---
title: "30 kW 240 V generator, 2.5% drop"
subject: "Electric Machine Design"
chapter: 5
key: "semester_5/electric_machine_design:dc-generator-30kw-240v-2-5pct"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-30kw-240v-2-5pct
site: BCT Engineering
---

# 30 kW 240 V generator, 2.5% drop

Electric Machine Design · Chapter 5: DC Machine Design

- Asked in: 2078 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-30kw-240v-2-5pct
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5

## Question

A design is required for a 30 kW, 4 pole, 900 rpm dc shunt generator, the full load terminal voltage being 240 V. Assume that the full load armature voltage drop is 2.5% of terminal voltage. Calculate main dimension of machine. Given that: Maximum gap flux density = 0.85 Wb/m²; Specific electric loading = 20000 ampere conductor per metre; Field resistance = 120 Ω; Ratio of pole arc to pole pitch = 0.7; Field form factor = 0.7

## Short answer

Ia=127 A, E=246 V, Pa=31.24 kW. Co=117.4, D²L=0.01773 m³. With square pole face, D=0.318 m, L=0.175 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).

The full-load armature voltage drop $I_aR_a$ is 2.5% of terminal voltage. Field form factor $K_f = B_{av}/B_m = 0.7$ (equal to $\psi$). Since no length ratio is given, the standard assumption of a **square pole face** (L = pole arc) is used.

#### Step 1: Armature power from currents and emf
$$
\begin{aligned}
I_L &= \frac{30000}{240} = 125.0\ \text{A} \\
I_f &= \frac{240}{120} = 2.0\ \text{A} \\
I_a &= I_L + I_f = 127.0\ \text{A} \\
E &= V + I_aR_a = 240 + 0.025\times240 = 246.0\ \text{V} \\
P_a &= E I_a = 246.0\times127.0 = 31.24\ \text{kW}
\end{aligned}
$$

#### Step 2: Loadings and $D^2L$
$$
\begin{aligned}
B_{av} &= \psi B_m = 0.7\times0.85 = 0.595\ \text{Wb/m}^2 \\
C_0 &= \pi^2\times0.595\times20000\times10^{-3} = 117.4 \\
n &= 900/60 = 15\ \text{rps} \\
D^2L &= \frac{31.24}{117.4\times15} = 0.01773\ \text{m}^3
\end{aligned}
$$

#### Step 3: Square pole face
$$
L = \psi\frac{\pi D}{p} = \frac{0.7\pi D}{4} = 0.5498D
$$

$$
\begin{aligned}
D^3 &= \frac{0.01773}{0.5498} = 0.03226\ \text{m}^3 \\
D &= 0.3183\ \text{m} \\
L &= 0.5498\times0.3183 = 0.1750\ \text{m}
\end{aligned}
$$

#### Check
- Pole pitch $\tau = \pi D/4 = 0.250$ m; pole arc $= 0.7\tau = 0.1750$ m $= L$.
- Peripheral speed $= \pi Dn = 15.0$ m/s; frequency $= pn/2 = 30$ Hz. Both acceptable.

**Answer: Armature diameter $D \approx 0.318$ m, armature core length $L \approx 0.175$ m.**

## Check yourself

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

1. With armature voltage drop 2.5% of 240 V, what is the generated emf?
   A. 234 V
   B. 240 V
   C. 246 V
   D. 247.2 V
   Answer: C. E = V + IaRa = 240 + 0.025 x 240 = 246 V. (Key point: Armature emf)

2. Compared with the 3% case, how does the 2.5% drop change the design?
   A. Pa rises, so D increases
   B. Pa falls slightly to 31.24 kW, so D is slightly smaller (0.318 m)
   C. D and L are unchanged exactly
   D. L becomes larger than the pole arc
   Answer: B. Lower drop gives lower E and Pa (246 x 127 = 31.24 kW), so D^2L and D reduce slightly. (Key point: Effect of smaller drop)

3. Why is Ia = 127 A rather than 125 A?
   A. Armature reaction adds 2 A
   B. The 2.5% drop adds 2 A
   C. The load current is 127 A
   D. The field current 240/120 = 2 A is supplied by the armature in addition to the 125 A load current
   Answer: D. For a shunt generator Ia = IL + If = 125 + 2 = 127 A. (Key point: Armature current)

4. What does the square pole face condition mean for the final result?
   A. Core length equals the diameter
   B. Core length equals pole pitch
   C. Core length equals 0.7 x diameter
   D. Core length equals the pole arc, 0.7τ = 0.175 m
   Answer: D. L = ψπD/p = 0.5498D = 0.175 m, equal to the pole arc 0.7 x 0.250. (Key point: Square pole face)

## Close variants set in other exams

- [30 kW 240 V generator, 3% drop](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-30kw-240v-3pct) — Electric Machine Design, chapter 5, asked 3×
- [Main dimensions of 30 kW 220 V shunt generator](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-30kw-220v) — Electric Machine Design, chapter 5, asked 4×

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