---
title: "Armature size, 7.5 kW 1000 rpm motor"
subject: "Electric Machine Design"
chapter: 5
key: "semester_5/electric_machine_design:dc-motor-7-5kw-1000rpm"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-motor-7-5kw-1000rpm
site: BCT Engineering
---

# Armature size, 7.5 kW 1000 rpm motor

Electric Machine Design · Chapter 5: DC Machine Design

- Asked in: 2079 Bhadra, 8 marks; 2076 Chaitra, 8 marks; 2075 Chaitra, 8 marks (asked 3 times)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-motor-7-5kw-1000rpm
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5

## Question

Calculate the diameter (D) and length (L) of armature for a 7.5 kW, 4 pole, 1000 rpm, 220 V shunt motor. Given: full load efficiency = 0.83; maximum gap flux density = 0.9 Wb/m²; specific electric loading = 30000 ampere conductor per meter; field form factor = 0.7. Assume that the maximum efficiency occurs at full load and the field current is 2.5% of rated current. The pole face is square.

## Short answer

Armature power Pa = 8.042 kW (using equal constant and variable loss at maximum efficiency). Output coefficient C0 = 186.54. D^2L = 2.587×10^-3 m^3. For square pole face, L = 0.5498 D, giving D ≈ 16.8 cm, L ≈ 9.2 cm.

## Answer

**Data:** 7.5 kW, 220 V shunt motor, 4 poles, 1000 rpm ($n = 16.67$ rps), $\eta = 0.83$, $B_g = 0.9$ Wb/m², $ac = 30000$ A/m, field form factor $\psi = 0.7$, $I_f = 2.5\%$ of rated current, maximum efficiency at full load, square pole face.

#### Armature power $P_a$
Input and losses:
$$
\begin{aligned}
P_{in} &= \frac{7500}{0.83} = 9036.1\ \text{W}, \qquad \text{Losses} = 9036.1 - 7500 = 1536.1\ \text{W}
\end{aligned}
$$
Maximum efficiency at full load means variable (armature copper) loss = constant loss:
$$
I_a^2R_a = \frac{1536.1}{2} = 768.1\ \text{W}
$$
Currents:
$$
\begin{aligned}
I &= \frac{9036.1}{220} = 41.07\ \text{A}, \qquad I_f = 0.025 \times 41.07 = 1.03\ \text{A} \\
I_a &= 41.07 - 1.03 = 40.05\ \text{A}
\end{aligned}
$$
Back emf:
$$
\begin{aligned}
I_aR_a &= \frac{768.1}{40.05} = 19.18\ \text{V} \\
E_b &= 220 - 19.18 = 200.82\ \text{V} \\
P_a &= E_bI_a = 200.82 \times 40.05 = 8042\ \text{W} = 8.042\ \text{kW}
\end{aligned}
$$

#### Output coefficient
$$
\begin{aligned}
B_{av} &= \psi B_g = 0.7 \times 0.9 = 0.63\ \text{Wb/m}^2 \\
C_0 &= \pi^2 B_{av}\,ac \times 10^{-3} = \pi^2 \times 0.63 \times 30000 \times 10^{-3} = 186.54
\end{aligned}
$$

#### $D^2L$
$$
D^2L = \frac{P_a}{C_0 n} = \frac{8.042}{186.54 \times 16.67} = 2.587 \times 10^{-3}\ \text{m}^3
$$

#### Square pole face
Pole arc $= L$, and pole arc $= \psi\tau = 0.7 \times \pi D/4$:
$$
L = 0.5498D
$$
$$
D^3 = \frac{2.587\times10^{-3}}{0.5498} = 4.705\times10^{-3} \Rightarrow D = 0.1676\ \text{m}
$$
$$
L = 0.5498 \times 0.1676 = 0.0921\ \text{m}
$$
Check: peripheral speed $= \pi \times 0.1676 \times 16.67 = 8.8$ m/s (acceptable).

**Answer:** $D \approx 16.8$ cm, $L \approx 9.2$ cm.

## Check yourself

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

1. What does maximum efficiency at full load imply?
   A. Armature copper loss equals the output
   B. Iron loss is zero
   C. Field loss equals armature loss
   D. Variable (armature copper) loss equals constant loss, each half of the total
   Answer: D. Total loss 1536 W gives Ia^2 Ra = 768 W. (Key point: Maximum efficiency condition)

2. How is Pa found?
   A. Pa = Eb Ia with Eb = V - Ia Ra
   B. Pa = V Ia
   C. Pa = output / efficiency
   D. Pa = V x line current
   Answer: A. Eb = 220 - 19.18 = 200.8 V and Pa = 200.8 x 40.05 = 8.04 kW. (Key point: Armature power)

3. How is Bav found?
   A. Bav = Bg / psi
   B. Bav = psi x Bg = 0.7 x 0.9 = 0.63
   C. Bav = Bg = 0.9
   D. Bav = psi + Bg
   Answer: B. Field form factor is Bav/Bg. (Key point: Bav from form factor)

4. What relation gives L for a square pole face?
   A. L = psi tau = 0.7 pi D/4 = 0.5498 D
   B. L = pi D/4
   C. L = 0.7 D
   D. L = D/0.7
   Answer: A. A square face means pole arc = L, giving D = 0.168 m and L = 0.092 m. (Key point: Square pole face)

## Close variants set in other exams

- [Armature size, 10 kW 900 rpm motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-motor-10kw-900rpm) — Electric Machine Design, chapter 5, asked 5×
- [Armature size, 7.5 kW 1800 rpm motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-motor-7-5kw-1800rpm) — Electric Machine Design, chapter 5, asked 5×

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