EMC II.pptx

ARYA INSTITUTE OF ENGINEERING TECHNOLOGYAND
MANAGEMENT
SUBJECT : Electrical Machine –II (4EE4-05)
FACULTY : SUBHASH CHANDER SWAMI
B.Tech. II YEAR, IV SEMESTER
1
Department of Electrical Engineering
ELECTRICAL MACHINE II

1.1.0 Definition of AC Machines (Rotating)
AC machines are generators that convert mechanical energy to ac electrical energy and motors
that convert ac electrical energy to mechanical energy.
The fundamental principles of ac machines are very simple, but unfortunately, they are somewhat
obscured by the complicated construction of real machines. There are two major classes of ac
machines
a) Synchronous Machines
b) Induction Machines.
Synchronous machines are motors and generators whose magnetic field current is supplied by a
separate de power source.
Induction machines are motors and generators whose field current is supplied by magnetic
induction (transformer action) into their field windings.
2

1.1.1 Introduction of Rotating machines
1. Electromechanical energy conversion takes place whenever a change in flux is associated with mechanical motion.
2. Speed voltage is generated in a coil when there is relative movement between the coil and magnetic field.
3. Alternating emf is generated if the change in flux linkage of the coil is cyclic.
4. The field windings which are the primary source of flux in a machine are, therefore, arranged to produce cyclic north-south
space distribution of poles.
5. A cylindrical structure is a natural choice for such a machine.
6. Coils which are the seats of induced emf’s are several in number in practical machines and are suitably connected in
series/parallels and in star/delta
7. 3-phase connection to give the desired voltage and to supply the rated current. This arrangement is called the armature
winding. When the armature coils carry currents they produce their own magnetic field which interacting with the magnetic
field of the field winding produces electro-magnetic torque tending to align the two magnetic fields.
8. The field winding and armature winding are appropriately positioned on a common magnetic circuit composed of two parts—
the stator (stationary member) and the rotor (rotating member). The stator is the annular portion of a cylinder in which rotates
a cylindrical rotor; there being an appropriate clearance (air-gap) between the two. The rotor axle is carried on two bearings
which are housed in two end-covers bolted on the two sides of the stator .
3

1.1.1 Introduction of Rotating machines CONTD…
8. The stator and rotor are made of high permeability magnetic material—silicon steel. Further, the member in
which the flux rotates is built up of thin insulated laminations to reduce eddy-current loss.
9. Since electromechanical energy conversion requires relative motion between the field and armature
winding, either of these could be placed on the stator or rotor. Because of practical convenience, field
windings are normally placed on the rotor in the class of machines called the synchronous machines; the
cross-sectional view of one such machine .The armature winding is housed in suitably
shaped slots cut out in the stator. The field winding is supplied with dc from an external source, called the
exciter, through a pair of slip-rings .The exciter is generally coupled directly to the
rotor shaft of the synchronous machine.
10. In an induction machine the stator has a 3-phase winding which draws a component of current from the
mains to set up a cyclic flux pattern in the air-gap which rotates at a speed corresponding to supply frequency
(synchronous speed ) and the rotor is either properly wound and the winding is short-circuited or is merely
a set of copper (or aluminium) bars placed in rotor slots short-circuited at each end by means of end-rings.
4

1.2.1 EMF Equation in a simple Rotating
Loop
ab
ab vBl
e 
sin


sin
).
( vBl
l
B
v
ein 


cd
cd vBl
e 
sin

S
N
r
Vcd
Vab
a b
b
a
d
c
l
+ -
t
vBL
ein 


 sin
sin
2 max


c d
B
0

 da
bc e
e
 B
r
F
ab

 B
 B
5

Induce Torque in current
Carrying Loop
cd
cd
cd rilB
r
F 

 sin
)
sin
)(
( 


 sin
2rilB
ind 
i
d
F 


ilB
B
l
i
F 

 )
(
ab
ab
ab rilB
r
F 

 sin
)
sin
)(
( 

l
b
a
c
d
0

 da
bc 

s
loop
ind B
kB 




a b
c d
Bloop
BS
6

Rotating Magnetic Field
• If three set of currents each of equal magnitude
and differing in phase by 120º flow in 3 phase
winding, then it will produce a rotating magnetic
field of constant magnitude.















90
5
.
1
240
)
240
sin(
120
)
120
sin(
0
sin
'
'
'
M
net
M
cc
M
bb
M
aa
B
B
T
t
B
B
T
t
B
B
T
t
B
B



a’
a
0

t

Resulting net
magnetic field
Apply three set of currents to the stator will
produce magnetic field intensity H and
magnetic flux B as follows:
Baa’
Bbb’
Bcc’
b
b’


c’
c
Bnet
7

The relation between the Electrical
Frequency and Mechanical Speed
• The rotating magnetic flux in stator (Bnet or Bs) can be
represented by one North and one South pole (2 pole machine).
These magnetic poles complete one mechanical rotation around
the stator for each electrical rotation.
m
e
m
e f
f 
 
 ,
For 4-pole machine, the
mechanical pole move halfway
around the stator in one
electrical cycle:
m
e
m
e
m
e f
f 


 2
,
2
,
2 


P
f
n
P
n
f
P
f
P
f
P
m
m
e
m
e
m
e
m
e
120
120
2
,
2
,
2
/ 




 



