Electrical machines 2 AC Machines

ELECTRICAL
MACHINES – II
(AC MACHINES)
Presented by
C.GOKUL
AP/EEE
Velalar College of Engg & Tech,Erode
EMAIL: gokulvlsi@gmail.com

Syllabus
EE6502 Electrical Machines -II

BOOKS Reference
LOCAL AUTHORS: {For THEORY use this books}
1.Electrical Machines-II by “Gnanavadivel” – Anuradha Publication
2. Electrical Machines-II by “Godse” – Technical Publication
For Problems:
 Electric Machines by Nagrath & Kothari {Refer Solved Problems}
 Electric Machinery by A.E.Fitgerald {Refer Solved Problems}

Important Website Reference
 Electrical Machines-II by S. B.
Sivasubramaniyan -MSEC, Chennai
 http://yourelectrichome.blogspot.in/
 http://www.electricaleasy.com/p/electri
cal-machines.html

NPTEL Reference
• Electrical Machines II by Dr. Krishna
Vasudevan & Prof. G. Sridhara Rao
Department of Electrical Engineering , IIT
Madras.
• Basic Electrical Technology by Prof. L.
Umanand - IISc Bangalore {video}

Electrical Machine?
Electrical machine is a device which
can convert
 Mechanical energy into electrical
energy (Generators/alternators)
 Electrical energy into mechanical
energy (Motors)
 AC current from one voltage level to
other voltage level without changing its
frequency (Transformers)
Presented by C.GOKUL,AP/EEE Velalar College of Engg & Tech , Erode

Fundamental Principle..
 Electrical Machines (irrespective of
AC or DC) work on the fundamental
principle of Faraday’s law of
Electromagnetic Induction.

Faraday’s Law
 Faraday’s Law of Electromagnetic
Induction states that an EMF is
induced in a coil when the magnetic
flux linking this coil changes with time
or
 The EMF generated is proportional to
the rate at which flux is changed.
d d
e N
dt dt
ψ ϕ
=− =−

Faraday’s Law – Illustration

Two forms of Induced EMF !
 The effect is same if the magnet is
moved and the coil is made stationery
 We call it as statically induced EMF
 The previous case is referred to as
Dynamically induced EMF

Governing Rules
 It becomes evident that there exists a
relationship between mechanical energy,
electrical energy and magnetic field.
 These three can be combined and precisely
put as governing rules each for generator
and for motor

Fleming’s Right hand rule
 For Generator

Fleming's Right hand rule(for Generator)

Fleming’s Left hand rule
 For Motor

Fleming's left hand rule (for motors)
 First finger - direction of magnetic field (N-S)
 Second finger - direction of current
(positive to negative)
 Thumb - movements of the wire

Maxwell’s Corkscrew rule
 If the electric current is moving away from the
observer, the direction of lines of force of the
magnetic field surrounding the conductor is
clockwise and that if the electric current is
moving towards an observer, the direction of
lines of force is anti-clockwise

Corkscrew (Screw driver) rule -
Illustration

Coiling of Conductor
 To augment the effect of flux, we coil the conductor
as the flux lines aid each other when they are in the
same direction and cancel each other when they are
in the opposite direction
 Many a times, conductor is coiled around a magnetic
material as surrounding air weakens the flux
 We refer the magnetic material
as armature core

Electromagnet
 The magnetic property of current carrying
conductor can be exploited to make the
conductor act as a magnet – Electromagnet
 This is useful because it is very difficult to
find permanent magnets with such high field
 Also permanent magnets are prone to ageing
problems

Whenever current passes through
a conductor…
 Opposition to flow of current
 Opposition to sudden change in current
 Opposition to sudden change in voltage
 Flux lines around the conductor

Inductive Effect
 Reactance EMF
 Lenz Law
An induced current is always in such a
direction as to oppose the motion or
change causing it

Capacitive effect
( ) 1
( ) ( )
q t
V t i t dt
C C
= = ∫
( ) ( )
( )
dq t dv t
i t C
dt dt
⇒ = =
Q
C
V
=

Resistive Network – Vector diagram

Inductive Network – Vector Diagram

Capacitive Network – Vector Diagram

Inductive & Capacitive effects -
combined

Pure L & C networks – not at all
possible!
 R-L network

Pure L & C networks – not at all
possible! – contd.
 R-C network

Current & Flux
 As already mentioned,
As the current, so the flux

Star connection
3
L
ph
L ph
V V
I I
=
=

Delta Connection
3
L
ph
L ph
V V
I I
=
=

Maxwell's Right Hand Grip
Rule

Generators
 The Generator converts mechanical power into
electrical power.
 Synchronous generators (Alternator) are
constant speed generators.
 The conversion of mechanical power into
electrical power is done through a coupling field
(magnetic field).
Magnetic
Mechanical
ElectricalInput Output

Electric Generator
G
Mechanical
Energy
Electrical
Energy
Stationary magnets - rotating magnets - electromagnets

Motor
 The Motor converts electrical power into
mechanical power.
Magnetic
Mechanical
Electrical
Input Output
M
Electrical
Energy
Mechanical
Energy

Basic Construction
Parts
Stator
Mechanical
Rotor
Armature
Electrical
Field
Rotating Part
Stationary Part

AC MACHINES
 Two categories:
1.Synchronous Machines:
 Synchronous Generators(Alternator)
 Primary Source of Electrical Energy
 Synchronous Motor
2.Asynchronous Machines(Induction Machines)

UNIT-1
Synchronous
Generator
(Alternator)

Synchronous Generators
Generator
Exciter
View of a two-pole round rotor generator and exciter.
(Westinghouse)

Synchronous Machines
• Synchronous generators or alternators are used to convert
mechanical power derived from steam, gas, or hydraulic-turbine
to ac electric power
• Synchronous generators are the primary source of electrical
energy we consume today
• Large ac power networks rely almost exclusively on synchronous
generators
• Synchronous motors are built in large units compare to induction
motors (Induction motors are cheaper for smaller ratings) and
used for constant speed industrial drives

Construction
 Basic parts of a synchronous generator:
• Rotor - dc excited winding
• Stator - 3-phase winding in which the ac emf is generated
 The manner in which the active parts of a synchronous machine
are cooled determines its overall physical size and structure

Armature Windings (On Stator)
• Armature windings connected are 3-phase and are
either star or delta connected
• It is the stationary part of the machine and is built up of
sheet-steel laminations having slots on its inner
periphery.
• The windings are 120 degrees apart and normally use
distributed windings

Field Windings (on Rotor)
• The field winding of a synchronous machine is always
energized with direct current
• Under steady state condition, the field or exciting
current is given
Ir = Vf/Rf
Vf = Direct voltage applied to the field winding
Rf= Field winding Resistance

Rotor
• Rotor is the rotating part of the machine
• Can be classified as: (a) Cylindrical Rotor and (b) Salient
Pole rotor
• Large salient-pole rotors are made of laminated poles
retaining the winding under the pole head.

Various Types of ROTOR
 Salient-pole Rotor
 Cylindrical or round rotor

1. Most hydraulic turbines have to turn at low speeds
(between 50 and 300 r/min)
2. A large number of poles are required on the rotor
Hydrogenerator
Turbine
Hydro (water)
D ≈ 10 m
Non-uniform
air-gap
N
S S
N
d-axis
q-axis
a. Salient-Pole Rotor

• Salient pole type rotor is used in low and medium speed
alternators
• This type of rotor consists of large number of projected
poles (called salient poles)
• Poles are also laminated to minimize the eddy current
losses.
• This type of rotor are large in diameters and short in
axial length.

Salient-Pole Synchronous Generator
Stator

L ≈ 10 m
D ≈ 1 mTurbine
Steam
Stato
r
Uniform air-
gap
Stator winding
Roto
r
Rotor winding
N
S
 High speed
 3600 r/min ⇒ 2-pole
 1800 r/min ⇒ 4-pole
 Direct-conductor cooling (using
hydrogen or water as coolant)
 Rating up to 2000 MVA
Turbogenerator
d-axis
q-axis
b. Cylindrical-Rotor(Non-Salient Pole)

• Cylindrical type rotors are used in high
speed alternators (turbo alternators)
• This type of rotor consists of a smooth and
solid steel cylinder having slots along its
outer periphery.
• Field windings are placed in these slots.

Cylindrical-Rotor Synchronous Generator
Stator
Cylindrical rotor

Working of Alternator &
frequency of Induced EMF

Working Principle
• It works on the principle of Electromagnetic induction
• In the synchronous generator field system is rotating and armature
winding is steady.
• Its works on principle opposite to the DC generator
• High voltage AC output coming from the armature terminal

Working Principle
• Armature Stator
• Field Rotor
• No commutator is
required {No need for
commutator because
we need AC only}

Every time a complete pair of poles crosses the conductor, the
induced voltage goes through one complete cycle. Therefore, the
generator frequency is given by
12060
.
2
pnnp
f ==
Frequency of Induced EMF
N=Rotor speed in r.p.m
P=number of rotor poles
f=frequency of induced EMF in Hz
No of cycles/revolution = No of pairs of poles = P/2
No of revolutions/second = N/60
No of cycles/second {Frequency}= (P/2)*(N/60)=PN/120

Advantages of stationary
armature
• At high voltages, it easier to insulate
stationary armature winding(30 kV or more)
• The high voltage output can be directly
taken out from the stationary armature.
• Rotor is Field winding. So low dc voltage
can be transferred safely
• Due to simple construction High speed of
Rotating DC field is possible.

