华北电力大学电力系统分析14年真题及答案

j0.125j0.105j0.0452)I(dkj0.105j0.125j0.105U(2)

负序网：

X(2)?[1/2?(XG(2)?XT1(2))?XL(2)]//XT3(2)?[1/2?(0.125?0.105)?0.045]//0.105?0.063j0.105j0.1351,1)I(dk

j0.105U(0)

零序网：X(0)?(XT1(0)?XL(0))//XT3(0)?(j0.105?j0.135)//j0.105?0.073

Ifa(1)?Uf0jX(1)?jX(2)?X(0)X(2)?X(0)X(0)X(2)?X(0)X(2)X(2)?X(0)1aa2?1??j10.31

0.063?0.073j0.063?j0.063?0.0730.063?j4.770.073?0.0630.073?j5.540.073?0.063Ifa(2)??Ifa(1)Ifa(0)??Ifa(1)?Ifa??1?I???a2?fb????Ifc????a??j10.31???j10.31?

1??Ifa(1)???21??Ifa(2)??Ifb?a?Ifa(1)?a?Ifa(2)?Ifa(0)???1????Ifa(0)??

1313Ifb?(??j)?(?j10.31)?(??j)?(j4.77)?j5.54??13.06?j8.312222?Ifb?Ifc?15.48

（2） 变压器T2中性点三相对称，因此正序、负序电压为0，只有零序电压；因此在

零序网中求得：

UT2(0)?Ifa(0)?5.54?XT3(0)XT3(0)?XT1(0)?XL:(0)?XT1(0)?U'B

0.105115?0.105??11.75(kV)0.105?0.135?0.0153（3） 由于中性线上只流过零序电流，因此在零序网中求得：

IT3?3?Ifa(0)??5.54?

XT3(0)XT3(0)?XT1(0)?XL:(0)?IB

0.10515??0.13(kA)0.105?0.135?0.0151153