某110kv变电站短路电流计算书一、短路电流计算取基准容量S j=100MV A，略去“*”，U j=115KV，I j=0.502A富兴变：地区电网电抗X 1=S j/S dx=I j/I dx=0.502/15.94=0.0315km线路电抗X2=X*L*(S j/Up2)=0.4*5*(100/1152)=0.015发电机电抗X3=(Xd’’%/100)*(S j/Seb)=(24.6/100)*(100/48)=0.51216km线路电抗X4=X*L*(S j/Up2)=0.4*16*(100/1152)=0.0495.6km线路电抗X5=X*L*(S j/Up2)=0.4*5.6*(100/1152)=0.01731.5MV A变压器电抗X6=X7=(Ud%/100)*(S j/Seb)=(10.5/100)*(100/31.5)=0.33350MV A变压器电抗X=(Ud%/100)*(Sj/Seb)=0.272X8=X3+X4+X5=0.578 X9=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1.1求富兴变供给d1’点(即d1点)的短路电流I x″=I j/(X1+X2)=0.502/(0.031+0.015)=10.913kAS x″=S j/(X1+X2)=100/(0.031+0.015)≈2173.913MV Ai chx1=√2 *K ch*I x″=√2 *1.8*10.913=27.776kAI ch=I x″√1+2(K ch-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X js=X8*S rg/S j=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:I ef=S rg/( 3U p)=48/(1.732x115)=0.241求得:I f’’= I*’’*I ef=3.993x0.241=0.962kAI f0.2’’= I*0.2’’*I ef=3.096x0.241=0.746kAI f0.4’’= I*4’’*I ef=3.043x0.241=0.732kAi chf=√2 *K ch*I f″=√2 *1.8*0.962=2.448kA1.3 求得d1’点的短路电流I x″=10.913+0.962=11.875kAi ch=27.776+2.448=30.224kAI ch=11.875√1+2*(1.8-1)2 =17.93kA2、求d2点的短路电流I x″=I j/(X1+X2+X6)=5.50/(0.031+0.015+0.333)=14.512kAi chx=2* K ch*I x2″=2*1.8*14.512=36.936kAI ch=I x″√1+2(K ch-1)2 =14.512*√1+2(1.8-1)2=21.913kA3、求d2’点的短路电流3.1求富兴变供给d2’点的短路电流I x″=I j/X13=5.5/0.402=13.68kAi chx1=√2 *K ch*I x″=√2 *1.8*13.68=34.82kAI ch=I x″√1+2(K ch-1)2 =13.68*√1+2(1.8-1)2=20.656kA3.2 求沙县城关水电站供给d2’点的短路电流将X14换算到以发电机容量为基准容量时的标幺值X js=X14*S rg/S j=5.08*48/100=2.438 查表得I *’’=0.425 I*0.2=0.431 I*4=0.431 换算到115kV下发电机的额定电流:I ef=S rg/( 3U p)=48/(1.732x10.5)=2.64 求得:I f’’= I*’’*I ef=0.425x2.64=1.122kAI f0.2’’= I*0.2’’*I ef=0.431x2.64=1.138I f4’’= I4’’*I ef=0.431x2.64=1.138kA3.3 求得d2’点的短路电流I x″=13.68+1.122=14.802kAi ch=1.414x1.8x14.802=37.674kAI ch=14.802√1+2*(1.8-1)2 =22.35kA同理：求得终期d2点的短路电流I x2″= I j/(X1+X2+X6)=5.50/(0.031+0.015+0.272)=17.3kAi chx= √2*1.8*17.3≈44kAI ch=I x″*√1+2(K ch-1)2=17.3*√1+1.28 =26.122kA求得终期d2’点的短路电流I x″=16.32+1.344=17.664kAi ch=1.414x1.8x17.664=44.96kAI ch= I x″√1+2*(1.8-1)2 =26.655kA二、10KV母线选择(铜13720N/cm2，铝6860N/cm2)1、据最大长期工作电流选择TMY-2(100*10)的母线水平放置，环境温度为２５℃时，载流量Ｉ=3248*0.9=2923A>1.05*2749=2886A (系数取0.9)2、检验热稳定√Q/C=√I2t/c=√17.6642*1.5/171=126.5mm2<(2*1000)mm23、检验动稳定短路电动力 f=17.248*(l/a)*ich2*B*10-2=17.248*[(1.3*102)/(0.25*102)]*44.962*10-2=1809.76N产生应力σx-x=M/W=fl/10w=(1809.76*130)/(10*33.3)=707N/cm2<13720N/cm2[ 若是单片矩形导体的机械应力σ= M/W=fl/10w=(1809.76*130)/(10*16.7)=1408.8 N/cm2<13720N/cm2 ] 求得绝缘子最大允许跨距l=(7.614/ich)*√aωσ=(7.614/44.96)*√40*33.3*13720≈754cm求导体片间作用力σx=f x2*l c2/hb2其中fx =9.8*kx*(ich2/b)*10-2=9.8*0.12*(44.962/1)*10-2=23.77N导体片间临界跨距 lef =1.77* *b*4√h/fx=1.77*65*4√10/23.77=92cm本工程取40cm则σx=(23.772*402)/(102*1)=9040.2N/cm2<铜 13720N/cm2σ=σx-x + σx =707+9040.2=9747.2 N/cm2<铜 13720N/cm2按机械共振条件确定最大允许跨距(共振35-155HZ) l2=(112*r i*ε)/f=(112*2.89*11400)/155=23800=>l=154cm本工程取l=1300mm三、支柱绝缘子选择手册P25510KV选ZS-35/8 ( 8*0.6=4.8kN)Fc=0.173*(l c/a)*i ch2=0.173*(1.3/0.4)*44.962=1135.9N<4.8KN四、穿墙套管选择CWWL-10 3150/2 ，额定弯曲破坏负荷8KN动稳定检验8.62*(0.6+1)/0.4*44.962*10-2=697N<0.6*8=4.8kN五、接地网110KV为有效接地系统，接地电阻要求≤0.5Ω（1）现有接地装置计算土壤电阻率ρ=φρ0令ρ=3*104*1.2Ω.cm则ρ=360Ω.cm设人工接地体,采用垂直接地体与水平接地体组成的复式接地装置的电阻原地网Rt =1/(n*ηc/Rc+ηs/Rs)其中Rc=[ρ/(2πl)]*ln*(4L/0.84b)=[3.6*104/(2π*250)]*ln[(4*250)/(0.84*5)]=23*5.5=126.5n=100根Rs=[ρ/(2πl)] *ln(8L2/πbh)=360/(2π*800)* ln[(8*8002)/(π*0.04*0.8)]=1.24查表ηc =0.58，ηs=0.25则R=1/(100*0.58/126.5+0.25/1.24) ≈1.5Ωt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h x=10m，则r x=(1.5h-2h x)p=(1.5x30-2x10)=25h0AB=h-D/7p=30-60/7=21.43m则A、B两针间高度为h x=10m水平面上保护范围的一侧b xAB=1.5h0-2h x=1.5*21.43-2*10=12.1 同理h0BC=h-D/7=30-50/7=22.85m则b XBC=1.5h0-2h x=1.5*22.85-2*10=14.28mh0CD=h-D/7=30-50/7=22.85m则b XCD=1.5*22.85-2*10=14.28mh0AD=h-D/7=30-22.5/7=26.79m则b XCD=1.5*26.79-2*10=25.19m110kV变电站升压改造计算书（电气部分）。