---
title: "Main dimensions and turns of 5 HP cage motor"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:main-dimensions-5hp-cage"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/main-dimensions-5hp-cage
site: BCT Engineering
---

# Main dimensions and turns of 5 HP cage motor

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

- Asked in: 2071 Chaitra, 10 marks
- Video: [Design of 5 HP Induction Motor, Main dimension core stack length L and airgap diameter D calculation](https://www.youtube.com/watch?v=j6kdvRUuUqk) — ShreeDevi Classes
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/main-dimensions-5hp-cage
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4

## Question

Find the main dimensions, number of stator turns and size of conductor of a 5 HP, 400 V, 3 phase, 4 pole, 50 Hz squirrel cage induction motor with star delta starting. Use the following data: Average flux density in the air gap = 0.4 Wb/m²; Ampere conductor per meter of armature periphery = 22000; Full load efficiency = 83%; Full load p.f. = 0.84 (lagging)

## Short answer

From output equation with L=τ: D=14.3 cm, L=11.3 cm. Stator turns per phase 372. Conductor area 1.115 mm² (bare diameter ≈1.19 mm).

## Answer

**Data:** 5 HP $= 5 \times 746 = 3730$ W, 400 V, 4 poles, 50 Hz ($n_s = 25$ rps), $B_{av} = 0.4$ Wb/m², $ac = 22000$ A/m, $\eta = 0.83$, pf $= 0.84$.
**Assumptions:** star-delta starting, so the motor runs in **delta** ($E_{ph} = 400$ V); $K_w = 0.955$; $L = \tau$ (good overall design); $\delta = 4$ A/mm².

#### Main dimensions
$$
\begin{aligned}
Q &= \frac{3.73}{0.83 \times 0.84} = 5.35\ \text{kVA} \\
C_0 &= 11 \times 0.4 \times 22000 \times 0.955 \times 10^{-3} = 92.44 \\
D^2L &= \frac{5.35}{92.44 \times 25} = 2.315\times10^{-3}\ \text{m}^3
\end{aligned}
$$
With $L = \tau = \pi D/4$:
$$
D^3 = \frac{4 \times 2.315\times10^{-3}}{\pi} = 2.948\times10^{-3} \Rightarrow D = 0.1434\ \text{m}, \quad L = 0.1126\ \text{m}
$$
Peripheral speed $= \pi \times 0.1434 \times 25 = 11.3$ m/s (acceptable).

#### Stator turns per phase
$$
\begin{aligned}
\Phi &= B_{av}\tau L = 0.4 \times 0.1126 \times 0.1126 = 5.07\times10^{-3}\ \text{Wb} \\
T_s &= \frac{400}{4.44 \times 50 \times 5.07\times10^{-3} \times 0.955} = 372
\end{aligned}
$$
With 36 slots ($q = 3$, slot pitch $= \pi \times 143.4/36 = 12.5$ mm): conductors per slot $= 6 \times 372/36 = 62$. So **$T_s = 372$**, 62 conductors per slot.

#### Size of conductor
$$
\begin{aligned}
I_{ph} &= \frac{5350}{3 \times 400} = 4.458\ \text{A} \\
a_s &= \frac{4.458}{4} = 1.115\ \text{mm}^2 \\
d &= \sqrt{\frac{4 \times 1.115}{\pi}} = 1.19\ \text{mm}
\end{aligned}
$$

**Answer:** $D = 14.3$ cm, $L = 11.3$ cm, $T_s = 372$ turns per phase, conductor 1.115 mm² (about 1.19 mm bare diameter).

## Check yourself

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

1. Why is Eph taken as 400 V for the star-delta started 5 HP motor?
   A. It runs in star, so phase voltage is 400 V
   B. Star-delta starting makes phase voltage 400/sqrt(3)
   C. It runs in delta, where phase voltage equals line voltage
   D. Phase voltage is always the line voltage in any connection
   Answer: C. Star-delta starting means the motor runs in delta, so Eph = 400 V. (Key point: Star-delta starting)

2. Which expression gives D^2L here?
   A. D^2L = Q x C0 x ns
   B. D^2L = C0 / (Q ns)
   C. D^2L = Q / (C0 x P)
   D. D^2L = Q / (C0 ns), with C0 = 11 Bav ac Kw x 10^-3
   Answer: D. From Q = C0 D^2 L ns, D^2L = 5.35/(92.44 x 25) = 2.315e-3 m^3. (Key point: D^2L from output equation)

3. How are D and L separated for the good-design condition used?
   A. Take L = 2D
   B. Take D = 2L
   C. Take L = tau/2
   D. Take L = tau = pi D / P and solve D^3
   Answer: D. With L = tau = pi D/4, D^3 = 4 D^2L / pi, giving D = 0.1434 m and L = 0.1126 m. (Key point: Separating D and L)

4. How is the stator conductor area found?
   A. Line current of 5.35 kVA divided by 3 x delta
   B. Iph multiplied by the current density
   C. Phase current Iph = Q/(3 x Eph) divided by the current density
   D. Iph divided by the number of slots
   Answer: C. Iph = 4.458 A and a = Iph/delta = 4.458/4 = 1.115 mm^2. (Key point: Conductor size)

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