---
title: "Stator design of 120 kW 750 rpm motor"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:stator-120kw-750rpm"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/stator-120kw-750rpm
site: BCT Engineering
---

# Stator design of 120 kW 750 rpm motor

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

- Asked in: 2072 Chaitra, 12 marks; 2071 Shrawan, 8 marks (asked 2 times)
- Video: [Video4 11 Part4 Numerical example on Main dimensions of IM](https://www.youtube.com/watch?v=OfdCYDSBQsA) — S.D Hirekodi
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/stator-120kw-750rpm
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4

## Question

Calculate (i) diameter (ii) length (iii) number of turns per phase (iv) full load current and cross section of conductors and (v) total I²R loss of stator of 3 phase, 120 kW, 2200 volts, 50 Hz, 750 rpm (synchronous speed), star connected slip ring induction motor from the following particulars: Bav = 0.48 tesla, ac = 26000 ampere conductor per metre, efficiency = 92%, power factor = 0.88, L = 1.25τ, Kw = 0.955, current density = 5 A/mm², mean length of stator conductors = 75 cm, ρ = 0.021 ohm per metre and mm² section.

## Short answer

C0=131.10, D2L=0.09045 m3 with L=1.25τ yields D=0.569 m, L=0.279 m. Ts=200 turns/phase, full-load current 38.9 A, conductor area 7.78 mm2; total stator I2R loss ≈3.68 kW.

## Answer

**Data:** 120 kW, 2200 V, 50 Hz, $N_s = 750$ rpm, star connected. $P = 120f/N_s = 8$ poles, $n_s = 12.5$ rps.

#### (i) and (ii) Diameter and length
$$
\begin{aligned}
Q &= \frac{120}{0.92 \times 0.88} = 148.22\ \text{kVA} \\
C_0 &= 11 \times 0.48 \times 26000 \times 0.955 \times 10^{-3} = 131.10 \\
D^2L &= \frac{148.22}{131.10 \times 12.5} = 0.09045\ \text{m}^3
\end{aligned}
$$
$L = 1.25\tau = 1.25 \times \pi D/8 = 0.4909D$:
$$
D^3 = \frac{0.09045}{0.4909} = 0.1843 \Rightarrow D = 0.569\ \text{m}, \quad L = 0.279\ \text{m}
$$

#### (iii) Turns per phase
$$
\begin{aligned}
\tau &= \frac{\pi \times 0.569}{8} = 0.2235\ \text{m} \\
\Phi &= B_{av}\tau L = 0.48 \times 0.2235 \times 0.2793 = 0.02996\ \text{Wb} \\
E_{ph} &= \frac{2200}{\sqrt3} = 1270.2\ \text{V} \\
T_s &= \frac{1270.2}{4.44 \times 50 \times 0.02996 \times 0.955} = 199.97 \approx 200
\end{aligned}
$$

#### (iv) Full-load current and conductor section
$$
\begin{aligned}
I_s &= \frac{Q \times 10^3}{\sqrt3 \times 2200} = \frac{148221}{3810.5} = 38.90\ \text{A} \\
a_s &= \frac{38.90}{5} = 7.78\ \text{mm}^2
\end{aligned}
$$

#### (v) Total stator $I^2R$ loss
Conductors per phase $= 2T_s = 400$, each 0.75 m long, so length per phase $= 400 \times 0.75 = 300$ m.
$$
\begin{aligned}
R_s &= \frac{\rho\, l}{a} = \frac{0.021 \times 300}{7.78} = 0.810\ \Omega \\
\text{Loss} &= 3 I_s^2 R_s = 3 \times 38.90^2 \times 0.810 = 3676\ \text{W}
\end{aligned}
$$
($\rho = 0.021$ Ω per m per mm² is copper at about 75 °C.)

**Answer:** $D = 56.9$ cm, $L = 27.9$ cm, $T_s = 200$, $I_s = 38.9$ A, $a_s = 7.78$ mm², stator copper loss $\approx 3.68$ kW.

## Check yourself

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

1. For a 750 rpm synchronous speed at 50 Hz, how many poles does the motor have?
   A. 8
   B. 4
   C. 6
   D. 10
   Answer: A. P = 120f/Ns = 120 x 50/750 = 8 poles, so ns = 12.5 rps. (Key point: Number of poles)

2. How is the kVA rating Q used in the output equation obtained from the 120 kW output?
   A. Q = kW x efficiency x power factor
   B. Q = kW/(efficiency x power factor) = 148.22 kVA
   C. Q = kW/efficiency only
   D. Q = kW/power factor only
   Answer: B. Input kVA must supply the output plus losses and reactive power, so divide kW by eta x cos phi. (Key point: Input kVA)

3. With L = 1.25 tau and 8 poles, how is D found from D^2L = 0.09045 m^3?
   A. Take D^2 = 0.09045 directly, so D = 0.30 m
   B. Put L = 1.25D, so D^3 = 0.09045/1.25
   C. Put L = 1.25 x pi D/8 = 0.4909D, so D^3 = 0.09045/0.4909 and D = 0.569 m
   D. Use D = sqrt(0.09045 x 8)
   Answer: C. The ratio L/tau with tau = pi D/P turns L into a multiple of D, so D^3 can be solved. (Key point: Separation of D and L)

4. In a star-connected stator, which voltage is used to find turns per phase?
   A. Line voltage 2200 V
   B. Line voltage divided by 3 = 733 V
   C. Line voltage times sqrt3
   D. Phase voltage 2200/sqrt3 = 1270 V
   Answer: D. In star, Eph = VL/sqrt3, and Ts = Eph/(4.44 f Phi Kw) gives about 200 turns. (Key point: Turns per phase)

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