Principle of operation – Equivalent circuit and phasor diagram – Starting methods – V and inverted V Curves – Power transforming and capability curves – Effect of varying excitation – Hunting – Synchronous condenser – Speed control of synchronous motor.

1. UNIT – II
SYNCHRONOUS MOTOR
Principle of operation – Equivalent circuit and phasor
diagram – Starting methods – V and inverted V Curves
– Power transforming and capability curves – Effect of
varying excitation – Hunting – Synchronous condenser
– Speed control of synchronous motor.

2. Features
• Operates at constant speed
• Construction is similar to AC generator
• By changing the field excitation, the PF can be
varied from lagging to leading
• Its efficiency is high low initial cost
• Improve the power factor

3. Principle of Operation
• When a three phase voltage is applied to a three
phase winding, the flux produced will be the
resultant of all the three fields has magnitude of
1.5𝜑𝑚.
• The field is rotating is space at a speed is given by,
𝑁𝑠 =
120×𝑓
𝑝
synchronous speed.
Interlocking between the stator and rotor poles, the
motor runs only at one speed.

5. Cross sectional view of rotor

6. PRINCIPLE OF OPERATION

7. Starting Methods of Synchronous Motor
1. By using DC source:
• Synchronous motor is coupled and started by
means of a DC compound motor. The speed of DC
motor is adjusted by the speed regulator.
• The synchronous motor is then excited and
synchronized with AC supply mains.

8. 2. By means of Damper Grids in the pole faces
• The synchronous motor is made self starting by using a
special winding on the rotor poles, known as damper
winding. The winding S.C copper bars which are placed
in the face of field poles.
• At the starting, motor provided with damper starts as a
squirrel cage induction motor, when AC supply is given
to the stator.
• Once the motor is given excitation, the motor starts to
runs at sub synchronous speed
• DC suppy is given to the field winding, motor gets
locked into synchronism and runs at synchronous
speed.
• The relative speed between damper winding and RMF
is zero. So, it disconnected from the circuit.

9. 3. As a Slip Ring Induction Motor:
• In order to obtain high starting torque, Syn. motor
can be stat as SRIM.
• By forming damper winding as star or delta
connected instead of short circuited.
• other end of the winding if connected via slip rings
also, external resistance can added in series with
rotor circuit
• When three phase supply is given, the motor runs
as SRIM and high starting torque can be achieved by
adding external resistance.
• The DC excitation is given to rotor when the motor
attains neared to synchronous speed.

10. By means of Pony Motor(Small AC Motor)
• A small direct coupled induction motor is called
pony motor used for starting.
• Before switching on the AC supply it must be
synchronized by AC bus bars.
• Once rotor is attains synchronous speed, DC
excitation is given.

11. Effect of Load on Synchronous Motor
• Synchronous motor, the speed is always constant
irrespective of the load, by increasing the load, to
supply extra torque and power, synchronous motor
draws more current.
• The method of drawing increased current from
mains on syn. Motor can not be same as that of a
DC shunt motor.
• The motor can be started by using some external
arrangement, then the motor is synchronized with
supply.
• At the instant of synchorozing an emf is induced in
the stator.

12. • This emf is called back emf.

13. • The angular displacement between the stator and
rotor poles is called torque angle or load angle or
coupling angle.
• Under light load condition, the torque angle is less
• The resultant voltage 𝐸𝑅 causes 𝐼𝑎 =
𝐸𝑅
𝑍𝑠
.
• 𝐼𝑎 lags 𝐸𝑅 by an angle 𝜃 = tan−1 𝑋𝑠
𝑅𝑎
internal angle.

