某110kv變電站短路電流計算書一、短路電流計算取基準容量Sj=100MVA，略去“*”，Uj=115KV，Ij=0.502A富興變：地區電網電抗X1=Sj/Sdx=Ij/Idx=0.502/15.94=0.0315km線路電抗X2=X*L*(Sj/Up2)=0.4*5*(100/1152)=0.015發電機電抗X3=(Xd’’%/100)*(Sj/Seb)=(24.6/100)*(100/48)=0.51216km線路電抗X4=X*L*(Sj/Up=0.4*16*(100/1152)=0.0495.6km線路電抗X5=X*L*(Sj/Up2=0.4*5.6*(100/1152)=0.01731.5MVA變壓器電抗X6=X7=(Ud%/100)*(Sj/Seb)=(10.5/100)*(100/31.5)=0.33350MVA變壓器電抗X=(Ud%/100)*(Sj/Seb)=0.272X8=X3+X4+X5=0.578X9=X1+X2=0.046X10=(X8*X9)/(X8+X9)X11=X10+X6=0.046地區電網支路的分布系數C1=X10/X9=0.935發電機支路的分布系數C2=X10/X8=0.074則X13=X11/C1=0.376/0.935=0.402X14=X11/C2=0.376/0.074=5.081、求d1’點的短路電流求富興變供給d1’點(即d1點)的短路電流Ix″=Ij/(X1+X2)=0.502/(0.031+0.015)=10.913kASx″=Sj/(X1+X2)=100/(0.031+0.015)≈2173.913MVAichx1=√2*Kch*Ix″=√2*1.8*10.913=27.776kAIch=Ix″√1+2(Kch-1)2=10.913*√1+2(1.8-1)2=10.913*1.51=16.479kA1.2求沙縣城關水電站供給d1’點的短路電流將發電機支路的等值電抗換算到以發電機容量為基準容量時的標幺值Xjs=X8*Srg/Sj=0.578*48/100=0.277查表得I*’’=3.993I*0.2=3.096I*4=3.043換算到115kV下發電機的額定電流:Ief=Srg/(3Up)=48/(1.732x115)=0.241求得:If’’=I*’’*Ief=3.993x0.241=0.962kAIf0.2’’=I*0.2’’*IIf0.4’’=I*4’’*Iichf=√2*Kch*If″=√2*1.8*0.962=2.448kA1.3求得d1’點的短路電流Ix″=10.913+0.962=11.875kAich=27.776+2.448=30.224kAIch=11.875√1+2*(1.8-1)2=17.93kA2、求d2點的短路電流Ix″=Ij/(X1+X2+X6)=5.50/(0.031+0.015+0.333)=14.512kAichx=2*Kch*Ix2″=2*1.8*14.512=36.936kAIch=Ix″√1+2(Kch-1)2=14.512*√1+2(1.8-1)2=21.913kA3、求d2’3.1求富興變供給d2’點的短路電流Ix″=Ij/X13=5.5/0.402=13.68kAichx1=√2*Kch*Ix″=√2*1.8*13.68=34.82kAIch=Ix″√1+2(Kch-1)2=13.68*√1+2(1.8-1)2則Rt=1/(100*0.58/126.5+0.25/1.24)≈1.5Ω六、現有避雷針保護范圍計算現下洋變有四支等高避雷針（相對站內地面標高），位置詳見B992C-D0101-03。令hx=10m，則rx=(1.5h-2hx)p=(1.5x30-2x10)=25h0AB=h-D/7p=30-60/7=21.43m則A、B兩針間高度為hx=10m水平面上保護范圍的一側bxAB=1.5h0-2hx=1.5*21.43-2*10=12.1同理h0BC=h-D/7=30-50/7=22.85m則bXBC=1.5