1. UNIT II
SYNCHRONOUS MOTOR

3. Ideal Condition on No Load
The ideal condition on no load can be assumed by neglecting various losses in the motor.
       Vph  = Ebph
Under this condition, the magnetic locking between stator and rotor is in such a way that the magnetic axes of both coincide with each other.
Eph
Vph
As magnitude of Ebph  and Vph  is same and opposes the phasor diagram for this condition

4. Synchronous Motor on No Load (With Losses)
various losses practically present on no load, the magnetic locking exists between stator and rotor but in such a way that there exists a small angle difference between the axes of two magnetic fields C B
δ is very small
Eph
Vph – Eph = IaZs
IaZs
OB = IaZs = ERph δ δ A
Eph O
Vph

5. Synchronous Motor on Load

6. C B
ER1ph
OB = ER1ph
Ebph
LIGHT LOAD δ1
Vph A
Eph O B C
δ2 > δ1
OB = ER2ph
ER2ph
ER2ph > ER1ph
Ebph
HEAVY LOAD δ2
Vph A
Eph O

7. Operation of Synchronous motor at constant load Variable Excitation
Field Excitation (Field Current) is changed
Keeping Load Constant,
The Synchronous motor reacts by changing its power factor of operation
Normal Excitation
Excitation is adjusted to get
Eb = Vph
Motor draws a certain current Ia from supply
PF of Motor is lagging
If changes Eb changes
If changes Eb also changes
Eb α If

8. B C
Constant Ia Cos ф
Normal Excitation
Excitation is adjusted
Eb = Vph ER
Ebph
Constant Vph ϴ δ A O
Vph ф Ia B
Under Excitation
Excitation is adjusted
Eb < Vph
Ebph ER δ ϴ A
Vph ф Ia B
Ebph ER Ia
Over Excitation
Excitation is adjusted
Eb > Vph ϴ δ ф A
Vph B
Critical Excitation
Excitation is adjusted
Eb > Vph
Ebph ER ϴ δ A Ia
Vph

9. Synchronous condensers
A synchronous motor takes a leading current when over-excited and, therefore, behaves as a capacitor.
An over-excited synchronous motor running on no-load in known as Synchronous condenser.
Synchronous Motor is over-excited it takes leading P.F. current
If Synchronous Motor is no NO-LOAD, where load angle δ
Over Excitation
Excitation is adjusted
Eb > Vph Ia Ia
Ebph ER
Ф = 90 δ 90
Vph

10. Advantages
By varying the field excitation, the magnitude of current drawn by the motor can be changed by any amount. This helps in achieving stepless control of power factor.
The motor windings have high thermal stability to short circuit currents.
The faults can be removed easily.
Disadvantages
(i) There are considerable losses in the motor.
(ii) The maintenance cost is high.

11. Hunting When Synchronous motor is on NO LOAD
The stator and rotor pole axis coincide with each other.
When Synchronous motor is LOAD the rotor pole fall back with respect to stator.
ER2 ER
Ebph
ER1 R δ
Load angle by which the Synchronous motor rotor falls back from the aim of stator is called load angle(δ) O
Vph
Ia1 Ia
Ia2

12. Causes of Hunting in Synchronous Motor
Sudden change in load.
Sudden change in field current.
A load containing harmonic torque.
Fault in supply system
Effects of Hunting in Synchronous Motor
It may lead to loss of synchronism.
Produces mechanical stresses.
Increases machine losses and cause temperature rise.
Cause greater surges in current and power flow

13. Reduction of Hunting in Synchronous Motor
Two techniques should be used to reduce hunting. These are –
Use of Damper Winding 
It consists of low electrical resistance copper / aluminium brush embedded in slots of pole faces in salient pole machine. Damper winding damps out hunting by producing torque opposite to slip of rotor. The magnitude of damping torque is proportional to the slip speed.
Use of Flywheels :
The prime mover is provided with a large and heavy flywheel. This increases the inertia of prime mover and helps in maintaining the rotor speed constant.
Designing synchronous machine with suitable synchronizing power coefficients

14. V curves and Inverted V curves

15. V curves and Inverted V curves
Unity PF
Armature Current I PF
Lagging PF
Leading PF A
Normal Excitation If

16. Synchronous Machines on Infinite Bus-Bars
Infinite bus-bars are those whose frequency and the phase and magnitude of potential differences are not affected by changes in the condition of any one machine connected to it.
The variation in excitation of a synchronous machine connected to an infinite bus-bar will cause a large change in the reactive component supplied by the alternator and so change its power.
The increase in torque of the prime-mover of one alternator, it is further loaded and equivalent load is removed from the other unit (s) with which the machine is paralleled. If the output of the alternator, whose prime-mover torque has been increased, becomes more than the total load being supplied, then the other machine (s) will operate as synchronous motor (s).
The active and reactive power loading of an alternator operating on an infinite bus-bar is controlled by controlling the input power to it and ecitation respectively.

