簡單電力系統如圖16-8所示，各元件的參數及初始運行情況均按照例16-1和例16-2給定的條件。假定在輸電線路之一的始端發生了兩相接地短路，線路兩側開關經0.1s同時切除，試計算極限切除角5，并用分段計算法計算轉子搖擺曲線和極限切除時間r，判斷系c-limc?lim統能否保持暫態穩定性。圖1x；=0.25，x2=0.2，發電機：Sgn=352.5x；=0.25，x2=0.2，Tjn=8s。變壓器：T-1S^1=360MVA，Vsti%=14，%1=10.5/242；T-2S^2=360MVA，匕晝%=14，k‘2=220/110。線路：l=250km，Xl=0.41。/km，匕=220kV。線路：運行條件：V0=115kV，P0=250MW，cos中運行條件：V0=115kV，P0=250MW，cos中=0.9500解題思路：5=cos-1P)(5“-50)+PmIIIC0S5"-Pm.〃Cos5°C-limPm〃廣PmII(17-6)1.求對應三種情況的等值電路和等值參數XXX圖22.對應的功率特性，得到PmII和PmIII50噸3.求極限切除角4.分段法求5(t)，直到5=5c.lim解:解:X=xWd=0.2.迪x冬=0.19G22SgnV；())352.59.072XLo=5Xl=5x0.586=2.93由例16-1的計算已知原始運行參數及網絡的參數:(一)計算功角特性(1)正常運行時。專穹1。馬小7,80=80=(一)計算功角特性(1)正常運行時。在此情況下可作系統的等值電路，如圖2(a)所示。+X+XT2=XL769功角特性方程為(2)短路故障時:147x1一一sin8=1.912sin80.769EV工sin8Xi輸電線路始端短路時的負序和零序等值網絡的等值電抗分別為PiX2廣((2)短路故障時:147x1一一sin8=1.912sin80.769EV工sin8Xi輸電線路始端短路時的負序和零序等值網絡的等值電抗分別為PiX2廣(XG2+X附加電抗為0￡XA=X2￡〃X0￡得到短路時的等值網絡如圖2(b)，于是XII=Xd+XT1+2XL+XT2(0.19+0.13)x(0.293+0.108)=0178

0.19+0.13+0.293.0.108.0.13x(1.465+0.108)_0120.13+1.465+0.108=.°』2x0.178=0.0720.12+0.178MM”)+XA=0.238+0.13+0.293+0.108+(0.238+013)x心293+0108)=2.820.07211X[[2.82(3)故障切除后的系統等值電路如圖2(c)。X山=X'd+Xt1+Xl+Xt2=0.238+0.13+0.586+0.108=1.062功角特性方程為P一一*sin8='.47x'sin8=1.384sin8111X[h1.062(二)計算極限切除角8c-lim先求＜5cr8=兀一sin-1%=180一sin-1=133.74。世PmIII1.384按式(17-6)有3=cos-1P0(廣D0)"miffCg「Pmffeg0c-lim―mUfP一PmIffmff=cos-1n1.0x面(133.74。-31.54。)+1.384cos133.74。-0.52cos31.54。1.384-0.5263.64。(三)根據分段計算法求tc-lim發電機慣性時間常數T=T亳=8x竺25=11.28(s)JJSB250△t取為0.05s，K=MAt2=18000x0.052=3.99T11.28J第一個時間段AP0)=P0-P"sin80=1-0.52xsin31.54。=0.728A8⑴=2KAP(0)=2x3.99x0.728=1.45。8⑴=80+A8⑴=31.54+1.45=32.99。第二個時間段AP^=P-P"sin8⑴=1-0.52xsin32.990=0.717A8(2)=A8⑴+KAP⑴=1.45+3.99x0.717=4.31。8(2)=8⑴+A8(2)=32.99+4.31=37.3。|AP，(2)=P0-P-sin8mff(2)=1-0.52xsin37.3。=0.685AP"(2)=P0-P-sin8miff(2)=1-1.384xsin37.3。=0.16A8(3)=A8+K](AP，(2)2(2)+△%；))第三個時間段開始瞬間，故障被切除，故=4.31+3.99x2x(0.685+0.16)=6.0。繼續進行計算，按照時間與功角的對應關系，可以作出轉子搖擺曲線，如圖8-25所示。從搖擺曲線上可以找到與極限切除角8.對應的極限切除時間t=0.329s。因為c-limc-limtc<tc-lim，所以系統能保持暫態穩定。圖3轉子搖擺曲線非線性方程組迭代求解：高斯賽德爾法，牛頓-拉夫遜法高斯賽德爾法迭代格式為：x（k+i）=_L（y—ax（k）—ax（k））TOC\o"1-5"\h\za1122133iix（k+1）=——（y—ax（k+1）—ax（k））a221123322,—1，，，、x（k+1）=——（y—ax（k+1）—ax（k+1））a331123233若式中的與?對于與、與對應&，片對應YM=YM=則對于第汁節點:j=lu:j=lpi~JQs

