250 HP 4-pole slip ring motor

Data: 250 HP $= 250 \times 746 = 186.5$ kW; 400 V, 50 Hz, 4 poles ($n_s = 1500/60 = 25$ rps); delta ($E_{ph} = 400$ V); $\eta = 0.9$, pf $= 0.9$, $B_{av} = 0.5$ Wb/m²; $\delta = 3.5$ A/mm²; $L/\tau = 1.2$; space factor 0.4. Assumptions: "ac = 3000" is taken as 30000 A/m (3000 is far below the practical 5000–45000 A/m range, so it is a printing error); $K_w = 0.955$.

Main dimensions

$$\begin{aligned} Q &= \frac{186.5}{0.9 \times 0.9} = 230.25\ \text{kVA} \\ C_0 &= 11B_{av}\,ac\,K_w\times10^{-3} = 11 \times 0.5 \times 30000 \times 0.955 \times 10^{-3} = 157.58 \\ D^2L &= \frac{Q}{C_0n_s} = \frac{230.25}{157.58 \times 25} = 0.05845\ \text{m}^3 \end{aligned}$$

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

$$D^3 = \frac{0.05845}{0.9425} = 0.0620 \Rightarrow D = 0.396\ \text{m}, \quad L = 0.373\ \text{m}$$

Turns per phase

$$\begin{aligned} \tau &= \pi \times 0.396/4 = 0.311\ \text{m} \\ \Phi &= B_{av}\tau L = 0.5 \times 0.311 \times 0.373 = 0.0580\ \text{Wb} \\ T_s &= \frac{E_{ph}}{4.44f\Phi K_w} = \frac{400}{4.44 \times 50 \times 0.0580 \times 0.955} = 32.5 \end{aligned}$$

Number of slots

Choose $q = 4$: $S_s = 3 \times 4 \times 4 = 48$ slots; slot pitch $= \pi \times 396/48 = 25.9$ mm (acceptable for this size). Conductors per slot $= 6 \times 32.5/48 = 4.07 \approx 4$, so $Z_s = 192$ and $T_s = 32$ (gap density becomes 0.508 Wb/m², acceptable).

Conductor cross-section

$$I_{ph} = \frac{Q\times10^3}{3E_{ph}} = \frac{230247}{3 \times 400} = 191.9\ \text{A}, \qquad a_s = \frac{191.9}{3.5} = 54.8\ \text{mm}^2$$

(A rectangular strip of about $4.5 \times 12.2$ mm.)

Slot area

$$A_{slot} = \frac{4 \times 54.8}{0.4} = 548\ \text{mm}^2$$

Answer: $D = 39.6$ cm, $L = 37.3$ cm, $T_s = 32$, 48 slots with 4 conductors each, conductor area 54.8 mm², slot area $\approx 548$ mm².