Example: Step Down Transformer: Difference between revisions

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



<math>I_2~=~\frac{s^*}{v_2^*}~=~200\angle -46^\circ </math>
<math>I_2~=~\frac{s^*}{v_2^*}~=~200\angle -46^\circ </math>



<math> R_H~+~a^2~R_l~=~1.6</math>
<math> R_H~+~a^2~R_l~=~1.6</math>

<math>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)</math>

<math>\vec V_1~=~893\angle 10.7\circ /text{instead of} 480 V </math>
<math>P_core~=~ P_c~=~250 w = \frac{v_1^2/R_{cH}=\frac{893^2}{3200}</math>
<math>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\vec V_2\right)</math>

<math>P_{cu}=16000 w </math>

<math> n~=~\frac{Re(s)}{Re(s)~+~P_{cu}~+~P_c}</math>
<math>\vec V_1~=~893\angle 10.7\circ ~~{instead~ of}~~ 480 V </math>


<math>P_{core}~=~ P_c~=~250 w = \frac{v_1^2}{R_{cH}}=\frac{893^2}{3200}</math>


<math> P_{cu}~=~16000 w </math>


<math> n~=~\frac {Re(s)}{Re(s)~+~P_{cu}~+~P_c}</math>


<math>VR~=~\frac{V_1-a~V_2}{a~V_2}</math>



Conclusion:

<math>n~=~67~% </math>

<math> VR~=~86~%</math>

Latest revision as of 02:55, 29 January 2010

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:

  • 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










Conclusion: