Three phase output equation and circular windings

Output equation of a 3-phase core type transformer

Let $A_i$ = net core area, $A_w$ = window area, $B_m$ = maximum flux density, $\delta$ = current density, $K_w$ = window space factor, $T$ = turns per phase, $I$ = phase current.

EMF per phase:

$$V \approx E = 4.44fB_mA_iT$$

There are three limbs and two windows. Each window holds one primary and one secondary of two adjacent limbs, so the copper area in one window is

$$K_wA_w = 2\left(\frac{I_pT_p}{\delta} + \frac{I_sT_s}{\delta}\right) = \frac{4AT}{\delta} \;\Rightarrow\; AT = \frac{K_wA_w\delta}{4}$$

Output:

$$\begin{aligned} Q &= 3VI\times10^{-3} = 3\times4.44fB_mA_i\,(IT)\times10^{-3}\\ &= 3\times4.44fB_mA_i\times\frac{K_wA_w\delta}{4}\times10^{-3}\\ Q &= 3.33\,fB_m\delta K_wA_wA_i\times10^{-3}\ \text{kVA} \end{aligned}$$

Why windings are made circular

1. Short-circuit forces: under a short circuit the radial forces act outward on the HV and inward on the LV winding. A circular coil takes them as uniform hoop stress and keeps its shape; a rectangular coil would bend at the flat sides.
2. Least copper: for a given enclosed area, a circle has the smallest perimeter, so the mean length of turn, copper weight, cost and $I^2R$ loss are least.
3. Ease of winding: circular coils are easily wound on formers on a lathe, with uniform tension.
4. Uniform insulation and cooling: clearances between core, LV and HV are uniform all round, and oil ducts are easy to provide.

Because the winding is circular, the core is made stepped (cruciform or multi-stepped) to fill the circle as fully as possible.