In general:
8

MMF and flux distribution
on AC Machine
• The flux in a real machine doesn't behave in a simple manner
assumed above since there is ferromagnetic rotor in the center of
machine with small air gap between rotor and stator.
– The reluctance of air gap is much higher than the reluctance of either the rotor and
stator. So the flux density vector B takes the shortest possible path across the air gap
and jumps perpendicularly between the rotor and stator.
How to produce sinusoidal voltage ? The flux density
must vary in sinusoidal manner.
The most straight foreword way to achieve a
sinusoidal variations of m.m.f along the surface of air
gap is to distribute the turns of the windings that
produce the m.m.f in closely spaced slots and to vary
the number of conductors in each slot in sinusoidal
manner.
360

cos
c
c N
n 


180
Fractional pitch winding is also
used to reduce harmonics and get
sinusoidal waves
9

Induced voltage in AC Machine
• The magnitude of flux density at a
point around the rotor is given by :

cos
M
B
B 
)
cos( 
 
 t
B
B M
• The magnitude of flux density
at a point around the stator is
given by :
BM


a-b

c-d
l
B
v
ein ).
( 

Airgap
a
b
c
d
t
t
vBL
e
e
e m
dc
ba
in 

 cos
cos
2 



t
N
e c
in 
cos

For N number of coil in each slot Department of Electrical Engineering
10

Induced volt in 3 phase coils
)
120
sin(
)
(
' 
 t
N
t
e c
bb 

t
N
e c
t
aa 
 sin
)
(
' 
)
240
sin(
)
(
'


 t
N
e c
t
cc 

RMS voltage in three phase stator:


 K
f
N
E c
A 
 2
11

Induce torque in AC Machine





sin
sin
2
net
R
ind
net
R
ind
S
R
S
ind
B
kB
B
kB
B
kB
rliB






12

AC Machine Power Flow
and Losses
Sync. Generator
Induction Motor
13

Induction Motors
14

Introduction
• Three-phase induction motors are the most common
and frequently encountered machines in industry
– 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
• not easy to have variable speed control
• requires a variable-frequency power-electronic drive for optimal
speed control
15

Construction
• An induction motor has two main parts
– a stationary stator
• consisting of a steel frame that supports a hollow,
cylindrical core
• core, constructed from stacked laminations (why?),
having a number of evenly spaced slots, providing the
space for the stator winding
Stator of IM
16

Construction
– a revolving rotor
• composed of punched laminations, stacked to create a series of rotor
slots, providing space for the rotor winding
• one of two types of rotor windings
• conventional 3-phase windings made of insulated wire (wound-rotor) »
similar to the winding on the stator
• aluminum bus bars shorted together at the ends by two aluminum rings,
forming a squirrel-cage shaped circuit (squirrel-cage)
• Two basic design types depending on the rotor design
– squirrel-cage: conducting bars laid into slots and shorted at both
ends by shorting rings.
– wound-rotor: complete set of three-phase windings exactly as the
stator. Usually Y-connected, the ends of the three rotor wires are
connected to 3 slip rings on the rotor shaft. In this way, the rotor
circuit is accessible.
17

Construction
Squirrel cage rotor
Wound rotor
Notice the
slip rings
18

Construction
Cutaway in a
typical wound-
rotor IM.
Notice the
brushes and
the slip rings
Brushes
Slip rings
19

• Balanced three phase windings, i.e.
mechanically displaced 120 degrees
form each other, fed by balanced three
phase source
• A rotating magnetic field with constant
magnitude is produced, rotating with a
speed
Where fe is the supply frequency and
P is the no. of poles and nsync is called the
synchronous speed in rpm (revolutions
per minute)
120 e
sync
f
n rpm
P

20

Synchronous speed
P 50 Hz 60 Hz
2 3000 3600
4 1500 1800
6 1000 1200
8 750 900
10 600 720
12 500 600
21

22

23

( ) ( ) ( ) ( )
net a b c
B t B t B t B t
  
sin( ) 0 sin( 120 ) 120 sin( 240) 240
M M M
B t B t B t
  
         
ˆ
sin( )
3
ˆ ˆ
[0.5 sin( 120 )] [ sin( 120 )]
2
3
ˆ ˆ
[0.5 sin( 240 )] [ sin( 240 )]
2
M
M M
M M
B t
B t B t
B t B t

 
 

     
     
x
x y
x y
24

1 3 1 3
ˆ
( ) [ sin( ) sin( ) cos( ) sin( ) cos( )]
4 4 4 4
3 3 3 3
ˆ
[ sin( ) cos( ) sin( ) cos( )]
4 4 4 4
net M M M M M
M M M M
B t B t B t B t B t B t
B t B t B t B t
    
   
    
    
x
y
ˆ ˆ
[1.5 sin( )] [1.5 cos( )]
M M
B t B t
 
 
x y
25

26

Principle of operation
• This rotating magnetic field cuts the rotor windings and
produces an induced voltage in the rotor windings
• Due to the fact that the rotor windings are short circuited, for
both squirrel cage and wound-rotor, and induced current
flows in the rotor windings
• The rotor current produces another magnetic field
• A torque is produced as a result of the interaction of those
two magnetic fields
Where ind is the induced torque and BR and BS are the magnetic
flux densities of the rotor and the stator respectively
ind R s
kB B
  
27

Induction motor speed
• At what speed will the IM run?
– Can the IM run at the synchronous speed, why?
– If rotor runs at the synchronous speed, which is the
same speed of the rotating magnetic field, then the
rotor will appear stationary to the rotating magnetic
field and the rotating magnetic field will not cut the
rotor. So, no induced current will flow in the rotor and
no rotor magnetic flux will be produced so no torque
is generated and the rotor speed will fall below the
synchronous speed
– When the speed falls, the rotating magnetic field will
cut the rotor windings and a torque is produced
28

