1、 天津渤海职业技术学院毕业设计论文220kV 变电所电气部分一次系统设计专业：电力系统自动化技术姓名：李佳鱼学校：天津渤海职业技术学院天津渤海职业技术学院 电气工程系- 0 -设计计算书短路电流计算1、 计算电路图和等值电路图 TS90/26-3QFS30-2P-360/PL-4/ 1KM150KMII2V5Vd2d3X143X7891056349202ddKV.51系统阻抗标幺值：设：SJ=100MVAX1=X2=X3=0.2X4=X5=X6=(Ud100 )*(Sj/Se)=(14.1/100)*(100/240)=0.59X7=X8=X9=X10=Xd*”*(Sj/Se)=0.167*(
2、100/300/0.85)=0.0473X7=X8=X9=X10= ( Ud% / 100 )*(Sj/Se)=(14.6/100)*(100/360)=0.0406X15=X16=X* Sj / Up = 0.4*150*( 100 / 230 ) = 0.1134X17=X18=X* Sj / Up = 0.4*100*( 100 / 230 ) = 0.0756根据主变的选择 SFPSLO-240000 型变压器，可查出：天津渤海职业技术学院 电气工程系- 1 -UdI-II% =14.6、UdI-III% =6.2、UdII-III% =9.84X19=X22=1/200*( UdI-
3、II%+ UdI-III%- UdII-III%)*(Sj/Se)=1/200*(14.6+6.2-9.84)*(100/240)=0.0228X20=X23=1/200*( UdI-II%+ UdII-III%- UdI-III%)*(Sj/Se)=1/200*(14.6+9.84-6.2)*(100/240)=0.0379X20=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： d1X28675930KV2X25=（X1+X4）/3=
4、0.0863X26=（X7+X11）/4=0.02198X27=X15/2=0.0567X28=X17/2=0.0378X29=X25+ X27=0.143X30=X26+ X28=0.05978用个别法求短路电流 水电厂 S1:Xjss1= X29*( SN1/ Sj )=0.143 * ( 3*200/0.875/100 ) = 0.98 水电厂 H1:XjsH1= X30*( SN1/ Sj )=0.0598 *( 4*300/0.85/100 ) = 0.844查运算曲线 ：t=0”时天津渤海职业技术学院 电气工程系- 2 -I*S-1”=1.061I*H-1”=1.242IS-1”=
5、 ( I*S-1” * SNS-1)(3 * Uj )=1.061*( 3*200/0.875)(3 * 230 )=1.826KAIchS-1= IS-1”*1+2*(Kch-1) =1.826*1+2*(1.85-1) =2.855KAIH-1”= (I*H-1”* SNH-1)/(3 * Uj )=1.242*(4*300/0.85)/(3 * 230 )=4.402KAIchH-1= IH-1”*1+2 * (Kch-1) =4.402*1+2 * (1.85-1) =6.883KAI”= IS-1”+ IH-1”=1.826+4.402=6.288KAIch1= IchS-1+ Ic
6、hH-1=2.855+6.833=9.738KAt=2”时I*t=2s-1”=1.225I*t=2H-1”=1.36It=2s-1”= (I*t=2s-1”* SNS-1)/ (3 * Uj )=1.225*(3*200/0.875)/ (3 * 230 )=2.109KAIt=2H-1”=(I*t=2H-1”*SNH-1)/(3 * Uj )=1.36*(4*300/0.85)/( 3 * 230 )=4.8198KAIt=2”= It=2s-1”+ It=2H-1”=2.109+4.8198=6.928KAT=4”时I*t=4s-1”=1.225I*t=4H-1”=1.375It=4s-1
7、”= (I*t=4s-1”* SNS-1)/ (3 * Uj )=1.225*(3*200/0.875)/ (3 * 230 )=2.109KAIt=4H-1”=(I*t=4H-1”*SNH-1)/(3 * Uj )=1.375*(4*300/0.85)/( 3 * 230)=4.873KAIt=4”= It=4s-1”+ It=4H-1”=2.109+4.873=6.982KA天津渤海职业技术学院 电气工程系- 3 -、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*0.0303
8、5/0.0598=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 X30d2X3d231X2930d2X320129水电厂 S1:Xjss1= X32*( SN1/ Sj )=0.246 *( 3*200/0.875/100 ) = 1.687水电厂 H1:XjsH1= X33*( SN1/ Sj )= 0.103*( 4*300/0.85/100 ) = 1.454查运算曲线 ：t=0”时I*S-1”=0.616I*H-1”=0.71IS-1”= ( I*S-1” * SNS-
9、1)(3 * Uj )=0.616*( 3*200/0.875)(3 * 230 )=1.06KAIchS-1= IS-1”*1+2*(Kch-1) =1.06*1+2*(1.85-1) =1.657KAIH-1”= (I*H-1”* SNH-1)/(3 * Uj )天津渤海职业技术学院 电气工程系- 4 -=0.71*(4*300/0.85)/(3 * 230 )=2.516KAIchH-1= IH-1”*1+2 * (Kch-1) =2.516*1+2 * (1.85-1) =3.934KAI”= IS-1”+ IH-1”=1.06+2.516=3.576KAIch1= IchS-1+ I
