88电力系统分析练习题及其答案

[例2-1]一条220kV的输电线,长180km,导线为LGJ-400(直径2.8cm),水平排列,相间距7m,求该线路的R,X,B,并画等值电路.解:电阻:kmSr/08.04005.3114.1418008.01lrR电抗:cmDeq88270027007003kmrDxeq/42.04.18.0882lg1445.08.0lg1445.016.7518042.01lxX电纳:kmSrDbeq/107.2104.1882lg58.710lg58.76661SlbB66110486180107.2等值电路:[例2-2]220kV架空线,水平排列,相间距7m，每相为2402LGJQ分裂导线，计算直径21.88mm，分裂间距400mm，求每相单位长度的电阻、电抗和电纳。解:电阻:kmSr/660.040225.311＝电抗:757.62400288.219.0dDDssbmmDeq882070002700070003kmDDxsbeq/31.0757.628820lg1445.0lg1445.01电纳:151.66400288.21rdreqkmSrDbeqeq/10567.310151.668820lg58.710lg58.76661[例2-3]一长度为600km的500kV架空线路，使用4×LGJQ-400四分裂导线，611110.0187,0.275,4.0510,0rkmxkmbSkmg。试计算该线路的形等值电路参数。解（1）精确计算。66(0.01870.275)0.2756(86.11)4.0510906000.27564.0510(83.1190)20.633988.060.021460.6335()0.5()0.01730.59220.592488.330.5924llZzrjxjkmkmyjbSkmlzyljshleejshlKl88.330.93450.270.633988.06()0.5()0.80610.01272(1)0.3886176.321.0351.07()0.3755176.39llYchleejchlKlshl计算形等效电路参数：6330.93450.270.275686.11600154.53860382(2)1.0350.074.0510903001.2581089.931.25810ZYZKzlYKjblSSjS（2）使用近似算法计算。2262226311110.2754.05600100.8663311()0.9336111.03312(9.72153.9)24.05103001.0331.25510rxbrxkxblrkbxlxkxblZkrljkljYjSjS与准确计算相比，电阻误差-0.4%，电抗误差-0.12%，电纳误差-0.24%，本例线路长度小于1000km，用实用近似公式计算已能够满足精确要求。如果直接取1,)(11.22165)ZYKKlj11则Z=（r+jx这时，电阻误差达15%，电抗误差7%，电纳误差-3.4%，误差已较大。例2－4330kV架空线路的参数为,/0579.00kmr,/316.00kmx,00g./105.360kmsb试分别计算长度为100，200，300，400和500线路的π型等值参数的近视值，修正值和精确值。解首先计算100km线路的参数（一）)6.3179.5(100)316.00579.0()(00`jjljxrZsjsjljbgY4600`1055.3100)1055.30()((二)修正参数计算9963.01001055.3316.031131126200lbxkr9982.0100]316.0/1055.30579.01055.3316.0[611)06112626200200(lxbrbxkx0009.11001055.3316.01211121126200lbxkb)5431.317686.5(100)316.09982.00579.09963.0()(00`jjljxkrkZxrssjljbkYb460`105533.31001055.30009.1（三）精确参数计算060000192.58255.300)2201.275914.299()1055.3/()316.00579.0()/()(jjjjjbgxrZckmbgxrjjjjj146000010)6355.09663.0()1055.3)(316.00579.0())((2410)6355.09663.0(10010)63555.109663.0(jjl计算双曲线函数。利用公式sh(x+jy)=shxcosy+jchxsinych(x+jy)=chxcosy+jshxsiny将l之值代入，便得222222210)6160.109609.0()106355.10sin()109633.0()106355.10cos()109633.0()106355.10109633.0(jjchshjshlsh2222222101026.09944.0)106355.10sin()109633.0()106355.10cos()109633.0()106355.10109633.0(jjshshjchlchII型电路的精确参数为sjsjjlshlchYjjjlshZZZcc422`10)5533.30006.0(5429.317684.5)1101026.09944.0(2)1(2`)5429.317684.5(10)6160.109609.0()2201.275914.299([例2-5]有一台SFL120000/110型的向10kV网络供电的降压变压器，铭牌给出的实验数据为：试计算归算到高压侧的变压参数。