东北电力学院毕业设计论文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：学校：东北电力学院设计计算书短路电流计算1、计算电路图和等值电路图TS900/296-32QFS300-2SSP-360/220 SSPSL-240/220100KM150KMI II III IIIIII230KV115KVKVKVd1d2d3X1 X4X2X3X7X8X9X10 X5X6X11X12X13X14X15X19X20X16X17X18X22X23d1d2d3230KV10.5KV115KV X21X24系统阻抗标幺值：设：SJ=100MVAX1=X2=X3=0.2X4=X5=X6=(Ud／100 )*(S j/S e)=(14.1/100)*(100/240)=0.59X7=X8=X9=X10=X d*”*(S j/S e)=0.167*(100/300/0.85)=0.0473X7=X8=X9=X10= ( Ud% / 100 )*(S j/S e)=(14.6/100)*(100/360) =0.0406X15=X16=X* S j / U p²= 0.4*150*( 100 / 230²) = 0.1134X17=X18=X* S j / U p²= 0.4*100*( 100 / 230²) = 0.0756根据主变的选择SFPSLO-240000型变压器，可查出： U dI-II % =14.6、U dI-III % =6.2、U dII-III % =9.84 X 19=X 22=1/200*( U dI-II %+ U dI-III %- U dII-III %)*(S j /S e )=1/200*(14.6+6.2-9.84)*(100/240)=0.0228X 20=X 23=1/200*( U dI-II %+ U dII-III %- U dI-III %)*(S j /S e )=1/200*(14.6+9.84-6.2)*(100/240)=0.0379X 20=X 23=1/200*( U dI-III %+ U dII-III %- U dI-II %)*(S j /S e )=1/200*(6.2+9.84-14.6)*(100/240)=0.003(1)、d 1点短路电流的计算：d1X28X26X27X25X29X30d1230KV230KVX 25=（X 1+X 4）/3=0.0863 X 26=（X 7+X 11）/4=0.02198 X 27=X 15/2=0.0567 X 28=X 17/2=0.0378 X 29=X 25+ X 27=0.143 X 30=X 26+ X 28=0.05978 用个别法求短路电流 ① 水电厂 S –1:X jss –1= X 29*( S N ∑1/ S j )=0.143 * ( 3*200/0.875/100 ) = 0.98②水电厂 H–1:X js H–1= X30*( S N∑1/ S j )=0.0598 *( 4*300/0.85/100 ) = 0.844 查运算曲线：t=0”时I*S-1”=1.061I*H-1”=1.242I S-1”= ( I*S-1” * S NS-1)／(√3 * U j )=1.061*( 3*200/0.875)／(√3 * 230)=1.826KAI ch S-1= I S-1”*√[1+2*(K ch-1)²]=1.826*√[1+2*(1.85-1)²]=2.855KAI H-1”= (I*H-1”* S NH-1)/(√3 * U j )=1.242*(4*300/0.85)/(√3 * 230 )=4.402KAI ch H-1= I H-1”*√[1+2 * (K ch-1)²]=4.402*√[1+2 * (1.85-1)²]=6.883KAI”= I S-1”+ I H-1”=1.826+4.402=6.288KAI ch1= I ch S-1+ I ch H-1=2.855+6.833=9.738KAt=2”时I*t=2s-1”=1.225I*t=2H-1”=1.36I t=2s-1”= (I*t=2s-1”* S NS-1)/ (√3 * U j )=1.225*(3*200/0.875)/ (√3 * 230)=2.109KAI t=2H-1”=(I*t=2H-1”*S NH-1)/(√3 * U j )=1.36*(4*300/0.85)/( √3 * 230)=4.8198KA I t=2”= I t=2s-1”+ I t=2H-1”=2.109+4.8198=6.928KAT=4”时I*t=4s-1”=1.225I*t=4H-1”=1.375I t=4s-1”= (I*t=4s-1”* S NS-1)/ (√3 * U j )=1.225*(3*200/0.875)/ (√3 * 230)=2.109KA I t=4H-1”=(I*t=4H-1”*S NH-1)/(√3 * U j )=1.375*(4*300/0.85)/( √3 * 230)=4.873KAI t=4”= I t=4s-1”+ I t=4H-1”=2.109+4.873=6.982KA⑵、d2点短路电流的计算：X31=（X19+X20）/2=0.03035X32=X29+X31+ X29*X31/ X30=0.143+0.03035+0.143*0.03035/0.0598=0.246X33=X30+X31+ X30*X31/ X29=0.0598+0.03035+0.0598*0.03035/0.143=0.103用个别法求短路电流d2d2①水电厂 