电力系统分析 苏小林 阎晓霞主编 课后答案 中国电力出版社 118页

'欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/阳光大学生网【欢迎光临】阳光大学生网：）最专业的课后答案，期末试题网站，一切成功源于积累，谨以此站献给所有奋斗路上的我们。关注大学生成长。一起分享大学里的学习资料和生活感悟，免费提供:大学生课后答案，大学考试题及答案，经典好书推荐及理由。第一章1-1解：105发电机G额定电压：1010.5KV100变压器额定变比：220110/100242110110T：，T：220:(110):(35)220:121:38.51210105/10010.51001001101103535T：，T：34101.11161.16.61-2解：(1)：发电机G：10.5KV变压器高低压侧额定电压：1211101103535T：，T：，T：12310.5351.138.5101.111(2)变压器实际变比：121(12.5%)12411035(15%)33.25T：，T：，T：12310.510.538.5101.1111-3解：日用电量：W702504802100480290412047022040(MWh)W2040日平均负荷：Pav85(MW)2424Pav85负荷率：0.708P120max 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/P50min最小负荷率：0.417P120maxW11-4解：T(1002000604000302760)5228(h)maxP100max第二章2-1解：线路长度为65km，属短线路，不计分布性参数31.5r0.105(/km)1S30033DDDD60006000120007559.5(mm)mabacbcDmx0.1445lg0.01570.4215(/km)1r7.5866b102.710(S/km)1DmlgrRrl0.105656.825()1Xxl0.42156527.3975()164Bbl2.710651.75510(S)1等效电路：2-2解：(1)不计分布性参数 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/31.5r0.0197(/km)1S440033DDDD12000120002400015119.1(mm)mabacbc44rrddd13.624004004002187.38(mm)eq121314Dm0.015715119.10.0157x0.1445lg0.1445lg0.279(/km)1rn187.384eq7.5866b103.97510(S/km)1DmlgreqRrl0.01971202.364()1Xxl0.27912033.48()164Bbl3.975101204.7710(S)1等效电路：(2)不计分布性参数31.5r0.0197(/km)1S440033DDDD14000140001400014000(mm)mabacbc44rrddd13.624504504502204.68(mm)eq121314Dm0.0157140000.0157x0.1445lg0.1445lg0.269(/km)1rn204.684eq7.5866b104.13110(S/km)1DmlgreqRrl0.01971202.364()1Xxl0.26912032.28()164Bbl4.131101204.957210(S)12-3解： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/Rrl0.120020()13XLl250210200125.66()164Bbl2500.01102006.2810(S)122l36200k1xb12502102500.01100.974r1133222r1b1llk1(xb)0.987，k1xb1.007x11b11x6121ZkRjkX0.97420j0.987125.6619.48j124.026()rx44YjkBj1.0076.2810j6.32410(S)b2-4解：221PSUN160110R3.63()T2221000S2100010N221Uk%UN110.5110X63.525()T2100S210010NP018-6G222.97610(S)T221000U1000110NI0%SN0.910-5B221.48810(S)T22100U100110N型等效电路：2-5解： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/1P(PPP)93(kW)S1S12S13S2321P(PPP)52(kW)S2S12S23S1321P(PPP)65(kW)S3S13S23S12222PS1UN93121R3.404()T1221000S100020N22PS2UN52121R1.903()T2221000S100020N22PS3UN65121R2.379()T3221000S100020N1U%(U%U%U%)11S1S12S13S2321U%(U%U%U%)0.5S2S12S23S3121U%(U%U%U%)7S3S31S23S12222US1%UN11121X80.526()T1100S10020N22US2%UN-0.5121X3.66()T2100S10020N22US3%UN7121X51.244()T3100S10020NP043.3-6G2.95710(S)T221000U1000121NI0%SN3.4620-5B4.72610(S)T22100U100121N等效电路图：2-6解： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/SN2P()P4P1080(kW)，P4P1380(kW)S23S23S23S13S13S3N1P(PPP)312.5(kW)S1S12S13S2321P(PPP)12.5(kW)S2S12S23S1321P(PPP)1067.5(kW)S3S13S23S12222PS1UN312.5220R1.867()T1221000S100090N22PS2UN12.5220R0.075()T2221000S100090N22PS3UN1067.5220R6.379()T3221000S100090NSNU%()U%2U%24.2，U%2U%37.2S23S23S23S31S31S3N1U%(U%U%U%)11.5S1S12S31S2321U%(U%U%U%)1.5S2S12S23S3121U%(U%U%U%)25.7S3S31S23S12222US1%UN11.5220X61.844()T1100S10090N22US2%UN-1.5220X8.067()T2100S10090N22US3%UN25.7220X138.209()T3100S10090NP0104-6G2.14910(S)T221000U1000220NI0%SN0.6590-5B1.20910(S)T22100U100220N等效电路图： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/2-7解：（1）以220kv电压作为基本级22PU620242SNT：R0.63()1T221000S1000240N22U%U14242SNX34.162()T100S100240NP01856G3.15910(S)T1221000U1000242NI0%SN0.52405B2.04910(S)T122100U100242NSN2T：P()P4P592(kW)，P4P880(kW)2S23S23S23S13S13S3N1P(PPP)483.5(kW)S1S12S13S2321P(PPP)195.5(kW)S2S12S23S1321P(PPP)396.5(kW)S3S13S23S12222PU483.5220S1NR0.722()T1221000S1000180N22PU195.5220S2NR0.292()T2221000S1000180N22PU396.5220S3NR0.592()T3221000S1000180N1U%(U%U%U%)14.95S1S12S31S2321U%(U%U%U%)0.95S2S12S23S312 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/1U%(U%U%U%)9.05S3S31S23S12222US1%UN14.95220X61.844()T1100S100180N22US2%UN0.95220X2.554()T2100S100180N22US3%UN9.05220X24.334()T3100S100180NP02006G4.13210(S)T2221000U1000220NI0%SN0.81805B2.97510(S)T222100U100220NL：Rrl0.08604.8()111Xxl0.4176025.02()1164Bbl2.9110601.74610(S)112202L：R0.13250()21.818()221212202X0.40550()66.942()2121622025B2.761050()4.174510(S)2121等效电路：2USBB(2)选取S1000MVA，U220kV，Z48.4()，Y0.02066(S)BBBB2SUBB*RT*XT*GT14*BT14T1：RT0.013，XT0.706，GT11.52910，BT19.91810ZBZBYBYB*RT1*RT2*RT3T2：RT10.015，RT20.006，RT30.012ZBZBZB*XT1*XT2*XT3XT10.831，XT20.053，XT30.503ZBZBZB 