200 kW 1420 rpm slip ring motor (star)

Data: 200 kW, 400 V, 50 Hz, 1420 rpm, so 4 poles and $n_s = 25$ rps; star connected; $\eta = 0.9$, pf $= 0.9$; $B_{av} = 0.5$, $ac = 30000$; $\delta = 3.5$ A/mm²; $L/\tau = 1.2$. Assume $K_w = 0.955$.

Main dimensions

$$\begin{aligned} Q &= \frac{200}{0.9 \times 0.9} = 246.9\ \text{kVA} \\ C_0 &= 11 \times 0.5 \times 30000 \times 0.955 \times 10^{-3} = 157.58 \\ D^2L &= \frac{246.9}{157.58 \times 25} = 0.06268\ \text{m}^3 \end{aligned}$$

$L = 1.2 \times \pi D/4 = 0.9425D$, so $D^3 = 0.0665$:

$$D = 0.405\ \text{m}, \qquad L = 0.382\ \text{m}$$

Turns per phase

$$\begin{aligned} \Phi &= B_{av}\tau L = 0.5 \times 0.318 \times 0.382 = 0.0608\ \text{Wb} \\ E_{ph} &= 400/\sqrt3 = 230.9\ \text{V} \\ T_s &= \frac{230.9}{4.44 \times 50 \times 0.0608 \times 0.955} = 17.9 \approx 18 \end{aligned}$$

Number of slots

Total conductors $= 6 \times 18 = 108$ with one path. With 48 slots this gives 2.25 per slot (not a whole number) and 36 slots gives a large slot pitch (35 mm). Use 2 parallel paths per phase and $q = 6$: $S_s = 3 \times 4 \times 6 = 72$, slot pitch $= \pi \times 405/72 = 17.7$ mm. Conductors $= 2 \times 108 = 216$, i.e. 3 per slot, still giving 18 series turns per phase.

Conductor size

$$\begin{aligned} I_{ph} &= \frac{246914}{\sqrt3 \times 400} = 356.4\ \text{A} \\ I_c &= \frac{356.4}{2} = 178.2\ \text{A per path} \\ a_s &= \frac{178.2}{3.5} = 50.9\ \text{mm}^2 \end{aligned}$$

(e.g. a strip about $4 \times 12.7$ mm).

Slot area

$$A_{slot} = \frac{3 \times 50.9}{0.4} = 381.8\ \text{mm}^2$$

Answer: $D = 40.5$ cm, $L = 38.2$ cm; $T_s = 18$; 72 slots (3 conductors per slot, 2 parallel paths); conductor 50.9 mm²; slot area $\approx 382$ mm². (With a single path, 36 slots × 3 conductors would also give $T_s = 18$, but with a large slot pitch.)