Production of Rotating Magnetic Field in Three Phase Induction Motor

PRODUCTION OF ROTATING MAGNETIC FIELD

In a three phase system, three phase coils are displaced by 120° with each other. The principle of a 3‒phase, two‒pole stator having three identical windings placed 120 degrees apart are shown in Fig. 5.5.

The assumed positive directions of the fluxes are shown in Fig. 5.6. Let the maximum value of flux due to any one of the three phases be ϕ_m. The resultant flux ϕ_r, at any instant, is equal to the phasor sum of the fluxes due to three phases. We will consider values of ϕ_r, at four instants 1/6th time‒period apart corresponding to points marked 0, 1, 2, and 3 is Fig. 5.6.

1. When ωt=0°

Let

ϕ₁ = ϕ_m sin ωt                   ………(5.1)

ϕ₂ = ϕ_m sin (ωt ‒ 120°)             ………(5.2)

ϕ₃ = ϕ_m sin (ωt + 120°)             ………(5.3)

Substituting ωt=0 in Eqn (5.1), (5.2) and (5.3), we get,

 ϕ₁ = ϕ_m sin 0 = 0

ϕ₂ = ϕ_m sin (‒ 120°) = ‒ √3/2 ϕ_m

ϕ₃ = ϕ_m sin (120°) = √3/2 ϕ_m

The vector for ϕ_r, in Fig. 5.7 shows the phasor sum of OB and OC. Therefore,

 ϕ_r = OD = OB cos 30° + OC cos 30°

But |OC|=|OB|

∴ ϕ_r = 20C cos 30°

= 2 × √3/2 × ϕ_m × √3/2

ϕ_r = (3/2) ϕ_m

2. When ωt = 60°

This instant corresponding to point 1 in Fig. 5.6.

Substituting ωt=60° in Eqn (5.1), (5.2) and (5.3), we get,

ϕ₁ = ϕ_m sin 60° = √3/2 ϕ_m

ϕ₂ = ϕ_m sin (60°‒ 120°) = ϕ_m sin (‒60°) = ‒ √3/2 ϕ_m

ϕ₃ = ϕ_m sin (60°+120°) = ϕ_m sin 180° = 0

The resultant vector ϕ_r is the vector sum of ϕ₁ and ϕ₂. It is found that ϕ_r is again 3/2 ϕ_m but rotated clockwise through an angle of 60° as shown in Fig. 5.8.

 ϕ_r = OB cos 30 + OA cos 30°

| OB | = |OA|

ϕ_r = 2OA cos 30° = 2 × √3/2 ϕ_m × √3/2  = 3/2 ϕ_m

3. When ωt = 120°

This instant corresponds to position 2. in Fig. 5.6.

Substituting ωt = 120° in Eqn (5,1), (5,2) and (5.3),

We get,

ϕ₁ = ϕ_m sin 120° = √3/2 ϕ_m

ϕ₂ = ϕ_m sin (120° – 120°) = 0

ϕ₃ = ϕ_m sin (120° + 120°)

= ϕ_m sin (240) = – √3/2 ϕ_m

The resultant vector ϕ_r is the vector sum of ϕ₁ and ϕ₂. (Ref Fig. 5.9). Therefore,

 ϕ_r = OC cos 30 + OA cos 30

 |OA|=|OC|

 ϕ_r = 2 OA cos 30°

 ϕ_r = 2 × √3/2 ϕ_m × √3/2 = 3/2 ϕ_m

Hence the resultant flux is again 3/2ϕ_m but has further rotated clockwise through an angle of 60° from position at instant 1 in Fig. 5.9.

4. When ωt = 180°

This instant corresponds to position 3 in Fig. 5.6. Substituting ωt = 180° in eqn (5.1), (5.2) and (5.3) we get,

ϕ₁ = ϕ_m sin 180° = 0

ϕ₂ = ϕ_m sin (180° – 120°) = √3/2 ϕ_m

ϕ₃ = ϕ_m sin (180° + 120°) = ‒ √3/2 ϕ_m

The resultant vector ϕ_r is the vector sum of ϕ₂ and ϕ₃ (Ref Fig. 5.9. Therefore,

ϕ_r = OC cos 30° + OB cos 30°

|OC| = |OB |

ϕ_r = 20B cos 30°

= 2× √3/2 ϕ_m × √3/2

= 3/2 ϕ_m

The resultant flux is 3/2 ϕ_m but has further rotated clockwise through an angle of 60° from position at instant 2 in Fig. 5.10.