15. Expression for Power and Torque
developed in Synchronous Motor
• OL – Supply voltage / Phase
• Ia – Armature current
• LM – Back emf at a load angle of 𝛿
• OM – Resultant Voltage 𝐸𝑅
• 𝐸𝑟 − 𝐼𝑎𝑍𝑠
• 𝐼𝑎 lags or leads V by an angle 𝜑 and lags behind 𝐸𝑟
by an angle 𝜃
𝜃 = tan−1
𝑋𝑠
𝑅𝑎
Line NS is drawn at an angle 𝜃 to LM

16. • LN and QS are perpendicular to NS and QL
• Mechanical power developed per phase in the rotor
𝑃𝑚𝑒𝑐ℎ = 𝐸𝑏𝐼𝑎 cos ψ ------------(1)
Consider ΔOMS, MS = 𝐼𝑎𝑍𝑠 cos ψ
MS = NS –NM = LQ –NM
𝐼𝑎𝑍𝑠 cos ψ = V cos 𝜃 − 𝛿 − 𝐸𝑏 cos 𝜃
𝐼𝑎 cos ψ =
𝑉
𝑍𝑠
cos 𝜃 − 𝛿 −
𝐸𝑏
𝑍𝑠
cos 𝜃 -------(2)
Substitute (2) in (1)
𝑃𝑚𝑒𝑐ℎ/𝑝ℎ = 𝐸𝑏
𝑉
𝑍𝑠
cos 𝜃 − 𝛿 −
𝐸𝑏
𝑍𝑠
cos 𝜃
𝑃𝑚𝑒𝑐ℎ/𝑝ℎ =
𝐸𝑏𝑉
𝑍𝑠
cos 𝜃 − 𝛿 −
𝐸𝑏
2
𝑍𝑠
cos 𝜃

17. This is the expression for mechanical power
developed in terms of load angle 𝛿 and the internal
angle 𝜃 of the motor for a constant voltage V &𝐸𝑏
If, 𝑍𝑠 = 𝑋𝑠, Ra is negligible, 𝜃 = 90°
𝑃𝑚𝑒𝑐ℎ/𝑝ℎ =
𝐸𝑏𝑉
𝑋𝑠
cos 90° − 𝛿
𝑃𝑚𝑒𝑐ℎ/𝑝ℎ =
𝐸𝑏𝑉
𝑋𝑠
sin 𝛿
If T is the gross torque developed by the motor, then,
𝑃𝑚𝑒𝑐ℎ = 𝑇 × 𝜔𝑛
T =
𝑃𝑚𝑒𝑐ℎ
2𝜋𝑁
60
=
9.55𝑃𝑚𝑒𝑐ℎ
𝑁

18. Condition for maximum power developed
Differentiating equ(4) w.r.t load angle
𝛿 and then equating in to zero
𝑑𝑃𝑚𝑒𝑐ℎ
𝑑𝛿
= −
𝐸𝑏𝑉
𝑋𝑠
sin 𝜃 − 𝛿 = 0
sin 𝜃 − 𝛿 = 0 𝜃 = 𝛿
Value of maximum power,
𝑃𝑚𝑒𝑐ℎ max =
𝐸𝑏𝑉
𝑍𝑠
−
𝐸𝑏
2
𝑍𝑠
cos 𝜃
Since running at constant speed, the maximum power
and hence torque depends on V and excitation
voltage 𝐸𝑏

19. • Maximum value of 𝜃 is 90°. For all values of V and
𝐸𝑏, maximum torque will be proportional to the
maximum power developed,
𝑃𝑚𝑒𝑐ℎ max =
𝐸𝑏𝑉
𝑍𝑠
when (𝛿 = 90°)
This corresponds to pull out torque,

20. • To determine the value of excitation or induced
emf𝐸𝑏 to give maximum power developed possible,
diff, w.r.t 𝐸𝑏 and equate to zero.
𝑑𝑃𝑚𝑒𝑐ℎ
𝑑𝐸𝑏
=
𝑉
𝑍𝑠
−
2𝐸𝑏
𝑍𝑠
cos 𝜃 = 0
𝐸𝑏 =
𝑉
2 cos 𝜃
Substitute 𝐸𝑏 in 𝑃𝑚𝑒𝑐ℎ max,
𝑃𝑚𝑒𝑐ℎ max =
𝑉2
2𝑍𝑠 cos 𝜃
−
𝑉2
4𝑍𝑠 cos 𝜃
=
𝑉2
4𝑍𝑠 cos 𝜃
=
𝑉2
4𝑅𝑠

21. Torque Developed in Synchronous Motor:
Starting Torque:
The torque developed by the motor at starting condition
while the rated voltage is applied to the stator. It is also
sometimes called “break away torque”.
Running Torque:
Running torque is the torque developed by the motor
under running condition.
It is determined by the output power rating and speed of
the driven machine.
Peak output power determines the maximum torque that
would be required .
The breakdown or maximum running torque of a motor
must be greater than this value in order to avoid stalling
of the machine.

22. Pull in torque
The amount of torque developed by synchronous
motor to pull into synchronism when changing from
induction to synchronous motor operation.
Pull out Torque
It is the maximum torque that the synchronous motor
will develop without pulling out of synchronism. Its
value ranges from 1.25 to 3.5 times the full load
torque. At pull load torque, load angle 𝛿is equal to
90°. The motor runs out of synchronism for 𝛿 greater
than 90°

23. Effect of Varying Excitation on Armature
current and power factor

25. V Curve and Inverted V Curves

27. Equivalent Circuit

30. Synchronous Condenser
• An over excited synchronous motor running on no
load.
• It takes a leading current when over excited and
therefore behaves as a capacitor
• When such a machine is connected in parallel with
induction motor or other devices that operate at
lagging PF, the leading kVAR supplied by the
synchronous motor partially neutralizes the lagging
kVAR of the loads.
• Power factor of the system is improved

31. • Impacts of Power Factor
P = VIcos ∅ I = 𝑃
V cos ∅
If PF gets lowered then, current will increase
dramatically.
• For withstanding high value of current, large size of
armature conductor must be used
• Cost of the machine is high
• More losses will occur hence efficiency gets
lowered
• Poor voltage regulation
Synchronous capacitors are installed in substations
solely for power factor improvement. They are
economical in large sizes than the static capacitors

32. V – applied voltage/ph
∅1 − angle by which the current 𝐼𝑎1 lags behind V
In order to improve the PF, the synchronous motor is
connected across the AC mains.
After that, the motor will draw the current 𝐼𝑎2 which
leads the applied voltage V by an angle of ∅𝑏.
The net effective current is the phasor sum of 𝐼𝑎1 and
𝐼𝑎2. This total current 𝐼𝑒𝑓𝑓 lags V by an angle of very
low value and the power factor improved.

34. Current locus for constant power lines:
In synchronous motor, the input power per phase is,
𝑃𝑖𝑛
3
If V be the input voltage/phase, the power component of
the current is,
Icos ∅ =
𝑃𝑖𝑛
3
×
1
𝑉
A vertical is first erected representing the axis for the
voltage V. A horizontal line is drawn at a point A such that
OA =
𝑃𝑖𝑛
3𝑉
The current locus is varies on the straight line for various
powers. For under excitation, it takes a lagging current
i.e., the current vector lying on the right side of the
voltage vector V.
For over excitation, it takes a leading current, the current
vector lying on the left side of the voltage vector V.

36. Current locus for Constant Power Developed
P = VIcos ∅ - 𝐼𝑎
2
𝑅𝑎
𝐼𝑎
2
-
𝑉
𝑅𝑎
Icos ∅ = -
𝑃
𝑅𝑎
𝐼𝑎
2
+
𝑉
2𝑅𝑎
2
- 2
𝑉
2𝑅𝑎
Icos ∅ =
𝑉
2𝑅𝑎
2
-
𝑃𝑚
𝑅𝑎
𝑉
2𝑅𝑎
and 𝐼𝑎 are taken as two sides of a triangle, shown
by OA and OB respectively. The third side is given by
𝑉
2𝑅𝑎
2
−
𝑃𝑚
𝑅𝑎
.
When Pm remains constant both
𝑉
2𝑅𝑎
and
𝑉
2𝑅𝑎
2
−
𝑃𝑚
𝑅𝑎
remains constant.

