電力工程習(xí)題集

1、X10.01570.0145 lg-DjPrC1b1 31.5S 116.327 kM60.0241 102 fc1100（查表LGJ-120的截面為116.3mm2）0.0241lg 663.168.54 10 9答：該線路每公里的電阻為_62.68X10610 68.54 10 9 F kM268 10 6SkM0.27 kM,電抗為0.423kM,電納為電力工程習(xí)題集1. 有一回110千伏架空輸電線路，導(dǎo)線型號(hào)為 LGJ-120,導(dǎo)線為水平排列并 經(jīng)過完全換位，線間距離為4米，試計(jì)算每公里線路的電阻，電抗和電容。查表得：LGJ-120的導(dǎo)線直徑為：15.2mm, r =7.6mm解：已知

2、線間距離為水平4米，故幾何均距離Djp 3 2D 3 2 4 1.25992 4 1.26 4 5.04M5 04/O.。157 o.01450.423 kM02、已知一臺(tái)SSPSOL型三相三線圈自耦變壓器,變壓器的額定容量比為300000/300000/15000（kVar，lo=O.5, UiN=242kV，U2N=121kV;U3N=13.8kV，Pk（i 2） 950kWPKd 3） 500kW; PK（2 3） 620kW; P0123kW;Uk（1 2）%=13.73； Uk（1 3）%=11.9； Uk（2 3）%=18.64求以高壓側(cè)為基準(zhǔn)的該變壓器的基本參數(shù).解:1進(jìn)行各Pk

3、的歸算.PK (12)950kW;Pk (1Pk(13)(SN)2=500X 4=2000kWPK (23)PK(2 3)(學(xué))2S3N =620 X 4=2480kW于是:PK1 = 2 Pk(1 2) + Pk(1 3) Pk(2 3) = 2 950+2000 2480=235kWGm2.3.4.Fk2 =PK3 BmPK (1PK(12) Pk1 =950 235=715 kW3) P1 =2000235=1765kW,103 UN10 % SnuN123 10 32.1 10 6S242210 50.530000010 5242252.56 10 S計(jì)算各繞組的電阻:sN(3105)

4、2PK2uN1037152422 103sN(3105)2PK3uN10317652422 1 03sN(3105)20.4651.149RK2RK2RK1 =235 2422 103Pk1 uN 1030.15295. 進(jìn)行Uk%的歸算:u K（1 2）% =13.73S1NUk（i 3）%=Uk（i 3）%x（s3n）=ii,9x2=23.8S2 Nu K（2 3）%=U K（2 3）%（ S3N ）=18.64 X 2=37.28于是u K1% = 2 U K（1 2）% + U K（1 3） % u K（2 3） %2=2 13.73+23.8 37.28=0.125Uk2%=U K（

5、1 2）% Uk1%=13.73 0.125=13.605u K3% =Uk（1 3）% Uk1%=23.8 0.125=23.6756. 求電抗:Xk110uNuK1 % =1024220.125=0.244SN:3105X K210u N102422SnuK2 % =310513.605=26.56X K310uN102422K3 % =Snu二 310523.675=46.223、某330kV輸電線路，采用LGJQ-300X 2的雙分裂導(dǎo)線，導(dǎo)線為水平排列并經(jīng)完全換位，線間距離為9m,分裂導(dǎo)線的距離為400mm。試求每公里線路的 電抗和電容值。解：已知線間距離D=9m，導(dǎo)線為水平排列，故

6、幾何平均距離Djp 3 2 D 1.26 9 11.34m查表得：LGJQ-300的導(dǎo)線直徑為23.7mmd 23 7r 11.85mm 0.01185m故半徑 22X1電抗:啤。.侶/卩.0157rd0.1445lg114 26.885 10 2子導(dǎo)線的幾何均距九400mm04m故等值半徑：rdr .dQW 0.4 10 46.885 10 2 口=0.00785+0.3203=0.328 kMC1電容:0.0241D jplgrd1060.0241 10 611.34lg26.885 10 21.087 10 8b1=0.01087FkM2 fc17.58D jplgrd10 67.581

