Six-stepped core area and turns (1000 kVA)

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: