电力系统暂态分析课后

1、电力系统暂态分析课后答案(整理版)最新电力系统暂态分析课后答案(整理版)最新电力系统暂态分析课后答案(整理版)最新第一章1-2-1对例1-2，取UB2110kV，SB30MVA,用正确和近似计算法计算参数标幺值。解：正确计算法：采纳第二段为基本段，取UB2110kV，SB30MVA，则其他两段的电压基准值分别为：UB1k1UB210.5110kV9.5kV121UB3UB2110k26.6kV1106.6电流基准值：IB1SB301.8kA3UB139.5IB2SB300.16kA3UB23110IB3SB302.62kA3UB336.6各元件的电抗标幺值分别为：发电机：x10.2610.52

2、300.32309.52变压器T1：x20.1051212300.121110231.52输电线路：x30.480300.0791102变压器T2：x40.1051102300.211521102电抗器：x50.0562.620.46.60.3电缆线路：x60.082.5300.146.6211电源电动势标幺值：E1.169.5近似算法：取SB30MVA，各段电压电流基准值分别为：UB110.5kV，IB1301.65kA310.5UB2115kV，IUB36.3kV，I2B33030.15kA1153032.75kA6.3各元件电抗标幺值：发电机：x10.2610.52300.263010.

3、52变压器T1：x2300.10.10531.5输电线路：x30.480300.0731152变压器T2：x4300.210.10515电抗器：x50.0562.750.446.30.3电缆线路：x60.082.5300.1516.3211电源电动势标幺值：E1.0510.5习题2解：（1）正确计算：UB3UB(110)115kVUUB2kB32115220209.1kV121UB1UB311522010.59.1kVk2k1121242各段的电流基准值为：IB1SB22014.0kA3UB139.1IB2SB2200.6kA3UB23209.1SB2201.1kAIB331153UB3各元件

4、的电抗标幺值分别为：发电机：x10.3010.522200.292400.89.12变压器T1：x20.1410.522200.1430029.12输电线路：x30.422302200.49209.12变压器T2：x40.1422022200.122802209.12(2)近似算法：UB110.5kV，IB122012.10kA310.5UB2231kV，IB22200.55kA3231UB3121kV，IB32201.05kA3121各元件电抗标幺值：发电机：x10.302200.22240/0.8变压器T1：x20.142200.10300220输电线路：x30.422300.402312

5、220变压器T2：x40.140.11280习题3重点：以下摘自国家电网公司电力系统安全稳固计算规定：暂态稳固是指电力系统遇到大扰动后，各同步电机保持同步运转并过渡到新的或恢复到本来稳态运转方式的能力，平常指保持第一、第二摇摆不失步的功角稳固，是电力系统功角稳固的一种形式。动向稳固是指电力系统遇到小的或大的扰动后，在自动调理和控制装置的作用下，保持较长过程的运转稳固性的能力，平常指电力系统遇到扰动后不发生发散振荡或连续的振荡，是电力系统功角稳固的另一种形式。二者均是系统受扰动后恢复的能力，均属于功角的稳固，暂态是偏移正常运转状态很小的临时的状态，能很快达到正常状态，而动向稳固更多依靠于自动调理

6、和控制装置的作用，时间较长，颠簸较大。(1)ImUm2*6.39.45KAarctanxarctan0.79757.64oz0.943r0.505TaL0.7970.005R(2)314*0.505(3)Im|0|0ia9.45cos(t27.64o)oe200t或ia9.45cos(t27.64o)8.37e200tib9.45cos(t147.64o)oe200tic9.45cos(t92.36o)oe200t或：将ua改写成ua2*6.3sin(t90o)，带入公式得ia9.45sin(t62.36o)9.45sin62.36oe200tib9.45sin(t57.64o)9.45sin

7、57.64oe200tic9.45sin(t182.36o)oe200t（3）a,b,c相初始相角分别|为o,o，o，故a相瞬时电流最大0.01iMImIm*e0.00510.73KA由|90|90得=或CIm|0|sin(|0|)Imsin()相2*0.18sin(309030)9.45sin(309057.64)8.12KAB相c相取基准值UB13.8kV,SB240MVA，则SB240IB10.04kA3*UB3*13.8发电机次暂态电抗标幺值xd*xdUN2*SB0.216SN/cosUB2变压器的电抗标幺值xT*US(%)*UTN2*SB0.13100SNUB2电流标幺值Im*12.

