1. Under load-simple calculations
Equivalent circuit represents one phase of a wye-connected motor. It is identical to the equivalent circuit of an ac generator.
The flux  created by the rotor induces a voltage EO in the stator. It depends on the dc exciting current Ix. EO varies with excitation.
Rotor and stator poles are lined up at no-load. Induced voltage EO is in phase with the line-to-neutral voltage E.
If, we adjust the excitation so that EO = E, the motor "floats" on the line and the line current I is practically zero.
The only current needed is to supply the small windage and friction losses in the motor, and so it is negligible.
Lecture 35
Electro Mechanical System

2. Under load-simple calculations
If we apply a mechanical load. The motor will begin to slow down, so the rotor poles fall behind the stator poles by an angle α.
Due to this mechanical shift, EO reaches its maximum value a little later than before. EO is now δ electrical degrees behind E.
The mechanical displacement α produces an electrical phase shift δ between EO and E, which produces a difference of potential Ex across the synchronous reactance Xs given by:
Ex = E – EO
A current I must flow in the circuit, given by: jIXS = Ex
I = – j Ex /XS = – j(E – EO)/Xs
I is nearly in phase with E, so the motor absorbs active power
Lecture 35
Electro Mechanical System

3. Electro Mechanical System
Example
A 500 hp. 720 r/min synchronous motor connected to a 3980 V, 3-phase line generates excitation voltage EO of 1790 V (line-to-neutral) when the dc exciting current is 25 A. The synchronous reactance is 22 Ω and the torque angle between EO and E is 30°.
Calculate;
Value of Ex
ac line current
Power factor of the motor
Approximate horsepower developed by the motor
Approximate torque developed at the shaft
Solution: This problem can best be solved by using vector notation.
a. The voltage E (line-to-neutral) applied to the motor has a value
E = EL /√3 = 3980/√ 3 = 2300 V
Let us select E as the reference phasor, whose angle with respect to the horizontal axis is assumed to be zero. Thus: E = 2300  0°
It follows that EO is given by the phasor: EO = 1790  – 30°
Ex = E – EO = (2300  0°) – (1790  – 300) = 1168  50°
Phasor Ex has a value of 1168 V and it leads phasor E by 50°.
Lecture 35
Electro Mechanical System

4. Electro Mechanical System
Example
The line current I is given by
jIXS = Ex so I = (1168  50°) / (22  90°) = 53  – 40°
Phasor I has a value of 53 A and it lags 40° behind phasor E.
c) The power factor of the motor is given by:
power factor = cos  = cos 40° = 0.766, or 76.6%
The power factor is lagging.
d) Total active power input to the stator:
Pi = 3 x ELNIL cos 
= 3 x 2300 x 53 x cos 40°
= W = kW
Neglecting the I2R losses and iron losses,
Electrical power transmitted is kW.
Approximate horsepower developed:
P = X 103/746 = 375 hp
e) Approximate torque:
T = (9.55 x P)/n
= (9.55 X X 103)/720
= 3715 N-m
Lecture 35
Electro Mechanical System

5. Electro Mechanical System
Reluctance Torque
If we gradually reduce the excitation when synchronous motor is running at no load.
Motor continues to run at synchronous speed, even when the excitation current is zero.
Flux produced by the stator prefers to cross the short gap between the poles and stator, rather than longer gap between the poles.
Reluctance of magnetic circuit is less in the axis of salient pole, so the flux is concentrated as shown in the diagram.
If a mechanical load is applied to the motor, the rotor poles will fall behind the stator, a considerable reluctance torque is developed, without any dc excitation.
The reluctance torque will become zero when the rotor poles are mid way between stator poles.
The reason is that N and S poles on the stator attracts the salient poles in the opposite direction
Lecture 35
Electro Mechanical System

6. Electro Mechanical System
Losses and Efficiency
Comparison between 200 hp and 2000 hp motor
The excitation power to excite 2000 hp is 4.2kW while 200 hp is 2.1kW, i.e. only twice more excitation power. Larger the synchronous motor, smaller per-unit excitation power needed
The total losses of 2000 hp is 38kW while 200 hp 9.5kW, More horse power they develop , smaller the relative losses. Efficiency of 2000 hp is 97.5% while 200 hp 94%.
Synchronous reactance is much larger than resistance. For hp XS is 122 times larger than RS where as for 200 hp XS is only 24 times larger than RS. As a result we can easily neglect resistance of larger motors.
Lecture 35
Electro Mechanical System

7. Excitation and reactive power
Consider a wye-connected synchronous motor connected to a 3-phase source with line voltage EL
The line current I produces mmf force Ua in the stator.
Rotor produces a dc mmf Ur
Total flux Φ is therefore combined action of Ua and Ur
Flux Φ induces line to neutral voltage in the Ea in the stator
Neglecting small IR drop in the stator, we can say that Ea = E
Like transformer E is fixed so flux Φ is also fixed
The flux Φ may be produced either by the stator or rotor or both
Lecture 35
Electro Mechanical System

8. Excitation and reactive power
If the rotor exciting current Ix is zero all the flux has to be produced by the stator.
The stator absorbs considerable reactive power from the three phase line.
If the rotor exciting current is increased the rotor mmf helps produce part of the flux, and less power is drawn from the system.
Raising the rotor exciting current, gradually the rotor produces all of the required flux, the stator circuit draws no reactive power (unity power factor)
If the exciting current exceeds this critical level, the stator delivers reactive power to the ac power system.
Lecture 35
Electro Mechanical System