Basic Electrical Engineering (3110005)

BEE-3110005

Winter-2019

Question-4c-OR

BE | Semester-1 Winter-2019 | 11-01-2020

Q4) (c)

Explain Generation of Rotating Magnetic Field in 3-phase Induction Motor with diagrams and equations.

Generation of Rotating Magnetic Field (RMF)

When stationary three phase winding coils are supplied by an alternating 3-phase supply then uniform Rotating Magnetic Field (or flux) of constant value is produced.
The principle of 3-phase, 2-pole stator having three identical winding coils are placed by 120° electrical apart. The sinusoidal flux due to three phase windings is shown in Fig. 2.
The directions of the positive fluxes are shown individually below at different positions.
Let us say that the maximum value of the flux due to any one of the three phases be $\emptyset_{m}$ . The resultant flux $\emptyset_{r}$ , at any instant is given by the resultant sum of the individual fluxes $\emptyset_{1}$ , $\emptyset_{2}$ , and $\emptyset_{3}$ , due to three phases.
We have considered the $\frac{1}{6}$ th time period apart corresponding to points marked 0, 1, 2 and 3 in Fig. 1.

When $θ = 0 °$ (at point 0),the Resultant flux,

We have,

\emptyset_{1} = 0

\emptyset_{2} = - \frac{\sqrt{3}}{2} \emptyset_{m}

\emptyset_{3} = \frac{\sqrt{3}}{2} \emptyset_{m}

Now,

\emptyset_{r} = \sqrt{{\emptyset_{2}}^{2} + {\emptyset_{3}}^{2} - 2 \emptyset_{2} \emptyset_{3} \cos θ}

\emptyset_{r} = \sqrt{{(- \frac{\sqrt{3}}{2} \emptyset_{m})}^{2} + {(\frac{\sqrt{3}}{2} \emptyset_{m})}^{2} - 2 (- \frac{\sqrt{3}}{2} \emptyset_{m}) (\frac{\sqrt{3}}{2} \emptyset_{m}) \cos 60 °}

\emptyset_{r} = \sqrt{\frac{3}{4} {\emptyset_{m}}^{2} + \frac{3}{4} {\emptyset_{m}}^{2} + \frac{3}{4} {\emptyset_{m}}^{2}}

\emptyset_{r} = \sqrt{[\frac{3}{4} + \frac{3}{4} + \frac{3}{4}] {\emptyset_{m}}^{2}}

\emptyset_{r} = \sqrt{[\frac{3 + 3 + 3}{4}] {\emptyset_{m}}^{2}}

\emptyset_{r} = \sqrt{[\frac{9}{4}] {\emptyset_{m}}^{2}}

\emptyset_{r} = \frac{3}{2} \emptyset_{m}

When $θ = 60 °$ (at point 1),the Resultant flux,

We have,

\emptyset_{1} = \frac{\sqrt{3}}{2} \emptyset_{m}

\emptyset_{2} = - \frac{\sqrt{3}}{2} \emptyset_{m}

\emptyset_{3} = 0

Now,

\emptyset_{r} = \sqrt{{\emptyset_{1}}^{2} + {\emptyset_{2}}^{2} - 2 \emptyset_{1} \emptyset_{2} \cos θ}

\emptyset_{r} = \sqrt{{(\frac{\sqrt{3}}{2} \emptyset_{m})}^{2} + {(- \frac{\sqrt{3}}{2} \emptyset_{m})}^{2} - 2 (\frac{\sqrt{3}}{2} \emptyset_{m}) (- \frac{\sqrt{3}}{2} \emptyset_{m}) \cos 60 °}

\emptyset_{r} = \sqrt{\frac{3}{4} {\emptyset_{m}}^{2} + \frac{3}{4} {\emptyset_{m}}^{2} + \frac{3}{2} (\frac{1}{2}) {\emptyset_{m}}^{2}}

\emptyset_{r} = \sqrt{[\frac{3}{4} + \frac{3}{4} + \frac{3}{4}] {\emptyset_{m}}^{2}}

\emptyset_{r} = \sqrt{[\frac{3 + 3 + 3}{4}] {\emptyset_{m}}^{2}}

\emptyset_{r} = \sqrt{[\frac{9}{4}] {\emptyset_{m}}^{2}}

\emptyset_{r} = \frac{3}{2} \emptyset_{m}

When $θ = 120 °$ (at point 2),the Resultant flux,

We have,

\emptyset_{1} = \frac{\sqrt{3}}{2} \emptyset_{m}

\emptyset_{2} = 0

\emptyset_{3} = - \frac{\sqrt{3}}{2} \emptyset_{m}

Now,

\emptyset_{r} = \sqrt{{\emptyset_{1}}^{2} + {\emptyset_{3}}^{2} - 2 \emptyset_{1} \emptyset_{3} \cos θ}

