Rotating Magnetic Field in Three-Phase Induction Motor

A Rotating Magnetic Field (RMF) is created when a 3-phase supply is fed to the stator winding of a 3-phase Induction motor. Due to the nature of this Magnetic field, its poles Continuously move around the stator rather than remaining in a fixed position on it. It is called Rotating Magnetic field (RMF) or RMF for this reason.

The Magnitude of this Rotating Magnetic field can be Demonstrated Mathematically to be constant and equal to 1.5 times the maximum flux (m) brought on by 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,

Synchronous speed, NS=(120f/P) RPM

Mathematical Analysis of Rotating Magnetic Field

Think about three Identical coils that are 120 degrees apart from one another in space. Let a Balanced 3-phase supply power these three coils. Therefore, along its own axis, each coil will Generate an Alternating flux. The three Instantaneous fluxes should now be given by,

φ1=φmsinωt…(1)

φ2=φmsin(ωt−120°)…(2)

φ3=φmsin(ωt+120°)…(3)

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

Resolve each flux into its Horizontal and Vertical Components, then find the phasor sum to determine the size of the Resulting flux.

Thus, the Resultant Horizontal Component of flux is given by,

φh=φ1−φ2cos60°−φ3cos60°=φ1−(φ2+φ3)cos60°

⇒φh=φ1−(1/2)(φ2+φ3)

⇒φh=(φmsinωt)−(1/2)[φmsin(ωt−120°)+φmsin(ωt+120°)]

⇒φh=(φmsinωt)−(φm/2)(sinωtcos120°−cosωtsin120°+sinωtcos120°+cosωtsin120°)

⇒φh=φmsinωt−[(φm/2)×(2sinωt)×(−1/2)]

⇒φh=(3/2)φmsinωt…(4)

The Resultant Vertical Component of the flux is given by,

φv=0−φ2cos30°+φ3cos30°=(−φ2+φ3)cos30°

⇒φv=[−φmsin(ωt−120°)+φmsin(ωt+120°)]cos30°

⇒φv=(√3/2)φm[−(sinωtcos120°−cosωtsin120°)+(sinωtcos120°+cosωtsin120°)]

⇒φv=(√3/2)φm(2cosωtsin120°)=(√3/2)φm×(2cosωt)×(√3/2)

⇒φv=(3/2)φmcosωt…(5)

Therefore, the resultant flux is given by,

φr=√(φ^2h+φ^2v)=√[{(3/2)φmsinωt}^2+{(3/2)φmcosωt}^2]

⇒φr=(3/2)φm(√(sin2ωt+cos2ωt))=(3/2)φm…(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 ( ϕ_m) per phase. Also, the resultant flux (ϕ_r) is independent of time, i.e., it is constant flux.

Again,

tanθ=φv/φh={((3/2)φmcosωt)}/{((3/2)φmsinωt)}=cotωt=tan(90°−ωt)

∴θ=(90°−ωt)…(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.

As a result, it is clear that the resulting flux rotates through space in a clockwise direction at a speed of radians per second. Consequently, for a P-pole machine,

ω=2πf; and f=PNS/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 ϕ_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.