電力網(wǎng)潮流電壓計(jì)算例題與matlab程序

1、9電力網(wǎng)潮流電壓計(jì)算例題與MATLAB程序 編寫 佘名寰 本文介紹了兩個(gè)電力網(wǎng)潮流計(jì)算例題。一例為5個(gè)節(jié)點(diǎn)5條支路；另一例有6個(gè)節(jié)點(diǎn)7條支路，有PQ負(fù)荷節(jié)點(diǎn)也有PV發(fā)電機(jī)節(jié)點(diǎn)，變壓器支路標(biāo)么變比不為1。本文給出了完整的計(jì)算程序和計(jì)算結(jié)果。程序包括計(jì)算網(wǎng)絡(luò)節(jié)點(diǎn)導(dǎo)納矩陣和潮流電壓兩部分，程序中引用的M函數(shù)見(jiàn)參考文獻(xiàn)1。例題選自研究生教材，比較典型，計(jì)算結(jié)果與教材核對(duì)基本一致。本文可供電力專業(yè)師生和基層技術(shù)人員潮流計(jì)算時(shí)參考?！纠?.3】如圖 1-4所示，五個(gè)節(jié)點(diǎn)三條母線簡(jiǎn)單系統(tǒng)，系統(tǒng)參數(shù)均用標(biāo)么值（pu.）表示。節(jié)點(diǎn)4、5通過(guò)變壓器與節(jié)點(diǎn)2、3相連，變比為1.05:1，節(jié)點(diǎn)1、2、3為PQ負(fù)荷節(jié)點(diǎn)

2、，節(jié)點(diǎn)4為PV發(fā)電機(jī)節(jié)點(diǎn)，節(jié)點(diǎn)5為參考節(jié)點(diǎn)，節(jié)點(diǎn)4、5電壓均為1.05，圖中箭頭表示各支路的正方向，各個(gè)節(jié)點(diǎn)功率如表2.1所示,計(jì)算各節(jié)點(diǎn)電壓、發(fā)電機(jī)無(wú)功功率和系統(tǒng)潮流分布：4Z23Z13Z21YC2YC3YC1Z42Z531:1.051.05:142315S3S2S1Z21=0.04+J0.25Z23=0.08+J0.30Z13=0.1+J0.35Z42=J0.015Z53=J0.03YC1=J0.25YC2=J0.50YC3=J0.25S1=1.6+J0.8S2=2+J1S3=3.7+J1.3P4=5.0V4=1.05V5=1.05P4=5.0V4=1.05vV5=1.05 圖1-4 例1

3、.3網(wǎng)絡(luò)接線圖表 1-1母線有功負(fù)載無(wú)功負(fù)載發(fā)電機(jī)有功發(fā)電機(jī)無(wú)功1PD1=1.6QD1=0.82PD2=2.0QD2=1.03PD3=3.7QD3=1.345PG4=5.0PG5=？QG4=未定QG5=？【例1.3】 程序說(shuō)明： 本例題與參考文獻(xiàn)2的【例2.3.1】不同之處在于有兩個(gè)變比不為1的變壓器支路，并且引入了PV發(fā)電機(jī)節(jié)點(diǎn)。由于網(wǎng)絡(luò)結(jié)構(gòu)與例圖1-1完全相同，網(wǎng)絡(luò)節(jié)點(diǎn)導(dǎo)納矩陣已在文獻(xiàn)1之1.3節(jié)求出不再贅述。潮流計(jì)算程序參考【例2.3.1】節(jié)點(diǎn)電壓符號(hào)矩陣：一共五個(gè)節(jié)點(diǎn)，獨(dú)立節(jié)點(diǎn)數(shù)N1=4，PV節(jié)點(diǎn)數(shù)N2=1 節(jié)點(diǎn)電壓幅值符號(hào)矩陣為 u=sym(u1,u2,u3,u4,u5); 節(jié)點(diǎn)電壓

4、相角符號(hào)矩陣為 delt=sym(d1,d2,d3,d4,d5);節(jié)點(diǎn)功率給定值為：三個(gè)PQ節(jié)點(diǎn)，一個(gè)PV 節(jié)點(diǎn)p=-1.6,-2,-3.7,5;q=-0.8,-1.0,-1.3;【例1.3】 源程序 NU114N.m%* NU114N.m , example 2-3-2, fig 2-2 * % %The following Program for load flow calculation is based on MATLAB2007clear% bus 1,2,3 is PQ bus,bus 4 is PV bus,bus 5 is slack busglobal Np Nb bt p

