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\ | |||
<math>\vec V_1~=~893\angle 10.7\circ | |||
<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: