1、电力系统合环电流的分析及控制对策s g 王伟灿 周昱甬K1ss u H, F, V ?/ V !K-, ? T,Y abYV110 kVsM 9 s,4 e +n , L=Bilb1oMs u; ; en ms |:TM71DS M :BcI|:1006-6357(2002)04-0026-031sXy500 kV + D M wg,13V220 kVM 1v,s g uaE( g) 61V110 kVM b u “d 5v1 p,s220kVM ,ss u,rE )HarK e “bu?v?,s u H oB“d/aM a20% =p Hqb LlV , TI n Hq b/ 110kVsM
2、110kV ,s vlY# V |n bm1“d 2“dL 2. 1“dLm1 U, 220 kV 2421La:2422L220 kV :Mi, :Ms1197L 1181Ls sM110 kVa bY, g ,119 kVP,7 :M113kV /byN,sM110a W6kVPb 220kVL :Mi B“d, op?1 pb2.2a | i :M“d B L H s,X m1(:MW,Mvar)bsM110 kV L:Uc= 111.9- 6.21,S Uc* = 0.973- 6.21;Uc= 121.40.69,S Uc* = 1.0560.69b9 M UM500kV Lv, |
3、1 L H 9 b2619 4 2002 M839 s3. 1 p # = q , pqE F(9 V ),e , m2 U,E FS b |SB= 100MW,UB=115kVbpE F: Z= 6ni= 1z i* = z1+ z2+ z3+ z4+z5 + z6 + z 7 = 0.00888+ j0.12045= 0.1207885.8m2 # sS1= 66.53- j21.3S1= - (75.63- j12.33)S2= 53.6- j19.3S3= 89.0- j19.9S4= 11.1- j5.3S5= 116.4- j27.3S6= - (195.94- j60.58)3.
4、 29 qV110 kVsMa LvlM,110 kV -110kV, H _ b$D=0.69+ 6.21= 6.9,y$Dl,#T 9 :$U*1.056- 0.973= 0.083,MA=90- 6.92 = 86.55b5$Ic* = $Uc*Zc*=0.08386.550.1207885.8= 0.68720.75,$I= $Ic*S B3cUB103= 0.68721003c115103= 345.01 AS$Sc* = $Ic* = 0.6872, q68.7MWb3.39 ss“ E, p 1181L,S12b | :220 kV L,m2$ , Z_ m2, o+p, o-p
5、bX m R, R V bps“ Ci:C1 = 0.00731+ 0.01339+ 0.030960.12045= 0.4289,C1= - (1- 0.4289)= - 0.5711 ,C2= 0.044350.12045= 0.3682,C3= C4= 0.030960.12045= 0.2570C5= 0.118950.12045= 0.9875,C5= - 0.0125,C6= 0.077840.12045= 0.6462,C6= - 0.3538S12 = 6ni= 1CiS i = (S1 + S1) C1+ S2 C2 +S3 C3 + S4 C4+ S5C5+ S6 C6 =
6、 118.52-j29.55272002 M4 Ic12= S123cUB103= 118.52- j29.553c115103= 595.04- j148.4= 613.314A3. 4YV1181L q qs I n1 qs q:S1181= S12+ $S= 118.52+ 68.7=187.22 MWI1181= I12+ $I= 613.3+ 345.01= 958.3A1181L K-680A,-Kv q135.4MWb W qV-b|Y“d, H V ? byN, L= = , 1b4h +n V - s9 V AS0 = S+ S,S= 6ni= 1CiSi:S ,Cis“ ,
7、Si,A ,Y = vly :a. M;b.s“;c.vlb1, y F a e, Vr e = rTb4. 1M“d, e B“dBZ Ta ,WM MbyN, e m 1 e b4sM110 kV , VYVM :MMsh9F :M (g )bsM110 kV , VYVh ? ,V7110 kV L, P $Ub :M ,|MshVB=, 110 kV L V 6114.56 kVb9 HqM f /, V p:$I= 248.7 A,5I1181= 248.7+ 613.3=862A - 958.3 AbA ,e evTb4.2M1 sss“ 1,7s“ | % E FvlbM“d E
8、 F VrMs“ “b7h =, V P1 s qb110 kVsM -, TMZ T,|220 kV :M110 kVa35 kV Ls ,s :M110 kV L#35 kVLbYVs“ E9 p:S12 = 98.36- j25.09 MW, I12 = 493.8-j126.0= 509.614.31 1181L 1 s q1Z T-Ab H, :Ma =E F9v,m2 = U,V p:Zc* = 0.09968+ j0.15140= 0.1517286.3,$Ic* = 0.08386.550.1517286.3= 0.54710.25 :$I= 274.7 A,$S54.7MW5YV1181L , I1181= I12+ $I=509.6+ 274.7= 784.3 AZ T - 958.3AbVn :Ms , 9 v/,r e “b5 s u T H,1 ?:(1)B“d,(2)MM,(3)20% =b H I n = Mb ? ! K-? 1p H, VYVa#M ZE, e s, 1 pbX110kVsM L= T bl :2001 M1215 s g s gyS 397|3100092819 4 2002 M8
