电力系统习题答案

SACSZ?C'?BCA?SB?'Z?Z?AB38.3885?38.111??50.002??48.234??30.806?35.754??19.972??57.265??MV?A74.76??51.244??33.7824?38.3639??(26.49?j20.967)MV?A

SAB?SC?SB?SAC?(30.2046?j23.693?25?j18?26.49?j20.9672)MV?A?(28.7146?j20.7258)MV?A

''SBC?SAB?SB'?(28.7146?j20.7258?25?j18)MV?A?(3.7416?j2.7258)MV?A（c）精确功率分布计算

?SAC?P2AC?Q2C2ACV(RAC?jXAC)26.492?20.96722?(13.5?j21)MV?A?(1.321?j2.055)MV?A1082

SAC1?SAC??SAC?(26.49?j20.9672?1.321?j2.055)MV?A?(27.811?j23.0222)MV?A

?SBC?P2BC?Q2C2BCV(RBC?jXBC)3.74162?2.72582?(22.5?j20.5)MV?A2108?(0.04134?j0.03766)MV?A SBC1?SBC??SBC?[(3.7416?0.04134)?j(2.7258?0.03766)MV?A?(3.7829?j2.7635)MV?A

?VBC?(?PR?BCBC?QCBCXBCVPX?jBCBC?QCBCRBCV3.7416?22.5?2.7258?20.53.7416?20.5?2.7258?22.5?j)kV108108?(1.2969?j0.1423)kV22?(??)?(?)VBVCVBCVBC?(108?1.2969)2?(0.1423)2kV?109.297KV

SAB2?SBC1?SB'?(3.7829?25)?j(2.7635?18)MV?A?(28.7829?j20.7635)MV?A

?SAB?P2AB2?Q2B2AB2V(RAB?jXAB)28.78292?20.76352?(10.8?j16.8)MV?A2109.297?(1.1388?j1.7714)MV?A SAB1?SAB??SAB?[(28.7829?j20.7635)?(1.1388?j1.7714)]MV?A?(29.9217?j22.5349)MV?A

SCD1?SC?(30.2046?j23.693)MV?A

'SCD2?SLD?(30?j20)MV?A

?VAB?(?PR?AB2AB?QCAB2XABVPX?jAB2AB?QCABRABV（2）A，B，C，D各点电压

28.7829?10.8?20.7635?16.828.7829?16.8?20.7635?10.8?j)kV109.297109.297?(6.0357?j2.3725)KV

V或

22?(??)?(?)VVVABABAB?(109.297?6.0357)2?(2.3725)2kV?115.357KV

22?(??)?(?)VAVCVACVAC20.49?13.5?20.9672?21226.49?21?20.9672?13.52)?()kV?115.4159KV108108115.4159?115.357?10000?0.05115.357?109.297KV,VC?108KV误差 VB ?(108??VT??(?P'CRT?QVC'CXT?jP'CXT?QVC'CRT30.2046?1.513?23.693?31.76330.2046?31.763?23.693?1.513?j)kV108108?(7.3913?j8.5513)KV

V'D?(VC??VT)2?(?VT)2（3）功率分点。

从计算结果可以看出有功功率和无功功率的分点均在节点C 5-9 题图5-9示一多端直流系统，已知线路电阻和节点功率的标么值如下：R12=0.02, R23=0.04, R34=0.04, R14=0.01, S1=0.3, S2= －0.2,S3=0.15。节点4为平衡节点，V４＝1.0。试用牛顿－拉夫逊法作潮流计算。

?(108?7.3913)2?(8.5513)2kV?100.97KV

11'1??100.91?KV?10.1kvVDVD110kT

图5-9 多端直流系统

解 （1）形成节点电导矩阵，各元素计算如下

RG?G12G11?112?1R?11411??1500.020.01????

??21R123121??500.02

G14?G41??1R141??1000.01G22?23112RRG?G??32??111??750.020.04??1??250.04R13423

G33?34123RG?GR(0)?R?111??500.040.04??1??250.04??43?R3434G44??P11141R?11??1250.010.04

(0)i(0)2（2）按给定的节点功率，设节点电压初值

?P1S?V1(0)(G11V1(0)V?GV12?1.0，计算节点不平衡量

?G14V4)

??0.3?1?(150?1?50?1?100?1)??0.3?P2?P3(0)(0)?P2S?V2(0)(0)(G21V1(0)(0)?G22V2(0)(0)?G23V3)

(0) ?0.2?1?(?50?1?75?1?25?1)?0.2?P3S?V3(G32V2?G33V3?G34V4)

??0.15?1?(?25?1?50?1?25?1)??0.15（3）计算雅可比矩正元素，形成修正方程式

J（0）11???P1?V1|0??2G11V1?G12V2?G14V4

（0）（0）??2?150?1?50?1?100?1??150（0）??P1（0）???|J12?0G12V1?50?1?50V2

J（0）21???P2?V1|0??G21V2?50（0）（0）

（0）（0）J（0）22???P2?V2|0??2G21V1?2G22V2?G23V3?50?1?2?75?1?25?1??75（0）??P2（0）???|J23?0G23V2?25V3

J（0）32???P3?V2|0??G32V3?25（0）（0）

（0）J（0）33???P3?V3|0??G32V2?2G33V3?G34V4

?25?1?2?50?1?25?1??50所得修正方程式如下

（4）求解修正方程式。

（a）用三角分解法求解休整方程，先对-丁阵作Crout分解 c11?J11??150

(0)?u12J12/c11?50/(?150)??1/3

(0)??(0)?0??V1???0.3???15050(0)??????0.2???50?7525???V2???(0)?25?50????0???V??0.15??3??

J?50

u?(J?cu)/c?(25?0)/(?175/3)??3/7 c?0,c?J?25

3????50?25?(?)??275/7cJcu7

21u13?0,c21?(0)23233132(0)2113(0)3222(0)33333223（b）进行消元演算，将修正方程左端的负号移到右端并计入

?Pi中，即