# HW 06 - ELEN3441FundamentalsofPowerEngineering Spring2011...

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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

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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
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## 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.

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HW 06 - ELEN3441FundamentalsofPowerEngineering Spring2011...

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