5、m k% Np is number of node point,Nb is number of braches,% p the transformer turns ratio,with off-nominal tap-setting,fig 2-4 ,PSCLF p6, % bt is two-dimensional array ,line one is point 1,line two is point 2,% m is number of node point, k is number of transformers Np=5;Nb=5; bt=2,3;4,5;p=1/1.05,1/1.0

6、5;m=5;k=2;% nstart-the start point of branches ,nend - the end point,% mm - network incidence matrix nstart=2,2,1,4,5; nend=1,3,3,2,3; mm=ffm(nstart,nend);% zb1,the series impedances of transmission line % yb1,the series admittances of transmission line zb1r=0.04,0.08,0.1,0.0,0.0; zb1i=0.25,0.30,0.3

7、5,0.015,0.03; zb1=zb1r+zb1i*j; yb1=zb1.(-1); yb=diag(conj(yb1); y=mm*yb*(mm); % yb0 ,the shunt admittances of transmission line yb0i=0.25,0.25,0.0,0.0,0.0; yb0=0+yb0i*j; y0=diag(conj(yb0); yg=mm*y0*(mm); yn=diag(diag(yg); yy=yn-yg; y=y+yn; % yn0,the node-admittance of shunt capacitor %yn0i=; %yn0=0+

8、yn0i*j; %ync=diag(conj(yn0); %y=y+ync; Y,YY=fdt1(y,yy); G=real(Y);B=imag(Y);% YY is the shund admittances of the line and transformers at each end,% Y is node-admittance matrix for network%*u=sym(u1,u2,u3,u4,u5);delt=sym(d1,d2,d3,d4,d5);% u - node voltage magnitude,delt - anglep=-1.6,-2,-3.7,5;q=-0.

9、8,-1.0,-1.3;k=0;precision=1;N1=4;%the N1 is the amount of the PQ and PV busN2=1;%the N2 is the amount of the PV bus for m=1:N1 for n=1:N1+1 pt(n)=u(m)*u(n)*(G(m,n)*cos(delt(m)-delt(n)+B(m,n)*sin(delt(m)-delt(n); end pp(m)=p(m)-sum(pt); end for m=1:N1-N2 for n=1:N1+1 qt(n)=u(m)*u(n)*(G(m,n)*sin(delt(

10、m)-delt(n)-B(m,n)*cos(delt(m)-delt(n); end qq(m)=q(m)-sum(qt); end J1=jacobian(pp,d1,d2,d3,d4,u1,u2,u3); J2=jacobian(qq,d1,d2,d3,d4,u1,u2,u3); J=vertcat(J1,J2); uu=1.0,1.0,1.0,1.05,1.05;dd=0,0,0,0,0; while precision0.00001 u1=uu(1);u2=uu(2);u3=uu(3);u4=uu(4);u5=uu(5); d1=dd(1);d2=dd(2);d3=dd(3);d4=d

11、d(4);d5=dd(5); for m=1:N1 PP(m)=eval(pp(m); end for m=1:N1-N2 PP(N1+m)=eval(qq(m); end JJ=eval(J); du=-inv(JJ)*PP;precision=max(abs(du); for n=1:N1 dd(n)=dd(n)+du(n); endfor n=1:N1-N2 uu(n)=uu(n)+du(N1+n); end k=k+1;endk-1,dd*180/pi,uu%*% the following program is used to calculate the Sm and Smnfor

12、n=1:N1+1 U(n)=uu(n)*(cos(dd(n)+j*sin(dd(n);end Um=conj(U); I=Y*Um;Sm=diag(Um)*conj(I)for m=1:N1+1 for n=1:N1+1 Smn(m,n)=U(m)*(conj(U(m)-conj(U(n)*conj(-Y(m,n)+U(m)*conj(U(m)*YY(m,n); endendSmn【例1.3】 潮流計(jì)算結(jié)果：迭代次數(shù)ans = 4節(jié)點(diǎn)1至5電壓相角（度）ans = -4.7785 17.8535 -4.2819 21.8433 0節(jié)點(diǎn)1至5電壓幅值（pu）uu = 0.8622 1.0779

