Presentation on Synchronous Machine.pptx

1
Synchronous Machines
ELEN 3441 Fundamentals of Power Engineering Spring 2008

2
Construction of Synchronous Machines
Synchronous machines are AC machines that have a field
circuit supplied by an external DC source.
In a synchronous generator, a DC current is applied to the rotor winding
producing a rotor magnetic field. The rotor is then turned by external means
producing a rotating magnetic field, which induces a 3-phase voltage within
the stator winding.
In a synchronous motor, a 3-phase set of stator currents produces a
rotating magnetic field causing the rotor magnetic field to align with it. The
rotor magnetic field is produced by a DC current applied to the rotor
winding.
Field windings are the windings producing the main magnetic field (rotor
windings for synchronous machines); armature windings are the windings
where the main voltage is induced (stator windings for synchronous
machines).

3
The rotor of a synchronous machine is a large electromagnet. The magnetic poles
can be either salient (sticking out of rotor surface) or non-salient construction.
Non-salient-pole rotor: usually two- and four-pole rotors.
Salient-pole rotor: four
and more poles.
Rotors are made laminated to reduce eddy current losses.

4
Salient pole with field
windings
Salient pole without
field windings –
observe laminations
A synchronous rotor with 8 salient poles

5
Two common approaches are used to supply a DC current to the field circuits on
the rotating rotor:
1. Supply the DC power from an external
DC source to the rotor by means of
slip rings and brushes;
2. Supply the DC power from a special
DC power source mounted directly on
the shaft of the machine.
Slip rings are metal rings completely encircling the shaft of a machine but insulated
from it. One end of a DC rotor winding is connected to each of the two slip rings on
the machine’s shaft. Graphite-like carbon brushes connected to DC terminals ride on
each slip ring supplying DC voltage to field windings regardless the position or speed
of the rotor.

6
Slip rings
Brush

7
Slip rings and brushes have certain disadvantages: increased friction and wear
(therefore, needed maintenance), brush voltage drop can introduce significant
power losses. Still this approach is used in most small synchronous machines.
On large generators and motors, brushless exciters are used.
A brushless exciter is a small AC generator whose field circuits are
mounted on the stator and armature circuits are mounted on the rotor
shaft. The exciter generator’s 3-phase output is rectified to DC by a 3-
phase rectifier (mounted on the shaft) and fed into the main DC field
circuit. It is possible to adjust the field current on the main machine by
controlling the small DC field current of the exciter generator (located on
the stator).
Since no mechanical contact occurs between the rotor and the stator, exciters of
this type require much less maintenance.

8
A brushless exciter: a
low 3-phase current is
rectified and used to
supply the field circuit
of the exciter (located
on the stator). The
output of the exciter’s
armature circuit (on the
rotor) is rectified and
used
current
as the field
of the main
machine.

9
To make
of
the
a
excitation
generator completely
independent
external
of any
power
source, a small pilot
exciter is often added
to the circuit. The pilot
exciter is an AC
with a
generator
permanent magnet
mounted on the rotor
shaft and a 3-phase
winding on the stator
producing the power
for the field circuit of
the exciter.

10
A rotor
of large
synchronous machine
with a brushless exciter
mounted on the same
shaft.
Many
generators
synchronous
having
brushless exciters also
include slip rings and
provide
brushes to
emergency source of
the field DC current.

11
A large
synchronous
machine with
the exciter
and salient
poles.

12
Rotation speed of Synchronous Generator
120
e
By the definition, synchronous generators produce electricity whose
frequency is synchronized with the mechanical rotational speed.
f 
nmP
(7.11.1)
Where fe is the electrical frequency, Hz;
nm is mechanical speed of magnetic field (rotor speed for synchronous
machine), rpm;
P is the number of poles.
Steam turbines are most efficient when rotating at high speed; therefore,
to generate 60 Hz, they are usually rotating at 3600 rpm and turn 2-pole
generators.
Water turbines are most efficient when rotating at low speeds (200-300
rpm); therefore, they usually turn generators with many poles.

