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=0.031 5km 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)=0.512 16km 线路电抗 X4=X*L*(Sj/Up2) =0.4*16*(100/1152)=0.049 5.6km 线路电抗 X5=X*L*(Sj/Up2) =0.4*5.6*(100/1152)=0.017 31.5MVA 变压器电抗 X6=X7= (Ud%/100)*(Sj/Seb)=(10.5/100)*(100/31.5)=0.333 50MVA 变压器电抗 X=(Ud%/100)*(Sj/Seb)=0.272 X8=X3+X4+X5=0.578 X9=X1+X2=0.046 X10=(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=0.402 X14=X11/C2=0.376/0.074=5.08 1、求 d1’点的短路电流 1.1 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)=10.913kA Sx″=Sj/(X1+X2)=100/(0.031+0.015) ≈2173.913MVA ichx1=√2 *Kch*Ix″=√2 *1.8*10.913=27.776kA Ich=Ix″√1+2(Kch-1)2 =10.913*√1+2(1.8-1)2 =10.913*1.51=16.479kA 1.2 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*’’=3.993 I*0.2=3.096 I*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=0.962kA If0.2’’= I*0.2’’*Ief=3.096x0.241=0.746kA If0.4’’= I*4’’*Ief=3.043x0.241=0.732kA ichf=√2 *Kch*If″=√2 *1.8*0.962=2.448kA 1.3 求得d1’点的短路电流 Ix″=10.913+0.962=11.875kA ich=27.776+2.448=30.224kA Ich=11.875√1+2*(1.8-1)2 =17.93kA 2、求d2 点的短路电流 Ix″=Ij/(X1+X2+X6)=5.50/(0.031+0.015+0.333) =14.512kA ichx=2* Kch*Ix2″=2*1.8*14.512=36.936kA Ich=Ix″√1+2(Kch-1)2 =14.512*√1+2(1.8-1)2 =21.913kA 3、求d2’点的短路电流 3.1 求富兴变供给d2’点的短路电流 Ix″=Ij/X13=5.5/0.402=13.68kA ichx1=√2 *Kch*Ix″=√2 *1.8*13.68=34.82kA Ich=Ix″√1+2(Kch-1)2 =13.68*√1+2(1.8-1)2 =20.656kA 3.2 求沙县城关水电站供给d2’点的短路电流 将X14 换算到以发...