Winding
Factors(Kp,Kd)
cos
2
sin
2
sin
2
p
d
K
m
K
m
α
β
β
=
 
 
 =
 
 
 

Pitch factor (Kp)
 Consider 4 pole, 3 phase machine having 24
conductors
 Pole pitch = 24 / 4 = 6 slots
 If Coil Pitch or Coil Span = pole pitch, then it
is referred to as full-pitched winding
 If Coil Pitch < pole pitch, it is referred to as
short-pitched winding

 Coil Span = 5 / 6 of pole pitch
 If falls short by 1 / 6 of pole pitch
or
 180 / 6 = 30 degrees

This is done primarily to
 Save copper of end connections
 Improve the wave-form of the generated emf
(sine wave)
 Eliminate the high frequency harmonics
There is a disadvantage attached to it
 Total voltage around the coil gets reduced
because, the emf induced in the two sides of
the coil is slightly out of phase
 Due to that, their resultant vectorial sum is less
than the arithmetic sum
 This is denoted by a factor Pitch factor, Kp or Kc

Pitch factor – Kp
p
Vectorsum
K
Arithmaticsum
=

Pitch factor – contd.
 Arithmatic sum

 Vector sum

_
_
2 cos
2
2
cos
2
p
s
s
Vector sum
K
Arithmatic sum
E
E
α
α
=
=
=

Distribution factor (Kd)
 As we know, each phase consists of
conductors distributed in number of slots to
form polar groups under each pole
 The result is that the emf induced in the
conductors constituting the polar group are
not in phase rather differ by an angle equal
to angular displacement of the slots

 For a 3 phase machine with 36 conductors, 4 pole,
no. of slots (conductors) / pole / phase is equal to 3
 Each phase consists of 3 slots
 Angular displacement between any two adjacent
slots = 180 / 9 = 20 degrees
 If the 3 coils are bunched in 1 slot, emf induced is
equal to the arithmetic sum (3Es)
 Practically, in distributed winding, vector sum has to
be calculated
 Kd = Vector sum / Arithmetic sum
_ _ _
_ _ _
d
emf with distributed winding
K
emf with concentrated winding
=

0 0
180 180
. _ _ _no of slots per pole n
β =

 For calculating Vector sum

2 sin
2
2 sin
2
sin
2
sin
2
d
d
m
r
K
m r
m
K
m
β
β
β
β
 
 
 =
 
 
 
 
 
 =
 
 
 

Problem:
Distribution factor /Breadth factor

Equation of Induced EMF
 Average emf induced per conductor = dφ / dt
Here, dφ = φP
 If P is number of poles and flux / pole is φ Weber
dt = time for N revolution = 60 / N second
Therefore,
 Average emf = dφ / dt = φP / (60 / N)
60
NPϕ
=

Equation of Induced EMF – contd.
We know,
 N = 120 f / P
Substituting, N we get
 Avg. emf per conductor = 2 f φ Volt
 If there are Z conductors / ph, then
Avg. emf induced / ph = 2 f φ Z Volt
 Ave emf induced (in turns) / ph = 4 f φ T Volt

Equation of Induced EMF – contd.
 We know, RMS value / Avg. Value = 1.11
 Therefore,
 RMS value of emf induced / ph = 1.11 (4 f φ T) V
= 4.44 f φ T Volt
 This is the actual value, but we have two other
factors coming in the picture, Kc and Kd
 These two reduces the emf induced
 RMS value of emf induced = (Kd) (Kc) 4.44 f φ T Volt

Armature
Reaction of
Alternator

Armature Reaction
 Main Flux Field Winding
 Secondary Flux Armature Winding
 Effect of Armature Flux on the Main Flux is
called Armature Reaction

Armature Reaction in alternator
I.) When load p.f. is unity
II.) When load p.f. is zero lagging
III.) When load p.f. is zero leading

Armature Reaction in alternator
I.) When load p.f. is unity
 distorted but not weakened.- the average flux in the
air-gap practically remains unaltered.
II.) When load p.f. is zero lagging
 the flux in the air-gap is weakened- the field
excitation will have to be increased to compensate
III.) When load p.f. is zero leading
the effect of armature reaction is wholly
magnetizing- the field excitation will have to be
reduced

1. Unity Power Factor Load
 Consider a purely resistive load connected to the
alternator, having unity power factor. As induced
e.m.f. Eph drives a current of Iaph and load power
factor is unity, Eph and Iph are in phase with each
other.
 If Φf is the main flux produced by the field
winding responsible for producing Eph then Eph lags
Φf by 90o .
 Now current through armature Ia, produces the
armature flux say Φa. So flux Φa and Ia are always in
the same direction.

• Phase difference of 90o between the armature flux and the main flux
• the two fluxes oppose each other on the left half of each pole while assist
each other on the right half of each pole.
• Average flux in the air gap remains constant but its distribution gets
distorted.
• Due to such distortion of the flux, there is small drop in the terminal voltage

2. Zero Lagging Power Factor Load
 Consider a purely inductive load connected to the
alternator, having zero lagging power factor.
 Iaph driven by Eph lags Eph by 90o which is the power
factor angle Φ.
 Induced e.m.f. Eph lags main flux Φf by 90o while
Φa is in the same direction as that of Ia.
 the armature flux and the main flux are exactly in
opposite direction to each other.

• As this effect causes reduction in the main flux, the terminal voltage
drops. This drop in the terminal voltage is more than the drop
corresponding to the unity p.f. load.

3. Zero Leading Power Factor Load
 Consider a purely capacitive load connected to the
alternator having zero leading power factor.
 This means that armature current Iaph driven by Eph,
leads Eph by 90o, which is the power factor angle Φ.
 Induced e.m.f. Eph lags Φf by 90o while Iaph and
Φa are always in the same direction.
 the armature flux and the main field flux are in the
same direction

• As this effect adds the flux to the main flux, greater
e.m.f. gets induced in the armature. Hence there is
increase in the terminal voltage for leading power factor
loads.

Phasor Diagram
for Synchronous
Generator/Alternator

Phasor Diagram of loaded
Alternator
Ef which denotes excitation voltage
Vt which denotes terminal voltage
Ia which denotes the armature current
θ which denotes the phase angle between Vt and Ia
ᴪ which denotes the angle between the Ef and Ia
δ which denotes the angle between the Ef and Vt
ra which denotes the armature per phase resistance
Two important points:
(1) If a machine is working as a synchronous generator then
direction of Ia will be in phase to that of the Ef.
(2) Phasor Ef is always ahead of Vt.

Lagging PF Unity PF Leading PF

a. Alternator at Lagging PF
 Ef by first taking the component of the Vt in the
direction of Ia
 Component of Vt in the direction of Ia is Vtcosθ ,
Total voltage drop is (Vtcosθ+Iara) along the Ia.
 we can calculate the voltage drop along the direction
perpendicular to Ia.
 The total voltage drop perpendicular to Ia is
(Vtsinθ+IaXs).
 With the help of triangle BOD in the first phasor
diagram we can write the expression for Ef as

b. Alternator at Unity PF
 Ef by first taking the component of the Vt in
the direction of Ia.
 θ = 0 hence we have ᴪ=δ.
 With the help of triangle BOD in the second
phasor diagram we can directly write the
expression for Ef as

c. Alternator at Leading PF
 Component in the direction of Ia is Vtcosθ.
 As the direction of Ia is same to that of the Vt thus
the total voltage drop is (Vtcosθ+Iara).
 Similarly we can write expression for the voltage
drop along the direction perpendicular to Ia.
 The total voltage drop comes out to be (Vtsinθ-IaXs).
 With the help of triangle BOD in the first phasor
diagram we can write the expression for Ef as

Determination of the parameters of
the equivalent circuit from test data
The equivalent circuit of a synchronous generator
that has been derived contains three quantities that
must be determined in order to completely
describe the behaviour of a real synchronous
generator:
The saturation characteristic: relationship between
If and φ (and therefore between If and Ef)
The synchronous reactance, Xs
The armature resistance, Ra

VOLTAGE
REGULATION
Voltage regulation of an alternator is
defined as the rise in terminal voltage of the
machine expressed as a fraction of
percentage of the initial voltage when
specified load at a particular power factor is
reduced to zero, the speed and excitation
remaining unchanged.

Voltage
Regulation
A convenient way to compare the voltage
behaviour of two generators is by their
voltage regulation (VR). The VR of a
synchronous generator at a given load,
power factor, and at rated speed is defined
as
%
V
VE
VR
fl
flnl
100×
−
=

Voltage
Regulation
Case 1: Lagging power factor:
A generator operating at a lagging power factor has a
positive voltage regulation.
Case 2: Unity power factor:
A generator operating at a unity power factor has a small
positive voltage regulation.
Case 3: Leading power factor:
A generator operating at a leading power factor has a
negative voltage regulation.

Voltage
Regulation
This value may be readily determined from
the phasor diagram for full load operation.
If the regulation is excessive, automatic
control of field current may be employed to
maintain a nearly constant terminal voltage
as load varies

Methods of
Determination of
voltage regulation

Methods of Determination of
voltage regulation
Synchronous Impedance Method / E.M.F.
Method
Ampere-turns method / M.M.F. method
ZPF(Zero Power Factor) Method / Potier
ASA Method

1. Synchronous Impedance
Method / E.M.F. Method
The method is also called E.M.F. method of determining
the regulation. The method requires following data to
calculate the regulation.
1. The armature resistance per phase (Ra).
2. Open circuit characteristics which is the graph of open
circuit voltage against the field current. This is possible by
conducting open circuit test on the alternator.
3. Short circuit characteristics which is the graph of short
circuit current against field current. This is possible by
conducting short circuit test on the alternator.

The alternator is coupled to a prime mover capable
of driving the alternator at its synchronous speed.
The armature is connected to the terminals of a
switch. The other terminals of the switch are short
circuited through an ammeter. The voltmeter is
connected across the lines to measure the open
circuit voltage of the alternator.
 The field winding is connected to a suitable d.c.
supply with rheostat connected in series. The field
excitation i.e. field current can be varied with the
help of this rheostat. The circuit diagram is shown
in the Fig.

Circuit Diagram for OC & SC test

a. Open Circuit Test
Procedure to conduct this test is as follows :
i) Start the prime mover and adjust the speed to the synchronous
speed of the alternator.
ii) Keeping rheostat in the field circuit maximum, switch on the d.c.
supply.
iii) The T.P.S.T switch in the armature circuit is kept open.
iv) With the help of rheostat, field current is varied from its
minimum value to the rated value. Due to this, flux increasing
the induced e.m.f.
Hence voltmeter reading, which is measuring line value of open
circuit voltage increases. For various values of field current,
voltmeter readings are observed.

Open-circuit test Characteristics
The generator is turned at the rated speed
The terminals are disconnected from all loads, and
the field current is set to zero.
Then the field current is gradually increased in
steps, and the terminal voltage is measured at each
step along the way.
It is thus possible to obtain an open-circuit
characteristic of a generator (Ef or Vt versus If)
from this information

Connection for Open Circuit Test

Short-circuit
test
Adjust the field current to zero and short-
circuit the terminals of the generator
through a set of ammeters.
Record the armature current Isc as the field
current is increased.
Such a plot is called short-circuit
characteristic.

Short-circuit test
 After completing the open circuit test observation, the field
rheostat is brought to maximum position, reducing field
current to a minimum value.
 The T.P.S.T switch is closed. As ammeter has negligible
resistance, the armature gets short circuited. Then the field
excitation is gradually increased till full load current is
obtained through armature winding.
 This can be observed on the ammeter connected in the
armature circuit. The graph of short circuit armature
current against field current is plotted from the observation
table of short circuit test. This graph is called short circuit
characteristics, S.C.C.

