Rotating Magnetic Field (R.M.F.)

Rotating Magnetic Field (R.M.F.)

The rotating magnetic field can be defined as the field or flux having constant amplitude but whose axis is continuously rotating in a plane with a certain speed. So if the arrangement is made to rotate a permanent magnet, then the resulting field is a rotating magnetic field. But in this method, it is necessary to rotate a magnet physically to produce rotating magnetic field.

But in three phase induction motors such a rotating magnetic field is produced by supplying currents to a set of stationary windings, with the help of three phase a.c. supply. The current carrying windings produce the magnetic field or flux. And due to interaction of three fluxes produced due to three phase supply, resultant flux has a constant magnitude and its axis rotating in space, without physically rotating the windings. This type of field is nothing but rotating magnetic field. Let us study how it happens.

1. Production of R.M.F.

A three phase induction motor consists of three phase winding as its stationary part called stator. The three phase stator winding is connected in star or delta. The three phase windings are displaced from each other by 120°. The windings are supplied by a balanced three phase a.c. supply. This is shown in the Fig. 5.2.1. The three phase windings are denoted as R-R', Y-Y' and B-B'.

The three phase currents flow simultaneously through the windings and are displaced from each other by 120° electrical. Each alternating phase current produces its own flux which is sinusoidal. So all three fluxes are sinusoidal and are separated from each other by 120°. If the phase sequence of the windings is R-Y-B, then mathematical equations for the instantaneous values of the three fluxes ϕ_R, ϕ_Y md ϕ_B can be written as,

ϕ_R = ϕ_m sin  (wt) = ϕ_m  sin θ     ... (5.2.1)

ϕ_Y =   ϕ_m sin          (wt - 120°) = ϕ_m sin (θ  -120°) ...(5.2.2)

ϕ_B = ϕ_m sin  (wt - 240°) = ϕ_m sin (θ  - 240°)          ... (5.2.3)

As windings are identical and supply is balanced, the magnitude of each flux is 0 m. Due to phase sequence R-Y-B, flux ϕ_Y lags behind ϕ_R by 120° and ϕ_B lags ϕ_R by 120°. So ϕ_B ultimately 1ags ϕ_R by 240° The flux ϕ_R is taken as reference while writing the equations.

The Fig. 5.2.2 (a) shows the waveforms of three fluxes in space. The Fig. 5.2.2 (b) shows the phasor diagram which clearly shows the assumed positive directions of each flux. Assumed positive direction means whenever the flux is positive it must be represented along the direction shown and whenever the flux is negative it must be represented along the opposite direction to the assumed positive direction.

Let ϕ_R, ϕ_Y  and ϕ_B be the instantaneous values of three fluxes. The resultant flux ϕ_RT is the phasor addition of ϕ_R, ϕ_Y  and ϕ_B.

Let us find 0 T at the instants 1, 2, 3 and 4 as shown in the Fig. 5.2.2 (a) which represents the values of 0 as 0°, 60°, 120° and 180° respectively. The phasor addition can be performed by obtaining the values of ϕ_R, ϕ_Y  and ϕ_B by substituting values of θ in the equations (5.2.1), (5.2.2) and (5.2.3).

Case 1: θ = 0°

Substituting in the equations (5.2.1), (5.2.2) and (5.2.3) we get,

ϕ_R = ϕ_m sin 0° = 0

ϕ_Y = ϕ_m sin (- 120^o )= -0.866 ϕ_m

ϕ_B = ϕ_m sin (- 240^o ) = + 0.866 ϕ_m

The phasor addition is shown in the Fig. 5.2.3 (a).

The positive values are shown in assumed positive directions while negative values are shown in opposite direction to the assumed positive directions of the respective fluxes. Refer to assumed positive directions shown in the Fig. 5.2.2 (b).

BD is drawn perpendicular from

So magnitude of ϕ_T is 1.5 ϕ_m and its position is vertically upwards at θ = 0°.

