电力系统解析理论课本习题matlab做.doc

PAGE PAGE 7 %例题3-1 l1=80 r1=0.21 x1=0.416 b=2.74/1000000 vn=110 S1=15 dp0=40.5 dps=128 vs=10.5 i0=3.5 sldb=30+12i sldc=20+15i %（1）计算参数并作出等值电路 rl=0.5*l1*r1 xl=0.5*l1*x1 bc=2*l1*b dqb=-0.5*bc*vn*vn rt=0.5*dps*vn^2/1000/S1^2 xt=0.5*vs*vn^2/100/S1 dq0=i0*S1/100 dp1=2*dp0/1000 dq1=2*dq0*i s0=dp1+dq1 sb=sldb+s0+dqb*i sc=sldc %（2)计算由母线A输出的功率 dst=(abs(sc)/vn)^2*(rt+xt*i) %变压器绕组中的功率损耗为 sc1=sc+dst s11=sc1+sb dsl=(abs(s11)/vn)^2*(rl+xl*i) %线路中的功率损耗为 s1=s11+dsl sa=s1+dqb*i %由母线A输出的功率为 %（3）计算各节点电压 p1=52.54 q1=32.1 va=117 %线路中电压降落的纵分量和横分量分别为： dvl=(p1*rl+q1*xl)/va pvl=(p1*xl-q1*rl)/va vb=sqrt((va-dvl)^2+(pvl)^2) %b点电压为 pc=20.18 qc=17.19 vcc=11 %变压器中电压降落的纵分量和横分量分别为： dvt=(pc*rt+qc*xt)/vb pvt=(pc*xt-qc*rt)/vb vc1=sqrt((vb-dvt)^2+(pvt)^2) %归算到高压侧的c点电压 vc=vc1*vcc/vn %变电所低压母线c的实际电压 l1 =80 r1 =0.2100 x1 =0.4160 b =2.7400e-006 vn =110 S1 =15 dp0 =40.5000 dps =128 vs =10.5000 i0 =3.5000 sldb =30.0000 +12.0000i sldc =20.0000 +15.0000i rl =8.4000 xl =16.6400 bc =4.3840e-004 dqb =-2.6523 rt = 3.4418 xt =42.3500 dq0 = 0.5250 dp1 =0.0810 dq1 =0 + 1.0500i s0 =0.0810 + 1.0500i sb =30.0810 +10.3977i sc =20.0000 +15.0000i dst =0.1778 + 2.1875i sc1 =20.1778 +17.1875i s11 =50.2588 +27.5852i dsl =2.2818 + 4.5201i s1 =52.5406 +32.1053i sa =52.5406 +29.4530i p1 =52.5400 q1 =32.1000 va =117 dvl =8.3374 pvl =5.1677 vb =108.7854 pc =20.1800 qc =17.1900 vcc =11 dvt =7.3305 pvt =7.3122 vc1 =101.7180 vc =10.1718 %例题3-2 s2=0.3+0.2i s3=0.5+0.3i s4=0.2+0.15i z12=1.2+2.4i z23=1.0+2.0i z24=1.5+3.0i vn=10 %(1)求线路始端功率 ds23=(abs(s3)^2/vn^2)*z23 ds24=(abs(s4)^2/vn^2)*z24 s23=s3+ds23 s24=s4+ds24 s112=s23+s24+s2 ds12=(abs(s112)/vn)^2*z12 s12=s112+ds12 %（2）求线路各点电压 dv12=(real(s12)*real(z12)+imag(s12)*imag(z12))/(1.05*vn) v2=1.05*vn-dv12 dv24=(real(s24)*real(z24)+imag(s24)*imag(z24))/v2 dv23=(real(s23)*real(z23)+imag(s23)*imag(z23))/v2 v3=v2-dv23 v4=v2-dv24 %（3）根据上述求得的线路各点电压，重新计算各线路的功率损耗和线路始端损耗 ds23=(abs(s3)/v3)^2*z23 ds24=(abs(s4)/v4)^2*z24 s23=s3+ds23 s24=s4+ds24 s112=s23+