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Half-load efficiency of 200 kVA transformer

Short answer

Using full-load efficiency, total losses are 3.2653 kW; maximum efficiency at 3/4 load gives copper loss 2.0898 kW and iron loss 1.1755 kW. At half load copper loss is 0.5224 kW and efficiency = 80/(80 + 1.1755 + 0.5224) = 97.92%.

A 200 kVA transformer has an efficiency of 98 % at full load if the maximum efficiency occurs at three quarters of full load, calculate efficiency at half load. Assume negligible magnetizing current and 0.8 pf at all loads.

Maximum efficiency occurs when copper loss equals iron loss. Use this with the full-load efficiency to find both losses.

Losses at full load

Pout,fl=200×0.8=160 kWTotal losses=1600.98−160=3.2653 kWPi+Pcu,fl=3.2653\begin{aligned} P_{out,fl} &= 200\times0.8 = 160\ \text{kW} \\ \text{Total losses} &= \frac{160}{0.98} - 160 = 3.2653\ \text{kW} \\ P_i + P_{cu,fl} &= 3.2653 \end{aligned}

Condition for maximum efficiency at 3/4 load

Pi=(34)2Pcu,fl=0.5625 Pcu,fl1.5625 Pcu,fl=3.2653⇒Pcu,fl=2.0898 kWPi=3.2653−2.0898=1.1755 kW\begin{aligned} P_i &= \left(\tfrac34\right)^2 P_{cu,fl} = 0.5625\,P_{cu,fl} \\ 1.5625\,P_{cu,fl} &= 3.2653 \Rightarrow P_{cu,fl} = 2.0898\ \text{kW} \\ P_i &= 3.2653 - 2.0898 = 1.1755\ \text{kW} \end{aligned}

Efficiency at half load

Pout=0.5×160=80 kWPcu=(0.5)2×2.0898=0.5224 kWη=8080+1.1755+0.5224=8081.698=97.92%\begin{aligned} P_{out} &= 0.5\times160 = 80\ \text{kW} \\ P_{cu} &= (0.5)^2\times2.0898 = 0.5224\ \text{kW} \\ \eta &= \frac{80}{80 + 1.1755 + 0.5224} = \frac{80}{81.698} = 97.92\% \end{aligned}

Answer: efficiency at half load ≈ 97.92%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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