17. (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ - ф)
Expression for Back EMF or Induced EMF per phase in Synchronous Motor
Case i : Under Excitation Eb < Vph Power factor is Lagging
Zs = Ra + jXs
ERph = IaZs
Ebph B
= │Zs│∟ ϴ
Ebph x δ ϴ δ A ф
Vph O
ERph ᴧ Iaph = ϴ Ia
Vph ᴧ Iaph = ф
X = ϴ - ф
Δ OAB
ERph = Iaph * Zs
(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ - ф)
Ebph = ERph
Sin x Sin δ
Sin δ = ERph Sin (ϴ - ф)
E bph

18. (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф)
Expression for Back EMF or Induced EMF per phase in Synchronous Motor
Case ii : Over Excitation Eb > Vph Power factor is Leading
Zs = Ra + jXs
ERph
Ebph Ia B
= │Zs│∟ ϴ
Ebph ϴ δ ф δ A
Vph O
ERph ᴧ Iaph = ϴ
Vph ᴧ Iaph = ф
X = ϴ + ф
Δ OAB
ERph = Iaph * Zs
(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф)
Sin δ = ERph Sin (ϴ + ф)
E bph

19. (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф)
Expression for Back EMF or Induced EMF per phase in Synchronous Motor
Case iii : Critical Excitation Eb ≡ Vph Power factor is Unity
Zs = Ra + jXs
ERph
Ebph B
= │Zs│∟ ϴ
Ebph ϴ δ ф δ A
Vph O
ERph ᴧ Iaph = ϴ Ia
Vph ᴧ Iaph = ф
X = ϴ + ф
Δ OAB
ERph = Iaph * Zs
(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф)
Ebph = ERph
Sin ϴ Sin δ
Sin δ = ERph Sin ϴ
E bph

20. (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф)
ERph = Iaph * Zs
+ sign for lagging
Zs = Ra + jXs
- sign for leading

21. A 400 V, 3 phase, star connected synchronous motor has an armature resistance of 0.2 ohms per phase and synchronous reactance of 2 ohms per phase. While driving a certain load, it takes 25 A from the supply. Calculate the back EMF induced in the motor if it is working with
0.8 Lagging
0.8 leading
Unity power factor conditions

22. (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф)
Given Data
VL = 400 V Star connected
ERph = Iaph * Zs
Ra = 0.2 Ohms
Zs = Ra + jXs
Xs = 2 Ohms
Zs = j 2 Ω
IL = Iaph = 25 A
Vph = VL / √3
= ∟ 84.29
ERph = Iaph * Zs
Vph = / √3
ERph = 25 * 2.001
Vph = V
ERph = V
(i) Cos ф = 0.8 Lagging
Ф = °
ϴ = 84.29°
(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф)
(Ebph )2 = (230.94)2 + (50.025)2 - 2x(230.94)x(50.025) * Cos (84.29 – 36.86)
Ebph = V
Ebph < Vph
<

23. (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф)
Vph = V
ERph = V
ϴ = 84.29°
(ii) Cos ф = 0.8 Leading
Ф = °
(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф)
(Ebph )2 = (230.94)2 + (50.025)2 - 2x(230.94)x(50.025) * Cos ( )
Ebph >Vph
<
Ebph = V
(iii) Cos ф = 1Unity PF
Ф = 0°
(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф)
(Ebph )2 = (230.94)2 + (50.025)2 - 2x(230.94)x(50.025) * Cos ( )
Ebph = V
Ebph ≡ Vph ≡

24. A 3 phase, 500 V, synchronous motor draws a current of 50 A from the supply while driving a certain load. The stator is star connected with armature resistance of 0.4 ohms per phase and a synchronous reactance of 4 ohms per phase. Find the power factor at which the motor would operate when the field current is adjusted to give the line values of generated emf as (i) 600 V (ii) 380 V

