HW 03 - ELEN3441FundamentalsofPowerEngineering Spring2011...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
ELEN 3441 – Fundamentals of Power Engineering Spring 2011 Page | 1 Homework 3 – Solutions 3.1. (25 pt.) The turns ratio of this transformer is a = 8000/277 = 28.89. Therefore, the secondary impedances referred to the primary side are c) To simplify the calculations, we use the simplified equivalent circuit referred to the primary side of the transformer:
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
ELEN 3441 – Fundamentals of Power Engineering Spring 2011 Page | 2 3.2. (20 pt.) We refer the circuit to the secondary (low-voltage) side. The feeder’s impedance referred to the secondary side is The secondary current I s is given by
Background image of page 2
ELEN 3441 – Fundamentals of Power Engineering Spring 2011 Page | 3 90 000 46.04 ; 46.04 31.79 2300 0.85 s IA == = ° s I a) The voltage at the power source of this system (referred to the secondary side) is () ( ) ( ) ' 2 300 0 46.04 31.79 1.12 4.11 46.04 31.79 0.12 0.5 2461.17 3.6 sl i n e E Q ZZ j jV ++= ° + ° + + ° + = source s s V' = V I I Therefore, the voltage at the power source is 14 000 2461.17 3.6 14.36 3.6 2 400 kV =∠ ° = ° sourse V b)
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/21/2011 for the course EE ee taught by Professor Ee during the Spring '10 term at Istanbul Technical University.

Page1 / 7

HW 03 - ELEN3441FundamentalsofPowerEngineering Spring2011...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online