Induction motor speed
• So, the IM will always run at a speed lower than
the synchronous speed
• The difference between the motor speed and the
synchronous speed is called the Slip
Where nslip= slip speed
nsync= speed of the magnetic field
nm = mechanical shaft speed of the motor
slip sync m
n n n
 
29

The Slip
sync m
sync
n n
s
n


Where s is the slip
Notice that : if the rotor runs at synchronous speed
s = 0
if the rotor is stationary
s = 1
Slip may be expressed as a percentage by multiplying the above
eq. by 100, notice that the slip is a ratio and doesn’t have units
30

Induction Motors and Transformers
• Both IM and transformer works on the principle of
induced voltage
– Transformer: voltage applied to the primary windings
produce an induced voltage in the secondary windings
– Induction motor: voltage applied to the stator windings
produce an induced voltage in the rotor windings
– The difference is that, in the case of the induction motor,
the secondary windings can move
– Due to the rotation of the rotor (the secondary winding of
the IM), the induced voltage in it does not have the same
frequency of the stator (the primary) voltage
31

Frequency
• The frequency of the voltage induced in the
rotor is given by
Where fr = the rotor frequency (Hz)
P = number of stator poles
n = slip speed (rpm)
120
r
P n
f


( )
120
120
s m
r
s
e
P n n
f
P sn
sf
 


 
32

Frequency
• What would be the frequency of the rotor’s
induced voltage at any speed nm?
• When the rotor is blocked (s=1) , the frequency
of the induced voltage is equal to the supply
frequency
• On the other hand, if the rotor runs at
synchronous speed (s = 0), the frequency will be
zero
r e
f s f

33

Torque
• While the input to the induction motor is electrical
power, its output is mechanical power and for that
we should know some terms and quantities
related to mechanical power
• Any mechanical load applied to the motor shaft
will introduce a Torque on the motor shaft. This
torque is related to the motor output power and
the rotor speed
and
.
out
load
m
P
N m


 2
/
60
m
m
n
rad s

 
34

Horse power
• Another unit used to measure mechanical
power is the horse power
• It is used to refer to the mechanical output
power of the motor
• Since we, as an electrical engineers, deal with
watts as a unit to measure electrical power,
there is a relation between horse power and
watts
746
hp watts

35

Example
A 208-V, 10hp, four pole, 60 Hz, Y-connected
induction motor has a full-load slip of 5
percent
1. What is the synchronous speed of this motor?
2. What is the rotor speed of this motor at rated
load?
3. What is the rotor frequency of this motor at
rated load?
4. What is the shaft torque of this motor at rated
load?
36

Solution
1.
2.
3.
4.
120 120(60)
1800
4
e
sync
f
n rpm
P
  
(1 )
(1 0.05) 1800 1710
m s
n s n
rpm
 
   
0.05 60 3
r e
f sf Hz
   
2
60
10 746 /
41.7 .
1710 2 (1/ 60)
out out
load
m
m
P P
n
hp watt hp
N m

 

 

 
 
37

Equivalent Circuit
• The induction motor is similar to the transformer
with the exception that its secondary windings are
free to rotate
As we noticed in the transformer, it is easier if we can
combine these two circuits in one circuit but there are
some difficulties
38

Equivalent Circuit
• When the rotor is locked (or blocked), i.e. s =1,
the largest voltage and rotor frequency are
induced in the rotor, Why?
• On the other side, if the rotor rotates at
synchronous speed, i.e. s = 0, the induced
voltage and frequency in the rotor will be equal
to zero, Why?
Where ER0 is the largest value of the rotor’s induced
voltage obtained at s = 1(loacked rotor)
0
R R
E sE

39

Equivalent Circuit
• The same is true for the frequency, i.e.
• It is known that
• So, as the frequency of the induced voltage in the
rotor changes, the reactance of the rotor circuit
also changes
Where Xr0 is the rotor reactance
at the supply frequency
(at blocked rotor)
r e
f s f

2
X L f L
 
 
0
2
2
r r r r r
e r
r
X L f L
sf L
sX
 

 


40

Equivalent Circuit
• Then, we can draw the rotor equivalent circuit as
follows
Where ER is the induced voltage in the rotor and RR is
the rotor resistance
41

Equivalent Circuit
• Now we can calculate the rotor current as
• Dividing both the numerator and denominator by
s so nothing changes we get
Where ER0 is the induced voltage and XR0 is the rotor
reactance at blocked rotor condition (s = 1)
0
0
( )
( )
R
R
R R
R
R R
E
I
R jX
sE
R jsX




0
0
( )
R
R
R
R
E
I
R
jX
s


42

Equivalent Circuit
• Now we can have the rotor equivalent circuit
43

Equivalent Circuit
• Now as we managed to solve the induced
voltage and different frequency problems, we
can combine the stator and rotor circuits in
one equivalent circuit
Where
2
2 0
2
2
2
1 0
eff R
eff R
R
eff
eff R
S
eff
R
X a X
R a R
I
I
a
E a E
N
a
N





44

Power losses in Induction machines
• Copper losses
– Copper loss in the stator (PSCL) = I1
2R1
– Copper loss in the rotor (PRCL) = I2
2R2
• Core loss (Pcore)
• Mechanical power loss due to friction and
windage
• How this power flow in the motor?
45

