Unequal stator/rotor slots and output equation

Why stator slots should never equal rotor slots

If $S_s = S_r$, all rotor teeth face stator teeth at the same time. At these positions the reluctance is minimum, and a strong alignment force holds the rotor there. This cogging (magnetic locking) can stop the motor from starting, and it also produces noise and vibration. So $S_r \ne S_s$ is always used.

Output equation in terms of main dimensions

Let $D$ = stator bore, $L$ = core length, $P$ = poles, $n_s$ = synchronous speed (rps).

1. kVA input: $Q = 3E_{ph}I_{ph}\times10^{-3}$
2. Emf: $E_{ph} = 4.44f\Phi T_{ph}K_w$, with $f = Pn_s/2$
3. Specific magnetic loading:

$$B_{av} = \frac{P\Phi}{\pi DL} \Rightarrow \Phi = \frac{\pi DL B_{av}}{P}$$

4. Specific electric loading (total conductors $6T_{ph}$):

$$ac = \frac{6I_{ph}T_{ph}}{\pi D} \Rightarrow I_{ph}T_{ph} = \frac{\pi D\,ac}{6}$$

5. Substitute into $Q$:

$$\begin{aligned} Q &= 3 \times 4.44 \times \frac{Pn_s}{2} \times \frac{\pi DLB_{av}}{P} \times K_w \times \frac{\pi D\,ac}{6} \times 10^{-3} \\ &= 1.11\pi^2 B_{av}\,ac\,K_w\times10^{-3}\, D^2Ln_s \\ &= 11B_{av}\,ac\,K_w\times10^{-3}\, D^2Ln_s \end{aligned}$$

Hence

$$Q = C_0D^2Ln_s, \qquad C_0 = 11B_{av}\,ac\,K_w\times10^{-3}$$

This shows the output (kVA) is proportional to:

• the main dimensions through $D^2L$ (active volume),
• the specific loadings $B_{av}$ and $ac$ through $C_0$,
• the speed $n_s$.

The kW rating is related by $Q = \text{kW}/(\eta\cos\phi)$. For example, with $B_{av} = 0.44$, $ac = 21000$, $K_w = 0.955$: $C_0 = 97.07$, so a 3.235 kVA, 4-pole motor needs $D^2L = 3.235/(97.07 \times 25) = 1.33 \times 10^{-3}$ m³.