Short-circuit
test
Adjust the field current to zero and short-circuit
the terminals of the generator through a set of
ammeters.
Record the armature current Isc as the field current
is increased.
Such a plot is called short-circuit characteristic.

Connection for Short
Circuit Test

Open and short circuit
characteristic

Curve feature
The OCC will be nonlinear due to the
saturation of the magnetic core at higher
levels of field current. The SCC will be
linear since the magnetic core does not
saturate under short-circuit conditions.

Determination of Xs
 For a particular field current IfA, the internal voltage Ef (=VA) could be found from
the occ and the short-circuit current flow Isc,A could be found from the scc.
 Then the synchronous reactance Xs could be obtained using
IfA
Ef or Vt (V) Air-gap line
OCC Isc (A)
SCC
If (A)
Vrated
VA
Isc,B
Isc, A
IfB
( )
scA
fA
unsat,saunsat,s
I
EV
XRZ
=
=+= 22
22
aunsat,sunsat,s RZX −=
scA
oc,t
scA
f
unsat,s
I
V
I
E
X =≈
: Ra is known from the DC test.
Since Xs,unsat>>Ra,

Xs under saturated condition
( )
scB
frated
sat,sasat,s
I
EV
XRZ
=
=+= 22
At V = Vrated,
22
asat,ssat,s RZX −= : Ra is known from the DC test.
IfA
Ef or Vt (V)
Air-gap line
OCC Isc (A)
SCC
If (A)
Vrated
VA
Isc,B
Isc, A
IfB

Advantages and Limitations of
Synchronous Impedance Method
 The value of synchronous impedance Zs for any load
condition can be calculated. Hence regulation of the
alternator at any load condition and load power factor can
be determined. Actual load need not be connected to the
alternator and hence method can be used for very high
capacity alternators.
 The main limitation of this method is that the method
gives large values of synchronous reactance. This leads to
high values of percentage regulation than the actual results.
Hence this method is called pessimistic method

Equivalent circuit & phasor diagram under
condition
Ia
Ef
Vt=0
jXs Ra
+
+
EfVt=0
jIaXs
IaRa
Ia

Short-circuit Ratio
 Another parameter used to describe synchronous generators is the
short-circuit ratio (SCR). The SCR of a generator defined as the ratio
of the field current required for the rated voltage at open circuit to the
field current required for the rated armature current at short circuit.
SCR is just the reciprocal of the per unit value of the saturated
synchronous reactance calculated by
[ ].u.pinX
I
I
SCR
sat_s
Iscrated_f
Vrated_f
1
=
=
Ef or Vt (V) Air-gap line
OCC
Isc (A)
SCC
If (A)
Vrated
Isc,rated
If_V rated If_Isc rated

Synchronous Generator Capability
Curves
 Synchronous generator capability curves are used to
determine the stability of the generator at various points of
operation. A particular capability curve generated in Lab
VIEW for an apparent power of 50,000W is shown in Fig.
The maximum prime-mover power is also reflected in it.

2. MMF method (Ampere turns method)
Tests: Conduct tests to find
 OCC (up to 125% of rated voltage)  refer diagram EMF
 SCC (for rated current)  refer diagram EMF

3. ZPF method (Potier method)
 ZPF (for rated current and rated voltage)
 Armature Resistance (if required)

4. ASA method
 ZPF (for rated current and rated voltage)
 Armature Resistance (if required)

Losses and
Efficiency
The losses in synchronous generator include:
1. Copper losses in
a) Armature
b) Field winding
c) The contacts between brushes
2. Core losses, Eddy current losses and
Hysteresis losses

Losses
3. Friction and windage losses,the brush
friction at the slip rings.
4. Stray load losses caused by eddy currents in
the armature conductors and by additional
core loss due to the distribution of magnetic
field under load conditions.

synchronous generator power flow
diagram
The three-phase synchronous generator power flow diagram

Synchronization
& Parallel
operation of
Alternator

Parallel operation of synchronous generators
There are several major advantages to operate generators in
parallel:
• Several generators can supply a bigger load than one machine
by itself.
• Having many generators increases the reliability of the power
system.
• It allows one or more generators to be removed for shutdown
or preventive maintenance.

Before connecting a generator in parallel with another
generator, it must be synchronized. A generator is said to be
synchronized when it meets all the following conditions:
• The rms line voltages of the two generators must be
equal.
• The two generators must have the same phase sequence.
• The phase angles of the two a phases must be equal.
• The oncoming generator frequency is equal to the
running system frequency.
Synchronization
Load
Generator 2
Generator 1
Switch
a
b
c
a/
b/
c/

Parallel operation of
synchronous generators
Most of synchronous generators are operating in parallel with other
synchronous generators to supply power to the same power system.
Obvious advantages of this arrangement are:
1. Several generators can supply a bigger load;
2. A failure of a single generator does not result in a total power loss to the load
increasing reliability of the power system;
3. Individual generators may be removed from the power system for maintenance
without shutting down the load;
4. A single generator not operating at near full load might be quite inefficient.
While having several generators in parallel, it is possible to turn off some of
them when operating the rest at near full-load condition.

Conditions required for
paralleling
A diagram shows that Generator 2
(oncoming generator) will be connected
in parallel when the switch S1 is closed.
However, closing the switch at an
arbitrary moment can severely
damage both generators!
If voltages are not exactly the same in both lines (i.e. in a and a’, b and b’ etc.), a
very large current will flow when the switch is closed. Therefore, to avoid this,
voltages coming from both generators must be exactly the same. Therefore, the
following conditions must be met:
1. The rms line voltages of the two generators must be equal.
2. The two generators must have the same phase sequence.
3. The phase angles of two a phases must be equal.
4. The frequency of the oncoming generator must be slightly higher than the
frequency of the running system.

Conditions required for
paralleling
If the phase sequences are different,
then even if one pair of voltages
(phases a) are in phase, the other two
pairs will be 1200 out of phase creating
huge currents in these phases.
If the frequencies of the generators are different, a large power transient may occur
until the generators stabilize at a common frequency. The frequencies of two
machines must be very close to each other but not exactly equal. If frequencies
differ by a small amount, the phase angles of the oncoming generator will change
slowly with respect to the phase angles of the running system.
If the angles between the voltages can be observed, it is possible to close the
switch S1 when the machines are in phase.

General procedure for
paralleling generators
When connecting the generator G2 to the running system, the following steps
should be taken:
1. Adjust the field current of the oncoming generator to make its terminal voltage
equal to the line voltage of the system (use a voltmeter).
2. Compare the phase sequences of the oncoming generator and the running
system. This can be done by different ways:
1) Connect a small induction motor to the terminals of the oncoming generator
and then to the terminals of the running system. If the motor rotates in the
same direction, the phase sequence is the same;
2) Connect three light bulbs across the
open terminals of the switch. As the phase
changes between the two generators, light
bulbs get brighter (large phase difference)
or dimmer (small phase difference). If all
three bulbs get bright and dark together,
both generators have the same phase
sequences.

General procedure for
paralleling generators
If phase sequences are different, two of the conductors on the oncoming
generator must be reversed.
3. The frequency of the oncoming generator is adjusted to be slightly higher than
the system’s frequency.
4. Turn on the switch connecting G2 to the system when phase angles are equal.
The simplest way to determine the moment when two generators are in phase is by
observing the same three light bulbs. When all three lights go out, the voltage
across them is zero and, therefore, machines are in phase.
A more accurate way is to use a synchroscope – a meter
measuring the difference in phase angles between two a
phases. However, a synchroscope does not check the
phase sequence since it only measures the phase
difference in one phase.
The whole process is usually automated…

Synchronization
LoadGenerat
or
Rest of the
power system
Generato
r
Xs1
Ef1
Xs2
Ef2
Xsn
Efn
Infinite bus
V, f are
constant
Xs eq = 0
G

Concept of the infinite bus
When a synchronous generator is connected to a power
system, the power system is often so large that nothing, the
operator of the generator does, will have much of an effect
on the power system. An example of this situation is the
connection of a single generator to the power grid. Our
power grid is so large that no reasonable action on the part
of one generator can cause an observable change in
overall grid frequency. This idea is idealized in the concept
of an infinite bus. An infinite bus is a power system so large
that its voltage and frequency do not vary regardless of
how much real or reactive power is drawn from or supplied
to it.

Steady-state power-
angle characteristics

Active and reactive power-angle characteristics
• P>0: generator operation
• P<0: motor operation
• Positive Q: delivering inductive vars for a generator action or
receiving inductive vars for a motor action
• Negaive Q: delivering capacitive vars for a generator action or
receiving capacitive vars for a motor action
Pm
Pe, Qe
Vt
Fig. Synchronous generator connected to an infinite bus.

• The real and reactive power delivered by a synchronous
generator or consumed by a synchronous motor can be
expressed in terms of the terminal voltage Vt, generated voltage
Ef, synchronous impedance Zs, and the power angle or torque
angle δ.
• Referring to Fig. 8, it is convenient to adopt a convention that
makes positive real power P and positive reactive power Q
delivered by an overexcited generator.
• The generator action corresponds to positive value of δ, while
the motor action corresponds to negative value of δ.
Pm
Pe, Qe
Vt

The complex power output of the generator in volt-
amperes per phase is given by
*
at
_
IVjQPS =+=
where:
Vt = terminal voltage per phase
Ia
* = complex conjugate of the armature current per phase
Taking the terminal voltage as reference
0jVV tt
_
+=
the excitation( at stator in case of motor) or the generated voltage,
( )δ+δ= sinjcosEE ff
_
Pm
Pe, Qe
Vt

Pm
Pe, Qe
Vt
and the armature current,
( )
s
ftf
s
t
_
f
_
a
_
jX
sinjEVcosE
jX
VE
I
δ+−δ
=
−
=
where Xs is the synchronous reactance per phase.
( )
s
tft
s
ft
s
tft
s
ft
s
ftf
t
*
a
_
t
_
X
VcosEV
Q
&
X
sinEV
P
X
VcosEV
j
X
sinEV
jX
sinjEVcosE
VIVjQPS
2
2
−δ
=
δ
=∴
−δ
+
δ
=








−
δ−−δ
==+=

Pm
Pe, Qe
Vt
s
tft
s
ft
X
VcosEV
Q&
X
sinEV
P
2
−δ
=
δ
=∴
• The above two equations for active and reactive powers hold
good for cylindrical-rotor synchronous machines for negligible
resistance
• To obtain the total power for a three-phase generator, the above
equations should be multiplied by 3 when the voltages are line-to-
neutral
• If the line-to-line magnitudes are used for the voltages, however,
these equations give the total three-phase power

Steady-state power-angle or torque-angle characteristic of a
cylindrical-rotor synchronous machine (with negligible
armature resistance).
+δ
Real power or
torque
generato
r
motor
+π+π/2
−π/2
0
−π
Pull-out
torque as a
generator
Pull-out
torque as a
motor
−δ

Steady-state stability limit
Total three-phase power: δ= sin
X
EV
P
s
ft3
The above equation shows that the power produced by a
synchronous generator depends on the angle δ between the Vt and
Ef. The maximum power that the generator can supply occurs when
δ=90o.
s
ft
X
EV
P
3
=
The maximum power indicated by this equation is called steady-state
stability limit of the generator. If we try to exceed this limit (such as by
admitting more steam to the turbine), the rotor will accelerate and lose
synchronism with the infinite bus. In practice, this condition is never
reached because the circuit breakers trip as soon as synchronism is
lost. We have to resynchronize the generator before it can again pick
up the load. Normally, real generators never even come close to the
limit. Full-load torque angle of 15o to 20o are more typical of real
machines.