Power flow in induction motor
46

Power relations
3 cos 3 cos
in L L ph ph
P V I V I
 
 
2
1 1
3
SCL
P I R

( )
AG in SCL core
P P P P
  
2
2 2
3
RCL
P I R

conv AG RCL
P P P
 
( )
out conv f w stray
P P P P

   conv
ind
m
P



47

Equivalent Circuit
• We can rearrange the equivalent circuit as
follows
Actual rotor
resistance
Resistance
equivalent to
mechanical load
48

Power relations
3 cos 3 cos
in L L ph ph
P V I V I
 
 
2
1 1
3
SCL
P I R

( )
AG in SCL core
P P P P
  
2
2 2
3
RCL
P I R

conv AG RCL
P P P
 
( )
out conv f w stray
P P P P

  
conv RCL
P P
  2 2
2
3
R
I
s

2 2
2
(1 )
3
R s
I
s


RCL
P
s

(1 )
RCL
P s
s


(1 )
conv AG
P s P
 
conv
ind
m
P



(1 )
(1 )
AG
s
s P
s 



49

Power relations
AG
P
RCL
P
conv
P
1
s
1-s
: :
1 : : 1-
AG RCL conv
P P P
s s
50

Example
A 480-V, 60 Hz, 50-hp, three phase induction motor
is drawing 60A at 0.85 PF lagging. The stator
copper losses are 2 kW, and the rotor copper
losses are 700 W. The friction and windage
losses are 600 W, the core losses are 1800 W,
and the stray losses are negligible. Find the
following quantities:
1. The air-gap power PAG.
2. The power converted Pconv.
3. The output power Pout.
4. The efficiency of the motor.
51

Solution
1.
2.
3.
3 cos
3 480 60 0.85 42.4 kW
in L L
P V I 

    
42.4 2 1.8 38.6 kW
AG in SCL core
P P P P
  
   
700
38.6 37.9 kW
1000
conv AG RCL
P P P
 
  
&
600
37.9 37.3 kW
1000
out conv F W
P P P
 
  
52

Solution
4.
37.3
50 hp
0.746
out
P  
100%
37.3
100 88%
42.4
out
in
P
P
  
  
53

Example
A 460-V, 25-hp, 60 Hz, four-pole, Y-connected induction motor
has the following impedances in ohms per phase referred to
the stator circuit:
R1= 0.641 R2= 0.332
X1= 1.106  X2= 0.464  XM= 26.3 
The total rotational losses are 1100 W and are assumed to be
constant. The core loss is lumped in with the rotational
losses. For a rotor slip of 2.2 percent at the rated voltage and
rated frequency, find the motor’s
1. Speed
2. Stator current
3. Power factor
4. Pconv and Pout
5. ind and load
6. Efficiency
54

Solution
1.
2.
120 120 60
1800 rpm
4
e
sync
f
n
P

  
(1 ) (1 0.022) 1800 1760 rpm
m sync
n s n
     
2
2 2
0.332
0.464
0.022
15.09 0.464 15.1 1.76
R
Z jX j
s
j
   
     
2
1 1
1/ 1/ 0.038 0.0662 1.76
1
12.94 31.1
0.0773 31.1
f
M
Z
jX Z j
 
     
    
  
55

Solution
3.
4.
0.641 1.106 12.94 31.1
11.72 7.79 14.07 33.6
tot stat f
Z Z Z
j
j
 
     
     
1
460 0
3 18.88 33.6 A
14.07 33.6
tot
V
I
Z

 
     
 
cos33.6 0.833 lagging
PF   
3 cos 3 460 18.88 0.833 12530 W
in L L
P V I 
     
2 2
1 1
3 3(18.88) 0.641 685 W
SCL
P I R
   
12530 685 11845 W
AG in SCL
P P P
    
56

Solution
5.
6.
(1 ) (1 0.022)(11845) 11585 W
conv AG
P s P
    
& 11585 1100 10485 W
10485
= 14.1hp
746
out conv F W
P P P
    

11845
62.8 N.m
1800
2
60
AG
ind
sync
P

 
  

10485
56.9 N.m
1760
2
60
out
load
m
P

 
  

10485
100% 100 83.7%
12530
out
in
P
P
     
57

Torque, power and Thevenin’s Theorem
• Thevenin’s theorem can be used to transform
the network to the left of points ‘a’ and ‘b’
into an equivalent voltage source VTH in series
with equivalent impedance RTH+jXTH
58

1 1
( )
M
TH
M
jX
V V
R j X X


 
1 1
( )//
TH TH M
R jX R jX jX
  
2 2
1 1
| | | |
( )
M
TH
M
X
V V
R X X


 
59

• Since XM>>X1 and XM>>R1
• Because XM>>X1 and XM+X1>>R1
1
M
TH
M
X
V V
X X



2
1
1
1
M
TH
M
TH
X
R R
X X
X X
 
  

 

60

Then the power converted to mechanical (Pconv)
2 2
2
2
2
( )
TH TH
T
TH TH
V V
I
Z R
R X X
s
 
 
  
 
 
2 2
2
(1 )
3
conv
R s
P I
s


And the internal mechanical torque (Tconv)
conv
ind
m
P



(1 )
conv
s
P
s 


2 2
2
3
AG
s s
R
I
P
s
 
 
61

2
2
2
2
2
2
3
( )
TH
ind
s
TH TH
V R
s
R
R X X
s


 
 
   
  
 
 
 
 
  
 
 
 
