Rotating Magnetic Field in Three-Phase Induction Motor

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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 ( _m) 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,

$$\mathrm{\text{Synchronous Speed, } N_{S} \:=\: \frac{120 f}{P} \: RPM}$$

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,

$$\mathrm{\varphi_{1} \:=\: \varphi_{m} \: \sin \omega t \:\: \dotso \: (1)}$$

$$\mathrm{\varphi_{2} \:=\: \varphi_{m} \: \sin(\omega t \: - \: 120°) \:\: \dotso \: (2)}$$

$$\mathrm{\varphi_{3} \:=\: \varphi_{m} \: \sin(\omega t \:+\: 120°) \:\: \dotso \: (3)}$$

Here, _m 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,

$$\mathrm{\varphi_{h} \:=\: \varphi_{1} \: - \: \varphi_{2} \: \cos 60° \: -\: \varphi_{3} \: \cos 60° \:=\: \varphi_{1} \: - \: (\varphi_{2} \:+\: \varphi_{3}) \cos 60°}$$

$$\mathrm{\Rightarrow \: \varphi_{h} \: = \: \varphi_{1} \:-\: \frac{1}{2} \: (\varphi_{2} \:+\: \varphi_{3})}$$

$$\mathrm{\Rightarrow \: \varphi_{h} \: = \: (\varphi_{m} \:\sin \omega t) \: - \: \frac{1}{2} \: [\varphi_{m} \: \sin(\omega t \: - \: 120°) \:+\: \varphi_{m} \: \sin(\omega t \:+\: 120°)]}$$

$$\mathrm{\Rightarrow \: \varphi_{h} \:=\: (\varphi_{m} \: \sin \omega t) \: - \: \frac{\varphi_{m}}{2}(\sin \omega t \: \cos 120° \:-\: \cos \omega t \: \sin 120° \:+\: \sin \omega t \: \cos 120° \:+\: \cos \omega t \: \sin 120°)}$$

$$\mathrm{\Rightarrow \: \varphi_{h} \:=\: \varphi_{m} \: \sin \omega t \:-\: \left[\frac{\varphi_{m}}{2} \:\times \: (2 \sin \omega t) \: \times \: \left(\frac{-1}{2}\right)\right]}$$

$$\mathrm{\Rightarrow \: \varphi_{h} \:=\: \frac{3}{2} \: \varphi_{m} \: \sin \omega t \:\: \dotso \: (4)}$$

The resultant vertical component of the flux is given by,

$$\mathrm{\varphi_{v} \:=\: 0 \:-\: \varphi_{2} \: \cos 30° \:+\: \varphi_{3} \: \cos 30° \:=\: (- \varphi_{2} \:+\: \varphi_{3}) \: \cos 30°}$$

$$\mathrm{\Rightarrow \: \varphi_{v} \:=\: [- \varphi_{m} \: \sin(\omega t \:-\: 120°) \:+\: \varphi_{m} \:\sin(\omega t \:+\: 120°)]\: \cos 30°}$$

$$\mathrm{\Rightarrow \: \varphi_{v} \:=\: \frac{\sqrt{3}}{2} \: \varphi_{m} [-(\sin \omega t \: \cos 120° \: -\: \cos \omega t \: \sin 120°) \:+\: (\sin \omega t \: \cos 120° \:+\: \cos \omega t \: \sin 120°)]}$$

$$\mathrm{\Rightarrow \: \varphi_{v} \:=\: \frac{\sqrt{3}}{2} \: \varphi_{m}(2 \cos \omega t \: \sin 120°) \:=\: \frac{\sqrt{3}}{2} \: \varphi_{m} \: \times \: (2 \cos \omega t) \: \times \: \frac{\sqrt{3}}{2}}$$

$$\mathrm{\Rightarrow \: \varphi_{v} \:=\: \frac{3}{2} \: \varphi_{m} \: \cos \omega t \:\: \dotso \: (5)}$$

Therefore, the resultant flux is given by,

$$\mathrm{\varphi_{r} \:=\: \sqrt{\varphi_{h}^{2} \:+\: \varphi_{v}^{2}} \:=\: \sqrt{\left(\frac{3}{2} \: \varphi_{m} \: \sin \omega t \right)^{2} \:+\: \left(\frac{3}{2} \: \varphi_{m} \: \cos \omega t \right)^{2}}}$$

$$\mathrm{\Rightarrow \: \varphi_{r} \:=\: \frac{3}{2} \: \varphi_{m} \: (\sqrt{\sin 2 \omega t \:+\: \cos 2 \omega t}) \:=\: \frac{3}{2} \: \varphi_{m} \:\: \dotso \: (6)}$$

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 ( _m) per phase. Also, the resultant flux (_r) is independent of time, i.e., it is constant flux.

Again,

$$\mathrm{\tan \theta \:=\: \frac{\varphi_{v}}{\varphi_{h}} \:=\: \frac{\left(\frac{3}{2} \: \varphi_{m} \cos \omega t\right)}{\left(\frac{3}{2} \: \varphi_{m} \sin \omega t \right)} \:=\: \cot \omega t \:=\: \tan(90° \: - \: \omega t)}$$

$$\mathrm{\therefore \: \theta \:=\: (90° \: -\: \omega t) \:\: \dotso \: (7)}$$

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,

$$\mathrm{\omega \:=\: 2\pi f; \:\: and \:\: f \:=\: \frac{PN_{S}}{120};}$$

Case I - When ωt = 0°

This instant is corresponds to the point 0 in the waveform. Putting ωt = 0° in the equations (1), (2) and (3), we get,

$$\mathrm{\varphi_{1} \: = \: \varphi_{m} \: \sin 0° \: = \: 0}$$

$$\mathrm{\varphi_{2} \: = \: \varphi_{m} \: \sin(0° \: - \: 120°) \: = \: -\frac{\sqrt{3}}{2}\: \varphi_{m}}$$

$$\mathrm{\varphi_{3} \: = \: \varphi_{m} \: \sin(0° \: + \: 120°) \: = \: \frac{\sqrt{3}}{2} \: \varphi_{m}}$$

Refer Figure-2, the phasor for ₂ is shown along OY and the phasor for ₃ is shown along OB. Thus the resultant flux _r is the phasor sum of OY and OB which is shown along OA. The magnitude of the resultant flux is given by,

$$\mathrm{\varphi_{r} \: = \: 0A \: = \: 20E \: = \: 2 \: 0B \: \cos 30° \: = \: 2 \: \times \: \frac{\sqrt{3}}{2} \: \varphi_{m} \: \times \: \frac{\sqrt{3}}{2} \: = \: \frac{3}{2} \: \varphi_{m}}$$

Case II - When ωt = 60°

This instant is corresponds to point 1. Putting ωt = 60° in the equations (1), (2) and (3), we get,

$$\mathrm{\varphi_{1} \: = \: \varphi_{m} \: \sin 60° \: = \: frac{\sqrt{3}}{2} \: \varphi_{m}}$$

$$\mathrm{\varphi_{2} \: = \: \varphi_{m} \: \sin(60° \: - \: 120°) \: = \: - frac{\sqrt{3}}{2} \: \varphi_{m}}$$

$$\mathrm{\varphi_{3} \: = \: \varphi_{m} \: \sin(60° \: + \: 120°) \: = \: 0}$$

The phasors ₁, ₂ and _r are shown in Figure-3. The value of resultant flux is given by,

$$\mathrm{\varphi_{r} \: = \: 0A \: = \: 2 \: 0R \: \cos 30° \: = \: 2 \: \times \: \frac{\sqrt{3}}{2} \: \varphi_{m} \: \times \: \frac{\sqrt{3}}{2} \: = \: \frac{3}{2} \: \varphi_{m}}$$

Hence, it can be seen that the resultant flux is again $(\frac{3}{2} \varphi_m)$ but has rotated through an angle of 60° in the clockwise direction.

Case III - When ωt = 120°

This instant is corresponds to point 2. Putting ωt = 120° in the equations (1), (2) and (3), we get,

$$\mathrm{\varphi_{1} \: = \: \varphi_{m} \: \sin 120° \: = \: \frac{\sqrt{3}}{2} \: \varphi_{m}}$$

$$\mathrm{\varphi_{2} \: = \: \varphi_{m} \: \sin(120° \: - \: 120°) \: = \: 0}$$

$$\mathrm{\varphi_{3} \: = \: \varphi_{m} \: \sin(120° \: + \: 120°) \: = \: - \frac{\sqrt{3}}{2} \: \varphi_{m}}$$

Thus, the resultant flux is given by,

$$\mathrm{\varphi_{r} \: = \: 0A \: = \: 2 \: 0R \: \cos 30° \: = \: 2 \: \times \: \frac{\sqrt{3}}{2} \: \varphi_{m} \: \times \: \frac{\sqrt{3}}{2} \: = \: \frac{3}{2} \: \varphi_{m}}$$

Hence, the resultant flux is again $(\frac{3}{2} \varphi_m)$ obtained but has further rotated by an angle of 60° form point 1 in the clockwise direction (see Figure-4).

Case IV - When ωt = 180°

This instant is corresponds to point 3. Putting ωt = 180° in the equations (1), (2) and (3), we get,

$$\mathrm{\varphi_{1} \: = \: \varphi_{m} \: \sin 180° \: = \: 0}$$

$$\mathrm{\varphi_{2} \: = \: \varphi_{m} \: \sin(180° \: - \: 120°) \: = \: \frac{\sqrt{3}}{2} \: \varphi_{m}}$$

$$\mathrm{\varphi_{3} \: = \: \varphi_{m} \: \sin(180° \: + \: 120°) \: = \: -\frac{\sqrt{3}}{2} \: \varphi_{m}}$$

Thus, the resultant flux is given by,

$$\mathrm{\varphi_{r} \: = \: 0A \: = \: 2 \: 0Y \: \cos 30° \: = \: 2 \: \times \: \frac{\sqrt{3}}{2} \: \varphi_{m} \: \times \: \frac{\sqrt{3}}{2} \: = \: \frac{3}{2} \: \varphi_{m}}$$

Again, the resultant flux is equal to $(\frac{3}{2} \varphi_m)$ but has further rotated by an angle of 60° form the point 2 in the clockwise direction (see Figure-5).

From the above discussion, it is clear that a three phase balanced supply produces a rotating magnetic field.

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 _m.
• 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.