Unit 2

Synchronous Motor

INTRODUCTION
A synchronous machine is an ac rotating machine whose speed under steady
state condition is proportional to the frequency of the current in its armature. The
magnetic field created by the armature currents rotates at the same speed as that
created by the field current on the rotor, which is rotating at the synchronous
speed, and a steady torque results.
Synchronous machines are commonly used as generators especially for large
power systems, such as turbine generators and hydroelectric generators in the grid
power supply. Because the rotor speed is proportional to the frequency of
excitation, synchronous motors can be used in situations where constant speed
drive is required. Since the reactive power generated by a synchronous machine
can be adjusted by controlling the magnitude of the rotor field current, unloaded
synchronous machines are also often installed in power systems solely for power
factor correction or for control of reactive kVA flow. Such machines, known as
synchronous condensers, may be more economical in the large sizes than static
capacitors.
With power electronic variable voltage variable frequency (VVVF) power
supplies, synchronous motors, especially those with permanent magnet rotors, are
widely used for variable speed drives. If the stator excitation of a permanent
magnet motor is controlled by its rotor position such that the stator field is always
90o (electrical) ahead of the rotor, the motor performance can be very close to the
conventional brushed dc motors, which is very much favored for variable speed
drives. The rotor position can be either detected by using rotor position sensors or
deduced from the induced emf in the stator windings. Since these types of motors
do not need brushes, they are known as brushless dc motors.
In this chapter, we concentrate on conventional synchronous machines whereas
the brushless dc motors will be discussed later in a separate chapter.

Principle of operation of the synchronous motor

➢ In order to understand the principle of operation of a synchronous
motor, let us examine what happens if we connect the armature winding
(laid out in the stator) of a 3- phase synchronous machine to a suitable
balanced 3-phase source and the field winding to a D.C source of
appropriate voltage.
➢ The current flowing through the field coils will set up stationary magnetic
poles of alternate North and South. (For convenience let us assume a salient
pole rotor, as shown in figure below). On the other hand, the 3-phase
currents flowing in the armature winding produce a rotating magnetic field
rotating at synchronous speed.
➢ In other words there will be moving North and South poles established in
the stator due to the 3- phase currents i.e at any location in the stator there
will be a North Pole at some instant of time and it will become a South
Pole after a time period corresponding to half a cycle. (After a time = 1/2f,
where f = frequency of the supply).
➢ Let us assume that the stationary South Pole in the rotor is aligned with the
North Pole in the stator moving in clockwise direction at a particular instant
of time, as shown in figure 2.1below.
 Fig 2.1 Principle of Operation of Synchronous Motor

➢ These two poles get attracted and try to maintain this alignment (as per
Lenz’s law) and hence the rotor pole tries to follow the stator pole as the
conditions are suitable for the production of t o r q u e in the clockwise
direction.
➢ However t h e r o t o r c a n n o t move instantaneously due to its mechanical
inertia, and so it needs some time to move. In the mean time, the stator pole
would quickly (a time duration corresponding to half a cycle) change its
polarity and becomes a South Pole.
➢ So the force of attraction will no longer be present and instead the like
poles experience a force of repulsion as shown in the figure 2.1. In other
words, the conditions are now suitable for the production of torque in the
production of torque in anticlockwise direction anticlockwise direction.
➢ Even this condition will not last longer as the stator pole would again
change to North Pole after a time of ½ f. Thus the rotor will experience an
alternating force which tries to move it clockwise and anticlockwise at
twice the frequency of the supply, i.e. at intervals corresponding to ½ f
 seconds. As this duration is quite small compared to the mechanical time
constant of the rotor, the rotor cannot respond and move in any direction.
The rotor continues to be stationary only.
➢ On the contrary if the rotor is brought to near synchronous speed by some
external means say a small motor (known as pony motor-which could be a
D.C or AC induction rotor) mounted on the same shaft as that of the rotor,
the rotor poles get locked to the unlike poles in the stator and the rotor
continues to run at the synchronous speed even if the supply to the pony
motor is disconnected.
➢ Thus the synchronous rotor cannot start rotating on its own or usually we
say that the synchronous rotor has no starting torque.
➢ So, some special provision has to be made either inside the machine or
outside of the machine so that the rotor is brought to near about its
synchronous speed.
➢ At that time, if the armature is supplied with electrical power, the rotor can
pull into step and continue to operate at its synchronous speed. Some of the
commonly used methods for starting synchronous rotor are described in the
following section.
Torque equation of a synchronous motor
When a synchronous machine is operated as a generator, a prime mover is
required to drive the generator. In steady state, the mechanical torque of the
prime mover should balance with the electromagnetic torque produced by the
generator and the mechanical loss torque due to friction and windage, or
Tpm = T + Tlosses
Multiplying the synchronous speed to both sides of the torque equation,
we have the power balance equation as
 Ppm=Pem+Plosses

Where Ppm=Tpmwsyn is the mechanical power supplied by the prime

mover, Pem=Twsynthe electromagnetic power of the generator, and Ploss=Tlosswsyn
the mechanical power loss of the system. The electromagnetic power is the
power being converted into the electrical power in the three phase stator
windings. That is
Pem=Tωsyn= 3EaIacosφEaIa

where EaIa is the angle between phasor Ea and Ia.

