Question

A three phase 300 kVA, 30 kV/1 kV $\left(\Delta /Y\right)$ 50 Hz transformer has primary (hv side) resistance per phase of $0.4\Omega$ and secondary (Iv side) resistance per phase of $0.1\Omega$. Iron loss of the transformer is 5000 W. The full load percentage efficiency of the transformer operated at unity power factor will be _____.

• A. 95.14

Answer 1

The correct option is A 95.14
$\text{For given}3-\varphi \text{transformer}$



$\text{Given,}{R}_1=0.4\Omega /phase$

$R_2=0.1\Omega /phase$

$\text{Iron loss = 5000 W}$

$\text{Primary line current,}{I}_{L1}=\frac{300\times 1000}{\sqrt{3}\times 3000}=\frac{100}{\sqrt{3}}A$

$\text{Primary phase current,}{I}_{ph,1}=\frac{I_{L1}}{\sqrt{3}}=\frac{100}{3}A$

$\text{Secondary line current,}{I}_{L2}=\frac{300\times 1000}{\sqrt{3}\times 1000}=\frac{300}{\sqrt{3}}A$

$\therefore \text{Secondary phase current}=\frac{300}{\sqrt{3}}A\text{(Star connection)}$

$\text{Copper loss on hv side}=3I_1^2{R}_1=3{\left(\frac{100}{3}\right)}^2\times \left(0.4\right)$ $=\frac{3\times 100\times 100}{9}\times 0.4=13333.33W$

$\text{Copper loss on lv side}=3I_2^2{R}_2=3{\left(\frac{300}{\sqrt{3}}\right)}^2\times \left(0.1\right)=9000W$

$\text{At unity power factor,}$

$\text{Efficiency,}\eta =\frac{(300\times 1000\times 1)}{(300\times 1000\times 1)+5000+1333.33+9000}$

$\eta =0.951374$

%$\eta =95.14\%$