37. The locus of the current phasor B, as q changes, is a
circle. The radius of the circle is
𝑉
2𝑅𝑎
2
−
𝑃𝑚
𝑅𝑎
remains
constant. and the co-ordinate of the center is given
by,
𝑉
2𝑅𝑎
, 0

38. The power developed by the motor is increases, the
radius AB on increasing.
Hence, the current locus becomes a circle of small
diameter. For a development of maximum power, the
radius of the circle must be zero.
𝑉
2𝑅𝑎
2
−
𝑃𝑚
𝑅𝑎
= 0
𝑃𝑚,𝑚𝑎𝑥
𝑅𝑎
=
𝑉2
4𝑅𝑎
2
𝑃𝑚,𝑚𝑎𝑥 =
𝑉2
4𝑅𝑎
When the mechanical power developed in the motor is
maximum, the current vector OB falls on the voltage
vector having magnitude of
𝑉
2𝑅𝑎
i.e., 𝐼𝑎 =
𝑉
2𝑅𝑎

39. Power input/ph = V ×
𝑉
2𝑅𝑎
=
𝑉2
2𝑅𝑎
Armature copper loss/ph = 𝐼𝑎
2
𝑅𝑎 =
𝑉2
4𝑅𝑎
Power developed 𝑃𝑚 = 𝑃𝑖𝑛 - Armature copper loss
Pm =
𝑉2
2𝑅𝑎
-
𝑉2
4𝑅𝑎
=
𝑉2
4𝑅𝑎
The motor efficiency at maximum output becomes
50%. Losses become the half of the input power

40. Current locus for constant Excitation
Consider the sides of the triangle to the left of voltage
vector V, impedance drop 𝐼𝑎𝑍𝑠 and the induced emf
E.
All the three sides of this triangle are divided by 𝑍𝑠 ,
we get triangle with sided,
𝑉
𝑍𝑠
, 𝐼𝑎 ,
𝐸
𝑍𝑠
.
If the field excitation remains constant
𝑉
𝑍𝑠
,
𝐸
𝑍𝑠
remains
constant and the current locus 𝐼𝑎 is a circle drawn for
a radius
𝐸
𝑍𝑠
from the point A.

42. Form fig.,
𝐼𝑎
2
=
𝑉
𝑍𝑠
2
+
𝐸𝑓
𝑍𝑠
2
- 2
𝐸𝑓
𝑍𝑠
𝑉
𝑍𝑠
cos 𝛿
𝐼𝑎 consists of two values. One value gives stable
operation and another value gives unstable operation.
The unstable operating point is represented by 𝐼𝑎. The
angle 𝛿 between
𝐸𝑓
𝑍𝑠
and
𝑉
𝑍𝑠
for unstable operating point is
more than ∅.
The maximum value of angle 𝛿 for the stable operation is
impedance angle 𝜑. When the synchronous motor is
loaded condition, the rotor field falls back from the stator
RMF by an angle 𝛿 and continuous to run in synchronous
speed.
the motor becomes unstable it may not be able to rotate
along with the rotating field if 𝛿 goes beyond ∅.

43. A six pole synchronuous motor has synchronous impedance of 10 ohm
and a resistance of 0.5ohm.when running on 2kV, 50Hz supply mains.
Its field excitation is such that the induced emf in the machine is 1.6kV.
Calculate the maximum torque developed by the machine
•

44. A 5kW, three phase star-connected 50HZ, 400V,
cylindrical rotor synchronous motor operates at
rated condition with 0.8pf lagging. The motor
efficiency excluding field and stator copper losses is
95% and Xs = 2.5Ohm. Calculate (i) Mechanical
power developed (ii) Armature current (iii) Back emf
(iv) Power angl (v) Maximum torque

48. A 1000kVA, 11000V, 3-ph, star connected motor has
armature resistance and synchronous reactance per phase
are 35ohm and 40 ohm respectively. Determine the induced
emf and angular retardation of the rotor when full load
0.8PF lagging and 0.8p.f leading condition

52. 1. A Star connected synchronous motor rated at 187KVA, 3-Ph, 300V, 47A, 50Hz,
187.5RPM has an effective resistance of 1.5 Ohm and a synchronous reactance of 20
ohm per phase. Determine the internal power developed by the motor when it is
operating at rated current and 0.8 power factor leading.(i/p = 149.78kW,
Pcu=9.94kW, Pdev = 139.84kW)
2. A 2.2kV, 3-ph, star connected synchronus motor has Zs=0.2+j2.2 ohms/ph. The
motor is operating at 0.5 power factor leading with a line current of 200A. Determine
the generated emf per phase.