Case 2 :      θ = 60°

Equations (5.2.1), (5.2.2) and (5.2.3) give us,

ϕ_R = ϕ_m sin 60° = + 0.866 ϕ_m

ϕ_Y = ϕ_m sin (- 60^o ) = -0.866 ϕ_m

ϕ_B = ϕ_m sin (- 180^o ) =  0

So ϕ_R is positive and ϕ_Y  is negative and hence drawing in appropriate directions we get phasor diagram as shown in the Fig. 5.2.3 (b).

Doing the same construction, drawing perpendicular from B on ϕ_T at D we get the same result as,

ϕ_T = 1.50 ϕ_m

But it can be seen that though its magnitude is 1.5 ϕ_m it has rotated through 60° in space, in clockwise direction, from its previous position.

Case 3 :      θ = 120°

Equations (5.2.1), (5.2.2) and (5.2.3) give us,

ϕ_R = ϕ_m sin 120 = + 0.866 ϕ_m

ϕ_Y = ϕ_m sin 0 = 0

ϕ_B = ϕ_m sin (- 120) = - 0.866 ϕ_m

So ϕ_R is positive and 0 B is negative. Showing ϕ_R and ϕ_B in the appropriate directions, we get the phasor diagram as shown in the Fig. 5.2.3 (c).

After doing the construction same as before i.e. drawing perpendicular from B on ϕ_T , it can be proved again that,

ϕ_T = 1.5 ϕ_m  

But the position of ϕ_T is such that it has rotated further through 60° from its previous position, in clockwise direction. And from its position at 0 = 0°, it has rotated through 120° in space, in clockwise direction.

Case 4 :      θ = 180°

From the equations (5.2.1), (5.2.2) and (5.2.3),

ϕ_R = ϕ_m sin (180°) = 0

ϕ_Y = ϕ_m sin (60°) = + 0.866 ϕ_m

ϕ_B  = ϕ_m sin (- 60°) = - 0.866 ϕ_m

So ϕ_R = 0, ϕ_Y is positive and ϕ_B is negative. Drawing ϕ_Y and ϕ_B in the appropriate directions, we get the phasor diagram as shown in the Fig. .2.3 (d)

From phasor diagram, it can be easily proved that,

ϕ_T = 1.5 ϕ_m

Thus the magnitude of ϕ_T once again remains same. But it can be seen that it has further rotated through 60° from its previous position in clockwise direction.

So for an electrical half cycle of 180°, the resultant ϕ_T has also rotated through 180°. This is applicable for the windings wound for 2 poles.

From the above discussion we have following conclusions :

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

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

Key Point This shows that when a three phase stationary windings are excited by balanced three phase a.c. supply then the resulting field produced is rotating magnetic field. Though nothing is physically rotating, the field produced is rotating in space having constant amplitude. 

2. Speed of R.M.F.

There exists a fixed relation between frequency f of a.c. supply to the windings, the number of poles P for which winding is wound and speed N r.p.m. of rotating magnetic field. For a standard frequency whatever speed of R.M.F. results is called synchronous speed, in case of induction motors. It is denoted as N_s.

N_s = 120f / P = Speed of R.M.F.

where          f = Supply frequency in Hz

P = Number of poles for which winding is wound

This is the speed with which R.M.F. rotates in space. Let us see how to change direction of rotation of R.M.F.

3. Direction of R.M.F

The direction of the R.M.F. is always from the axis of the leading phase of the three phase winding towards the lagging phase of the winding. In a phase sequence of R-Y-B, phase R leads Y by 120° and Y leads B by 120°. So R.M.F. rotates from axis of R to axis of Y and then to axis of B and so on. So its direction is clockwise as shown in the Fig. 5.2.4 (a). This direction can be reversed by interchanging any two terminals of the three phase windings while connecting to the three phase supply. The terminals Y and B are shown interchanged in the Fig. 5.2.4 (b). In such case the direction of R.M.F. will be anticlockwise.

As Y and B of windings are connected to B and Y from winding point of view the phase sequence becomes R-B-Y. Thus R.M.F. axis follows the direction from R to B to Y which is anticlockwise.

Key Point Thus by interchanging any two terminals of three phase winding while connecting it to three phase a.c. supply, direction of rotation of R.M.F. gets reversed.

Review Question

1.  Explain how a three phase supply produces a rotating magnetic field of constant value at constant speed with vector diagrams.