13、1.0364 1.0500 1.0500節(jié)點(diǎn)1至5功率（pu）Sm = -1.6000 - 0.8000i -2.0000 - 1.0000i -3.7000 - 1.3000i 5.0000 + 1.8131i 2.5794 + 2.2994i支路功率（行列編號(hào)均為節(jié)點(diǎn)1至5）Smn = 0 -1.4662 - 0.4091i -0.1338 - 0.3909i 0 0 1.5845 + 0.6726i 0 1.4155 - 0.2443i -5.0000 - 1.4282i 0 0.1568 + 0.4713i -1.2774 + 0.2032i 0 0 -2.5794 - 1.9745i

14、 0 5.0000 + 1.8131i 0 0 0 0 0 2.5794 + 2.2994i 0 0 【例1.4】如圖 1-5所示，六個(gè)節(jié)點(diǎn)網(wǎng)絡(luò)。節(jié)點(diǎn)2、1間和節(jié)點(diǎn)4、3間通過(guò)變壓器相連，變比分別為0.909:1和0.976:1，節(jié)點(diǎn)1、2、3、4為PQ負(fù)荷節(jié)點(diǎn)，節(jié)點(diǎn)5為PV發(fā)電機(jī)節(jié)點(diǎn)，節(jié)點(diǎn)電壓為1.10;節(jié)點(diǎn)6為參考節(jié)點(diǎn),節(jié)點(diǎn)電壓為1.05。輸電線路和變壓器數(shù)據(jù)見(jiàn)表2.2,各個(gè)節(jié)點(diǎn)功率如表2.3所示,計(jì)算各節(jié)點(diǎn)電壓、發(fā)電機(jī)無(wú)功功率和系統(tǒng)潮流分布：GG87563214910V5=1.10V6=1.05 圖1-5 例1.4網(wǎng)絡(luò)接線圖表1-2 輸電線路和變壓器數(shù)據(jù)匯編線路首端節(jié)點(diǎn)編號(hào)線路末端節(jié)點(diǎn)編

15、號(hào)充電功率換算導(dǎo)納BC/2線路阻抗換算電導(dǎo)G線路阻抗換算電納B變壓器變比 620.00700.2791-1.2910 640.00900.2170-0.9137 510.00.2224-0.3230 530.00.2383-0.6542 210.00.0-3.75940.909 240.00760.2771-1.1625 4 30.00.0-1.66670.976表 1-3 各節(jié)點(diǎn)功率母線有功負(fù)載無(wú)功負(fù)載發(fā)電機(jī)有功發(fā)電機(jī)無(wú)功1PD1=0.275QD1=0.0652PD2=0.0QD2=0.03PD3=0.15QD3=0.09456PD4=0.25QD4=0.025PG5=0.25PG6=？QG

16、4=未定QG6=？【例1.4】 程序說(shuō)明： 本例題計(jì)算程序可分為形成節(jié)點(diǎn)導(dǎo)納矩陣和計(jì)算節(jié)點(diǎn)電壓潮流兩部分。計(jì)算節(jié)點(diǎn)導(dǎo)納矩陣參考文獻(xiàn)1第1.3節(jié)。這里有6個(gè)節(jié)點(diǎn)7條支路，節(jié)點(diǎn)編號(hào)原則是先PQ節(jié)點(diǎn)，后PV節(jié)點(diǎn)，平衡節(jié)點(diǎn)編號(hào)最大，支路序號(hào)如圖所示。各支路的首末端節(jié)點(diǎn)編號(hào)用數(shù)組nstart和nend表示，各支路導(dǎo)納用數(shù)組yb1r,yb1i表示，各支路兩端充電電容導(dǎo)納為數(shù)組yb0i:Np=6;Nb=7; nstart=6,6,2,3,5,5,1;nend=4,2,4,4,3,1,2; yb1r=0.2170,0.2791,0.2771,0.0,0.2383,0.2224,0.0; yb1i=-0.913

