电力系统分析理论课本习题MATLAB做.doc

%例题3-1l1=80r1=0.21x1=0.416b=2.74/1000000vn=110S1=15dp0=40.5dps=128vs=10.5i0=3.5sldb=30+12isldc=20+15i%（1）计算参数并作出等值电路rl=0.5*l1*r1xl=0.5*l1*x1bc=2*l1*bdqb=-0.5*bc*vn*vnrt=0.5*dps*vn2/1000/S12xt=0.5*vs*vn2/100/S1dq0=i0*S1/100dp1=2*dp0/1000dq1=2*dq0*is0=dp1+dq1sb=sldb+s0+dqb*isc=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+dsts11=sc1+sbdsl=(abs(s11)/vn)2*(rl+xl*i) %线路中的功率损耗为s1=s11+dslsa=s1+dqb*i %由母线A输出的功率为%（3）计算各节点电压p1=52.54q1=32.1va=117%线路中电压降落的纵分量和横分量分别为：dvl=(p1*rl+q1*xl)/vapvl=(p1*xl-q1*rl)/vavb=sqrt(va-dvl)2+(pvl)2) %b点电压为pc=20.18qc=17.19vcc=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)/vbpvt=(pc*xt-qc*rt)/vbvc1=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 =80r1 =0.2100x1 =0.4160b =2.7400e-006vn =110S1 =15dp0 =40.5000dps =128vs =10.5000i0 =3.5000sldb =30.0000 +12.0000isldc =20.0000 +15.0000irl =8.4000xl =16.6400bc =4.3840e-004dqb =-2.6523rt = 3.4418xt =42.3500dq0 = 0.5250dp1 =0.0810dq1 =0 + 1.0500is0 =0.0810 + 1.0500isb =30.0810 +10.3977isc =20.0000 +15.0000idst =0.1778 + 2.1875isc1 =20.1778 +17.1875is11 =50.2588 +27.5852idsl =2.2818 + 4.5201is1 =52.5406 +32.1053isa =52.5406 +29.4530ip1 =52.5400q1 =32.1000va =117dvl =8.3374pvl =5.1677vb =108.7854pc =20.1800qc =17.1900vcc =11dvt =7.3305pvt =7.3122vc1 =101.7180vc =10.1718 %例题3-2s2=0.3+0.2is3=0.5+0.3is4=0.2+0.15iz12=1.2+2.4iz23=1.0+2.0iz24=1.5+3.0ivn=10%(1)求线路始端功率ds23=(abs(s3)2/vn2)*z23ds24=(abs(s4)2/vn2)*z24s23=s3+ds23s24=s4+ds24s112=s23+s24+s2ds12=(abs(s112)/vn)2*z12s12=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-dv12dv24=(real(s24)*real(z24)+imag(s24)*imag(z24)/v2dv23=(real(s23)*real(z23)+imag(s23)*imag(z23)/v2v3=v2-dv23v4=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*z23ds24=(abs(s4)/v4)2*z24s23=s3+ds23s24=s4+ds24s112=s23+s24+s2ds12=(abs(s112)/v2)2*z12%从而可得线路始端功率s12=s112+ds12s2 =0.3000 + 0.2000is3 =0.5000 + 0.3000is4 =0.2000 + 0.1500iz12 =1.2000 + 2.4000iz23 =1.0000 + 2.0000iz24 =1.5000 + 3.0000ivn =10ds23 =0.0034 + 