202302151214

15/02/23 12:15:07

@power-systems @study @py7002

If the frequency of the voltage in the windings of the motor is $f$ . Then the revolutions per minute of the magnetic field (at constant magnitude) is $n = f \times 60 = \frac{2 ω}{p 2 π} \times 60$ . Where p is the number of poles.
Slip is $ω_{s} - ω_{r} = s ω_{s}$ .
$ω_{s l} = ω_{s} - ω_{r}$
If s is positive, the moment of the coil is ‘chasing’ the rotating magnetic field. The greater this ‘tug’ on the magnetic moment, the greater the torque.
There is a changing magnetic field in the windings. The back emf induced in the windings is dependent on loops of the coil and possibly other factors. The emf induced in the ‘air gap’ or just the middle of the stator is $E_{a g} = k_{3} f ϕ_{a g}$ .
The voltage or back emf in the rotor is then $k_{2} f_{s l} ϕ_{a g}$ where k2 is a rotor dependent constant. This is a relative voltage though right, because of $f_{s l}$ ?
I might just have to take the derivations for granted.
Ultimately, the power loss in the rotor, as a percentage of the output electromagnetic power is $P_{r} = \frac{f _{s l}}{f - f _{s l}}$ .
At small values of $f_{s l}$ the motor speed varies approximately linearly with source frequency.
Slip frequency $f_{s l}$ is some portion of source frequency $f$ , determined by the relative speed of the rotating magnetic field relative to the rotor. $f_{s l} = \frac{ω _{s l}}{ω _{s}}$ .