13
Internal generated voltage of a
Synchronous Generator
The magnitude of internal generated voltage induced in a given stator is
EA  2NC f  K
where K is a constant representing the construction of the machine,  is flux in it
and  is its rotation speed.
Since flux in the
machine depends
on the field current
it, the
generated
through
internal
voltage
function
is a
of the
rotor field current.
Magnetization curve (open-circuit characteristic) of a
synchronous machine

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Equivalent circuit of a
Synchronous Generator
The internally generated voltage in a single phase of a
synchronous machine EA is not usually the voltage appearing
at its terminals. It equals to the output voltage V only when
there is no armature current in the machine. The reasons
that the armature voltage EA is not equal to the output
voltage V are:
1. Distortion of the air-gap magnetic field caused by the
current flowing in the stator (armature reaction);
2. Self-inductance of the armature coils;
3. Resistance of the armature coils;
4. Effect of salient-pole rotor shapes.

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Equivalent circuit of a Synchronous
Generator
Assuming that the load reactance is X, the armature reaction voltage is
Estat   jXIA (7.17.1)
The phase voltage is then V  EA  jXIA (7.17.2)
Armature reactance can be modeled by the following
circuit…
However, in addition to armature reactance effect,
the stator coil has a self-inductance LA (XA is the
corresponding reactance) and the stator has
resistance RA. The phase voltage is thus
V  EA  jXIA  jX AIA  RIA (7.17.3)

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Generator
Often, armature reactance and self-inductance are combined into the synchronous
reactance of the machine:
(7.18.1)
V  EA  jXS IA  RIA
XS  X  XA
Therefore, the phase voltage is
(7.18.2)
The equivalent circuit of a 3-phase
synchronous generator is shown.
The adjustable resistor Radj controls the
field current and, therefore, the rotor
magnetic field.

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Generator
A synchronous generator can be Y- or -connected:
The terminal voltage will be
VT  3V  for Y
VT V  for 
(7.19.1) (7.19.2)

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Equivalent circuit of a synchronous
generator
Note: the discussion above assumed a balanced load on the generator!
Since – for balanced loads – the three phases of a synchronous generator are
identical except for phase angles, per-phase equivalent circuits are often used.

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Phasor diagram of a Synchronous
Generator
Since the voltages in a synchronous generator are AC voltages, they are usually
one phase is
expressed as phasors. A vector plot of voltages and currents within
called a phasor diagram.
A phasor diagram of a synchronous generator
with a unity power factor (resistive load)
Lagging power factor (inductive load): a larger
than for leading PF internal generated voltage
EA is needed to form the same phase voltage.
Leading power factor (capacitive load).
For a given field current and magnitude of
load current, the terminal voltage is lower for
lagging loads and higher for leading loads.

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torque in Synchronous
Power &
Generators
A synchronous generator needs to be connected to a prime mover whose speed is
reasonably constant (to ensure constant frequency of the generated voltage) for
various loads.
The applied mechanical power
Pin appm (7.22.1)
(7.22.2)
is partially converted to electricity
Pconv indm  3EAIA cos
Where  is the angle between
EA and IA.
The power-flow diagram of a
synchronous generator.

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Power &
Generators
The real output power of the synchronous generator is
Pout  3VTIL cos  3VIA cos
The reactive output power of the synchronous generator is
Qout  3VTIL sin  3VIA sin
(7.23.1)
(7.23.2)
Recall that the power factor angle  is the angle between V and IA and not the
angle between VT and IL.
In real synchronous machines of any size, the
armature resistance RA << XS and, therefore,
the armature resistance can be ignored. Thus,
a simplified phasor diagram indicates that
S
X
EA sin
IA cos  (7.23.3)

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Power &
Generators
Then the real output power of the synchronous generator can be approximated as
out
S
P
X

3VEA sin
(7.24.1)
We observe that electrical losses are assumed to be zero since the resistance is
neglected. Therefore:
(7.24.2)
Pconv  Pout
Here  is the torque angle of the machine – the angle between V and EA.
The maximum power can be supplied by the generator when  = 900:
S
X
3VEA
Pmax  (7.24.3)

