Synchronous Motor

1. Introduction
If a three phase supply is given to the stator of a three phase alternator, it can work as a motor. As is is driven at synchronous speed, it is called synchronous generator. So if alternator is run as a motor. It will rotate at a synchronous speed. Such a device which converts an electrical energy into a mechanical energy running at synchronous speed is called synchronous motor. Synchronous motor works only at synchronous speed and can not work at a speed other than the synchronous speed. Its speed is constant irrespective of load, no doubt, its speed changes for an instant at the time of loading.

2. Types

The two types of synchronous motor are,

1. Three phase synchronous motors

2. Single phase synchronous motor

The single phase synchronous motor are further classified as reluctance motor and hysteresis motor.

The three phase synchronous motor works on the concept of rotating magnetic field. The field produced by stationary three phase winding, which rotates in space is called rotating magnetic field. Its speed is always synchronous and given by,
Ns = 120f/P

Where P = Number of poles for which winding is wound

f = Frequency of the supply.

3. Rotating magnetic field (R.M.F.)

The rotating magnetic field can be defined as the field or flux having constant amplitude but whose axis rotates in a plane at a certain speed.e.g. permanent magnet rotating in a space produces a rotating magnetic field. Similarly if an arrangement is made to rotate the poles, with constant excitation supplied, the resulting field is rotating magnetic field. So a field produced in an air gap of a rotating field type alternator is of rotating type. But this is all about production of R.M.F. by physically rotating poles or magnet. In practice such a rotating magnetic field can be produced by exciting a set of stationary coils or wi9nding with the help of polyphase a.c. supply. The resultant flux produced in such a case has constant magnitude and its axis rotates in space without physically rotating the winding. Let us study how it happens.

3.1 Production of rotating magnetic field

Consider a three phase windings displaced in space by 120^o, supplied by a three phase a.c. supply. The three phase currents are also displaced from each other by 120^o. the flux produced by each phase current is also sinusoidal in nature and all three fluxes are separated from each other by120^o. If the phase sequence of the windings is 1-2-3, then the mathematical equation for the instantaneous values of the fluxes Φ₁, Φ₂ and Φ₃can be given as,

Φ₁= Φ_m sin(ωt) = Φ_m sin θ ...........(1)

Φ₂= sin (ωt - 120^o) = Φ_m sin (θ - 120^o) ............(2)

Φ₃= Φ_m sin (ωt - 240^o) = Φ_m sin (θ - 240^o) .............(3)

As windings are identical and supply is balanced the amplitude of each flux is same i.e.Φ_m . The waveform of three fluxes are shown in the Fig.1(a) while the assumed positive directions of these fluxes in space are shown in the Fig.1(b). Assumed positive direction means whenever the instantaneous value of flux is positive, in vector diagram it must be represented along its assumed positive direction. And if flux has negative instantaneous value then must be represented in opposite direction to assumed positive direction, in the vector diagram.

Fig. 1

Let Φ₁, Φ₂ and Φ₃ be the instantaneous values of the fluxes. The resultant flux Φ_Tat any instant is given by phasor combination of Φ₁, Φ₂ and Φ₃at that instant. Let us find out at four different instant 1, 2, 3 and 4 as shown in the Fig. 1(a) i.e. respectively at θ = 0^o, 60^o, 120^o and 180^o.

Case i) θ = 0^o
       Substituting in equations (1), (2) and (3) we get,
                                         Φ₁= Φ_m sin 0^o = 0
                                         Φ₂ = Φ_m sin(-120^o ) = -0.866 Φ_m
                                          Φ₃= Φ_m sin (-240^o) = + 0.866 Φ_m

Show positive values in assumed positive directions and negative in opposite directions to assumed positive directions.

Hence vector diagram looks like as shown in The Fig. 2(a).
BD is perpendicular drawn from B on ' Φ_T '.

