Rotating Magnetic Field in Three-Phase Induction Motor

Digital ElectronicsElectronElectronics & Electrical

When 3-phase supply is fed to the stator winding of the 3-phase induction motor, a rotating magnetic field (RMF) is produced. This magnetic field is such that its poles do not remain in a fixed position on the stator but go on shifting their positions around the stator. For this reason, it is known as rotating magnetic field (RMF) or RMF.

Mathematically, it can be shown that the magnitude of this rotating magnetic field is constant and is equal to 1.5 times of the maximum flux ( ϕm) due to current in any phase.

The speed of the rotating magnetic field is known as synchronous speed (NS). The value of synchronous speed depends upon the number poles (P) on the stator and the supply frequency (f). Therefore,

$$\mathrm{Synchronous\:speed, 𝑁_𝑆 =\frac{120 𝑓}{𝑃}RPM}$$

Mathematical Analysis of Rotating Magnetic Field

Consider three identical coils which are displaced 120° apart from each other in space. Let these three coils are energised from a balanced 3-phase supply. Hence, each coil will produce an alternating flux along its own axis. Now, let the three instantaneous fluxes are given by,

$$\mathrm{\varphi_1 = \varphi_m sin \omega t … (1)}$$

$$\mathrm{\varphi_2 = \varphi_m sin(\omega t − 120°) … (2)}$$

$$\mathrm{\varphi_3 = \varphi_m sin(\omega t + 120°) … (3)}$$

Here, ϕm is the maximum value of flux due to current in any phase. The phasor diagram shows the three fluxes.

To determine the magnitude of the resultant flux, resolve each flux into horizontal and vertical components and then find their phasor sum.

Thus, the resultant horizontal component of flux is given by,

$$\mathrm{\varphi_h = \varphi_1 − \varphi_2 cos 60° − \varphi_3 cos 60° = \varphi_1 − (\varphi_2 + \varphi_3) cos 60°}$$

$$\mathrm{⇒ \varphi_h = \varphi_1 −\frac{1}{2}(\varphi_2 + \varphi_3)}$$

$$\mathrm{⇒ \varphi_h = (\varphi_m sin \omega t) −\frac{1}{2}[\varphi_m sin(\omega t − 120°) + \varphi_m sin(\omega t + 120°)]}$$

$$\mathrm{⇒ \varphi_h = (\varphi_m sin \omega t) −\frac{\varphi_m}{2}(sin \omega t\:cos 120° − cos \omega t\:sin 120°+ sin \omega t\:𝑐𝑜𝑠120° + cos \omega t\:sin 120°)}$$

$$\mathrm{⇒ \varphi_h = \varphi_m sin \omega t − [\frac{\varphi_m}{2}× (2 sin \omega t) × (\frac{-1}{2})]}$$

$$\mathrm{⇒ \varphi_h =\frac{3}{2}\varphi_m sin \omega t … (4)}$$

The resultant vertical component of the flux is given by,

$$\mathrm{\varphi_v = 0 − \varphi_2 cos 30° + \varphi_3 cos 30° = (−\varphi_2 + \varphi_3) cos 30°}$$

$$\mathrm{⇒ \varphi_v = [−\varphi_m sin(\omega t − 120°) + \varphi_m sin(\omega t + 120°)] cos 30°}$$

$$\mathrm{⇒ \varphi_v =\frac{\sqrt{3}}{2}\varphi_m [−(sin \omega t\:cos 120° − cos \omega t\:sin 120°)+(sin \omega t \:𝑐𝑜𝑠120° + cos \omega t\:sin 120°)]}$$

$$\mathrm{⇒ \varphi_v =\frac{\sqrt{3}}{2}\varphi_m(2 cos \omega t\:sin 120°) =\frac{\sqrt{3}}{2}\varphi_m × (2 cos \omega t) ×\frac{\sqrt{3}}{2}}$$

$$\mathrm{⇒ \varphi_v =\frac{3}{2}\varphi_m cos \omega t … (5)}$$

Therefore, the resultant flux is given by,

$$\mathrm{\varphi_r = \sqrt{\varphi_h^2 + \varphi_v^2} = \sqrt{(\frac{3}{2}\varphi_m sin \omega t)^2+ (\frac{3}{2}\varphi_m cos \omega t)^2}}$$

$$\mathrm{⇒ \varphi_r =\frac{3}{2}\varphi_m (\sqrt{sin2 \omega t + cos2 \omega t}) =\frac{3}{2}\varphi_m … (6)}$$

Hence, from the eqn. (6) it is clear that the magnitude of the resultant rotating magnetic field is equal to 1.5 times of maximum value of the flux ( ϕm) per phase. Also, the resultant flux (ϕr) is independent of time, i.e., it is constant flux.

Again,

$$\mathrm{tan \theta =\frac{\varphi_v}{\varphi_h}=\frac{(\frac{3}{2}\varphi_m cos \omega t)}{(\frac{3}{2}\varphi_m sin \omega t)}= cot \omega t = tan(90° − \omega t)}$$

$$\mathrm{\therefore \theta = (90° − \omega t) … (7)}$$

Eqn. (7) shows that the angle is the function of time. Hence,

  • Case 1 – At ωt = 0°; θ = 90°. It is corresponding to position A in the above figure.
  • Case 2 – At ωt = 90°; θ = 0°. It is corresponding to position B.
  • Case 3 – At ωt = 180°; θ = -90°. It is corresponding to position C.
  • Case 4 – At ωt = 270°; θ = -180°. It is corresponding to position D.

Hence, it can be seen that the resultant flux rotates in space in the clockwise direction with an angular velocity of ω radians per second. Therefore, for a machine of P poles,

$$\mathrm{\omega = 2𝜋𝑓;\: and\: 𝑓 =\frac{𝑃𝑁_𝑆}{120};}$$

The following conclusions can be drawn from the above discussion −

  • The 3-phase currents of a balanced 3-phase supply system produce a resultant flux of constant magnitude in the motor. The magnitude of the flux at every instant is 1.5 ϕm.
  • The resultant flux is rotating in nature and it rotates at an angular velocity same as that of the supply currents.
  • The direction of rotation of the resultant flux depends upon the phase sequence of supply system.
raja
Published on 28-Aug-2021 13:57:12
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