---
title: "Main dimensions of 5 kW generator (Bav 0.42)"
subject: "Electric Machine Design"
chapter: 5
key: "semester_5/electric_machine_design:dc-generator-5kw-bav-042"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-5kw-bav-042
site: BCT Engineering
---

# Main dimensions of 5 kW generator (Bav 0.42)

Electric Machine Design · Chapter 5: DC Machine Design

- Asked in: 2076 Asoj, 10 marks; 2070 Asar, 8 marks (asked 2 times)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-5kw-bav-042
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5

## Question

Calculate main dimensions of a 5 kW, 250 V, 4 pole, 1500 rpm dc shunt generator having full load efficiency of 0.87 and designed to have a square pole face. Assume average flux density in gap = 0.42 Wb/m², ampere conductors per meter = 15,000 and ratio of pole arc to pole pitch = 0.66.

## Short answer

Armature power Pa = 5.249 kW, C0 = 62.18, D^2L = 3.377×10^-3 m^3. For square pole face, L = 0.5184 D, giving D ≈ 18.7 cm and L ≈ 9.68 cm. Peripheral speed and frequency check are within limits.

## Answer

**Data:** 5 kW, 250 V, 4 poles, 1500 rpm ($n = 25$ rps) shunt generator, $\eta = 0.87$, $B_{av} = 0.42$ Wb/m², $ac = 15000$ A/m, $\psi = 0.66$, square pole face.

#### Armature power
Taking armature copper loss as one-third of total losses (generator):
$$
P_a = P\,\frac{1+2\eta}{3\eta} = 5 \times \frac{1 + 2 \times 0.87}{3 \times 0.87} = 5 \times \frac{2.74}{2.61} = 5.249\ \text{kW}
$$

#### Output coefficient
$$
C_0 = \pi^2B_{av}\,ac\times10^{-3} = \pi^2 \times 0.42 \times 15000 \times 10^{-3} = 62.18
$$

#### $D^2L$
$$
D^2L = \frac{P_a}{C_0n} = \frac{5.249}{62.18 \times 25} = 3.377\times10^{-3}\ \text{m}^3
$$

#### Square pole face
Pole arc $= L$ and pole arc $= \psi\tau = 0.66 \times \pi D/4 = 0.5184D$, so $L = 0.5184D$:
$$
\begin{aligned}
D^3 &= \frac{3.377\times10^{-3}}{0.5184} = 6.514\times10^{-3}\ \text{m}^3 \\
D &= 0.1868\ \text{m} \\
L &= 0.5184 \times 0.1868 = 0.0968\ \text{m}
\end{aligned}
$$

#### Checks
- Pole pitch $\tau = \pi D/4 = 0.1467$ m; pole arc $= 0.66\tau = 0.0968$ m $= L$ (square face).
- Peripheral speed $= \pi Dn = 14.7$ m/s (below 30 m/s).
- Frequency $= Pn/2 = 50$ Hz (upper limit of the normal range).

**Answer:** $D \approx 18.7$ cm, $L \approx 9.68$ 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 generator?
   A. Pa = P(1 + 2 eta)/(3 eta) = 5.249 kW
   B. Pa = P(2 + eta)/(3 eta)
   C. Pa = P x eta
   D. Pa = P / (3 eta)
   Answer: A. Armature copper loss is taken as one-third of total losses. (Key point: Generator armature power)

2. What is C0 for Bav = 0.42 and ac = 15000?
   A. C0 = 11 Bav ac x 10^-3 with Bav 0.42
   B. C0 = pi^2 Bav ac x 10^-3 = 62.18
   C. C0 = pi Bav ac x 10^-3
   D. C0 = Bav ac x 10^-3
   Answer: B. Substituting gives C0 = 62.18. (Key point: Output coefficient)

3. What relation between L and D follows from the square face with psi = 0.66 and 4 poles?
   A. L = pi D/4
   B. L = 0.66 x pi D/4 = 0.5184 D
   C. L = 0.66 D
   D. L = D/0.5184
   Answer: B. Pole arc = psi tau = L. (Key point: Square face)

4. What are D and L?
   A. D and L swapped
   B. D about half of this with L twice
   C. D about 30 cm with L about 15 cm
   D. D about 18.7 cm, L about 9.68 cm
   Answer: D. D^3 = D^2L/0.5184 with D^2L = 3.377e-3. (Key point: Result)

## Close variants set in other exams

- [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×

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