電力系統(tǒng)暫態(tài)分析課后

1、電力系統(tǒng)暫態(tài)分析課后答案(整理版)最新電力系統(tǒng)暫態(tài)分析課后答案(整理版)最新電力系統(tǒng)暫態(tài)分析課后答案(整理版)最新第一章1-2-1對(duì)例1-2，取UB2110kV，SB30MVA,用正確和近似計(jì)算法計(jì)算參數(shù)標(biāo)幺值。解：正確計(jì)算法：采納第二段為基本段，取UB2110kV，SB30MVA，則其他兩段的電壓基準(zhǔn)值分別為：UB1k1UB210.5110kV9.5kV121UB3UB2110k26.6kV1106.6電流基準(zhǔn)值：IB1SB301.8kA3UB139.5IB2SB300.16kA3UB23110IB3SB302.62kA3UB336.6各元件的電抗標(biāo)幺值分別為：發(fā)電機(jī)：x10.2610.52

2、300.32309.52變壓器T1：x20.1051212300.121110231.52輸電線路：x30.480300.0791102變壓器T2：x40.1051102300.211521102電抗器：x50.0562.620.46.60.3電纜線路：x60.082.5300.146.6211電源電動(dòng)勢(shì)標(biāo)幺值：E1.169.5近似算法：取SB30MVA，各段電壓電流基準(zhǔn)值分別為：UB110.5kV，IB1301.65kA310.5UB2115kV，IUB36.3kV，I2B33030.15kA1153032.75kA6.3各元件電抗標(biāo)幺值：發(fā)電機(jī)：x10.2610.52300.263010.

3、52變壓器T1：x2300.10.10531.5輸電線路：x30.480300.0731152變壓器T2：x4300.210.10515電抗器：x50.0562.750.446.30.3電纜線路：x60.082.5300.1516.3211電源電動(dòng)勢(shì)標(biāo)幺值：E1.0510.5習(xí)題2解：（1）正確計(jì)算：UB3UB(110)115kVUUB2kB32115220209.1kV121UB1UB311522010.59.1kVk2k1121242各段的電流基準(zhǔn)值為：IB1SB22014.0kA3UB139.1IB2SB2200.6kA3UB23209.1SB2201.1kAIB331153UB3各元件

4、的電抗標(biāo)幺值分別為：發(fā)電機(jī)：x10.3010.522200.292400.89.12變壓器T1：x20.1410.522200.1430029.12輸電線路：x30.422302200.49209.12變壓器T2：x40.1422022200.122802209.12(2)近似算法：UB110.5kV，IB122012.10kA310.5UB2231kV，IB22200.55kA3231UB3121kV，IB32201.05kA3121各元件電抗標(biāo)幺值：發(fā)電機(jī)：x10.302200.22240/0.8變壓器T1：x20.142200.10300220輸電線路：x30.422300.402312

5、220變壓器T2：x40.140.11280習(xí)題3重點(diǎn)：以下摘自國(guó)家電網(wǎng)公司電力系統(tǒng)安全穩(wěn)固計(jì)算規(guī)定：暫態(tài)穩(wěn)固是指電力系統(tǒng)遇到大擾動(dòng)后，各同步電機(jī)保持同步運(yùn)轉(zhuǎn)并過渡到新的或恢復(fù)到本來穩(wěn)態(tài)運(yùn)轉(zhuǎn)方式的能力，平常指保持第一、第二搖擺不失步的功角穩(wěn)固，是電力系統(tǒng)功角穩(wěn)固的一種形式。動(dòng)向穩(wěn)固是指電力系統(tǒng)遇到小的或大的擾動(dòng)后，在自動(dòng)調(diào)理和控制裝置的作用下，保持較長(zhǎng)過程的運(yùn)轉(zhuǎn)穩(wěn)固性的能力，平常指電力系統(tǒng)遇到擾動(dòng)后不發(fā)生發(fā)散振蕩或連續(xù)的振蕩，是電力系統(tǒng)功角穩(wěn)固的另一種形式。二者均是系統(tǒng)受擾動(dòng)后恢復(fù)的能力，均屬于功角的穩(wěn)固，暫態(tài)是偏移正常運(yùn)轉(zhuǎn)狀態(tài)很小的臨時(shí)的狀態(tài)，能很快達(dá)到正常狀態(tài)，而動(dòng)向穩(wěn)固更多依靠于自動(dòng)調(diào)理

