Example: Step Down Transformer: Difference between revisions

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: $R_{L} = 0.2 Ω, X_{L} = 0.6 Ω, R_{H} = 0.8 Ω, X_{H} = 1.6 Ω, R_{c H} = 3.2 k Ω, X_{m H} = 1.2 k Ω$

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 $a = 2 a n d s = 48000 ∠ \cos^{- 1} (0.7)$

$I_{2} = \frac{s^{*}}{v_{2}^{*}} = 200 ∠ - 4 6^{\circ}$

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

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

${\vec{V}}_{1} = 893 ∠ 10.7 \circ i n s t e a d o f 480 V$

$P_{c o r e} = P_{c} = 250 w = \frac{v_{1}^{2}}{R_{c H}} = \frac{89 3^{2}}{3200}$

$P_{c u} = 16000 w$

$n = \frac{R e (s)}{R e (s) + P_{c u} + P_{c}}$

$V R = \frac{V_{1} - a V_{2}}{a V_{2}}$

Conclusion:

$n = 67 %$

$V R = 86 %$

@@ Line 9: / Line 9: @@
 From the given data of the step down transformer we can see that
-<math>\text a=2~and~s=48000~\angle cos^{-1}      \</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> 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\vec V_2\right)</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>

Example: Step Down Transformer: Difference between revisions

Latest revision as of 02:55, 29 January 2010

Problem Statement

Solution

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