---
title: "Main dimensions and poles of 40 kW shunt motor"
subject: "Electric Machine Design"
chapter: 5
key: "semester_5/electric_machine_design:dc-motor-40kw-poles"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-motor-40kw-poles
site: BCT Engineering
---

# Main dimensions and poles of 40 kW shunt motor

Electric Machine Design · Chapter 5: DC Machine Design

- Asked in: 2071 Chaitra, 8 marks
- Video: [DC Machine Main Dimension numerical problem 2 #electricalengineering #electricmotors](https://www.youtube.com/watch?v=u_HbAS-skR8) — EEE, SJBIT
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-motor-40kw-poles
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5

## Question

Calculate the main dimensions and the number of poles of a 40 kW, 240 V, 1450 rpm dc shunt motor so that a square pole face is obtained. The average gap density is 0.5 Wb/m² and ampere conductors per meter are 22000. Take full load efficiency of 92% and ratio of pole arc to pole pitch of 0.7.

## Short answer

Pa=(2+η)/(3η)×P=42.32 kW, Co=108.6, D²L=0.01613 m³. For square pole face, trying p=4 gives D=0.308 m, L=0.170 m, frequency 48.3 Hz (acceptable); p=6 exceeds 50 Hz, p=2 gives L>D. Hence 4 poles, D=0.308 m, L=0.170 m.

## Answer

The main dimensions come from the **output equation of a dc machine** (A.K. Sawhney):

$$
P_a = C_0 D^2 L n, \qquad C_0 = \pi^2 B_{av}\, ac \times 10^{-3}\ \text{kW/(m}^3\text{ rps)}
$$

where $P_a$ = armature power (kW), $D$ = armature diameter, $L$ = core length (m), $n$ = speed (rps).

For a motor, Sawhney takes the armature copper loss as about one-third of the total losses, which gives

$$
P_a = \frac{2+\eta}{3\eta}\,P
$$

#### Step 1: Armature power and $D^2L$
$$
\begin{aligned}
P_a &= \frac{2+0.92}{3\times0.92}\times40 = 42.32\ \text{kW} \\
C_0 &= \pi^2 \times 0.5 \times 22000\times10^{-3} = 108.6 \\
n &= 1450/60 = 24.17\ \text{rps} \\
D^2L &= \frac{42.32}{108.6\times24.17} = 0.01613\ \text{m}^3
\end{aligned}
$$

#### Step 2: Square pole face
Core length = pole arc: $L = \psi\pi D/p = 0.7\pi D/p$, so

$$
D^3 = \frac{D^2L\; p}{0.7\pi}
$$

#### Step 3: Choice of number of poles
Try common pole numbers and check frequency ($f = pn/2 \le 50$ Hz) and peripheral speed ($\le 30$ m/s):

| Poles $p$ | $D$ (m) | $L$ (m) | $\pi Dn$ (m/s) | $f$ (Hz) |
|---|---|---|---|---|
| 2 | 0.245 | 0.269 | 18.6 | 24.2 |
| 4 | 0.308 | 0.170 | 23.4 | 48.3 |
| 6 | 0.353 | 0.129 | 26.8 | 72.5 |

- $p = 6$ gives $f = 72.5$ Hz, too high (large iron loss).
- $p = 2$ gives a long, narrow machine ($L > D$) with heavy armature reaction per pole and large yoke; not used for 40 kW.
- $p = 4$ keeps $f$ below 50 Hz with a reasonable $L/D$. Armature current $\approx 40000/(0.92\times240) = 181$ A, so with a lap winding the current per path ($\approx 45$ A) and current per brush arm ($\approx 90$ A) are small.

**Choose 4 poles.**

#### Step 4: Main dimensions for $p = 4$
$$
\begin{aligned}
D^3 &= \frac{0.01613\times4}{0.7\pi} = 0.02934\ \text{m}^3 \\
D &= 0.308\ \text{m} \\
L &= \frac{0.7\pi\times0.308}{4} = 0.170\ \text{m} \\
\tau &= \frac{\pi D}{p} = 0.242\ \text{m}, \quad \text{pole arc} = 0.7\tau = 0.170\ \text{m}
\end{aligned}
$$

**Answer: Number of poles = 4, armature diameter $D \approx 0.308$ m, core length $L \approx 0.170$ m (pole face 0.170 m x 0.170 m).**

## Check yourself

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

1. For a dc motor, which relation gives the armature power Pa from the rated output P (Sawhney)?
   A. Pa = (1+2η)/(3η) x P
   B. Pa = (2+η)/(3η) x P
   C. Pa = η x P
   D. Pa = P/η
   Answer: B. For a motor, armature copper loss is taken as one-third of total losses, giving Pa = (2+η)/(3η) P; (1+2η)/(3η) is the generator form. (Key point: Armature power of a motor)

2. What condition gives a square pole face for p poles and pole-arc ratio ψ?
   A. L = D
   B. L = πD/p
   C. L = ψπD/p
   D. L = ψD
   Answer: C. A square pole face means core length equals pole arc, L = ψτ = ψπD/p. (Key point: Square pole face condition)

3. Why is p = 6 rejected for this motor?
   A. Flux frequency would be 72.5 Hz, above the 50 Hz limit, causing large iron loss
   B. Peripheral speed would exceed 30 m/s
   C. The armature diameter would be too small
   D. Current per brush arm would be too high
   Answer: A. f = pn/2 = 6 x 24.17/2 = 72.5 Hz, too high; peripheral speed (26.8 m/s) was still acceptable. (Key point: Why 6 poles is rejected)

4. Why is a 2-pole design not used for this 40 kW motor?
   A. Its frequency is too high
   B. It needs too many brushes
   C. Its peripheral speed is too high
   D. It gives a long narrow machine (L > D) with heavy armature reaction per pole and a large yoke
   Answer: D. With p = 2, L = 0.269 m exceeds D = 0.245 m, and armature reaction per pole and yoke size become large, so 4 poles is chosen. (Key point: Why 2 poles is rejected)

## Close variants set in other exams

- [Main dimensions and poles of 37 kW shunt motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-motor-37kw-poles) — Electric Machine Design, chapter 5, asked 3×

## Related in Electric Machine Design

- [Main dimensions of 5 kW generator (Bav 0.40)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-5kw-bav-040) — Electric Machine Design, chapter 5, asked 4×
- [30 kW 240 V generator, square pole](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-30kw-240v-square) — Electric Machine Design, chapter 5, asked 2×
- [600 kW generator: dimensions, poles, air gap](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-600kw-constraints) — Electric Machine Design, chapter 5
- [Main dimensions of 600 kW square-pole generator](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-600kw-square) — Electric Machine Design, chapter 5
- [1000 kW generator: dimensions, poles, air gap](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-1000kw-air-gap) — Electric Machine Design, chapter 5
- [Main dimensions of 5 kW generator (Bav 0.42)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-5kw-bav-042) — Electric Machine Design, chapter 5, asked 4×

## Related in other subjects

- [Hollow shaft 40 kW at 240 rpm: diameters](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ahollow-shaft-40kw-240rpm) — Strength of Materials, chapter 6
- [Shunt motor 480 V, 740 conductors](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Ashunt-motor-480v-740-conductors) — Electrical Machine, chapter 4
- [Reverse current (plugging) braking of 20 kW DC shunt motor](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Adc-shunt-reverse-current-braking-20kw) — Utilization of Electrical Energy, chapter 3