 
2 2
2
2
2
2
3
1
( )
TH
ind
s
TH TH
R
V
s
R
R X X
s


 
 
 

 
  
 
 
62

Torque-speed characteristics
Typical torque-speed characteristics of induction motor
63

Comments
1. The induced torque is zero at synchronous
speed. Discussed earlier.
2. The curve is nearly linear between no-load and
full load. In this range, the rotor resistance is
much greater than the reactance, so the rotor
current, torque increase linearly with the slip.
3. There is a maximum possible torque that can’t
be exceeded. This torque is called pullout torque
and is 2 to 3 times the rated full-load torque.
64

Comments
4. The starting torque of the motor is slightly
higher than its full-load torque, so the motor
will start carrying any load it can supply at
full load.
5. The torque of the motor for a given slip
varies as the square of the applied voltage.
6. If the rotor is driven faster than synchronous
speed it will run as a generator, converting
mechanical power to electric power.
65

Complete Speed-torque c/c
66

Maximum torque
• Maximum torque occurs when the power
transferred to R2/s is maximum.
• This condition occurs when R2/s equals the
magnitude of the impedance RTH + j (XTH + X2)
max
2 2
2
2
( )
TH TH
T
R
R X X
s
  
max
2
2 2
2
( )
T
TH TH
R
s
R X X

 
67

Maximum torque
• The corresponding maximum torque of an induction motor
equals
The slip at maximum torque is directly proportional to the rotor resistance R2
The maximum torque is independent of R2
2
max 2 2
2
3
1
2 ( )
TH
s TH TH TH
V
R R X X


 
 

 
  
 
68

Maximum torque
• Rotor resistance can be increased by inserting
external resistance in the rotor of a wound-
rotor induction motor.
The
value of the maximum torque remains
unaffected
but
the speed at which it occurs can be controlled.
69

Maximum torque
Effect of rotor resistance on torque-speed characteristic
70

Example
A two-pole, 50-Hz induction motor supplies 15kW
to a load at a speed of 2950 rpm.
1. What is the motor’s slip?
2. What is the induced torque in the motor in N.m
under these conditions?
3. What will be the operating speed of the motor if
its torque is doubled?
4. How much power will be supplied by the motor
when the torque is doubled?
71

Solution
1.
2.
120 120 50
3000 rpm
2
3000 2950
0.0167 or 1.67%
3000
e
sync
sync m
sync
f
n
P
n n
s
n

  
 
  
3
no given
assume and
15 10
48.6 N.m
2
2950
60
f W
conv load ind load
conv
ind
m
P
P P
P
 




  

  

72

Solution
3. In the low-slip region, the torque-speed curve is
linear and the induced torque is direct
proportional to slip. So, if the torque is doubled
the new slip will be 3.33% and the motor speed
will be
4. (1 ) (1 0.0333) 3000 2900 rpm
m sync
n s n
     
2
(2 48.6) (2900 ) 29.5 kW
60
conv ind m
P  


    
73

Example
A 460-V, 25-hp, 60-Hz, four-pole, Y-connected wound-
rotor induction motor has the following impedances
in ohms per phase referred to the stator circuit
R1= 0.641 R2= 0.332
X1= 1.106  X2= 0.464  XM= 26.3 
1. What is the maximum torque of this motor? At what
speed and slip does it occur?
2. What is the starting torque of this motor?
3. If the rotor resistance is doubled, what is the speed
at which the maximum torque now occur? What is
the new starting torque of the motor?
4. Calculate and plot the T-s c/c for both cases.
74

Solution
2 2
1 1
2 2
( )
460
26.3
3 255.2 V
(0.641) (1.106 26.3)
M
TH
M
X
V V
R X X


 

 
 
2
1
1
2
26.3
(0.641) 0.590
1.106 26.3
M
TH
M
X
R R
X X
 
  

 
 
  
 

 
1 1.106
TH
X X
  
75

Solution
1.
The corresponding speed is
max
2
2 2
2
2 2
( )
0.332
0.198
(0.590) (1.106 0.464)
T
TH TH
R
s
R X X

 
 
 
(1 ) (1 0.198) 1800 1444 rpm
m sync
n s n
     
76

Solution
The torque at this speed is
2
max 2 2
2
2
2 2
3
1
2 ( )
3 (255.2)
2
2 (1800 )[0.590 (0.590) (1.106 0.464) ]
60
229 N.m
TH
s TH TH TH
V
R R X X



 
 

 
  
 


    

77

Solution
2. The starting torque can be found from the
torque eqn. by substituting s = 1
 
2 2
2
1
2
2
2
1
2
2
2 2
2 2
2
2 2
3
1
( )
3
[ ( ) ]
3 (255.2) (0.332)
2
1800 [(0.590 0.332) (1.106 0.464) ]
60
104 N.m
TH
start ind s
s
TH TH
s
TH
s TH TH
R
V
s
R
R X X
s
V R
R R X X
 





 
 
 
 
 
  
 
 

  
 

    

78

Solution
3. If the rotor resistance is doubled, then the slip at
maximum torque doubles too
The corresponding speed is
The maximum torque is still
max = 229 N.m
max
2
2 2
2
0.396
( )
T
TH TH
R
s
R X X
 
 
(1 ) (1 0.396) 1800 1087 rpm
m sync
n s n
     
79

Solution
The starting torque is now 2
2 2
3 (255.2) (0.664)
2
1800 [(0.590 0.664) (1.106 0.464) ]
60
170 N.m
start


 

    

