---
title: "250 HP 1430 rpm slip ring motor"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:slip-ring-250hp-1430rpm"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/slip-ring-250hp-1430rpm
site: BCT Engineering
---

# 250 HP 1430 rpm slip ring motor

Electric Machine Design · Chapter 4: Three phase induction motor design

- Asked in: 2072 Kartik, 10 marks
- 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-1430rpm
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4

## Question

Determine the main dimensions, turns per phase, number of slots, conductor size and slot area of a 250 HP, 3-φ, 400 V, 50 Hz, 1430 rpm slip ring induction motor. Assume Bav = 0.5 Wb/m², ac = 30000 ampere conductors per meter, efficiency = 0.9, Power factor = 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. Appropriate values for additional data required may be assumed.

## 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, 1430 rpm, so 4 poles with $N_s = 1500$ rpm ($n_s = 25$ 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.
Given $ac = 30000$ A/m. **Assumed:** $K_w = 0.955$ (full-pitch, 60° spread), 1 HP = 746 W.

#### 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. What winding factor is assumed when none is given?
   A. 0.5
   B. 1.0
   C. 0.7
   D. 0.955 (full-pitch, 60 degree spread)
   Answer: D. A standard assumption for a three-phase distributed winding. (Key point: Assumed Kw)

2. With L = 0.9425D and D^2L = 0.05845 m^3, what is D?
   A. 0.396 m
   B. 0.62 m
   C. 0.242 m
   D. 0.0620 m
   Answer: A. D^3 = 0.0620, so D = 0.396 m and L = 0.373 m. (Key point: Diameter)

3. Which formula gives turns per phase?
   A. Ts = 4.44 f Phi Kw/Eph
   B. Ts = Eph/(4.44 f Phi Kw), with Eph = 400 V for delta
   C. Ts = Eph/(2.22 f Phi)
   D. Ts = 400/sqrt3 /(4.44 f Phi Kw)
   Answer: B. Delta gives Eph = line voltage; Ts = 32.5. (Key point: Turns formula)

4. With 4 conductors per slot, how many slots?
   A. 24
   B. 72
   C. 48, with q = 4
   D. 36
   Answer: C. 6 x 32.5/48 = 4.07, close to 4. (Key point: Slot choice)

## Close variants set in other exams

- [200 kW 1420 rpm slip ring motor (star)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aslip-ring-200kw-star) — Electric Machine Design, chapter 4, asked 5×
- [250 HP 1410 rpm slip ring motor (kw 0.955)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aslip-ring-250hp-kw0955) — Electric Machine Design, chapter 4, asked 6×

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