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torque in synchronous
Power &
generators
The maximum power specified by (7.24.3) is called the static stability limit
of the generator. Normally, real generators do not approach this limit: full-
load torque angles are usually between 150 and 200.
The induced torque is
ind  kBR  BS  kBR  Bnet  kBR Bnet sin (7.25.1)
Notice that the torque angle  is also the angle between the rotor magnetic field
BR and the net magnetic field Bnet.
Alternatively, the induced torque is
ind
m S

 X

3VEA sin
(7.25.2)

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Measuring parameters of
Synchronous Generator Model
The three quantities must be determined in order to describe the generator model:
1. The relationship between field current and flux (and therefore between the field
current IF and the internal generated voltage EA);
2. The synchronous reactance;
3. The armature resistance.
We conduct first the open-circuit test on the synchronous generator: the generator
is rotated at the rated speed, all the terminals are disconnected from loads, the
field current is set to zero first. Next, the field current is increased in steps and the
phase voltage (whish is equal to the internal generated voltage EA since the
armature current is zero) is measured.
Therefore, it is possible to plot the dependence of the internal generated voltage
on the field current – the open-circuit characteristic (OCC) of the generator.

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Synchronous Generator model
Since the unsaturated core of the machine has a
reluctance thousands times lower than the
reluctance of the air-gap, the resulting flux
increases linearly first. When the saturation is
reached, the core reluctance greatly increases
causing the flux to increase much slower with
the increase of the mmf.
We conduct next the short-circuit test on the synchronous generator: the generator
is rotated at the rated speed, all the terminals are short-circuited through
ammeters, the field current is set to zero first. Next, the field current is increased in
steps and the armature current IA is measured as the field current is increased.
The plot of armature current (or line current) vs. the field current is the short-circuit
characteristic (SCC) of the generator.

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The SCC is a straight line since, for the
short-circuited terminals, the magnitude of
the armature current is
A S
EA
R2
IA 
 X 2
(7.28.1)
The equivalent generator’s circuit during SC
The resulting
phasor diagram
The magnetic
fields during
short-circuit test
Since BS almost cancels BR, the
net field Bnet is very small.

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An approximate method to determine the synchronous reactance XS at a given
field current:
1. Get the internal generated voltage EA from the OCC at that field current.
2. Get the short-circuit current IA,SC at that field current from the SCC.
3. Find XS from
EA
XS 
I A,SC
Since the internal machine impedance is
2 2
S
EA
IA,SC
ZS  RA  X   XS since XS ? RA
(7.29.1)
(7.29.2)

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A drawback of this method is that the internal generated voltage EA is measured
during the OCC, where the machine can be saturated for large field currents, while
the armature current is measured in SCC, where the core is unsaturated.
Therefore, this approach is accurate for unsaturated cores only.
The approximate value of synchronous
reactance varies with the degree of
saturation of the OCC.
Therefore, the value of the synchronous
reactance for a given problem should be
estimated at the approximate load of the
machine.
The winding’s resistance can be
approximated by applying a DC voltage
to a stationary machine’s winding and
measuring the current. However, AC
resistance is slightly larger than DC
resistance (skin effect).

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Synchronous Generator model: Ex
Example 7.1: A 200 kVA, 480 V, 50 Hz, Y-connected synchronous generator with a
rated field current of 5 A was tested and the following data were obtained:
1. VT,OC = 540 V at the rated IF.
2. IL,SC = 300 A at the rated IF.
3. When a DC voltage of 10 V was applied to two of the terminals, a current of 25 A
was measured.
Find the generator’s model at the rated conditions (i.e., the armature resistance and
the approximate synchronous reactance).
Since the generator is Y-connected, a DC
voltage was applied between its two
phases. Therefore:
10
A
DC
A
DC
2R
I
R
2I

VDC

VDC
  0.2 
225

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Synchronous Generator model: Ex
The internal generated voltage at the rated field current is
3 3
A ,OC
E V 
VT

