---
title: "Main dimensions of 15 kW 2810 rpm cage motor"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:main-dimensions-15kw-2810rpm"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/main-dimensions-15kw-2810rpm
site: BCT Engineering
---

# Main dimensions of 15 kW 2810 rpm cage motor

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

- Asked in: 2074 Asoj, 8 marks; 2070 Chaitra, 8 marks (asked 2 times)
- Video: [Problem 1 based on Main Dimensions #Induction_Motor #sdyelgatte #Problem #designofelectricalmachines](https://www.youtube.com/watch?v=f2BOX58CWPg) — S D Yelgatte 📖 ⚡
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/main-dimensions-15kw-2810rpm
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4

## Question

Determine the main dimensions of a 15 kW, 3 phase, 400 V, 50 Hz, 2810 rpm squirrel cage induction motor having efficiency of 0.88 and a full load power factor of 0.9. Assume specific magnetic loading = 0.5 Wb/m² and specific electric loading = 25000 A/m. Take rotor peripheral speed as approximately 20 m/s at synchronous speed.

## Short answer

C0=131.31, D2L=2.885×10-3 m3. From peripheral speed 20 m/s, D=0.1273 m, then L=0.178 m (L/τ≈0.89).

## Answer

**Data:** 15 kW, 400 V, 50 Hz, 2810 rpm. Nearest synchronous speed is 3000 rpm, so $P = 2$ and $n_s = 50$ rps. Assume $K_w = 0.955$.

#### Output coefficient and $D^2L$
$$
\begin{aligned}
Q &= \frac{15}{0.88 \times 0.9} = 18.94\ \text{kVA} \\
C_0 &= 11 B_{av}\, ac\, K_w \times 10^{-3} = 11 \times 0.5 \times 25000 \times 0.955 \times 10^{-3} = 131.31 \\
D^2L &= \frac{Q}{C_0 n_s} = \frac{18.94}{131.31 \times 50} = 2.885 \times 10^{-3}\ \text{m}^3
\end{aligned}
$$

#### Diameter from peripheral speed
$$
v_a = \pi D n_s = 20\ \text{m/s} \Rightarrow D = \frac{20}{\pi \times 50} = 0.1273\ \text{m}
$$

#### Core length
$$
L = \frac{2.885\times10^{-3}}{0.1273^2} = 0.178\ \text{m}
$$

#### Check
Pole pitch $\tau = \pi D/2 = 0.200$ m, so $L/\tau = 0.89$. This lies in the usual range (about 0.6 to 2), so the proportions are acceptable.

**Answer:** $D \approx 12.7$ cm, $L \approx 17.8$ cm.

## Check yourself

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

1. For a 2810 rpm, 50 Hz induction motor, what pole number and synchronous speed are used?
   A. 2 poles, 3000 rpm (50 rps)
   B. 2 poles, 2810 rpm
   C. 4 poles, 1500 rpm
   D. 4 poles, 2810 rpm
   Answer: A. Rated speed is slightly below synchronous speed; the nearest synchronous speed is 3000 rpm, giving P = 2. (Key point: Synchronous speed)

2. How is C0 calculated?
   A. C0 = 11 x Bav x ac x 10^3 only, without Kw
   B. C0 = 11 x Bav x ac x Kw x 10^-3
   C. C0 = Bav x ac x Kw/11
   D. C0 = 11 x Bav x Kw/ac
   Answer: B. The output equation Q = C0 D^2 L ns has C0 = 11 Bav ac Kw x 10^-3, here 131.31. (Key point: Output coefficient)

3. How is the diameter found in this problem?
   A. From L/tau ratio, as no speed is given
   B. From D = sqrt(D^2L)
   C. From peripheral speed: D = va/(pi ns) = 20/(pi x 50) = 0.127 m
   D. From D = va x ns/pi
   Answer: C. Peripheral speed va = pi D ns is given, so D follows directly and L = D^2L/D^2. (Key point: Diameter from speed)

4. What does L/tau = 0.89 show?
   A. It is below 1, so the design must be rejected
   B. It shows peripheral speed is too high
   C. It gives the power factor of the motor
   D. It lies in the usual range of about 0.6 to 2, so proportions are acceptable
   Answer: D. L/tau is a sanity check on the shape of the machine after D and L are found. (Key point: Check of proportions)

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