---
title: "200 kW 4-pole slip ring motor (delta)"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:slip-ring-200kw-delta"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/slip-ring-200kw-delta
site: BCT Engineering
---

# 200 kW 4-pole slip ring motor (delta)

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

- Asked in: 2079 Bhadra, 12 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-200kw-delta
- 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 200 kW, 400 V, 4-pole, 50 Hz slip ring induction motor. Assume Bav = 0.5 Wb/m², ac = 3000 AC/m, η = 0.9, p.f. = 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=40.5 cm, L=38.2 cm. Stator turns per phase Ts=32, using 48 slots with 4 conductors per slot. Conductor cross-section is 58.8 mm², giving a slot area of about 588 mm².

## Answer

**Data:** 200 kW, 400 V, 4 poles, 50 Hz ($n_s = 25$ rps), delta, $\eta = 0.9$, pf $= 0.9$, $\delta = 3.5$ A/mm², space factor 0.4, $L/\tau = 1.2$.
**Assumptions:** "ac = 3000" is taken as **30000 A/m** (3000 is far below the practical range of 5000–45000 A/m); $K_w = 0.955$.

#### Main dimensions
$$
\begin{aligned}
Q &= \frac{200}{0.9 \times 0.9} = 246.9\ \text{kVA} \\
C_0 &= 11 \times 0.5 \times 30000 \times 0.955 \times 10^{-3} = 157.58 \\
D^2L &= \frac{246.9}{157.58 \times 25} = 0.06268\ \text{m}^3
\end{aligned}
$$
$L = 1.2\tau = 1.2\pi D/4 = 0.9425D$:
$$
D^3 = \frac{0.06268}{0.9425} = 0.0665 \Rightarrow D = 0.405\ \text{m}, \quad L = 0.382\ \text{m}
$$

#### Turns per phase
$$
\begin{aligned}
\tau &= \pi \times 0.405/4 = 0.318\ \text{m} \\
\Phi &= 0.5 \times 0.318 \times 0.382 = 0.0608\ \text{Wb} \\
T_s &= \frac{400}{4.44 \times 50 \times 0.0608 \times 0.955} = 31.06
\end{aligned}
$$

#### Number of slots
Try $q = 4$: $S_s = 3 \times 4 \times 4 = 48$, slot pitch $= \pi \times 405/48 = 26.5$ mm (acceptable for a large machine).
Conductors per slot $= 6 \times 31.06/48 = 3.88 \approx 4$, so $Z_s = 192$ and **$T_s = 32$** (gap density becomes 0.485 Wb/m²).

#### Conductor and slot area
$$
\begin{aligned}
I_{ph} &= \frac{246914}{3 \times 400} = 205.8\ \text{A} \\
a_s &= \frac{205.8}{3.5} = 58.8\ \text{mm}^2 \\
A_{slot} &= \frac{4 \times 58.8}{0.4} = 588\ \text{mm}^2
\end{aligned}
$$
(A strip of about $4.5 \times 13$ mm could be used.)

**Answer:** $D = 40.5$ cm, $L = 38.2$ cm, $T_s = 32$, 48 slots (4 conductors per slot), conductor 58.8 mm², slot area $\approx 588$ mm².

## Check yourself

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

1. Why is 'ac = 3000' treated as 30000 A/m?
   A. Because the motor is delta connected
   B. To keep Bav at 0.5
   C. To make Kw equal to 1
   D. 3000 is far below the practical range of 5000-45000 A/m, so it is a misprint
   Answer: D. A realistic ac value is needed to get sensible dimensions. (Key point: Correction of data)

2. What are D^2L and the resulting D when L = 1.2 tau for 4 poles?
   A. D^2L = 0.0627 m^3 and D = 0.405 m
   B. D^2L = 0.627 m^3 and D = 0.81 m
   C. D^2L = 0.0627 m^3 and D = 0.81 m
   D. D^2L = 0.0063 m^3 and D = 0.2 m
   Answer: A. D^3 = 0.0627/0.9425 = 0.0665, giving D = 0.405 m. (Key point: Main dimensions)

3. Why is Ts rounded to 32 rather than 31?
   A. Because delta needs even turns
   B. Conductors per slot must be a whole number: 4 x 48 = 192, so Ts = 192/6 = 32
   C. To increase the current
   D. To make q = 3
   Answer: B. 6 x 31.06/48 = 3.88, rounded to 4 per slot. (Key point: Turns per phase)

4. How is the slot area of 588 mm^2 obtained?
   A. 4 x 58.8 x 0.4
   B. 58.8/0.4
   C. 4 x 58.8/0.4
   D. 205.8/0.4
   Answer: C. Slot area equals conductors per slot times conductor area divided by space factor. (Key point: Slot area)

## Close variants set in other exams

- [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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