Question

A single phase transformer has the ratio $\frac{x_{e2}}{r_{e2}}=3$ and equivalent resistance is 1.5% and per unit core loss is 0.01. When voltage regulation of this transformer is 4%, then the full load efficiency of this transformer is ($x_{e2}and{r}_{e2}$ are equivalent reactance and resistance of transformer referred to secondary).

• A. 98.10%
• B. 97.56%
• C. 96.88%
• D. 90.73%

Answer 1

The correct option is C 96.88%

$\text{Let full load be 1 p.u.}$
$\text{Core loss will be 0.01 p.u.}$

$\text{Copper loss will be}\frac{1.5}{100}p.u.=0.015p.u.$
$r_{e2}=0.015p.u.$

$x_{e2}=3\times 0.015=0.045p.u.$

$\text{\% V.R = \% R cos}\varphi \text{+\% X sin}\varphi \text{(Lagging load)}$

$0.04=0.015cos\varphi +0.045sin\varphi$

$\frac{0.04}{0.045}=\frac{1}{3}cos\varphi +sin\varphi$

$\text{By multiplying and dividing by 1.054 on RHS}$
$0.888=(\sin 18.43\cos \varphi +\sin \varphi \cos 18.43)\times 1.054$

$sin(\varphi +18.43)=\frac{0.888}{1.054}=0.8425$

$\varphi +18.43=57.40$
$\varphi ={38.975}^{\circ }$
$\cos \varphi =0.777lag$
$\eta =\frac{1\times 0.777}{1\times 0.777+0.01+0.015}=96.88\%$

$\text{Alternate Solution:}$

$P_i=0.01p.u.$
$P_{cu\left(rated\right)}=0.015p.u.$
$X_{e2}=3\times 0.015=0.045$
$Z_{e2}={(X_{e2}^2+r_{e2}^2)}^{1/2}=0.474p.u.$

${\theta }_z=ta{n}^{-1}\left(\frac{X_{e1}}{r_{e2}}\right)={71.565}^{\circ }$

$\text{Voltage regulation =}{Z}_{pu}cos({\theta }_z-\varphi )$

$0.04=0.0474cos(71.565-\varphi )$

$\varphi ={39.05}^{\circ }$

$pf=cos39.05=0.777$
$\eta =\frac{1\times 0.777}{1\times 0.777+0.01+0.015}=96.88\%$