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# homework4solutions - Homework 4 Solutions 3.4 V £3 = 23(er...

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Unformatted text preview: Homework 4 Solutions 3.4 V £3 = 23(er v 24002240 N 2400 E =—'E,=——— 230 =23oov (a) ' N3‘ 240( ) — ___ _ — . l 3 _l D (b) S}: l§‘[2=[i) {30X Ozcos 08 El 23040° J = 347.84 - 36.87° _ El 230400 2 ..~__—_ 12 347.84 — 36.87° = 0.529 + j0.397 Q = 0661343637" 9 (c) 22(5) 17-2 =|00722 =66.l3z36.87°§2 (d) PI =P2 =80(0.8)=64kW Q, =Q2 =64tan(36.87°) =43 kVAR 3.10 Rated current magnitude on the 66-kV side is given by II = '5'000 =227.3A 66 I,2RH1,,=(227.3)2 RM, =100x10-‘ .-.Rw,=1.94o <— _Z_ :55le '4' 227.3 Then Xw,=,/—afl,—R3q, = (24.2)3—(194)2 =24.12\$2 <— = 24.2 Q 3.14 ‘ 340(0" V 24(X):340 5L 3 ‘21) 72= V z—cos-'(P.F.)=50x‘0 240 A—cos" (0.8) = 208.34—36.87°A v.) .__NZ_ , -———l =i(208.34—36.87°)=20.83z—36.87°A NI 10 N h) I = ._I 2 =10(24040°)= 240010°v N2 I = E] +<Rygl +jXrgl)l_l < , =240040°+(1+j2.5)(20.83z—36.87°) = 2400+ 56.095431.329° =2447.9+ j29.l66 = 2448.40.683°V f ‘3 3W3 4 V VS = El +(Rfrt’d +jxfeed + R rql + jqul = 240040° + (2.0 + j4.5)(20.83z — 36.87°) = 2400 +102.59429.168° . “ =2489.6+ 150.00 = 2490.41.1505ov N; l: UAW \ \/ (c) SS = VJ; =(249041.1505°)(20.832 36.87°) = m! was? as" PS : Re(§s) .37kw _ 7 delivered to the sending end of feeder. Q5 : [m ) 3.19 x 0300 x 10-‘1’1; 3.01103 x 10" PU i: p. = 1114-3037 11,7351 x 103111,I * [2.1701 x 10'3 17w, 1‘" P. z 1040" Zone I Zone 2 Shaw, = 50 kVA V5”: = 240V Vle = 2400V 4W, = (2400f/50x10‘ = 1 152 Q (a) VW=1.040°+(8.6803x10-3 + j2.l701x 10-2 )(1 .04 — 36.87°) = 1.0 +0.023373431.33° = 1.01997 + j0.0l 2157 = 1 .02040.683° pu 17] = 17 VW = (1 .02040.683°)(2400) = 2448.40.683° v | pu (b) 17 = 1.040°+(1.7361x10-2 +j3.9063x10‘2 )(1.04— 36.87°) um = l.0+0.042747429. 168° = 1.03733 + [0.020833 = 1.037 SA] .1505°pu vs = 141mm“ = (1 .037541.1505°)(2400) = 2490.41.1505°v ((2) PW + jQ,,,.. = 1;," =(1.037541.1505°)(1.0436.87°) =1.0375438.02° = 0.8173 + j0.6390 per unit P5 = (0.8173)(50) = 40.87 kW d3]. In d Qs = (0.6390)(50) = 3195 kVARS} L lure 3.23 Select a common base of lOOMVA and 22kV on the generator side; Base voltage at bus I is 22kV; this ﬁxes-the voltage bases for the remaining buses in accordance with the transformer turns ratios. Using Eq. 3.3.1 1, per-unit reactances on the selected base are given by G:X=0.18(ﬂ]=0.2;T, : X=O.l(@]=0.2 9O 50 T, :X=0.O6(ﬂj=0.15:7'1 :X=0.06 ﬂ =0.15 ‘ 40 ‘ 4O T1:X=O.O64[ﬂ)=0.l6;ﬁ:X=0.08[@]=0.2 ‘ 40 40 M : X=O.185(ﬂ)(l—O'£] =0.25 66.5 ll . (220)1 48.4 Foere 1. {BASE = 100 :4840 and X=4—84—=0.l 110 3 For Line 2, EMF =(—) =l2l§2 and X=§£=054 ’ 100 121 The load complex power at 0.6 Lagging pfis Ema, = 57453.13°MVA _ 10.45 I 2 The load impedance in OHMS is -Z-L = 57: 53130 = g’ ‘ T ‘ ' ‘ Lam = l.l495+jl.53267 Q The base impedance for the load is (l l)2/l00=l.2l Q l.l495+jl.53267 Load Impedance in pu = I 2] = 0.95 + jl .2667 The per-unit equivalent circuit is shown below: 1%“ 4 g 10.2 N! 10.15 7 [0. l6 [9.54 _/'0.2 W FW— J1m- .__._AT.T # 10.2 n95 7L... IL 4. jl.‘_’667 Per-unit impedance diagram Losses are minimum at 0 degrees = 16.606 MW (there can be a +/- 0.] variation in this value values of the power ﬂow solution tolerance). MW Losses 30 --o-- MW Losses 71!) ...
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## This note was uploaded on 11/17/2011 for the course ECE 476 taught by Professor Overbye,t during the Fall '08 term at University of Illinois, Urbana Champaign.

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homework4solutions - Homework 4 Solutions 3.4 V £3 = 23(er...

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