Pull-out torque
The maximum torque or pull-out torque per phase that a two-
pole round-rotor synchronous motor can develop is





π
=
ω
=
60
2 s
max
m
max
max
n
PP
T
where ns is the synchronous speed of the motor in rpm
P
δ
P or Q
Q
Fig. Active and reactive power as a function of the internal angle

BLONDELS TWO REACTION
THEORY

BLONDELS TWO REACTION
THEORY
In case of cylindrical pole machines, the direct-axis
and the quadrature axis mmfs act on the same magnetic
circuits, hence they can be summed up as complexors.
However, in a salient-pole machine, the two mmfs do not
act on the same magnetic circuit.
The direct axis component Fad operates over a
magnetic circuit identical with that of the field system,
while the q-axis component Faq is applied across the
interpole space, producing a flux distribution different
from that of Fad or the Field mmf.

The Blondel's two reaction theory hence
considers the results of the cross and direct-
reaction components separately and if saturation
is neglected, accounts for their different effects
by assigning to each an appropriate value for
armature-reaction "reactive" respectively Xaq and
Xad .
Considering the leakage reactance, the combined reactance
values becomes
Xad = X + X ad and X sq = X aq
Xsq < Xsd as a given current component of the q-axis gives rise
to a smaller flux due to the higher reluctance of the magnetic path.

• Let lq and Id be the q and d-axis components
of the current I in the armature reference to the
phasor diagram in Figure. We get the following
relationships
• Iq= I cos (σ+θ) Ia = I cosφ
• Id = I sin (σ+ φ) Ir = I sinφ
I = √(Id
2 + Iq
2)= = √(Id
2 + Ir
2)
• where Ia and Ir are the active and reactive
components of current I.

Slip Test (for salient pole machines only)

Short Circuit Transients
for Synchronous
Generator

Short Circuit Phenomenon
Consider a two pole elementary single phase alternator with concentrated
stator winding as shown in Fig. 4. Consider a two pole elementary single
phase alternator with concentrated stator winding as shown in Fig. 4.

The corresponding waveforms for stator and rotor currents are shown in the
Fig

Let short circuit occurs at position of rotor shown in Fig. 4(a)
when there are no stator linkages. After 1/4 Rev as shown Fig. 4(b), it
tends to establish full normal linkage in stator winding. The stator
opposes this by a current in the shown direction as to force the flux in
the leakage path. The rotor current must increase to maintain its flux
constant. It reduces to normal at position (c) where stator current is
again reduces to zero. The waveform of stator current and field current
shown in the Fig. 5. changes totally if the position of rotor at the instant
of short circuit is different. Thus the short circuit current is a function of
relative position of stator and rotor.
Using the theorem of constant linkages a three phase short
circuit can also be studied. After the instant of short circuit the flux
linking with the stator will not change. A stationary image of main pole
flux is produced in the stator. Thus a d.c. component of current is
carried by each phase.
The magnitude of d.c. component of current is different for each
phase as the instant on the voltage wave at which short circuit occurs is
different for each phase. The rotor tries to maintain its own poles

The rotor current is normal each time when rotor poles
occupy the position same as that during short circuit and the
current in the stator will be zero if the machine is previously
unloaded. After one half cycle from this position the stator and
rotor poles are again coincident but the poles are opposite. To
maintain the flux linkages constant, the current in rotor reaches to
its peak value.
The stationary field produced by poles on the stator
induces a normal frequency emf in the rotor. Thus the rotor
current is fluctuating whose resultant a.c. component develops
fundamental frequency flux which rotates and again produces in
the stator winding double frequency or second harmonic
currents. Thus the waveform of transient current consists of
fundamental, a.c. and second harmonic components of currents.
Thus whenever short circuit occurs in three phase generator
then the stator currents are distorted from pure sine wave and
are similar to those obtained when an alternating voltage is
suddenly applied to series R-L circuit.

Stator Currents during Short Circuit
• If a generator having negligible resistance, excited and
running on no load is suddenly undergoing short circuit at its
terminals, then the emf induced in the stator winding is used
to circulate short circuit current through it. Initially the
reactance to be taken into consideration is not the
synchronous reactance of the machine. The effect of armature
flux (reaction) is to reduce the main field flux.
• But the flux linking with stator and rotor can not change
instantaneously because of the induction associated with the
windings. Thus at the short circuit instant, the armature
reaction is ineffective. It will not reduce the main flux. Thus
the synchronous reactance will not come into picture at the
moment of short circuit. The only limiting factor for short
circuit current at this instant is the leakage reactance.

After some time from the instant of short circuit, the
armature reaction slowly shows its effect and the alternator then
reaches to steady state. Thus the short circuit current reaches to
high value for some time and then settles to steady value.
It can be seen that during the initial instant of short circuit
is dependent on induced emf and leakage reactance which is
similar to the case which we have considered previously of
voltage source suddenly applied to series R-L circuit. The
instant in the cycle at which short occurs also affects the short
circuit current. Near zero e.m.f. (or voltage) it has doubling
effect. The expressions that we have derived are applicable only
during initial conditions of short circuit as the induced emf also
reduces after some tome because of increased armature
reaction.
The short circuit currents in the three phases during short
circuit are as shown in the Fig(next slide)

Capability Curves of
Synchronous
Generators

• The rating of synchronous generators is specified in terms of
maximum apparent power in KVA and MVA load at a specified
power factor (normally 80, 85 or 90 percent lagging) and voltage for
which they are designed to operate under steady state conditions.
This load is carried by the alternators continuously without
overheating. With the help of automatic voltage regulators the
terminal voltage of the alternator is kept constant (normally within
±5% of rated voltage).
• The power factor is also important factor that must be specified.
This is because the alternator that is designed to operate at 0.95 p.f.
lagging at rated load will require more field current when operate at
0.85 p.f. lagging at rated load. More field current results in
overheating of the field system which is undesirable. For this
compounding curves of the alternators can be drawn.
• If synchronous generator is supplying power at constant
frequency to a load whose power factor is constant then curve
showing variation of field current versus armature current when
constant power factor load is varied is called compounding curve for
alternator.

• To maintain the terminal voltage constant the lagging power factors
require more field excitation that that required for leading power
factors. Hence there is limitation on output given by exciter and
current flowing in field coils because of lagging power factors.
• The ability of prime mover decides the active power output of the
alternator which is limited to a value within the apparent power
rating. The capability curve for synchronous generator specifies the
bounds within which it can operate safely.
• The loading on generator should not exceed the generator rating as it
may lead to heating of stator. The turbine rating is the limiting factor
for MW loading. The operation of generator should be away from
steady state stability limit (δ = 90o). The field current should not
exceed its limiting value as it may cause rotor heating.
• All these considerations provides performance curves which are
important in practical applications. A set of capability curves for an
alternator is shown in Fig. 2. The effect of increased Hydrogen
pressure is shown which increases the cooling.

• When the active power and voltage are fixed the allowable reactive
power loading is limited by either armature or field winding heating.
From the capability curve shown in Fig. 2, the maximum reactive
power loadings can be obtained for different power loadings with
the operation at rated voltage. From unity p.f. to rated p.f. (0.8 as
shown in Fig. 2), the limiting factor is armature heating while for
lower power factors field heating is limiting factor.
This fact can be derived as follows :
• If the alternator is operating is constant terminal voltage and
armature current which the limiting value corresponding to heating
then the operation of alternator is at constant value of apparent
power as the apparent power is product of terminal voltage and
current, both of which are constant.
• If P is per unit active power and Q is per unit reactive power
then per unit apparent power is given by,

• Similarly, considering the alternator to be operating at constant terminal
voltage and field current (hence E) is limited to a maximum value obtained
by heating limits.
• Thus induced voltage E is given by,
If Ra is assumed to be zero then
The apparent power can be written as,
Substituting value of Īa obtained from (1) in equation (2),
Taking magnitudes,
• This equation also represents a circle with centre at (0, -Vt
2/Xs). These two circles are
represents in the Fig. 3 (see next post as Fig. 1). The field heating and armature heating
limitation on machine operation can be seen from this Fig. 3 (see next post as Fig.1).
• The rating of machine which consists of apparent power and power factor is specified as
the point of intersection of these circles as shown in the Fig. 4. So that the machine operates
safely.

UNIT-2
SYNCHRONOUS MOTOR
Presented by
C.GOKUL
AP/EEE

Synchronous Motor
 3 phase AC supply is given to the stator and
mechanical energy is obtained from the rotor
 Reverse of alternator operation
 However, field poles are given electrical
supply to excite the poles (electromagnets !)
 Rated between 150kW to 15MW with speeds
ranging from 150 to 1800 rpm.
 Constant speed motor

Basics – Rotating Magnetic Field
 When 3 phase supply is given to the stator
winding, 3 phase current flows which
produces 3 phase flux
 The MMF wave of the stator will have
rotating effect on the rotor
 The effect of the field will be equal to that
produced by a rotating pole

Rotating Magnetic Field (R.M.F) –
contd.

