Stator bore, length and conductors of 70 hp motor

Data: 70 HP $= 70 \times 746 = 52.22$ kW, 415 V, star, 6 poles ($n_s = 1000/60 = 16.67$ rps), $ac = 32000$ A/m, $B_{av} = 0.51$ Wb/m², $\eta = 0.9$, pf $= 0.91$, $L = \tau$. Assume $K_w = 0.955$.

Diameter and core length

$$\begin{aligned} Q &= \frac{52.22}{0.9 \times 0.91} = 63.76\ \text{kVA} \\ C_0 &= 11 \times 0.51 \times 32000 \times 0.955 \times 10^{-3} = 171.44 \\ D^2L &= \frac{63.76}{171.44 \times 16.67} = 0.02232\ \text{m}^3 \end{aligned}$$

$L = \tau = \pi D/6$:

$$D^3 = \frac{6 \times 0.02232}{\pi} = 0.04262 \Rightarrow D = 0.349\ \text{m}, \quad L = 0.183\ \text{m}$$

Number of stator conductors

Line (= phase) current for star:

$$I = \frac{63760}{\sqrt3 \times 415} = 88.70\ \text{A}$$

With 2 parallel paths, current per conductor $I_z = 88.70/2 = 44.35$ A.

$$Z = \frac{ac \times \pi D}{I_z} = \frac{32000 \times \pi \times 0.3493}{44.35} = 792$$

Conductors per slot

Slot loading $\le 750$ A-conductors, so conductors per slot $\le 750/44.35 = 16.9$.

• Choose $q = 3$: $S_s = 3 \times 6 \times 3 = 54$ slots, slot pitch $= \pi \times 349.3/54 = 20.3$ mm.
• Conductors per slot $= 792/54 = 14.7$, take 14 (even, for a double-layer winding).
• Total conductors $= 54 \times 14 = 756$; slot loading $= 14 \times 44.35 = 621$ A $< 750$ A.

Check: series turns per phase $= 756/(3 \times 2 \times 2) = 63$; from emf, $T_s = 239.6/(4.44 \times 50 \times \Phi \times 0.955) = 66.2$ with $\Phi = 0.51 \times 0.1829 \times 0.1829 = 0.01706$ Wb, so the gap density rises slightly to 0.536 Wb/m², which is acceptable.

Answer: $D \approx 34.9$ cm, $L \approx 18.3$ cm; about 792 conductors needed, arranged as 54 slots × 14 conductors = 756 conductors, slot loading 621 A-conductors.