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电力系统三大计算例题
22页1、简单电力系统如图16-8所示,各元件的参数及初始运行情况均按照例16-1和例16-2给定的条件。假定在输电线路之一的始端发生了两相接地短路,线路两侧开关经0.1s同时切除,试计算极限切除角&时,并用分段计算法计算转子摇摆曲线和极限切除时间tcim ,判断系统能否保持暂态稳定性。G Jl T-2】 I 11 Tin K C055n发电机:Sn=352.5MVA Pgn=300MWVgn=10.5KV , Xd=1.0 ,xq=0.6 , xd = 0.25, x2 = 0.2 ,Tjn =8s。变压器:T -1 Stni =360MVA , Vsti% =14 , kT1 =10.5/242 ;T -2 STN2 =360MVA , VST2%=14, kT2 =220/110。线路:l = 250km , xL =0.41C/km, vn =220kV。运行条件:V=115kV, Po=250MW, cos%=0.95。解题思路:_cc-1 /( cr -; 0)PmIII cos cr 一 PmII cos 0c lim(17-6)cosPmIII PmII1.求对应三种情况的等值电
2、路和等值参数X Xi Xiit 涵 谒 吐莅 诉(8)j 用jrijXu0js(b)j 玉j*TLj*Ljn图22 .对应的功率特性,得到PmII和PnIII跖和6cr3 .求极限切除角4 .分段法求6( t ),直到6=%.lim解:SB VGNG2 - x2 -2-SGN VB(I)25010.52= 0.2 =0.19352.5 9.072XLo =5Xl =5 0.586 =2.93由例16-1的计算已知原始运行参数及网络的参数:Pt=P0=1.0, Eg =1.47, c.o =、.0 =31.54(一)计算功角特性(1)正常运行时。在此情况下可作系统的等值电路,XlX1 -Xd Xt1 Xt22如图2(a)所示。=Xd 7 =0.769功角特性方程为PiE0V 1.47 1sin c. =sin、. =1.912sin、.X 0.769(2)短路故障时:输电线路始端短路时的负序和零序等值网络的等值电抗分别为X2: =(XG2 Xt1) XL XT2一2(0.19 0.13)0.293 0.108=0.1780.19 0.13 0.293.0.108附加电抗为X 07 = X
3、T1 2X L0 XT2_ 0.13 (1.465 0.108) _ o 120.13 1.465 0.1080.12 0.178X . =X2:X0; = 0.072一 一 0.12 0.178得到短路时的等值网络如图2(b),于(Xd Xt1 ) X L Xt212Xii =XdXt1Xl Xt2 2X A(0238 013) (0.293 0108)= 0.238 0.13 0.293 0.108 - 二2820.072故障时的功角特性为EgV1.47 1Pii =sinsin c =0.52sin、X ii2.82(3)故障切除后的系统等值电路如图2(c)。XIII =Xd XT1 XL XT2 =0.238 0.13 0.586 0.108 =1.062功角特性方程为E0V .1.47 1 .Pm =sin、. =sin、. = 1.384sin、.Xm1.062(二)计算极限切除角dclim-1 P1 1.0先求 6 cr&r =冗-sin =180 sin=133.74Pmiii1.384按式(17-6)有aJ_ P0( Bcr Bo) + PmI11cos Bcr Pm
4、II cos 0Sc lim 二 CoSPm III -PmII1.0x (133.7431.54 j +1.384cos133.74 二0.52cos31.54 二 -1180= cos =63.641.384-0.52I(三)根据分段计算法求tclim发电机惯性时间常数TJ =Tjn SN =8 竺竺=11.28 (s)J JN SB250 t 取为 0.05s,K =犯&2 =18000 X0.052 =3.99 TJ11.28第一个时间段Pg-PmII sin、。=1-0.52 sin31.54 : 0.7 2811(1)=-KP(0)=- 3.99 0.728 =1.45 22、.=c.0 J.=31.54 1.45=32.99第二个时间段沪=P -PmII sin、=1-0.52 sin32.990 =0.717、.(2),:、(1) K =1.45 3.99 0.717=4.31、.=、.(1).:c.(2)=32.99 4.31 =37.3 I、C第三个时间段开始瞬间,故障被切除,故P(2)=P。PmII sin、(2)=1-0.52 sin37.3、=0.685P(2
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