---
title: "Main dimensions and poles of 37 kW shunt motor"
subject: "Electric Machine Design"
chapter: 5
key: "semester_5/electric_machine_design:dc-motor-37kw-poles"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-motor-37kw-poles
site: BCT Engineering
---

# Main dimensions and poles of 37 kW shunt motor

Electric Machine Design · Chapter 5: DC Machine Design

- Asked in: 2078 Bhadra, 10 marks; 2073 Shrawan, 8 marks (asked 2 times)
- 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-37kw-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 37 kW, 230 V, 1400 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 90% and the ratio of pole arc to pole pitch is 0.7.

## Short answer

Armature power Pa = 39.74 kW, C0 = 108.57, D^2L = 0.01569 m^3. Comparing pole numbers, 4 poles give acceptable frequency (46.7 Hz) and current per path. With square pole face L = 0.5498 D, D ≈ 30.6 cm, L ≈ 16.8 cm.

## Answer

**Data:** 37 kW, 230 V, 1400 rpm ($n = 23.33$ rps) shunt motor, $B_{av} = 0.5$ Wb/m², $ac = 22000$ A/m, $\eta = 0.9$, $\psi = 0.7$, square pole face.

#### Armature power
Taking armature copper loss as one-third of total losses (motor):
$$
P_a = P\,\frac{2+\eta}{3\eta} = 37 \times \frac{2.9}{2.7} = 39.74\ \text{kW}
$$

#### Output coefficient and $D^2L$
$$
\begin{aligned}
C_0 &= \pi^2B_{av}\,ac\times10^{-3} = \pi^2 \times 0.5 \times 22 = 108.57 \\
D^2L &= \frac{P_a}{C_0n} = \frac{39.74}{108.57 \times 23.33} = 0.01569\ \text{m}^3
\end{aligned}
$$

#### Square pole face
$L = \text{pole arc} = \psi\tau = 0.7\pi D/P$, so $D^3 = D^2L \cdot P/(0.7\pi)$.

#### Choice of number of poles
Full-load current $\approx 37000/(0.9 \times 230) = 178.7$ A.

| $P$ | $D$ (m) | $L$ (m) | $f = Pn/2$ (Hz) | Current per path (lap) |
|---|---|---|---|---|
| 2 | 0.243 | 0.267 | 23.3 | 89 A |
| 4 | 0.306 | 0.168 | 46.7 | 45 A |
| 6 | 0.350 | 0.128 | 70.0 | 30 A |

- 6 poles: frequency 70 Hz exceeds the usual 25–50 Hz limit (high iron loss).
- 2 poles: frequency 23.3 Hz is below the usual range; the core is long ($L > D$) and armature mmf per pole is high ($ac\,\pi D/2P \approx 4190$ A).
- 4 poles: frequency 46.7 Hz is within range, current per path 45 A, armature mmf per pole $= 22000 \times \pi \times 0.306/8 = 2640$ A, peripheral speed $= \pi \times 0.306 \times 23.33 = 22.4$ m/s (below 30 m/s).

**Choose $P = 4$.** Then
$$
\begin{aligned}
L &= \frac{0.7\pi D}{4} = 0.5498D \\
D^3 &= \frac{0.01569}{0.5498} = 0.02853 \Rightarrow D = 0.306\ \text{m} \\
L &= 0.5498 \times 0.306 = 0.168\ \text{m}
\end{aligned}
$$

**Answer:** 4 poles, $D \approx 30.6$ cm, $L \approx 16.8$ cm.

## Check yourself

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

1. Which formula gives Pa for the motor?
   A. Pa = P(2 + eta)/(3 eta)
   B. Pa = P(1 + 2 eta)/(3 eta)
   C. Pa = P / eta
   D. Pa = P x eta
   Answer: A. This assumes armature copper loss is one-third of total; Pa = 39.74 kW. (Key point: Armature power for motor)

2. Why is 6 poles rejected?
   A. Current per path is too high
   B. Pole pitch is too large
   C. Frequency of 70 Hz exceeds the 25-50 Hz limit and iron loss is high
   D. Peripheral speed exceeds 30 m/s
   Answer: C. f = Pn/2 = 6 x 23.33/2 = 70 Hz. (Key point: Why not 6 poles)

3. What are the final main dimensions?
   A. 4 poles, D about 16.8 cm, L about 30.6 cm
   B. 6 poles, D about 35 cm, L about 12.8 cm
   C. 2 poles, D about 24 cm, L about 27 cm
   D. 4 poles, D about 30.6 cm, L about 16.8 cm
   Answer: D. L = 0.5498 D with D^3 = 0.02853. (Key point: Result)

## Close variants set in other exams

- [Main dimensions and poles of 40 kW shunt motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-motor-40kw-poles) — Electric Machine Design, chapter 5, asked 3×

## Related in Electric Machine Design

- [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
- [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
- [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×
- [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×
- [Main dimensions of 200 kW shunt generator](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-200kw) — Electric Machine Design, chapter 5

## Related in other subjects

- [Shunt motor speed with field resistance added (250 V, 1400 rpm)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-250v-0p2ohm-1400rpm-50ohm) — Electrical Machines I, chapter 4, asked 5×
- [Torque of 4-pole induction motor at 1400 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-4pole-2ohm-8ohm-5nm-1400rpm) — Electrical Machines I, chapter 5
- [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