---
title: "1000 kW generator: dimensions, poles, air gap"
subject: "Electric Machine Design"
chapter: 5
key: "semester_5/electric_machine_design:dc-generator-1000kw-air-gap"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-1000kw-air-gap
site: BCT Engineering
---

# 1000 kW generator: dimensions, poles, air gap

Electric Machine Design · Chapter 5: DC Machine Design

- Asked in: 2071 Shrawan, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-1000kw-air-gap
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5

## Question

Find the main dimension, number of poles and length of air gap of a 1000 kW, 500 V, 30[?] rpm dc generator. Assume the specific magnetic loading Bav = 0.7 Wb/m², ampere conductor per meter = 40000, square pole face, ratio of pole arc to pole pitch is 0.7. Assume efficiency as 92% and gap contraction factor as 1.15.

## Short answer

With 300 rpm, armature power 1029 kW, D²L=0.7447 m³, p=10 chosen from brush current limit. Square pole face gives D≈1.50 m, L≈0.33 m. Air-gap mmf 50% of armature mmf yields length ≈5.1 mm.

## Answer

**Assumptions:** the speed is read as **300 rpm** (the digit after "30" is unclear; 300 rpm is the usual value for this problem). The air-gap mmf is taken as **50% of the armature mmf per pole** (usual design value, not stated), and the limits $f \le 50$ Hz and current per brush arm $\le 400$ A are used to choose the poles.

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 generator (Sawhney), $P_a = \dfrac{1+2\eta}{3\eta}P$.

#### Step 1: $D^2L$
$$
\begin{aligned}
P_a &= \frac{1+2\times0.92}{3\times0.92}\times1000 = 1029\ \text{kW} \\
C_0 &= \pi^2\times0.7\times40000\times10^{-3} = 276.3 \\
n &= 300/60 = 5\ \text{rps} \\
D^2L &= \frac{1029}{276.3\times5} = 0.7447\ \text{m}^3
\end{aligned}
$$

#### Step 2: Number of poles
Armature current $I_a \approx 10^6/500 = 2000$ A. Lap winding, current per brush arm $= 2I_a/p \le 400 \Rightarrow p \ge 10$. Square pole face gives $L = 0.7\pi D/p$, so $D^3 = D^2L\,p/(0.7\pi)$.

| $p$ | $D$ (m) | $L$ (m) | $f$ (Hz) | A/brush arm | $AT_a$/pole |
|---|---|---|---|---|---|
| 8 | 1.394 | 0.383 | 20 | 500 | 10 949 |
| 10 | 1.502 | 0.330 | 25 | 400 | 9 435 |
| 12 | 1.596 | 0.292 | 30 | 333 | 8 355 |

$p = 8$ fails the brush current limit. $p = 10$ is the smallest number that satisfies it, with low frequency and armature mmf below 10 000 A. **Choose $p = 10$.**

#### Step 3: Main dimensions ($p = 10$)
$$
\begin{aligned}
D^3 &= \frac{0.7447\times10}{0.7\pi} = 3.386\ \text{m}^3 \\
D &= 1.502\ \text{m} \\
L &= \frac{0.7\pi\times1.502}{10} = 0.330\ \text{m}
\end{aligned}
$$

Peripheral speed $= \pi Dn = \pi\times1.502\times5 = 23.6$ m/s (safe).

#### Step 4: Air-gap length
$$
\begin{aligned}
AT_a &= \frac{ac\,\pi D}{2p} = \frac{40000\times\pi\times1.502}{20} = 9435\ \text{A} \\
AT_g &= 0.5\times9435 = 4718\ \text{A} \\
B_g &= \frac{B_{av}}{\psi} = \frac{0.7}{0.7} = 1.0\ \text{Wb/m}^2 \\
l_g &= \frac{AT_g}{800000\,B_g k_g} = \frac{4718}{800000\times1.0\times1.15} = 5.13\ \text{mm}
\end{aligned}
$$

**Answer: Poles = 10, $D \approx 1.50$ m, $L \approx 0.33$ m, air gap $l_g \approx 5.1$ mm.**

## Check yourself

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

1. Why is p = 10 chosen over p = 8?
   A. p = 8 gives frequency above 50 Hz
   B. p = 8 gives current per brush arm 2 x 2000/8 = 500 A, over the 400 A limit; 10 is the smallest that satisfies it
   C. p = 8 gives peripheral speed above 40 m/s
   D. p = 10 gives a smaller diameter
   Answer: B. Brush arm current 2Ia/p <= 400 A requires p >= 10 for Ia = 2000 A. (Key point: Pole number from brush current)

2. For a square pole face with ψ = 0.7 and p = 10, how is D found?
   A. D^3 = D^2L x p/(0.7π), giving D = 1.502 m and L = 0.330 m
   B. D^3 = D^2L/p
   C. D = sqrt(D^2L/L) with L = D
   D. D^3 = D^2L x 0.7π/p
   Answer: A. L = 0.7πD/p, so D^3 = 0.7447 x 10/(0.7π) = 3.386 and D = 1.502 m. (Key point: Square pole face)

3. What are ATa and ATg?
   A. ATa = ac x D / p = 6008 A, ATg = ATa
   B. ATa = ac πD/2p = 9435 A and ATg = 50% of it = 4718 A
   C. ATa = ac πD = 188700 A, ATg = ATa/2
   D. ATa = 4718 A and ATg = 9435 A
   Answer: B. ATa = 40000 x π x 1.502/20 = 9435 A; the gap takes 50%, so 4718 A. (Key point: Armature mmf and gap mmf)

4. What are Bg and lg?
   A. Bg = 0.49, lg = 10 mm
   B. Bg = Bav x ψ = 0.49, lg = 7 mm
   C. Bg = Bav = 0.7, lg = 7.3 mm
   D. Bg = Bav/ψ = 1.0 Wb/m^2, lg = 4718/(800000 x 1.0 x 1.15) = 5.1 mm
   Answer: D. Peak gap density is Bav/ψ = 1.0, so lg = ATg/(800000 Bg kg) = 5.13 mm. (Key point: Gap density and air-gap length)

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