'Sc ?Sc??PT?j?QT?20?j15?0.18?j2.19MVA
?20.18?j17.19MVA'''S?S?Sb?20.18?j17.19?30.08?j10.4MVA c c?50.26?j27.59MVA线路中的功率损耗为
?S1''? ?SL?? ?V???RL?jXL??N?50.262?27.592?(8.4?j16.6)MVA?2.28?j4.51MVA11022于是可得 S'1?S1''??SL?50.26?j27.59?2.28?j4.51MVA
?52.54?j32.1MVA由母线A输出的功率为
SA?S1'?j?QB?52.54?j32.1?j2.65MVA?52.54?j29.45MVA
（3）计算各节点电压。
线路中电压降落的纵分量和横分量分别为
P1'RL?Q1'XL52.24?8.4?32.1?16.6?VL??kV?8.3kVVA117P1'XL?Q1'RL52.24?16.6?32.1?8.4?VL??kV?5.2kV
VA117
b点电压为
Vb??VA??VL????VL??22?117?8.3?2?5.22kV?108.8kV
变压器中电压降落的纵，横分量分别为
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?V?PR?QX?20.18?3.4?17.19?42.4kV?7.3kV
''cTcTTVb108.8 ?VT'Pc'XT?QCRT20.18?42.4?17.19?3.4 ??kV?7.3kVVb108.8归算到高压侧的c点电压
V??V??V????V???108.8?7.3??7.3kV?101.7kV
'2222cbTT变电所低压母线c的实际电压
Vc?Vc'?1111?101.7?kV?10.17kV 110110如果在上述计算中都不计电压降落的横分量，所得结果为 V?108.7kV， V?101.4kV， Vc?10.14kV
'bc与计及电压降落横分量的计算结果相比，误差很小。
3.3 某一额定电压为10kV的两端供电网，如图所示。线路L1、L2和线路长度分别为10km，4km和3km，线路L4L3导线型号均为LJ-185，
??10.5?0kV、为2km长的LJ-70导线；各负荷点负荷如图所示。试求V?A?B?10.4?0?kV时的初始功率分布，且找到电压最低点。（线路参数VLJ-185：z=0.17+j0.38Ω/km；LJ-70：z=0.45+j0.4Ω/km）
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解 线路等值阻抗
Z?10?(0.17?j0.38)?1.7?j3.8?
L1 ZL2?4?(0.17?j0.38)?0.68?j1.52? Z?3?(0.17?j0.38)?0.51?j1.14?
L3 ZL4?2?(0.45?j0.4)?0.9?j0.8? 求C点和D点的运算负荷，为 ?SCCE0.32?0.162?(0.9?j0.8)?1.04?j0.925kVA 210 S?2600?j1600?300?j160?1.04?j0.925
?2901.04?j1760.925kVASD?600?j200?1600?j1000?2200?j1200kVA
循环功率 Sc?V??A?10.5?10.4??10?VB??VN??Z?17??0.17?j0.38?
?339.43?0.17?j0.38?kVA?580?j129kVA1?2901.04?7?2200?3?j1760.925?7?j1200?3??Sc17?1582.78?j936.85?580?j129?2162.78?j1065.85kVASAC?1?2901.04?10?2200?14?j1760.925?10?j1200?14??Sc17?3518.26?j2024.07?580?j129?2938.26?j1895.07kVASBD?15
SAC?SBD?2162.78?j1065.85?2938.26?j1895.07?5101.04?j2960.92kVASC?SD?2901.04?j1760.925?2200?j1200
?5101.04?j2960.92kVA
SDC?SBD?SD?2938.26?j1895.07?2200?j1200 ?738.26?j695.07kVAC点为功率分点，可推算出E点为电压最低点。进一步可求得E点电压 ?SAC2.162?1.072?(1.7?j3.8)MVA?98.78?j220.8kVA 210'SAC?2162.78?j1065.85?98.78?j220.8?2261.56?j1286.65kVA
?V?2.26?1.7?1.29?3.8?0.8328kV
AC10.5VC?VA??VAC?10.5?0.8328?9.6672kV
?VCE?0.301?0.9?0.161?0.8?0.041kV
9.6672VE?VC??VCE?9.6672?0.041?9.6262kV
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