540
 311.8V
The synchronous reactance at the rated field current is precisely
2 2 2 2
311.82
3002
A
A A
A,SC
E2
I2
XS  ZS  R   R   0.2 1.02 
We observe that if XS was estimated via the approximate formula, the result would
be:
S
A,SC
EA
I 300
X  
311.8
 1.04 
Which is close to the previous result.
The error ignoring RA is much smaller
than the error due to core saturation.
The equivalent circuit

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The Synchronous Generator operating
alone
The behavior of a synchronous generator varies greatly under
load depending on the power factor of the load and on
whether the generator is working alone or in parallel with other
synchronous generators.
Although most of the synchronous generators in the world
operate as parts of large power systems, we start our
discussion assuming that the synchronous generator works
alone.
Unless otherwise stated, the speed of the generator is
assumed constant.

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The Synchronous Generator
operating alone
Effects of load changes
A increase in the load is an
increase in the real and/or
reactive power drawn from the
generator.
Since the field resistor is unaffected, the field current is constant and, therefore, the
flux  is constant too. Since the speed is assumed as constant, the magnitude of
the internal generated voltage is constant also.
Assuming the same power factor of the load, change in load will change the
magnitude of the armature current IA. However, the angle will be the same (for a
constant PF). Thus, the armature reaction voltage jXSIA will be larger for the
increased load. Since the magnitude of the internal generated voltage is constant
(7.34.1)
EA V  jXS IA
Armature reaction voltage vector will “move parallel” to its initial position.

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operating alone
Increase load effect on generators with
Lagging PF
Leading PF
Unity PF

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operating alone
Generally, when a load on a synchronous generator is added, the following
changes can be observed:
1. For lagging (inductive) loads, the phase (and terminal) voltage
decreases significantly.
2. For unity power factor (purely resistive) loads, the phase (and
terminal) voltage decreases slightly.
3. For leading (capacitive) loads, the phase (and terminal) voltage rises.
Effects of adding loads can be described by the voltage regulation:
VR 
Vnl Vfl
100%
Vfl
(7.36.1)
Where Vnl is the no-load voltage of the generator and Vfl is its full-load voltage.

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operating alone
A synchronous generator operating at a lagging power factor has a fairly large
positive voltage regulation. A synchronous generator operating at a unity power
factor has a small positive voltage regulation. A synchronous generator operating
at a leading power factor often has a negative voltage regulation.
Normally, a constant terminal voltage supplied by a generator is desired. Since the
armature reactance cannot be controlled, an obvious approach to adjust the
terminal voltage is by controlling the internal generated voltage EA = K. This
may be done by changing flux in the machine while varying the value of the field
resistance RF, which is summarized:
1. Decreasing the field resistance increases the field current in the generator.
2. An increase in the field current increases the flux in the machine.
3. An increased flux leads to the increase in the internal generated voltage.
4. An increase in the internal generated voltage increases the terminal voltage of
the generator.
Therefore, the terminal voltage of the generator can be changed by adjusting the
field resistance.

49
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.

50
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 both generators must be equal.
4. The frequency of the oncoming generator must be slightly higher than the
frequency of the running system.

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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.

59
Generators in parallel with other
generators of the same size
When a generator is working alone, its real and reactive power are fixed and
determined by the load.
When a generator is connected to an infinite bus, its frequency and the terminal
voltage are constant and determined by a bus.
When two generators of the same size
are connected to the same load, the
sum of the real and reactive powers
supplied by the two generators must
equal the real and reactive powers
demanded by the load:
Ptot  Pload
Qtot  Qload
 PG1  PG2
 QG1 QG2
(7.59.1)
(7.59.2)

68
Synchronous motors
The field current
produces
a steady-state
IF of the motor
rotor
magnetic field BR. A 3-phase set of
voltages applied to the stator produces
a 3-phase current flow in the windings.
A 3-phase set of currents in an
armature winding produces a uniform
rotating magnetic field Bs.
Two magnetic fields are present in the machine, and the rotor field tends to align
with the stator magnetic field. Since the stator magnetic field is rotating, the rotor
magnetic field will try to catch up pulling the rotor.
The larger the angle between two magnetic fields (up to a certain maximum), the
greater the torque on the rotor of the machine.

