一、短路電流計算取基準容量Sjj=100MMVA，略去“*”，Uj=115KKV，Ij=0.5002A富興變：地區(qū)電電網(wǎng)電抗X1=Sj/Sdx=Ij/Idx=0.5002/15..94=0..0315km線路電抗抗X2=X*L**(Sj/Up2)=0..4*5*((100/11152)=0.0015發(fā)電機電抗X3=(Xd’’%/1000)*(Sjj/Seb))=(224.6/1100)*((100/448)=0..51216km線路電電抗X4=X*L**(Sj/Up2)=00.4*166*(1000/11522)=0.00495.6km線路路電抗X5=X*L**(Sj/Up2)=00.4*5..6*(1000/11552)=0.001731.5MVAA變壓器電抗抗X6=X7=(Ud%%/100))*(Sj/Seb))=(10..5/1000)*(1000/31..5)=0..33350MVA變壓壓器電抗XX=(Ud%%/100))*(Sj//Seb)==0.2722X8=X3+X4+X5=0.5778X9=X1+X2=0.0446X10=(X8*X9)/(X8+X9)X11=X10+X6=0.0446地區(qū)電網(wǎng)支路的的分布系數(shù)C1=X10/X9=0.9335發(fā)電機支路的分分布系數(shù)C2=X10/X8=0.0744則X13=XX11/C1=0.3776/0.9935=0..402X114=X11/C2=0.3776/0.0074=5..081、求d1’點點的短路電流流求富興變供給dd1’點(即d1點)的短路電流流Ix″=Ijj/(X1+X2)=0.5502/(00.031++0.0155)=10..913kAASx″=Sj//(X1+X2)=1000/(0.0031+0..015)≈2173.9913MVAAichx11=√2*Kchh*Ix″=√2*1..8*10..913=227.7766kAIch=Ix″″√1+2((Kch-1)2=10.9913*√1+2(11.8-1))2=10.9133*1.511=16.4479kA1.2求沙縣城關(guān)關(guān)水電站供給給d1’點的短路電電流將發(fā)電機支路的的等值電抗換換算到以發(fā)電電機容量為基準容量量時的標幺值值Xjs=X8**Srg/Sj=0.5778*48//100=00.277查表得I**’’=3.9993I*0.2=3..096I*4=3.0043換算到115kkV下發(fā)電機機的額定電流流:Ief=Srg/(33Up)=48//(1.7332x1155)=0.2241求得:If’’’=I*’’*Ief=3.9993x0..241=00.962kkAIf00.2’’=I*0..2’’*Ief=3.0096x0..241=00.746kkAIf00.4’’=I*4’’*Ief=3.0043x0..241=00.732kkAicchf=√2*Kchh*If″=√2*1..8*0.9962=2..448kAA1..3求得d1’點的短路電電流Ix″=10..913+00.962==11.8775kAich=27..776+22.448==30.2224kAIch=111.875√√1+2*((1.8-11)2=17..93kA2、求d2點的的短路電流Ix″=Ij//(X1+X2+X6)=5.550/(0..031+00.015++0.3333)=14.5122kAichxx=2*KKch*Ix2″=2*1..8*14..512=336.9366kAIch=Ix″√1+2((Kch-1)2=14.5512*√1+2(11.8-1))2=21.9133kA3、求d2’點的短路電電流3.1求富興變供供給d2’點的短路電電流Ix″=Ijj/X13=5.55/0.4002=13..68kAichx1==√2*Kchh*Ix″=√2*1..8*13..68=344.82kAAIch=Ix″″√1+2((Kch-1)2=13.668*√1+2(11.8-1))2=20.6566kA3.2求沙縣城關(guān)關(guān)水電站供給給d2’點的短路電電流將X14換算到到以發(fā)電機容容量為基準容容量時的標幺值Xjss=X14*Srg/Sj=5.08*48//100=22.438查表得I**’’=0.4255I*0.2=0.4311I*4=0.4311換算到1155kV下發(fā)電電機的額定電電流:Ief=Srg/(33Up)=48//(1.7332x10..5)=2..64求得:If’’’=I*’’*Ief=0.4425x2..64=1..122kAAIf00.2’’=I*0..2’’*Ief=0.4431x2..64=1..138If44’’=I4’’*Ief=0.4431x2..64=1..138kAA3.3求得d2’點的短路電電流Ix″=13..68+1..122=114.8022kAich=