---
title: "Main dimensions of 600 kW square-pole generator"
subject: "Electric Machine Design"
chapter: 5
key: "semester_5/electric_machine_design:dc-generator-600kw-square"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-600kw-square
site: BCT Engineering
---

# Main dimensions of 600 kW square-pole generator

Electric Machine Design · Chapter 5: DC Machine Design

- Asked in: 2070 Chaitra, 8 marks
- Video: [Main dimensions calculation of DC Machines](https://www.youtube.com/watch?v=pGPKMK2SaTQ) — Dr.Ramesh P
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-600kw-square
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5

## Question

A 600 kW, 500 V, 900 rpm, dc shunt generator is designed to have a square pole face. The loadings are Bav = 0.6 Wb/m², ac = 35000 A/m, full load efficiency = 0.91, ratio of pole arc to pole pitch = 0.75. Calculate the main dimensions of the machine.

## Short answer

Armature power 619.8 kW, output coefficient 207.3, D²L=0.1994 m³. With square pole face (L=0.3927D) and p=6, D≈0.80 m, L≈0.31 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 generator (Sawhney), $P_a = \dfrac{1+2\eta}{3\eta}P$.

#### Step 1: $D^2L$
$$
\begin{aligned}
P_a &= \frac{1+2\times0.91}{3\times0.91}\times600 = 619.8\ \text{kW} \\
C_0 &= \pi^2\times0.6\times35000\times10^{-3} = 207.3 \\
n &= 900/60 = 15\ \text{rps} \\
D^2L &= \frac{619.8}{207.3\times15} = 0.1994\ \text{m}^3
\end{aligned}
$$

#### Step 2: Number of poles (assumed from usual limits)
- Frequency $f = pn/2 \le 50$ Hz $\Rightarrow p \le 6.67$.
- Armature current $\approx 600000/500 = 1200$ A. With a lap winding, current per brush arm $2I_a/p \le 400$ A $\Rightarrow p \ge 6$.

So **$p = 6$** ($f = 45$ Hz, 400 A per brush arm).

#### Step 3: Square pole face
$$
L = \psi\frac{\pi D}{p} = \frac{0.75\pi D}{6} = 0.3927D
$$

$$
\begin{aligned}
D^3 &= \frac{0.1994}{0.3927} = 0.5077\ \text{m}^3 \\
D &= 0.798\ \text{m} \\
L &= 0.3927\times0.798 = 0.313\ \text{m}
\end{aligned}
$$

#### Check
- Pole pitch $\tau = \pi\times0.798/6 = 0.418$ m; pole arc $= 0.75\tau = 0.313$ m $= L$.
- Peripheral speed $= \pi Dn = \pi\times0.798\times15 = 37.6$ m/s (below 40 m/s).
- Armature mmf per pole $= ac\,\pi D/2p = 35000\times\pi\times0.798/12 = 7310$ A (reasonable).

**Answer: $p = 6$, armature diameter $D \approx 0.80$ m, core length $L \approx 0.31$ m.**

## Check yourself

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

1. How is p = 6 chosen for the 600 kW, 900 rpm generator?
   A. Frequency pn/2 <= 50 Hz gives p <= 6.67 and brush current 2Ia/p <= 400 A gives p >= 6
   B. Only the peripheral speed limit decides it
   C. Because 6 is the smallest even number
   D. Because ψ = 0.75 requires 6 poles
   Answer: A. Ia is about 1200 A; the two limits leave only p = 6 (45 Hz, 400 A per brush arm). (Key point: Pole number selection)

2. With p = 6 and ψ = 0.75, how are D and L found from D^2L = 0.1994?
   A. L = D, so D = 0.585 m
   B. L = 0.75D, so D = 0.643 m
   C. L = 0.75πD/6 = 0.3927D, so D^3 = 0.5077 and D = 0.798 m, L = 0.313 m
   D. L = 0.75πD, so D = 0.35 m
   Answer: C. Square face gives L = ψπD/p = 0.3927D, hence D = 0.798 m and L = 0.313 m. (Key point: Square pole face dimensions)

3. Which statement describes the checks on the result?
   A. Peripheral speed 37.6 m/s is below 40 m/s and armature mmf per pole is about 7310 A, reasonable
   B. Peripheral speed exceeds 40 m/s but is accepted
   C. Armature mmf per pole is 73100 A
   D. Pole arc is 0.418 m, larger than L
   Answer: A. πDn = 37.6 m/s < 40 and ac πD/2p = 7310 A; pole arc 0.313 m equals L. (Key point: Checks on the result)

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