--- title: "250 HP 4-pole slip ring motor" subject: "Electric Machine Design" chapter: 4 key: "semester_5/electric_machine_design:slip-ring-250hp-4pole" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/slip-ring-250hp-4pole site: BCT Engineering --- # 250 HP 4-pole slip ring motor Electric Machine Design · Chapter 4: Three phase induction motor design - Asked in: 2074 Chaitra, 8 marks - Video: [Problem 1 based on Main Dimensions #Induction_Motor #sdyelgatte #Problem #designofelectricalmachines](https://www.youtube.com/watch?v=f2BOX58CWPg) — S D Yelgatte 📖 ⚡ - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/slip-ring-250hp-4pole - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4 ## Question Determine the main dimension, turns per phase, number of slots and slot area of a 250 HP, 400 V, 4-pole, 50 Hz slip ring induction motor. Assume Bav = 0.5 Wb/m², ac = 3000 Ampere conductor/m, efficiency (η) = 0.9, pf = 0.9, current density = 3.5 A/mm². The slot space factor is 0.4 and ratio of core length to pole pitch is 1.2. The machine is delta connected. ## Short answer Main dimensions: D=39.6 cm, L=37.3 cm. Stator turns per phase Ts=32, 48 slots with 4 conductors per slot. Conductor area is 54.8 mm², slot area about 548 mm². ## Answer **Data:** 250 HP $= 250 \times 746 = 186.5$ kW; 400 V, 50 Hz, 4 poles ($n_s = 1500/60 = 25$ rps); delta ($E_{ph} = 400$ V); $\eta = 0.9$, pf $= 0.9$, $B_{av} = 0.5$ Wb/m²; $\delta = 3.5$ A/mm²; $L/\tau = 1.2$; space factor 0.4. **Assumptions:** "ac = 3000" is taken as **30000 A/m** (3000 is far below the practical 5000–45000 A/m range, so it is a printing error); $K_w = 0.955$. #### Main dimensions $$ \begin{aligned} Q &= \frac{186.5}{0.9 \times 0.9} = 230.25\ \text{kVA} \\ C_0 &= 11B_{av}\,ac\,K_w\times10^{-3} = 11 \times 0.5 \times 30000 \times 0.955 \times 10^{-3} = 157.58 \\ D^2L &= \frac{Q}{C_0n_s} = \frac{230.25}{157.58 \times 25} = 0.05845\ \text{m}^3 \end{aligned} $$ With $L = 1.2\tau = 1.2\pi D/4 = 0.9425D$: $$ D^3 = \frac{0.05845}{0.9425} = 0.0620 \Rightarrow D = 0.396\ \text{m}, \quad L = 0.373\ \text{m} $$ #### Turns per phase $$ \begin{aligned} \tau &= \pi \times 0.396/4 = 0.311\ \text{m} \\ \Phi &= B_{av}\tau L = 0.5 \times 0.311 \times 0.373 = 0.0580\ \text{Wb} \\ T_s &= \frac{E_{ph}}{4.44f\Phi K_w} = \frac{400}{4.44 \times 50 \times 0.0580 \times 0.955} = 32.5 \end{aligned} $$ #### Number of slots Choose $q = 4$: $S_s = 3 \times 4 \times 4 = 48$ slots; slot pitch $= \pi \times 396/48 = 25.9$ mm (acceptable for this size). Conductors per slot $= 6 \times 32.5/48 = 4.07 \approx 4$, so $Z_s = 192$ and **$T_s = 32$** (gap density becomes 0.508 Wb/m², acceptable). #### Conductor cross-section $$ I_{ph} = \frac{Q\times10^3}{3E_{ph}} = \frac{230247}{3 \times 400} = 191.9\ \text{A}, \qquad a_s = \frac{191.9}{3.5} = 54.8\ \text{mm}^2 $$ (A rectangular strip of about $4.5 \times 12.2$ mm.) #### Slot area $$ A_{slot} = \frac{4 \times 54.8}{0.4} = 548\ \text{mm}^2 $$ **Answer:** $D = 39.6$ cm, $L = 37.3$ cm, $T_s = 32$, 48 slots with 4 conductors each, conductor area 54.8 mm², slot area $\approx 548$ mm². ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is ac = 3000 changed to 30000 A/m? A. Because HP is converted to kW B. Because the machine is delta connected C. Because of ns D. 3000 is far below the practical range (5000-45000 A/m) Answer: D. The value is taken as a misprint to give sensible dimensions. (Key point: Misprinted ac) 2. What is Q for 250 HP at eta = 0.9 and pf = 0.9? A. 186.5/(0.9 x 0.9) = 230.25 kVA B. 186.5/0.9 = 207 kVA C. 250/0.81 = 308.6 kVA D. 186.5 x 0.81 = 151 kVA Answer: A. Q = kW/(eta cos phi) with 250 HP = 186.5 kW. (Key point: Output kVA) 3. How is the slot area 548 mm^2 found? A. 4 x 54.8 x 0.4 B. 4 x 54.8/0.4 C. 54.8/0.4 D. 3.5 x 54.8 Answer: B. Copper per slot divided by space factor. (Key point: Slot area) 4. After Ts is rounded to 32, what happens to gap flux density? A. It falls to 0.45 B. It becomes 1.0 C. It rises slightly to 0.508 Wb/m^2 D. It is unchanged Answer: C. Fewer turns than 32.5 mean more flux for same voltage. 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