---
title: "Armature dimensions of 50 kW shunt generator"
subject: "Electric Machine Design"
chapter: 5
key: "semester_5/electric_machine_design:dc-generator-50kw-armature"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-50kw-armature
site: BCT Engineering
---

# Armature dimensions of 50 kW shunt generator

Electric Machine Design · Chapter 5: DC Machine Design

- Asked in: 2072 Kartik, 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-50kw-armature
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5

## Question

A design is required for a 50 kW, 4 pole, 6000 rpm, dc shunt generator. The full load terminal voltage is 220 V. If the maximum gap density is 0.83 Wb/m² and the armature ampere conductors per meter are 30000, calculate suitable dimensions of armature core to give a square pole face. Assume that the full load armature voltage drop is 3% of the rated terminal voltage and that the field current is 1% of rated full load current, ratio of pole arc to pole pitch is 0.67.

## Short answer

Assuming 600 rpm, armature power 52.02 kW, Bav=0.556 Wb/m², output coefficient 164.7 gives D²L=0.03159 m³. Square pole face yields D≈0.39 m, L≈0.21 m.

## Answer

**Assumption:** the speed is read as **600 rpm** (the standard textbook version of this problem). At 6000 rpm a 4-pole machine would have a flux frequency of 200 Hz and a peripheral speed near 57 m/s, which is not practical; the method is the same either way.

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).

#### Step 1: Armature power
$$
\begin{aligned}
I_L &= \frac{50000}{220} = 227.27\ \text{A} \\
I_f &= 0.01\times227.27 = 2.27\ \text{A} \\
I_a &= I_L + I_f = 229.55\ \text{A} \\
E &= 220 + 0.03\times220 = 226.6\ \text{V} \\
P_a &= E I_a = 226.6\times229.55 = 52.02\ \text{kW}
\end{aligned}
$$

#### Step 2: Loadings and $D^2L$
$$
\begin{aligned}
B_{av} &= \psi B_m = 0.67\times0.83 = 0.556\ \text{Wb/m}^2 \\
C_0 &= \pi^2\times0.556\times30000\times10^{-3} = 164.7 \\
n &= 600/60 = 10\ \text{rps} \\
D^2L &= \frac{52.02}{164.7\times10} = 0.03159\ \text{m}^3
\end{aligned}
$$

#### Step 3: Square pole face
$$
L = \psi\frac{\pi D}{p} = \frac{0.67\pi D}{4} = 0.5262D
$$

$$
\begin{aligned}
D^3 &= \frac{0.03159}{0.5262} = 0.06003\ \text{m}^3 \\
D &= 0.392\ \text{m} \\
L &= 0.5262\times0.392 = 0.206\ \text{m}
\end{aligned}
$$

#### Check
- Pole pitch $\tau = \pi\times0.392/4 = 0.308$ m; pole arc $= 0.67\tau = 0.206$ m $= L$.
- Peripheral speed $= \pi Dn = 12.3$ m/s; frequency $= 4\times10/2 = 20$ Hz. Both acceptable.

**Answer (600 rpm): $D \approx 0.39$ m, $L \approx 0.21$ m.** (If 6000 rpm is used literally: $D^2L = 0.00316$ m³, giving $D \approx 0.182$ m and $L \approx 0.096$ m.)

## Check yourself

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

1. Why is the speed read as 600 rpm instead of 6000 rpm?
   A. The answer requires lower efficiency
   B. At 6000 rpm a 4-pole machine has 200 Hz flux frequency and about 57 m/s peripheral speed, which is impractical
   C. 600 rpm gives a higher power
   D. The output equation does not work at 6000 rpm
   Answer: B. 6000 rpm is impractical (f = 200 Hz, high peripheral speed), so the standard 600 rpm version is used. (Key point: Speed assumption)

2. How is Ia found when field current is 1% of full-load current?
   A. Ia = IL + 0.01 IL = 227.27 + 2.27 = 229.55 A
   B. Ia = IL - 0.01 IL
   C. Ia = IL = 227.27 A
   D. Ia = 1.03 IL
   Answer: A. The armature supplies load plus field current: Ia = 229.55 A. (Key point: Armature current)

3. What is Bav used in C0 here?
   A. 0.83 Wb/m^2
   B. 1.24 Wb/m^2
   C. 0.67 Wb/m^2
   D. 0.556 Wb/m^2, from ψ x Bm = 0.67 x 0.83
   Answer: D. Bav = ψ Bm = 0.556, giving C0 = π^2 x 0.556 x 30 = 164.7. (Key point: Average gap density)

4. What are D and L for the square pole face?
   A. D = 0.182 m, L = 0.096 m
   B. L = 0.5262D, so D^3 = 0.03159/0.5262 and D = 0.392 m, L = 0.206 m
   C. D = 0.5 m, L = 0.5 m
   D. D = 0.308 m, L = 0.17 m
   Answer: B. L = 0.67πD/4 = 0.5262D, D^3 = 0.06003, D = 0.392 m, L = 0.206 m (equal to pole arc). (Key point: Square pole face dimensions)

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