69
Synchronous motor equivalent
circuit
A synchronous motor has the same
equivalent circuit as synchronous
generator, except that the direction of
power flow (and the direction of IA) is
reversed. Per-phase circuit is shown:
A change in direction of IA changes the Kirchhoff’s voltage law equation:
EA V  jXS IA  RAIA
V  EA  jXS IA  RAIA
Therefore, the internal generated voltage is
We observe that this is exactly the same equation as the equation for the generator,
except that the sign on the current terms is reversed.
(7.69.1)
(7.69.2)

70
Synchronous motor vs.
synchronous generator
Let us
diagram
suppose that a phasor
of synchronous
generator is shown. BR produces
Bnet
EA, produces V, and BS
Estat = -jXSIA. The
produces
rotation on both
counterclockwise
induced torque is
diagrams is
and the
ind  kBR  Bnet (7.70.1)
clockwise, opposing the direction of rotation. In other words, the induced torque in
generators is a counter-torque that opposes the rotation caused by external torque.
If the prime mover loses power, the rotor will slow down and the rotor field BR will
fall behind the magnetic field in the machine Bnet. Therefore, the operation of the
machine changes…

72
Steady-state operation of motor:
Torque-speed curve
Usually, synchronous motors are connected to large power systems (infinite bus);
therefore, their terminal voltage and system frequency are constant regardless
the motor load. Since the motor speed is locked to the electrical frequency, the
speed should be constant regardless the load.
The steady-state speed of the motor is
constant from no-load to the maximum torque
that motor can supply (pullout torque).
Therefore, the speed regulation of
synchronous motor is 0%.
The induced torque is
ind  kBR Bnet sin
or
ind
m S
 X
 
3VEA
sin
(7.72.1)
(7.72.2)

73
Torque-speed curve
The maximum pullout torque occurs when  = 900:
Normal full-load torques are much less than that (usually, about 3 times smaller).
When the torque on the shaft of a synchronous motor exceeds the pullout torque,
the rotor can no longer remain locked to the stator and net magnetic fields. It starts
to slip behind them. As the motor slows down, the stator magnetic field “laps” it
repeatedly, and the direction of the induced torque in the rotor reverses with each
pass. As a result, huge torque surges of alternating direction cause the motor
vibrate severely. The loss of synchronization after the pullout torque is exceeded is
known as slipping poles.
max
m S
3VEA
 X
  kBRBnet  (7.73.1)

74
Effect of torque changes
Assuming that a synchronous motor operates
initially with a leading PF.
If the load on the motor increases, the rotor
initially slows down increasing the torque angle
. As a result, the induced torque increases
speeding up the rotor up to the synchronous
speed with a larger torque angle .
Since the terminal voltage and frequency
supplied to the motor are constant, the
magnitude of internal generated voltage
must be constant at the load changes
(EA = K and field current is constant).

75
Effect of torque changes
Assuming that the armature resistance is negligible, the power converted from
electrical to mechanical form in the motor will be the same as its input power:
S
X
3VEA
sin
P  3VIA cos  (7.73.1)
Since the phase voltage is constant, the quantities IAcos and EAsin are directly
proportional to the power supplied by (and to) the motor. When the power
supplied by the motor increases, the distance proportional to power increases.
Since the internal generated voltage is
constant, its phasor “swings down” as
load increases. The quantity jXSIA has to
increase; therefore, the armature current
IA increases too.
Also, the PF angle changes too moving
from leading to lagging.