8、890.130.216第二章2-2-1一发电机、变压器组的高压侧断路器处于断开状态，发电机空载运转，其端电压为额定电压。试计算变压器高压侧忽然三相短路后短路电流交流重量初始值Im。发电机：SN200MW，UN13.8kV，cosN0.9，xd0.92，xd0.32，xd0.2变压器：SN240MVA，220kV/13.8kV，US(%)13解：取基准值UB13.8kV，SB240MVASB240电流基准值IB10.04kA3UB313.8则变压器电抗标幺值xTUS%UTN2SB1313.822400.13100SNUB210024013.82发电机次暂态电抗标幺值xdxdUN2SB0.213.

9、822402SN220020.216cosNUB0.913.8次暂态电流标幺值I112.86xTxd0.130.22有名值Im22.8610.0438.05kA2-3-1例2-1的发电机在短路前处于额定运转状态。（1）分别用E，E和Eq计算短路电流交流重量I，I和Id；（2）计算稳态短路电流I。?cos10.85解：（1）U010，I01132?短路前的电动势：E0U0jxdI01j0.167321.0977.4?E0U0jxdI01j0.269321.16611.3Id01sin(41.132)0.957Uq01cos41.10.754Eq0Uq0 xdId00.7540.2690.9571

10、.01Eq0Uq0 xdId00.7542.260.9572.92因此有：IE0 xd1.097/0.1676.57IE0 xd1.166/0.2694.33IdEq0 xd1.01/0.2693.75（2）IEq0/xd2.92/2.261.292-4-1解:对不计阻尼绕组的情况Udrid&dqUqriq&qdUfrfif&fdxdidxadifqxqiqfxadidxfif由定子三相开路：id=iq=0得：Ud&xaddifqddtUq&qdxadifUf0rfif&frfifxfdifdtUf0rftUf0可以解得：ifexfrfrfUf0rft带入得：UdxadexfrfrftrftU

11、f0exfUf0)xadUf0(1exfUqxad()rfrfrf可得：Uacossin1UdUbcos(120o)sin(120o)1gUqUccos(120o)sin(120o)1U0UdcostUqsintUdcos(t120o)Uqsin(t120o)Udcos(t120o)Uqsin(t120o)rfUf0sint(1erfttUaxadUf0exfcostxadxf)xfrfUbxadUf0exfUcxadUf0exfrftUf0rftxf120o)120o)exfcos(txadsin(t(1)rfrftUf0rftxf120o)120o)exfcos(txadsin(t(1)r

12、f2-5,6-1Eqmtid(1eTd)xdxLiq0uGdxL*iq0tuGqxL*idEqm*xL(1eTd)xdxL2-5,6-2励磁突增前：&UG1.317.5oI|0|=jxL&oEQUGJxqI|0|1.3550.3I&q|0|1.31cos(50.3o7.5o)50.3o0.9650.3oI&q|0|1.31sin(50.3o7.5o)(50.3o90o)o而&oEq|0|=UG+jxdIdjxqIq1.8450.3由Uf|0|0.1Uf|0|得：Eqm0.184EqmttidTd)(1e0.136(1e0.26)xdxqiq0iat0.136(1e0.26)costEqm*xL

13、ttuGqTd)0.027(1(1ee0.26)xdxLuGd0tuGa0.027(1e0.26)sint设励磁突增前:UGa|0|2cos(t15o)ia|0|2*1.31cos(t7.5o)tiaia|0|ia2*1.31cos(t7.5o)0.136(1e0.26)cost则有：t2cos(t15o)0.027(1UGae0.26)sint第三章3-1-1j0.15j0.075x1j0.075j0.1-1j0.1-1j0.1x2-1X11j0.15j13;1160j0.2j0.1X21j391j0.15;170013j0.075j601I39jj17.9487700?j0.075IG1I

14、f13j4.6154;jj0.07560?j13IG2IG30.5If1360j6.6667;j0.075j60IG14.61546015.22683*10.5IG26.66676021.99443*10.53-1-2x7X3+x4+x6XX-110.17826;8110.5(X1X2)X70.5(X3X4X6)10.2565;If1913.898561X9X1X2X8IG1X8X21000IfX257.758X1X2321IG2IG30.5IfX70.5(X3X4X6)X110000.5(X1X2)0.5(X3X4X6)X1X2X210.343213-1-3(1)SB1000MVA,均匀额定

