How is full-load efficiency at 0.8 pf obtained?

Output 240 kW; losses 4.3 kW; eta = 240/(240 + 4.3) = 98.24%

Output 300 kW; losses 1.8 kW only

Output 240 kW; losses 2.5 kW; eta = 98.97%

Output 300 kVA; losses 4.3 kW; eta = 98.6%

At what kVA is efficiency maximum?

300 x sqrt(1.8/2.5) = 254.56 kVA

2081 Kartik (new course) · 6 marks

300 kVA transformer efficiency calculation

Short answer

Full-load efficiency at 0.8 lagging pf is98.24%. Maximum efficiency occurs at254.56 kVA, yielding98.26% at0.8 pf.

A 300 kVA single phase transformer has iron losses of 1.8 kW. The full load copper losses is 2500 watts. Calculate, (i) Efficiency at full load, 0.8 lagging pf (ii) kVA supplied at maximum Efficiency (iii) Maximum Efficiency at 0.8 lagging pf.

Given: $S=300$ kVA, $P_i=1.8$ kW, full-load $P_{cu}=2.5$ kW.

(i) Full-load efficiency at 0.8 lagging pf.

\text{Output}=300\times0.8=240\ \text{kW}

\text{Total loss}=1.8+2.5=4.3\ \text{kW}

\eta=\frac{240}{240+4.3}\times100=98.24\%

(ii) kVA at maximum efficiency. Efficiency is maximum when copper loss = iron loss:

x=\sqrt{\frac{P_i}{P_{cu}}}=\sqrt{\frac{1.8}{2.5}}=0.8485

\text{kVA}=300\times0.8485=254.56\ \text{kVA}

(iii) Maximum efficiency at 0.8 pf. Output $=254.56\times0.8=203.65$ kW. At this load, the copper loss equals the iron loss $=1.8$ kW, so the total loss is $2\times1.8=3.6$ kW:

\eta_{max}=\frac{203.65}{203.65+3.6}\times100=98.26\%

Answer: (i) 98.24%, (ii) 254.56 kVA, (iii) 98.26%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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