HW 06 - ELEN3441FundamentalsofPowerEngineering Spring2011...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ELEN 3441 – Fundamentals of Power Engineering Spring 2011 Page | 1 Homework 6 – Solutions 6.1. (15 pt.) The speed of a synchronous machine is related to its frequency by the equation 120 e m f n P = To make a 50 Hz and a 60 Hz machine have the same mechanical speed so that they can be coupled together, we see that ,1 ,2 12 120 120 ee sync f f n PP == ,2 2 1, 1 60 12 50 10 e e f P Pf = = = Therefore, a 10-pole synchronous motor must be coupled to a 12-pole synchronous generator to accomplish this frequency conversion. 6.2. (20 pt.) (a) If the no-load terminal voltage is 480 V, the required field current can be read directly from the open-circuit characteristic. It is 4.55 A. (b) This generator is Y-connected, so I L = I A . At rated conditions, the line and phase current in this generator is
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 6.3. (20 pt) (a) The base phase voltage of this generator is Therefore, the base impedance of the generator is (a) The generator impedances in ohms are: (b) The rated armature current is The power factor is 0.8 lagging, so
Background image of page 2
Image of page 3
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 / 4

HW 06 - ELEN3441FundamentalsofPowerEngineering Spring2011...

This preview shows document pages 1 - 3. Sign up to view the full document.

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