Iron loss and maximum efficiency of 20 kVA 220/2200 V transformer

Given (referred to the 220 V primary): $R_0 = 500\ \Omega$, $X_0 = 160\ \Omega$, $R_{01} = 0.04\ \Omega$, $X_{01} = 0.1\ \Omega$. Rated primary voltage $V_1 = 220$ V.

(i) Iron loss

Iron loss is the power taken by the core-loss resistance $R_0$ at rated voltage:

$$P_i = \frac{V_1^2}{R_0} = \frac{220^2}{500} = 96.8\ \text{W}$$

($X_0$ only carries the magnetising current, which takes no real power.)

(ii) Current for maximum efficiency

Efficiency is maximum when copper loss equals iron loss:

$$\begin{aligned} I_1^2 R_{01} &= P_i \\ I_1 &= \sqrt{\frac{P_i}{R_{01}}} = \sqrt{\frac{96.8}{0.04}} = \sqrt{2420} = 49.19\ \text{A} \end{aligned}$$

Full-load primary current is $20000/220 = 90.91$ A, so maximum efficiency occurs at $49.19/90.91 = 0.541$, i.e. about 54% of full load.

Maximum efficiency

Assuming unity power factor (pf not given, and maximum efficiency is normally quoted at upf):

$$\begin{aligned} \text{Output} &= V_1 I_1 \cos\phi = 220 \times 49.19 \times 1 = 10822.6\ \text{W} \\ \text{Total loss} &= P_i + P_{cu} = 96.8 + 96.8 = 193.6\ \text{W} \\ \eta_{max} &= \frac{10822.6}{10822.6 + 193.6} \times 100 = 98.24\% \end{aligned}$$

Answer: Iron loss = 96.8 W; maximum efficiency occurs at a primary current of 49.19 A; $\eta_{max} \approx$ 98.24% at unity pf.