Efficiency and loading of 50 kVA transformer at 40 kW

Given: $R_0 = 600\ \Omega$, $X_0 = 200\ \Omega$ (primary, 200 V); $R_{02} = 2\ \Omega$, $X_{02} = 4\ \Omega$ (secondary, 2200 V). Load: 40 kW at 0.65 pf lagging, at rated voltage 2200 V.

Load kVA and current

$$\begin{aligned} S_{load} &= \frac{P}{\cos\phi} = \frac{40}{0.65} = 61.54\ \text{kVA} \\ I_2 &= \frac{61538}{2200} = 27.97\ \text{A} \end{aligned}$$

Rated secondary current $= 50000/2200 = 22.73$ A.

Losses

$$\begin{aligned} P_i &= \frac{V_1^2}{R_0} = \frac{200^2}{600} = 66.67\ \text{W} \\ P_{cu} &= I_2^2R_{02} = 27.97^2 \times 2 = 1564.87\ \text{W} \end{aligned}$$

Efficiency

$$\begin{aligned} \eta &= \frac{40000}{40000 + 66.67 + 1564.87}\times 100 \\ &= \frac{40000}{41631.54}\times 100 = 96.08\% \end{aligned}$$

Over-load or under-load?

The transformer is rated 50 kVA but supplies 61.54 kVA. The power (40 kW) is below 50 kW, but what matters for heating is the kVA (current), so it is over-loaded:

$$\%\text{Overload} = \frac{61.54 - 50}{50}\times 100 = 23.08\%$$

Same by current: $\frac{27.97 - 22.73}{22.73}\times 100 = 23.08\%$.

Answer: Efficiency = 96.08%; the transformer is over-loaded by 23.08% because of the low power factor.