--- title: "Rotor slots and turns, 90 kW motor" subject: "Electric Machine Design" chapter: 4 key: "semester_5/electric_machine_design:rotor-design-90kw-slip-ring" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/rotor-design-90kw-slip-ring site: BCT Engineering --- # Rotor slots and turns, 90 kW motor Electric Machine Design · Chapter 4: Three phase induction motor design - Asked in: 2075 Asoj, 6 marks - Video: [Problems on design of wound rotor](https://www.youtube.com/watch?v=ZBs2EMmNwik) — Anand Melinamane - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/rotor-design-90kw-slip-ring - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4 ## Question A 90 kW, 500 V, three phase, 8 pole slip ring induction motor having 0.9 efficiency and power factor of 0.86 has 63 stator slots with 6 conductors per slot. If the slip ring voltage on open circuit is to be about 400 V, find the number of rotor slots, rotor turns per phase, number of conductors per slot and appropriate full load rotor current per phase. Both stator and rotor are star connected. ## Short answer From voltage ratio and mmf balance: 72 rotor slots, 4 conductors per slot give 48 turns per phase, open-circuit line voltage 381 V; full-load rotor current ≈150 A per phase. ## Answer **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are stator turns per phase found? A. Ts = 378 B. Ts = 378/3 C. Ts = 63 x 6/2 D. Zs = 63 x 6 = 378, Ts = Zs/6 = 63 Answer: D. Total conductors are shared by three phases with 2 conductors per turn, so Ts = Zs/6. (Key point: Stator turns) 2. How is the rotor turn count estimated from open-circuit voltage? A. Tr = Ts x Er/Es with phase voltages, about 50 B. Tr = Ts x 500/400 C. Tr = Ts D. Tr = Ts x Is/Ir Answer: A. At standstill, voltage ratio equals turns ratio, using 230.9/288.7. (Key point: Rotor turns) 3. Why is 72 rotor slots chosen? A. To equal the 63 stator slots B. Integral slots per pole per phase (q = 3) for a wound rotor with Sr different from Ss C. Because it is 8 x 9 D. Because fractional slots are required Answer: B. A wound rotor needs an integral q, and unequal slots avoid cogging. (Key point: Rotor slots) 4. How is full-load rotor current estimated? A. Ir = Is Tr/Ts B. Ir = Is C. Ir = 0.85 Is Ts/Tr, about 150 A D. Ir = 0.85 Is Tr/Ts Answer: C. Rotor mmf is about 85% of stator mmf (rest is magnetising), so Ir = 0.85 x 134.3 x 63/48. 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