7、06lg 1134 66.885 10 63.4195 10 64、cos =0.88因此，負(fù)荷功率:S2P2 jP tan線路末端充電功率 1=150+j 150X tan(cos 0.88)=150+j 80.96 MVAB 232QC2 j U； j 0.243 10 3 2202j11.762MVAR通過線路阻抗的功率S2 S2QC2 =150+j 69.2 MVAUlP2 Rl Q2 X L150 14.4 69.2 75.24線路的電壓降:U2220試計(jì)算習(xí)題5所示輸電系統(tǒng)的線路及變壓器內(nèi)電壓降落的縱分量及橫分量;以及線路和變壓器的功耗(一)已知：P2=150MW,=33.48 k

8、VUlP 2 X l Q2Rl150 75.24 69.2 14.4U2220=46.77 kV從而得：U1； (22033.48)246.772257.76kVPl線路的功率損耗:2 2P?Q2U22Rl2 215069.2220214.4=8.12 MW2Ql蟲2Q2UTXl150269.22220275.24=42.43MVAR S1 S2Sl =150+ j 69.2+8.12+j 42.43=158.12+j 111.63 MVA線路首端的充電功率：B 232QC1 j U1 j 0.243 10257.76j16.642MVARI. 3 S1Qc1 =158.12+j 111.63

9、-j 16.14=158.12+j 95.49 MVAs是通過變壓器的功率.（二）變壓器上的壓降:UkRrkQ1 X kU1158122.0595.4935.1514.28kv257.76Uk(R Xk Q1 Rk)158.12 35.0595.49 2.05 20.80kVU1257.76從而:2Uk .(257.76 14.28)20.802272.83 kV變壓器上的功率損耗:Ptn Pon PkS1(1 Sn)n-S1158.12j95.49MVAPo96kWSn510 kVAPk700kWPt2 9622700型竺295.49 )4 1041.386MWQt(Io%SN )100Si

10、/k% Sn (100 n SN)21.9 10510012 105158.12295.492524.27MVAR4 10100（注：若Io%12 ,則Qt 44.47MVAR)SK S1SK 158.12j95.49 1.386 j 24.27159.506j119.76MVA199.4636.9 MVA(J139.96)(注：若 qk 為 139.96，則 Sk2122 41.27 MVA)Sk 212.2MVA兩臺(tái)變壓器的額定容量。2X Sn 200 MVA5、某110KV雙回輸電線路，線路長100公里，輸送功率為80MW，功率因素0.85,已知最大負(fù)荷利用小時(shí)數(shù)為Tmax=6000小時(shí)

11、，如果線路利用鋼芯鋁導(dǎo)線， 試選擇導(dǎo)線的截面和型號(hào)。解：因?yàn)榫€路為110k V,按經(jīng)濟(jì)電流密度選擇導(dǎo)線截面,線路輸送電流：IP.3U cos1I80 1 034943 110 0.85AI 494247A每回線的電流：2 2由Tmax=6000h,查得經(jīng)濟(jì)電流密度Jec=0.9,P88表2-11導(dǎo)線截面:IJ ec2470.9274mm2查附表3 P324得知:以上兩種型號(hào)均滿足要求,且兩回線的其中一條斷線時(shí),單回路仍能通過全型號(hào)溫度70 r80 r90 rLGJ-240420A520A600ALGJ-300511A638A740A部494A的電流。所選截面自然滿足按不發(fā)生電暈的條件和機(jī)械強(qiáng)度