S–1:X jss–1= X32*( S N∑1/ S j )=0.246 *( 3*200/0.875/100 ) = 1.687 ②水电厂 H–1:X js H–1= X33*( S N∑1/ S j )= 0.103*( 4*300/0.85/100 ) = 1.454 查运算曲线：t=0”时I*S-1”=0.616I*H-1”=0.71I S-1”= ( I*S-1” * S NS-1)／(√3 * U j )=0.616*( 3*200/0.875)／(√3 * 230)=1.06KAI ch S-1= I S-1”*√[1+2*(K ch-1)²]=1.06*√[1+2*(1.85-1)²]=1.657KAI H-1”= (I*H-1”* S NH-1)/(√3 * U j )=0.71*(4*300/0.85)/(√3 * 230 )=2.516KAI ch H-1= I H-1”*√[1+2 * (K ch-1)²]=2.516*√[1+2 * (1.85-1)²]=3.934KAI”= I S-1”+ I H-1”=1.06+2.516=3.576KAI ch1= I ch S-1+ I ch H-1=1.657+3.934=5.591KAt=2”时I*t=2s-1”=0.649I*t=2H-1”=0.74I t=2s-1”= (I*t=2s-1”* S NS-1)/ (√3 * U j )=0.649*(3*200/0.875)/ (√3 * 230)=1.117KA I t=2H-1”=(I*t=2H-1”*S NH-1)/(√3 * U j )=0.74*(4*300/0.85)/( √3 * 230)=2.623KAI t=2”= I t=2s-1”+ I t=2H-1”=1.117+2.623=3.74KAT=4”时I*t=4s-1”=0.649I*t=4H-1”=0.74I t=4s-1”= (I*t=4s-1”* S NS-1)/ (√3 * U j )=0.649*(3*200/0.875)/ (√3 * 230)=1.117KA I t=4H-1”=(I*t=4H-1”*S NH-1)/(√3 * U j )=0.74*(4*300/0.85)/( √3 * 230)=2.623KAI t=4”= I t=4s-1”+ I t=4H-1”=1.117+2.623=3.74KA⑶、d3点短路电流的计算：X34=（X19+X21）/2=0.0129X35=X29+X34+ X29*X34/ X30=0.143+0.0129+0.143*0.0129/0.0598=0.187X36=X30+X34+ X30*X34/ X29=0.0598+0.0129+0.0598*0.0129/0.143=0.078用个别法求短路电流①水电厂 S–1:X jss–1= X35*( S N∑1/ S j )=0.187 *( 3*200/0.875/100 ) = 1.282 ②水电厂 H–1:X js H–1= X36*( S N∑1/ S j )= 0.078*( 4*300/0.85/100 ) = 1.101 查运算曲线：t=0”时I*S-1”=0.810I*H-1”=0.94I S-1”= ( I*S-1” * S NS-1)／(√3 * U j )=0.810*( 3*200/0.875)／(√3 * 230)=1.394KAI ch S-1= I S-1”*√[1+2*(K ch-1)²]=1.394*√[1+2*(1.85-1)²]=12.18KAI H-1”= (I*H-1”* S NH-1)/(√3 * U j )=0.94*(4*300/0.85)/(√3 * 230 )=3.331KAI ch H-1= I H-1”*√[1+2 * (K ch-1)²]=3.331*√[1+2 * (1.85-1)²]=5.21KAI”= I S-1”+ I H-1”=1.394+3.331=4.725KAI ch1= I ch S-1+ I ch H-1=2.81+5.21=7.39KAt=2”时I*t=2s-1”=0.888I*t=2H-1”=1.011I t=2s-1”= (I*t=2s-1”* S NS-1)/ (√3 * U j )=0.888*(3*200/0.875)/ (√3 * 230)=1.529KA I t=2H-1”=(I*t=2H-1”*S NH-1)/(√3 * U j )=1.011*(4*300/0.85)/( √3 * 230)=3.583KA I t=2”= I t=2s-1”+ I t=2H-1”=1.529+3.583=5.112KAT=4”时I*t=4s-1”=0.888I*t=4H-1”=1.011I t=4s-1”= (I*t=4s-1”* S NS-1)/ (√3 * U j )=0.888*(3*200/0.875)/ (√3 * 230)=1.529KA I t=4H-1”=(I*t=4H-1”*S NH-1)/(√3 * U j )=1.011*(4*300/0.85)/(√3 * 230)=3.583KAI t=4”= I t=4s-1”+ I t=4H-1”=1.529+3.583=5.112KA电气设备的选择与校验一、断路器的选择与校验，隔离开关的选择与校验1、220KV电压等级断路器S n=240MVA最大工作电流：I max =1.05* S n/(√3 * U n )=1.05*240/(1.732*220)=661A选SW2-220型断路器假定主保护动作时间为0.05”，后备保护3.9”。