Rotor slots and turns, 90 kW motor

Data: 90 kW, 500 V, 8 poles, star/star, $\eta = 0.9$, pf $= 0.86$, 63 stator slots with 6 conductors per slot, rotor open-circuit (slip-ring) voltage about 400 V.

Stator turns per phase

$$Z_s = 63 \times 6 = 378, \qquad T_s = \frac{378}{6} = 63$$

Rotor turns per phase (approximate)

At standstill the voltage ratio equals the turns ratio (taking $K_{ws} \approx K_{wr}$):

$$\begin{aligned} E_s &= 500/\sqrt3 = 288.7\ \text{V}, \qquad E_r = 400/\sqrt3 = 230.9\ \text{V} \\ T_r &= T_s \times \frac{E_r}{E_s} = 63 \times \frac{230.9}{288.7} = 50.4 \end{aligned}$$

Rotor conductors $\approx 6 \times 50.4 = 302$.

Number of rotor slots

• Stator slots per pole per phase $= 63/(8 \times 3) = 2.625$ (fractional).
• For a wound rotor an integral value is used, and $S_r \ne S_s$.
• Take $q_r = 3$: $S_r = 3 \times 8 \times 3 =$ 72 slots.

Conductors per slot and rotor turns

$302/72 = 4.2$, so take 4 conductors per slot (double layer):

$$Z_r = 72 \times 4 = 288, \qquad T_r = 288/6 = 48$$

Slip-ring voltage on open circuit:

$$V_r = 500 \times \frac{48}{63} = 381\ \text{V} \ (\text{about 400 V, acceptable})$$

Full-load rotor current

Stator current:

$$I_s = \frac{90000}{\sqrt3 \times 500 \times 0.9 \times 0.86} = 134.3\ \text{A}$$

Rotor mmf is about 85% of stator mmf (the rest is magnetising), so:

$$I_r = 0.85 \times I_s \times \frac{T_s}{T_r} = 0.85 \times 134.3 \times \frac{63}{48} = 149.8\ \text{A}$$

Answer: 72 rotor slots, 4 conductors per slot, $T_r = 48$ turns per phase (open-circuit voltage $\approx 381$ V), full-load rotor current $\approx 150$ A per phase.