53. A 6600V, 3-ph, star connected synchronous motor draws a
full load current of 80A at 0.8pf leading. The armature
resistance is 2.2 ohm and reactance 22ohm per phase. If the
stray losses of the machine are 3200W find a) Induced emf
ii) Output power iii) Efficiency of the machine

54. A 400 V, 6-pole, 3-Ph, 50Hz, star connected synchronous motor has a
resistance and synchronous impedance of 0.5 ohm and 4 ohm
respectively. It takes a current of 15A at UPF when operating with a
certain field current. If the load torque is increased untill the line
current Is increased to 60A, the field current remain unchanged,
calculate the gross torque developed and new power factor.

55. Advantages:
1. The speed is constant and independent of the load
2. Operate at higher efficiencies
3. Electro magneitc power varies linearly with the voltage
4. These motors can be constructed with wider air gaps than induction motor, which
make them mechanically strong
5. An over excited synchronous motor having a leading power factor can be operated
in parallel with induction motors
Disadvantages:
1. It cannot be started under load
2. Requires DC excitation which must be supplied from the external source
3. It has a tendency to hunt
4. It cannot be used for variable speed jobs, as there is no possibility of speed
adjustment
5. Collector rings and brushes are required
Applications:
1. They are used in power houses and substations in parallel to the bus-bars to
improve power factor
2. Synchronous motors are also used to regulate the voltage at the end of
transmission lines
3. Fans, blowers, pumps, compressor, rupper mills and paper mills

59. A 12MVA, 5000V, 3-Ph, 4-Pole 50HZ alternator is
connected to infinite bus bar. The short circuit
current Is 4 times the normal full load current. And
the moment of inertia of the rotating system is
22000 kg-m2.determine the normal period of
oscillation

60. CAPABILITY CURVES

62. POWER ANGLE CHARACTERISITCS

64. A 6-pole, 3-ph, IM deverlopes a power of 22.38kW including
mech losses.which total of 1.492kw at a speed of 950RPM
on 550V, 50Hz supply with a PF of 0.88.caculate i) Slip ii)
Rotor Copper loss iii) total i/p if stator losses are 2000W iv)
Effi v)Line current vi) number of cycles completed by the
rotor frequency per minite

65. A 6-pole, 3-ph, 50Hz IM has a slip of 1% at no load
and 3% full load. Find i) synchronous speed ii) no
load speed iii) full load speed iv) frequency of rotor
current at stand still v) frequency of rotor current at
full load.

66. Types of Torque
Starting Torque:
The torque developed by the motor at starting condition
while the rated voltage is applied to the stator. It is also
sometimes called “break away torque”.
Running Torque:
Running torque is the torque developed by the motor
under running condition.
It is determined by the output power rating and speed of
the driven machine.
Peak output power determines the maximum torque that
would be required .
The breakdown or maximum running torque of a motor
must be greater than this value in order to avoid stalling
of the machine.

67. Pull in torque
The amount of torque developed by synchronous
motor to pull into synchronism when changing from
induction to synchronous motor operation.
Pull out Torque
It is the maximum torque that the synchronous motor
will develop without pulling out of synchronism. Its
value ranges from 1.25 to 3.5 times the full load
torque. At pull load torque, load angle 𝛿is equal to
90°. The motor runs out of synchronism for 𝛿 greater
than 90°

68. Hunting:
The oscillation of rotor about its final equilibrium
position is called hunting.
Causes:
1. Due to fault occurrence in the system that is
supplied by the generator.
2. Sudden changes of load
3. Sudden changes of field current
4. Cycle variations of load torque

69. • Motor is loaded, rotor falls back in phase by
the coupling angle or load angle
• Load increase- angle increase, more torque
• Sudden decrease in load- motor immediately
pulled up- new value of load angle

70. Effects of Hunting:
1. Variation in supply voltage
2. Increases the chance of loosing synchronism
3. Increases the possibility of resonance
4. Huge mechanical stress may be developed in the
rotor
5. Loss increases in the machine
6. Increase in temperature
Reduction of Hunting:
1. By providing damper winding
2. By using flywheels