RMF – contd.
( ) ( )
( ) ( )
sin sin .......................( )
sin 120 sin 120 ...................( )
sin 240 sin 240 ...................( )
R m m
Y m m
B m m
t a
t b
t c
φ φ ω φ θ
φ φ ω φ θ
φ φ ω φ θ
= =
= − = −
= − = −

RMF – contd.
 Looking back at the waveform again, we see
that at any instant, one waveform has zero
magnitude and one has a positive value and
the other, negative value
 Let us consider at the following instances –
0, 60, 120, 180 degrees

RMF – contd.
 Case (i) φ = 0 (look at the waveform)

RMF – contd.
 Simply substitute φ = 0 in equations a, b, c
( ) ( )
( ) ( )
sin sin0 0
3
sin 120 sin 0 120
2
3
sin 240 sin 0 240
2
R m m
Y m m m
B m m m
φ φ θ φ
φ φ θ φ φ
φ φ θ φ φ
= = =
= − = − =−
= − = − =+

RMF – contd.
 Case (i) - Phasor diagram

RMF – contd.
 Case (ii) φ = 60 (look at the waveform)

RMF – contd.
( ) ( )
( ) ( )
3
sin sin 60
2
3
sin 120 sin 60 120
2
sin 240 sin 60 240 0
R m m m
Y m m m
B m m
φ φ θ φ φ
φ φ θ φ φ
φ φ θ φ
= = =
= − = − =−
= − = − =

RMF – contd.
 Case (iii) φ = 120 (look at the waveform)

RMF – contd.
( ) ( )
( ) ( )
3
sin sin120
2
sin 120 sin 120 120 0
3
sin 240 sin 120 240
2
R m m m
Y m m
R m m m
φ φ θ φ φ
φ φ θ φ
φ φ θ φ φ
= = =
= − = − =
= − = − =−

RMF – contd.
 Case (iv) φ = 180 (look at the waveform)

RMF – contd.
( ) ( )
( ) ( )
sin sin180 0
3
sin 120 sin 180 120
2
3
sin 240 sin 180 240
2
R m m
Y m m m
B m m m
φ φ θ φ
φ φ θ φ φ
φ φ θ φ φ
= = =
= − = − =
= − = − =−

RMF – contd.
 It is found that the resultant flux line is
rotating at constant magnitude
 This we refer as rotating field or revolving
field
 The speed at which it rotates will be at
synchronous speed – Ns = (120 f / P )
 Direction of rotation will be in the clockwise
direction as shown in the previous slide

Operation
 We have a rotating field at the stator
 Rotor is another magnet
 If properly aligned (?!) these two magnets will
attract each other
 Since the stator field is rotating at
synchronous speed, it will carry the rotor
magnet along with it due to attraction
(magnetic locking)

Magnetic Locking - Illustration

Why - ?
 It is true that magnetic locking will make the
rotor run at synchronous speed
 Locking cannot happen instantly in a
machine (?)
 This makes synchronous motors not self
starting

Not self starting
 Due to inertia

How to make Syn. Motor self
starting
 If the rotor is moved by external means (to
overcome inertial force acting on it) then
there is a chance for the motor to get started

Procedure to make SM self start
 3 ph supply is given to the stator
 Motor is driven by external means
 Rotor is excited
 At an instant rotor poles will be locked with
the stator field and motor will run at syn.
speed

Back EMF &
V Curves ,
Inverted V Curves

EMF generation in a motor ? !
 We call it as back emf
 Similar to generated emf in an alternator
 Rotor rotating at synchronous speed will
induce emf in the stationary armature
conductors
 The ac voltage applied has to overcome this
back emf to circulate current through the
armature winding

Back emf
 As given, emf is proportional to flux
4.44b C dE K K fTφ=

Slight deviation from the topic (?)

Increase in Load…
 In a Synchronous motor with increase in load
δ increases

Increase in Load, o.k – What about
the speed ?
 The speed of the Synchronous motor speed
stays constant at synchronous speed even
when the load is increased
 Magnetic locking between the stator and
rotor (stiffness of coupling) keeps the rotor
run at synchronous speed
 But when the angle of separation (δ) is 90,
then stiffness (locking) is lost and the motor
ceases to run

At constant load, varying the
excitation…

Kindly see to it that
 In all the cases discussed above, magnitude
of current vector changes
 Power factor changes
 But the product Icosφ would be constant so
that active power drawn by the machine
remains constant

What actually happens ?
 The resultant air gap flux is due to ac
armature winding and dc field winding
 If the field is sufficient enough to set up the
constant air gap flux then the magnetizing
armature current required from the ac source
is zero – hence the machine operates at
unity power factor – this field current is the
normal field current or normal excitation

 If the field current is less than the normal
excitation – then the machine is under
excited
 This deficiency in flux must be made by the
armature mmf – so the armature winding
draws magnetizing current or lagging
reactive MVA – leaving the machine to
operate at lagging power factor

 In case the field current is made more than
its normal operation – then the machine is
over excited
 This excess flux must be neutralized by the
armature mmf – so the machine draws
demagnetizing current or leading reactive
MVA – leaving the machine to operate at
leading power factor

Better Illustration
 Similarly,

Synchronous motor in pf
improvement
 This feature of synchronous motor makes it
suitable for improving the power factor of the
system
 Motors are overexcited so that it draws
leading current from the supply
 The motor here is referred to as synchronous
condenser

Circle Diagrams
 This offers a quick graphical solution to many
problems

Circle Diagrams – contd.
 Excitation Circle diagram
 It gives the locus of armature current, as the
excitation voltage and load angle are varied

Excitation Circle Diagram
 It is based on the voltage equation of a motor
given by
 It can be expressed as
t f a sV E I Z= +
ft
a
s s
EV
I
Z Z
= −

Excitation Circle Diagram – contd.
 Each component in the above expression is
a current component
 It can be taken in such a way that they lag
from their corresponding voltage component
by power factor angle
ft
a
s s
EV
I
Z Z
= −

 Same result can be obtained mathematically
as follows
 With Vt as reference
ft
a
s s
EV
I
Z Z
= −
0 ft
a
s s
EV
I
Z Z
δ
φ φ
∠ −∠
= −
∠ ∠

ft
a
s s
EV
I
Z Z
φ δ φ= ∠ − − ∠ − −
( ) ( ) ( )( )cos sin cos sin
ft
a
s s
EV
I j j
Z Z
φ φ δ φ δ φ
 
= − − + − + 
 
( ) ( )
Re
cos cos sin sinf ft t
a
s s s s
arranging
E EV V
I j
Z Z Z Z
φ δ φ φ δ φ
   
= − + + − + +   
   

( ) ( )
2 2
2
cos cos sin sinf ft t
a
s s s s
Magnitude
E EV V
I
Z Z Z Z
φ δ φ φ δ φ
   
= − + + − + +   
   
( ) ( )
2 2
2
2 cos cos sin sin
f ft t
a
s s s s
E EV V
I
Z Z Z Z
δ φ φ δ φ φ
   
= + − + + +      
   
( ) ( )
2 2
2
2 cos cos sin sin cos sin cos cos sin sin
f ft t
a
s s s s
E EV V
I
Z Z Z Z
δ φ δ φ φ δ φ δ φ φ
   
= + − − + +      
   

( ) ( )
2 2
2
2 cos cos sin sin cos sin cos cos sin sin
f ft t
a
s s s s
E EV V
I
Z Z Z Z
δ φ δ φ φ δ φ δ φ φ
   
= + − − + +      
   
2 2
2 2 2
2 cos cos sin sin cos sin cos sin cos sin
f ft t
a
s s s s
E EV V
I
Z Z Z Z
δ φ δ φ φ δ φ φ δ φ
   
 = + − − + +     
   
2 2
2 2 2
2 cos cos cos sin
f ft t
a
s s s s
E EV V
I
Z Z Z Z
δ φ δ φ
   
 = + − +     
   
2 2
2
2 cos
f ft t
a
s s s s
E EV V
I
Z Z Z Z
δ
   
= + −   
   

 The above equation says that Vt / Zs is one
side of a triangle, whose other side is given
by Ef / Zs
 The third side is given by Ia
2 2
2
2 cos
f ft t
a
s s s s
E EV V
I
Z Z Z Z
δ
   
= + −   
   

 Coming back to our diagram (kindly verify the
sides)

 In the diagram, if Vt is assumed constant,
then Vt / Zs is a constant
 Now, if Ef (the excitation) is fixed, Ef / Zs
vector and Ia vector follow the path of a circle
as load is changed on the motor
 This locus is referred to as Excitation circle
 Excitation circle defines the magnitude and
power factor of Ia and the load angle δ, for
different shaft loads

 Same old diagram

Power Circle Diagram
 This again gives the locus of armature
current, as the mechanical power developed
and power factor is varied

 Power output per phase is given as
 P is the mechanical power developed
including iron and mechanical losses
2
cost a a aP V I I rφ= −

 The equation can be written as,
 Dividing the whole equation by ra and
rearranging it, we get
2
cos 0t
a a
a a
V P
I I
r r
φ− + =
2 2 2 2
cos sin cos 0t
a a a
a a
V P
I I I
r r
φ φ φ+ − + =

 Subsitituting x = Ia sinφ and y = Ia cosφ, the equation
becomes
 This is equation of circle with
2 2 2 2
cos sin cos 0t
a a a
a a
V P
I I I
r r
φ φ φ+ − + =
2 2
0t
a a
V P
x y y
r r
+ − + =
2
0, &
2 2
t t
a a a
V V P
centre radius
r r r
   
= = −   
   

 Alternatively,
 We know,
 Adding Vt / 2 ra on either side we get,
2
cos 0t
a a
a a
V P
I I
r r
φ− + =
2 2
2
cos
2 2
t t t
a a
a a a a
V V VP
I I
r r r r
φ
   
− + + =   
   

 Slight Modification, yields
2 2
2
Re ,
cos
2 2
t t t
a a
a a a a
arranging
V V V P
I I
r r r r
φ
   
+ − = −   
   
2 2
2
2 cos
2 2 2
t t t
a a
a a a a
V V V P
I I
r r r r
φ
   
+ − = −   
   

 The above expression shows that
is one side of a triangle whose other two
sides are Ia and Vt / 2ra seperated by φ
2 2
2
2 cos
2 2 2
t t t
a a
a a a a
V V V P
I I
r r r r
φ
   
+ − = −   
   
2
2
t
a a
V P
r r
 
− 
 

 Going back to the power circle diagram

Power Circle Diagram - Inference
 At Pmax, armature current is in phase with
Vt/2ra, hence the power factor is unity
 Magnitude of armature current is given by
Vt/2ra

Power Circle Diagram - Inference
 At Pmax, we know, radius of the power circle
is zero
 Substituting, radius = 0, we get
2
max
0
2
t
a a
V P
r r
 