\emptyset_{r} = \sqrt{{(\frac{\sqrt{3}}{2} \emptyset_{m})}^{2} + {(- \frac{\sqrt{3}}{2} \emptyset_{m})}^{2} - 2 (\frac{\sqrt{3}}{2} \emptyset_{m}) (- \frac{\sqrt{3}}{2} \emptyset_{m}) \cos 60 °}

\emptyset_{r} = \sqrt{\frac{3}{4} {\emptyset_{m}}^{2} + \frac{3}{4} {\emptyset_{m}}^{2} + \frac{3}{4} {\emptyset_{m}}^{2}}

\emptyset_{r} = \sqrt{[\frac{3}{4} + \frac{3}{4} + \frac{3}{4}] {\emptyset_{m}}^{2}}

\emptyset_{r} = \sqrt{[\frac{3 + 3 + 3}{4}] {\emptyset_{m}}^{2}}

\emptyset_{r} = \sqrt{[\frac{9}{4}] {\emptyset_{m}}^{2}}

\emptyset_{r} = \frac{3}{2} \emptyset_{m}

When $θ = 180 °$ (at point 3),the Resultant flux,

We have,

\emptyset_{1} = 0

\emptyset_{2} = \frac{\sqrt{3}}{2} \emptyset_{m}

\emptyset_{3} = - \frac{\sqrt{3}}{2} \emptyset_{m}

Now,

\emptyset_{r} = \sqrt{{\emptyset_{2}}^{2} + {\emptyset_{3}}^{2} - 2 \emptyset_{2} \emptyset_{3} \cos θ}

\emptyset_{r} = \sqrt{{(- \frac{\sqrt{3}}{2} \emptyset_{m})}^{2} + {(\frac{\sqrt{3}}{2} \emptyset_{m})}^{2} - 2 (- \frac{\sqrt{3}}{2} \emptyset_{m}) (\frac{\sqrt{3}}{2} \emptyset_{m}) \cos 60 °}

\emptyset_{r} = \sqrt{\frac{3}{4} {\emptyset_{m}}^{2} + \frac{3}{4} {\emptyset_{m}}^{2} + \frac{3}{4} {\emptyset_{m}}^{2}}

\emptyset_{r} = \sqrt{[\frac{3}{4} + \frac{3}{4} + \frac{3}{4}] {\emptyset_{m}}^{2}}

\emptyset_{r} = \sqrt{[\frac{3 + 3 + 3}{4}] {\emptyset_{m}}^{2}}

\emptyset_{r} = \sqrt{[\frac{9}{4}] {\emptyset_{m}}^{2}}

\emptyset_{r} = \frac{3}{2} \emptyset_{m}

FOR UNDERSTANDING ONLY

Note: Any above cases, any one vector has zero value and 60° is the angle between other two present vectors as per law of parallelogram, hence, in each above cases cos60 is considered. The value of θ is considered with sin function, as the waveforms are sinewaves.

Basic Electrical Engineering (3110005)

Q4) (c)

Explain Generation of Rotating Magnetic Field in 3-phase Induction Motor with diagrams and equations.

Generation of Rotating Magnetic Field (RMF)

When $θ = 0 °$ (at point 0),the Resultant flux,

When $θ = 60 °$ (at point 1),the Resultant flux,

When $θ = 120 °$ (at point 2),the Resultant flux,

When $θ = 180 °$ (at point 3),the Resultant flux,

Questions

Go to Question Paper

Basic Electrical Engineering (3110005)

Q4) (c)

Explain Generation of Rotating Magnetic Field in 3-phase Induction Motor with diagrams and equations.

Generation of Rotating Magnetic Field (RMF)

When θ = 0° (at point 0),the Resultant flux,

When θ = 60° (at point 1),the Resultant flux,

When θ = 120° (at point 2),the Resultant flux,

When θ = 180° (at point 3),the Resultant flux,

Questions

Go to Question Paper

When $θ = 0 °$ (at point 0),the Resultant flux,

When $θ = 60 °$ (at point 1),the Resultant flux,

When $θ = 120 °$ (at point 2),the Resultant flux,

When $θ = 180 °$ (at point 3),the Resultant flux,