---
title: "Air gap and rotor bars, 10 kW motor"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:cage-rotor-10kw-design-data"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/cage-rotor-10kw-design-data
site: BCT Engineering
---

# Air gap and rotor bars, 10 kW motor

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

- Asked in: 2073 Shrawan, 12 marks
- Video: [Air Gap Length Estimation |Design of Three Phase Induction Motor|](https://www.youtube.com/watch?v=gs9NO-ht6Q4) — Anish Vora
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/cage-rotor-10kw-design-data
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4

## Question

The following design data are provided for an induction motor: Diameter of stator bore (D) = 16 cm, Length of stator core (L) = 8.5 cm, Average flux density (Bav) = 0.44 Wb/m², Power factor = 0.85, Efficiency = 86%, Frequency = 50 Hz, Current density = 5 A/mm², Stator slots = 36, Rotor slots = 30, Length of rotor bar = 15 cm, Mean diameter of end ring = 12 cm, Resistivity of bar conductor = 0.020 Ohm-metre, Power output of 3-phase, 4-pole, 400 V, delta connected = 10 kW. Calculate no-load maximum flux, length of air gap, no. of turns per phase, rotor bar current and area, end ring current and area, losses in bars and end rings.

## Short answer

No-load flux 4.70 mWb, air gap 0.433 mm, stator turns 402. Rotor bar current 744 A (area 148.8 mm²); end-ring current 1776 A (area 355.2 mm²). Bar loss 334.8 W, end-ring loss 133.9 W.

## Answer

**Data:** 10 kW, 400 V, 3-phase, 4 poles, delta, 50 Hz; $D = 0.16$ m, $L = 0.085$ m, $B_{av} = 0.44$ Wb/m², $\eta = 0.86$, pf $= 0.85$, $\delta = 5$ A/mm², $S_s = 36$, $S_r = 30$, bar length 0.15 m, end-ring mean diameter 0.12 m.
**Assumptions:** $K_{ws} = 0.955$; resistivity taken as $0.02$ Ω per m per mm² (i.e. $0.02 \times 10^{-6}$ Ω·m; "0.020 Ω-m" is a misprint); same current density 5 A/mm² in bars and rings.

#### Flux per pole
$$
\begin{aligned}
\tau &= \frac{\pi D}{P} = \frac{\pi \times 0.16}{4} = 0.1257\ \text{m} \\
\Phi_m &= B_{av}\tau L = 0.44 \times 0.1257 \times 0.085 = 4.70\times10^{-3}\ \text{Wb}
\end{aligned}
$$

#### Length of air gap
$$
l_g = 0.2 + 2\sqrt{DL} = 0.2 + 2\sqrt{0.16 \times 0.085} = 0.433\ \text{mm}
$$

#### Turns per phase
$$
T_s = \frac{E_{ph}}{4.44f\Phi K_w} = \frac{400}{4.44 \times 50 \times 4.70\times10^{-3} \times 0.955} = 401.4
$$
Conductors per slot $= 6 \times 401.4/36 = 66.9 \approx 67$, so $Z_s = 2412$ and **$T_s = 402$**.

#### Stator phase current
$$
Q = \frac{10}{0.86 \times 0.85} = 13.68\ \text{kVA}, \qquad I_s = \frac{13680}{3 \times 400} = 11.40\ \text{A}
$$

#### Rotor bar current and area
Rotor mmf $\approx 0.85$ of stator mmf:
$$
\begin{aligned}
I_b &= 0.85 \times \frac{6 I_s T_s K_{ws}}{S_r} = 0.85 \times \frac{6 \times 11.40 \times 402 \times 0.955}{30} = 744.0\ \text{A} \\
a_b &= \frac{744.0}{5} = 148.8\ \text{mm}^2
\end{aligned}
$$

#### End-ring current and area
$$
\begin{aligned}
I_e &= \frac{S_r I_b}{\pi P} = \frac{30 \times 744.0}{\pi \times 4} = 1776\ \text{A} \\
a_e &= \frac{1776}{5} = 355.2\ \text{mm}^2
\end{aligned}
$$

#### Losses in bars
$$
\begin{aligned}
r_b &= \frac{\rho l_b}{a_b} = \frac{0.02 \times 0.15}{148.8} = 2.016\times10^{-5}\ \Omega \\
P_b &= S_r I_b^2 r_b = 30 \times 744.0^2 \times 2.016\times10^{-5} = 334.8\ \text{W}
\end{aligned}
$$

#### Losses in end rings
$$
\begin{aligned}
r_e &= \frac{\rho\,\pi D_e}{a_e} = \frac{0.02 \times \pi \times 0.12}{355.2} = 2.122\times10^{-5}\ \Omega \\
P_e &= 2I_e^2 r_e = 2 \times 1776^2 \times 2.122\times10^{-5} = 133.9\ \text{W}
\end{aligned}
$$

Total rotor copper loss $= 334.8 + 133.9 = 468.7$ W.

| Quantity | Value |
|---|---|
| Flux per pole | 4.70 mWb |
| Air gap | 0.433 mm |
| Turns per phase | 402 |
| Bar current, area | 744 A, 148.8 mm² |
| End-ring current, area | 1776 A, 355.2 mm² |
| Bar loss | 334.8 W |
| End-ring loss | 133.9 W |

## Check yourself

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

1. How is the flux per pole found for the 10 kW cage motor?
   A. Phi = Bav x tau x L with tau = pi D / P
   B. Phi = Bav x D x L
   C. Phi = Bav x pi D x L
   D. Phi = Bav x tau^2
   Answer: A. Flux per pole is the average density times the pole area, tau x L, where tau = pi D / P = 0.1257 m. (Key point: Flux per pole)

2. Which empirical relation gives the air gap length used in the answer?
   A. lg = 0.2 + 2 sqrt(DL) mm
   B. lg = 0.2 sqrt(DL) mm
   C. lg = 2 + 0.2 sqrt(DL) mm
   D. lg = 0.2 + 2 DL mm
   Answer: A. The answer uses lg = 0.2 + 2 sqrt(DL) mm, which gives 0.433 mm. (Key point: Air gap length)

3. In a delta-connected stator at 400 V, what phase emf is used in Ts = Eph/(4.44 f Phi Kw)?
   A. 400 V, since phase voltage equals line voltage in delta
   B. 400/sqrt(3) = 231 V
   C. 400 x sqrt(3) = 693 V
   D. 400/3 = 133 V
   Answer: A. In delta the phase voltage equals the line voltage, so Eph = 400 V and Ts = 401.4, rounded to 402 turns. (Key point: Turns per phase)

4. How is the rotor bar current obtained in the answer?
   A. Ib = Is x Ts / Sr
   B. From the rotor mmf taken as 0.85 of the stator mmf: Ib = 0.85 x 6 Is Ts Kws / Sr
   C. Ib = Is x Sr / Ts
   D. Ib = 6 Is Ts Kws x Sr
   Answer: B. Rotor mmf is about 0.85 of stator mmf, and this mmf is shared among the Sr rotor bars, giving 744 A. (Key point: Rotor bar current)

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