电力系统分析潮流计算例题

电力系统的潮流计算西安交通大学自动化学院2012.103．1电网结构如图3—11所示，其额定电压为10KV。已知各节点的负荷功率及参数：MVAjS)2.03.0(2，MVAjS)3.05.0(3，MVAjS)15.02.0(4)4.22.1(12jZ，)0.20.1(23jZ，)0.35.1(24jZ试求电压和功率分布。解：（1）先假设各节点电压均为额定电压，求线路始端功率。0068.00034.0)21(103.05.0)(22223232232323jjjXRVQPSN0019.00009.0)35.1(1015.02.0)(22224242242424jjjXRVQPSN则：3068.05034.023323jSSS1519.02009.024424jSSS6587.00043.122423'12jSSSS又0346.00173.0)4.22.1(106587.00043.1)(22212122'12'1212jjjXRVQPSN故：6933.00216.112'1212jSSS（2）再用已知的线路始端电压kVV5.101及上述求得的线路始端功率12S，求出线路各点电压。kVVXQRPV2752.05.104.26933.02.10216.1)(11212121212kVVVV2248.101212kVVVVkVVXQRPV1508.100740.0)(242422424242424kVVVVkVVXQRPV1156.101092.0)(232322323232323（3）根据上述求得的线路各点电压，重新计算各线路的功率损耗和线路始端功率。0066.00033.0)21(12.103.05.022223jjS0018.00009.0)35.1(15.1015.02.022224jjS故3066.05033.023323jSSS1518.02009.024424jSSS则6584.00042.122423'12jSSSS又0331.00166.0)4.22.1(22.106584.00042.122212jjS从而可得线路始端功率6915.00208.112jS这个结果与第（1）步所得计算结果之差小于0.3%，所以第（2）和第（3）的结果可作为最终计算结果；若相差较大，则应返回第（2）步重新计算，直道相差较小为止。3.2如图所示简单系统，额定电压为110KV双回输电线路，长度为80km，采用LGJ-150导线，其单位长度的参数为：r=0.21Ω/km，x=0.416Ω/km,b=2.74kmS/106。变电所中装有两台三相110/11kV的变压器，每台的容量为15MVA,其参数为：5.3%,5.10%,128P5.40s0osIVkWkWP，。母线A的实际运行电压为117kV，负荷功率：MVAjSMVAjSLDcLDb1520,1230。当变压器取主轴时，求母线c的电压。解（1）计算参数并作出等值电路。输电线路的等值电阻、电抗和电纳分别为4.821.08021LR6.16416.08021LXSSBc461038.41074.2802由于线路电压未知，可用线路额定电压计算线路产生的充电功率，并将其等分为两部分，便得var65.2var1101038.42121242MMVBQNcB将BQ分别接于节点A和b，作为节点负荷的一部分。两台变压器并联运行时，它们的等值电阻、电抗及励磁功率分别为4.3151000110128211000212222NNsTSVPR4.42151001105.1021100%21222NNsTSVVRMVAjMVAjQjPoo05.108.0)100155.30405.0(2变压器的励磁功率也作为接于节点b的负荷，于是节点b的负荷MVAjMVAjjjQjPQjSSBLDbb4.1008.3065.205.108.01230)(00节点c的功率即是负荷功率MVAjSc1520这样就得到图所示的等值电路（2）计算母线A输出的功率。先按电力网络的额定电压计算电力网络中的功率损耗。变压器绕组中的功率损耗为MVAjMVAjjXRVSSTTNcT19.218.0)4.424.3(11015202222由图可知MVAjMVAjjQjPSSTTcc19.1718.2019.218.01520'MVAjMVAjjSSSbcc59.2726.504.1008.3019.1718.20'''线路中的功率损耗为MVAjMVAjjXRVSSLLNL51.428.2)6.164.8(11059.2726.502222''1于是可得MVAjMVAjjSSSL1.3254.5251.428.259.2726.50''1'1由母线A输出的功率为MVAjMVAjjQjSSBA45.2954.5265.21.3254.52'1（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kVkVVXQRPVALLL3.81176.161.324.824.52'1'1kVkVVRQXPVALLL2.51174.81.326.1624.52'1'1b点电压为kVkVVVVVLLAb8.1082.