71. Damper windings:
• Damper windings are used in salient pole machine to provide
damping.
• The damper winding consists of low resistance copper or
aluminium bars embedded in notches carved in rotor pole
shoes and short circuited at both ends by rings.
• When the rotor is running at synchronous speed there is no
induced emf in the bars. When the speed is different from
synchronous speed, there is a relative motion between stator
magnetic field and rotor and voltage induced in the bars
• The bars are short circuited at the ends, the current flows,
and produces magnetic field
• The interaction of the two fields produces a torque, which
tend to make the rotor speed equal to synchronous speed.
• The addition of damper winding helps in damping out the
rotor oscillations

72. A 10HP, 400V, 3-ph, star connected synchronous
motor has Zs=0.35+j2.8 ohm/ph. Find the angle of
retard and the voltage at which it must be excited to
give a full load output at 0.866pf leading. Assume
efficieny as 0.88.(Pin=8.358kW, 𝞱=82.9, Φ=30,
V=231V, Er=39.3V, Eb=249V, δ=8.36)

73. Speed control
• A synchronous motor is constructionally same an alternator
• - It runs at synchronous speed or it remains stand still
• - Speed can be varied by varying supply frequency because
synchronous speed, Ns = (120f/p)
• - Due to unavailability of economical variable frequency
sources, this method of speed control was not used in past
& they were mainly used for constant speed applications
• - The development of semiconductor variable frequency
sources such as inverter & cycloconverter allowed the use
of synchronous motor in variable speed applications

74. Types of synchronous motors
• Commonly used synchronous motors are
1. Wound field synchronous motor (Cylindrical &
salient pole)
2. Permanent magnet synchronous motor
3. Synchronous reluctance motor
4. Hysteresis motor
• - All these motors have a stator with 3 phase
winding which is connected to an AC source
• - Fractional horse power synchronous reluctance
& hysteresis motors employ a 1 phase stator

75. True synchronous mode

76. Speed – Torque Characteristics

77. Self controlled mode

78. Self control mode using a rotor
position sensor or stator voltage
sensor

79. True synchronous mode where the speed of
motor is determined by
the external independent oscillator

80. Self controlled synchronous motor
drive employing a load
commutated thyristor inverter

84. `

85. Generator phasor diagram

87. Capability Curve of Synchronous
Generator:
• Capability Curve of Synchronous Generator defines the
bounds within which it can operate safely. Various
bounds imposed on the machine are:
• MVA-loading cannot exceed the generator rating. This
limit is imposed by the stator heating.
• MW-loading cannot exceed the turbine rating which is
given by MVA (rating) x pf (rating).
• The generator must operate a safe margin away from
the steady-state stability limit (δ = 90°). This can be laid
down as a maximum allowable value of δ
• The maximum field current cannot exceed a specified
value imposed by rotor

88. • To draw the capability curve of the
synchronous generator, its phasor diagram is
used which is redrawn in Fig. armature
resistance is neglected. After multiplying
voltage magnitude of each voltage phasor by
(3Vt/Xs), the phasor diagram is redrawn in Fig.
8.49. It is immediately recognized that OMN is
the complex power triangle (in 3-phase
values) wherein

91. Case 1 − Under-Excitation of
Synchronous Motor
• The synchronous motor is said to be under-
excited if the field excitation is such that
(Ef<V). Under this condition, the armature
current (Ia) lags behind the supply voltage (V)
and consumes lagging reactive power (Q)
i.e. the motor power factor is lagging as
shown in Figure-1.

92. Case 2 − Normal Excitation of
Synchronous Motor
• If (Ef=V), the synchronous motor is said to
be normally excited. Under this condition, the
reactive power of the motor is zero (i.e., Q =
0), that is the motor is neither absorbing nor
delivering reactive power. Thus, the power
factor of the motor is unity. For a given load,
at unity power factor, the resultant voltage (Er)
and hence, the armature current (Ia) are
minimum as shown in Figure-2.

93. Case 3 − Over-Excitation of
Synchronous Motor
• The synchronous motor is said to be over-
excited if the field excitation is such that
(Ef>V). Under such conditions, the armature
current (Ia) leads the supply voltage (V) and
the motor supplies lagging reactive power to
the system. Hence, motor power factor is
leading as shown in Figure-3.