10、chH-1=1.657+3.934=5.591KAt=2”时I*t=2s-1”=0.649I*t=2H-1”=0.74It=2s-1”= (I*t=2s-1”* SNS-1)/ (3 * Uj )=0.649*(3*200/0.875)/ (3 * 230 )=1.117KAIt=2H-1”=(I*t=2H-1”*SNH-1)/(3 * Uj )=0.74*(4*300/0.85)/( 3 * 230 )=2.623KAIt=2”= It=2s-1”+ It=2H-1”=1.117+2.623=3.74KAT=4”时I*t=4s-1”=0.649I*t=4H-1”=0.74It=4s-1”=
11、(I*t=4s-1”* SNS-1)/ (3 * Uj )=0.649*(3*200/0.875)/ (3 * 230 )=1.117KAIt=4H-1”=(I*t=4H-1”*SNH-1)/(3 * Uj )=0.74*(4*300/0.85)/( 3 * 230)=2.623KAIt=4”= It=4s-1”+ It=4H-1”=1.117+2.623=3.74KA、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+ X3
12、0*X34/ X29=0.0598+0.0129+0.0598*0.0129/0.143=0.078用个别法求短路电流天津渤海职业技术学院 电气工程系- 5 -水电厂 S1:Xjss1= X35*( SN1/ Sj )=0.187 *( 3*200/0.875/100 ) = 1.282水电厂 H1:XjsH1= X36*( SN1/ Sj )= 0.078*( 4*300/0.85/100 ) = 1.101查运算曲线 ：t=0”时I*S-1”=0.810I*H-1”=0.94IS-1”= ( I*S-1” * SNS-1)(3 * Uj )=0.810*( 3*200/0.875)(3 *
13、 230 )=1.394KAIchS-1= IS-1”*1+2*(Kch-1) =1.394*1+2*(1.85-1) =12.18KAIH-1”= (I*H-1”* SNH-1)/(3 * Uj )=0.94*(4*300/0.85)/(3 * 230 )=3.331KAIchH-1= IH-1”*1+2 * (Kch-1) =3.331*1+2 * (1.85-1) =5.21KAI”= IS-1”+ IH-1”=1.394+3.331=4.725KAIch1= IchS-1+ IchH-1=2.81+5.21=7.39KAt=2”时I*t=2s-1”=0.888I*t=2H-1”=1.0
14、11It=2s-1”= (I*t=2s-1”* SNS-1)/ (3 * Uj )=0.888*(3*200/0.875)/ (3 * 230 )=1.529KAIt=2H-1”=(I*t=2H-1”*SNH-1)/(3 * Uj )=1.011*(4*300/0.85)/( 3 * 230 )=3.583KAIt=2”= It=2s-1”+ It=2H-1”=1.529+3.583=5.112KAT=4”时I*t=4s-1”=0.888天津渤海职业技术学院 电气工程系- 6 -I*t=4H-1”=1.011It=4s-1”= (I*t=4s-1”* SNS-1)/ (3 * Uj )=0.8
15、88*(3*200/0.875)/ (3 * 230 )=1.529KAIt=4H-1”=(I*t=4H-1”*SNH-1)/(3 * Uj )=1.011*(4*300/0.85)/(3 * 230)=3.583KAIt=4”= It=4s-1”+ It=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Sn=240MVA最大工作电流：Imax =1.05* Sn/(3 * Un )=1.05*240/(1.732*220)=661A选 SW2-220 型断路器型号 额 定电 压额 定电 流额定断
16、开电流极限通过电流热稳定电流断 开容 量固有分闸时间SW2-220 220KV 1000A 15.8KA 59KA 21KA 5000MVA 0.05”假定主保护动作时间为 0.05”，后备保护 3.9”。则：断路计算时间：tk=0.05+0.39+0.05=4”短路点选主变压器与断路器之间，流经该断路器的短路电流为最大。1）热稳定校验;I”=6.228KA 、I2.0=6.928KA 、I4.0=6.56KAQP=(tK/12)*(I”+10Itk/2Itk )=(4/12)*(6.228+10*6.928+6.56)=187.26(KA).S由于 tK1 秒 故不计 QNP所以：QK= Q
17、P=187.26(KA).S天津渤海职业技术学院 电气工程系- 7 -It*t=21*4=1764(KA).SQKIt*t 热稳定满足。2）动稳定校验：ish=Ksh2 * I”=2.55* I”=2.55*6.228=15.8814KAies=59KA因为 ies ish ，所以动稳定满足。3） I”=6.228KAINbr=15.8KAish=15.8814KAINcl=59KA综上所述：选 SW2-220 型断路器隔离开关选 GW6-220G 型型 号 额定电压 额定电流 动稳定电流 热稳定电流GW6-220G 220KV 1000A 50KA 21KA(5 秒)1) 热稳定校验QK=187.26(KA).SIt*t=21*5=2205(KA).SQKIt*t 热稳定满足。2) 动稳定校验ish=15.8814KAies=50KA因为 ies ish ，所以动稳定满足。综上所述：选 GW6-220G 型隔离开关满足要求。2、 110KV 电压等级断路器Sn=240MVA最大工作电流：IMax=1.05*Sn/(3 * Un )=1.05*240/(1.732*110)=1323A
