## Introduction And Working Of 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 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

**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__

^{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 ?

_{1}, ?

_{2}and ?

_{3 }can be given as,

?_{1 }= ?_{m } sin(?t) = ?_{m } sin ? ………..(1)

_{2 }= sin (?t – 120

^{o}) = ?

_{m }sin (? – 120

^{o}) …………(2)

_{3 }= ?

_{m }sin (?t – 240

^{o}) = ?

_{m }sin (? – 240

^{o}) ………….(3)

_{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 |

_{1}, ?

_{2 }and ?

_{3 }be the instantaneous values of the fluxes. The resultant flux ?

_{T }at any instant is given by phasor combination of ?

_{1}, ?

_{2 }and ?

_{3 }at 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 ? = 0

^{o}, 60

^{o}, 120

^{o}and 180

^{o}.

**Case i) **? = 0^{o}

Substituting in equations (1), (2) and (3) we get,

?_{1 }= ?_{m } sin 0^{o} = 0

?_{2 } = ?_{m } sin(-120^{o} ) = -0.866 ?_{m }

?_{3 }= ?_{m } sin (-240^{o}) = + 0.866 ?_{m }

Hence vector diagram looks like as shown in The Fig. 2(a).

BD is perpendicular drawn from B on ‘ ?_{T } ‘.

Fig. 2 a and b |

**. ^{.}. **OD = DA = ?

_{T}/2

^{o}

**.**cos 30

^{.}.^{o}= OD/OB = (?

_{T}/2)/(0.866 ?

_{m })

**.**?

^{.}._{T }= 2 x 0.866 ?

_{m }x cos 30

^{o}

_{m }

So magnitude of resultant flux is 1.5 times the maximum value of an individual flux.

**Case ii) **? = 60^{o}

Substituting in equations (1), (2) and (3) we get,

?_{1 }= ?_{m } sin 60^{o }= +0.866 ?_{m }

_{2 }= ?

_{m }sin (-60

^{o}) = -0866 ?

_{m }

_{3 }= ?

_{m }sin (-180

^{o}) = 0

So ?_{1 } is positive and ?_{2 } 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,

?_{T }= 1.5 ?_{m }

^{o }in clockwise direction, from its previous position.

** Case iii) **? = 120^{o}

Substituting in equations (1), (2) and (3), we get,

?_{1 }= ?m sin 120^{o} = +0.866 ?m

_{2 }= ?m sin 0

^{o}= 0

_{3 }= ?m sin (-120

^{o}) = -0.866 ?m

_{1 }is positive, ?

_{2 }is zero and ?

_{3 }is negative. So vector diagram looks like as shown in the Fig. 2(c). From the Fig. 2(c), it can be proved easily that,

_{T }= 1.5 ?

_{m }

Fig. 2 c and d |

_{m}, same as before. While it is further rotated in space by from its previous position at ? = 60

^{o}

**Case iv)** ? = 180^{o}

Substituting in equations (1), (2) and (3), we get,

?_{1 }= ?m sin (180^{o}) = 0

_{2 }= ?m sin (60

^{o}) = +0.866 ?m

_{3 }= ?m sin (-60

^{o})

So ?_{1 }= 0, ?_{2 } is positive and ?_{3 } is negative. The vector diagram is as shown in the Fig. 2(d).

From the vector diagram, it can be proved that,

?_{T }= 1.5 ?m

^{o}in clockwise direction from its position for ? = 120

^{o}

^{o}. This is applicable for 2 pole winding. From this discussion we can have following conclusions :

^{o}, has a constant amplitude of 1.5 ?m where ?m is maximum amplitude of an individual flux due to any phase.

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.

**Key Point**: This is nothing but,

^{ o}mechanical =

^{o}electrical for 2 pole case.

^{o}electrical

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,

__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^{o }and 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).

Fig.3 Reversal of direction R.M.F. |

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