例：如圖所示三相電力系統，節點2上發電機無功出力范圍為35Mvar到0Mvar。試用高斯-塞德爾法計算潮流（阻抗為標么值形式）。選取功率的基準值為100MVA。節點節點電壓發電機注入功率MWMvar負荷MWMvar11.05+J0.0？？0021.0320？50203？06025分析:由已知條件可知：節點1為平衡節點，節點2為PV節點，節點3為PQ節點。解：（1）形成節點導納矩陣>23=1/Z23=1.667-j5.0y=1/Z=1.25-j3.75y=1/Z=5-j15.01313Y=y12+y13-y12%-yy+y-y12122323-孔義3EJ-1.25+j3.752.9167-j8.75-1.6667+j5.06.25-J18.75-1.25+j3.73-5.0+j15.0-5.0+j15.0-1.6667+j5.06.6667-j20.0=1.03Z0U3=1.0Z0TOC\o"1-5"\h\zU=1.05Z0=1.03Z0U3=1.0Z0一-3*220)=Im(U2zY2jj=1=Im[1.03/0x(-1.25—j3.73)x1.05Z0oo+1.03Z0x(2.9167+j8.75)x1.03Z0oo+1.03Z0x(-1.6667-j5.0)x1.0Z0]oo=0.07766P=20一50=-0.321001P-jQ(0)、U⑴二(2jQ2-YU-YU(0)).2*(0)21?123?322U2_1[-0.3-j0.07766-TOC\o"1-5"\h\z2.9167—j8.75103(-1.25+j3.73)x1.05Z0-(-1.6667+j5.0)x1.0/0]oo_1.018/-1.79oU⑴=1.03Z-1.7921P—jQo、U⑴_(3jQ3-YU-YU⑴).3*(0)31-132?233U3_1[-0.6—j0.25-6.6667-j2010(-5.0+j15)x1.05/0-(-1.6667+j5.0)x1.03/-1.79]oo_1.048Z-2.12o牛頓-拉夫遜法習題3-11：三節點電力系統，所有節點電壓幅值為1.0,求潮流近似解，并將發電功率、負荷功率和線路功率標在系統圖上。此外：(Qg3=0),Vi=V2=1,確定各節點類型，并進行潮流計算，迭代2步。加2有功負荷無功負荷有功注入無功注入1——2"1106??——~/|3|1.0|1.0|0|0解答：一、確定節點類型：節點1為平衡節點；節點2為PV節點；節點3為PQ節點；二、潮流計算：⑴節點導納矩陣中每個非對角元素均為-壽=5/，每個主對角線元素均為-10/，所'-IO；5j5J_以該電力系統的節點導納矩陣為Y=G+jB=5j-10;5j5j5j-10;(2)第一次迭代計算選擇5⑴=0，各節點電壓初始值為1,計算各節點2、3有功功率和節點3無功功率3=x比x((?2icos$21+sin<J2i)+^xGjj+*x(G23cos6公+B公sin6公)]_=4x(Hxsin以+匕xsin323)g=*x[Kx(G31cos^,+月3】sin$3】)+匕xG33+匕x(G32cos8^+B32sin&)]=匕x(Ksin%]+V^2xB,2sin%?)Q3=匕x[《x(G31sin^,-B31cos%)-止x%+V2x(G32sinJ32-Bi2cosJ32)]=Kxcos^,,-Kx5?-cos%)功率失配量(計算值一給定值)：注意：若計算功率失配量=給定值一計算值，則J—1.4pl1.401.4一1.0■-1.0—0-L0[-1.0-1.00[-L0雅可比矩陣：一明明明明墊oQ.明V七&1cos%+匕匕&3cos以-V2V3cos腿*B?sin323一匕匕B，3cos如3匕匕Lcos鳥]+匕匕B，3cos&匕sin、+V2B32sin3^一*七&32sin&2匕g831sin以+*^832sin肉2-^^31cos^j-2J^533-^532cos^210-50-51000010各變量修正量計算'?-10-50一■-0.12-由-5100場得0.04X0一00io]0.1_修正量二'0_■-0.12-■0.12_故—0—0.04——0.0410.1_0.9第一次迭代值：——~0~0+-0.12--0.04——0.12-0.04_1__-0.1_0.9備節點電壓為％=1，^=1，=0.9,功=0,S2=0J2,&=—0°4(3)第二次迭代計算(.