某110kv变电站短路电流计算书

                            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换算到以发电机容量为基准容量时的     

                                 标幺值 Xjs=X14*Srg/Sj=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:

                                      Ief=Srg/( 3Up)=48/(1.732x10.5)=2.

                              求得:If’’= I*’’*Ief=0.425x2.=1.122kA

                                  If0.2’’= I*0.2’’*Ief=0.431x2.=1.138

                                  If4’’= I4’’*Ief=0.431x2.=1.138kA

                            3.3 求得d2’点的短路电流 

                               Ix″=13.68+1.122=14.802kA

                               ich=1.414x1.8x14.802=37.674kA

                               Ich=14.802√1+2*(1.8-1)2 =22.35kA

                            同理：求得终期d2点的短路电流

                              Ix2″= Ij/(X1+X2+X6)

                                  =5.50/(0.031+0.015+0.272)=17.3kA

                              ichx= √2*1.8*17.3≈44kA

                              Ich=Ix″*√1+2(Kch-1)2  

                                                       =17.3*√1+1.28  =26.122kA

                                  求得终期d2’点的短路电流

                               Ix″=16.32+1.344=17.6kA

                               ich=1.414x1.8x17.6=44.96kA

                               Ich= Ix″√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=√17.62*1.5/171=126.5mm2<(2*1000)mm2

3、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=fx2*lc2/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/cm2

  σ=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*ri*ε)/f=(112*2.*11400)/155=23800=>l=154cm

       本工程取l=1300mm

三、支柱绝缘子选择   手册P255

     10KV选ZS-35/8    ( 8*0.6=4.8kN)

    Fc=0.173*(lc/a)*ich2=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ρ0=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=126.5

                   n=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)=25

   h0AB=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=1.5h0-2hx=1.5*22.85-2*10=14.28m

       h0CD=h-D/7=30-50/7=22.85m

       则 bXCD=1.5*22.85-2*10=14.28m

       h0AD=h-D/7=30-22.5/7=26.79m

       则 bXCD=1.5*26.79-2*10=25.19m

110kV变电站升压改造

计算书

（电气部分）