---
title: "250 HP 1410 rpm slip ring motor (kw 0.95)"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:slip-ring-250hp-1410rpm"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/slip-ring-250hp-1410rpm
site: BCT Engineering
---

# 250 HP 1410 rpm slip ring motor (kw 0.95)

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

- Asked in: 2078 Kartik, 10 marks
- Video: [Design of slip ring induction motor](https://www.youtube.com/watch?v=5iyvJ2Y-k1s) — Praful Nandankar 
- 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-1410rpm
- 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 average flux density = 0.5 Wb/m²; Ampere conductor per meter = 30000; Efficiency = 0.9 and power factor = 0.9; Winding factor = 0.95; Current density = 3 A/mm². The slot space factor is 0.4 and the ratio core length to pole pitch is 1.2. The machine is delta connected.

## Short answer

Main dimensions: D=39.65 cm, L=37.37 cm. Stator turns per phase Ts=32, using 48 slots with 4 conductors each. Conductor area is 64 mm², slot area approximately 640 mm².

## Answer

**Data:** 250 HP $= 250 \times 746 = 186.5$ kW; 400 V, 50 Hz, 1410 rpm, so 4 poles and $n_s = 25$ rps; delta ($E_{ph} = 400$ V); $\eta = 0.9$, pf $= 0.9$, $K_w = 0.95$, $\delta = 3$ A/mm², $L/\tau = 1.2$.

#### Main dimensions
$$
\begin{aligned}
Q &= \frac{186.5}{0.9 \times 0.9} = 230.25\ \text{kVA} \\
C_0 &= 11 \times 0.5 \times 30000 \times 0.95 \times 10^{-3} = 156.75 \\
D^2L &= \frac{230.25}{156.75 \times 25} = 0.05876\ \text{m}^3
\end{aligned}
$$
$L = 1.2\pi D/4 = 0.9425D$:
$$
D^3 = \frac{0.05876}{0.9425} = 0.06234 \Rightarrow D = 0.3965\ \text{m}, \quad L = 0.3737\ \text{m}
$$

#### Turns per phase
$$
\begin{aligned}
\tau &= \pi \times 0.3965/4 = 0.3114\ \text{m} \\
\Phi &= 0.5 \times 0.3114 \times 0.3737 = 0.05819\ \text{Wb} \\
T_s &= \frac{400}{4.44 \times 50 \times 0.05819 \times 0.95} = 32.6
\end{aligned}
$$

#### Number of slots
$q = 4$: $S_s = 3 \times 4 \times 4 = 48$, slot pitch $= \pi \times 396.5/48 = 25.95$ mm.
Conductors per slot $= 6 \times 32.6/48 = 4.07 \approx 4$ → $Z_s = 192$, **$T_s = 32$**.

#### Conductor cross-section
$$
I_{ph} = \frac{230247}{3 \times 400} = 191.9\ \text{A}, \qquad a_s = \frac{191.9}{3} = 63.96\ \text{mm}^2
$$

#### Slot area
$$
A_{slot} = \frac{4 \times 63.96}{0.4} = 639.6\ \text{mm}^2
$$

**Answer:** $D = 39.65$ cm, $L = 37.37$ cm, $T_s = 32$, 48 slots, conductor $\approx 64$ mm², slot area $\approx 640$ mm².

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the rating converted?
   A. 250 x 1000 = 250 kW
   B. 250/746 kW
   C. 250 x 0.746 x 3 kW
   D. 250 x 746 = 186.5 kW
   Answer: D. 1 HP = 746 W. (Key point: HP to kW)

2. What are the poles and ns for 1410 rpm, 50 Hz?
   A. 4 poles, 25 rps
   B. 2 poles, 50 rps
   C. 6 poles, 16.7 rps
   D. 4 poles, 23.5 rps
   Answer: A. Nearest synchronous speed above 1410 rpm is 1500 rpm. (Key point: Poles)

3. With Kw = 0.95, what is the output coefficient?
   A. 157.58
   B. C0 = 11 x 0.5 x 30000 x 0.95 x 10^-3 = 156.75
   C. 165
   D. 148.2
   Answer: B. Using Kw = 0.95 rather than 0.955 gives 156.75. (Key point: Effect of Kw)

4. Delta phase current is 191.9 A and delta = 3 A/mm^2; what is the area?
   A. 54.8 mm^2
   B. 575.7 mm^2
   C. 63.96 mm^2
   D. 6.4 mm^2
   Answer: C. as = 191.9/3 = 63.96 mm^2. (Key point: Conductor area)

## Close variants set in other exams

- [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×
- [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×

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