---
title: "250 HP 1410 rpm slip ring motor (kw 0.955)"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:slip-ring-250hp-kw0955"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/slip-ring-250hp-kw0955
site: BCT Engineering
---

# 250 HP 1410 rpm slip ring motor (kw 0.955)

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

- Asked in: 2073 Chaitra, 8 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-kw0955
- 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, conductor cross section and slot area of a 250 HP, 3 phase, 50 Hz, 400 V, 1410 rpm slip ring induction motor. Assume Bav = 0.5 Wb/m², Ampere conductor per meter = 30,000, Efficiency = 0.9 and Power factor = 0.9, Winding factor = 0.955, Current density = 3.5 A/mm². The slot space factor is 0.4 and the 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, 1410 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, $K_w = 0.955$.

#### 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. How is Iph found for this delta machine?
   A. Iph = Q x 10^3/(sqrt3 x 400) = 332 A
   B. Iph = Q x 10^3/400 = 575.6 A
   C. Iph = 186.5 x 10^3/400
   D. Iph = Q x 10^3/(3 x 400) = 191.9 A
   Answer: D. For delta, Eph = VL and Iph = Q/(3 Eph). (Key point: Delta phase current)

2. With q = 4, how many slots are there?
   A. 48 (3 x 4 x 4)
   B. 24
   C. 36
   D. 96
   Answer: A. Slots = phases x poles x q. (Key point: Slots)

3. Why Ts = 32 when calculated is 32.5?
   A. Rounded down only to reduce loss
   B. 4 conductors per slot x 48 slots = 192, so Ts = 32
   C. Because Kw = 0.955
   D. Because delta requires multiples of 8
   Answer: B. Conductors per slot is an integer, which fixes Ts at 32 and slightly raises flux density to 0.508. (Key point: Ts check)

4. What area results with delta = 3.5 A/mm^2?
   A. 63.96 mm^2
   B. 191.9 mm^2
   C. 54.8 mm^2
   D. 5.48 mm^2
   Answer: C. 191.9/3.5 = 54.8 mm^2. (Key point: Conductor area)

## Close variants set in other exams

- [250 HP 1410 rpm slip ring motor (kw 0.95)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aslip-ring-250hp-1410rpm) — Electric Machine Design, chapter 4, asked 3×
- [250 HP 4-pole slip ring motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aslip-ring-250hp-4pole) — Electric Machine Design, chapter 4, asked 5×
- [200 kW 4-pole slip ring motor (delta)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aslip-ring-200kw-delta) — Electric Machine Design, chapter 4, asked 4×
- [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 1430 rpm slip ring motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aslip-ring-250hp-1430rpm) — Electric Machine Design, chapter 4, asked 3×

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