Main dimensions and turns of 5 HP cage motor

Data: 5 HP $= 5 \times 746 = 3730$ W, 400 V, 4 poles, 50 Hz ($n_s = 25$ rps), $B_{av} = 0.4$ Wb/m², $ac = 22000$ A/m, $\eta = 0.83$, pf $= 0.84$. Assumptions: star-delta starting, so the motor runs in delta ($E_{ph} = 400$ V); $K_w = 0.955$; $L = \tau$ (good overall design); $\delta = 4$ A/mm².

Main dimensions

$$\begin{aligned} Q &= \frac{3.73}{0.83 \times 0.84} = 5.35\ \text{kVA} \\ C_0 &= 11 \times 0.4 \times 22000 \times 0.955 \times 10^{-3} = 92.44 \\ D^2L &= \frac{5.35}{92.44 \times 25} = 2.315\times10^{-3}\ \text{m}^3 \end{aligned}$$

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

$$D^3 = \frac{4 \times 2.315\times10^{-3}}{\pi} = 2.948\times10^{-3} \Rightarrow D = 0.1434\ \text{m}, \quad L = 0.1126\ \text{m}$$

Peripheral speed $= \pi \times 0.1434 \times 25 = 11.3$ m/s (acceptable).

Stator turns per phase

$$\begin{aligned} \Phi &= B_{av}\tau L = 0.4 \times 0.1126 \times 0.1126 = 5.07\times10^{-3}\ \text{Wb} \\ T_s &= \frac{400}{4.44 \times 50 \times 5.07\times10^{-3} \times 0.955} = 372 \end{aligned}$$

With 36 slots ($q = 3$, slot pitch $= \pi \times 143.4/36 = 12.5$ mm): conductors per slot $= 6 \times 372/36 = 62$. So $T_s = 372$, 62 conductors per slot.

Size of conductor

$$\begin{aligned} I_{ph} &= \frac{5350}{3 \times 400} = 4.458\ \text{A} \\ a_s &= \frac{4.458}{4} = 1.115\ \text{mm}^2 \\ d &= \sqrt{\frac{4 \times 1.115}{\pi}} = 1.19\ \text{mm} \end{aligned}$$

Answer: $D = 14.3$ cm, $L = 11.3$ cm, $T_s = 372$ turns per phase, conductor 1.115 mm² (about 1.19 mm bare diameter).