Core, turns and conductors of 5 kVA transformer

Data: $Q$ = 5 kVA, $f$ = 50 Hz, $B_m$ = 1 Wb/m², $\delta$ = 1.4 A/mm², $K_w$ = 0.2, $H_w = 3W_w$, square core, and copper area in window $K_wA_w = 0.6A_i$.

Core area

Single-phase output equation:

$$Q = 2.22fB_m\delta\,(K_wA_w)\,A_i\times10^{-3}$$

With $K_wA_w = 0.6A_i$:

$$\begin{aligned} Q &= 2.22fB_m\delta\,(0.6A_i)A_i\times10^{-3}\\ 5 &= 2.22\times50\times1\times1.4\times10^6\times0.6\times10^{-3}\,A_i^2 = 93240\,A_i^2\\ A_i &= \sqrt{\frac{5}{93240}} = 0.00732\ \text{m}^2 = 73.2\ \text{cm}^2 \end{aligned}$$

Core dimensions (square core)

For a square core with stacking factor 0.9: $A_i = 0.9\times0.5d^2 = 0.45d^2$, side $a = d/\sqrt2 = 0.707d$.

$$\begin{aligned} d &= \sqrt{\frac{0.00732}{0.45}} = 127.6\ \text{mm}\\ a &= 0.707d = 90.2\ \text{mm (side of square limb)} \end{aligned}$$

Window

$$\begin{aligned} K_wA_w &= 0.6A_i = 0.00439\ \text{m}^2\\ A_w &= \frac{0.00439}{0.2} = 0.02197\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{3}} = 85.6\ \text{mm},\quad H_w = 3W_w = 256.7\ \text{mm}\\ D &= d + W_w = 213.1\ \text{mm (centre distance of limbs)} \end{aligned}$$

Turns

$$\begin{aligned} E_t &= 4.44fB_mA_i = 4.44\times50\times1\times0.00732 = 1.626\ \text{V}\\ T_{LV} &= \frac{400}{1.626} = 246.0 \approx 246\\ T_{HV} &= 246\times\frac{11000}{400} = 6765.0 \approx 6765 \end{aligned}$$

Conductor sections

$$\begin{aligned} I_{HV} &= \frac{5000}{11000} = 0.4545\ \text{A},\quad a_{HV} = \frac{0.4545}{1.4} = 0.325\ \text{mm}^2\\ I_{LV} &= \frac{5000}{400} = 12.5\ \text{A},\quad a_{LV} = \frac{12.5}{1.4} = 8.93\ \text{mm}^2 \end{aligned}$$

The HV conductor ≈ 0.33 mm² (about 0.65 mm diameter round wire); the LV conductor ≈ 8.9 mm² (round wire about 3.4 mm diameter, or a small strip).

Answer: $A_i$ ≈ 73.2 cm², square limb 90.2 mm (circle 127.6 mm), window 85.6 × 256.7 mm; HV 6765 turns of 0.325 mm², LV 246 turns of 8.93 mm².