15、电压为基准值；1000X12500.3020.274;5252XXXXXXif2345861410000.19472010000.2170.32556009121000112010.179411X1X2X3X5X20.373410.17411X3X2X515.75;if5.751000158.084；X6321i2ifX3X83.072;i2f3.072100084.46X8321X3X11000i3ifX3X8X1X2X30.9247;i3f0.924732125.423;X3X2X31000i系ifX3X8X1X2X31.7533;i系1.753332148.203;(2)f2点短路X71

16、11X1（）0.5X2X3ifX711.579;if0.5X40.133310001.57924.639；321i2i30.5ifX10.4053;iG20.405310006.324X10.5(X2X3)321iB0.5ifX10.7684;iG30.7684100011.99X10.5(X2X3)3213-2-1应用例3-4已求得Y矩阵因子计算3-1-1,并与已有的计算结果比较。j26.6666j10j10解：Yj10j33.3333j10j10j10j20.0因子表中内容为j0.03750.3750.375j0.0338030.464791j0.1014200节点2处注入单位电流，则电流

17、向量I1;利用已求得的R和D1计算电压向量，获取20节点自阻抗和互阻抗。1W10W10-0.3751W2I1W210.3750.4647911W30W30.464791X1j0.0375W10X2j0.033803W2j0.033803X3j0.101420W3j0.0471391-0.375-0.375U10U1Z12j0.0385691-0.464791U2j0.033803U2Z22j0.0557131U3j0.047139U3Z32j0.047139j0.0729j0.0386j0.0557也许用MATLAB对Y矩阵求逆获取阻抗矩阵Zj0.0386j0.0557j0.0471从而获取j

18、0.0557j0.0471j0.1014Z12j0.038569Z22j0.055713；Z32j0.0471392的短路电流：1j17.95；节点IfZ22U1Z120.692各节点电压：U2Z22If1；U3Z320.845U1U10.307U21U20；U3U30.155?U1j0.462；发电机电流：IG1j0.15?1*U2IG2IG3j6.67。2j0.075第四章4-1-1如有三相不对称电流流入一用电设备，试问：1）改用电设备在什么状况下，三相电流中零序电流为零2）当零序电流为零时，用电设备端口三相电压中有无零序电压?I(0)Ia?用电?Ib设备U(0)Z(0)?Ic答：（1）负

19、载中性点不接地；三相电压对称；负载中性点接地，且三相负载不对称时，端口三相电压对称。（2）因为零序电流为零，故无零序电压。4-1-2由不对称重量法变换知：111&Za111&Ua(1)Ia(1)a2a1&Zba2a1&Ua(2)Ia(2)aa21&Zcaa21&Ua(0)Ia(0)得：U&a(1)1ZaZbZcZaa2ZbaZcZaaZba2Zc&ZaaZbaZcZaZb2ZcZaa2ZbZcUa(2)=a&3Za2aZcZaaZb2ZaZbZcUa(0)aZbaZc即没法获取三序独立的电压降方程。ggE(1)11aa2Ea4-2-1ga2g解：E(2)aaEbg3111gE(0)Ec11aa

20、21aa2a13111jI&a(1)I&a(2)I&a(0)0.7887j0.45530.2113j0.1220j0.3333gI(0)0ggE(1)0.7887j0.4553I(1)0.2276j0.3943j2j2ggE(2)0.2113j0.1220I(2)0.0610j0.1056j2j2ggIa111I(1)0.1666j0.4999gga2Iba1I(2)0.1667j0.4999gaa21g0.333IcI(0)gggUngEaIa(j2)j0.33324-5-1解：1）三绕组开路直接接地xIxIIx110.62336.410.014I2100 xII110.636.42310.

21、122100零序电压标幺值为：U(0)1030.143121则一侧电流标幺值：II0.1431.3490.0140.12则实质电流为：II1.34912030.772kA121则二侧电流标幺值：III0.1431.3490.0140.12则实质电流为：III1.34912030.4251kA220公共绕组电流：IIIIII0.7720.4251.197kA中性点电流：In3(0.7720.425)1.041kA中性点电压：Un0中性点经阻抗接地xIxIIxI0.124xII0.044零序电压标幺值保持不变，为：U(0)0.143则一侧电流标幺值：II0.1430.1240.8510.044实质

22、电流为：II1.8511200.487kA3121二侧电流标幺值：IIIII0.851则实质电流为：120III0.8510.268kA3220公共绕组电流：IIIIII0.4870.2680.755kA中性点电流：In3(0.4870.268)0.657kA中性点电压：Un12.50.6578.21kV4-6-1图4-37所示的系统中一回线路停运，另一回线路发生接地故障，试做出其零序网络图。1L2LG1T133T2G2xn1xn2解：画出其零序等值电路U(0)第五章不对称故障的分析计算5-1-1B、C相分别经阻抗接地的等值图：图1图1表示f点发生两相短路接地，其界限条件为?Ifa0，UfbU

23、fc0变换为对称重量的表示形式为：?(0)0If(1)If(2)If?Uf(1)Uf(2)Uf(0)复合序网：5-1-2图5-33示出系统中节点f处不对称的情况。若已知xf1、Uf01，由f点看入系统的x(1)x(2)1，系统内无中性点接地。试计算?Ifa、b、c。afbcxfxfxf?Uf0 x(1)f(1)?Uf(1)n(1)x(2)f(2)?Uf(2)n(2)x(1)f(0)?Uf(2)n(0)a）x(1)/xff(1)戴维南等值xf?Uf0Uf(1)n(1)x(2)/xff(2)xf?Uf(2)n(2)xff(0)xf?Uf(2)n(0)（b）x(1)/xff(1)?Uf0Uf(1)n

24、(1)x(2)/xff(2)?Uf(2)n(2)xff(0)?Uf(2)n(0)（c）解：正负零三序网如图（a），各序端口的戴维南等值电路如图（b）（a）单相短路，复合序网图如图（c）则：I(1)I(2)I(0)Uf010.5x(1)/xfx(2)/xfxf0.50.51（b）5-1-3图5-34示出一简单系统。若在线路始端处丈量Za?、Zb?、UagIaUbgIb?Za、Zb、Zc和ZcUcgIc。试分别作出f点发生三相短路和三种不对称短路时（可取0、1）的关系曲线，并分析计算结果。fGTxn1l解：其正序等值电路：EaxGxTlxl5-2-1已知图3-35所示的变压器星形侧?B、C相短路的

25、If。试以If为参照向量绘制出三角形侧线路上的三相电流相量：（1）对称重量法；（2）相重量法。?xIaaA?IbybB?IcczIfC1、对称重量法?IA(1)1aa2IA1aa20?1a2?11a2?IA(2)1aIBaIf?3?3?111111IA(0)ICIf?IA(1)Ia(1)?3Ic(2)?IaIc3If?Ia(2)?Ic(1)IA(2)三角侧零序无通路，不含零序重量，则：?IaIa(1)Ia(2)?IbIb(1)Ib(2)?IcIc(1)Ic(2)2、相重量法?Ib(2)?Ib(1)3?If33?If33?If3?23?IbIf3?If?电流向量图：此中相电流Ia与相电流IA同相

26、位，Ib与IB、Ic与IC同相位。?1?1?1?N1：N23：1。且IaIA、IbIB、IcIC。原副边匝数比333I?aIA?Ia?IcIBIbIC化为矩阵形式为：?Ia110Ia110IA1100?1?1?Ib011I011IB011Ifb3?101?3101?101?IcIcICIf5-2-2由正序增广网络看法得：XXX(2)(2)XX(0)(0)0.0950.1290.0950.0540.129111If(1)X0.0950.0546.711X(1)0.149IfbIfcMIf(1)X31X(2)X(0)310.0950.12930.8691.506X(0)2(0.0950.129)2

27、(2)XUf(1)Uf(2)Uf(0)If(1)X(2)(2)XX(0)6.7110.2441.639(0)Uf(a)Uf(1)Uf(2)Uf(0)3Uf(1)31.6394.917Uf(b)Uf(c)05-3-1解：（1）中间性点接地时因为由公式Ia3I(1)3Uqk0z(1)z(2)z(0)3zqk因此Ia31.1j0.49j(2.252.252.25)2）中间性点不接地时因为中间性点不接地时，则没有零序电流因此Ia2I(1)2Uqk0z(1)j0.326z(2)z(0)3zqk第六章电力系统稳固性问题归纳和各元件的机电特征6-2-1（1）TJ2.74GD2n28.74s1000SB（2）

