Prelim_exam_sol_sp03 - Copy

Prelim_exam_sol_sp03 - Copy - Ph.D. Preliminary Examination...

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Ph.D. Preliminary Examination Spring 2003 Solutions
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2 Problem 1 (Core) Code Number:__________
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3 Problem 2 (Core) Code Number:__________
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4 Problem 3 (Core) Code Number:__________
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5 Problem 4 (Core) Code Number:__________
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6 Problem 5 (Core) Code Number:__________
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7
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8 Problem 6 (Core) Code Number:__________
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9 Problem 7 (Core) Code Number:__________
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10 Problem 8 (Core) Code Number:__________
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12 Problem 9 (Core) Code Number:__________ SOLUTION (a) The output frequency of a synchronous generator is proportional to the speed. Therefore since the frequency at 1800 rpm is 60 Hz, the frequency at 1200 rpm is 60 (1200/1800) = 40 Hz. (b) The no load voltage is directly proportional to the frequency and the field current. Therefore, V V load no 200 8 6 1800 1200 400 _ = = (c) The no load voltage is the internal voltage, E af . The terminal voltage, V t , is the internal voltage minus the drop across X s . The value of X s at 40 Hz is 0.8(40/60) = 0.533 Ω. V V V j I jX E V t t a s af t 4 . 195 cos 200 31 . 12 ) 80 )( 533 . 0 ( sin 200 0 0 ) 0 80 )( 533 . 0 ( 200 ~ ~ ~ = = ° = = = = = δ
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Problem 10 (Core) Code Number:__________ Let n V ~ be the voltage at the neutral (common point) of the electric load. Kirchoff’s voltage law provides: n a c b a c b a V I I I j I j I I j I j ~ ~ 20 ) ~ ~ ( 5 . 1 ~ 0 . 4 ) ~ ~ ( 0 . 1 ~ 5 . 3 2 . 7 + + + + + + = n b c a b c a b j V I I I j I j I I j I j e ~ ~ 20 ) ~ ~ ( 5 . 1 ~ 0 . 4 ) ~ ~ ( 0 . 1 ~ 5 . 3 2 . 7 0 120 + + + + + + = n c a b c a b c j V I I I j I j I I j I j e ~ ~ 20 ) ~ ~ ( 5 . 1 ~ 0 . 4 ) ~ ~ ( 0 . 1 ~ 5 . 3 2 . 7 0 240 + + + + + + = c b a I I I ~ ~ ~ 0 . 0 + + = Solution of above equations yields kA e I j a 0 64 . 15 2773 . 0 ~ = kA
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This note was uploaded on 09/23/2010 for the course ECE 0000 taught by Professor Ddaa during the Fall '10 term at Georgia Institute of Technology.

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Prelim_exam_sol_sp03 - Copy - Ph.D. Preliminary Examination...

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