电力系统分析试卷A答案

GР ''E=1.05Р X''d=0.26Р V=10.5kVS=31.5MVA10.5/121kVVs%=10.5x1=0.4?/kmS=15MVA80km110/6.6kVVs%=10.5XR=5%0.3kA6kVx2=0.08?2.5kmT1IIT2LIIIk(3)Р - 3 -Р 解:Р SB?100MVA,VB1?10.5kV,VB2?115kV,VB3?6.3kV (1分)Р IB1?Р ??5.5kA (1分)Р ??0.5021kA (1分)Р ??9.1646kA (1分) IB2?Р IB3?Р 各元件标幺值电抗如下: 发电机XSB1?X''Р dР S?0.26?100Р ?0.87 N30Р 变压器XSB2?Р Vs%100S?0.105?100Р ?0.33 N31.5线路XSB3?x1lР V2?0.4?80?100Р 2?0.242 B2115Р 变压器Xs%SB4?Р V100S?0.105?100Р ?0.7 N15Р 电抗器X5?Р ?0.05100Р 6.3Р 2Р ?1.4547 电缆XSB6?x2lР V2?0.08?2.5?100Р ?0.481 B36.32Р 电源电动势标幺值E?E''Р ?1.05 短路点的总电抗X??0.87?0.33?0.242?0.7?1.4547?0.481?4.0777 短路电流标幺值I*?Р EX?1.05Р 4.0777Р ?0.2575 ?所以,发电机电流I?0.2575IB1?0.2575?5.5?1.41625kA 短路点电流Ik?0.2575IB3?0.2575?9.1646?2.36kA - 4 -Р (1分) 1分)Р 1分) 1分)Р 1分) (1分) (1分) (1分) (1分) (1分) (1分) ( ( ( (