28、TJNTJSN14.98sSB6-2-2若在例6-2中的发电机是一台凸极机。其参数为：SN300MW，UN18kV，cosN0.875，xd1.298，xq0.912，xd0.458试计算发电机分别保持Eq0，Eq0，Uq0为常数时，发电机的功率特征。P0,cos0G1T1LT2U115kVq?Eq?jId(xdxq)jIxq?EQ?jIxd?E?jIxejI(xTxL)EqUG?jIqxq?UqUG?IqI?dIdUd解：（1）取基准值SB250MVA，UB(110)115kV，UB(220)115220209kV，则121阻抗参数以下：2502xd1.2892421.2602093000.

29、8752502xq0.9122420.8922093000.8752502xd0.4582420.4482093000.8752502xT12420.1300.142093602502xT22200.1080.14209360 xL10.412002500.23522092系统的综合阻抗为：xexT1xLxT20.1300.1080.2350.473xdxdxe1.2600.4731.733xqxqxe0.8920.4731.365xdxdxe0.4480.4730.921（2）正常运转时的UG0，E0，Eq0，Eq0：P02501，Q01tg(cos10.98)0.2，U1151250115

30、由凸极机向量图得：?令US10，则：I(P0jQ0)US(1j0.2)101.019811.3099?EQ0USjxqI10j1.365(1j0.2)1.866546.9974IdIsin()1.0198sin(46.997411.3099)0.8677?Eq0EQ0Id(xdxq)1.86650.8677(1.7331.365)2.1858?E0UsjxdI10j0.921(1j0.2)1.500237.8736Eq0Ecos()1.5002cos(47.0037.8736)1.4812?UG0USjxeI10j0.473(1j0.2)1.192423.3702与例题6-2UG0(UQ0 x

31、e)2(P0 xe)2(10.20.473)2(0.473)21.193UUE0(10.20.921)20.92121.5EQ0(10.21.365)21.36521.8665Eq0EQ0Id(xdxq)1.86650.31932.18580tg11.36546.9910.21.365Eq0Uq0Id0 xdUq0EQ0Uq0 xqxd1.866cos46.991.4809cos46.991.3650.921(3)各电动势、电压分别保持不变时发电机的功率特征：Eq0UU2xdxPEqsinqsin2xd2xdxqPEEq0UU2xqxdsin2xdsin2xdxqqPEEUsinEUsin1U

32、xd)sinxdsin(1xdxdEPUqUGUsinGUGUsinsin1U(1xe)sinxexeUGxq6-2-3由公式EqUqIdxdEqUqIdxd0UdIqxq和消去Id可得0UdIqxqEqxdEq(xdxd)Ucos1.882C0.882cosxd6-2-4EUPsinxdxTxL/x并联电抗器使系统中xd减小，因此发电机的功率极限将增大。第七章7-2-1(1)POUGUsin1.0510.837.5550XT1GsinGGXLXT20.8?00?UGU1.0537.555100I0.82714.67j(XT1XLXT2)j0.8?0.82714.67066.610EqUjIX

33、d1j1.81.569900时，PMEqU1.56910.8717X1.8dPMP08.963%10%;因此不切合要求KpPM（2）POUGUsin1.0510.822.40XT1XLXT2G0.5sinGG?00?UGU1.0522.4100I0.84.29j(XT1XLXT2)j0.8?4.29048.3560EqUjIXd1j0.81.31.3878故048.3560；Id（0-4.290)；0.8sin48.3560.5564EqEQId(xdxq)1.38780.5564(10.8)1.499PEq1.499sin10.2sin2sin0.0513sin2;1.521.51.3dPEqcos0.103cos2084.210;PEqm1.令dKP10.825%;0.87-3-1判据一：K10判据二：K4KeK5K40判据三：K1K2K4）Ke(K1K6K2K5)K1K2K40K1K2K3K4（K3K3第八章电力系统暂态稳固8-2-2在例8-1中若扰动是忽然断开一回线路，是判断系统能否保持暂态稳固。P0220MWcos00.98U115kV300MW200kM18kV360MW360MWcos0.85x10.41/kMxdxq