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

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

$R_H+a^2{R}_l=1.6$ Failed to parse (unknown function "\vecV"): {\displaystyle 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^2X_L))+a\vecV_2\right)} ${\overrightarrow{V}}_1=893\mathrm{\angle }10.7\circ /textinsteadof480V$ Failed to parse (syntax error): {\displaystyle P_core~=~ P_c~=~250 w = \frac{v_1^2/R_{cH}=\frac{893^2}{3200}} $P_{cu}=16000w$ $n=\frac{Re(s)}{Re(s)+P_{cu}+P_c}$