250 HP 1410 rpm slip ring motor (kw 0.95)

Data: 250 HP $= 250 \times 746 = 186.5$ kW; 400 V, 50 Hz, 1410 rpm, so 4 poles and $n_s = 25$ rps; delta ($E_{ph} = 400$ V); $\eta = 0.9$, pf $= 0.9$, $K_w = 0.95$, $\delta = 3$ A/mm², $L/\tau = 1.2$.

Main dimensions

$$\begin{aligned} Q &= \frac{186.5}{0.9 \times 0.9} = 230.25\ \text{kVA} \\ C_0 &= 11 \times 0.5 \times 30000 \times 0.95 \times 10^{-3} = 156.75 \\ D^2L &= \frac{230.25}{156.75 \times 25} = 0.05876\ \text{m}^3 \end{aligned}$$

$L = 1.2\pi D/4 = 0.9425D$:

$$D^3 = \frac{0.05876}{0.9425} = 0.06234 \Rightarrow D = 0.3965\ \text{m}, \quad L = 0.3737\ \text{m}$$

Turns per phase

$$\begin{aligned} \tau &= \pi \times 0.3965/4 = 0.3114\ \text{m} \\ \Phi &= 0.5 \times 0.3114 \times 0.3737 = 0.05819\ \text{Wb} \\ T_s &= \frac{400}{4.44 \times 50 \times 0.05819 \times 0.95} = 32.6 \end{aligned}$$

Number of slots

$q = 4$: $S_s = 3 \times 4 \times 4 = 48$, slot pitch $= \pi \times 396.5/48 = 25.95$ mm. Conductors per slot $= 6 \times 32.6/48 = 4.07 \approx 4$ → $Z_s = 192$, $T_s = 32$.

Conductor cross-section

$$I_{ph} = \frac{230247}{3 \times 400} = 191.9\ \text{A}, \qquad a_s = \frac{191.9}{3} = 63.96\ \text{mm}^2$$

Slot area

$$A_{slot} = \frac{4 \times 63.96}{0.4} = 639.6\ \text{mm}^2$$

Answer: $D = 39.65$ cm, $L = 37.37$ cm, $T_s = 32$, 48 slots, conductor $\approx 64$ mm², slot area $\approx 640$ mm².