---
title: "Main dimensions of 200 kW shunt generator"
subject: "Electric Machine Design"
chapter: 5
key: "semester_5/electric_machine_design:dc-generator-200kw"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-200kw
site: BCT Engineering
---

# Main dimensions of 200 kW shunt generator

Electric Machine Design · Chapter 5: DC Machine Design

- Asked in: 2081 Bhadra, 8 marks
- 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-generator-200kw
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5

## Question

Find the main dimensions of a 200 kW, 250 V, 6 pole, 1000 rpm dc shunt generator. The maximum value of flux density in the gap is 0.87 Wb/m² and ampere conductors per meter of armature periphery are 31,000. The ratio of pole arc to pitch is 0.67 and the efficiency is 91 percent. Assume the ratio of length of core to pole pitch is 0.75.

## Short answer

Pa=(1+2η)/(3η)×P=206.6 kW, Co=178.3, D²L=0.0695 m³. Using L/τ=0.75 gives L=0.3927D, thus D=0.561 m, L=0.220 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, taking armature copper loss as about one-third of total losses (Sawhney),

$$
P_a = \frac{1+2\eta}{3\eta}\,P
$$

#### Step 1: Armature power
$$
P_a = \frac{1+2\times0.91}{3\times0.91}\times200 = \frac{2.82}{2.73}\times200 = 206.6\ \text{kW}
$$

#### Step 2: Loadings and output coefficient
$$
\begin{aligned}
B_{av} &= \psi B_m = 0.67\times0.87 = 0.583\ \text{Wb/m}^2 \\
C_0 &= \pi^2\times0.583\times31000\times10^{-3} = 178.3 \\
n &= 1000/60 = 16.67\ \text{rps} \\
D^2L &= \frac{206.6}{178.3\times16.67} = 0.0695\ \text{m}^3
\end{aligned}
$$

#### Step 3: Use $L/\tau = 0.75$
$$
\tau = \frac{\pi D}{p} = \frac{\pi D}{6}, \qquad L = 0.75\tau = \frac{0.75\pi D}{6} = 0.3927D
$$

$$
\begin{aligned}
D^3 &= \frac{0.0695}{0.3927} = 0.1770\ \text{m}^3 \\
D &= 0.561\ \text{m} \\
L &= 0.3927\times0.561 = 0.220\ \text{m}
\end{aligned}
$$

#### Check
- Pole pitch $\tau = \pi\times0.561/6 = 0.294$ m; pole arc $= 0.67\tau = 0.197$ m.
- Peripheral speed $= \pi D n = \pi\times0.561\times16.67 = 29.4$ m/s (within the usual 30 m/s limit).
- Frequency $= pn/2 = 6\times16.67/2 = 50$ Hz (acceptable).

**Answer: Armature diameter $D \approx 0.561$ m, core length $L \approx 0.220$ m.**

## Check yourself

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

1. How is the average gap flux density found from the given maximum value?
   A. Bav = Bm / ψ
   B. Bav = ψ x Bm = 0.67 x 0.87
   C. Bav = Bm
   D. Bav = Bm x L/τ
   Answer: B. Bav = ψ Bm = 0.583 Wb/m^2, which is then used in C0. (Key point: Average gap density)

2. With L/τ = 0.75 and 6 poles, how is D obtained from D^2L?
   A. L = 0.75D, so D^3 = D^2L/0.75
   B. L = 0.75 x 0.67 x D, so D^3 = D^2L/0.503
   C. L = 0.75πD/6 = 0.3927D, so D^3 = D^2L/0.3927
   D. L = 0.75πD, so D^3 = D^2L/2.356
   Answer: C. L = 0.75τ with τ = πD/p gives L = 0.3927D, so D^3 = 0.0695/0.3927 = 0.177 and D = 0.561 m. (Key point: Using L/τ to find D)

3. Which check confirms the design is acceptable?
   A. Frequency pn/2 = 50 Hz and peripheral speed 29.4 m/s, within the usual limits
   B. Frequency 100 Hz and peripheral speed 15 m/s
   C. Frequency 25 Hz and peripheral speed 60 m/s
   D. Frequency pn = 100 Hz, which is acceptable
   Answer: A. f = 6 x 16.67/2 = 50 Hz and πDn = 29.4 m/s, both within the usual 50 Hz and 30 m/s limits. (Key point: Design check)

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