---
title: "Output equation and separation of D, L"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:output-equation-separation-dl"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/output-equation-separation-dl
site: BCT Engineering
---

# Output equation and separation of D, L

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

- Asked in: 2074 Chaitra, 8 marks
- 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/output-equation-separation-dl
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4

## Question

Derive the expressions for output equation of three phase induction machine. Explain the separation of D and L.

## Short answer

Output equation Q = C0 D2 L ns with C0 = 11 Bav ac Kw × 10-3 is derived from EMF and specific loadings. D and L are separated by choosing a ratio L/τ (e.g., L/τ=1 for overall design, 1.5–2 for minimum cost, or τ=√(0.18L) for best power factor), solving D3 = (D2L·P)/(kπ) and checking peripheral speed.

## Answer

#### Output equation
Let $D$ = bore diameter, $L$ = core length, $P$ = poles, $n_s$ = synchronous speed (rps), $T_{ph}$ = turns per phase.

kVA input:
$$
Q = 3E_{ph}I_{ph}\times10^{-3}, \qquad E_{ph} = 4.44 f\Phi T_{ph}K_w, \qquad f = \frac{Pn_s}{2}
$$
Specific magnetic loading:
$$
B_{av} = \frac{P\Phi}{\pi DL} \Rightarrow \Phi = \frac{B_{av}\pi DL}{P}
$$
Specific electric loading (6$T_{ph}$ conductors):
$$
ac = \frac{6I_{ph}T_{ph}}{\pi D} \Rightarrow I_{ph}T_{ph} = \frac{ac\,\pi D}{6}
$$
Substituting:
$$
\begin{aligned}
Q &= 3 \times 4.44 \times \frac{Pn_s}{2} \times \frac{B_{av}\pi DL}{P} \times K_w \times \frac{ac\,\pi D}{6} \times 10^{-3} \\
&= 11 B_{av}\, ac\, K_w \times 10^{-3}\, D^2 L n_s = C_0 D^2 L n_s
\end{aligned}
$$
where $C_0 = 11B_{av}\,ac\,K_w\times10^{-3}$ is the output coefficient and $Q = \text{kW}/(\eta\cos\phi)$.

#### Separation of $D$ and $L$
The output equation gives only the product $D^2L$. To get $D$ and $L$ separately, a ratio $L/\tau$ (with $\tau = \pi D/P$) is chosen according to the design aim:

| Design aim | Ratio chosen |
|---|---|
| Minimum cost | $L/\tau = 1.5$ to $2$ |
| Good power factor | $L/\tau = 1.0$ to $1.25$ |
| Good efficiency | $L/\tau = 1.5$ |
| Good overall design | $L/\tau = 1$ |
| Best power factor | $\tau = \sqrt{0.18L}$ |

With $L = k\tau = k\pi D/P$:
$$
D^3 = \frac{D^2L \cdot P}{k\pi}, \qquad L = \frac{D^2L}{D^2}
$$

#### Checks after separation
- **Peripheral speed**: $v_a = \pi D n_s$ must not exceed about 30 m/s for normal rotors (higher only with special construction).
- **Ventilating ducts**: if $L$ exceeds about 100–125 mm, radial ducts (about 10 mm wide) are provided.
- **Practical proportions**: very long, thin machines are hard to cool and the shaft deflects; very short, large-diameter machines have a high end-winding share.

#### Example
For $D^2L = 0.0110$ m³, 6 poles, $L = \tau$: $D^3 = 0.0110 \times 6/\pi = 0.0210$, so $D = 0.276$ m and $L = 0.1445$ m, with $v_a = \pi \times 0.276 \times 16.67 = 14.4$ m/s (acceptable).

## Check yourself

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

1. In the derivation, how is ac related to phase current and turns?
   A. ac = 6 Iph Tph/(pi D), since there are 6 Tph conductors in total
   B. ac = 3 Iph Tph/(pi D)
   C. ac = 6 Iph/(pi D Tph)
   D. ac = Iph Tph/(6 pi D)
   Answer: A. Three phases each have 2 Tph conductors, giving 6 Tph conductors spread over the periphery pi D. (Key point: Specific electric loading)

2. What is the final output equation?
   A. Q = C0 D L^2 ns
   B. Q = C0 D^2 L ns with C0 = 11 Bav ac Kw x 10^-3
   C. Q = C0 D^2 L/ns
   D. Q = C0 D L ns
   Answer: B. Substituting flux and ampere-conductors gives 1.11 pi^2 = 11 and output proportional to D^2 L ns. (Key point: Result of output equation)

3. Why must D and L be separated by a chosen ratio?
   A. The equation gives D only, so L is free
   B. Two equations are available for D and L already
   C. The equation gives only the product D^2L, so one extra relation (L/tau) is required
   D. The speed is unknown
   Answer: C. One equation with two unknowns needs an additional design-based ratio. (Key point: Why separation is needed)

4. Which ratio of L/tau is suggested for good power factor?
   A. 1.5 to 2
   B. 3 to 4
   C. 0.2 to 0.4
   D. 1.0 to 1.25
   Answer: D. Power factor is best with a compact, nearly square shape, while minimum cost uses 1.5 to 2. (Key point: Choice of L/tau)

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