Question

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating. 2- the speed of the rotor when the slip is 0.05. 3- the frequency of the rotor currents when the slip is 0.04. 4- the frequency of the rotor currents at standstill.

Answer 1

Answer:

The answer is below

Explanation:

1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:

$n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm$

2) The speed of the rotor is the motor speed. The slip is given by:

$Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm$

3) The frequency of the rotor is given as:

$f_r=slip*f_s\\f_r=0.04*60=2.4\ Hz$

4) At standstill, the speed of the motor is 0, therefore the slip is 1.

The frequency of the rotor is given as:

$f_r=slip*f_s\\f_r=1*60=60\ Hz$