80

Determination of motor parameters
• Due to the similarity between the induction
motor equivalent circuit and the transformer
equivalent circuit, same tests are used to
determine the values of the motor parameters.
– DC test: determine the stator resistance R1
– No-load test: determine the rotational losses and
magnetization current (similar to no-load test in
Transformers).
– Locked-rotor test: determine the rotor and stator
impedances (similar to short-circuit test in
Transformers).
81

DC test
– The purpose of the DC test is to determine R1. A
variable DC voltage source is connected
between two stator terminals.
– The DC source is adjusted to provide
approximately rated stator current, and the
resistance between the two stator leads is
determined from the voltmeter and ammeter
readings.
82

DC test
– then
– If the stator is Y-connected, the per phase stator
resistance is
– If the stator is delta-connected, the per phase
stator resistance is
DC
DC
DC
V
R
I

1
2
DC
R
R 
1
3
2
DC
R R

83

No-load test
1. The motor is allowed to spin freely
2. The only load on the motor is the friction and
windage losses, so all Pconv is consumed by
mechanical losses
3. The slip is very small
84

No-load test
4. At this small slip
The equivalent circuit reduces to…
2 2
2 2
(1 ) R (1 )
&
R s s
R X
s s
 
85

No-load test
5. Combining Rc & RF+W we get……
86

No-load test
6. At the no-load conditions, the input power
measured by meters must equal the losses in the
motor.
7. The PRCL is negligible because I2 is extremely
small because R2(1-s)/s is very large.
8. The input power equals
Where
&
2
1 1
3
in SCL core F W
rot
P P P P
I R P
  
 
&
rot core F W
P P P
 
87

No-load test
9. The equivalent input impedance is thus
approximately
If X1 can be found, in some other fashion, the
magnetizing impedance XM will be known
1
1,
eq M
nl
V
Z X X
I

  
88

Blocked-rotor test
• In this test, the rotor is locked or blocked so
that it cannot move, a voltage is applied to
the motor, and the resulting voltage, current
and power are measured.
89

Blocked-rotor test
• The AC voltage applied to the stator is
adjusted so that the current flow is
approximately full-load value.
• The locked-rotor power factor can be found as
• The magnitude of the total impedance
cos
3
in
l l
P
PF
V I

 
LR
V
Z
I


90

Blocked-rotor test
Where X’1 and X’2 are the stator and rotor
reactances at the test frequency respectively
'
cos sin
LR LR LR
LR LR
Z R jX
Z j Z
 
 
 
1 2
' ' '
1 2
LR
LR
R R R
X X X
 
 
2 1
LR
R R R
 
'
1 2
rated
LR LR
test
f
X X X X
f
  
91

Blocked-rotor test
X1 and X2 as function of XLR
Rotor Design X1 X2
Wound rotor 0.5 XLR 0.5 XLR
Design A 0.5 XLR 0.5 XLR
Design B 0.4 XLR 0.6 XLR
Design C 0.3 XLR 0.7 XLR
Design D 0.5 XLR 0.5 XLR
92

Example
The following test data were taken on a 7.5-hp, four-pole, 208-V, 60-Hz,
design A, Y-connected IM having a rated current of 28 A.
DC Test:
VDC = 13.6 V IDC = 28.0 A
No-load Test:
Vl = 208 V f = 60 Hz
I = 8.17 A Pin = 420 W
Locked-rotor Test:
Vl = 25 V f = 15 Hz
I = 27.9 A Pin = 920 W
(a) Sketch the per-phase equivalent circuit of this motor.
(b) Find the slip at pull-out torque, and find the value of the pull-out
torque.
93

94
Learning Outcomes
• At the end of the lecture, student should to:
– Understand the principle and the nature of 3 phase
induction machines.
– Perform an analysis on induction machines which is
the most rugged and the most widely used machine in
industry.

95
Contents
– Overview of Three-Phase Induction Motor
– Construction
– Principle of Operation
– Equivalent Circuit
• Power Flow, Losses and Efficiency
• Torque-Speed Characteristics
– Speed Control
– Overview of Single-Phase Induction Motor

96
Overview of Three-Phase Induction Motor
• Induction motors are used worldwide in many residential,
commercial, industrial, and utility applications.
• Induction Motors transform electrical energy into
mechanical energy.
• It can be part of a pump or fan, or connected to some other
form of mechanical equipment such as a winder, conveyor,
or mixer.

97
Introduction
General aspects
• A induction machine can be used as either a
induction generator or a induction motor.
• Induction motors are popularly used in the
industry
• Focus on three-phase induction motor
• Main features: cheap and low maintenance
• Main disadvantages: speed control is not easy

98

99
Construction
• The three basic parts of an AC motor are the rotor, stator, and
enclosure.
• The stator and the rotor are electrical circuits that perform as
electromagnets.

Squirrel Cage Rotor
100

101
Construction (Stator construction)
• The stator is the stationary electrical part of the motor.
• The stator core of a National Electrical Manufacturers Association
(NEMA) motor is made up of several hundred thin laminations.
• Stator laminations are stacked together forming a hollow cylinder. Coils
of insulated wire are inserted into slots of the stator core.
• Electromagnetism is the principle behind motor operation. Each
grouping of coils, together with the steel core it surrounds, form an
electromagnet. The stator windings are connected directly to the power
source.

102
Construction (Rotor construction)
• The rotor is the rotating part of the
electromagnetic circuit.
• It can be found in two types:
– Squirrel cage
– Wound rotor
• However, the most common type of rotor is
the “squirrel cage” rotor.