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/*GT24*BT24GT2210，BT21.4410YBYB*R1*X1*B13L1：R10.099，X10.517，B18.45110ZBZBYB*R2*X2*B23L2：R20.451，X21.383，B22.02110ZBZBYB等效电路：(3)选取S1000MVA，UUBBav 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/*PSSB6201000T：R0.0111T221000S1000240N*US%SB141000X0.583T100S100240N*P01854G1.8510T11000S10001000B*I0%SN0.52403B1.210T1100S1001000B*PS1SB483.51000T：R0.0152T1221000S1000180N*PS2SB195.51000R0.006()T2221000S1000180N*PS3SB396.51000R0.012()T3221000S1000180N*US1%SB14.951000X0.831T1100S100180N*US2%SB0.951000X0.053T2100S100180N*US3%SB9.051000X0.503T3100S100180N*P02004G210T21000S10001000B*I0%SN0.81803BT21.4410100S1001000B*SB1000L1：R1r1l20.086020.091Uav230*SB1000X1x1l20.4176020.473Uav23022*Uav62303B1b1l2.9110609.23610SB1000*SB1000L2：R2r2l20.1325020.499Uav115*SB1000X2x2l20.4055021.531Uav11522*Uav61153B2b2l2.7610501.82510SB1000 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/等效电路为：2-8解：(1)计算线路参数：22PSUN410121T：R0.417()1T221000S1000120N22US%UN10.5121X12.811()T100S100120NP099.46G6.78910(S)T1221000U1000121NI0%SN0.61205B4.91810(S)T122100U100121N22PSUN300110T：RRR0.457()2T1T2T3221000S200063N1U%(U%U%U%)10S1S12S31S2321U%(U%U%U%)0.5S2S12S23S3121U%(U%U%U%)6S3S31S23S122 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/22US1%UN10110X19.206()T1100S10063N22US2%UN0.5110X0.96()T2100S10063N22US3%UN6110X11.524()T3100S10063NP084.76G710(S)T2221000U1000110NI0%SN1.2635B6.24810(S)T222100U100110NL：Rrl0.4228033.76()1111Xxl0.4298034.32()11164Bbl2.6610802.12810(S)111L：Rrl0.4225021.1()2222Xxl0.4295021.45()22264Bbl2.6610501.3310(S)222L：Rrl0.4226025.32()1333Xxl0.4296025.74()33364Bbl2.6610601.59610(S)33322PSUN148110T：R1.805()3T221000S100031.5N22US%UN10.5110X40.333()T100S10031.5NP038.56G3.18210(S)T3221000U1000110NI0%SN0.831.55B2.08310(S)T322100U100110N等效电路图： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/(2)标幺制下，按精确计算法计算，选取2UBSBSB1000MVA，UB110kV，ZB12.1()，YB20.083(S)SBUB*RT*XT*GT15*BT14T1：RT0.034，XT1.059，GT18.1810，BT15.92510ZBZBYBYB***RT1*XT1**T2：RT1RT2RT30.038，XT11.587，XT20.079，XT30.952ZBZB*GT25*BT24GT28.43410，BT27.52810YBYB*RT*XT*GT35*BT34T3：RT0.149，XT3.333，GT33.83410，BT32.5110ZBZBYBYB*R1*X1*B13L1：R12.79，X12.836，B12.56410ZBZBYB*R2*X2*B23L2：R21.744，X21.773，B21.60210ZBZBYB*R3*X3*B33L3：R32.093，X32.127，B31.92310ZBZBYB2-9解：选取S1000MVA，UUBBav*SB100G：xdxd1.6280.1628SN1000*US1%SB10.51000T1：XT10.875100SN100120*SB1000L1：x1x1l120.410023.025Uav115*US2%SB10.51000T2：XT21.167100SN10090*UNSB61000R：XRXR20.0521.0913INUav346.3*SB1000L2：R2r2l220.45252283.447Uav6.3*SB1000x2x2l220.0825250.391Uav6.3等效电路图： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/第三章3-1解：(1)计算线路参数：111RR1r1l0.131006.5()222111XX1x1l0.41210021.05()22264B2B12b1l22.7101005.410(S)~12142SY2jBU2j5.410210j11.907(MVA)22~~~S2S2SY240j30j11.90740j18.093(MVA)2222~P2Q24018.093SZ(RjX)(6.5j21.05)0.284j0.92(MVA)22U2210~~~S1S2SZ40j18.0930.284j0.9240.284j19.013(MVA)P2RQ2X406.518.09321.05U3.052(kV)U2210P2XQ2R4021.0518.0936.5U3.45(kV)U221022U1(U2U)U213.08(kV)~12142SY1jBU1j5.410213.08j12.259(MVA)22~~~S1S1SY140.284j19.013j12.25940.284j6.754(MVA)电压向量图为：~(2)空载运行时：S20~~12142S2SY2jBU2j5.410210j11.907(MVA)22114UBU2X5.41021021.051.194(kV)22 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/114UBUR5.4102106.50.369(kV)22222U1(2101.194)0.369208.806(kV)222~P2Q2(11.907)SZ(RjX)(6.5j21.05)0.021j0.068(MVA)22U2210~~~S1S2SZj11.9070.021j0.0680.021j11.839(MVA)~12142SY1jBU1j5.410208.806j11.772(MVA)22~~~S1S1SY10.021j11.839j11.7720.021j23.611(MVA)5003-2解：计算变压器参数(归算到高压侧)：U35500(kV)35SN232502P()P()94.91647.569(kW)S1-3S1-3S3603NSN232502P()P()1616.319(kW)S2-3S2-3S3603N1P(PPP)218.675(kW)S1S1-2S1-3S2-321P(PPP)187.425(kW)S2S1-2S2-3S1-321P(PPP)1428.894(kW)S3S1-3S2-3S1-221U%(U%U%U%)13.1S1S1-2S1-3S2-321U%(U%U%U%)2.3S2S1-2S2-3S1-321U%(U%U%U%)31.5S3S1-3S2-3S1-2222PS1UN218.675500R0.097()T1221000S1000(3250)N22PS2UN187.425500R0.083()T2221000S1000(3250)N22PS3UN1428.894500R0.635()T3221000S1000(3250)N 