6、和控制裝置的作用，時(shí)間較長(zhǎng)，顛簸較大。(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：將ua改寫成ua2*6.3sin(t90o)，帶入公式得ia9.45sin(t62.36o)9.45sin62.36oe200tib9.45sin(t57.64o)9.45sin

7、57.64oe200tic9.45sin(t182.36o)oe200t（3）a,b,c相初始相角分別|為o,o，o，故a相瞬時(shí)電流最大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相取基準(zhǔn)值UB13.8kV,SB240MVA，則SB240IB10.04kA3*UB3*13.8發(fā)電機(jī)次暫態(tài)電抗標(biāo)幺值xd*xdUN2*SB0.216SN/cosUB2變壓器的電抗標(biāo)幺值xT*US(%)*UTN2*SB0.13100SNUB2電流標(biāo)幺值Im*12.

8、890.130.216第二章2-2-1一發(fā)電機(jī)、變壓器組的高壓側(cè)斷路器處于斷開狀態(tài)，發(fā)電機(jī)空載運(yùn)轉(zhuǎn)，其端電壓為額定電壓。試計(jì)算變壓器高壓側(cè)忽然三相短路后短路電流交流重量初始值Im。發(fā)電機(jī)：SN200MW，UN13.8kV，cosN0.9，xd0.92，xd0.32，xd0.2變壓器：SN240MVA，220kV/13.8kV，US(%)13解：取基準(zhǔn)值UB13.8kV，SB240MVASB240電流基準(zhǔn)值IB10.04kA3UB313.8則變壓器電抗標(biāo)幺值xTUS%UTN2SB1313.822400.13100SNUB210024013.82發(fā)電機(jī)次暫態(tài)電抗標(biāo)幺值xdxdUN2SB0.213.

9、822402SN220020.216cosNUB0.913.8次暫態(tài)電流標(biāo)幺值I112.86xTxd0.130.22有名值Im22.8610.0438.05kA2-3-1例2-1的發(fā)電機(jī)在短路前處于額定運(yùn)轉(zhuǎn)狀態(tài)。（1）分別用E，E和Eq計(jì)算短路電流交流重量I，I和Id；（2）計(jì)算穩(wěn)態(tài)短路電流I。?cos10.85解：（1）U010，I01132?短路前的電動(dòng)勢(shì)：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:對(duì)不計(jì)阻尼繞組的情況Udrid&dqUqriq&qdUfrfif&fdxdidxadifqxqiqfxadidxfif由定子三相開路：id=iq=0得：Ud&xaddifqddtUq&qdxadifUf0rfif&frfifxfdifdtUf0rftUf0可以解得：ifexfrfrfUf0rft帶入得：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勵(lì)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設(shè)勵(lì)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則有：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,均勻額定

15、電壓為基準(zhǔn)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點(diǎn)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應(yīng)用例3-4已求得Y矩陣因子計(jì)算3-1-1,并與已有的計(jì)算結(jié)果比較。j26.6666j10j10解：Yj10j33.3333j10j10j10j20.0因子表中內(nèi)容為j0.03750.3750.375j0.0338030.464791j0.1014200節(jié)點(diǎn)2處注入單位電流，則電流

17、向量I1;利用已求得的R和D1計(jì)算電壓向量，獲取20節(jié)點(diǎn)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也許用MATLAB對(duì)Y矩陣求逆獲取阻抗矩陣Zj0.0386j0.0557j0.0471從而獲取j

