Stator design of 120 kW 750 rpm motor

Data: 120 kW, 2200 V, 50 Hz, $N_s = 750$ rpm, star connected. $P = 120f/N_s = 8$ poles, $n_s = 12.5$ rps.

(i) and (ii) Diameter and length

$$\begin{aligned} Q &= \frac{120}{0.92 \times 0.88} = 148.22\ \text{kVA} \\ C_0 &= 11 \times 0.48 \times 26000 \times 0.955 \times 10^{-3} = 131.10 \\ D^2L &= \frac{148.22}{131.10 \times 12.5} = 0.09045\ \text{m}^3 \end{aligned}$$

$L = 1.25\tau = 1.25 \times \pi D/8 = 0.4909D$:

$$D^3 = \frac{0.09045}{0.4909} = 0.1843 \Rightarrow D = 0.569\ \text{m}, \quad L = 0.279\ \text{m}$$

(iii) Turns per phase

$$\begin{aligned} \tau &= \frac{\pi \times 0.569}{8} = 0.2235\ \text{m} \\ \Phi &= B_{av}\tau L = 0.48 \times 0.2235 \times 0.2793 = 0.02996\ \text{Wb} \\ E_{ph} &= \frac{2200}{\sqrt3} = 1270.2\ \text{V} \\ T_s &= \frac{1270.2}{4.44 \times 50 \times 0.02996 \times 0.955} = 199.97 \approx 200 \end{aligned}$$

(iv) Full-load current and conductor section

$$\begin{aligned} I_s &= \frac{Q \times 10^3}{\sqrt3 \times 2200} = \frac{148221}{3810.5} = 38.90\ \text{A} \\ a_s &= \frac{38.90}{5} = 7.78\ \text{mm}^2 \end{aligned}$$

(v) Total stator $I^2R$ loss

Conductors per phase $= 2T_s = 400$, each 0.75 m long, so length per phase $= 400 \times 0.75 = 300$ m.

$$\begin{aligned} R_s &= \frac{\rho\, l}{a} = \frac{0.021 \times 300}{7.78} = 0.810\ \Omega \\ \text{Loss} &= 3 I_s^2 R_s = 3 \times 38.90^2 \times 0.810 = 3676\ \text{W} \end{aligned}$$

($\rho = 0.021$ Ω per m per mm² is copper at about 75 °C.)

Answer: $D = 56.9$ cm, $L = 27.9$ cm, $T_s = 200$, $I_s = 38.9$ A, $a_s = 7.78$ mm², stator copper loss $\approx 3.68$ kW.