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100 kVA transformer efficiency at full and half load

Short answer

(i) Full load, unity pf efficiency = 97.89%. (ii) Half load, 0.8 pf efficiency = 96.95%.

A transformer is rated at 100 kVA. At full load its copper loss is 1200 W and its iron loss is 960 W. Calculate: (i) Efficiency at full load, unity power factor (ii) Efficiency at half load, 0.8 power factor

Given: 100 kVA, Pcu,fl=1200P_{cu,fl} = 1200 W, Pi=960P_i = 960 W.

The efficiency at a load fraction xx and power factor cos⁡ϕ\cos\phi is

η=xScos⁡ϕxScos⁡ϕ+Pi+x2Pcu,fl\eta = \frac{xS\cos\phi}{xS\cos\phi + P_i + x^2P_{cu,fl}}

(i) Full load, unity power factor (x=1x = 1)

Pout=100×1=100 kWLosses=0.96+1.2=2.16 kWη=100100+2.16×100=97.89%\begin{aligned} P_{out} &= 100\times1 = 100\ \text{kW} \\ \text{Losses} &= 0.96 + 1.2 = 2.16\ \text{kW} \\ \eta &= \frac{100}{100 + 2.16}\times100 = 97.89\% \end{aligned}

(ii) Half load, 0.8 power factor (x=0.5x = 0.5)

Pout=0.5×100×0.8=40 kWPcu=0.52×1.2=0.3 kWLosses=0.96+0.3=1.26 kWη=4040+1.26×100=96.95%\begin{aligned} P_{out} &= 0.5\times100\times0.8 = 40\ \text{kW} \\ P_{cu} &= 0.5^2\times1.2 = 0.3\ \text{kW} \\ \text{Losses} &= 0.96 + 0.3 = 1.26\ \text{kW} \\ \eta &= \frac{40}{40 + 1.26}\times100 = 96.95\% \end{aligned}

Answer: (i) 97.89%; (ii) 96.95%

✓ Answer checked by Claude Opus 5.5 (high effort)

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