76
Effect of field current changes
Assuming that a synchronous motor operates
initially with a lagging PF.
If, for the constant load, the field current on the
motor increases, the magnitude of the internal
generated voltage EA increases.
Since changes in IA do not affect the shaft
speed and the motor load is constant, the
real power supplied by the motor is
unchanged. Therefore, the distances
proportional to power on the phasor
diagram (EAsin and IAcos) must be
constant.
Notice that as EA increases, the magnitude of the armature current IA first
decreases and then increases again. At low EA, the armature current is lagging and
the motor is an inductive load that consumes reactive power Q. As the field current
increases , IA eventually lines up with V, and the motor is purely resistive. As the
field current further increases, IA becomes leading and the motor is a capacitive
load that supplies reactive power Q to the system (consumes –Q).

77
A plot of armature current vs. field current is
called a synchronous motor V curve. V
curves for different levels of real power
have their minimum at unity PF, when only
real power is supplied to the motor. For field
currents less than the one giving the
minimum IA, the armature current is lagging
and the motor consumes reactive power.
For field currents greater than the one
giving the minimum IA, the armature current
is leading and the motor supplies reactive
power to the system.
Therefore, by controlling the field current of a synchronous
motor, the reactive power consumed or supplied to the power
system can be controlled.

78
When the projection of the phasor EA onto
V (EAcos) is shorter than V, a
synchronous motor has a lagging current
and consumes Q. Since the field current is
small in this situation, the motor is sais to
be under-excited.
When the projection of the phasor EA
onto V (EAcos) is longer than V, a
synchronous motor has a leading
current and supplies Q to the system.
Since the field current is large in this
situation, the motor is sais to be over-
excited.

79
power factor correction
Assuming that a load contains a
synchronous motor (whose PF
can be adjusted) in addition to
motors of other types. What
does the ability to set the PF of
one of the loads do for the
power system?
Let us consider a large power system operating at 480 V. Load 1 is an induction
motor consuming 100 kW at 0.78 PF lagging, and load 2 is an induction motor
consuming 200 kW at 0.8 PF lagging. Load 3 is a synchronous motor whose real
power consumption is 150 kW.
a. If the synchronous motor is adjusted to 0.85 PF lagging, what is the line current?
b. If the synchronous motor is adjusted to 0.85 PF leading, what is the line current?
c. Assuming that the line losses are PLL = 3IL
2RL, how du these losses compare in
the two cases?

80
a. The real power of load 1 is 100 kW, and the reactive power of load 1 is
 
1 1
1
Q  P tan 100tan cos 0.78  80.2 kVAR
The real power of load 2 is 200 kW, and the reactive power of load 2 is
 
2 2
1
Q  P tan  200tan cos 0.8 150 kVAR
The real power of load 3 is 150 kW, and the reactive power of load 3 is
 
1
Q  P tan 150tan cos 0.85  93kVAR
3 3
The total real load is Ptot  P
1  P2  P3 100 200 150  450 kW
The total reactive load is Qtot  Q1  Q2 Q3  80.2 150  93  kVAR
The equivalent system PF is
 
Q  323.2 
PF  cos  cos tan1
  cos tan1
  0.812 lagging
  450 
The line current is
P 
450 000
L
L
I
P
3V cos
 667 A
3 4800.812
 tot


81
b. The real and reactive powers of loads 1 and 2 are the same. The reactive power
of load 3 is
 
3 3 93kVAR
1
Q  P tan 150tan cos 0.85  
The total real load is Ptot  P
1  P2  P3 100 200 150  450 kW
The total reactive load is Qtot  Q1 Q2 Q3  80.2150 93 kVAR
The equivalent system PF is
137.2
Q
1 1
   
PF  cos  cos tan  cos tan  0.957 lagging
 P  
450 
   
The line current is
450 000
L
L
I
P
3V cos
 566 A
3 4800.957
 tot


82
c. The transmission line losses in the first case are
P  3I 2
R 1344 700 R
LL L L L
The transmission line losses in the second case are
P  3I 2
R  96170 R
LL L L L
We notice that the transmission power losses are 28% less in the second
case, while the real power supplied to the loads is the same.

