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 |
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<math> |
<math>~~~a=2~and~s=48000~\angle \cos^{-1}(0.7)</math> |
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<math>I_2~=~\frac{s^*}{v_2^*}~=~200\angle -46^\circ </math> |
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<math> R_H~+~a^2~R_l~=~1.6</math> |
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<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> |
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<math>\vec V_1~=~893\angle 10.7\circ /text{instead of} 480 V </math> |
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<math>P_core~=~ P_c~=~250 w = \frac{v_1^2/R_{cH}=\frac{893^2}{3200}</math> |
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<math>P_{cu}=16000 w </math> |
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<math> n~=~\frac{Re(s)}{Re(s)~+~P_{cu}~+~P_c}</math> |
Revision as of 01:38, 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
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)} Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_core~=~ P_c~=~250 w = \frac{v_1^2/R_{cH}=\frac{893^2}{3200}}