− = 
 
2
max
4
t
a
V
P
r
⇒ =

Power Circle Diagram- Inference
 Maximum power input,
 Efficiency is given by
2
,max cos .1
2 2
t t
in t a t
a a
V V
P V I V
r r
φ
 
⇒ = = = 
 
( )
( )
2
max
2
,max
/ 4
50%
/ 2
t a
in t a
V rP
P V r
η= = =

 As we see, 50 % efficiency is too low a value
for synchronous motor
 At this efficiency, since the losses are about
half of that of the input, temperature rise
reaches the permissible limit
 As such, maximum power output presented
earlier cannot be met in practice

V – curves (again?!)
 We know, excitation circle diagram shows
locus of armature current as a function of
excitation voltage
 Power circle diagram shows locus of
armature current as a function of power
 When these two circles are super imposed…

TORQUE EQUATION
& POWER
EQUATION

Power Developed by Synchronous
Motor
 Consider the phasor diagram

Motor
 In a motor power developed can be given as
 Looking at the phasor diagram again
cosm b aP E I ψ=

Motor
 We need to manipulate the vector diagram to
arrive at the expression

Motor

Torque Developed by Synchronous
Motor
 We know(e), T (2π Ns) = P if Ns is in rps
 So, T = P / (2π Ns)
 or T = P / (2π Ns) if Ns is in rpm

Maximum power developed
 Condition for maximum power developed can
be found by differentiating the power
expression by δ and equating it to zero (as
usual)
( )
2
cos cosb b
m
s s
E V E
P
Z Z
θ δ θ= − −
( )
,
sin 0m b
s
Differentiating
dP E V
d Z
θ δ
δ
=− − =

Maximum power developed -
condition
( )sin 0b
s
E V
Z
θ δ− − =
( )sin 0θ δ− =
0θ δ⇒ − =
θ δ⇒ =

 Substituting θ = δ, in the power expression,
we get,
2
,max cosb b
m
s s
E V E
P
Z Z
δ= −
2
,max cosb b
m
s s
or
E V E
P
Z Z
θ= −

 If
 Substituting, cos θ = Ra / Zs
,max
0a
b
m
s
R
E V
P
Z
≈
=
2
,max
b b a
m
s s s
E V E R
P
Z Z Z
 
= −  
 

2
,max
b b a
m
s s s
E V E R
P
Z Z Z
 
= −  
 
( )2
,max
,
4
2
s
b a m
a
Solving
Z
E V V R P
R
 = ± −  

Maximum power developed –
condition
 As the equation says, Power developed
depends on excitation
 Differentiating with respect to Eb
( )
2
cos cosb b
m
s s
E V E
P
Z Z
θ δ θ= − −
( )
2
cos cos 0m b b
b b s s
dP E V Ed
dE dE Z Z
θ δ θ
 
= − −= 
 

condition
( )
2
cos cos 0m b b
b b s s
dP E V Ed
dE dE Z Z
θ δ θ
 
= − −= 
 
2
s
b
a
VZ
E
R
=

condition
 This is the value of Eb which will make
developed power to be maximum
 The maximum power is given by substituting
the condition (Eb) in Pm expression
2
s
b
a
VZ
E
R
=
2 2
,max
2 4
m
a a
V V
P
R R
= −

Operation of
infinite bus
bars

Operation of AC Generators in Parallel
with Large Power Systems
• Isolated synchronous generator supplying its own load is very
rare (emergency generators)
• In general applications more than one generator operating in
parallel to supply loads
• In Iran national grid hundreds of generators share the load on
the system
• Advantages of generators operating in parallel:
1- several generators can supply a larger load
2- having many generators in parallel increase the
reliability of power system
3- having many generators operating in parallel allows
one or more of them to be removed for shutdown &
preventive maintenance
4- if only one generator employed & not operating near full load, it
will be relatively inefficient

INFINITE BUS
• When a Syn. Gen. connected to power system,
power sys. is so large that nothing operator of
generator does, have much effect on pwr. sys.
• Example: connection of a single generator to a
large power grid (i.e. Iran grid), no reasonable
action on part of one generator can cause an
observable change in overall grid frequency
• This idea belong to definition of “Infinite Bus”
which is: a so large power system, that its
voltage & frequency do not vary, (regardless of
amount of real and reactive power load)

• When a syn. Gen.
connected to a
power system:
1-The real power
versus frequency
characteristic of
such a system
2-And the reactive
power-voltage
characteristic

• Behavior of a generator
connected to a large
system
A generator connected in
parallel with a large
system as shown
• Frequency & voltage of
all machines must be the
same, their real power-
frequency (& reactive
power-voltage)
characteristics plotted
back to back 

• Assume generator just been paralleled with
infinite bus, generator will be “floating” on the
line, supplying a small amount of real power
and little or no reactive power
• Suppose generator paralleled, however its
frequency being slightly lower than system’s
operating frequency
 At this frequency power supplied by
generator is less than system’s operating
frequency, generator will consume energy and
runs as motor

• In order that a generator comes on line and
supply power instead of consuming it, we
should ensure that oncoming machine’s
frequency is adjusted higher than running
system’s frequency
• Many generators have “reverse-power trip”
system
• And if such a generator ever starts to consume
power it will be automatically disconnected from
line

• As seen earlier, synchronous motor is not self
starting. It is necessary to rotate the rotor at a
speed very near to synchronous speed. This is
possible by various method in practice. The
various methods to start the synchronous motor
are,
1. Using pony motors
2. Using damper winding
3. As a slip ring induction motor
4. Using small d.c. machine coupled to it.

1. Using pony motors
• In this method, the rotor is brought to the
synchronous speed with the help of some
external device like small induction motor. Such
an external device is called 'pony motor'.
• Once the rotor attains the synchronous
speed, the d.c. excitation to the rotor is switched
on. Once the synchronism is established pony
motor is decoupled. The motor then continues to
rotate as synchronous motor.

3. As a Slip Ring Induction Motor
Refer Unit 3 for detail understanding

4. Using Small D.C. Machine
• Many a times, a large synchronous motor are provided
with a coupled d.c. machine. This machine is used as a
d.c. motor to rotate the synchronous motor at a
synchronous speed. Then the excitation to the rotor is
provided. Once motor starts running as a synchronous
motor, the same d.c. machine acts as a d.c. generator
called exciter. The field of the synchronous motor is then
excited by this exciter itself.

Current loci for constant
power input, constant
excitation and constant power
developed
Refer Book for
detail study

Current loci for constant power input

Current loci for constant power
developed(PM)

Current locus for constant Excitation

Natural
frequency of
oscillations
Refer Book

Damper
windingsRefer Book for
detail study

Synchronous motors are not self starting machines. These
machines are made self starting by providing a special winding in the
rotor poles, known as damper winding or squirrel cage windings. The
damper winding consists of short circuited copper bars embedded in the
face of the rotor poles
When an ac supply is provided to stator of a 3-phase
synchronous motor, stator winding produces rotating magnetic field.
Due to the damper winding present in the rotor winding of the
synchronous motor, machine starts as induction motor (Induction
machine works on the principle of induction. Damper windings in
synchronous motor will carryout the same task of induction motor rotor
windings.
Therefore due to damper windings synchronous motor starts as
induction motor and continue to accelerate). The exciter for synchronous
motor moves along with rotor. When the motor attains about 95% of the
synchronous speed, the rotor windings is connected to exciter terminals
and the rotor is magnetically locked by the rotating magnetic field of
stator and it runs as a synchronous motor.

Functions of Damper Windings:
• Damper windings helps the synchronous motor to start
on its own (self starting machine) by providing starting
torque
• By providing damper windings in the rotor of
synchronous motor "Hunting of machine“ can be
suppressed.
When there is change in load, excitation or change in
other conditions of the systems rotor of the synchronous
motor will oscillate to and fro about an equilibrium
position. At times these oscillations becomes more
violent and resulting in loss of synchronism of the motor
and comes to halt.

• When synchronous motor is over excited it takes leading
p.f. current. If synchronous motor is on no load, where
load angle δ is very small and it is over excited (Eb > V)
then power factor angle increases almost up to 90o. And
motor runs with almost zero leading power factor
condition.

• This characteristics is similar to a normal capacitor
which takes leading power factor current. Hence over
excited synchronous motor operating on no load
condition is called as synchronous condenser or
synchronous capacitor. This is the property due to which
synchronous motor is used as a phase advancer or as
power improvement device.
Disadvantage of Low Power Factor
• In various industries, many machines are of induction motor
type. The lighting and heating loads are supplied through
transformers. The induction motors and transformers draw lagging
current from the supply. Hence the overall power factor is very low
and lagging in nature.
• The power is given by,
P = VI cosΦ .............. single phase
... I = P/(VcosΦ)

The high current due to low p.f. has following disadvantages
:
1. For higher current, conductor size required is more which
increases the cost.
2. The p.f. is given by
cosΦ = Active power/ Apparent = (P in KW)/ (S in KVA)
Thus for fixed active power P, low p.f. demands large KVA
rating
alternators and transformers. This increases the cost.
3. Large current means more copper losses and poor
efficiency.
4. Large current causes large voltage drops in transmission
lines, alternators and other equipments. This results into poor
regulation.

Unit-3
Three phase Induction
Motor
Presented By
C.GOKUL
AP/EEE

Construction of
Induction Motor

Principle of
Operation

Compare
Induction motor &
Transformer

356
Losses - Summary
Efficiency (η) =
Poutput
Pinput

357
Motor Torque
Tm =
9.55 Pm
n
9.55 (1 – s) Pr
ns (1 – s)
=
= 9.55 Pr / ns
Tm = 9.55 Pr / ns

358
I2R losses in the rotor
Pjr = s Pr
Pjr = rotor I2R losses [W]
s = slip
Pr = power transmitted to the rotor [W]
Mechanical Power
Pm = Pr - Pjr
= Pr - s Pr
= (1 – s) Pr

LOAD TEST ON THREE PHASE INDUCTION MOTOR

No Load Test or Running Light Test or
Open Circuit Test
This test gives
1. Core loss
2. F & W loss
3. No load current I0
5. Ic, Rc, Iμ, Xm
6. Mechanical faults, noise
Rated per voltage V0, with
rated freq is given to stator.
Motor is run at NO LOAD
STATOR
A
I0
VV0
R
YB
ROTOR
N
W0
P0, I0 and V0 are recorded
P0 = I0
2r1+Pc+Pfw
4. No load power factor

No load power factor is small,
0.05 to 0.15
1. Ic=I0cosθ0 2. Iμ=I0sinθ0
3.
On No load, Motor runs near to syn speed
So, s ≈ zero 1/s=α or open circuit
4.
00
0
0
IV
P
θC =os
)(, 11000
c
0
c jxrIVE
I
E
R +−==
µI
E
X 0
m =
r1
r2/s
jx1 jx2
I2
jXm
Rc
I0
I0
Ic IΦ
V0
opencircuit
provided x1 is known

The F & W loss Pfw, can be obtained from this
test.
Vary input voltage and note input power
Input Power
Input Voltage
Pfw
Thus Pc=P0 - I0
2r1 - Pfw

Rotor is blocked, Speed = 0, slip = 1
Blocked Rotor test or Short Circuit Test
A
Isc
V Vsc
R
YB
N
Wsc
3-ph Variac
I M
Rotor is blocked or held stationary by
belt pulley or by hand
Low voltage is applied upto rated stator current
Voltage Vsc, Current Isc and Power Psc are measured.

