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: $R_L=0.2\mathrm{\Omega },X_L=0.6\mathrm{\Omega },R_H=0.8\mathrm{\Omega },X_H=1.6\mathrm{\Omega },R_{cH}=3.2k\mathrm{\Omega },X_{mH}=1.2k\mathrm{\Omega }$

• 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 $a=2ands=48000\mathrm{\angle }{\cos }^{-1}(0.7)$

$I_2=\frac{s^{*}}{v_2^{*}}=200\mathrm{\angle }-46^{\circ }$

$R_H+a^2{R}_l=1.6$

$j(X_H+A^2{X}_L)=J4=\frac{1}{a}{\overrightarrow{I}}_2((R_H+a^2{R}_L)+j(X_H+a^2{X}_L))+a{\overrightarrow{V}}_2)$

${\overrightarrow{V}}_1=893\mathrm{\angle }10.7\circ insteadof480V$

$P_{core}=P_c=250w=\frac{v_1^2}{R_{cH}}=\frac{893^2}{3200}$

$P_{cu}=16000w$

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

$VR=\frac{V_1-a{V}_2}{a{V}_2}$

Conclusion:

$n=67\mathrm{\%}$

$VR=86\mathrm{\%}$