同上)計算各節點2、3有功功率和節點3無功功率4=匕x比x(G21cos站+珥1sin瓦i)+《xG22+匕乂爐君co$&+^23sin<3^)]=匕乂sin%+匕又夕為sin^n)13155R=匕xBx(G31cos再i+鳥]sin%)+*xG33+匕x(0,cos%+電sin&筮)]=WM*%sin鳥]+匕x馬?sin%)-0.8969Q3-V2x[^x(G3Isin為一電(:崩如)一匕x旦3+匕x(G32sinS拿一為cos晶)]=匕x(—咒尻1COS&I-gX&3—八夕32COS<5^2)1.4-1.0△00-1.00.8389功率失配量（給定值一計算值）、-0.08450.10310.1611雅可比矩陣:明832四832況ds.明

就

明明9.4066-4.44250.7169-4.44258.9389-0.89690.7966-0.99658.0679-或?明88.dP1dS2明，

豹5rr3。匕9.4066444250.7169-4.4425071698.9389-0.8969*0.89697.2611各變伺修正竟計算■9.4066-4.44250.79661-0.0055-由=444258.9389-0.9965得=0.0112△00一7169-0.89698.0679JK.0.0217_-9.4066-4.44250.71691r叫_場-0.0055-或由-4.44258.9389-0.8969AS,得=0.0112_Afi_0.7169-0.89697.2611J化一△火_0.0241%-0.12"■-0.0055--0.1255_~0.12--0.0055^3=-0.04—0.0112———0.0512或=-0.04—0.0112A-°?9一0.0217_0.8783_0.90.0241x0.90.1255-0.05120.8783備節點電壓為匕=耳,V2=\.匕=0.8783,少=0,如二01255,&=—0.0512一、計算發電功率、負荷功率和線路功率節點1的總的有功、無功注入功率，以及節點2的總的無功注入功率：*=Kx[Kx(GL2cos+%sinS「)十KxGu+Kx(G”cos功③+晶sin^J]=Kx(K.xsin島十W*B】3sin6i3)=5xsin(-0.1255)+0.8783x5xsin(0.0512)=-0.4012Q.=Kx[Kx(G?&in(J17一晶cos%)-、xBV]+Kx(Gsin^?-B..cos<5,)1■*^■11JlkZvIZ1Z1Z1ZzJ11JvJJ.BJ1JzJ=Vxx(—cos<5]2—、——句3cos%)=0.6536Q2-Kx[I^x(G2]sin如]—B2lcos8^-V2^B12+*x(G23sin為—B23cosS2i)]=V2x(―咒B2icosSr-當*-匕均3cos曷③)=0.7163由節點1負荷為1.0+0.6j,可知節點1發電功率為-0.4012+0.6536j十l.0+0.6j=0.59X8+1.2536j由節點2無功負荷為0,可知節點2無功發電功率為0.7163各線路功率：S12=V11\2=Ki匕]Mz=(1—IN—0.1255)x5j=-0.6259+0.0393i私-KiZb二討山―-(1-0.8783Z0.0512)x5;=0.2247+0.61431§23=V2113或厄-六也=1^0.1255(1Z-0.1255-0.8783Z0.0512)x5;=0.7719十0.6769iS2I=V2Z21=1X0.1255(12—0.1255—1)x57=0.6259+0.0393i?*■*■申8u=FaZ31=尸，［*一人03］=08783N-0,0512(0,87834),0512-l)x5j=-0,22470.5287i????+■+&=r3hi=皿［匕一*］乃1=0.87832：—0.0512(0.8783/0.0512—1匕一0.1255)x5/-0.7719-0.466li復雜電網短路電流計算線性方程組求解任一網絡用節點阻抗矩陣表示的節點電壓方程為U“1U2U34氣U“1U2U34氣^7"8U91-10」Z11Z14Z...14Z22Z33Z44Z55Z66Z77Z88Z99Z00I1-10」,11則任一點三相短路時的電流I*=—=——，其中Zf是短路支路阻抗，取決于故障回路，如果是金屬性短路，fZ+ZZ+Z故障回路，如果是金屬性短路，■ZZ...