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/22US1%UN13.1500X43.667()T1100S1003250N22US2%UN2.3500X7.667()T2100S1003250N22US3%UN31.5500X105()T3100S1003250NP098.97G3.95610(S)T221000U1000500NI0%SN0.0832506B2.410(S)T22100U100500N2222~PQ9056S(RjX)(0.635j105)0.029j4.719(MVA)ZT32T3T32U500~~SS90j5690.029j60.719(MVA)2ZT3PRT3QXT3900.63556105U11.874(kV)T3U500PXT3QRT390105560.635U18.829(kV)T3U50022U(50011.874)18.829521.22(kV)222~400194S(0.083j7.667)0.066j6.061(MVA)ZT22500~~SS400j194400.066j187.939(MVA)2ZT2400.0660.083187.939(7.667)U2.748(kV)T2512.22400.066(7.667)187.9390.083U6.019(kV)T2512.2222U[512.22(2.748)](6.019)515.003(kV)2230(12.5%)中压侧电压：UU230.979(kV)中3500~~~SSS400.066j187.93990.029j60.719490.095j248.658(MVA)222490.0950.097248.65843.667U21.291(kV)T1512.22490.09543.667248.6580.097U41.734(kV)T1512.2222U(512.2221.291)41.734535.141(kV)1高压侧电压：UU535.141(kV)高1 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/3-3解：计算线路参数：Rr1l0.1310013()Xx1l0.410040()64Bb1l2.8101002.810(S)220电压归算到高压侧：UUC200(kV)38.522~8040SZT(3.025j50.4)0.605j10.08(MVA)2200~~~S2S2SZT80j400.605j10.0880.605j50.08(MVA)PRTQXT803.0254050.4UZT11.29(kV)U200PXTQRT8050.4403.025UZT19.555(kV)U20022U2(UUZT)UZT212.193(kV)~2662SYT(GTjBT)U2(2.0710j29.710)212.1930.093j1.337(MVA)~12142SY2jBU2j2.810212.193j6.304(MVA)22~~~~S2S2SYTSY280.698j45.113(MVA)P2RQ2X80.6981345.11340UZ13.448(kV)U2212.193P2XQ2R80.69840-45.11313UZ12.448(kV)U2212.19322U1(U2UZ)UZ225.984(kV)2222~P2Q280.69845.113SZ(RjX)(13j40)2.468j7.593(MVA)22U2212.193~12142SY1jBU2j2.810225.984j7.15(MVA)22~~~~S1S2SZSY183.166j45.556(MVA)即始端电压为225.984kV，始端功率为(83.166+j45.556)MVA 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/3-4解：计算线路参数：AB:R0.08655.2()ABX0.416526.65()AB64B2.710651.75510(S)ABB:R0.108808.64()1B1X0.428033.6()B164B2.6910802.15210(S)B1设全网电压为额定电压UN220kV，计算功率分布： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/22~~~2010662SSS(13.95j218.2)(1.72910j5.8610)220TBZTBYTB22200.228j2.538(MVA)~~SS20j1020.228j12.538(MVA)BTB22~4030S(1.46j0)0.075(MVA)ZT2222022~3020S(2.92j117.7)0.078j3.161(MVA)ZT32220~~SS40j3040.075j30(MVA)2ZT2~~SS30j2030.078j23.161(MVA)3ZT3~~~SSS70.153j53.161(MVA)22322~70.15353.161S(1.46j48.9)0.234j7.828(MVA)ZT12220~~~SSS70.387j60.989(MVA)12ZT1~2662S(GjB)U(0.84710j3.5310)2200.041j0.171(MVA)YTTTN~~~~142SSS70.428j61.16(MVA)Sj2.15210220j5.208(MVA)11YTYB112~~~SSS70.428j55.952(MVA)11YB1122~70.42855.952S(8.64j33.6)1.444j5.617(MVA)ZB12220~142Sj2.15210220j5.208(MVA)YB122~~~SSS71.872j61.569(MVA)B11ZB1~~~SSS71.872j56.361(MVA)B1B1YB12~~~SSS71.872j56.36120.228j12.53892.1j68.899(MVA)B1B1B~142Sj1.75510220j4.247(MVA)YAB12~~~SSS92.1j64.652(MVA)ABB1YAB122~92.164.652S(5.2j26.65)1.36j6.972(MVA)ZAB2220~~~SSS93.46j71.624(MVA)ABABZAB~142Sj1.75510242j5.139(MVA)YAB22~~~SSS93.46j66.458(MVA)AABYAB2用给定的始端电压和求得的功率分布，计算各段中的电压和变电所高压母线电压： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/PABRABQABXAB93.465.271.62426.65U9.896(kV)ABU242APABXABQABRAB93.4626.6571.6245.2U8.753(kV)ABU242A22U(2429.896)8.753232.269(kV)BPB1RB1QB1XB171.8728.6461.56933.6U11.58(kV)B1U232.269BPB1XB1QB1RB171.87233.661.5698.64U8.107(kV)B1U232.269B22U(232.26911.58)8.107220.838(kV)1P1RT1Q1XT170.3871.4660.98948.9U13.97(kV)ZT1U220.8381P1XT1Q1RT170.38748.960.9891.46U15.183(kV)ZT1U220.838122U(220.83813.97)15.183207.424(kV)0P2RT2Q2XT240.0751.46300U0.282(kV)ZT2U207.4240P2XT2Q2RT240.0750301.46U0.211(kV)ZT2U207.424022U(207.4240.282)(0.211)207.142(kV)2P3RT3Q3XT330.0782.9223.161117.7U13.566(kV)ZT3U207.4240P3XT3Q3RT330.078117.723.1612.92U16.741(kV)ZT3U207.424022U(207.42413.566)16.741194.58(kV)3121∴110kV侧的实际电压为：U110kVU2113.928(kV)22038.535kV侧的实际电压为：U35kVU334.052(kV)220对于B点，~0~SBSBSYTB20.228j12.538(0.084j0.284)20.144j12.254(MVA) 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/38.5变压器T1低压侧实际电压：UB35kVUBT38.552(kV)220将计算结果标于图上：3-5解：设全网电压为额定电压U110kV，计算功率分布：N2222~P2Q2120151S(RjX)(2.84j1.92)0.073j0.05(MVA)Z22T2T22U21102N~~SS20j1520.073j15.05(MVA)2Z22222~P3Q311081S(RjX)(5.68j24)0.038j0.163(MVA)Z32T3T32U21102N~~SS10j810.038j8.163(MVA)3Z3~~~SSS30.111j23.213(MVA)12322~30.11121.2131S(2.84j41.29)0.17j2.466(MVA)Z121102 