Single Phase Transformer Solved Problems

1)A single phase 90 kVA, 3.2 kV/220 V, 50Hz transformer has 89\% efficiency at unity power factor both at full load and half load. Determine the efficiency at 70\% of full load and 0.8 power factor.

\eta_{100\%}=\eta_{50\%}

\dfrac{VIcos\phi}{VIcos\phi+P_i+P_{cu}}=\dfrac{xVIcos\phi}{xVIcos\phi+P_i+x^2P_{cu}}

\dfrac{VIcos\phi}{VIcos\phi+P_i+P_{cu}}=\dfrac{0.5VIcos\phi}{0.5VIcos\phi+P_i+0.5^2P_{cu}}

P_i+(0.5)^2P_{cu}=0.5P_i+0.5P_{cu}

P_i-0.5P_i=0.5P_{cu}-(0.5)^2P_{cu}

P_{cu}=2P_i

Efficiency at 100\%

\eta_{100\%}= \dfrac{VIcos\phi}{VIcos\phi+P_i+P_{cu}}

\eta_{100\%}= \dfrac{VIcos\phi}{VIcos\phi+3P_i}

\eta_{100\%}= \dfrac{90\times 10^3}{90\times 10^3+3P_i}

P_i = 3707.8651 W

P_{cu} =2 P_i = 7415.7302 W

Now efficiency at 70\% of full load

\eta_{70\%} = \dfrac{xVIcos\phi}{xVIcos\phi+P_i+x^2P_{cu}}

\eta_{70\%} = \dfrac{0.7\times 90\times 10^3 \times 0.8}{0.7\times 90\times 10^3 \times 0.8+3707.8651+0.7^2 \times 7415.7302}

\eta_{70\%} = 87.28 \%