Mechanical loss =0
Rc and Xm >> r2+jx2
Therefore, Zsc = Vsc / Isc
=Rsc+jXsc
This test gives copper loss
Since slip is 1, secondary is short circuited
jx1 jx2
jXm
Rc
r1 Isc
I0
Ic IΦ
Vsc
r2





 −
s
s1
r2
Core loss negligible
Hence omitted scscIV
P
cosθ sc
sc = =0.8 to 0.9

Class of motor x1 x2
= r1+r2
1. Class A (normal Tst and Ist) 0.5 0.5
For wound rotor motor, x1 = x2 = Xsc /2
Rsc= Psc/Isc
2
22
scscsc RZX −=
r2= Rsc – r1
21 xx +=
For squirrel cage motor,
2. Class B (normal Tst and low Ist) 0.4 0.6
3. Class C (high Tst and low Ist) 0.3 0.7
4. Class D (high Tst and high slip) 0.5 0.5

But the advantage of circle diagram is that
torque and slip can be known from circle diagram
The circle diagram is constructed with the help of
Graphical representation
The equivalent ckt., operating ch. can be obtained
by computer quickly and accurately
1. No load test (I0 & θ0)
2. Blocked rotor test (Isc & θsc)
Circle Diagram of Ind Motor
extremities or Limits of stator current, Power,

x
y
I0
θ0
Isc
θsc
1. Draw x and y axes(V1 on y axis)
2. Draw I0 and Isc(=V1/Zsc)
3. Draw parallel line to x axis from I0.
This line indicates constant loss vertically
V1
Line I0Isc is
output line
4. Join I0 and Isc
Output line
O

x
y
I0
θ0
θsc
C
Output line
L1
T
V1
5. Draw perpendicular bisector to output line
6. Draw circle with C as a centre
7. Draw perpendicular from Isc on x axis..
8. Divide IscL1 in such a way that.
LossCuStator
LossCuRotor
r
'r
LT
TI
1
2
1
sc
==
Isc
L2
O

x
y
I0
θ0
θsc
C
Output line
9. Join I0T. This is called as Torque Line.
10. Suppose 1cm=Xamp, so 1cm=V1.X= power scale
Rated output power/V1X = Total cm for rated o/p power
Torque line
V1
Total cm for rated output power=IscR
Isc
T
R
L1
L2
O
rated output power

x
y
I0
θ0
θsc
C
Output line
11. From R, draw line parallel to output line crossing at P & P’.
P is operating point
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
12. Join O and P. Cosθ1 is operating pf.
θ1
Lebel O’, T’ , L1’ and L2’
13. From P draw perpendicular on x axis
O

x
y
I0
θ0
θsc
C
Output line
14. Determine the following
1. Constant Losses and copper losses
=Core loss + F & W loss
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
L1L2=L1’L2’=constant losses
α no load current I0
θ1
O

x
y
I0
θ0
θsc
C
Output line
At standstill, input power = IscL2 L1L2=Constant Loss
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
Constant loss= Stator core loss +rotor core loss (f)
F & W loss=0
θ1
O

x
y
I0
θ0
θsc
C
Output line
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
At operating point P, input power = PL2’,
θ1
L1’L2’=Constant Loss
Constant loss = Stator core loss + F & W loss
Rotor core loss ≈ 0 (sf)
Thus L1L2=L1’L2’= Constant loss
O

x
y
I0
θ0
θsc
C
Output line
At standstill, Stator Cu loss=TL1
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
rotor Cu loss = IscT
At P, stator Cu loss =T’L1’ and
θ1
rotor Cu loss = O’T’
O

x
y
I0
θ0
θsc
C
Output line
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
2. Output Power and Torque
θ1
Output Power = O’P
The gap betn output line and circle is OUTPUT Power.
At I0, o/p=0, at Isc, o/p=0
O”
Pmax
Max output power=PmaxO”
Slip1 0
Ns0 Speed
Pmax
T”
L1”
L2”O

x
y
I0
θ0
θsc
C
Output line
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
θ1
Output Torque = T’P
The gap betn torque line and circle is OUTPUT torque.
At I0, torque=0, but at
Isc, torque=T Isc
Pmax
=Starting torque
Tmax
Max output torque=TmaxT”’
Slip1 0
Ns0 Speed
Tmax
O”
T”
L1”
L2”
O”’
T”’
L1”’
L2”’O

x
y
I0
θ0
θsc
C
Output line
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
θ1
Max Power and Max Torque are not occurring at same time
Contradiction to max power transfer theorem
Pmax
Tmax
O”
T”
L1”
L2”
O”’
T”’
L1”’
L2”’O

x
y
I0
θ0
θsc
C
Output line
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
θ1
Pmax
Tmax
O”
T”
L1”
L2”
O”’
T”’
L1”’
L2”’
Air gap power Pg = Input power – Stator Cu loss- core loss
=PL2’-T’L1’-L1’L2’
3. Slip, Power factor and Efficiency
= PT’
s = rotor Cu loss/Pg =O’T’/PT’
"
""
TP
TO
smp
max
=
'"
'"'"
TT
TO
smt
max
=
O

x
y
I0
θ0
θsc
C
Output line
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
θ1
Pmax
Tmax
O”
T”
L1”
L2”
O”’
T”’
L1”’
L2”’
3. Slip, Power factor and Efficiency
O
Power factor cosθ1 = PL2’/OP
PO’/PL2’Efficiency=

x
y
I0
θ0
θsc
C
Output line
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
θ1
s=0
The gap betn circle and T & s=α is braking torque
O”
Pmax
Tmax
T”
4. Braking Torque
O
s=1
s=α
braking torque
Slip 0
Ns
0 Speed
Te
1α

x
y
I0
θ0
θsc
C
Output line
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
θ1
s=0
O”
Pmax
Tmax
T”
5. Induction Generator
O
s=1
s=α
braking torque

x
y
I0
θ0
θsc
C
Output line
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
θ1
s=0
(Generator)
O”
Pmax
Tmax
T”
O
s=1
s=α
braking torque
s= -ve
θG
G
OG=Gen Current
O’G=Mech I/p
L2’G=Active power
OL2’=reactive powerPGmax

x
y
I0
θ0
θsc
C
Output line
Torque line
V1
Isc
T
R
P
P’
L2’
L1’
T’
O’
L1
L2
θ1
s=0
(Generator)
O”
Pmax
Tmax
T”
O
s=1
s=α
braking torque
s= -ve
θG
G
OG=Gen Current
O’G=Mech I/p
L2’G=Active power
OL2’=reactive powerPGmax
Slip 0 -1
Speed
2Ns
Ns
0
Slip
Speed
Te
1α

CIRCLE DIAGRAM OF AN INDUCTION MOTOR- Summary
H
T
Fig. 3.3

Separation of
Losses

SEPARATION OF NO LOAD LOSSES
The separation of core loss and mechanical loss (windage and friction) can be obtained by
no load test conducted from variable voltage, rated frequency supply. Step by step reduce
the voltage till the machine slip suddenly start to increase and the motor tends to rest
(stall). The core loss decrease almost square of the voltage and windage and friction loss
remains almost constant. Plot the curve between applied voltage (V) and power (Po),
extended to V=0 which gives mechanical loss.
Mechanical loss will be obtained from graph
Magnetic loss + mechanical loss = output power
Therefore., magnetic loss = output power – mechanical loss

Formulae for calculating the equivalent circuit parameters:
Z0 = Voc /(Ioc / √3)
R0 = Woc / (Ioc) 2
X0 = √[( Z0)2 - (R0)2
ϕ0 = cos-1 [Woc / (√3 * Voc * Ioc )]
RBR = Wsc / (Isc)2
ZBR = Vsc / (Isc/ √3)
XBR = √[( ZBR)2 - (RBR)2]
RiWF – Resistance accounting for rotational losses
R1 = 1.2 * stator winding resistance (dc)
Pr = Woc – Ioc
2 * R1 (since Pr = P0 – 3 * (Ioc / √3)2 * R1)
RiWF = Voc
2 / Pr
Xm – Magnetizing reactance
IiWF = Voc / Riwf
Im = (Ioc
2 - IiWF
2)1/2
Xm = Voc / Im

DOUBLE CAGE ROTOR
Double Cage Rotor has two independent cages on the same rotor slots,
one inside the other for the production of high starting torque. The
outer cage (alloy) in the rotor has high resistance and low reactance
which is used for starting purpose. The inner cage (copper) has a low
resistance and high reactance which is used for running purpose. The
constructional arrangement and torque-speed characteristics as shown
in fig. 3.5.
Advantages:
 High starting torque.
 Low I2R loss under running conditions and high efficiency.

Fig. 3.5
Double Cage construction
Torque-Slip Characteristics
Slip

Equivalent Circuit:
If the magnetising current is neglected, then the equivalent circuit is reduced to
Rotor
‘ ‘

Principle of operation
Induction generators and motors produce electrical power when
their rotor is rotated faster than the synchronous speed. For a four-
pole motor operating on a 50 Hz will have synchronous speed equal
to 1500 rpm.
In normal motor operation, stator flux rotation is faster than the
rotor rotation. This is causing stator flux to induce rotor currents,
which create rotor flux with magnetic polarity opposite to stator. In
this way, rotor is dragged along behind stator flux, by value equal to
slip.
In generator operation, a prime mover (turbine, engine) is driving
the rotor above the synchronous speed. Stator flux still induces
currents in the rotor, but since the opposing rotor flux is now cutting
the stator coils, active current is produced in stator coils and motor
is now operating as a generator and sending power back to the
electrical grid.
INDUCTION GENERATOR

Fig. 3.4 current Locus for Induction Machine
a. Sub-synchronous (motor) b. Super-synchronous (generator)

Fig.3.5 Phasor Diagram
Fig. 3.6 Torque-Slip Characteristics
When the machine runs as induction generator, the vector diagram shown in fig.3.5. This is
possible only if the machine is mechanically driven above the synchronous speed.
OA-no load current
AB-stator current to overcome rotor mmf
OB-total stator current

Fig.3.4b the point P in the lower half of the circle shows operating point as an induction
generator.
PT-stator electrical output
ST-Core, friction and windage losses
RS-Stator copper loss
QR-Rotor copper loss
PQ-Mechanical input
PR-Rotor input
Slip
Efficiency
PR
QR
inputrotor
losscopperrotor
==
PQ
PT
input
output
==
The torque-slip curve is shown in fig.3.6.Torque will become zero at synchronous speed. If the
speed increases above the synchronous speed, the slip will be negative.
Induction generator differs from the synchronous generator as
 Dc current excitation is not required.
 Synchronisation is not required.

Advantages:
 It does not hunt or drop out of synchronism
 Simple in construction
 Cheaper in cost
 Easy maintenance
 Induction regulators provide a constant voltage adjustment depending on the
loading of the lines.
Disadvantages:
 Cannot be operated independently.
 Deliver only leading current.
 Dangerously high voltages may occur over long transmission lines if the
synchronous machines at the far end become disconnected and the line capacitance
excites the induction machines.
 The induction generator is not helpful in system stability.
Applications:
 For installation in small power stations where it can be operated in parallel and
feeding into a common mains without attendant.
 For braking purpose in railway work.

SYNCHRONOUS INDUCTION MOTOR
It is possible to make the slip ring induction motor to run at synchronous speed when its
secondary winding is fed from a dc source. Such motors are then called as synchronous
induction motor.
3Φ
Supply
Stator
Fig. 3.3

Rotor connections for dc excitation:

Heating will always occur with normal three phase rotor winding as in fig.3.4. The two phase
windings (e and f) gives uniform heating but produce large harmonics and noise. In those
machines primary chording is commonly employed to reduce the effect of harmonics.
The synchronous induction motor is generally built for outputs greater than 30HP because of its
higher cost of the dc exciter. These motors are employed in applications where a constant speed
is desirable such as compressors, fans, pumps, etc., If load torque is high and the machines goes
out of synchronism, it continues to run as an induction motor. As soon as the load torque falls
sufficiently low, the machines will automatically synchronize.
Fig 3.4

Advantages:
 It will start and synchronise itself against heavy loads.
 No separate damper winding is required.
 The exciter may be small unit due to smaller air-gap.

Example 5.1 A 3-phase, 460 V, 100 hp, 60 Hz, four-pole 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 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.
Solution:
rpm
p
f
ns 1800
4
60*120120
===
( ) ( ) rpmnsn s 17101800*05.011 =−=−=
(b) 1800 (same as synchronous speed)

Example 4.2 A no-load test conducted on a 30 hp, 835 r/min, 440 V, 3-phase, 60 Hz squirrel-cage
induction motor yielded the following results:
No-load voltage (line-to-line): 440 V
No-load current: 14 A
No-load power: 1470 W
Resistance measured between two terminals: 0.5 Ω
The locked-rotor test, conducted at reduced voltage, gave the following results:
Locked-rotor voltage (line-to-line): 163 V
Locked-rotor power: 7200 W
Locked-rotor current: 60 A
Determine the equivalent circuit of the motor.
Solution:
Assuming the stator windings are connected in way, the resistance per phase is:
Ω== 25.02/5.01R
From the no-load test:
PhaseV
V
V LL
/254
3
440
3
1 ===
Ω=== 143.18
14
254
1
1
I
V
ZNL

Ω=== 5.2
14*3
1470
3 22
1I
P
R NL
NL
97.175.2143.18 2222
=−=−= NLNLNL RZX
Ω==+ 97.171 NLm XXX
Ω=== 6667.0
60*3
7200
3
22
1
BL
BL
BL
I
P
R
From the blocked-rotor test
The blocked-rotor reactance is:
( ) Ω=−=−= 42.16667.05685.1 2222
BLBLBL RZX
Ω=′+≅ 42.121 XXX BL

Ω=′=∴ 71.021 XX
Ω=−=−= 26.1771.097.171XXX NLm
Ω=−=−= 4167.025.06667.01RRR BL
Ω=




 +
=






 +′
=′∴ 4517.04167.0*
26.17
26.1771.0
22
2
2 R
X
XX
R
m
m

Example 5.3 The following test results are obtained from a three-phase 60 hp, 2200
V, six-pole, 60 Hz squirrel-cage induction motor.
(1) No-load test:
Supply frequency = 60 Hz, Line voltage = 2200 V
Line current = 4.5 A, Input power = 1600 W
(2) Blocked-rotor test:
Frequency = 15 Hz, Line voltage = 270 V
Line current = 25 A, Input power = 9000 W
(3) Average DC resistance per stator phase: 2.8 Ω
(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 of
Fig.5.16.

PhaseVV /2.1270
3
2200
1 == Ω=== 27.282
5.4
2.1270
1
1
I
V
ZNL
Ω=== 34.26
5.4*3
1600
3 22
1I
P
R NL
NL

(a) No-Load equivalent Circuit (b) Locked rotor equivalent circuit
Ω=−=−= 28134.2627.282 2222
NLNLNL RZX
Ω==+ 2811 NLm XXX
=281.0 Ω.
Ω=== 8.4
25*3
9000
3 22
1I
P
R BL
BL
Ω=−=−=′ 28.28.412 RRR BL

impedance at 15 Hz is:
Ω=== 24.6
25*3
270
1
1
I
V
ZBL
The blocked-rotor reactance at 15 Hz is
( ) Ω=−= 98.38.424.6 22
BLX
Its value at 60 Hz is
Ω== 92.15
15
60
*98.3BLX
21 XXXBL ′+≅
Ω==′=∴ 96.7
2
92.15
21 XX at 60 Hz
Ω=−= 04.27396.7281mX
Ω=−=−= 28.28.41RRR BL
Ω=




 +
=′ 12.22
04.273
04.27396.7
2
2R

)c (
11 97.0
04.27396.7
04.273
VVVth =
+
≅
Ω==≅ 63.28.2*97.097.0 2
1
2
RRth
Ω=≅ 96.71XXth

Example 4.4 A three-phase, 460 V, 1740 rpm, 60 Hz, four-pole
wound-rotor induction motor has the following parameters per
phase:
1R = 0.25 Ω, 2.02 =′R Ω, 5.021 =′= XX Ω, 30=mX Ω
The rotational losses are 1700 watts. With the rotor terminals
short-circuited, find
(a) (i) Starting current when started direct on full voltage.
(ii) Starting torque.
(b) (i) Full-load slip.
(ii) Full-load current.
(iii) Ratio of starting current to full-load current.
(iv) Full-load power factor.
(v) Full-load torque.
(iv) Internal efficiency and motor efficiency at full load.
(c) (i) Slip at which maximum torque is developed.
(ii) Maximum torque developed.
(d) How much external resistance per phase should be
connected in the rotor circuit so that maximum torque occurs at
start?

%5.87100*
4.32022
3.28022
==motorη
( ) ( ) %7.96100*0333.01100*1int =−=−= sernalη
(c) (i)
(c) (ii)

Note that for parts (a) and (b) it is not necessary to use Thevenin
equivalent circuit. Calculation can be based on the equivalent circuit of
Fig.5.15 as follows:

A three-phase, 460 V, 60 Hz, six-pole wound-rotor induction motor
drives a constant load of 100 N - m at a speed of 1140 rpm when
the rotor terminals are short-circuited. It is required to reduce the
speed of the motor to 1000 rpm by inserting resistances in the
rotor circuit. Determine the value of the resistance if the rotor
winding resistance per phase is 0.2 ohms. Neglect rotational
losses. The stator-to-rotor turns ratio is unity.

Example The following test results are obtained from three
phase 100hp,460 V, eight pole star connected induction machine
No-load test : 460 V, 60 Hz, 40 A, 4.2 kW. Blocked rotor test is
100V, 60Hz, 140A 8kW. Average DC resistor between two stator
terminals is 0.152 Ω
(a)Determine the parameters of the equivalent circuit.
(b)The motor is connected to 3ϕ , 460 V, 60 Hz supply and runs
at 873 rpm. Determine the input current, input power, air
gap power, rotor cupper loss, mechanical power developed,
output power and efficiency of the motor.
(c) Determine the speed of the rotor field relative to stator
structure and stator rotating field

Solution: From no load test:
( ) Ω== 64.6
40
3/460
NLZa
Ω=== 875.0
40*3
4200
*3 22
1I
P
R NL
NL
Ω=−= 58.6875.064.6 22
NLX
Ω=+ 58.61 mXX
From blocked rotor test:
Ω== 136.0
140*3
8000
2BLR Ω== 076.0
2
152.0
1R
Ω== 412.0
140
3/100
BLZ

Ω=−= 389.0136.0412.0 22
BLX
Ω==′= 1945.0
2
389.0
21 XX
3855.61945.058.6 =−=mX
Ω=−=−= 06.0076.0136.01RRR BL
0637.006.0*
3855.6
3855.61945.0
2
2 =




 +
=′R
0.076 j0.195
j6.386
j0.195
s
0637.0
Ω Ω
Ω
Ω
389.021 =′+ XX

( ) rpm
P
f
nb s 900
8
60*120120
===
03.0
900
873900
=
−
=
−
=
s
s
n
nn
s
123.2
03.0
0637.02
==
′
s
R
Input impedance ( )( )
( )
Ω∠=
++
+
++= o
j
jj
jZ 16.27121.2
195.0386.6123.2
195.0123.2386.6
195.0076.01
o
Z
V
I 16.2722.125
16.2712.2
3/460
1
1
1 −∠=
∠
==
Input power:
( ) kWP o
in 767.8816.27cos22.125*
3
460
*3 ==

Stator CU losses:
kWPst 575.3076.0*22.125*3 2
==
Air gap power
kWPag 192.85575.3767.88 =−=
Rotor CU losses
kWsPP ag 556.2192.85*03.02 ===
Mechanical power developed:
( ) ( ) kWPsP agmech 636.82192.85*03.011 =−=−=
rotmechout PPP −=
From no load test: WRIPP NLrot 2.3835076.0*40*34200*3 2
1
2
1 =−=−=
kWPout 8.782.383510*636.82 3
=−=
%77.88100*
767.88
8.78
100* ===
in
out
P
P
η

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