0.0068ids24 =0.0009 + 0.0019is23 =0.5034 + 0.3068is24 =0.2009 + 0.1519is112 =1.0043 + 0.6587ids12 =0.0173 + 0.0346is12 =1.0216 + 0.6933idv12 =0.2752v2 =10.2248dv24 =0.0740dv23 =0.1092v3 =10.1155v4 =10.1507ds23 =0.0033 + 0.0066ids24 =0.0009 + 0.0018is23 =0.5033 + 0.3066is24 =0.2009 + 0.1518is112 =1.0042 + 0.6585ids12 =0.0166 + 0.0331is12 =1.0208 + 0.6916i %例题3-3vn=10l1=10l2=4l3=3l4=2va=10.5vb=10.4z1=0.17+0.38*iz2=0.45+0.4*i%的线路等值阻抗zl1=10*z1zl2=4*z1zl3=3*z1zl4=2*z2%求C和D点的运算负荷，为sc1=2600+1600*ise=300+160*isd1=600+200*isd2=1600+1000*idsce=(abs(se/1000)/vn)2*zl4*1000sc=sc1+se+dscesd=sd1+sd2%循环功率z11=0.17-0.38*iscc=1000*(va-vb)*vn/(17*z11)sac=(real(sc)*7+imag(sc)*7*i+real(sd)*3+imag(sd)*3*i)/17+sccsbd=(real(sc)*10+imag(sc)*10*i+real(sd)*14+imag(sd)*14*i)/17-sccs1=sac+sbds2=sc+sdsdc=sbd-sd%C点位功率分点，可推测出E点电压最低点，进一步可求得E点电压dsac=(abs(sac/1000)/vn)2*zl1*1000s1ac=sac+dsacdvac=(real(s1ac/1000)*real(zl1)+imag(s1ac/1000)*imag(zl1)/vavc=va-dvacsce=se+dscedvce=(real(sce/1000)*real(zl4)+imag(sce/1000)*imag(zl4)/vcve=vc-dvcevn =10l1 =10l2 =4l3 =3l4 =2va =10.5000vb =10.4000z1 =0.1700 + 0.3800iz2 =0.4500 + 0.4000izl1 =1.7000 + 3.8000izl2 =0.6800 + 1.5200izl3 =0.5100 + 1.1400izl4 =0.9000 + 0.8000isc1 =2.6000e+003 +1.6000e+003ise =3.0000e+002 +1.6000e+002isd1 =6.0000e+002 +2.0000e+002isd2 =1.6000e+003 +1.0000e+003idsce =1.0404 + 0.9248isc =2.9010e+003 +1.7609e+003isd =2.2000e+003 +1.2000e+003iz11 =0.1700 - 0.3800iscc =5.7703e+001 +1.2898e+002isac =1.6405e+003 +1.0658e+003isbd =3.4606e+003 +1.8951e+003is1 =5.1010e+003 +2.9609e+003is2 =5.1010e+003 +2.9609e+003isdc =1.2606e+003 +6.9509e+002idsac =6.5062e+001 +1.4543e+002is1ac =1.7055e+003 +1.2113e+003idvac =0.7145vc =9.7855sce =3.0104e+002 +1.6092e+002idvce =0.0408ve =9.7447 %例题3-4r1=0.27x1=0.423 b1=2.69/1000000r2=r1x2=x1b2=b1r3=0.45x3=0.44b3=2.58/1000000l1=60l2=50l3=40vn=110snb=20ds0b=0.05+0.6*irtb=4.84xtb=63.5snc=10ds0c=0.03+0.35*irtc=11.4xtc=127sldb=24+18*isldc=12+9*i%（1）计算网络参数及制定等值电路Z1=l1*(r1+x1*i) %线路1 