12、條件的要求6、某500千伏遠(yuǎn)距離輸電線路，線路長為 550kM，送功率80萬kW,功率因素0.95,最大負(fù)荷利用小時(shí)Tmax=6500小時(shí)，試選擇導(dǎo)線的截面和型號(hào)解：線路輸送的電流:P3U cos800 1033 500 0.95972.38A按經(jīng)濟(jì)電流密度查得 Jec=0.9I 972.382F1080.42mmJ ec 0.9按電暈條件要求500kV,線路至少要用LGJQ-400X 390 r829A單根LGJQ-400的載流容量為：70 T80E573A716A用三根分裂導(dǎo)線后，足以通過工作電流。7、圖中d點(diǎn)發(fā)生三相短路，求次暫態(tài)短路電流和穩(wěn)態(tài)短路電流值。F:汽輪發(fā)電機(jī)、有自動(dòng)電壓調(diào)整器

13、：Sf =60 MVA, Uf=10.5 kV, Xd =0.125;B1:Sb1=60 MVA.Uk%=10.5變比：10.5/121 kVB2:Sb2=60 MVA.Uk%=10.5變比：110/38.5 kL :X=0.4 Q /kML=80Km、用近似法計(jì)算，變壓器變比用平均額定電壓Uav解:1.將發(fā)電機(jī)的次暫態(tài)電抗Xd折換成有名值:X X uN X d X dSn0.125警0.23(取 k仁 115/10.5, k2=37/115)2.將各元件電抗歸算至短路側(cè):（短路點(diǎn)uav=37KV , Sj=60MVA)X d(37)1152Xd荷37211520.2337210.522.85

14、6Xl(37)2.396 2.396 3.318.1Uk%uN37210.51152372100Sn11521006021152Uk%uN10.53722.396100Sn1006020.4 8037203.312.396115X L(37)X B1(37)X B2(37)X1 X B1(37) X B2(37)次暫態(tài)短路電流:E。Id Xd X1XdX1Uav3 )1.08 10.5115371153(2.8568.1)10.51.08 37,3 10.9562.1kA根據(jù)x fXd有名值:E。1.083723.07kV .3=0.05 . I LC =4.5.查 P158,圖4-13:E

15、=3.753.75K0 I LC0.47 4.51.773X d(37)Xd XjU1.773 SNF2NF 代21 21.77310.526011523722210.521152E (37)I (37)3721.77340.4660E Uj3.753780.11kV、3(.37)80.11X d (37)X140.46 8.11.65kA判強(qiáng)勵(lì)是否動(dòng)作:I (37) X11.658.113.365kV發(fā)電機(jī)母線電壓:Uf10.51153710.51153721.36kV(37) X1 U F強(qiáng)勵(lì)動(dòng)作，I(37)所求值。、用標(biāo)么值計(jì)算：(取Sj=60MVA, Uj=Uav)1 求各元件電抗標(biāo)么

16、值:Xd =0.125Uk%Sj10.560X B10.105100Sn10060Uk%Sj10.560X B20.105100SN10060Sj60XlXlj2u20.48011520.145XdEf K 0 1 LC3.750.47 4.51.773IIId11E01.081.082.250.48XdX10.125 0.105 0.105 0.,145實(shí)際值:IIISjI602.252.1kAJ3 37I dd 3Uj按 X fk =0.05 . Ilc =4.5 .查得 E =3.75也375!.76Xd X,1.773 0.105 0.105 0.,1452.1281.76Sj3U j

17、1.76一601.65kA.3 37X11.760.3550.625強(qiáng)勵(lì)動(dòng)作，丨即為所求。、用精確法求解（有名值）Xd =0.125 Xd =0.23 Q1.求出短路點(diǎn)側(cè)各元件電抗值：Xd(38.5) = xdk12k；=.23121210.5238.5211023.74X B1(38.5)Uk% uN100 Sn10.51001212602進(jìn) 3.14110X B2(38.5)Uk%uN100 SN10.5 38.52100 602.59X L(38.5)0.4 8038.5211023.92X1E。IdXdXdI Xd X13.14 2.591.08Un.3 k1 k2E0xd X=0.0