1.4414x1..8x14..802=337.6744kAIch=114.8022√1+2*((1.8-11)2=22..35kA同理：求得得終期d2點的短路路電流Ix2″=IIj/(X1+X2+X6)=5.50/((0.0311+0.0115+0.2272)=117.3kAAichxx=√2*1.88*17.33≈44kAIch=Ix″″*√1+2(KKch-1)2=17.3*√√1+1.228=226.1222kA求得終期d2’點的短路電電流Ix″=16..32+1..344=117.6644kAich=1.4414x1..8x17..664=444.96kkAIch=Ixx″√1+2**(1.8--1)2=26..655kAA二、10KV母母線選擇(銅137220N/cmm2，鋁68600N/cm2)據(jù)最大長期工作作電流選擇TMY-22(100*110)的母線線水平放置，環(huán)環(huán)境溫度為２２５℃時，載流量Ｉ=32248*0..9=29223A>1..05*27749=28886A(系數(shù)取0.9)檢驗熱穩(wěn)定√Q/C=√II2t/c=√17.66642*1.5//171=1126.5mmm2<(2*11000)mmm2檢驗動穩(wěn)定短路電動力f=177.248**(l/a))*ich2*B*100-2=177.248**[(1.33*102)/(0..25*1002)]*444.962*10-2=18009.76NN產(chǎn)生應力σx--x=M/WW=fl/110w=(11809.776*1300)/(100*33.33)=7077N/cm22<137220N/cmm2[若是單單片矩形導體體的機械應力力σ=M/WW=fl/110w=(1809..76*1300)/(100*16.77)=14008.8NN/cm2<137220N/cmm2]]求得絕緣子最大大允許跨距l(xiāng)=(7.6614/icch)*√aωσ=(7.6614/444.96)**√40*333.3*133720≈754cmm求導體片間作用用力σx=fx2*lc2/hb2其中fx=9..8*kx*(ich2/b)*10-2=9.88*0.122*(44.9662/1)*10-2=23..77N導體片間臨界跨跨距l(xiāng)ef=1.777***b*4√h/fx=1.777*65*44√10/233.77=992cm本工程取40ccm則σx=(233.772*402)/(1002*1)=99040.22N/cm22<銅137200N/cm22σ=σx-x+σx=7077+90400.2=97747.2N/cm22<銅137200N/cm22按機械共振條件件確定最大允允許跨距(共振35-1555HZ)l2=(1122*ri*ε)/f=((112*22.89*111400))/155==238000=>l=1154cm本工程取l==1300mmm三、支柱絕緣子子選擇手冊P255100KV選ZS-355/8(8*0..6=4.88kN)Fc==0.1733*(lc/a)*iich2=0.1773*(1..3/0.44)*44..962=11355.9N<44.8KN四、穿墻套管選選擇CWWL--1031150/2，額定彎曲曲破壞負荷8KN動穩(wěn)定檢驗8.662*(0..6+1)//0.4*444.962*10-2=697N<0..6*8=44.8kN五、接地網(wǎng)1110KV為有效接地地系統(tǒng)，接地地電阻要求≤0.5Ω（1）現(xiàn)現(xiàn)有接地裝置置計算土壤電阻率ρ=φρ0令ρ0=3*1004*1.2Ω.cm則ρ=360ΩΩ.cm設人工接接地體,采用垂直接接地體與水平平接地體組成成的復式接地地裝置的電阻阻原地網(wǎng)Rt=1/(nn*ηc/Rc+ηs/Rs)其中Rc=[ρ/(2πl(wèi))]*ln*((4L/0..84b)=[3.6*1104/(2π*250))]*ln[[(4*2550)/(00.84*55)]=223*5.55=126..5n=1100根Rs==[ρ/(2πl(wèi))]**ln(8LL2/πbh)==360/((2π*800))*ln[[(8*80002)/(π*0.044*0.8))]=1.224查表ηc=0..58，ηs=0.25則Rt=1/(1000*0.558/1266.5+0..25/1..24)≈1.5Ω六、現(xiàn)有避雷針針保護范圍計計算現(xiàn)下洋變有四支支等高避雷針針（相對站內(nèi)內(nèi)地面標高），位位置詳見B9992C-DD0101--03。令hx=100m，則rx=(1.55h-2hxx)p=(11.5x300-2x100)=25h0ABB=h-D//7p=300-60/77=21.443m則A、B兩針間間高度為hx=10m水平平面上保護范范圍的一側(cè)bxABB=1.5hh0-2hx