83
The ability to adjust the power factor of one or more loads in a power system can
significantly affect the efficiency of the power system: the lower the PF, the greater
the losses in the power lines. Since most loads in a typical power system are
induction motors, having one or more over-excided synchronous motors (leading
loads) in the system is useful for the following reasons:
1. A leading load supplies some reactive power to lagging loads in the system.
Since this reactive power does not travel along the transmission line,
transmission line current is reduced reducing power losses.
2. Since the transmission line carries less current, the line can be smaller for a
given power flow reducing system cost.
3. The over-excited mode of synchronous motor increases the motor’s maximum
torque.
Usage of synchronous motors or other equipment increasing the overall system’s
PF is called power-factor correction. Since a synchronous motor can provide PF
correction, many loads that can accept constant speed are driven by over-excited
synchronous motors.

84
Starting synchronous motors
Consider a 60 Hz synchronous motor.
When the power is applied to the stator windings, the rotor (and,
therefore its magnetic field BR) is stationary. The stator magnetic field
BS starts sweeping around the motor at synchronous speed.
Note that the induced torque on the shaft
ind  kBR  BS (7.84.1)
is zero at t = 0 since both magnetic fields are aligned.
At t = 1/240 s the rotor has barely moved but the stator
magnetic field BS has rotated by 900. Therefore, the torque
on the shaft is non-zero and counter-clockwise.

85
At t = 1/120 s the rotor and stator magnetic fields point in opposite
directions, and the induced torque on the shaft is zero again.
At t = 3/240 s the stator magnetic fields point to the
right, and the induced torque on the shaft is non-
zero but clockwise.
Finally, at t = 1/60 s the rotor and stator magnetic fields are aligned
again, and the induced torque on the shaft is zero.
During one electrical cycle, the torque was counter-clockwise
and then clockwise, and the average torque is zero. The
motor will vibrate heavily and finally overheats!

86
Three basic approaches can be used to safely start a synchronous motor:
1. Reduce the speed of the stator magnetic field to a low enough value
that the rotor can accelerate and two magnetic fields lock in during one
half-cycle of field rotation. This can be achieved by reducing the
frequency of the applied electric power (which used to be difficult but
can be done now).
2. Use an external prime mover to accelerate the synchronous motor up
to synchronous speed, go through the paralleling procedure, and bring
the machine on the line as a generator. Next, turning off the prime
mover will make the synchronous machine a motor.
3. Use damper windings or amortisseur windings – the most popular.

87
Motor starting by amortisseur or
damper windings
Amortisseur (damper) windings are special bars
laid into notches carved in the rotor face and then
shorted out on each end by a large shorting ring.

91
Motor starting by amortisseur or
damper windings
We observe that the torque is either counter-clockwise or zero, but it is always
unidirectional. Since the net torque is nonzero, the motor will speed up.
However, the rotor will never reach the synchronous speed! If a rotor was running
at the synchronous speed, the speed of stator magnetic field BS would be the same
as the speed of the rotor and, therefore, no relative motion between the rotor and
the stator magnetic field. If there is no relative motion, no voltage is induced and,
therefore, the torque will be zero.
Instead, when the rotor’s speed is close to synchronous, the regular field current
can be turned on and the motor will operate normally. In real machines, field circuit
are shorted during starting. Therefore, if a machine has damper winding:
1. Disconnect the field windings from their DC power source and short them out;
2. Apply a 3-phase voltage to the stator and let the rotor to accelerate up to near-
synchronous speed. The motor should have no load on its shaft to enable motor
speed to approach the synchronous speed as closely as possible;
3. Connect the DC field circuit to its power source: the motor will lock at
synchronous speed and loads may be added to the shaft.

92
Relationship between synchronous
generators and motors
Synchronous generator and synchronous
motor are physically the same machines!
A synchronous machine can supply real
power to (generator) or consume real
power (motor) from a power system. It
can also either consume or supply
reactive power to the system.
1. The distinguishing characteristic of a
synchronous generator (supplying P)
is that EA lies ahead of V while for a
motor EA lies behind V.
2. The distinguishing characteristic of a
machine supplying reactive power Q
is that Eacos > V (regardless
whether it is a motor or generator).
The machine consuming reactive
power Q has Eacos < V .

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