---
title: "Output equation of three phase induction motor"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:induction-motor-output-equation"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/induction-motor-output-equation
site: BCT Engineering
---

# Output equation of three phase induction motor

Electric Machine Design · Chapter 4: Three phase induction motor design

- Asked in: 2082 Baishakh, 6 marks; 2078 Kartik, 8 marks; 2075 Asoj, 5 marks; 2073 Shrawan, 8 marks; 2073 Chaitra, 8 marks; 2069 Asar, 8 marks (asked 6 times)
- Video: [Output equation of three phase Induction Motor | Design of Electrical Machines by S. D.Yelgatte](https://www.youtube.com/watch?v=m8vAfFiVWzQ) — S D Yelgatte 📖 ⚡
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/induction-motor-output-equation
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4

## Question

Derive the output equation for three phase induction motor.

## Short answer

The output equation is Q = C0 D² L ns with C0 = 11 Bav ac Kw ×10⁻³, relating kVA input to stator bore diameter D, core length L, synchronous speed ns, and specific loadings Bav and ac, derived from emf and mmf expressions.

## Answer

The **output equation** relates the kVA input rating $Q$ of a 3-phase induction motor to its main dimensions ($D$ = stator bore diameter, $L$ = core length), its specific loadings ($B_{av}$, $ac$) and its synchronous speed $n_s$.

#### Symbols
- $P$ = number of poles, $f$ = frequency, $n_s$ = synchronous speed (rps)
- $\Phi$ = flux per pole (Wb), $T_{ph}$ = turns per phase, $K_w$ = winding factor
- $E_{ph}$, $I_{ph}$ = phase voltage and phase current
- $B_{av}$ = specific magnetic loading (Wb/m²), $ac$ = specific electric loading (A/m)

#### Step 1: kVA input
Neglecting the small drop, $V_{ph} \approx E_{ph}$:
$$
Q = 3E_{ph}I_{ph}\times 10^{-3}\ \text{kVA}
$$

#### Step 2: Induced emf
$$
E_{ph} = 4.44\, f\, \Phi\, T_{ph} K_w, \qquad f = \frac{P n_s}{2}
$$

#### Step 3: Magnetic loading
Total flux around the gap is $P\Phi$ over area $\pi D L$:
$$
B_{av} = \frac{P\Phi}{\pi D L} \;\Rightarrow\; \Phi = \frac{B_{av}\pi D L}{P}
$$

#### Step 4: Electric loading
Total stator conductors $= 3 \times 2T_{ph} = 6T_{ph}$, each carrying $I_{ph}$:
$$
ac = \frac{6 I_{ph} T_{ph}}{\pi D} \;\Rightarrow\; I_{ph}T_{ph} = \frac{ac\,\pi D}{6}
$$

#### Step 5: Substitute
$$
\begin{aligned}
Q &= 3 \times 4.44 \times \frac{P n_s}{2} \times \frac{B_{av}\pi D L}{P} \times K_w \times \frac{ac\,\pi D}{6} \times 10^{-3} \\
&= 1.11\,\pi^2 B_{av}\, ac\, K_w\, D^2 L\, n_s \times 10^{-3} \\
&= 11\, B_{av}\, ac\, K_w \times 10^{-3}\, D^2 L\, n_s
\end{aligned}
$$

So
$$
Q = C_0 D^2 L n_s, \qquad C_0 = 11 B_{av}\, ac\, K_w \times 10^{-3}
$$

$C_0$ is called the **output coefficient**. The kVA input is found from the rated shaft output:
$$
Q = \frac{\text{kW output}}{\eta \cos\phi}
$$

#### Meaning of the result
- $D^2L = \dfrac{Q}{C_0 n_s}$, so the volume of the active part is directly proportional to rating and inversely proportional to speed and specific loadings.
- Higher $B_{av}$ and $ac$ give a smaller, cheaper machine, but power factor, temperature rise and overload capacity limit them.
- A high-speed motor is smaller than a low-speed motor of the same rating.
- Typical ranges: $B_{av} = 0.3$–$0.6$ Wb/m², $ac = 5000$–$45000$ A/m, $K_w \approx 0.955$ for a full-pitch, 60° phase-spread winding.

The product $D^2L$ is then split into $D$ and $L$ using a chosen ratio $L/\tau$ (for example $L/\tau = 1$ for good overall design) and a check on peripheral speed ($\pi D n_s \le 30$ m/s for normal designs).

## Check yourself

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

1. What is the starting expression for the kVA input of a 3-phase induction motor?
   A. Q = E*I x 10^-3
   B. Q = 3 Eph Iph x 10^-3, taking Vph approximately equal to Eph
   C. Q = sqrt3 Eph Iph x 10^-3 with Eph per phase
   D. Q = 3 Eph Iph cos(phi) x 10^-3
   Answer: B. Input kVA is 3 Vph Iph x 10^-3, with Vph approximated by Eph. (Key point: kVA input)

2. How is ac related to phase current and turns?
   A. ac = 3 Iph Tph/(pi D)
   B. ac = Iph Tph/(pi D)
   C. ac = 6 Iph Tph/(pi D), since there are 6Tph conductors in total
   D. ac = 2 Iph Tph/D
   Answer: C. Three phases with 2Tph conductors each give 6Tph conductors, each carrying Iph, spread over pi D. (Key point: Specific electric loading)

3. How is flux per pole related to Bav?
   A. Phi = Bav pi D L
   B. Phi = Bav pi D L/P, since P poles share the gap area pi D L
   C. Phi = Bav D L
   D. Phi = Bav pi D/(P L)
   Answer: B. Bav = P Phi/(pi D L), so Phi = Bav pi D L/P. (Key point: Flux per pole)

4. What does the final output equation Q = C0 D^2 L ns say about size?
   A. D^2L is proportional to speed
   B. D^2L = Q/(C0 ns), so a higher speed or higher loadings give a smaller machine
   C. D^2L is independent of speed and loadings
   D. D^2L increases with Bav and ac
   Answer: B. With C0 = 11 Bav ac Kw x 10^-3, volume D^2L is inversely proportional to ns and the loadings. (Key point: Final result and meaning)

## Related in Electric Machine Design

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