Chap_11_hmk_prob

Chap_11_hmk_prob - 202 (11.3) Determine the symmetrical...

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Unformatted text preview: 202 (11.3) Determine the symmetrical components of the three currents Ia= 1042 , Ib =10fim, and Ic = 1011.10: A. Solution: l“=;—(1040: + 10135031013102) = 31,110 +9.848 ~j1.736 +9.848 +J'1.736) = 9399/0; A 112l=§<10/_0: +10a410: +10a501) = %(10 - 3.420 +0397 ~ 3.420 #9397) = 1053401 A Agogéqom‘; + 101230“ + 10113.02) = 3—00 -6.428 -f7.66 -6.428 +1766) =0.952/180° A The components of Ib and Ic are easily found from )2”, If) and 120). Check: I3 = 9.899 + 1.053 - 0.952 = 10.00. (11.4) The currents flowing in the lines toward a balanced load connected in A are Ia = 100g); , Ib = 141.4111ng , and Ic = 100%. Find the symmetrical components of the given line currents and draw phasor diagrams of the positive— and negative- sequence line and phase currents. What is 131-, in amperes? 203 Solution: The phasor diagrams for positive- and negative-sequence currents in the lines and in the A—connected load show the desired relations, namely: a a '3 '3 For the given currents, we find 1311):? 100 + 14141335" + 100131312) 2 107.7 —}28.9 = 11155.5: A 1;”: .134 100 + 141.4Lm51+ 100121;?) = —7.73 +1283 = 299/105" A 11% :4 100 — 100 4100 +j100) = 0 (since zero-sequence cannot flow into the A). and, 1 Iib’ = 1%5 /—15° +30° = 64.4/15" = 62.2 +j16.66 1;?) = 23:9 4193’ ao" = 1726/75" = 4.47 +jl6.67 3 13;, = 66.67 +1333?) = 74.5/26.6° A (1 1.12) A Y-connected synchronous generator has sequence reactances X0 = 0.09, X1 -—- 0.22, and X2 = 0.36, all in per unit. The neutral point of the machine is grounded through a reactance of 0.09 per unit. The machine is running on no load with rated terminal voltage when it suffers and unbalanced fault. The fault currents out of the machine are L1 = 0, L5 = 3.‘l5£,,1_fl2’a_a , and Ic = 3.751311 , all in per unit with respect to phase a line-to-neutral voltage. Determine (a) The terminal voltages in each phase of the machine with respect to ground, (b) The voltage of the neutral point of the machine with respect to ground, and (c) The nature (type) of the fault from the results of part (a). Solution: (5:) Z; = jO.22 p.u., 22 = 0.36 p.u., 2'0 = Zgo + 32” =j0.09 + 3xj0.09 = 0.36 p.u. I3 = 0,11; = 3.7540: p.u., and LE = 3.75,_/_Qip.u. 1(0) 1 1 1 0 11.25 3.75513? = 12.5 3.75/30° 11.25 .30) = 49% ~_— 41.25 x j0.36 = 0.45 p.u. a”) -_- Em*§1)21= 1101—0125 x 10.22) = 0.45 p.u. v9) = 4922 = 41.25 x j0.36 = 0.45 p.u. a and, V3 1 1 1 0.45m: 1.35m: Vb = 1 32 a 045$ = 0 p.u. Vc 1 a 32 DAB/g 0 (b) VI] = —3I§°’ x 10.09 p.u. = —3 x 11.25 X p.09 p.u. = 0.3375 p.u. (:3) since Vb = V; = 0, it is a double~line-to-ground fault. (1 1.13) Solve Prob. 11.12 if the fault currents in per unit are I; = 0, I}; = -2.986LQE , and Ic = 2.986.LO_a . Solution: (a) 13(0) 1 1 1 0 0 13(1) =13: 1 a az -2.986 = -j1.724 13(2) 1 22 a 2.986 11.724 £0] 2 _I;0}ZO = 0 am = Em 4921 = 11.0: -(-;‘1.724)(10.22) =0.621p.u. v.52) = 4922 = -—(-jl.724)(j0.36) =0.621p.u- Va 1 1 1 0 124ng Vb = 1 32 a 0.621 = —0.621fl 13-11. Vc 1 a 32 0.621 —0.621[_Q: (b) Since 1‘30) 3 0, V” = 0. (c) Since Vb = Vc, it is a line—to-line fault. ...
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Chap_11_hmk_prob - 202 (11.3) Determine the symmetrical...

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