17、7,-1.2910,-1.1625,-1.6667,-0.6542,-0.3230,-3.7594; yb0i=0.0099,0.0070,0.0076,0.0,0.0,0.0,0.0;兩臺(tái)變壓器跨接在節(jié)點(diǎn)2-1和4-3之間，參見(jiàn)圖1-2，i節(jié)點(diǎn)對(duì)應(yīng)2、4，j節(jié)點(diǎn)對(duì)應(yīng)1、3，變比為1：1/0.909和1:1/0.976,程序中用數(shù)組bt 和p 表示，m=6,k=2表明用6個(gè)節(jié)點(diǎn)兩條變壓器支路： bt=2,4;1,3;p=1/0.909,1/0.976;m=6;k=2; 計(jì)算潮流電壓參照例題1.3。這里獨(dú)立節(jié)點(diǎn)數(shù)N1=5，PV節(jié)點(diǎn)數(shù)N2=1，節(jié)點(diǎn)電壓初始給定值幅值為數(shù)組uu,相角為dd，節(jié)點(diǎn)有功

18、功率為p,無(wú)功功率為q:uu=1.000,1.000,1.000,1.000,1.100,1.050;dd=0,0,0,0,0,0;p=-.275,0,-0.150,-0.250,0.250;q=-0.065,0,-0.090,-0.02【例1.4】源程序 NU117N.m%*NU117N.M example 2.3.3, fig 2-3 * % %The following Program for load flow calculation is based on MATLAB R2007clear% bus 1,2,3,4 is PQ bus,bus 5 is PV bus,bus 6 i

19、s slack busu=sym(u1,u2,u3,u4,u5,u6);delt=sym(d1,d2,d3,d4,d5,d6); global Np Nb bt p m k Np=6;Nb=7; bt=2,4;1,3;p=1/0.909,1/0.976;m=6;k=2; nstart=6,6,2,3,5,5,1;nend=4,2,4,4,3,1,2; mm=ffm(nstart,nend); % yb1,the series admittance of transmission line yb1r=0.2170,0.2791,0.2771,0.0,0.2383,0.2224,0.0; yb1i

20、=-0.9137,-1.2910,-1.1625,-1.6667,-0.6542,-0.3230,-3.7594; yb1=yb1r+yb1i*j; %yb1=zb1.(-1); yb=diag(conj(yb1); y=mm*yb*(mm); % yb0 ,the shunt admittances of transmission line yb0i=0.0099,0.0070,0.0076,0.0,0.0,0.0,0.0; yb0=0+yb0i*j; y0=diag(conj(yb0); yg=mm*y0*(mm); yn=diag(diag(yg); yy=yn-yg; y=y+yn;

21、Y,YY=fdt1(y,yy);%*G=real(Y);B=imag(Y);p=-.275,0,-0.150,-0.250,0.250;q=-0.065,0,-0.090,-0.025;k=0;precision=1;N1=5;%the N1 is the amount of the PQ and PV busN2=1;%the N2 is the amount of the PV bus for m=1:N1 for n=1:N1+1 pt(n)=u(m)*u(n)*(G(m,n)*cos(delt(m)-delt(n)+B(m,n)*sin(delt(m)-delt(n); end pp(

22、m)=p(m)-sum(pt); end for m=1:N1-N2 for n=1:N1+1 qt(n)=u(m)*u(n)*(G(m,n)*sin(delt(m)-delt(n)-B(m,n)*cos(delt(m)-delt(n); end qq(m)=q(m)-sum(qt); end J1=jacobian(pp,d1,d2,d3,d4,d5,u1,u2,u3,u4); J2=jacobian(qq,d1,d2,d3,d4,d5,u1,u2,u3,u4); J=vertcat(J1,J2); uu=1.000,1.000,1.000,1.000,1.100,1.050;dd=0,0,

23、0,0,0,0; while precision0.0000001 u1=uu(1);u2=uu(2);u3=uu(3);u4=uu(4);u5=uu(5);u6=uu(6); d1=dd(1);d2=dd(2);d3=dd(3);d4=dd(4);d5=dd(5);d6=dd(6); for m=1:N1 PP(m)=eval(pp(m); end for m=1:N1-N2 PP(N1+m)=eval(qq(m); end JJ=eval(J); du=-inv(JJ)*PP;precision=max(abs(PP); for n=1:N1 dd(n)=dd(n)+du(n); endfor n=1:N1-N2 uu(n)=uu(n)+du(N1+n); end k=k+1;endk-1,dd*180/pi,uu%*% the following program is used to calculate the Sm and Smnfor n=1:N1+1 U(n)=uu(n)*(cos(dd(n)+j*sin(dd(n);end Um=co