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Test_3_solutions

# Test_3_solutions - Page 1 0f 6 NORTH CAROLINA STATE...

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Unformatted text preview: Page 1 0f 6 NORTH CAROLINA STATE UNIVERSITY DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING Lecturer: John W. Gajda, P.E. ECE451 Test 3 Fall 2010 NAME » '53 5 W7 .' Problem No. 1 (possible 25 pts.) Problem No. 2 (possible 25 pts.) Problem No. 3 (possible 25 pts.) Problem No. 4 (possible 25 pts.) TOTAL (possible 100 pts.) NOTE: THIS IS A STRICTLY CLOSED BOOK EXAM. ALL WORK MUST BE SHOWN. DO ALL FOUR PROBLEMS. Honor pledge: I certify that I have neither given nor received unauthorized aid while taking this test. Signature: Problem Page 2 of 6 No. 1 (25 points) A circuit breaker is closed in to a faulted circuit. - The source is 60 Hz, 22.86 kV line-to-line RMS nominal voltage, infinite bus - The circuit beyond the breaker is faulted one mile from the substation with a three phase bolted fault. - That mile of line has a resistance of 1.0 ohms, and a reactance of 6.0 ohms. fl. Parts A & B (15’ points): What is the (a) the angle in the cycle (alpha) and (b) the instantaneous value of the voltage (line-to-neutral) which will produce the maximum DC / \ _ .J " W. A; ’7. X I / g ,5” 1 / P 0/ L If“. r A T f b - I! I‘lfth.“ " \l r— ’— " v — 1V. -r—w‘ M L 4. / j I“, ‘I L / ,.. yd -- .1 __ \ " . v i // ‘ “ r— I ’7 'i' _ m" 1‘ :\-"L', “JV. ,an’r—l-‘fm _’- - ca r“ r' .77 H 7 j} 0 IL: ,ﬁ i if Part C ()1 points): Assuming alpha = 90 degrees, find the value of the DC component at 2.5 cycles after the breaker closing. ‘_ _ , . 7— s . . ‘ f __ r ‘I J /. Am, I weee-nw—— . . 90‘ r .m. — -ir-a‘f‘ ,{i‘ f .ILJ-ﬁfli fir ’t J” / i j: F A 0 )c If A F r ' { 7 r A — ‘— r ‘1 I I 7 .- n g I * ~9§H2~w~i-e /cx%§, :;«brzr“ 6.“ A =3 R 3 L : 3.: é- : | I 1’ " Part D (2 points): Wthical tripping time for a modern vacuum-bottle circuit breaker? Circle one: (1 cycle I w/ 15 cycles 1 30 cycles) Page 3 of 6 Problem No. 2 (25 points) A 60 Hz generator is rated 2.8 MVA, 480 V (line-to-line), with X”d = 0.25 per unit. It is controlling its bus voltage to 1.0 per unit, and it is sewing a 2.0 MW load running at 0.9 lagging power factor. if the generator suddenly undergoes a three phase bolted fault at its terminals, find the magnitude of the initial symmetrical rms current in the generator. First find the answer in per-unit, then calculate the answer in real amperes, using the machine ratings for your per-unit bases. ,uJ ) .\_) T— x. Page 3 of 6 Problem No. 3 (25 points) Page 4 of 6 The following currents are measured at a feeder circuit breaker monitoring system: A phase = 150 amps, ang 90 deg B phase = 160 amps, ang 320 deg C phase = 170 amps, ang 220 deg Assume this is an A-B-C sequence system. Part A (20 points): Calculate the positive, negative, a and the zero sequence quantities for B phase and for each worth 4 points). f i r r -* - !. '27., , L ' " n‘ 3‘ a , 1 r. Part B (3 points): This circuit breaker serves the only nd zero sequence quantities for A phase, C phase (you will have five answers, feeder out of a substation with a transformer that has a wye-grounded secondary. How many amps will the feeder breaker ground relay see? (don't worry about CT ratio here...assume a CT ratio of 1:1, so the relay sees ) Another way to ask this question: What is the number of amps that will be flowing in the transformer secondary neutral? l 1 Part C (2 points): In generators, what type of sequence current is not produced by the generator but can have dama in heating effe_cts for the generator rotor? Circle one: (positive sequence current [\negative sequence—[email protected] zero sequence current) Page 4 of6 Page 5 of 6 Problem No. 4 (25 points) An isolated generator (not connected to the grid) is rated 419 kVA, 480V, and has X”d = 25%, X2 = 20%, and X0 = 5%. It has a wye~grounded I wye-grounded step-up transformer with an impedance of X = 10%. For each question below, assume a bolted fault at the transformer’s high voltage terminals, and assume that all faults are subtransient. Part A (4 points): Find the magnitude of a three phase fault in amperes per-unit. X247 , Xe , .t \ \ ' ‘-1 Part B (7 points): Assume the generator is wye-grounded. Draw the appropriate sequence network diagram used to find the magnitude of a single line-to-ground fault {with all impedances labeled), and calculate the per-unit value of the single-line-to-ground fault in amperes per-unit. Page 5 of 6 Page 6 of 6 Part C (7 points): Assume the generator is grounded with a 1 ohm neutral reactor. Draw the appropriate sequence network diagram used to find the magnitude of a single line~to-ground fault (with all impedances labeled), and calculate the per—unit value of the single-line-to- ground fault in amperes per-unit. .f- I g (a an? ’ ,3 ,T’i" iii ‘ ' x 1‘ ‘~\ ‘ i i f r ‘ r I bi rt l ‘5 J l c- : _ :r ’— - C ’3 /V . t _ 5X»! ".-r ‘ fir .: \ \ ’ l i l r——_ \ r ; A c a? i z D, f‘ _ r; ' I Part C (7 points): Assume the generator is ungrounded. Draw the appropriate sequence network diagram used to find the magnitude of a single line-to-ground fault (with all impedances labeled), and calculate the per-unit value of the single-line-to-ground fault in amperes peraunit. ._< ,‘J ‘I /,r _ ,1 A”; 7 ~—; p i 1 ~ - "e .__ ;\ 3,. 'jr, ‘f / / _. Kl J _/!/l-x I, git? - / N 7 J9 \A '2 ‘y' '1 r f I ._ 1 l , r44, , __ AM. . 5 j ' l E . x; {a , Page 6 of 6 ...
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Test_3_solutions - Page 1 0f 6 NORTH CAROLINA STATE...

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