Fig
2.2. A synchronous machine operated as generator
For l a rg e r s yn c h ro no us g en e r at o rs , the wi nd in g resistance is
generally much smaller than the synchronous reactance, and thus the per
phase Circuit equation can be approximately written as

Va = Ea – jXsIa
 2.3. Phasor diagram for synchronous motor
The corresponding phasor diagram is shown in fig 2.3 right hand side.
From the phasor diagram, we can readily obtain
Easin = XsIscosφ
When the phase winding resistance is ignored, the output electrical
power equals the electromagnetic power, or

Pem= Pout = 3VaIacosφ

Therefore
3𝑉𝑎 𝐼𝑎
𝑃𝑒𝑚 = sin 𝛿
𝑋𝑠
and
𝑃𝑒𝑚 3𝑉𝑎 𝐼𝑎
𝑇= sin 𝛿 = sin 𝛿
𝜔𝑠𝑦𝑛 𝜔𝑠𝑦𝑛 𝑋𝑠

Fig 2.4 working of synchronous motor

 Where d is the angle between the phasor of the voltage and the emf, known
as the load angle. When the stator winding resistance is ignored, d can also be
regarded as the angle between the rotor and stator rotating magnetic fields.
The electromagnetic torque of a synchronous machine is proportional to the sine
function of the load angle, as plotted in the diagram above, where the curve in the
third quadrant is for the situation when the machine is operated as a motor, where
the electromagnetic torque is negative because the armature current direction is
reversed.

Starting Methods of synchronous motor and explain?

Basically there are three methods that are used to start a synchronous motor:
➢ To reduce the speed of the rotating magnetic field of the stator to a low
enough value that the rotor can easily accelerate and lock in with it
during one half-cycle of the rotating magnetic field’s rotation. This is
done by reducing the frequency of the applied electric power.
➢ This method is usually followed in the case of inverter-fed synchronous
motor operating under variable speed drive applications.
➢ To use an external prime mover to accelerate the rotor of synchronous motor
near to its synchronous speed and then supply the rotor as well as stator.
➢ Of course care should be taken to ensure that the direction of rotation of the
rotor as well as that of the rotating magnetic field of the stator is the same.
➢ This method is usually followed in the laboratory- the synchronous machine
is started as a generator and is then connected to the supply mains by
following the synchronization or paralleling procedure. Then the power
supply to the prime mover is disconnected so that the synchronous machine
will continue to operate as a motor.
➢ To use damper windings or amortisseur windings if these are provided in
the machine.
➢ The damper windings or amortisseur windings are provided in most of the
large synchronous motors in order to nullify the oscillations of the rotor
whenever the synchronous machine is subjected to a periodically varying
load.

Connecting a synchronous machine to a infinite bus bar.

➢ In order to simplify the ideas as much as possible the resistance of the

generator will be neglected; in practice this assumption is usually reasonable.
Figure 2.5 (a) shows the schematic diagram of a machine connected to an
infinite bus bar along with the corresponding phasor diagram.

Figure 2.5

➢ If losses are neglected the power output from the turbine is equal to the
power output from the generator. The angle  between the E and V phasors
is known as the load angle is dependent on the power input from the turbine
shaft.
➢ With an isolated machine supplying its own load the latter dictates the power
 required and hence the load angle.
➢ when connected to an infinite-busbar system, however, the load delivered by
the machine is no longer directly dependent on the connected load.
➢ By changing the turbine output and hence  the generator can be made to take
on any load the operator desires, subject to economic and technical limits.

From the phasor diagram in Figure 2.5 (b), the power delivered to the
infinite busbar = VI cos  per phase but,

E IX s

sin(90   ) sin 



hence 

E
I cos   sin 
Xs

VE
 sin 
 power delivered
Xs
This expression is of extreme importance as it governs to a large extent the
operation of a power system.
➢ Equation is shown plotted in Figure 2.6. The maximum power is obtained
at  = 90o.

