系统如图所示,已知T1、T2额定容量100MVA,UK%=10.5,系统S1的短

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问题 系统如图所示,已知T1、T2额定容量100MVA,UK%=10.5,系统S1的短路容量为1000MVA,系统S2的短路容量为800MVA。当f点发生三相短路时,短路点的短路容量(MVA)及短路电流(kA)为(取SB=100MVA,线路电抗标幺值为x=0.05)(  )。A.859.0MVA,4.55kAB.844.9MVA,4.55kAC.859.0MVA,4.24kAD.847.5MVA,4.26kA

选项 A.859.0MVA,4.55kA
B.844.9MVA,4.55kA
C.859.0MVA,4.24kA
D.847.5MVA,4.26kA

答案 D

解析 输电线路各参数元件计算如下:系统S1电抗标幺值为:X*S1=SB/SS1=100/1000=0.1。系统S2电抗标幺值为:X*S2=SB/SS2=100/800=0.125。变压器T1、T2电抗标幺值为:X*T1=X*T2=(Uk%/100)×SB/SrT=(10.5/100)×(100/100)=0.105。短路点等效总电抗为:X*∑=(0.1+0.105)∥(0.125+0.105+0.05)=0.118。短路电流有名值为:短路容量为:
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