25. (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф)
Given Data
VL = 500 V Star connected
ERph = Iaph * Zs
Ra = 0.4 Ohms
Zs = Ra + jXs
Xs = 4 Ohms
Zs = j 4 Ω
IL = Iaph = 50 A
Vph = VL / √3
= ∟ 84.89
ϴ = 84.89°
ERph = Iaph * Zs
Vph = / √3
(i) Cos ф = ?
Ф = ?
Vph = V
ERph = 50 * 4.019
(i) Eb LINE = 600 v
ERph = V
Ebph = Eb Line / √3
Ebph = / √3
= V
(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф)
( )2 = (288.67)2 + (200.95)2 - 2x(288.67)x(200.95) * Cos ( Ф )
(i) Cos ф = Leading
Ф =
Ф = 3.877
Ebph > Vph
>

26. (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф)
Given Data
ERph = Iaph * Zs
Vph= VL / √3
Vph = / √3
Zs = Ra + jXs
Vph = V
Zs = j 4 Ω
= ∟ 84.89
(ii) Eb LINE = 380 v
ϴ = 84.89°
ERph = Iaph * Zs
Cos ф = ?
Ф = ?
ERph = 50 * 4.019
ERph = V
Ebph = Eb Line / √3
Ebph = / √3
= V
(Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ ± ф)
( )2 = (288.67)2 + (200.95)2 - 2x(288.67)x(200.95) * Cos ( Ф )
Cos ф = Lagging
Ф = 34.93
Ф = 34.93
Ebph < Vph
<

27. A 3 phase, 6600 V, star connected synchronous motor delivers 500 kW power to the full load efficiency is 83%. Its armature resistance of 0.3 ohms per phase and a synchronous reactance of 3.2 ohms per phase. Its works with 0.8 leading power factor on full load. Calculate (i) Generated EMF on full load
(ii) The load angle
Given Data
VL = 6600 V Star connected
Ra = 0.3Ohms
Ƞ = Output
Input
Xs = 3.2 Ohms
Efficiency = 83 %
Pin = VLILCOS ф
Pout = 500 kW
Cos ф = 0.8 leading

28. (Ebph )2 = (Vph )2 + (ERph )2 - 2Vph ERph * Cos (ϴ + ф)
Sin δ = ERph Sin ϴ
E bph
(Ebph )2 = ( )2 + (211.71)2 -2x( )x(211.71)*Cos ( )
Ebph = V
Sin δ = ERph Sin ϴ
E bph
Sin δ = (211.71)2 Sin ( )
Sin δ = 2.63

29. Equivalent Circuit and torque equationZs
Equivalent circuit for one armature phase Eb δ V Eb O δ
IaZs
-Eb
IaXs
IaRa Ia
V is the Vector sum of
Reversed back EMF and
Impedance drop
V = -Eb +Ia Ra
Load Angle δ

30. V O A δ
90 - ф ф ф
IaXs Ia Eb 90 B C
AB = Eb Sin δ
Since Stator copper losses neglected
Pin also represents gross mechanical
Power(Pm)
Cos ф = AB / IaXs
AB = IaXs Cos ф
Pm= (3 V Eb Sin δ ) / Xs
AB = Eb Sin δ
T = Pm / ɷm
ɷm = (2πN )/ 60
IaXs Cos ф = Eb Sin δ
Ia Cos ф = Eb Sin δ / Xs
T = (3 V Eb Sin δ )
((2πN ) / 60 )Xs
P = V Ia Cos ф
T = 9.55 Pm / N
P = V Eb Sin δ / Xs
Pin =(3 V Eb Sin δ ) / Xs
For 3 phase

31. Torque in Synchronous Motor
Starting Torque :
This is the torque developed by the synchronous motor at start when rated voltage is applied to the stator.
It is also called BREAK AWAY torque.
It is necessary to overcome friction and inertia
Running Torque :
It is the torque developed by the synchronous motor under running conditions.
It is decided by the OUTPUT rating of the motor and speed
Pull in Torque :
Initially Synchronous motor is rotated at a speed slightly less than the synchronous speed. The amount of torque developed by the motor at the time of pulling into synchronous speed is called PULL in Torque
Pull our Torque :
When the Synchronous motor is loaded, the rotor falls back with respect to stator
by an angle called load angle δ. As δ increases magnetic locking between stator and rotor decreases.
The maximum torque developed by the synchronous motor without pulling out of synchronous speed is called Pull out Torque

32. Applications of Synchronous Motors
Synchronous motor are used drive load which are at constant speed applications in industries.
They are pumps, compressor, blowers, rolling mills, paper mills, crushers etc.
Synchronous motor are used to drive an AC generator to generate electricity at frequency
other than supply frequency
Synchronous motor are used to improve power factor (lagging)
Synchronous motor are used to regulate the Transmission line voltage