103
• Induction motor types:
 Squirrel cage type:
Rotor winding is composed of copper bars embedded in
the rotor slots and shorted at both end by end rings
Simple, low cost, robust, low maintenance
 Wound rotor type:
Rotor winding is wound by wires. The winding terminals
can be connected to external circuits through slip rings
and brushes.
Easy to control speed, more expensive.

104
Wound Rotor
Squirrel-Cage Rotor
/rotor winding
Short circuits all
rotor bars.

105
Construction (Enclosure)
• The enclosure consists of a frame (or yoke) and two end
brackets (or bearing housings). The stator is mounted inside the
frame. The rotor fits inside the stator with a slight air gap
separating it from the stator. There is NO direct physical
connection between the rotor and the stator.
Stator
Rotor
Air gap
• The enclosure also protects the electrical
and operating parts of the motor from
harmful effects of the environment in which
the motor operates. Bearings, mounted on
the shaft, support the rotor and allow it to
turn. A fan, also mounted on the shaft, is
used on the motor shown below for cooling.

106
Construction (Enclosure)

107
Nameplate

ELECTRICAL MACHINE II 108
• When a 3 phase stator winding is connected to a 3 phase voltage
supply, 3 phase current will flow in the windings, which also will
induced 3 phase flux in the stator.
• These flux will rotate at a speed called a Synchronous Speed, ns.
The flux is called as Rotating magnetic Field
• Synchronous speed: speed of rotating flux
• Where; p = is the number of poles, and
f = the frequency of supply
p
f
ns
120


a Fc
-93 10 113 216
-1.5
-1
-0.5
0
0.5
1
1.5
a’
c’
b’
b c
a
a’
c’
b’
b c
a
a’
c’
b’
b c
a
a’
c’
b’
b c
Fb
Fa F
Fb
Fc
F
Fa
F
Fb
Fc Fc Fb
F
Space angle () in degrees
F
Fa Fc
Fb
t = t0= t4
t = t1
t = t2 t = t3
t = t0= t4
RMF(Rotating Magnetic Field)
109

AC Machine Stator
110

Axis of phase a
a’
a’
-90 -40 10 60 110 160 210 260
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Fa
Space angle (theta) in degrees
t0
t01
t12
t2
a
MMF Due to ‘a’ phase current
111

1 Cycle
Amp
time
t0
t1 t2 t3 t4
t01 t12
Currents in different phases of AC Machine
112

Slip Ring Rotor
•The rotor contains windings similar to stator.
•The connections from rotor are brought out using slip rings that
are rotating with the rotor and carbon brushes that are static.
113

114
Slip and Rotor Speed
1. Slip s
– The rotor speed of an Induction machine is different from the speed
of Rotating magnetic field. The % difference of the speed is called slip.
– Where; ns = synchronous speed (rpm)
nr = mechanical speed of rotor (rpm)
– under normal operating conditions, s= 0.01 ~ 0.05, which is very
small and the actual speed is very close to synchronous speed.
– Note that : s is not negligible
)
1
( s
n
n
OR
n
n
n
s s
r
s
r
s





115
Slip and Rotor Speed
• Rotor Speed
– When the rotor move at rotor speed, nr (rps), the stator flux will circulate
the rotor conductor at a speed of (ns-nr) per second. Hence, the
frequency of the rotor is written as:
• Where; s= slip
f = supply frequency
sf
p
n
n
f r
s
r


 )
(
f
s
f
i
ii
ii
p
n
n
f
n
n
Rotor
At
i
p
n
f
n
stator
At
Note
r
r
s
r
p
f
r
s
s
p
f
s
.
:
)
(
)
(
)
.....(
120
)
(
:
)
.....(
120
:
:
120
120











116
Principle of Operation
• Torque producing mechanism
When a 3 phase stator winding is connected to a 3
phase voltage supply, 3 phase current will flow in the
windings, hence the stator is energized.
A rotating flux Φ is produced in the air gap. The flux
Φ induces a voltage Ea in the rotor winding (like a
transformer).
The induced voltage produces rotor current, if rotor
circuit is closed.
The rotor current interacts with the flux Φ, producing
torque. The rotor rotates in the direction of the
rotating flux.

117
Direction of Rotor Rotates
• Q: How to change the direction of
• rotation?
• • A: Change the phase sequence of the
• power supply.

118
• Conventional equivalent circuit
 Note:
● Never use three-phase equivalent circuit. Always use
per- phase equivalent circuit.
● The equivalent circuit always bases on the Y
connection regardless of the actual connection of the
motor.
● Induction machine equivalent circuit is very similar to
the single-phase equivalent circuit of transformer. It is
composed of stator circuit and rotor circuit
Equivalent Circuit of Induction Machines

119
• Step1 Rotor winding is open
(The rotor will not rotate)
• Note:
– the frequency of E2 is the same as that of E1 since the rotor is at
standstill. At standstill s=1.
f f

120

121
• Step2 Rotor winding is shorted
(Under normal operating conditions, the rotor winding is shorted. The slip is s)
• Note:
– the frequency of E2 is fr=sf because rotor is rotating.
f fr

122
• Step3 Eliminate f2
Keep the rotor current same:

123
• Step 4 Referred to the stator side
• Note:
– X’2 and R’2 will be given or measured. In practice, we do not have to
calculate them from above equations.
– Always refer the rotor side parameters to stator side.
– Rc represents core loss, which is the core loss of stator side.