18、5 .K01 LC3.929.651.0826.410.512138.510.51103.74 9.651.97kALC =4.5 .查得 E =3.753.750.47 4.51.77326.4kVST k12k； = 1.773UjE k1k23XdX13.7510.5260121210.5210.5 竺10.538.5110.3(53 9.65)158.83(53 9.65)1.46kA8、當(dāng)短路電流發(fā)生在圖中的d點(diǎn)時(shí)，試求次暫態(tài)短路電流和沖擊短路電流。已知發(fā)電機(jī) F: Xd =0.125, Sn=30MVA, U n =6.3kV變壓器 B: Sn=30MVA, Uk%=io.5電抗器

19、 P： 1 N =400A, U n =6kV, Xp%=4、用標(biāo)幺值求解：S U -設(shè) Sj =30MVA, u j =6.3kV求各元件阻抗:Xd =0.125Uk% 10.5Xb = 100 = 100 =0.105Xp% Un 弐Xp%= 100 X 3 I N X U j4630= 100 3 0.4 6.32 =0.26、等值電路XE XdEcXb0.0571.080.1250.0571.03650.105Xd XbiiXd Xb125 O。50.0570.125 0.105Id1.0365IdId I jXpIdi ch、2kch I d= 0.057Sj.3U j3.270.2

20、6303.27 8.99kAV3 6.3、2 1.9 8.99 24.273kA若還要求I*,則可以用運(yùn)算曲線法:、由等值電路求轉(zhuǎn)移電抗0.105X2X5 X2 X3X2X3X10.1050.260.105 0.260.1250.58330、求計(jì)算電抗：XX4js =0.695X 30 =0.695對(duì)電源X5js不需歸算：X5js = X5=0.583查表 P326,附圖 7,丨1 =1.5;P327附圖 9, I1 =1.72對(duì)系統(tǒng)：I2 =0.583 =1.72 I2 =1.72爐電源的周期分量不衰減）、求電流實(shí)際值:Sn30丨1= Il x 3U j =1.5X 3 6.3=4.124

21、kASn30|2= |2 x 3U j =1.72X 3 6.3=4.729 kA 1 = |i + |2=4.124+4.729=8.853kAich 2 kchld 2 1.9 8.853 23.79kA若kch =1 8 則 ichw2 1.8 8.853 22.54kA30301 =1.72X 3 6.3 + 1.72X 3 6.3 =4.729+4.729=9.458AUj26.32Zj1.323校驗(yàn)：Sj 30Uf3X3I3 x 0.26X 1.323X 9.458=5.635VUNI =9.458kA9、Uc=85Z 175 V,求其正序、負(fù)序和零序電壓。有一三相不平衡系統(tǒng)，已知

22、 Ua=80/ 10 V, Ub=70/ 135 V,Ua1111a2 aUa1Ub1 1aa28010Ua22 aa2 aa 7013533Ua0111Uc111 8517580 107025585 55109.42j15.921180 1070585 295182.28j453380 107013585 17555.42j70.491110.578.2836.858.28187.8513.8662.5813.86389.9 128.0629.97128.06所以:Ua1=36.85/ 8.28Ua2=62.85/ -13.86U a0 U b0 U c0Ub1=36.85/ -111.72

23、U b2=62.85/ 106.14=29.97/ 128.Uc1=36.85/ 128.28Uc2=62.85/ -133.8610、有一三相不平衡系統(tǒng)，已知Ua=80/ 10。伏,Ub=70/ 135伏,Uc=85/ 175伏，試求其正序、負(fù)序和零序電壓。解：1Ua1 -(Ua3aUBa2Uc)1 (80 10370 2558555 )1丄(78.783j13.8918.12j67.61 48.75 j69.63)1-(109.413j15.91)36.47j5.3)36.85 8.27 伏1Ua2 (Ua3a2UBa Uc)1 (80 10370 158565 )1-(78.783j 1