B1=l1*b1dqb1=-B1*vn2/2Z2=l2*(r2+x2*i) %线路2 B2=l2*b2dqb2=-B2*vn2/2Z3=l3*(r3+x3*i) %线路3 B3=l3*b3dqb3=-B3*vn2/2ztb=(rtb+xtb*i)/2 %变电所bdS0b=2*ds0bztc=(rtc+xtc*i)/2 %变电所cdS0c=2*ds0c%（2）计算节点b和c得运算负荷dstb=(abs(sldb)/vn)2*ztbsb=sldb+dstb+dS0b+dqb1*i+dqb3*idstc=(abs(sldc)/vn)2*ztcsc=sldc+dstc+dS0c+dqb3*i+dqb2*i%（3)计算闭式网络的功率分布s1=(sb*(conj(Z2)+conj(Z3)+sc*conj(Z2)/(conj(Z1)+conj(Z2)+conj(Z3)s2=(sc*(conj(Z1)+conj(Z3)+sb*conj(Z1)/(conj(Z1)+conj(Z2)+conj(Z3)s12=s1+s2sbc=sb+scs3=sb-s1%（4）计算电压损耗dsl1=(abs(s1)/vn)2*Z1sa1=s1+dsl1va=117dv1=(real(sa1)*real(Z1)+imag(sa1)*imag(Z1)/vavb=va-dv1r1 =0.2700x1 =0.4230b1 =2.6900e-006r2 =0.2700x2 =0.4230b2 =2.6900e-006r3 =0.4500x3 =0.4400b3 =2.5800e-006l1 =60l2 =50l3 =40vn =110snb =20ds0b =0.0500 + 0.6000irtb =4.8400xtb =63.5000snc =10ds0c =0.0300 + 0.3500irtc =11.4000xtc =127sldb =24.0000 +18.0000isldc =12.0000 + 9.0000iZ1 =16.2000 +25.3800iB1 =1.6140e-004dqb1 =-0.9765Z2 =13.5000 +21.1500iB2 =1.3450e-004dqb2 =-0.8137Z3 =18.0000 +17.6000iB3 =1.0320e-004dqb3 =-0.6244ztb =2.4200 +31.7500idS0b =0.1000 + 1.2000iztc =5.7000 +63.5000idS0c =0.0600 + 0.7000idstb =0.1800 + 2.3616isb =24.2800 +19.9607idstc =0.1060 + 1.1808isc =12.1660 + 9.4427is1 =18.6410 +15.7972is2 =17.8050 +13.6062is12 =36.4460 +29.4034isbc =36.4460 +29.4034is3 =5.6390 + 4.1635idsl1 =0.7993 + 1.2523isa1 =19.4404 +17.0495iva =117dv1 =6.3902vb =110.6098 %例题3-5k1=110/11k2=115.5/11zt1=1*izt2=1*ivb=10%（1）计算变压器的功率分布sld=16+12*is1ld=sld/2s2ld=s1ld%（2）求循环功率dE=vb*(k2/k1-1)%故循环功率为sc=(vb*dE)/(conj(zt1)+conj(zt2)%（3）计算俩台变压器的实际功率分布st1=s1ld+scst2=s2ld-sc%（4）计算高压侧电压vat1=(vb+(imag(st1)*imag(zt1)/vb)*k1 %按变压器T-1计算vat2=(vb+(imag(st2)*imag(zt2)/vb)*k2 %按变压器T-2计算%计及电压降落的横分量，按T-1和T-2计算可分别得vat1=108.79vat2=109%（5）计算从高压母线输入变压器T-1和T-2的功率st11=st1+(abs(st1)/vb)2*zt1st12=st2+(abs(st2)/vb)2*zt2%输入高压母线的总功率为s=st11+st12k1 =10k2 =10.5000zt1 =0 + 1.0000izt2 =0 + 1.0000ivb =10sld =16.0000 +12.0000is1ld =8.0000 + 6.0000is2ld =8.0000 + 6.0000idE =0.5000sc =0 + 2.5000ist1 =8.0000 + 8.5000ist2 =8.0000 + 3.5000ivat1 =108.5000vat2 =108.6750vat1 =108.7900vat2 =109st11 =8.0000 + 9.8625ist12 =8.0000 + 4.2625is =16.0000 +14.1250i %例题3-6s=40+30*icosa=0.8Tmax=4500r=0.17x=0.409b=2.82/1000000dp0=86dps=200I0=2.7Vs=10.5sn=31.5%最大负荷时变压器的绕组功率损耗为dst=2*(dps+1000*Vs*sn*i/100)*(real(s)/0.8/2/sn)2ds0=2*(dp0+1000*I0*sn*i/100) %变压器的铁芯功率损耗为l=100v=110qb2=-2*b*l*v2/2 %线路末端充电功率%等值电路中流过线路等值阻抗的功率为s1=s+dst/1000+ds0/1000+qb2*irt=r*l/2dpl=(abs(s1)/v)2*rt %线路上的有功功率损耗T=8760t=3150dwt=2*dp0*T+real(dst)*t %变压器全年的电能损耗dwl=1000*dpl*t %线路全年的电能损耗dw=dwt+dwl %输电系统全年的总电能损耗s =40.0000 +30.0000icosa =0.8000Tmax = 4500r =0.1700x =0.4090b =2.8200e-006dp0 =86dps =200I0 =2.7000Vs =10.5000sn =31.5000dst =2.5195e+002 +4.1667e+003ids0 =1.7200e+002 +1.7010e+003il =100v =110qb2 =-3.4122s1 = 40.4240 +32.4555irt = 8.500dpl =1.8879T =8760t =3150dwt =2.3004e+006dwl =5.9468e+006dw =8.2472e+006 %例题5-2z1=4.32+10.5*iv1=35va=36vb=10Smax=8+5*iSmin=4+3*izt=0.69+7.84*i%变压器阻抗与线路阻抗合并得等值阻抗z=z1+zt%线路首端输送功率为Samax=Smax+(abs(Smax)/v1)2*zSamin=Smin+(abs(Smin)/v1)2*z%B点折算到高压侧电压为V1bmax=va-(real(Samax)*real(z)+imag(Samax)*imag(z)/vaV1bmin=va-(real(Samin)*real(z)+imag(Samin)*imag(z)/va%最大和最小负荷时对应的分接头电压Vbmax=0.95*vbVbmin=1.05*vbV2n=10.5Vtmax=V1bmax*V2n/VbmaxVtmin=V1bmin*V2n/Vbmin%取平均值Vt=(Vtmax+Vtmin)/2%选择变压器最接近的分接头a=(Vt/v1-1)*100%所以取-2.5%的分接头，即Vt1=(1-0.025)*v1%按所选分接头校验10KV母线的实际电压Vbmax1=V1bmax*V2n/Vt1dvmax=(Vbmax1-10)/10Vbmin1=V1bmin*V2n/Vt1dvmax=(Vbmin1-10)/10z1 =4.3200 +10.5000iv1 =35va =36vb =10Smax =8.0000 + 5.0000iSmin =4.0000 + 3.0000izt =0.6900 + 7.8400iz = 5.0100 +18.3400iSamax =8.3640 + 6.3325iSamin =4.1022 + 3.3743iV1bmax =31.6100V1bmin =33.7101Vbmax =9.5000Vbmin =10.5000V2n =10.5000Vtmax =34.9373Vtmin =33.7101Vt =34.3237a =-1.9322Vt1 =34.1250Vbmax1 =9.7261dvmax =-0.0274Vbmin1 =10.3723dvmax =0.0372 %例题5-3Smax=25+18iSmin=14+10iZt=3+30iV1max=120V1min=114Pmax=real(Smax)Qmax=imag(Smax)Pmin=real(Smin)Qmin=imag(Smin)R=real(Zt)X=imag(Zt)%最大负荷时变压器的电压降为dVmax=(Pmax*R+Qmax*X)/V1max%规算至高电压侧的低电压为V2max=V1max+dVmax%最小负荷时变压器的电压降为dVmin=(Pmin*R+Qmin*X)/V1min%规算至高电压侧的低电压为V2min=V1min+dVmin%假定最大负荷时发电机电压为6.6kv，最小负荷时电压为6kvV1tmax=V2max*6.3/6.6V1tmin=V2min*6.3/6V1t=(V1tmax+V1tmin)/2%选择最近的分接头121kv%校验：最大负荷时发电机端实际电压为V2max*6.3/121%最大负荷时发电机端实际电压为V2min*6.3/121Smax =25.0000 +18.0000iSmin =14.0000 +10.0000iZt =3.0000 +30.0000iV1max =120V1min =114Pmax =25Qmax =18Pmin =14Qmin =10R =3X =30dVmax =5.1250V2max =125.1250dVmin =3V2min =117V1tmax =119.437V1tmin =122.8500V1t =121.1438ans =6.5148ans =6.0917 %例题5-4V1n=110V2n=35V3n=6Vn3=6.6P1=12.8Q1=9.6R1=2.94X1=65P2=6.4Q2=4.8R2=4.42X2=-1.5P3=6.4Q3=4.8R3=4.42X3=37.7V1max=112V1min=115%（1）求最大、最小负荷时各绕组的电压损耗dV1max=(P1*R1+Q1*X1)/V1max %最大负荷时dV2max=(P2*R2+Q2*X2)/(V1max-dV1max)dV3max=(P3*R3+Q3*X3)/(V1max-dV1max)dV1min=(0.5*P1*R1+0.5*Q1*X1)/V1min %最小负荷时 dV2min=(0.5*P2*R2+0.5*Q2*X2)/(V1min-dV1min)dV3min=(0.5*P3*R3+0.5*Q3*X3)/(V1min-dV1min)%（2）求最大、最小负荷时个母线电压V1max=112 %最大负荷时V2max1=V1max-dV1max-dV2maxV3max1=V1max-dV1max-dV3maxV1min=115 %最小负荷时V2min1=V1min-dV1min-dV2minV3min1=V1min-dV1min-dV3min%（3)选择高压绕组分接头V3max=V3n*(1+0)V3min=V3n*(1+0.075)Vt1max=V3max1*Vn3/V3maxVt1min=V3min1*Vn3/V3min Vt1=(Vt1max+Vt1min)/2Vt1=115.5 %选用110+5%的分接头%（4）校验低压母线电压V3max=V3max1*Vn3/Vt1 %最大负荷时V3min=V3min1*Vn3/Vt1 %最小负荷时%低压母线电压偏移dV3max=(V3max-V3n)/V3n %最大负荷时dV3min=(V3min-V3n)/V3n %最小负荷时%（5）根据中压母线的调压要求，又高、中压俩侧，选择中压绕组的分接头%最大、最小负荷时中压母线调压要求电压为V2max=V2n*(1+0)V2min=V2n*(1+0.075)%最大、最小负荷时中压绕组的分接头电压为Vt2max=V2max*Vt1/V2max1Vt2min=V2min*Vt1/V2min1Vt2=(Vt2max+Vt2min)/2%于是就选电压为38.5kV的主抽头Vt2=38.5%（6）校验中压侧母线电压V2max=V2max1*Vt2/Vt1 %最大负荷时V2min=V2min1*Vt2/Vt1 %最小负荷时%中压母线电压偏移：dV2max=(V2max-V2n)/V2ndV2min=(V2min-V2n)/V2nV1n =110V2n =35V3n =6Vn3 =6.6000P1 =12.8000Q1 =9.6000R1 =2.9400X1 =65P2 =6.4000Q2 =4.8000R2 =4.4200X2 =-1.5000P3 =6.4000Q3 =4.8000R3 =4.4200X3 =37.7000V1max =112V1min =115dV1max =5.9074dV2max =0.1988dV3max =1.9723dV1min =2.8767dV2min =0.0940dV3min =0.9331V1max =112V2max1 =105.8938V3max1 =104.1203V1min =115V2min1 =112.0293V3min1 =111.1902V3max =6V3min =6.4500Vt1max =114.5323Vt1min =113.7760Vt1 =114.1542Vt1 =115.5000V3max =5.9497V3min =6.3537dV3max =-0.0084dV3min =0.0590V2max =35V2min =37.625Vt2max =38.1750Vt2min =38.7906Vt2 =38.4828Vt2 =38.5000V2max =35.2979V2min = 37.3431dV2max =0.0085dV2min =0.0669 %例题7-3Sg1=100,X11d=0.183,cosa1=0.85 %发电机G-1Sg2=50,X22d=0.141,cosa2=0.8 %发电机G-2St1=120,Vs1=14.2 %变压器T-1 St2=63,Vs2=14.5 %变压器T-2 l1=170,x1=0.427 %线路L-1l2=120,x2=0.432 %线路L-2l3=100,x3=0.432 %线路L-3 Sld=160,vn1=230%（1）各电抗标幺值SB=100, VB=230,xl=0.35,E3=0.8X1=X11d*SB/(Sg1/cosa1) %发电机G-1X2=X22d*SB/(Sg2/cosa2) %发电机G-2X3=xl*SB/Sld %负荷LDX4=Vs1*SB/100/St1 %变压器T-1X5=Vs2*SB/100/St2 %变压器T-2X6=x1*l1*SB/vn12 %线路L-1X7=x2*l2*SB/vn12 %线路L-2X8=x3*l3*SB/vn12 %线路L-3E1=1.08 %取发电机的次暂态电势E2=1.08%(2)简化网络X9=X1+X4X10=X2+X5%将X6 X7 X8构成的三角形化为星型X11=X6*X7/(X6+X7+X8)X12=X6*X8/(X6+X7+X8)X13=X7*X8/(X6+X7+X8)%将E1、E2俩条有源支路并联 X14=(X9+X11)*(X10+X12)/(X9+X11+X10+X12)+X13E12=1.08 %（3）计算起始暂态电流%有发电机提供的起始次暂态电流为 Ib=E12/X14%由负荷LD提供的起始次暂态电流为 ILDb=E3/X3%短路点总的起始次暂态电流为 Ifb=Ib+ILDb%基准电流IB=SB/sqrt(3)/VB%有起始次暂态电流有名值为Ifb=8.63*IB%（4）计算冲击电流%有发电机冲击系数kim=1.8，综合负荷LD冲击系数kimLD=1，短路点的冲击电流为kim=1.8,kimLD=1iim=(kim*Ib*sqrt(2)+kimLD*ILDb*sqrt(2)*IBSg1 =100X11d =0.1830cosa1 =0.8500Sg2 =50X22d =0.1410cosa2 =0.8000St1 =120Vs1 =14.2000St2 =63Vs2 =14.5000l1 =170x1 =0.4270l2 =120x2 =0.4320l3 =100x3 =0.4320Sld =160vn1 =230SB =100VB =230xl =0.3500E3 =0.8000X1 =0.1555X2 =0.2256X3 =0.2188X4 =0.1183X5 =0.2302X6 =0.1372X7 =0.0980X8 =0.0817E1 =1.0800E2 =1.0800X9 =0.2739X10 =0.4558X11 =0.0424X12 =0.0354X13 =0.0253X14 =0.2177E12 =1.0800Ib =4.9620ILDb =3.6571Ifb =8.6191IB =0.2510Ifb =2.1663kim =1.8000kimLD =1iim = 4.4690 %例题8-1SG=50,COSg=0.8,Xdb=0.15,X2=0.18,E1=1.08 %发电机GST1=60,VS=10.5,xn=22 %变压器T-1、T-2SLD=15,X1D=1.2,X2D=0.35 %负荷LDL=50,x1=0.4 ,x0=3*x1 %输电线路L%（1）各元件参数标幺值计算SB=100XG1=Xdb*SB/(SG/COSg),XG2=X2*SB/(SG/COSg) %发电机XT1=(VS/100)*(SB/ST1) %变压器T-1变压器T-2XT2=XT1v=37Xn=xn*SB/v2 %中性点接地电阻 XLD1=X1D*SB/SLD,XLD2=X2D*SB/SLD, %负荷LDXL1=L*x1*100/v2 ,XL0=3*XL1 %输电线路L%（2）制定各序网络(见课本222页）%（3）网络化简，求组合电势和各序组合电抗E=E1*XLD1/(XG1+XLD1)Xa=XG1*XLD1/(XG1+XLD1)+XT1+XL1 %正序电抗Xb=XG2*XLD2/(XG2+XLD2)+XT1+XL1 %负序电抗Xc=(XT1+3*xn+XL0)*XT2/(XT1+3*xn+XL0+XT2) %零序电抗SG =50COSg =0.8000Xdb =0.1500X2 =0.1800E1 =1.0800ST1 =60VS =10.5000xn =22SLD =15X1D =1.2000X2D =0.3500L =50x1 =0.4000x0 =1.2000SB =100XG1 =0.2400XG2 =0.2880XT1 =0.1750XT2 =0.1750v =37Xn =1.6070XLD1 =8XLD2 =2.3333XL1 =.4609XL0 =4.3828E =1.0485Xa =1.8689Xb =1.8923Xc =0.1746 %例题8-2Sg1=100cos1=0.85Xd1=0.183X12=0.223Sg2=50cos2=0.8Xd2=0.141Xx22=0.172St1=120Vs1=14.2St2=63Vs2=14.5L1=120Xl=0.432X0=5*XlVb=230Sb=100Xg11=Xd1*Sb/(Sg1/cos1)Xt11=Vs1*Sb/(100*St1)Xl11=0.5*L1*Xl*Sb/Vb2Xt21=Vs2*Sb/(100*St2)Xg21=Xd2*Sb/(Sg2/cos2)Xg12=X12*Sb/(Sg1/cos1)Xt12=Vs1*Sb/(100*St1)Xl12=0.5*L1*Xl*Sb/Vb2Xt22=Vs2*Sb/(100*St2)Xg22=Xt22*Sb/(Sg2/cos2)Xl0=5*Xl11Xt10=Vs1*Sb/(100*St1)Xt20=Vs2*Sb/(100*St2)%（1）制定各序等值电路X1=(Xg11+Xt11+Xl11)*(Xt21+Xg21)/(Xg11+Xt11+Xl11)+(Xt21+Xg21)X2=(Xg12+Xt12+Xl12)*(Xt22+Xg22)/(Xg12+Xt12+Xl12)+(Xt22+Xg22)X0=(Xl0+Xt10)*Xt20/(Xl0+Xt10)+Xt20)%（2）计算各种不对称短路时的短路电流Xd1=X2+X0 %单相接地短路m1=3Ed=1Ia11=Ed/(X1+Xd1)Ib=Sb/(sqrt(3)*Vb) %基准电流If1=m1*Ia11*IbXd2=X2 %俩相短路m2=sqrt(3)Ia12=Ed/(X1+Xd2)If2=m2*Ia12*IbXd11=X2*X0/(X2+X0) %；俩相短路接地m11=sqrt(3)*sqrt(1-(X2*X0/(X2+X0)2)Ia111=Ed/(X1+Xd11)If11=m11*Ia111*IbSg1 =100cos1 =0.8500Xd1 =0.1830X12 =0.2230Sg2 = 50cos2 = 0.8000Xd2 =0.1410Xx22 =0.1720St1 =120Vs1 =14.2000St2 =63Vs2 =14.5000L1 =120Xl =0.4320X0 =2.1600Vb =230Sb =100Xg11 =0.1555Xt11 =0.1183Xl11 =0.0490Xt21 =0.2302Xg21 =0.2256Xg12 =0.1896Xt12 =0.1183Xl12 =0.0490Xt22 =0.2302Xg22 =0.3683Xl0 =0.2450Xt10 =0.1183Xt20 =0.2302X1 =0.1890X2 =0.2236X0 =0.1409Xd1 =0.3645m1 =3Ed =1Ia11 =1.8069Ib =0.2510If1 =1.3607Xd2 =0.2236m2 =1.7321Ia12 =2.4240If2 =1.0539Xd11 =0.0864m11 =1.5128Ia111 =3.6308If11 =1.3788 %例题8-3V1=230a=-0.5+1i*(sqrt(3)/2)A=a*aX1=0.1890X2=0.2092X0=0.1409Ia1=1.855Va1=i*(1-X1*Ia1)Va2=-1i*X2*Ia1Va0=-1i*X0*Ia1a=-0.5+1i*(sqrt(3)/2)A=-0.5-1i*(sqrt(3)/2)Vb1=A*Va1+a*Va2+Va0Vb11=abs(Vb1)Vc1=a*Va1+A*Va2+Va0%b、c相电压有名值为Vb=Vb11*V1/sqrt(3)Vc=VbV1 =230a = -0.5