220kV变电所电气部分一次系统设计论文.doc

220kV变电所电气部分一次系统设计 设计计算书 专业：电力系统及其自动化 姓名：宋铁成 学校：吉林电子信息职业技术学院 设计计算书 短路电流计算 计算电路图和等值电路图 系统阻抗标幺值：设：SJ=100MVA X1=X2=X3=0.2 X4=X5=X6=(Ud／100 )*(Sj/Se)=(14.1/100)*(100/240)=0.59 X7=X8=X9=X10=Xd*”*(Sj/Se)=0.167*(100/300/0.85)=0.0473 X7=X8=X9=X10= ( Ud% / 100 )*(Sj/Se)=(14.6/100)*(100/360) =0.0406 X15=X16=X* Sj / Up2 = 0.4*150*( 100 / 2302 ) = 0.1134 X17=X18=X* Sj / Up2 = 0.4*100*( 100 / 2302 ) = 0.0756 根据主变的选择SFPSLO-240000型变压器，可查出： UdI-II% =14.6、UdI-III% =6.2、UdII-III% =9.84 X19=X22=1/200*( UdI-II%+ UdI-III%- UdII-III%)*(Sj/Se) =1/200*(14.6+6.2-9.84)*(100/240)=0.0228 X20=X23=1/200*( UdI-II%+ UdII-III%- UdI-III%)*(Sj/Se) =1/200*(14.6+9.84-6.2)*(100/240)=0.0379 X20=X23=1/200*( UdI-III%+ UdII-III%- UdI-II%)*(Sj/Se) =1/200*(6.2+9.84-14.6)*(100/240)=0.003 (1)、d1点短路电流的计算： X25=（X1+X4）/3=0.0863 X26=（X7+X11）/4=0.02198 X27=X15/2=0.0567 X28=X17/2=0.0378 X29=X25+ X27=0.143 X30=X26+ X28=0.05978 用个别法求短路电流 水电厂 S–1: Xjss–1= X29*( SN∑1/ Sj )=0.143 * ( 3*200/0.875/100 ) = 0.98 水电厂 H–1: XjsH–1= X30*( SN∑1/ Sj )=0.0598 *( 4*300/0.85/100 ) = 0.844 查运算曲线 ： t=0”时 I*S-1”=1.061 I*H-1”=1.242 IS-1”= ( I*S-1” * SNS-1)／(√3 * Uj ) =1.061*( 3*200/0.875)／(√3 * 230 )=1.826KA IchS-1= IS-1”*√[1+2*(Kch-1) 2] =1.826*√[1+2*(1.85-1) 2]=2.855KA IH-1”= (I*H-1”* SNH-1)/(√3 * Uj ) =1.242*(4*300/0.85)/(√3 * 230 )=4.402KA IchH-1= IH-1”*√[1+2 * (Kch-1) 2] =4.402*√[1+2 * (1.85-1) 2]=6.883KA I”= IS-1”+ IH-1”=1.826+4.402=6.288KA Ich1= IchS-1+ IchH-1=2.855+6.833=9.738KA t=2”时 I*t=2s-1”=1.225 I*t=2H-1”=1.36 It=2s-1”= (I*t=2s-1”* SNS-1)/ (√3 * Uj ) =1.225*(3*200/0.875)/ (√3 * 230 )=2.109KA It=2H-1”=(I*t=2H-1”*SNH-1)/(√3 * Uj ) =1.36*(4*300/0.85)/( √3 * 230 )=4.8198KA It=2”= It=2s-1”+ It=2H-1”=2.109+4.8198=6.928KA T=4”时 I*t=4s-1”=1.225 I*t=4H-1”=1.375 It=4s-1”= (I*t=4s-1”* SNS-1)/ (√3 * Uj ) =1.225*(3*200/0.875)/ (√3 * 230 )=2.109KA It=4H-1”=(I*t=4H-1”*SNH-1)/(√3 * Uj ) =1.375*(4*300/0.85)/( √3 * 230)=4.873KA It=4”= It=4s-1”+ It=4H-1”=2.109+4.873=6.982KA ⑵、d2点短路电流的计算： X31=（X19+X20）/2=0.03035 X32=X29+X31+ X29*X31/ X30 =0.14