电力系统潮流上机计算实验报告 1 1. 手算过程 已知: 节点1:PQ 节点, s(1)= -0.5000-j0.3500 节点2:PV 节点, p(2)=0.4000 v(2)=1.0500 节点3:平衡节点,U(3)=1.0000∠0.0000 网络的连接图: 0.0500+j0.2000 1 0.0500+j0.2000 23 1)计算节点导纳矩阵 由2 0 0 0.00 5 0 0.01 2jZ 7 1.41 8.11 2jy; 2 0 0 0.00 5 0 0.01 3jZ 7 1.41 8.11 3jy; 导纳矩阵中的各元素: 4 2.93 6.27 1.41 8.17 1.41 8.11 31 21 1jjjyyY; 7 1.41 8.11 21 2jyY; 7 1.41 8.11 31 3jyY; 2 1Y7 1.41 8.11 21 2jyY; 7 1.41 8.12 12 2jyY; 002 32 3jyY; 3 1Y7 1.41 8.11 31 3jyY; 3 2Y002 32 3jyY; 7 1.41 8.13 13 3jyY; 形成导纳矩阵BY : 7 1.41 8.1007 1.41 8.1007 1.41 8.17 1.41 8.17 1.41 8.17 1.41 8.14 2.93 6.2jjjjjjjjjYB 2)计算各 PQ、PV 节点功率的不平衡量,及PV 节点电压的不平衡量: 取:0 0 0.00 0 0.1)0(1)0(1)0(1jjfeU 0 0 0.00 0 0.1)0(2)0(2)0(2jjfeU 节点3 是平衡节点,保持0 0 0.00 0 0.1333jjfeU为定值。 njjjijjijijijjijiieBfGffBeGeP1)0()0()0()0()0()0()0(; 电力系统潮流上机计算实验报告 2 njjjijjijijijjijiieBfGefBeGfQ1)0()0()0()0()0()0()0(; );(2)0(2)0(2)0(iiifeU )0.142.90.036.2(0.0)0.042.90.136.2(0.1)0(1P )0.171.40.018.1(0.0)0.071.40.118.1(0.1 )0.171.40.018.1(0.0)0.071.40.118.1(0.1 0.0; )0.142.90.036.2(0.1)0.042.90.136.2(0.0)0(1Q )0.171.40.018.1(0.1)0.071.40.118.1(0.0 )0.171.40.018.1(0.1)0.071.40.118.1(0.0 0.0; )0.171.40.018.1(0.0)0.071.40.118.1(0.1)0(2P )0.171.40.018.1(0.0)0.071.40.118.1(0.1 )0.00.00.00.0(0.0)0.10.00.10.0(0.1 0.0; 101)(222)0(22...