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/~~~SSS30.281j25.679(MVA)11Z1~1262SjBUj218.810110j2.647(MVA)Y1N2~~~SSS30.281j23.032(MVA)A1Y122~30.28123.0321S(16.86j33.25)1.008j1.989(MVA)Z21102~~~SSS31.289j25.021(MVA)AA1Z~1262SjBUj218.810115j2.894(MVA)Y2A2~~~SSS31.289j22.127(MVA)AAY2根据给定电压U115kV计算各段电压：A11PRQXAA231.2890.516.8625.0210.533.252U5.911(kV)ZU115APA0.5XQA0.5R31.2890.533.2525.0210.516.86U2.689(kV)ZU115A22U(1155.911)2.689109.122(kV)1P10.5RT1Q10.5XT130.2810.52.8425.6790.541.29U5.252(kV)Z1U109.1221PA0.5XT1QA0.5RT130.2810.541.2925.6790.52.84U5.395(kV)Z1U109.122A22U(109.1225.252)5.395104.01(kV)0P20.5RT2Q20.5XT220.0730.52.8415.050.51.92U0.413(kV)Z2U104.010P20.5XT2Q20.5RT220.0730.51.9215.050.52.84U0.02(kV)ZT2U104.01022U(104.010.413)(0.02)103.597(kV)2P30.5RT3Q30.5XT310.0380.55.688.1630.524U1.216(kV)Z3U104.010P30.5XT3Q30.5RT310.0380.5248.1630.55.68U0.935(kV)Z3U104.01022U(104.011.216)0.935102.798(kV)3变压器中、低压侧实际电压为： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/38.5UU36.259(kV)中211011UU10.28(kV)低3110将计算结果标于图上：3-6解：计算线路参数：L:Rrl0.0541307.02()111Xxl0.30213039.26()1164Bbl3.75101304.87510(S)11L:Rrl0.054703.78()212Xxl0.3027021.14()1264Bbl3.7510702.62510(S)1222PSUN860242T:R0.389()T221000S1000360N22US%UN14.3242X23.263()T100S100360NP01906G3.24410(S)T221000U1000242NIS%SN0.283605B1.72110(S)T22100U100242N(1)设全网电压为额定电压UN220kV 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/~~42SSj4.87510220j23.595(MVA)Yl12Yl11~~~SSS270j106.405(MVA)BBYl112222~PBQB270106.4051SZ[(7.02j39.26)]6.108j34.159(MVA)ZAB2AB2U2202N~~~SSS276.108j140.564(MVA)ABZAB~~~SSS276.108j116.969(MVA)ABAYl121PARQAX1276.1087.02140.56439.26U16.21(kV)2U2230A1PXQR1276.10839.26140.5647.02AAU21.42(kV)2U2230A22U(23016.21)21.42214.86(kV)B~~42SSj2.62510220j12.705(MVA)Yl22Yl21~~~SSS180j74.295(MVA)CCYl212222~PQ18074.2951CCSZ[(3.78j21.14)]1.481j8.281(MVA)ZAC2AC2U2202N~~~SSS181.481j82.576(MVA)ACZAC~~~SSS181.481j69.871(MVA)ACAYl221PRQX1181.4813.7882.57621.14AAU5.286(kV)2U2230A1PXQR1181.48121.1482.5763.78AAU7.662(kV)2U2230A22U(2305.286)7.662224.845(kV)C即B点电压为214.86kV，C点电压为224.845kV。~300(2)厂用电负荷：S28%56.471(MVA)10.85~~~SPjQScosjSsin48j29.748(MVA)厂厂厂厂厂~~~SSS276.108j116.969181.481j69.871(MVA)AABAC22~~~P0I0%SNPS(0.5SA)US%(0.5SA)SSS2(j)2[j]TBYTBZTB210001001000S100SNN221900.28360860494.26414.3494.2642(j)2(j)1.191j50.536(MVA)21000100100043601004360~~~~SSSS506.78j267.124(MVA)厂ATB即发电机的发电负荷为(506.78+j267.124)MVA。3-7解：(1)不计功率损耗 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/~(3030)(20j15)30(10j10)S15j12(MVA)a403030~(4030)(10j10)40(20j15)S15j13(MVA)a403030~~~~SS30j25SSaaBC正常运行时，最大电压损耗出现在B点：U110kVN150.3340120.42940U3.672(kV)B110U1103.672106.328(kV)B110106.328最大电压损耗为：100%3.338%110AB故障切除：~~~SSS20j1510j1030j25(MVA)ABC30300.3325300.429U5.625(kV)C110U1105.625104.375(kV)C20300.3315300.429U3.747(kV)B104.375U104.3753.747100.628(kV)B110100.628电压损耗：100%8.52%110AC故障切除： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/~~~SSS20j1510j1030j25(MVA)ABC30400.3325400.429U7.5(kV)B110U1107.5102.5(kV)B10300.3310300.429U2.221(kV)C102.5U102.52.221100.279(kV)C110100.279电压损耗：100%8.837%110即切除AC线路时的电压损耗最大，为8.837%(2)计及功率损耗的功率分布：由(1)知S15j12MVA；S5j3MVA，取U110kVBBN22~1512S(0.33j0.429)400.403j0.523(MVA)AB2110~~~SSS15.043j12.523(MVA)ABAB22~53S(0.33j0.429)300.028j0.036(MVA)BC2110~~~SSS5.028j3.036(MVA)CBBC~~~SSS10j105.028j3.03615.028j13.036(MVA)CCC22~15.02813.036S(0.33j0.429)300.324j0.421(MVA)AC2110~~~SSS15.352j13.457(MVA)ACAC正常运行时，最大电压损耗出现在B点：15.4030.334012.5230.42940U3.637(kV)115U1153.637111.363(kV)B115111.363最大电压损耗为：100%3.306%115 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/AB故障切除：设全网为额定电压U110kVN22~2015S(0.33j0.429)300.511j0.665(MVA)BC2110~~~SSS20.511j15.665(MVA)BBBC~~~SSS10j1020.511j15.66530.511j25.665(MVA)CCB22~30.51125.665S(0.33j0.429)301.301j1.691(MVA)AC2110~~~SSS31.812j27.356(MVA)ACAC31.8120.333027.3560.42930U5.8(kV)C115U1155.8109.2(kV)C20.5110.333015.6650.42930U3.706(kV)B109.2U109.23.706105.494(kV)B115105.494电压损耗为：100%8.642%110AC故障切除：设全网为额定电压U110kVN22~1010S(0.33j0.429)300.164j0.213(MVA)BC2110~~~SSS10j100.164j0.21310.164j10.213(MVA)CCBC~~~SSS20j1510.164j10.21330.164j25.213(MVA)BBC22~30.16425.213S(0.33j0.429)401.686j2.192(MVA)AB2110~~~SSS31.85j27.405(MVA)ABAB31.850.334027.4050.42940U7.745(kV)B115U1157.745107.255(kV)B10.1640.333010.2130.42930U2.164(kV)C107.255U107.2552.164105.091(kV)C115105.091电压损耗为：100%9.008%110即切除AC线路时的电压损耗最大，为9.008%(3)不计功率损耗： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/~(20j15)(13.5j13.29.9j12.87)(10j10)(9.9j12.87)S15.093j12.36(MVA)a13.2j17.1613.5j13.29.9j12.87~(10j10)(13.2j17.1613.5j13.2)(20j15)(13.2j17.16)S14.906j12.64(MVA)a13.2j17.1613.5j13.29.9j12.87~~~~SS30j25SSaaBC正常运行时，最大电压损耗出现在B点：15.0930.334012.360.42940U3.739(kV)B110U1103.739106.261(kV)B110106.261最大电压损耗为：100%3.399%110AB故障切除：~~~SSS20j1510j1030j25(MVA)ABC30300.3325300.429U5.625(kV)C110U1105.625104.375(kV)C20300.4515300.44U4.484(kV)B104.375U104.3754.48499.891(kV)B11099.891电压损耗：100%9.19%110AC故障切除： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/~~~SSS20j1510j1030j25(MVA)ABC30400.3325400.429U7.5(kV)B110U1107.5102.5(kV)B10300.4510300.44U2.605(kV)C102.5U102.52.60599.895(kV)C11099.895电压损耗：100%9.186%110即切除AB线路时的电压损耗最大，为9.19%∴阻抗增大后，其最大电压损耗也将增大。3-8解：UU1150∴等效电路图如下：12不计功率损耗，计算功率分布：15151515(40j20)(3j12j)(80j50)(j)~8282S50.182j27.818(MVA)115152j83j12j82~(80j50)(3j122j8)(40j20)2j8S69.818j42.182(MVA)415152j83j12j82~5j20~SS43.636j26.364(MVA)443j125j20~3j12~SS26.182j15.818(MVA)443j125j20~~~~~SSS120j70SS14423功率分点在3点，分解网络进行计算 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/~S69.818j42.182(MVA)3~~~SSS50.182j27.818(40j20)10.182j7.818(MVA)312~5j20~SS43.636j26.364(MVA)433j125j20~3j12~SS26.182j15.818(MVA)443j125j202222~P4Q443.63626.364S(RjX)(3j12)0.644j2.578(MVA)34234342U110N~~~SSS43.636j26.3640.644j2.57844.28j28.942(MVA)44342222~P4Q426.18215.818S(RjX)(5j20)0.387j1.547(MVA)13213132U110N~~~SSS26.182j15.8180.387j1.54726.569j17.365(MVA)134132222~P3Q310.1827.818S(RjX)(3j12)0.041j0.163(MVA)23223232U110N~~~SSS10.182j7.8180.041j0.16310.223j7.981(MVA)2323~~~SSS40j2010.223j7.98150.223j27.981(MVA)2222222~P2Q250.22327.981S(RjX)(2j8)0.546j2.185(MVA)12212122U110N~~~SSS50.769j30.166(MVA)12212按给定电压UU1150计算各段电压：14P4R34Q4X3444.28328.94212U4.175(kV)34U1154U1154.175110.825(kV)3P12R12Q12X1250.769230.1668U2.981(kV)12U1151U1152.981112.019(kV)3将结果标于图上： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/3-9解：参数计算：L:Rrl0.081008()，Xxl0.410040()1111111L:Rrl0.06804.8()，Xxl0.428033.6()222222L:Rrl0.08705.6()，Xxl0.427029.4()23333322P1SU1N800242T:R2.082()1T1221000S10001501N22U1S%U1N14242X54.66()T1100S1001501N22P2SU2N800230T:R1.881()2T2221000S10001502N22U2S%U2N14230X49.373()T2100S1001502N22P3SU3N300220T、T、T:RRR2.269()345T3T4T5221000S1000803N22U3S%U3N10.5220XXX63.525()T3T4T5100S100803N(1)断路器QF合上时：222222~PSS5US%S5300(4025)10.5(4025)Sjj0.104j2.92(MVA)T5221000S100S1008010080NN~~~SSS40j250.104j2.9240.104j27.92(MVA)25T522~PS(S4/2)US%(S4/2)S2[j]T3421000S100SNN2222300(2010)10.5(2010)2[j]0.002j0.328(MVA)2410080100480~~~SSS20j100.002j0.32820.002j10.328(MVA)34T34不计网损及并联导纳，计算功率分布： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/~(40.104j27.92)(8j405.6j29.4)(20.002j10.328)(8j40)S128j405.6j29.44.8j33.634.5j23.508(MVA)~(20.002j10.328)(4.8j33.65.6j29.4)(40.104j27.92（)4.8j33.6）S138j405.6j29.44.8j33.625.606j14.739(MVA)~~~~SS60.106j38.247(MVA)SS1213232点为功率分点，设全网为额定电压U220kVN~~S34.5j23.508(MVA)，S25.606j14.739(20.002j10.328)5.604j4.411(MVA)222222~P2Q234.523.508S(4.8j33.6)(4.8j33.6)0.173j1.21(MVA)l222U220N~~~SSS34.673j24.718(MVA)12l22222~P2Q25.6044.411S(5.6j29.4)(5.6j29.4)0.006j0.031(MVA)l322U220N~~~SSS5.61j4.442(MVA)32l3~~~SSS20.002j10.3285.61j4.44225.612j14.77(MVA)3332222~P3Q325.61214.77S(8j40)(8j40)0.144j0.722(MVA)l122U220N~~~SSS25.756j15.492(MVA)13l1计算结果标于图上： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/(2)断路器QF断开后，因变压器变比不等，产生循环功率，为顺时针方向，dU24223012kV不计网损及并联导纳，计算功率分布：~(20.002j10.328)(4.8j33.65.6j29.41.881j49.373)S18j405.6j29.44.8j33.61.881j49.3732.082j54.66(40.104j27.92（)4.8j33.61.881j49.373）8j405.6j29.44.8j33.61.881j49.3732.082j54.66122208j405.6j29.44.8j33.61.881j49.3732.082j54.6628.542j28.944(MVA)~(40.104j27.92)(2.082j54.668j405.6j29.4)S18j405.6j29.44.8j33.61.881j49.3732.082j54.66(20.002j10.328)(8j402.082j54.66)8j405.6j29.44.8j33.61.881j49.3732.082j54.66122208j405.6j29.44.8j33.61.881j49.3732.082j54.6631.564j9.304(MVA)~~~~SS60.106j38.247(MVA)SS11232点为功率分点，设全网为额定电压U220kVN 