
• At heavy loads:
– As load increase, the slip increase, rotor speed
falls down,
– thus, more relative motion appears and produce
stronger ER,
– larger rotor current IR and so rotor magnetic
field BR will be seen.
– The angle of the rotor current will be also
changed.
– The increase in BR tend to increase in the
torque.
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• Starting torque: is 200-250% of the full load
torque (rated torque).
• Pullout torque: Occurs at the point where
for an incremental increase in load the
increase in the rotor current is exactly
balanced by the decrease in the rotor power
factor. It is 200-250 % of the full load
torque.
27
A typical induction motor torquespeed characteristic curve
28
Torque-Speed Characteristic
Curve Regions
Low-slip region:
In this region the motor slip increases approximately
linearly with increase load & rotor mechanical speed
decreases approximately linearly with load.
• In this region rotor reactance is negligible, so rotor PF is
approximately unity, while rotor current increases linearly
with slip.
• The entire normal steady-state operating range of an
induction motor is included in this linear low-slip region.
• Moderate-slip region
In moderate-slip region rotor frequency is higher than
before, & rotor reactance is on the same order of magnitude
as rotor resistance.
- In this region rotor current, no longer increases as rapidly
as before and the P.F. starts to drop
- peak torque (pullout torque) of motor occurs at point
where, for an incremental increase in load, increase in
rotor current is exactly balanced by decrease in rotor P.F.
High-slip region:
• In high- slip region, induced torque actually
decreases with increased load, since the increase
in rotor current is completely overshadowed by
decrease in rotor P.F.
• For a typical induction motor, pullout torque is
200 to 250 % of rated full-load torque
• And starting torque (at zero speed) is about 150%
of full-load torque
• Unlike synchronous motor, induction motor can
start with a full-load attached to its shaft
Comments on IM torque speed curve
1- Induced torque of motor is zero at Syn. Speed.
2- Torque-speed curve is nearly linear between no
load and full load. In this range rotor resistance is
much larger than its reactance so rotor current,
rotor magnetic field & induced torque increase
linearly with increasing slip
3- There is a maximum possible torque that cannot be
exceeded (pullout torque) is 2 to 3 times rated fullload torque of motor.
Comments on IM torque speed curve (con.)
4- Starting torque on motor is slightly larger than its
full-load torque, so this motor will start carrying
any load that it can supply at full power
5- Note: that torque on motor for a given slip varies
as square of applied voltage. This is useful in one
form of induction motor speed control that will be
described.
6- If rotor of induction motor driven faster than sync.
Speed, direction of Tind reverses & machine
become Gen. converting Pmech to Pelec.
Induction Motor/Generator Mode
Induction Motor Maximum Torque
Using Thevenin Equivalent
• Thevenin equivalent voltage of induction motor:
ZTH=RTH+jXTH = jXM(R1+jX1)/[R1+j(X1+XM)]
• Since XM>>X1 and XM+X1>>R1 , Thevenin
resistance & reactance can be approximated as:
RTH ≈ R1 ( XM/ [X1+XM] ) ^2
XTH ≈ X1
•
V TH
I2 
( R TH 
R2
s
)  ( X TH  X 2 )
2
2
Induction Motor Maximum Torque
In IM the maximum power transfer occurs when:
R2/s=√RTH^2 + (XTH+X2)^2
So maximum slip Smax is:
Smax=R2 / √RTH^2 + (XTH+X2)^2
By applying this value of slip to torque equation
in slide (8):
 ind  m  3 I
2
R2
(1  s )
s
2
 m ax 
3V T H
2 syn c
R

TH


R
2
TH

Note: smax ~ R2 , Maximum torque is independent of R2
 X TH
 X
2

2



Effect of Varying Rotor Resistance
in wound rotor IM
Effect of varying rotor resistance on T-ω of wound
rotor
Example (2)
• A 2 pole, 50 Hz induction motor supplies
15kW to a load at a speed of 2950 r/min.
Determine:
- The motor’s slip?
- The induced torque in the motor in Nm under these
conditions?
- The operating speed of the motor be if its torque is
doubled?
- The power will be supplied by the motor when the
torque is doubled?
Solution:
(a) nsync= 120fe/p= 120x50/2=3000 r/min
s= 3000-2950/3000=0.0167 or 1.67%
(b) Tind=Pconv/ωm=15 / (2950)(2πx1/60)=48.6 N.m.
(c) In the low slip region, the torque-speed is
linear & induced torque ~ s doubling Tind slip would be
3.33 % 
nm=(1-s)nsync =(1-0.0333)(3000)=2900 r/min
(d) Pconv=Tind ωm=97.2 x 2900 x 2πx1/60=29.5 kW
Example (3)
• A 460V, 25hp, 60Hz, 4-pole, Y-connected wound
rotor induction motor has the following impedances
in ohms per-phase referred to the stator circuit:
• R1 = 0.641 Ω
R2 = 0.332 Ω
• X1 = 1.106 Ω
X2 = 0.464 Ω Xm = 26.3 Ω
a)What is the max torque of this motor? At what
speed and slip does it occur?
b) What is the starting torque?
c) When the rotor resistance is doubled, what is the
speed at which the max torque now occurs?
d) What is the new starting torque?
Solution : using Thevenin equivalent :
V TH  V
Xm
R   X1  X m 
2
1
2
= 266/ √(0.641)^2+(1.106+26.3)^2= 255.2 V
=(0.641)(26.3/[1.106+26.3])^2=0.59 Ω
XTH≈X1=1.106 Ω
(a) Smax = R2 / √RTH^2 + (XTH+X2)^2
=0.332/√(0.59)^2+(1.106+0.464)^2=0.198
-
This corresponds to a mechanical speed of :
Nm=(1-s)Nsync=(1-0.198)(1800)=1444 r/min
- The maximum torque at this speed :
2
 m ax 
3V T H
2 syn c  R T H 


RTH 
2
 X TH
 X
2 
2



= 3(255.2)^2 / {2x188.5x[0.59+√0.59^2+(1.106+0.464)^2]} =229
N.m.
(b) Starting torque of motor found by s=1
2
 start 
3V TH R 2
 sync [( R TH  R 2 )  ( X TH  X 2 ) ]
2
2

= 3x255.2^2 x 0.332 /
{188.5x[(0.59+0.332)^2+(1.106+0.464)^2]}=104 N.m.
c) Rotor resistance is doubled,  s at Tmax doubles
smax=0.396 , and the speed at Tmax is:
nm=(1-s)Nsync=(1-0.396)(1800)=1087 r/min
Maximum torque is still:
Tmax=229 N.m. and starting torque is :
Tstart=3(255.2)(0.664) /
{(188.5)[(0.59+0.664)^2+(1.106+0.464)^2]} =170 N.m.