---
title: "Stator bore, length and conductors of 70 hp motor"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:stator-bore-70hp-motor"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/stator-bore-70hp-motor
site: BCT Engineering
---

# Stator bore, length and conductors of 70 hp motor

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

- Asked in: 2069 Asar, 10 marks
- Video: [Video 4 4 Main Dimensions of Induction motor](https://www.youtube.com/watch?v=xgDkTfRR-p4) — S.D Hirekodi
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/stator-bore-70hp-motor
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4

## Question

Determine the diameter of stator bore and core length of a 70 h.p., 415 V, 3-phase, 50 Hz star connected, 6 pole induction motor for which the specific electric and magnetic loadings are 32000 A/m and 0.51 Wb/m² respectively. Take the efficiency as 90 percent and power factor as 0.91. Assume pole pitch equal to core length. Estimate the number of stator conductors required for a winding in which the conductors are connected in two parallel paths. Choose a suitable number of conductors per slot so that slot loading does not exceed 750 ampere conductors.

## Short answer

Stator bore 34.9 cm, core length 18.3 cm. With 2 parallel paths, about 792 conductors needed. Using 54 slots × 14 conductors gives 756 total conductors, slot loading 621 A-conductors.

## Answer

**Data:** 70 HP $= 70 \times 746 = 52.22$ kW, 415 V, star, 6 poles ($n_s = 1000/60 = 16.67$ rps), $ac = 32000$ A/m, $B_{av} = 0.51$ Wb/m², $\eta = 0.9$, pf $= 0.91$, $L = \tau$. Assume $K_w = 0.955$.

#### Diameter and core length
$$
\begin{aligned}
Q &= \frac{52.22}{0.9 \times 0.91} = 63.76\ \text{kVA} \\
C_0 &= 11 \times 0.51 \times 32000 \times 0.955 \times 10^{-3} = 171.44 \\
D^2L &= \frac{63.76}{171.44 \times 16.67} = 0.02232\ \text{m}^3
\end{aligned}
$$
$L = \tau = \pi D/6$:
$$
D^3 = \frac{6 \times 0.02232}{\pi} = 0.04262 \Rightarrow D = 0.349\ \text{m}, \quad L = 0.183\ \text{m}
$$

#### Number of stator conductors
Line (= phase) current for star:
$$
I = \frac{63760}{\sqrt3 \times 415} = 88.70\ \text{A}
$$
With 2 parallel paths, current per conductor $I_z = 88.70/2 = 44.35$ A.
$$
Z = \frac{ac \times \pi D}{I_z} = \frac{32000 \times \pi \times 0.3493}{44.35} = 792
$$

#### Conductors per slot
Slot loading $\le 750$ A-conductors, so conductors per slot $\le 750/44.35 = 16.9$.
- Choose $q = 3$: $S_s = 3 \times 6 \times 3 = 54$ slots, slot pitch $= \pi \times 349.3/54 = 20.3$ mm.
- Conductors per slot $= 792/54 = 14.7$, take **14** (even, for a double-layer winding).
- Total conductors $= 54 \times 14 = 756$; slot loading $= 14 \times 44.35 = 621$ A $< 750$ A.

Check: series turns per phase $= 756/(3 \times 2 \times 2) = 63$; from emf, $T_s = 239.6/(4.44 \times 50 \times \Phi \times 0.955) = 66.2$ with $\Phi = 0.51 \times 0.1829 \times 0.1829 = 0.01706$ Wb, so the gap density rises slightly to 0.536 Wb/m², which is acceptable.

**Answer:** $D \approx 34.9$ cm, $L \approx 18.3$ cm; about 792 conductors needed, arranged as 54 slots × 14 conductors = 756 conductors, slot loading 621 A-conductors.

## Check yourself

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

1. How is the kVA Q obtained from the 70 HP rating?
   A. Q = 52.22 x efficiency x power factor
   B. Q = 52.22 / efficiency only
   C. Q = (70 x 0.746)/(efficiency x power factor)
   D. Q = 70 x 0.746 x pf
   Answer: C. Q = 52.22/(0.9 x 0.91) = 63.76 kVA. (Key point: Power rating to kVA)

2. With L = tau for a 6 pole machine, what is D^3 in terms of D^2L?
   A. D^3 = pi D^2L / 6
   B. D^3 = 3 D^2L / pi
   C. D^3 = 6 D^2L / pi
   D. D^3 = D^2L / (6 pi)
   Answer: C. tau = pi D/6 = L, so D^2L = pi D^3/6, giving D = 0.349 m and L = 0.183 m. (Key point: Pole-pitch condition)

3. Which formula gives the total stator conductors Z?
   A. Z = ac x D / Iz
   B. Z = ac x pi D / Iz, where Iz is the current per conductor
   C. Z = ac x Iz / (pi D)
   D. Z = Iz x pi D / ac
   Answer: B. ac = Iz Z/(pi D), so Z = 32000 x pi x 0.3493/44.35 = 792. (Key point: Total conductors)

4. With two parallel paths in a star winding, what is the current per conductor?
   A. The full line current, 88.7 A
   B. Half the line current, 44.35 A
   C. Twice the line current
   D. Line current / sqrt(3)
   Answer: B. Star line current equals phase current (88.7 A), and 2 parallel paths split it to 44.35 A. (Key point: Current per conductor)

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