Stator design of 120 kW 1480 rpm motor

Data: 120 kW, 2200 V, 50 Hz, 1480 rpm, star connected. Nearest synchronous speed 1500 rpm, so $P = 4$ and $n_s = 25$ rps. $B_{av} = 0.8$ T is used as given (it is higher than the usual 0.3–0.6 Wb/m²).

i) and ii) Diameter and length

$$\begin{aligned} Q &= \frac{120}{0.92 \times 0.88} = 148.22\ \text{kVA} \\ C_0 &= 11B_{av}\,ac\,K_w\times10^{-3} = 11 \times 0.8 \times 26000 \times 0.955 \times 10^{-3} = 218.50 \\ D^2L &= \frac{148.22}{218.50 \times 25} = 0.02713\ \text{m}^3 \end{aligned}$$

$L = 1.25\tau = 1.25 \times \pi D/4 = 0.9817D$:

$$D^3 = \frac{0.02713}{0.9817} = 0.02764 \Rightarrow D = 0.302\ \text{m}, \quad L = 0.297\ \text{m}$$

iii) Turns per phase $T_s$

$$\begin{aligned} \tau &= \pi \times 0.3024/4 = 0.2375\ \text{m} \\ \Phi &= B_{av}\tau L = 0.8 \times 0.2375 \times 0.2968 = 0.05639\ \text{Wb} \\ E_{ph} &= 2200/\sqrt3 = 1270.2\ \text{V} \\ T_s &= \frac{1270.2}{4.44 \times 50 \times 0.05639 \times 0.955} = 106.2 \approx 106 \end{aligned}$$

iv) Full-load current and conductor area $a_s$

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

v) Stator $I^2R$ loss

Conductors per phase $= 2T_s = 212$, each 0.75 m long, so length per phase $= 212 \times 0.75 = 159$ m.

$$\begin{aligned} R_s &= \frac{0.021 \times 159}{7.78} = 0.429\ \Omega \\ \text{Loss} &= 3I_s^2R_s = 3 \times 38.90^2 \times 0.429 = 1948\ \text{W} \end{aligned}$$

Answer: $D = 30.2$ cm, $L = 29.7$ cm, $T_s = 106$, $I_s = 38.9$ A, $a_s = 7.78$ mm², stator copper loss $\approx 1.95$ kW.