➢ If  becomes larger than 90o due to an attempt to obtain more than

Pmax, increase in  results in less power output and the machine
becomes unstable and loses synchronism.
➢ Loss of synchronism results in the interchange of current surges
between the generator and network as the poles of the machine pull into
synchronism and then out again.

Figure 2.6

➢ If the power output of the generator is increased by small increments with the

no- load voltage kept constant, the limited of stability occurs at  = 90o and
is known as the steady-state stability limit.
➢ There is another limit of stability due to a sudden large change in
conditions such as caused by a fault, known as the transient stability

limit, and it is possible for the rotor to oscillate beyond 90o a number of
times.
➢ If these oscillations diminish, the machine is stable. The load angle  has a
physical significance; it is the angle between like radial marks on the end of
the rotor shaft of the machine and on an imaginary rotor representing the
system.
➢ The marks are in identical physical positions when the machine is on no-
load. The synchronizing power coefficient = dP/d watts per radian and
the synchronizing torque coefficient = (1/ws)/(dP/d).

➢ In figure 2.7(a) the phasor diagram for the limiting steady-state condition is
shown. It should be noted that in this condition current is always leading.
The following figures, 2.7(b), (c) and (d), show the phasor diagrams for
various operational conditions.

➢ Another interesting operating condition is variable power and constant

excitation. This is shown in Figure 4.
➢ In this case as V and E are constant when the power from the turbine is
increased  must increase and the power factor changes.
➢ It is convenient to summarize the above types of an operation in a single
diagram or chart which will enable an operator to see immediately whether
the machine is operating within the limits of stability and rating.
Figure 2.7
 Figure 2.8

The performance chart of a synchronous generator

Consider figure 2.9(a), the phasor diagram for a round-rotor machine ignoring
resistance. The locus of constant IXs, I, and hence MVA is a circle and the locus of
constant E a circle. Hence,

0s is proportional to VI or MVA
ps is proportional to VI sin  or MVAr
sq is proportional to VI cos or MW

To obtain the scaling factor for MVA, MVAr and MW the fact that at zero

excitation, E = 0 and IXs = V, is used, from which I is V/Xs at 90o leading to 00’,
corresponding to VAr/phase.

Figure 2.9(b) represents the construction of a chart for a 60 MW machine.

Figure 2.9
 Machine data 60 MW, 0.8 pf, 75 MVA
11.8 kV, SCR 0.63, 3000
rev/min
Maximum exciter current
500 A

The chart will refer to complete three-phase values of MW and MAVr.

When the excitation and hence E are reduced to zero, the current leads V by 90o
and is equal to (V/Xs), ie 11,800/3 x 2.94. The leading vars correspond to this =

11,8002/2.94 = 47 MVAr.
➢ With centre 0 a number of semicircles are drawn of radii equal to various
MVA loadings, the most important being the 75 MVA circle.
➢ Arcs with 0’ as centre are drawn with various multiples of 00’ (or V) as radii
to give the loci for constant excitation. Lines may also be drawn from 0
corresponding to various power factors, but for clarity only 0.8 pf lagging is
shown.
➢ The operational limits are fixed as follows. The rated turbine output gives a
60MW limit which is drawn as shown, ie line efg, which meets the 75 MVA
line in g. The MVA arc governs the thermal loading of the machine, ie the
stator temperature rise.
➢ so that over portion gh the output is decided by the MVA rating. At point h the
rotor heating becomes more decisive and the arc hj is decided by the
maximum excitation current allowable, in this case assumed to be 2.5 p.u.
The remaining limit is that governed by loss of synchronism at leading
power factors. The theoretical limit is the line perpendicular to 00'’at 0'’(ie 
 = 90o), but in practice a safety margin is introduced to allow a further
increase in load of either 10 or 20 per cent before instability.
➢ In Figure 5 a 10 per cent margin is used and is represented by ecd: it
is constructed in the following manner.
➢ Considering point ‘a’ on the theoretical limit on the E = 1 p.u. arc, the power
0’a is reduced by 10 per cent of the rated power (ie by 6MW) to 0’b; the
operating point must, however, still be on the same E arc and b is projected
to c which is a point on the new limiting curve.
➢ This is repeated for several excitations giving finally the curve ecd.
The complete operating limit is shown shaded and the operator should
normally work within the area bounded by this line.

As an example of the use of the chart, the full-load operating point g (60
MW, 0.8 p.f.lagging) will require an excitation E of 2.3 p.u. and the measured

load angle  is 33o. This can be checked by using, power = VE/Xs sin , ie

11,8002 ×2.3
60×106 = sin 𝛿
2.94

From which

=33.4
 1. Name the important characteristics of a synchronous motor not found in
an induction motor. (QU: V and Inverted V curves) (8) [APL/MAY- 2008]

➢ V curve is the graph showing the relation of armature current as a function of

field current in synchronous machines.
➢ The purpose of the curve is to show the variation in the magnitude of the
armature current as the excitation voltage of the machine is varied.

