Example: Step Down Transformer

Problem Statement

Given a 48 kVA, 480/240 V/V step down transformer operating at rated load with a power factor of 0.7 (lag), determine the efficiency and voltage regulation.

Given: $R_{L}~=~0.2~\Omega ,~~X_{L}~=~0.6~\Omega ,~~R_{H}~=~0.8~\Omega ,~~X_{H}~=~1.6~\Omega ,~~R_{cH}~=~3.2~k\Omega ,~~X_{mH}~=~1.2~k\Omega$

• this problem is from EMEC EXAM 1, Winter 08 by legendary Dr. Cross

Solution

From the given data of the step down transformer we can see that $~~~a=2~and~s=48000~\angle \cos ^{-1}(0.7)$

$I_{2}~=~{\frac {s^{*}}{v_{2}^{*}}}~=~200\angle -46^{\circ }$

$R_{H}~+~a^{2}~R_{l}~=~1.6$

$j~(X_{H}~+A^{2}~X_{L})~=~J~4~=~{\frac {1}{a}}~{\vec {I}}_{2}~\left((R_{H}~+~a^{2}~R_{L})~+~j(X_{H}~+~a^{2}X_{L}))+a{\vec {V}}_{2}\right)$

${\vec {V}}_{1}~=~893\angle 10.7\circ ~~{instead~of}~~480V$

$P_{core}~=~P_{c}~=~250w={\frac {v_{1}^{2}}{R_{cH}}}={\frac {893^{2}}{3200}}$

$P_{cu}~=~16000w$

$n~=~{\frac {Re(s)}{Re(s)~+~P_{cu}~+~P_{c}}}$

$VR~=~{\frac {V_{1}-a~V_{2}}{a~V_{2}}}$

Conclusion:

$n~=~67~\%$

$VR~=~86~\%$