Iron loss and maximum efficiency of 20 kVA 250/2500 V transformer

Given (referred to the 250 V primary): $R_0 = 600\ \Omega$, $X_0 = 180\ \Omega$, $R_{01} = 0.05\ \Omega$, $X_{01} = 0.15\ \Omega$; $V_1 = 250$ V.

(i) Iron loss

Only the core-loss resistance $R_0$ absorbs real power in the shunt branch, so at rated voltage:

$$P_i = \frac{V_1^2}{R_0} = \frac{250^2}{600} = 104.17\ \text{W}$$

(ii) Primary current for maximum efficiency

Maximum efficiency occurs when copper loss = iron loss:

$$\begin{aligned} I_1^2R_{01} &= P_i \\ I_1 &= \sqrt{\frac{104.17}{0.05}} = \sqrt{2083.3} = 45.64\ \text{A} \end{aligned}$$

Full-load primary current $= 20000/250 = 80$ A, so this is $45.64/80 = 0.571$, i.e. about 57% of full load.

Maximum efficiency

Taking unity power factor (not stated; maximum efficiency is normally quoted at upf):

$$\begin{aligned} \text{Output} &= 250 \times 45.64 \times 1 = 11410.9\ \text{W} \\ \text{Losses} &= P_i + P_{cu} = 104.17 + 104.17 = 208.33\ \text{W} \\ \eta_{max} &= \frac{11410.9}{11410.9 + 208.33}\times 100 = 98.21\% \end{aligned}$$

Answer: Iron loss = 104.17 W; maximum efficiency at a primary current of 45.64 A; $\eta_{max} \approx$ 98.21% (upf).