18、0.0557j0.0471j0.1014Z12j0.038569Z22j0.055713；Z32j0.0471392的短路電流：1j17.95；節(jié)點(diǎn)IfZ22U1Z120.692各節(jié)點(diǎn)電壓：U2Z22If1；U3Z320.845U1U10.307U21U20；U3U30.155?U1j0.462；發(fā)電機(jī)電流：IG1j0.15?1*U2IG2IG3j6.67。2j0.075第四章4-1-1如有三相不對(duì)稱電流流入一用電設(shè)備，試問：1）改用電設(shè)備在什么狀況下，三相電流中零序電流為零2）當(dāng)零序電流為零時(shí)，用電設(shè)備端口三相電壓中有無零序電壓?I(0)Ia?用電?Ib設(shè)備U(0)Z(0)?Ic答：（1）負(fù)

19、載中性點(diǎn)不接地；三相電壓對(duì)稱；負(fù)載中性點(diǎn)接地，且三相負(fù)載不對(duì)稱時(shí)，端口三相電壓對(duì)稱。（2）因?yàn)榱阈螂娏鳛榱?，故無零序電壓。4-1-2由不對(duì)稱重量法變換知：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即沒法獲取三序獨(dú)立的電壓降方程。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）三繞組開路直接接地xIxIIx110.62336.410.014I2100 xII110.636.42310.

21、122100零序電壓標(biāo)幺值為：U(0)1030.143121則一側(cè)電流標(biāo)幺值：II0.1431.3490.0140.12則實(shí)質(zhì)電流為：II1.34912030.772kA121則二側(cè)電流標(biāo)幺值：III0.1431.3490.0140.12則實(shí)質(zhì)電流為：III1.34912030.4251kA220公共繞組電流：IIIIII0.7720.4251.197kA中性點(diǎn)電流：In3(0.7720.425)1.041kA中性點(diǎn)電壓：Un0中性點(diǎn)經(jīng)阻抗接地xIxIIxI0.124xII0.044零序電壓標(biāo)幺值保持不變，為：U(0)0.143則一側(cè)電流標(biāo)幺值：II0.1430.1240.8510.044實(shí)質(zhì)

22、電流為：II1.8511200.487kA3121二側(cè)電流標(biāo)幺值：IIIII0.851則實(shí)質(zhì)電流為：120III0.8510.268kA3220公共繞組電流：IIIIII0.4870.2680.755kA中性點(diǎn)電流：In3(0.4870.268)0.657kA中性點(diǎn)電壓：Un12.50.6578.21kV4-6-1圖4-37所示的系統(tǒng)中一回線路停運(yùn)，另一回線路發(fā)生接地故障，試做出其零序網(wǎng)絡(luò)圖。1L2LG1T133T2G2xn1xn2解：畫出其零序等值電路U(0)第五章不對(duì)稱故障的分析計(jì)算5-1-1B、C相分別經(jīng)阻抗接地的等值圖：圖1圖1表示f點(diǎn)發(fā)生兩相短路接地，其界限條件為?Ifa0，UfbU

23、fc0變換為對(duì)稱重量的表示形式為：?(0)0If(1)If(2)If?Uf(1)Uf(2)Uf(0)復(fù)合序網(wǎng)：5-1-2圖5-33示出系統(tǒng)中節(jié)點(diǎn)f處不對(duì)稱的情況。若已知xf1、Uf01，由f點(diǎn)看入系統(tǒng)的x(1)x(2)1，系統(tǒng)內(nèi)無中性點(diǎn)接地。試計(jì)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)戴維南等值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：正負(fù)零三序網(wǎng)如圖（a），各序端口的戴維南等值電路如圖（b）（a）單相短路，復(fù)合序網(wǎng)圖如圖（c）則：I(1)I(2)I(0)Uf010.5x(1)/xfx(2)/xfxf0.50.51（b）5-1-3圖5-34示出一簡(jiǎn)單系統(tǒng)。若在線路始端處丈量Za?、Zb?、UagIaUbgIb?Za、Zb、Zc和ZcUcgIc。試分別作出f點(diǎn)發(fā)生三相短路和三種不對(duì)稱短路時(shí)（可取0、1）的關(guān)系曲線，并分析計(jì)算結(jié)果。fGTxn1l解：其正序等值電路：EaxGxTlxl5-2-1已知圖3-35所示的變壓器星形側(cè)?B、C相短路的

25、If。試以If為參照向量繪制出三角形側(cè)線路上的三相電流相量：（1）對(duì)稱重量法；（2）相重量法。?xIaaA?IbybB?IcczIfC1、對(duì)稱重量法?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)三角側(cè)零序無通路，不含零序重量，則：?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?電流向量圖：此中相電流Ia與相電流IA同相

26、位，Ib與IB、Ic與IC同相位。?1?1?1?N1：N23：1。且IaIA、IbIB、IcIC。原副邊匝數(shù)比333I?aIA?Ia?IcIBIbIC化為矩陣形式為：?Ia110Ia110IA1100?1?1?Ib011I011IB011Ifb3?101?3101?101?IcIcICIf5-2-2由正序增廣網(wǎng)絡(luò)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）中間性點(diǎn)接地時(shí)因?yàn)橛晒絀a3I(1)3Uqk0z(1)z(2)z(0)3zqk因此Ia31.1j0.49j(2.252.252.25)2）中間性點(diǎn)不接地時(shí)因?yàn)橹虚g性點(diǎn)不接地時(shí)，則沒有零序電流因此Ia2I(1)2Uqk0z(1)j0.326z(2)z(0)3zqk第六章電力系統(tǒng)穩(wěn)固性問題歸納和各元件的機(jī)電特征6-2-1（1）TJ2.74GD2n28.74s1000SB（2）

28、TJNTJSN14.98sSB6-2-2若在例6-2中的發(fā)電機(jī)是一臺(tái)凸極機(jī)。其參數(shù)為：SN300MW，UN18kV，cosN0.875，xd1.298，xq0.912，xd0.458試計(jì)算發(fā)電機(jī)分別保持Eq0，Eq0，Uq0為常數(shù)時(shí)，發(fā)電機(jī)的功率特征。P0,cos0G1T1LT2U115kVq?Eq?jId(xdxq)jIxq?EQ?jIxd?E?jIxejI(xTxL)EqUG?jIqxq?UqUG?IqI?dIdUd解：（1）取基準(zhǔn)值SB250MVA，UB(110)115kV，UB(220)115220209kV，則121阻抗參數(shù)以下：2502xd1.2892421.2602093000.

29、8752502xq0.9122420.8922093000.8752502xd0.4582420.4482093000.8752502xT12420.1300.142093602502xT22200.1080.14209360 xL10.412002500.23522092系統(tǒng)的綜合阻抗為：xexT1xLxT20.1300.1080.2350.473xdxdxe1.2600.4731.733xqxqxe0.8920.4731.365xdxdxe0.4480.4730.921（2）正常運(yùn)轉(zhuǎn)時(shí)的UG0，E0，Eq0，Eq0：P02501，Q01tg(cos10.98)0.2，U1151250115

30、由凸極機(jī)向量圖得：?令US10，則：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與例題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)各電動(dòng)勢(shì)、電壓分別保持不變時(shí)發(fā)電機(jī)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并聯(lián)電抗器使系統(tǒng)中xd減小，因此發(fā)電機(jī)的功率極限將增大。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時(shí)，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判據(jù)一：K10判據(jù)二：K4KeK5K40判據(jù)三：K1K2K4）Ke(K1K6K2K5)K1K2K40K1K2K3K4（K3K3第八章電力系統(tǒng)暫態(tài)穩(wěn)固8-2-2在例8-1中若擾動(dòng)是忽然斷開一回線路，是判斷系統(tǒng)能否保持暫態(tài)穩(wěn)固。P0220MWcos00.98U115kV300MW200kM18kV360MW360MWcos0.85x10.41/kMxdxq