---
title: "Six-stepped core area and turns (1000 kVA)"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:six-stepped-core-1000kva"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/six-stepped-core-1000kva
site: BCT Engineering
---

# Six-stepped core area and turns (1000 kVA)

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2069 Asar, 10 marks
- Video: [Three Phase Transformer complete design core, window, winding, space factor, specific loading, turns](https://www.youtube.com/watch?v=O7FdTj5wCV0) — ShreeDevi Classes
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/six-stepped-core-1000kva
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

A 3 phase core type power transformer has the following design specification: 1000 kVA, 50 Hz, 6600/433 volt, delta/star connected. Estimate: i) Net iron area of the core (with 6 steps core) ii) Diameter of the circumscribing circle iii) Number of HV and LV turns. Assume: Ratio magnetic loading (φm) / electrical loading (AT) = 1.62×10⁻⁴ Wb/AT; Flux density = 1.57 T; Net iron area = 0.6 × Gross iron area (for 6 stepped); Stacking factor = 0.89

## Short answer

Using the given ratio 1.62×10⁻⁴ Wb/AT, net iron area Ai = 0.314 m², circumscribing circle diameter d = 0.724 m. LV turns = 2, HV turns =53.

## Answer

Data: 1000 kVA, 3-phase, 50 Hz, 6600 V (delta) / 433 V (star), $r = \phi_m/AT = 1.62\times10^{-4}$ Wb/AT, $B_m$ = 1.57 T, 6-stepped core, stacking factor 0.89.

Reading of the data: the net iron area of a 6-stepped core is taken as $A_i = 0.6d^2$, where $d$ is the diameter of the circumscribing circle (the factor 0.6 already includes the stacking factor; the gross stepped area is $A_i/0.89$).

#### Flux from the loading ratio
Per phase kVA:

$$
Q_{ph} = 4.44f\phi_m(AT)\times10^{-3} \;\Rightarrow\; \phi_m\cdot AT = \frac{(1000/3)\times10^3}{4.44\times50} = 1501.5
$$

With $AT = \phi_m/r$:

$$
\begin{aligned}
\phi_m^2 &= r\times1501.5 = 1.62\times10^{-4}\times1501.5\\
\phi_m &= 0.4932\ \text{Wb}\\
E_t &= 4.44f\phi_m = 109.49\ \text{V per turn}
\end{aligned}
$$

#### i) Net iron area
$$
A_i = \frac{\phi_m}{B_m} = \frac{0.4932}{1.57} = 0.3141\ \text{m}^2
$$

Gross area of the stepped core $= A_i/0.89 = 0.3530$ m².

#### ii) Diameter of circumscribing circle
$$
d = \sqrt{\frac{A_i}{0.6}} = \sqrt{\frac{0.3141}{0.6}} = 0.724\ \text{m}
$$

#### iii) Turns
LV phase voltage (star) $= 433/\sqrt3 = 250.0$ V; HV phase voltage (delta) = 6600 V.

$$
\begin{aligned}
T_{LV} &= \frac{250.0}{109.49} = 2.28 \approx 2\\
T_{HV} &= T_{LV}\times\frac{6600}{250.0} = 52.8 \approx 53
\end{aligned}
$$

**Answer (with the ratio as printed): $A_i$ ≈ 0.314 m², $d$ ≈ 0.724 m, LV 2 turns, HV 53 turns.**

#### Note on the data
The printed ratio gives about 110 V per turn, far above normal for 1000 kVA ($E_t = K\sqrt{Q}$ with $K \approx 0.6$ gives ≈ 19 V). The ratio was most likely meant to be $1.62\times10^{-6}$ Wb/AT. The same steps then give:

| Quantity | $r = 1.62\times10^{-6}$ |
|---|---|
| $\phi_m$ | 0.0493 Wb |
| $E_t$ | 10.95 V |
| Net iron area $A_i$ | 0.0314 m² (314 cm²) |
| Circle diameter $d$ | 229 mm |
| LV turns | 22.8 → 23 |
| HV turns | 607 |

## Check yourself

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

1. How is the flux phi_m obtained from the ratio r = phi_m/AT?
   A. phi_m = r*Bm
   B. From per-phase kVA = 4.44 f phi_m AT x 10^-3 with AT = phi_m/r, so phi_m^2 = r*(phi_m*AT) and phi_m = 0.493 Wb
   C. phi_m = Et/(4.44 f) with Et assumed 0.6*sqrt(Q)
   D. phi_m = AT*Bm/r
   Answer: B. The product phi_m*AT is fixed by the per-phase kVA (1501.5), so with AT = phi_m/r, phi_m = sqrt(r*1501.5) = 0.493 Wb. (Key point: Flux from the loading ratio)

2. How is the net iron area found once phi_m = 0.4932 Wb is known?
   A. Ai = phi_m*Bm = 0.774 m^2
   B. Ai = phi_m/(Bm*0.89) = 0.353 m^2 is the net area
   C. Ai = phi_m/Bm = 0.314 m^2
   D. Ai = 0.6*phi_m = 0.296 m^2
   Answer: C. Net iron area is the flux divided by the maximum flux density: 0.4932/1.57 = 0.314 m^2. Dividing by 0.89 gives the gross area. (Key point: Net iron area)

3. For the 6-stepped core the diameter of the circumscribing circle is found from:
   A. d = sqrt(Ai/0.56)
   B. d = sqrt(Ai/0.6) = 0.724 m
   C. d = Ai/0.6
   D. d = sqrt(Ai/0.785)
   Answer: B. The answer uses Ai = 0.6 d^2 for the 6-stepped core, so d = sqrt(0.3141/0.6) = 0.724 m. (Key point: Circumscribing circle)

4. Why does the answer suspect the printed ratio 1.62x10^-4 Wb/AT?
   A. It gives a negative flux
   B. It gives about 110 V per turn, far above the normal about 19 V for 1000 kVA; 1.62x10^-6 was probably meant
   C. It makes the core area smaller than the gross area
   D. It gives more LV turns than HV turns
   Answer: B. Et = 109.5 V per turn is unrealistic; with 1.62x10^-6, Et = 10.95 V and the results become sensible (d = 229 mm, 23 LV turns). (Key point: Data check on the ratio)

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