2.10 V curves
➢ The synchronous motor “V Curves” shown above illustrate the effect of
excitation (field amps) on the armature (stator) amps and on system power
factor.
➢ There are separate “V” Curves for No-Load and Full-Load and sometimes the
motor manufacturer publishes curves for 25%, 50%, and 75% load.
➢ Note that the Armature Amperage and Power Factor “V” Curves are actually
inverted “V’s”.
➢ Assume it is desired to determine the field excitation which will produce unity
power factor operation at full motor load.
➢ Project across from the unity power factor (100%) operating point on the Y-
Axis to the peak of the inverted Power Factor “V” Curve (blue line). From this
intersection, project down (red line) from the full-load unity power factor
(100%) operating point to determine the required field current on the X-Axis.
➢ In this example the required DC field current is shown to be just over 10 amps.
Note at unity power factor operation the armature (stator) full-load amps are at
the minimum value.
➢ Increasing the field amps above the value required for unity power factor
operation will cause the machine to run with a leading power factor, while
field weakening caused the motor power factor to become lagging.
➢ When the motor runs either leading or lagging, the armature (stator) amps
increases above the unity power factor value.

2.11 Inverted V curves

Expression for the power developed by the synchronous motor when
connected to infinite bus.
Consider an under-excited star-connected synchronous motor driving
a mechanical load. Fig. (2.12 (i)) shows the equivalent circuit for one phase, while
Fig. (2.12 (ii)) shows the phasor diagram.

2.13 Power Flow diagrams

Except for very small machines, the armature resistance of a synchronous
motor is relatively insignificant compared to its synchronous reactance, so that
Equation below to be approximated to
𝑉𝑇 = 𝐸𝑓 + 𝑗𝐼𝑎 𝑋𝑠
The equivalent-circuit and phasor diagram corresponding to this relation are
shown in Fig. 2.14 and Fig.2.15. These are normally used for analyzing the
behavior of a synchronous motor, due to changes in load and/or changes in field
excitation. From this phasor diagram, we have,
IaXs cos θi = −Ef sin δ
Multiplying through by VT and rearranging terms we have,
𝑉𝑇 𝐼𝑎
𝑉𝑇 𝐼𝑎 cos 𝜑𝑖 = − sin 𝛿
𝑋𝑠
Since the left side of Eqn above is an expression for active power input
and as the winding resistance is assumed to be negligible this power input
will also represent the electromagnetic power developed, per phase, by the
synchronous motor.
Pin,ph = VT Ia cos φi
Or
𝑉𝑇 𝐸𝑓
𝑃𝑖𝑛,𝑝ℎ = − sin 𝛿
𝑋𝑠

Thus, for a three-phase synchronous motor,

Pin = 3 x VT Ia cos φi
Or
−𝑉𝑇 𝐸𝑓
𝑃𝑖𝑛,𝑝ℎ = 3× sin 𝛿
𝑋𝑠
Equation Pin,ph, called the synchronous-machine power equation, expresses
the electro magnetic power developed per phase by a cylindrical-rotor motor, in
terms of its excitation voltage and power angle. Assuming a constant source
voltage and constant supply frequency, the above equations may be expressed as
proportionalities that are very useful for analyzing the behavior of a synchronous-
motor:
P ∝ Ia cos θ
P ∝ Ia cos θ

2.14 equivalent circuit of a synchronous motor assuming negligible armature

resistance

2.15 Phasor diagram model for synchronous motor

Hunting in a synchronous motor
➢ Sometimes an alternator will not operate satisfactorily with others due to
hunting. If the driving torque applied to an alternator is pulsating such as that
produced by a diesel engine.
➢ the alternator rotor may be pulled periodically ahead of or behind its normal
position as it rotates. This oscillating action is called hunting.
➢ Hunting causes the alternators to shift load from one to another.
➢ In some cases, this oscillation of power becomes cumulative and violent enough
to cause the alternator to pull out of synchronism.
➢ In salient-pole machines, hunting is reduced by providing damper winding. It
consists of short-circuited copper bars embedded in the pole faces as shown in
Fig. 2.16 When hunting occurs, there is shifting of armature flux across the pole
faces, thereby inducing currents in the damper winding.
➢ Since any induced current opposes the action that produces it, the hunting
action is opposed by the flow of induced currents.
➢ The following points may be noted:

2.16 Hunting

Prepared by, Approved by,

N.Jegan AP/EEE HOD/EEE