04_SynchMotors

ECE 3650 Electrical Machines Athula Rajapakse

1

Synchronous Motors A synchronous motor is represented using the same equivalent circuit as a generator. However, in the synchronous motor equivalent circuit, the current is marked in the reversed direction, following the motor convention.

Fig. 1

For the equivalent circuit shown in Fig. 1,

asatf IjXRVE )( (48)

The phasor diagrams of the synchronous motor under lagging, unity and leading power factors are shown in Fig. 49.

Lagging pf Unity pf

Leading pf

Fig. 2

In the motor operation, the internal voltage Ef always lags behind the terminal voltage Vt. Under lagging power factor conditions, | Ef |<| Vt| and under leading power factor conditions | Ef |>| Vt|.

j Xs Ia

+ Ef _

+ Vt _

Ra

Te m

TL If

Ia

Vt

Ef

-RaIa

-jXsIa

Ia Vt

Ef

-RaIa

-jXsIa

Ia

Vt

Ef -RaIa

-jXsIa


2

Torque-Speed Characteristics Using the equivalent circuit, it is possible to calculate the converted power (from electrical to mechanical). This is equal to the real power input to the ideal source in the equivalent source. If the armature winding resistance is assumed negligible, the converted power is equal to the input real power:

)sin(||||

3)cos(||||3 s

ftatcon X

EVIVP

(49)

If the losses are neglected, the mechanical power output is equal to the converted power. The torque can be calculated by

)sin(||||

3

sm

ftind X

EV

(50)

Since a synchronous machine can operate only at its synchronous speed, its torque-speed characteristic is represented by a vertical straight line as shown in Fig 3. The pull out torque (maximum torque) is given by

sm

ftind X

EV

||||3max_ (occurs at = 90o ) (51)

Fig. 3 If the load torque greater than the pull out torque, pole slipping can occur

Stator magnetic field “laps” rotor repeatedly Torque surges in one way and the other way Huge vibration that may lead to mechanical damage

synspeed

Torque

pull out

rated


3

Effect of Load Change with Constant Excitation If a synchronous motor is connected to an infinite bus and its field excitation current is held constant, the magnitudes of the terminal voltage |Vt| and the internal voltage |Ef| remain constant. If the mechanical load on the machine is increased, increased load is met by increasing the load angle (according to (49). The change in load angle results in changes to the armature current magnitude and power factor (or more specifically to the product (Ia cos()) . This is illustrated in the phasor diagrams in Fig. 4. In drawing these phasor diagrams, Ra has been assumed negligible.

Fig. 4

1

Ia1

Vt

Ef

jXsIa1

3 P1

P2

P3

Ia2

Ia3

P1

P2

P3

Ia

Vt

Ef

Iacos() P

jXsIa

Xs Ia cos()=Ef sin() P


4

Effect of Changing Excitation at a Constant Load Again consider a synchronous motor connected to an infinite bus so that Vt is constant. If the field excitation current is changed while keeping the mechanical load driven by the motor constant, causes the machine power factor (or the reactive power) to change. This is illustrated in the phasor diagram shown in Fig. 5 (assume Ra is negligible). Since out and m are constant, Pout remains constant therefore, the quantities “Ia cos()” and “Ef sin()” must remain constant during the change of |Ef|. As a result,

When |Ef | increases, |Ia| first decreases and then increases The machine power factor changes from lagging to leading

Fig. 5 Synchronous Machine V – curves Plot of Ia Vs If under constant load and constant Vt (shown in Fig. 53) are called synchronous motor V-curves.

Fig. 6

1

Ia3

Vt

Ef1

Iacos() P

jXsIa3

3 P

Ia2

Ia1

P

Ef If

Ef2 Ef3

P1

P2

P3

P3> P2> P1 Unity pf line

Leading pf Lagging pf

If

Ia

Over excited Q- generated

Under excited Q- consumed


5

Synchronous Condenser (Capacitor) Consider a synchronous motor connected to an infinite bus as before. If there is no mechanical load attached to the motor, it will not draw any real power from the supply (except for the power required to overcome the friction and windage). If the field is “over excited” (i.e. |Ef| >|Vt|) the synchronous motor can supply reactive power like a capacitor. When a synchronous motor is run at no load, just to provide reactive power, the synchronous motor is called a “synchronous condenser” or “synchronous capacitor”.

Starting of synchronous motors A Synchronous motor operates only at the synchronous speed. Therefore starting needs special mechanism. Some possible methods are:

Reduce the speed of stator magnetic field initially by using a variable frequency supply. o Need to keep V/f constant to avoid core saturation

Use an external prime mover to accelerate under no load, bring the machine online as a generator and then disconnect the prime mover.

o Exciters can be used as starting motors Use damper windings (amortisseur windings) and start as an induction motor.

o Large starting currents – may not be suitable for large machines

Fig. 7


6

Example A 208 V, 45 kVA, 0.8 pf leading, -connected, 60 Hz synchronous motor has a synchronous reactance of 2.5 per phase and a negligible armature resistance. Its friction and windage losses are 1.5 kW, and its core losses are 1.0 kW. Initially the shaft is supplying a 15 hp load and the motors power factor is 0.85 leading.

a) Sketch the phasor diagram of this motor and find Ef, the line current IL and Ia.

b) Assume that the shaft load is increased to 30 hp. Sketch the new phasor diagram of the motor.

c) Find Ia , IL, and Ef after the load change Example The synchronous motor in the previous example is supplying a 15 hp load with an initial power factor of 0.85 lagging. The field current If at this condition is 4.0 A.

d) Sketch the initial phasor diagram of this motor.

e) If the motor flux is increased by 25%, sketch the new phasor diagram of the motor. What are the Ef, Ia and pf of the motor now?

f) Assume that the flux in the motor varies linearly with the field current If. Make a plot of Ia Vs If for the synchronous motor with a 15 hp load.

04_SynchMotors

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