解由型号知，.110,20000kVVAkVSNN高压侧额定电压各参数如下：08.4102000011013510322322NNSTSVPR53.6310200001001105.1010100%32322NNSTSVVXssVPGNT6333201082.11011022101011110102.1310110100200008.010100%21632320NNTNNTVVssVSIBk8.0%,22,5.10%,13500IkWPVkWPss例2-6三相三绕组降压变压器的型号为SFPSL-120000/220，额定容量为120MVA/120MVA/60MVA，额定电压为：220kV/121kV/11kV,kWPS601)21(，kWPS5.182)31(，kWPS5.132)32(，85.14%)21(SV，25.28%)31(SV，96.7%)32(SV，kWP1350，663.0%0I，求变压器归算到220kV侧的参数，并作出等值电路。解：（1）求各绕组的电阻kWkWSSPSSPPPNNSNNSSS5.400])60120(5.132)60120(5.182601[5.0])()([212223)32(23)31()21(1同理可得：kWPkWPSS5.3295.20032电阻计算如下：346.112010002005.4001000222211NNSTSVPR107.1674.032TTRR（2）求各绕组电抗68.10%72.2%57.17%)%%(21%32)32()31()21(1SSSSSVVVVVV电抗计算：86.7012010022057.17100%2211NNSTSVVX08.4397.1032TTXX变压器阻抗参数：)68.70346.1(111jjXRZTTT)08.43107.1()97.10674.0(32jZjZTT（3）求导纳SjBGYSSVSIBSSVPGTTTNNTNT66220622010)4.1679.2(104.16220100120663.0100%1079.222010001351000例2—7试计算2—15（a）所示输电系统各元件电抗的标幺值。已知各元件的参数如下：发电机：26.0,5.10,30)()()(NGNGNGXkVVMVAS，变压器T-1：;121/5.10,5.10%,5.31)(TISNTikVMVAS变压器T-2：;6.6/110,5.10%,15)(TISNTikVMVAS电抗器:5%,3.0,6)()(RNRNRXkAIkVV；架空线路长80km，每公里电抗为4.0；电缆线路长2.5km,每公里电抗为08.0。解首先选择基准值。取全系统的基准功率MVASB100。为了使标幺值参数的等值电－路中不出现串联的理想变压器，选取相邻段的基准电压比kkkkTIIIIIBTB2)(1)III(,。这样，只要选出三段中的某一段的基准电压，其余的基准电压就可以由基准变比确定了。选第I段的基准电压,5.10)(kVVIB于是kVkVkVVIIIBIBIIB1211215.1015.101)()()(kVkVkkVkVVIIIIIBIIIBIBIIIIIBIIBIIIB26.7)6.6110()1215.10(15.1011)()()()()()(各元件电抗的标幺值为87.05.10100305.1026.0222)()(2)()*(1IBBNGNGBGVSSVXX33.05.101005.315.101005.10100%222)()(12)(18)(12IBBNTNITSBTVSSVVXX22.0100804.012122*)(3)(VSXXXIIBBLBL58.0100151005.10100%121110222)(22*)(24)()1(2VSSVVXXIIBNTBNTsBT09.11003.031005100%26.76322)()*(5)()(VSIVVXXIIBNRBNRRBR38.01005.208.026.722*)(6)(VSXXXIIBBCBC[例2-8]给定基准功率MVASB100，基准电压等于各级平均额定电压。假定发电机电势标幺值等于1.0。试计算例2-7的输电系统在电缆末端短路的短路电流（分别按元件标幺参数的近似值和精确值计算）。解按题给条件，各级基准电压应为.3.6,115,5.10)()()(kVkVkVVVVIIIBIIBIB各元件电抗的标幺值计算如下：87.01003026.05.105.10222)(2*)(1)()(VSSVXXIBNGBNGNG33.01005.311005.10100%5.105.10222)(122)()1(1VSSVVXIBNTBNTs24.0100