# HW 04 - Spring2011 Homework 4 Solutions 4.1(25 pt a At...

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ELEN 3441 – Fundamentals of Power Engineering Spring 2011 Page | 1 Homework 4 – Solutions 4.1. (25 pt.) a) At no-load conditions, E A = V T = 240 V. The field current is given by 240 0.96 175 75 T F adj F V I A RR == = ++ From Figure 5-1, this field current would produce an internal generated voltage E Ao of 277 V at a speed n o of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be 0 00 0 240 1200 1040 277 A A AA En E n nr p m E = = = = b) The minimum speed will occur when R adj = 100 , and the maximum speed will occur when R adj = 400 . The field current when R adj = 100 is: 240 1.37 100 75 T F adj F V I A = From the magnetization curve, this field current would produce an internal generated voltage E Ao of 289 V at a speed n o of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be 0 0 240 1200 997 289 A A E n p m E = = = = The field current when R adj = 400 is: 240 0.505 400 75 T F adj F V I A = From the magnetization curve, this field current would produce an internal generated voltage E Ao of 207 V at a speed n o of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be 0 0 240 1200 1391 207 A A E n p m E = = = = c) The starting current of this machine (ignoring the small field current) is , 240 1263 0.19 T Ls tar t A V I A R = The rated current is 110 A, so the starting current is 11.5 times greater than the full-load current.

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HW 04 - Spring2011 Homework 4 Solutions 4.1(25 pt a At...

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