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Reasons: slot numbers and speed vs size

Short answer

i) Equal stator and rotor slots cause magnetic locking (cogging), preventing starting and causing vibration and noise. ii) Higher speed reduces D2L for the same output because volume is inversely proportional to speed, yielding a smaller, lighter machine.

Give reasons for the followings: i) Stator slots should never be equal to the number of rotor slots ii) The size of induction machine will be small if it is designed with higher speed for the same output.

i) Stator slots should never equal rotor slots

  • If Ss=SrS_s = S_r, every rotor tooth lines up with a stator tooth at the same time. The reluctance of the magnetic path is then minimum at these positions and the rotor tends to stay there.
  • This strong alignment torque is called cogging or magnetic locking. The locking torque can exceed the starting torque, so the motor may fail to start, especially at reduced voltage.
  • Equal slot numbers also give strong slot-harmonic torques, which cause noise and vibration.
  • Therefore SrS_r is always made different from SsS_s (and differences like Ss−Sr=±PS_s - S_r = \pm P, ±2P\pm 2P, ±5P\pm 5P are also avoided).

ii) A higher-speed machine is smaller for the same output

From the output equation:

Q=C0D2Lns  ⇒  D2L=QC0nsQ = C_0 D^2 L n_s \;\Rightarrow\; D^2 L = \frac{Q}{C_0 n_s}
  • For the same output QQ and the same specific loadings (C0C_0 fixed), the active volume D2LD^2L is inversely proportional to the synchronous speed nsn_s.
  • A higher speed means fewer poles. The same power is produced with less torque (T=P/ωT = P/\omega), and torque decides the size of the machine.
  • Example: a 4-pole motor (ns=25n_s = 25 rps) needs only half the D2LD^2L of an 8-pole motor (ns=12.5n_s = 12.5 rps) of the same rating.
  • Hence a high-speed machine is smaller, lighter and cheaper.

✓ Answer checked by Claude Sonnet 5.5

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