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/~~S31.564j9.304(MVA)，S28.542j28.944(20.002j10.328)8.54j18.616(MVA)222222~P2Q231.5649.304S(4.8j33.6)(4.8j33.6)0.107j0.752(MVA)L222U220N~~~SSS31.671j10.056(MVA)L22L222~31.67110.056S(1.881j49.373)0.043j1.126(MVA)T22220~~~SSS31.714j11.182(MVA)1L2T222~8.5418.616S(5.6j29.4)0.049j0.255(MVA)L32220~~~SSS8.589j18.871(MVA)32L3~~~SSS20.002j10.3288.589j18.87128.591j29.199(MVA)33322~28.59129.199S(8j40)0.276j1.38(MVA)L12220~~~SSS28.867j30.579(MVA)L13L122~28.86730.579S(2.082j54.66)0.076j1.997(MVA)T12220~~~SSS28.943j32.576(MVA)1L1T1计算结果标于图上：3-10解：因变压器变比不等，产生循环功率，为顺时针方向，dU23120922kV不计功率损耗，计算基本功率分布： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/~(39j25)(39j251j256j504j40)SA2j2038j8039j251j256j504j40(150j100)(6j504j40)(50j30)(4j40)2j2038j8039j251j256j504j40222202j2038j8039j251j256j504j4098.306j55.594(MVA)~(50j30)(6j501j2539j2538j802j20)SA2j2038j8039j251j256j504j40(150j100)(1j2539j2538j802j20)(39j25)(38j802j20)2j2038j8039j251j256j504j40222202j2038j8039j251j256j504j40140.694j99.407(MVA)~~~~~SS239j155(MVA)SSSAAabc~~~SSS98.306j55.594(39j25)59.306j30.594(MVA)CAC~~~SSS140.694j99.407(50j30)90.694j69.407(MVA)aAa计算结果标于图上： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/第四章 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/4-1解：111（1）Yyyy3.667j11111213140.1j0.30.15j0.450.05j0.151YY1j312210.1j0.31YY0.667j213310.15j0.451YY2j614410.05j0.1511Yyy2.724j7.312221230.1j0.30.08j0.21YY1.724j4.3123320.08j0.2YY02442111Yyyy3.391j9.31333132340.15j0.450.08j0.20.1j0.31YY1j334430.1j0.311Yyy3j94441430.05j0.150.1j0.3得到节点导纳矩阵：3.667j111j30.667j22j61j32.724j7.311.724j4.310YB0.667j21.724j4.313.391j9.311j32j601j33j9(2)Y1j311Y1j322YY1j31221修改后的导纳矩阵中：Y3.667j111j32.667j811YY1j31j301221Y2.724j7.311j31.724j4.3122其余元素不变。 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/2.667-j800.667j22j601.724j4.311.724j4.310YB0.667j21.724j4.313.391j9.311j32j601j33j94-2解：11（1）Yyy0.935j4.2951112130.12j0.50.08j0.41YY0.454j1.89112210.12j0.51YY0.481j2.40413310.08j0.4YY01441YY0155111Yyy1.042j4.2442221250.12j0.50.1j0.4YY02332YY024421YY0.588j2.35325520.1j0.411Yyyyj250.481j30.737333031340.08j0.4j0.31YYj3.3333443j0.3YY03553111Yyyj6.6664434452j0.3kj0.3*11Yj3.33345kj0.3*11Y0.588j5.68655j0.30.1j0.4得到节点导纳矩阵：0.935j4.2950.454j1.8910.481j2.404000.454j1.8911.042j4.244000.588j2.353Y0.481j2.40400.481j30.737j3.3330B00j3.333j6.666j3.33300.588j2.3530j3.3330.588j5.686 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/111(2)Y()j0.5844221.11.0j0.3Y055111YY()j0.30345541.11.0j0.3修改后的矩阵中：Yj6.666j0.58j6.08644YYj3.333j0.303j3.034554修改后的导纳矩阵为：0.935j4.2950.454j1.8910.481j2.404000.454j1.8911.042j4.244000.588j2.353Y0.481j2.40400.481j30.737j3.3330B00j3.333j6.086j3.0300.588j2.3530j3.030.588j5.6864-3解：（1）有名制下： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/38.5k0.303T1121(15%)38.5k0.326T2121(12.5%)11Y0.016j0.473(S)112k0.8j23T1YY0122111YY0.005j0.143(S)13310.3030.8j23YY01441YY0155111Y0.032j0.209(S)222k6.8j44T2YY0233211YY0.011j0.068(S)24420.3266.8j44YY0255211Yyyyy20.014j0.089(S)333435303110.5j400.8j231YY0.006j0.023(S)344310.5j401YY0.006j0.023(S)355310.5j4011Yyyyy20.015j0.068(S)443445402410.5j406.8j441YY0.006j0.023(S)455410.5j401Yyy20.012j0.046(S)55354510.5j40∴有名制下的节点导纳矩阵为：0.016j0.47300.005j0.1430000.032j0.2900.011j0.0680Y0.005j0.14300.014j0.0890.006j0.0230.006j0.023B00.011j0.00680.006j0.0230.015j0.0680.006j0.023000.006j0.0230.006j0.0230.012j0.046 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/（2）标幺制下：取S1000MVAUU115KVBBav37kk0.332T1T2115211UavY0.255j5.485112k0.8j23ST1BYY01221211115YY0.082j1.76613310.3220.8j231000YY01441YY01551211UavY0.383j2.806222k6.8j44ST2BYY02332211115YY0.123j0.90424420.3226.8j441000YY025522115Y(0.014j0.089)0.185j1.1773310002115YY(0.006j0.023)0.079j0.304344310002115YY(0.006j0.023)0.079j0.304355310002115Y(0.015j0.068)0.198j0.8994410002115YY(0.006j0.023)0.079j0.304455410002115Y(0.012j0.046)0.159j0.608551000标幺制下导纳矩阵：0.255j5.48500.082j1.7660000.383j2.80600.123j0.9040Y0.082j1.76600.185j1.1770.079j0.3040.079j0.304B00.123j0.9040.079j0.3040.198j0.8990.079j0.304000.079j0.3040.079j0.3040.159j0.6084-4解：先计算系统的节点导纳矩阵： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/1Y1j3110.1j0.31YY1j312210.1j0.31Y1j3220.1j0.31j31j3YB1j31j3又P0.5Q0.31S1S(0)(0)(0)(0)给定电压初值：ee1.0ff01212计算P,Q（只有一个PQ节点和一个平衡节点）ii22(0)(0)(0)(0)(0)(0)(0)P1P1S[e1(G1jejB1jfj)f1(G1jfjB1jej)]0.5j1j122(0)(0)(0)(0)(0)(0)(0)Q1Q1S[f1(G1jejB1jfj)e1(G1jfjB1jej)]0.3j1j1计算雅可比矩阵各元素： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/P21(G1jejB1jfj)G11e1B11f11e1j1P21(G1jfjB1jej)B11e1G11f13f1j1Q21(G1jfjB1jej)B11e1G11f13e1j1Q21(G1jejB1jfj)G11e1B11f11f1j1P13e(0)11(0)Q131f1(0)1310.50.14e1(0)f1310.30.12(1)(0)(0)eee0.86111(1)(0)(0)fff0.12111(1)P0.51(1)Q0.31P10.72e1P12.76f1Q12.16e1Q11.72f1∴第一次迭代后修正方程式为：(1)P10.722.76e1(1)Q12.161.72f1(1)0.50.722.76e1即(1)0.32.161.72f14-5解（1）节点1为平衡节点，节点2为PV节点（2）节点导纳矩阵中各元素为：1Yyyj0.1j9.9111012j0.11YYyj10122112j0.1 