-1MT:1.TAU-——_2AU；AU%4...JAU:5?AU:6」AU:7"AU8iAU；-9...J-11Z14...Z22...Z3314Z44Z55Z66Z77Z88Z99「Z-00.0;0.，0"「二二y-1*；?…f■■0?10i+0:;0:;00.7690；其值為0；則則AUn是各個結點故障狀態下疊加的電壓量，故障后的電壓應為j，其中zij為ij兩結點U=U+AU=U-1Z，則任一支路電流I=AUi~AUnnl0lnnl0lfj，其中zij為ij兩結點阻抗矩陣:j0.0935j0.0210j0.0177j0.0384j0.0497j0.0747j0.0654j0.0642j0.0712j0.0210j0.0641j0.0324j0.0417j0.0374j0.0281j0.0316j0.032j0.0294j0.0177j0.0324j0.0505j0.035j0.0314j0.0236j0.0265j0.0269j0.0247j0.0384j0.0417j0.035j0.0759j0.0682j0.0512j0.0575j0.0584j0.0536j0.0497j0.0374j0.0314j0.0682j0.0883j0.0663j0.0744j0.0755j0.0694j0.0747j0.0281j0.0236j0.0512j0.0663j0.0996j0.0873j0.0856j0.0949j0.0654j0.0316j0.0265j0.0575j0.0744j0.0873j0.1354j0.1134j0.1104j0.0642j0.032j0.0269j0.0584j0.0755j0.0856j0.1134j0.1603j0.1104j0.0712j0.0294j0.0247j0.0536j0.0694j0.0949j0.096j0.1104j0.1667因子表的標準形式為:j0.03750000-0.750000j0.0375-0.375-0.375000000j10j0.0338-0.465000000j10j13.75j0.0232-0.7720000000j33.333j0.0385-0.3855-0.193-0.12850j20000j10j0.0321-0.383-0.0413-0.3210000j5j11.927j0.0514-0.5719-0.19670000j3.333j1.285j11.34j0.087-0.661900000j10j3.829j7.6023j0.1665對U求解的回代運算:TOC\o"1-5"\h\zU=19=j0.16661+j0.1103I+j0.09591+j0.09481+j0.06921+j0.0534199987654+j0.0248I+j0.0293I+j0.0711IU=18-Y（8）U=j0.1103I+j0.1601I+j0.1133I+j0.08551+j0.0753I8889998765+j0.05821+j0.0271+j0.03191+j0.06121U=17-Y⑺U-Y⑺U=j0.09591+j0.1133I+j0.1351I+j0.08721777997889876+j0.07421+j0.0573I+j0.02661+j0.03141+j0.0651U=16-Y（6）U-Y（6）U-Y（6）U=j0.0948I+j0.08541+j0.0872166699688677987+j0.0995I+j0.0661I+j0.0511I+j0.02371+j0.0281+j0.0743I654321U=15-Y（5）U-Y（5）U-Y（5）U-Y（5）U=j0.0692I+j0.0753I+j0.0742155599588577566987+j0.0661I+j0.0881I+j0.06791+j0.03151+j0.03721+j0.049154321U=14-Y⑷U-Y⑷U-Y（4）U-Y（4）U-Y（4）U=j0.05341+j0.0581I4449948847746645598+j0.0573I+j0.051I+j0.068I+j0.07561+j0.0351I+j0.04151+j0.03781654321U=13-Y⑶U-Y⑶U-Y⑶U-Y⑶U-Y⑶U-Y⑶U=j0.0248I333993883773663553449+j0.0271+j0.02661+j0.02371+j0.03161+j0.0351I+j0.0501I+j0.0321+j0.01761U=12-Y⑵U-Y⑵U-Y⑵U-Y（2）U-Y⑵U-Y⑵U-Y⑵U=j0.0293I222992882772662552442339+j0.03191+j0.03151+j0.0281+j0.03741+j0.041