Fig. 2 a and b

.^.. OD = DA = Φ_T/2

In triangle ∟OBD = 30^o

.^.. cos 30^o = OD/OB = (Φ_T/2)/(0.866 Φ_m )

.^.. Φ_T = 2 x 0.866 Φ_m x cos 30^o

= 1.5 Φ_m

       So magnitude of resultant flux is 1.5 times the maximum value of an individual flux.
Case ii) θ = 60^o
       Substituting in equations (1), (2) and (3) we get,
                              Φ₁= Φ_m sin 60^o= +0.866 Φ_m

Φ₂ = Φ_m sin (-60^o) = -0866 Φ_m

Φ₃= Φ_m sin (-180^o) = 0

       So Φ₁ is positive and Φ₂ is negative so vector diagram looks like as shown in the Fig. 2(b).
       It can be seen that from the Fig. 2(b), that,
                                 Φ_T= 1.5 Φ_m

So magnitude of the resultant is same as before but is is rotated in space by 60^o in clockwise direction, from its previous position.

Case iii) θ = 120^o
Substituting in equations (1), (2) and (3), we get,
Φ₁= Φm sin 120^o = +0.866 Φm

Φ₂ = Φm sin 0^o = 0

Φ₃= Φm sin (-120^o ) = -0.866 Φm

So Φ₁ is positive, Φ₂ is zero and Φ₃is negative. So vector diagram looks like as shown in the Fig. 2(c). From the Fig. 2(c), it can be proved easily that,

Φ_T= 1.5 Φ_m

Fig. 2 c and d

So magnitude of the resultant is once again 1.5 Φ_m, same as before. While it is further rotated in space by from its previous position at θ = 60^o

Case iv) θ = 180^o
Substituting in equations (1), (2) and (3), we get,
Φ₁= Φm sin (180^o) = 0

Φ₂ = Φm sin (60^o) = +0.866 Φm

Φ₃= Φm sin (-60^o)

= -0.866 Φm

       So Φ₁= 0, Φ₂ is positive and Φ₃ is negative. The vector diagram is as shown in the Fig. 2(d).
       From the vector diagram, it can be proved that,
                                  Φ_T= 1.5 Φm

So magnitude of resultant flux is once again 1.5 Φm but is further rotated by 60^o in clockwise direction from its position for θ = 120^o

So for a half cycle of the fluxes, the resultant has rotated through180^o. This is applicable for 2 pole winding. From this discussion we can have following conclusions :

a) The resultant of the three alternating fluxes, separated from each other by 120^o, has a constant amplitude of 1.5 Φm where Φm is maximum amplitude of an individual flux due to any phase.

b) The resultant always keeps on rotating with a certain speed in space.

This is nothing but satisfying the definition of a rotating magnetic field. Hence we can conclude that the three phase stationary winding when connected to a three phase a.c. supply produces a rotating magnetic field.

The speed of the resultant is in space, for electrical of the fluxes for a 2 pole winding as discussed above.

Key Point : This is nothing but,

^omechanical = ^oelectrical for 2 pole case.

If winding is wound for P poles, then resultant will complete 2/P revolution for 360^oelectrical of the fluxes. The relation is exactly similar to what we have discussed earlier in case of alternator. So resultant flux bears a fixed relation between speed of rotation, supply frequency and number of poles for which winding is wound. The relation is derived while studying an alternator. So for a standard supply frequency of f Hz of a three phase a.c. supply and 'P' poles of the three windings, the speed of the rotating magnetic field is Ns r.p.m.

Key Point : So for a rotating magnetic field,

Ns = 120f/P r.p.m.

3.2 Direction of rotating magnetic field

The direction of the rotating magnetic field is always from the axis of the leading phase of the three phase winding towards the lagging phase of the winding. In the example above the phase sequence is 1-2-3 i.e. phase 1 leads 2 by 120^oand phase 2 leads 3 by 120^o. So rotating magnetic field rotates from axis of 1 to axis of 2 and then to axis of e i.e. in the clockwise direction as seen above. This direction can be reversed by changing any two terminals of three phase winding while connecting them to the three phase supply. So in practice for a phase sequence of R-Y-B, the rotating magnetic field is rotating in clockwise direction, then by changing any two terminals of the winding it can be changed to anticlockwise, as shown in the Fig. 3(a) and (b).