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/1Yyyj0.1j9.9222120j0.1∴节点导纳矩阵为：j9.9j10YBj10j9.9(0)(0)(0)(0)（3）给定电压初值：ee1.0ff0又P1520512122S22(0)(0)(0)(0)(0)(0)(0)P2P2S[e2(G2jejB2jfj)f2(G2jfjB2jej)]5j1j122(0)2(0)2U2U2S(e2f2)0计算雅可比矩阵各元素：P22(G2jejB2jfj)G22e2B22f20e2j1P22(G2jfjB2jej)B22e2G22f210f2j12U22e22e22U22f20f2P010e(0)222(0)U220f2e(0)0.52(0)f20(1)(0)(0)eee0.5222(1)(0)(0)fff0222(1)P522U20.75P20e2P210f22U22e21.0e22U20f2 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/∴第一次迭代后修正方程式为：(1)P2010e22(1)U21.00f2(1)5010e2(1)0.751.00f2未迭代前的功率方程：平衡节点功率：~2**S1P1jQ1U1Y1jUj1.0[j9.91.0(j101.0)]j0.1j1线路功率：~*****S12P12jQ12U1[U1y10(U1U2)y12]1.0[1.0(j0.1)0(j10)]j0.1~*****S21P21jQ21U2[U2y20(U2U1)y21]j0.1线路上的损耗：~~~S12S12S21j0.2迭代后的功率方程：平衡节点功率：~2**S1P1jQ1U1Y1jUj1.0[j9.91.0(j100.5)]j4.9j1线路功率：~*****S12P12jQ12U1[U1y10(U1U2)y12]1.0[1.0(j0.1)(10.5)(j10)]j4.9~*****S21P21jQ21U2[U2y20(U2U1)y21]0.5[0.5(j0.1)(0.51)(j10)]j2.525线路上的损耗：~~~S12S12S21j4.9j2.525j2.3754-6解：1Y1120j400.01j0.02Y12Y2101Y2220j400.01j0.02Y23Y3220j402Y3340j800.01j0.02 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/20j40020j40YB020j4020j4020j4020j4040j801为平衡节点，2为PV节点，3为PQ节点(0)(0)(0)(0)(0)(0)e1ee0fff01231232计算Pi,Qi,Ui(P2S0.4,P3S-1,Q3S0.5)33(0)P2P2Se2(G2jejB2jfj)f2(G2jfjB2jej)]0.4j1j12222U2U2S(e2f2)033(0)PP3Se3(G3jejB3jfj)f3(G3jfjB3jej)]1j1j133Q3Q3Sf3(G3jejB3jfj)e3(G3jfjB3jej)]0.5j1j1P32(G2jejB2jfj)G22e2B22f220e2j1P32(G2jfjB2jej)B22e2G22f240f2j12U22e22e22U20f2P2G23e2B23f220e3P2B23e2G23f240f32U20e32U20f3P3G32e3B32f320e2P3B32e3G32f340f2 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/Q3B32e3G32f340e2Q3G32e3B32f320f2P33(G3jejB3jfj)G33e3B33f340e3j1P33(G3jfjB3jej)B33e3G33f380f3j1Q33(G3jfjB3jej)B33e3G33f380e3j1Q33(G3jfjB3jej)B33e3G33f340e3j10.420402040e202000f2120404080e30.540208040f3(0)2040204010.40e2(0)f200000.0052e(0)2040408010.013(0)f3402080400.50.01∴第一次迭代后节点3电压为：(1)(0)(0)(1)(0)(0)eee0.99fff0.01333333U30.99j0.010.990.5794-7解：导纳矩阵为：1Y122j0.025.882j23.490.02j0.081Y12Y212.941j11.7650.02j0.081Y13Y312.941j11.7650.02j0.081Y2222j0.025.882j23.490.02j0.081Y23Y322.941j11.7650.02j0.08Y335.882j23.49 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/5.882j23.492,941j11.7652,941j11.765YB2,941j11.7655.882j23.492,941j11.7652,941j11.7652,941j11.7655.882j23.491为平衡节点，2，3为PQ节点。(0)(0)(0)(0)(0)(0)e1.0ee1.0fff0123123计算Pi,Qi(P2S0.5,Q2S1.0,P3S-1.5,Q3S0.4)33(0)P2P2Se2(G2jejB2jfj)f2(G2jfjB2jej)]0.5j1j133Q2Q2Sf2(Gj2ejB2jfj)e2(G2jfjB2jej)]1.04j1j133(0)P3P3Se3(G3jejB3jfj)f3(G3jfjB3jej)]1.5j1j133Q3Q3Sf3(G3jejB3jfj)e3(G3jfjB3jej)]0.44j1j1P32(GeBf)GeBf5.8822jj2jj222222e2j1P32(GfBe)BeGf23.532jj2jj222222f2j1Q32(GfBe)BeGf23.452jj2jj222222e2j1Q32(GeBf)GeBf5.8822jj2jj222222f2j1P2G23e2B23f25.882e3P2B23e2G23f223.49f3Q2B23e2G23f223.49e3Q2G23e2B23f25.882f3P3G32e3B32f32.941e2P3B32e3G32f311.765f2 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/Q3B32e3G32f311.765e2Q3G32e3B32f32.941f2P33(G3jejB3jfj)G33e3B33f35.882e3j1P33(G3jfjB3jej)B33e3G33f323.53f3j1Q33(G3jfjB3jej)B33e3G33f323.45e3j1Q33(G3jfjB3jej)B33e3G33f35.882e3j10.55.88223.535.88223.49e21.0423.455.88223.495.882f21.52.94111.7655.88223.53e30.4411.7652.94123.455.882f3e(0)0.02942(0)f0.04932e(0)0.01733(0)f30.0398(1)(0)(0)eee1.0294222(1)(0)(0)fff0.0493222(1)(0)(0)eee1.0173333(1)(0)(0)fff0.398333迭代后节点功率方程：（平衡节点）~3**S1P1jQ1U1Y1jUjj11.0[(5.882j23.49)1.0(2.941j11.765)(1.0294j0.0493)(-2.941-j11.765)(1.0173j0.398)]3.9651-j1.615线路上：~*****S12P12jQ12U1[U1y10(U1U2)y12]1.0{1.0(j0.02)[1(1.0294j0.0493)](2.941j11.765)}0.6665j0.2209~*****S21P21jQ21U2[U2y20(U2U1)y21](1.0294j0.0493)[(1.0294-j0.0493)(-j0.02)(1.0294-j0.0493-1.0)(2.941j11.765)]0.6762j0.2184 