200 kW 4-pole slip ring motor (delta)

Data: 200 kW, 400 V, 4 poles, 50 Hz ($n_s = 25$ rps), delta, $\eta = 0.9$, pf $= 0.9$, $\delta = 3.5$ A/mm², space factor 0.4, $L/\tau = 1.2$. Assumptions: "ac = 3000" is taken as 30000 A/m (3000 is far below the practical range of 5000–45000 A/m); $K_w = 0.955$.

Main dimensions

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

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

$$D^3 = \frac{0.06268}{0.9425} = 0.0665 \Rightarrow D = 0.405\ \text{m}, \quad L = 0.382\ \text{m}$$

Turns per phase

$$\begin{aligned} \tau &= \pi \times 0.405/4 = 0.318\ \text{m} \\ \Phi &= 0.5 \times 0.318 \times 0.382 = 0.0608\ \text{Wb} \\ T_s &= \frac{400}{4.44 \times 50 \times 0.0608 \times 0.955} = 31.06 \end{aligned}$$

Number of slots

Try $q = 4$: $S_s = 3 \times 4 \times 4 = 48$, slot pitch $= \pi \times 405/48 = 26.5$ mm (acceptable for a large machine). Conductors per slot $= 6 \times 31.06/48 = 3.88 \approx 4$, so $Z_s = 192$ and $T_s = 32$ (gap density becomes 0.485 Wb/m²).

Conductor and slot area

$$\begin{aligned} I_{ph} &= \frac{246914}{3 \times 400} = 205.8\ \text{A} \\ a_s &= \frac{205.8}{3.5} = 58.8\ \text{mm}^2 \\ A_{slot} &= \frac{4 \times 58.8}{0.4} = 588\ \text{mm}^2 \end{aligned}$$

(A strip of about $4.5 \times 13$ mm could be used.)

Answer: $D = 40.5$ cm, $L = 38.2$ cm, $T_s = 32$, 48 slots (4 conductors per slot), conductor 58.8 mm², slot area $\approx 588$ mm².