124
• IEEE recommended equivalent circuit
• Note:
– Rc is omitted. The core loss is lumped with the
rotational loss.

125
• IEEE recommended equivalent circuit
Note: can be separated into 2 PARTS
• Purpose :
– to obtain the developed mechanical
I1 1
R
1
X
m
X
'
2
X '
2
R
s
s
R

1
'
2
1
V
s
R2
s
s
R
R
s
R )
1
(
2
2
2 



126
Analysis of Induction Machines
• For simplicity, let assume
Is=I1 , IR=I2
(s=stator, R=rotor)
 
R
m
s
Total
s
s
s
c
m
m
c
m
c
m
R
R
R
Z
Z
Z
Z
jX
R
Z
neglected
R
jX
Z
neglected
R
jX
R
Z
jX
s
R
Z
//
;
;
;
//
;
'
'










ZR
Zm
Zs
Vs1
Is1 Im1 IR1
T
s
s
Z
V
I 

1
1


127
Analysis of Induction Machines





















m
RM
m
R
RM
R
s
T
m
R
RM
Z
V
I
Z
V
I
Hence
V
Z
Z
Z
V
Rules
Dividing
Voltage






1
1
1
1
1
1
,
//
,
ZR
Zm
Zs
Vs1
Is1 Im1 IR1




1
1
1
1
,
s
R
m
m
R
s
R
m
R
m
I
Z
Z
Z
I
I
Z
Z
Z
I
Rules
Dividing
Current
















OR
Note : 1hp =746Watt

128
Power Flow Diagram
Pin (Motor)
Pin (Stator)
Pcore loss
(Pc)
Pair Gap
(Pag)
Pdeveloped
Pmechanical
Pconverted
(Pm)
Pout, Po
Pstator copper
loss, (Pscu)
Protor copper
loss (Prcu)
Pwindage, friction,
etc
(P - Given)

cos
3 s
s I
V
s
R
I R
R
'
'
3 2
2
3 







c
RM
R
V '
'
3 2
R
R R
I





 
s
s
R
I R
R
1
'
'
3 2
W
hp 746
1 
s
s R
I
2
3
Pin (Rotor)

129
Power Flow Diagram
• Ratio:
Pag Prcu Pm
s
R
I R
R
'
'
3 2
'
'
3 2
R
R R
I 




 
s
s
R
I R
R
1
'
'
3 2
s
1
1
1

s
1
1 s s

1
Ratio makes the analysis simpler to find the value of the particular power if we have
another particular power. For example:
s
s
P
P
m
rcu


1

130
Efficiency
Watt
x
W
hp
x
P
I
V
P
otherwise
P
P
P
P
P
P
given
are
P
if
P
P
out
s
s
in
m
o
losses
in
o
losses
in
out
746
746
cos
3
,
,
%
100














131
Torque-Equation
• Torque, can be derived from power equation in term of
mechanical power or electrical power.
n
P
T
Hence
s
rad
n
where
T
P
Power




2
60
,
)
/
(
60
2
,
,



r
o
o
r
m
m
n
P
T
Torque
Output
n
P
T
Torque
Mechanical
Thus


2
60
,
2
60
,
,



132
Torque-Equation
• Note that, Mechanical torque can written in terms of circuit
parameters. This is determined by using approximation method
...
...
...
)
1
(
'
'
3
)
1
(
'
'
3
2
2

















r
R
R
r
m
m
m
r
m
R
R
m
s
s
R
I
P
T
T
P
and
s
s
R
I
P



















 2
2
2
)
'
(
)
'
(
'
2
)
(
3
R
R
R
s
RM
m
sX
R
sR
n
V
T


Hence, Plot Tm vs s
Tm
ns
smax is the slip for Tmax to occur
s=1
Tst
Tmax
smax

133
Torque-Equation






























2
2
2
)
'
(
)
'
(
'
60
2
)
(
3
1
,
R
s
R
s
R
s
s
st
X
X
R
R
R
n
V
T
s
Torque
Starting

















































2
2
2
max
2
2
max
)
'
(
)
(
1
60
2
2
)
(
3
)
'
(
)
(
'
R
s
s
s
s
s
R
s
R
X
X
R
R
n
V
T
X
R
R
s



134
Speed Control
• There are 3 types of speed control of 3 phase
induction machines
i. Varying rotor resistance
ii. Varying supply voltage
iii. Varying supply voltage and supply frequency

135
Varying rotor resistance
• For wound rotor only
• Speed is decreasing
• Constant maximum torque
• The speed at which max
torque occurs changes
• Disadvantages:
– large speed regulation
– Power loss in Rext – reduce
the efficiency
T
ns~nNL
T
nr1
nr2
nr3
n
nr1< nr2< nr3
R1
R2
R3
R1< R2< R3

136
Varying supply voltage
• Maximum torque changes
• The speed which at max
torque occurs is constant (at
max torque, XR=RR/s
• Relatively simple method –
uses power electronics circuit
for voltage controller
• Suitable for fan type load
• Disadvantages :
– Large speed regulation since ~
ns
T
ns~nNL
T
nr1
nr2
nr3
n
nr1> nr2 > nr3
V1
V2
V3
V1> V2 > V3
V decreasing

137
• The best method since
supply voltage and supply
frequency is varied to keep
V/f constant
• Maintain speed regulation
• uses power electronics
circuit for frequency and
voltage controller
• Constant maximum torque
Varying supply voltage and supply frequency
T
nNL1
T
nr1
nr2
nr3
n
f decreasing
nNL2
nNL3

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