24、3.8967.61j18.12 35.92 j77.04)1(182.313j45.03)60.77j5.01)62.6013.87 伏1Ua0 -(Ua Ub Uc)31(80 1070 13585 175 )31-(78.78 j 13.89 49.5j49.5 84.86 j7.41)1(55.4 j 70.8)18.47j23.6329.97 128.05 伏11、如圖所示220KV單回路架空線路，裝有兩根良導(dǎo)體的架空地線,線路各LGJJ150解:查表得:LGJJLGJJ零序電抗:取Dd150300x0t)X1000MrD jP 3 DAB D BC DCA17.5225.682X；t

25、。XTOX。8.75mm12.84mm0.,434lgVrDjP0.81r0.81 12.84 10 310.4004 103.5 5 1 053 26.3M3M部分尺寸如圖，已知導(dǎo)線為LGJJ300型鋼芯鋁線，架空地線為 型鋼芯鋁線，求該線路單位長度的零序電抗和零序電容。X。0.434(lg000 10)1.36.kM7.445DnT6.(22 52)(32 52)(5282)6 29 34 896.67MXnTO0.434 Ig Dd 0.434lg10000.944 kMDnT6.672 jP3 rD2p 3 10.4 25 3 4 10 17.445 10 1MrTPjJtDt0.81

26、 rD/0.81 8.75 10 3 6 0.206MXto0.434lg-D-DrTPj0.434 嚴(yán)0.2061.6 kM(T)X0最終X。xIt。XTO1.3601.60.80 kM2,求零序電納:D AAdabDacH PjrPj7.5810 63 lg匕rPjDnrlg DnTH TPj lgrTPj6.5 213M DBBdCC.132 52.194 13.93M.132 10226916.4M:D aaDbBDcCDaBDBCDca93dBcdAbdAc9 133 194 194 269 14.115M9 12.843 2525 100 10 10.8MD nr6.(1822 2

27、2 )(1832)(18282)6 328 333 38818.67MrTPjG Dt8.7510 3 60.23MDnT6.85MH TPj4 232(23262)23.38M0.43545D nr lgDnrlg也rPj普 1.24666.85HTPj lgrTPj嚴(yán)0.232.007c0t)10 63 (1.244660.42545 )2.007b0T)2.193 10 66.981007.582.193 10 6S kM10 9 F kM12、某110kV單回架空線路，采用等邊三角形布置，如圖上方有一根鋼質(zhì)架空地導(dǎo)線：LGJ 240架空地線：GJ 120鍍鋅鋼鉸線，d 14mm求單位長

28、度的零序電感和零序電容解：LGJ240：GJ120：3.5M10.64mm2匕7mm2D jPr 0.81r0.8110.648.62mmX。O.,434 lg yd=VrDjp0.434lg310 10$8.62 3.51.52 kMXTO0.434 lg D1000沁廠看2237XnTo 0.434lg0.434lg10000.972 kMDnT5.75DnT6(422 21 )(5.752 22 )(7.512)6 17 37.0625 57.255.57 M(T)X 0X。X 2nTO1.520.97221.10 . kMXTO2.237L(T)L0仃）X03.4910 3H kM10

29、0H TPj2 1530MrPj9 r3DBCDABDAC9 (10.64 10 3)3 3.560.507MH PjIonn9 15 18.5.”(7.5 9.25)322G (11 9.25)232 )2rTPj9315 18.522 41.0625 289.5625lg匕rPjD nTC lg18.5M18.5F 156210 3m6 (22.521)(2621)(24.2524)H TPjrTPj30lgk3-6326 507.25677 592.062524.25 MB噲7.58b0T)lg6 507.25 677 592.06256 17 37.0625 57.25106I6 y1

