某110kv变电站短路电流计算书

某110kv变电站短路电流计算书一、短路电流计算取基准容量Sj=100MVA，略去“*”，Uj=115KV，Ij=0.502A富兴变：地区电网电抗X1=Sj/Sdx=Ij/Idx=0.502/15.94=0.0315km线路电抗X2=X*L*(Sj/Up2)=0.4*5*(100/1152)=0.015发电机电抗X3=(Xd’’%/100)*(Sj/Seb)=(24.6/100)*(100/48)=0.51216km线路电抗X4=X*L*(Sj/Up=0.4*16*(100/1152)=0.0495.6km线路电抗X5=X*L*(Sj/Up2=0.4*5.6*(100/1152)=0.01731.5MVA变压器电抗X6=X7=(Ud%/100)*(Sj/Seb)=(10.5/100)*(100/31.5)=0.33350MVA变压器电抗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地区电网支路的分布系数C1=X10/X9=0.935发电机支路的分布系数C2=X10/X8=0.074则X13=X11/C1=0.376/0.935=0.402X14=X11/C2=0.376/0.074=5.081、求d1’点的短路电流求富兴变供给d1’点(即d1点)的短路电流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求沙县城关水电站供给d1’点的短路电流将发电机支路的等值电抗换算到以发电机容量为基准容量时的标幺值Xjs=X8*Srg/Sj=0.578*48/100=0.277查表得I*’’=3.993I*0.2=3.096I*4=3.043换算到115kV下发电机的额定电流: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’点的短路电流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点的短路电流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求富兴变供给d2’点的短路电流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则Rt=1/(100*0.58/126.5+0.25/1.24)≈1.5Ω六、现有避雷针保护范围计算现下洋变有四支等高避雷针（相对站内地面标高），位置详见B992C-D0101-03。令hx=10m，则rx=(1.5h-2hx)p=(1.5x30-2x10)=25h0AB=h-D/7p=30-60/7=21.43m则A、B两针间高度为hx=10m水平面上保护范围的一侧bxAB=1.5h0-2hx=1.5*21.43-2*10=12.1同理h0BC=h-D/7=30-50/7=22.85m则bXBC=1.5