---
title: "200 kW 1420 rpm slip ring motor (star)"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:slip-ring-200kw-star"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/slip-ring-200kw-star
site: BCT Engineering
---

# 200 kW 1420 rpm slip ring motor (star)

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

- Asked in: 2078 Bhadra, 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-200kw-star
- 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 200 kW, 3φ, 400 V, 50 Hz, 1420 rpm slip ring induction motor. Assume Bav = 0.5 Wb/m², ac = 30000 ampere conductors per meter, efficiency = 0.9, p.f = 0.9, 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 star connected.

## Short answer

Main dimensions: D=40.5 cm, L=38.2 cm; with star connection, Ts=18. Using 72 slots, 3 conductors per slot and 2 parallel paths per phase. Conductor cross-section is 50.9 mm², slot area about 382 mm².

## Answer

**Data:** 200 kW, 400 V, 50 Hz, 1420 rpm, so 4 poles and $n_s = 25$ rps; star connected; $\eta = 0.9$, pf $= 0.9$; $B_{av} = 0.5$, $ac = 30000$; $\delta = 3.5$ A/mm²; $L/\tau = 1.2$. Assume $K_w = 0.955$.

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

#### Turns per phase
$$
\begin{aligned}
\Phi &= B_{av}\tau L = 0.5 \times 0.318 \times 0.382 = 0.0608\ \text{Wb} \\
E_{ph} &= 400/\sqrt3 = 230.9\ \text{V} \\
T_s &= \frac{230.9}{4.44 \times 50 \times 0.0608 \times 0.955} = 17.9 \approx 18
\end{aligned}
$$

#### Number of slots
Total conductors $= 6 \times 18 = 108$ with one path. With 48 slots this gives 2.25 per slot (not a whole number) and 36 slots gives a large slot pitch (35 mm).
Use **2 parallel paths per phase** and $q = 6$: $S_s = 3 \times 4 \times 6 = 72$, slot pitch $= \pi \times 405/72 = 17.7$ mm.
Conductors $= 2 \times 108 = 216$, i.e. **3 per slot**, still giving 18 series turns per phase.

#### Conductor size
$$
\begin{aligned}
I_{ph} &= \frac{246914}{\sqrt3 \times 400} = 356.4\ \text{A} \\
I_c &= \frac{356.4}{2} = 178.2\ \text{A per path} \\
a_s &= \frac{178.2}{3.5} = 50.9\ \text{mm}^2
\end{aligned}
$$
(e.g. a strip about $4 \times 12.7$ mm).

#### Slot area
$$
A_{slot} = \frac{3 \times 50.9}{0.4} = 381.8\ \text{mm}^2
$$

**Answer:** $D = 40.5$ cm, $L = 38.2$ cm; $T_s = 18$; 72 slots (3 conductors per slot, 2 parallel paths); conductor 50.9 mm²; slot area $\approx 382$ mm². (With a single path, 36 slots × 3 conductors would also give $T_s = 18$, but with a large slot pitch.)

## Check yourself

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

1. What phase voltage is used for the star connection?
   A. 400 V
   B. 400/3 V
   C. 400 x sqrt3 V
   D. 400/sqrt3 = 230.9 V
   Answer: D. In star, phase emf is line voltage divided by sqrt3. (Key point: Star phase voltage)

2. How is Iph found for star?
   A. Iph = Q x 10^3/(sqrt3 x 400) = 356.4 A
   B. Iph = Q x 10^3/(3 x 400)
   C. Iph = Q x 10^3/400
   D. Iph = Q x 10^3/(3 x 230.9)/sqrt3
   Answer: A. In star, phase current equals line current = Q/(sqrt3 VL). (Key point: Phase current)

3. Why are 2 parallel paths used?
   A. To halve the voltage
   B. With one path, 108 conductors do not divide into a whole number per slot with a suitable pitch; 72 slots with 3 conductors fixes it
   C. To double Kw
   D. To reduce the flux
   Answer: B. Two paths double the conductors to 216 (3 per slot in 72 slots) while keeping 18 series turns. (Key point: Parallel paths)

4. What current is used for conductor area with 2 paths?
   A. 356.4 A, giving 101.8 mm^2
   B. 356.4 A, giving 50.9 mm^2
   C. 178.2 A per path, giving 50.9 mm^2
   D. 89.1 A, giving 25.5 mm^2
   Answer: C. Each conductor carries half the phase current: 178.2/3.5 = 50.9 mm^2. (Key point: Conductor size)

## 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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