Design: 100 kVA single phase, square core

Standard method (A.K. Sawhney, A Course in Electrical Machine Design): $E_t = K\sqrt{Q}$, net core area ratio for the core type, output equation for the window, yoke depth = width of largest stamping. Assumed: yoke area 1.2 × core area (hot-rolled steel).

(i) Overall dimensions of core

Voltage per turn, $E_t = K\sqrt{Q}$ ($Q$ in kVA):

$$\begin{aligned} E_t &= 0.75\sqrt{100} = 7.500\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{7.500}{4.44\times50} = 0.03378\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.03378}{1.2} = 0.02815\ \text{m}^2 = 281.5\ \text{cm}^2 \end{aligned}$$

For a square core with stacking factor 0.9, $A_i = 0.45d^2$ (standard ratio, Sawhney):

$$\begin{aligned} d &= \sqrt{\frac{A_i}{0.45}} = \sqrt{\frac{0.02815}{0.45}} = 0.2501\ \text{m} = 250.1\ \text{mm}\\ a &= 0.71d = 177.6\ \text{mm} \end{aligned}$$

Gross core area $A_{gi} = A_i/0.9 = 0.03128$ m².

Output equation (single phase): $Q = 2.22fB_m\delta K_wA_wA_i\times10^{-3}$

$$\begin{aligned} A_w &= \frac{Q}{2.22fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{100}{2.22\times50\times1.2\times2.5\times10^6\times0.28\times0.02815\times10^{-3}}\\ &= 0.03810\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2}} = 138.0\ \text{mm},\quad H_w = 2W_w = 276.0\ \text{mm}\\ D &= d + W_w = 388.1\ \text{mm (distance between adjacent limb centres)} \end{aligned}$$

Yoke:

Yoke area 20% more than core area (hot-rolled steel, to reduce yoke loss and magnetising current). Yoke depth = width of largest stamping:

$$\begin{aligned} A_y &= 1.2A_i = 0.03378\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.03754\ \text{m}^2\\ D_y &= a = 177.6\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.03754}{0.1776} = 211.4\ \text{mm} \end{aligned}$$

(ii) Overall dimensions of frame

$$\begin{aligned} H &= H_w + 2H_y = 276.0 + 2(211.4) = 698.8\ \text{mm}\\ W &= D + a = 388.1 + 177.6 = 565.7\ \text{mm}\\ \text{Depth} &= D_y = a = 177.6\ \text{mm} \end{aligned}$$

 +-----------------------+  ---
 |       top yoke        |   ^
 +-----+-----------+-----+   |
 |limb |  window   |limb |   H
 |     |  Ww x Hw  |     |   |
 +-----+-----------+-----+   |
 |      bottom yoke      |   v
 +-----------------------+  ---
 |<--------- W --------->|
    |<----- D ----->|

(iii) Turns and conductor areas

Currents: $I_{LV} = \frac{Q}{V_{LV}} = 208.33$ A; $I_{HV} = 45.455$ A.

$$\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{480.0}{7.500} = 64.00 \approx 64\\ E_t\ (\text{revised}) &= \frac{480.0}{64} = 7.500\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 293.3 \approx 293\\ a_{LV} &= \frac{I_{LV}}{\delta} = 83.33\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 18.182\ \text{mm}^2 \end{aligned}$$

The LV conductor (83.3 mm²) would be a rectangular strip (for example several strips in parallel); the HV conductor is a round or small rectangular wire.

Answer: $d$ = 250.1 mm ($a$ = 177.6 mm), window 138 × 276 mm; frame 699 × 566 × 178 mm; HV 293 turns of 18.18 mm², LV 64 turns of 83.33 mm².