电力系统分析-第三章例题

第三章电力系统潮流分布计算3-2已知图3-2所示输电线路始末端电压分别为248kV、220kV，末端有功功率负荷为220MW，无功功率负荷为165MVAR。试求始端功率因数。3-2解：62.26105.5220422yQ(MVAR)83.33105.5248421yQ(MVAR)38.13822062.261652202jjjS(MVA)求Z12中的功率损耗：21.194165.23183.55165.1138.13822083.55165.11)408(22038.138220122212jjjSjjS38.160165.23183.3321.194165.2311jjjS(MVA)8216.038.160165.231165.231cos223-8额定电压110kV的辐射形电网各段阻抗及负荷如图3-8所示。已知电源A的电压为121kV，求功率分布和各母线电压。（注：考虑功率损耗，可以不计电压降落的横分量U）。3-8解：设01100NCUU8+j40248kV220kV①②S1ˊS1习题解图3-2C1j5.5×10-4(S)C2ˊS2···40+j30MVA20+j30ΩSA·ˊSB·SB·〞10+j8MVAABC习题解图3-820+j40ΩP2=220MWQ2=165MVARU2U18+j40ΩC1j5.5×10-4SC2j5.5×10-4S习题图3-2A20+j40ΩB20+j30ΩC40+j30MVA10+j8MVA习题图3-8083.27545.32676.4338.2407.22207.30676.4338.2)4020(110407.22207.30407.22207.30953.7793.93040593.7793.9407.0271.0)810(407.0271.0)3020(110810222222jjjSSSjjSjjjSSSjjjSSSjjSABBAABBBBBCCBBC已知UA=121kV332.1412140083.2720545.32ABUkV668.106332.14121ABABUUUkV972.3668.10630)593.7(20)793.9(BCUkV64.110972.3668.106BCBCUUUkV3-13由A、B两端供电的电力网，其线路阻抗和负荷功率等如图3-13示。试求当A、B两端供电电压相等（即UA=UB）时，各段线路的输送功率是多少？（不计线路的功率损耗）3-13解：1、4支路合并，等值电路如图3-13a所示。510)63()126(63)1530(1020)63()126(126)1530()510()1530(10201530168)84)(1530()126)(1020(1020168)42)(1020()84)(1530()(42)63(324111411414jjjjjZZZSSjjjjjZZZSSjjjSSSjjjjjjSjjjjjjSjZBBaAabBA4+j82+j42+j4SASaSbSabSBab114习题解图3-13a·····Z2=4+j8ΩZ3=2+j4ΩZ4=3+j6ΩBSa=30+j15MVASb=20+j10MVAaAZ1=6+j12ΩUAbUb习题图3-13··输送功率分布如图3-13b所示。3-16发电厂C的输出功率、负荷D及E的负荷功率以及线路长度参数如图3-16所示。当0112CUkV，BAUU，并计及线路CE上的功率损耗，但不计其它线路上功率损耗时，求线路AD、DE、及BE上通过的功率。3-16解：965.910112501002224222RUQPPCCCCE93.1920112501002224222XUQPQCCCCE)530()201050100(2560)(201095.19965.9jjjjSSSSjjSCECEECE如图3-16所示)1428()3040()1612(2510921612251092105)63)(530()42)(3040()(1612)424(212)42)(530()63)(3040()(321211321332jjjSSSjjjSSjSSjjjjjjZZZZZSZSSjjjjjjZZZZSZZSSDDADEBEADEDEDBEEDAD则110+j5220+j10a310+j5b4UA30+j1520+j10UB习题解图3-13b20+j10SD=40+j30SADSDESBEZ1=2+j4Z1=1+j2Z1=2+j4AB〞SE=-(30+j5)习题解图3-16····DE·～～～AEDZ1=2+j4ΩZ2=1+j2ΩZ3=2+j4ΩSD=40+j30MVACBL3=10kmL2=5kmL1=10kmSE=60+j25MVAZ4=10+j20ΩL4=50kmSC=100+j50MVAUC=112kV·习题图3-16···3-18如图3-18所示，已知闭式网参数如下：421jZ632jZ843jZ101Lkm152Lkm203Lkm负荷参数：BS=10+j5MVACS=30+j15MVA电源参数：UA=110kV试求闭式网上的潮流分布及B点电压值（计算时，不计线路上的功率损耗）。3-18解：如图3-18a等值电路.56.1011.21)56.511.11()510(56.511.11)44.989.18()1530(44.989.184510)510(25)1530(10152010)510()1015)(1530(jjjSSSjjjSSSjjjjjSBCBABACCBCAc▽C点为功率分点。于是网络从C点打开，变成两个辐射网如图3-18b所示。3.015.0)84(11044.989.182223jjSˊA4+j8ΩAZ3CZ2BZ118.89+j9.44MVA11.11+j5.5610+j5ˊSB·SA·习题解图3-18bˊSA·BAZ3Z2Z14+j8Ω203+j6Ω152+j4Ω1030+j15SC·10+j5ˊA习题指导图3-18aC▽SB·～AUA=110kVL3L1CBSCSBL2习题图3-18··1+j2.5Ω1+j3Ω10km5+j3MVA8+j6MVA10km10kmA123习题图3-216365.101483.210765.00383.056.511.115100765.00383.0)63(0127558.0)63(11056.511.1174.904.1930.015.044.989.18222223jjjjSSSSjjjSjjjSSSBCBBACA82.1024.218218.102409.211853.00926.06365.101483.211853.00926.0)42(0463.0)42(1106365.101483.2112221jjjjSSSjjjSBA78.07769.011028.4348.42110482.10224.211AUQXPRU22.10978.01101UUUABkV3-21如图3-21所示电力系统，已知Z12、Z23、Z31均为1+j3Ω，UA=37kV，若不计线路上的功率损耗及电压降落的横分量，求功率分布及最低点电压。3-21解：⑴功率分布：913)46()57(46)11()35(11)57()68(5710101010)35(20)68(12131232121332313jjjSSSjjjSSSjjjSSSjjjSA3点为功率分点，电压最低。⑵电压分布：43.356104.0041.366104.0041.363517041.36959.037959.037259113131313111UUUUUUUUAAA1+j2.5Ω1+j3Ω10km5+j3MVA8+j6MVA10km10kmA123习题图3-21功率分布如图3-21所示。36.04113+j9MVA6+j4MVA10km5+j3MVA8+j6MVA1+j1MVA7+j5MVAA12337kV35.43习题解图3-21▽