30、-(lgx Igy)60.6253 (A7.58 10 60.62523(1.562)3.6321.737 106 S kM仃）C0b0T)5.5310 9 F kM13、有如圖5-72所示電力網(wǎng),已知10.EKV o A|230KV(3)50IM30KM2+50MVAB35-T2F: 2X 50MW, cos 0.8, UN=10.5kV, Xd* 0.125 , X2* 0.15, X0* 0.08B1,B2: SE50000 型，S=50000kVA U%=12電壓比 242/10.5 kV,Y0/ 連接組B3,B4: SF-40000 型，SN=40000kVA UK%=12電壓比 2

31、42/6.6 kV, Y0/ 連接組線路：l=80kM,Xi=0.4Q /kM|X0= 3Xi1,試分別作出下列三種運(yùn)行方式下:1)B1|B2,B3,B4的中性點(diǎn)全部接地;2)B1 中性點(diǎn)不接地,B2,B3,B4中性點(diǎn)接地;3)B1,B4中性點(diǎn)不接地,B2,B3中性點(diǎn)接地;當(dāng)d點(diǎn)發(fā)生單相接地短路時(shí)的零序等值網(wǎng)絡(luò)2,求出上述三種下d點(diǎn)單相接地短路時(shí)的短路電流和非故障相電壓升高，并進(jìn)行比較.3,求出當(dāng)B1,B2,B3,B4全部接地時(shí)d點(diǎn)發(fā)生兩相接地短路時(shí)的短路電流.解:1,三種運(yùn)行方式下的零序網(wǎng)絡(luò)：0- 24XR20r xIQ(Q4 30. 3IB300. 1140.069XbioXB30XB20

32、X,cE402,各元件各序阻抗值(取Sj=100MVAUj=Upj=230 kV)XdSiXd氏F:1000.1250.250()0.8X2X2SjSN0.151000.24X。X。SSN0.810050()0.80.128X B1 X B2B:Uk%100SSN12型 0.2450Uk% Sj工型0.3100 40XB2D0-24Xbio址X B120. 114監(jiān)0XM )(Xii1X B32(0.1 0.120.258/0.1730.038)/(0.023 0.173)設(shè)變壓器的正,負(fù),零序阻抗相等.Xi1L:SjX1I12Uj1000.4 5022300.038 X12X ii 1SX1

33、I2U20.4 30j223020.023 XII2X103Xi13 0.0380.114X II 03Xii13 0.0230.069100 SnX B3 X B4根據(jù)序網(wǎng)圖求解X1E, X2刀,和X0刀,并求解短路電流與非故障相電壓升高。0.258 0.1730.258 0.1730.104X2負(fù)序網(wǎng)絡(luò)基本上和正序一樣，只是沒有E，且發(fā)電機(jī)負(fù)序阻抗為 TX2X B12XB3XH2 )/(XH2 尹(0.120.278/0.173Ij(1)0.120.278 0.1730.2780.1070.1730.038)/(0.023 0.15)Sj、3UX。Id1)Id1)X0X11003 :(X

34、B1(_2_(0.120.107X1Id1)0.251kA230Xi0)/(Xii00.144)/(0.0693EX。Sj3U j10.01他 1.0870.104非故障相電壓的升高系數(shù):dj(1.5 1.087)23:(1.087 2)24UbUc djXB3 )0.15)3 1.080.104 0.107 0.11310.011002.51kA、3 2301.014(Xb1(0.240.135Id1)2301.014Xi0)/(Xii0230134.69kVX B3 )2 )0.144)/(0.069 0.15)3EX1 X2 X3 1.080.104 0.107 0.1359.36Id1)Id1) I j 9.36 0.2512.35kAX。X10.1351.2980.104dj(1.5 1.298)2(1.2982)231.0484UbUc dj23032301.04 139.18kVV3(3)Xo(Xb1Xi0)/(Xii0 Xb3 )Id1)Id1)X。djUb3,在（1）已知:(n)I dX(1J)(1,1)1 d1(1,1) mi (1,1)I dI (1,1)I d(0.240.1