A 500-kVA, 3-phase, 50-Hz transformer has a voltage ratio (line voltage) of 33/11-kV and is delta/star connected. The resistances per phase are: high voltage 35 Ω, low voltage 0.876 Ω and the iron loss is 3050W. Calculate the value of efficiency at full-load and one-half of full-load respectively (a) at unity p.f. and (b) 0.8 p.f.

Transformation ratio $K =\frac{11,000}{\sqrt{3}\times 33,000}=\frac{1}{3\sqrt{3}}$

Per phase $R_{02} = 0.876 + {(1/3\sqrt{3})}^2 \times 35 = 2.172\Omega$

Secondary phase current $=\frac{500,000}{\sqrt{3}\times 11,000}$

$=\frac{500}{11\sqrt{3}}A$

Full-load condition

Full load total Cu loss,

$=3\times {\left(\frac{500}{11\sqrt{3}}\right)}^2\times 2.172$ = 4,490 W;

Iron loss = 3,050 W

Total full-load losses

= 4,490 +3,050 = 7,540 W;

Output at unity p.f. = 500 kW

∴ F.L. efficiency $=\frac{500,000}{507,540}$ = 0.9854 or 98.54%;

Output at 0.8 p.f.= 400 kW

∴ Efficiency $=\frac{400,000}{407.540}$ = 0.982 or 98.2%

Half-load condition

Output at unity p.f.= 250 kW

Cu losses $={\left(\frac{1}{2}\right)}^2\times 4,490$= 1,222 W

Total losses = 3,050 + 1,122 = 4,172 W

∴ $\eta =\frac{250,000}{254,172}$ = 0.9835 = 98.35%

Output at 0.8 p.f. = 200 kW

$\therefore \eta =\frac{200,000}{204,172}$ = 0.98 or 98%