53.81172222变压器中电压降落的纵，横分量分别为kVkVVXQRPVbTcTcT3.78.1084.4219.174.318.20''kVkVVRQXPVbTCTcT3.78.1084.319.174.4218.20''归算到高压侧的c点电压kVkVVVVVTTbc7.1013.73.78.1082222'变电所低压母线c的实际电压kVkVVVcc17.10110117.10111011'如果在上述计算中都不计电压降落的横分量，所得结果为kVVb7.108，kVVc4.101'，kVVc14.10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1L、2L和3L导线型号均为LJ-185，线路长度分别为10km，4km和3km，线路4L为2km长的LJ-70导线；各负荷点负荷如图所示。试求kVVA05.10、kVVB04.10时的初始功率分布，且找到电压最低点。（线路参数LJ-185：z=0.17+j0.38Ω/km；LJ-70：z=0.45+j0.4Ω/km）解线路等值阻抗8.37.1)38.017.0(101jjZL52.168.0)38.017.0(42jjZL14.151.0)38.017.0(33jjZL8.09.0)4.045.0(24jjZL求C点和D点的运算负荷，为kVAjjSCE925.004.1)8.09.0(1016.03.0222kVAjjjjSC925.176004.2901925.004.116030016002600kVAjjjSD1200220010001600200600循环功率kVAjkVAjjZVVVSNBAc12958038.017.043.33938.017.017104.105.10kVAjjjSjjScAC85.106578.216212958085.93678.1582312007925.176032200704.2901171kVAjjjSjjScBD07.189526.293812958007.202426.351814120010925.17601422001004.2901171kVAjjjSSBDAC92.296004.510107.189526.293885.106578.2162kVAjjjSSDC92.296004.510112002200925.176004.2901kVAjjjSSSDBDCD07.69526.7381200220007.189526.2938C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kVAjMVAjSAC8.22078.98)8.37.1(1007.116.2222kVAjjjSAC65.128656.22618.22078.9885.106578.2162'kVVAC8328.05.108.329.17.126.2kVVVVACAC6672.98328.05.10kVVCE041.06672.98.0161.09.0301.0kVVVVCECE6262.9041.06672.93.4图所示110kV闭式电网，A点为某发电厂的高压母线，其运行电压为117kV。网络各组件参数为：变电所bMVASN20，MVAjS6.005.00，84.4TR，5.63TX变电所cMVASN10，MVAjS35.003.00，4.11TR，127TX负荷功率MVAjSLDb1824，MVAjSLDc912试求电力网络的功率分布及最大电压损耗。解（1）计算网络参数及制定等值电路。线路Ⅰ：38.252.1660)423.027.0(jjZSSB461061.1601069.2var95.1var1101061.1224MMQB线路Ⅱ：15.215.1350)423.027.0(jjZSSB461035.1501069.2var63.1var1101035.1224MMQB线路Ⅱ：6.171840)44.045.0(jjZSSB461003.1401058.2var25.1var1101003.1224MMQB变电所b：75.3142.25.6384.421jjZTbMVAjMVAjSb2.11.06.005.020变电所b：5.637.51274.1121jjZTcMVAjMVAjSc7.006.035.003.020等值电路如图所示（2）计算节点b和c的运算负荷。MVAjMVAjSTb36.218.075.3124.21101824222MVAjMVAjjjjjQjQjSSSSBBIobTbLDbb96.1928.24623.0975.02.11.036.218.01824MVAjMVAjSTc18.1106.05.637.5110912222MVAjMVAjjjjjQjQjSSSSBBocTcLDcc44.917.12815.0623.07.006.018.1106.0912（3）计算闭式网络的功率分布。MVAjMVAjjjjjZZZZSZZSScb79.1564.1813.647.4715.215.1344.917.1275.385.3196.1928.24