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/~*****S13U1[U1y10(U1U3)y13]1.0[j0.02(1.01.0173j0.398)(2.941j11.765)]4.1737j3.2259~*****S31U3[U3y30(U3U1)y31](1.0173j0.398)[(1.0173j0.398)(j0.02)(1.0173j0.3981.0)(2.941j11.765)]2.9699j4.8987~*****S23U2[U2y20(U2U3)y23](1.0294j0.0493)[(1.0294j0.0493)(j0.02)(1.0294j0.04931.0173j0.398)(2.941j11.765)]5.396j1.4476~*****S32U3[U3y30(U3U2)y32](1.0173j0.398)[(1.0173j0.398)(j0.02)(1.0173j0.3981.0294j0.0493)(2.941j11.765)]5.8566j0.8913（2）１为平衡节点，２为PQ节点，3为PV节点(0)(0)(0)(0)(0)(0)e1.0e1.0e1.0fff01231232计算P2,Q2,P3,U3(P2S0.5,Q2S1.0,P3S-1.5,U3S1.0) 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/33(0)P2P2Se2(G2jejB2jfj)f2(G2jfjB2jej)]0.5j1j133Q2Q2Sf2(Gj2ejB2jfj)e2(G2jfjB2jej)]1.04j1j133(0)P3P3Se3(G3jejB3jfj)f3(G3jfjB3jej)]1.5j1j12222U3U3S(e3f3)0由（１）知：P2P2Q2Q25.882；23.53；23.45；5.882e2f2e2f2P2P2Q2Q25.882；23.49；23.49；5.882e3f3e3f222P3P3U3U35.882；23.53；2e30；0e3f3e3f322P3P3U3U32.941；11.765；0e2f2e2f20.55.88223.535.88223.49e21.0423.455.88223.495.882f21.52.94111.7655.88223.53e30002.00f3e(0)0.04672(0)f0.04492e(0)03(0)f30.0354(1)(0)(0)eee1.0467222(1)(0)(0)fff0.0449222(1)(0)(0)eee1.0333(1)(0)(0)fff0.0354333平衡节点功率：~3**S1P1jQ1U1Y1jUjj11.0[(5.882j23.49)1.0(2.941j11.765)(1.0467j0.0449)(-2.941-j11.765)(1.0j0.0354)]-0.2491-j0.5615线路上： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/~*****S12P12jQ12U1[U1y10(U1U2)y12]1.0{1.0(j0.02)[1(1.0467j0.0449)](2.941j11.765)}0.6656j0.4374~*****S21P21jQ21U2[U2y20(U2U1)y21](1.0467j0.0449)[(1.0467-j0.0449)(-j0.02)(1.0467-j0.0449-1.0)(2.941j11.765)]0.678j0.4449~*****S13U1[U1y10(U1U3)y13]1.0[j0.02(1.01.0j0.0354)(2.941j11.765)]0.4165j0.1241~*****S31U3[U3y30(U3U1)y31](1.0j0.0354)[(1.0j0.0354)(j0.02)(1.0j0.03541.0)(2.941j11.765)]0.4128j0.0988~*****S23U2[U2y20(U2U3)y23](1.0467j0.0449)[(1.0467j0.0449)(j0.02)(1.0467j0.04491.0j0.0354)(2.941j11.765)]1.1186j0.3546~*****S32U3[U3y30(U3U2)y32](1.0j0.0354)[(1.0j0.0354)(j0.02)(1.0j0.03541.0467j0.0449)(2.941j11.765)]1.0932j0.2954-8解：导纳矩阵为： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/11Y11y14y12j20j0.1j0.1Y12Y21y12j10Y13Y310Y14Y41j10Y15Y5101Y22y12y23y243j30j0.1Y23Y32j10Y24Y42j10Y25Y520Y33y32y35j20Y34Y430Y35Y53j101Y44y41y42y453j30j0.1Y45Y54j101Y552j20j0.1j20j100j100j10j30j10j100YB0j10j200j10j10j100j30j1000j10j10j201为平衡节点，2、3为PV节点，4、5为PQ节点 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/(0)(0)(0)(0)(0)U1U2U3U4U51.0(0)(0)(0)(0)(0)123450.0P2S0.9;P3S0.2;P4S1.72;P5S1.75Q4S0.6;Q5S0.555P2P2SU2Uj(G2jcos2jB2jsin2j)0.9j15P3P3SU3Uj(G3jcos3jB3jsin3j)0.2j15P4P4SU4Uj(G4jcos4jB4jsin4j)1.72j15P5P5SU5Uj(G5jcos5jB5jsin5j)1.75j15Q4Q4SU4Uj(G4jsin4jB4jcos4j)0.6j15Q5Q5SU5Uj(G5jsin5jB5jcos5j)0.55j1P52H22U2Uj(G2jsin2jB2jcos2j)302j1j2P2H23U2U3(G23sin23B23cos23)103H24U2U4(G24sin24B24cos24)10H25U2U5(G25sin25B25cos25)0 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/H32U3U2(G32sin32B32cos32)10P2H23U2U3(G23sin23B23cos23)35H33U3Uj(G3jsin3jB3jcos3j)20j1j3H34U3U4(G34sin34B34cos34)0H35U3U5(G35sin35B35cos35)10H42U4U2(G42sin42B42cos42)10H43U4U3(G43sin43B43cos43)05H44U4Uj(G4jsin4jB4jcos4j)30j1j4H45U4U5(G45sin45B45cos45)10H52U5U2(G52sin52B52cos52)0H53U5U3(G53sin53B53cos53)10H54U5U4(G54sin54B54cos54)105H55U5Uj(G5jsin5jB5jcos5j)20j1j5N24U2U4(G24cos24B24sin24)0N25U2U5(G25cos25B25sin25)0N34U3U4(G34cos34B34sin34)0N35U3U5(G35cos35B35sin35)052N44U4Uj(G4jcos4jB4jsin4j)2U4G440j1j4N45U4U5(G45cos45B45sin45)0N54U5U4(G54cos54B54sin54)052N55U5Uj(G5jcos5jB5jsin5j)2U5G550j1j5J42U4U2(G42cos42B42sin42)0J43U4U3(G43cos43B43sin43)05J44U4Uj(G4jcos4jB4jsin4j)0j1j4J45U4U5(G45cos45B45sin45)0J52U5U2(G52cos52B52sin52)0J53U5U3(G53cos53B53sin53)0J54U5U4(G54cos54B54sin54)0 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/5J55U5Uj(G5jcos5jB5jsin5j)0j1j552L44U4Uj(G4jsin4jB4jcos4j)2U4B4430j1j4L45U4U5(G45sin45B45cos45)10L54U5U4(G54sin54B54cos54)1052L55U5Uj(G5jsin5jB5jcos5j)2U5B5520j1j50.930101000020.210200100031.7210030100041.7501010200050.600003010U4/U40.5500001020U5/U5(0)0.21032(0)0.29413(0)0.24674(0)50.3579U4/U40.035U5/U50.045(0)(0)U40.034,U50.045修正各点电压：(1)(0)(0)UUU0.965444(1)(0)(0)U5U5U50.955(1)(0)(0)0.2103222(1)(0)(0)0.2941333(1)(0)(0)0.2467444(1)(0)(0)5550.3579U21.00.2103U31.00.2941U40.9650.2467U50.9550.3579 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/平衡节点功率：~5**S1U1Y1jUjj201.0(j10)10.21030(j10)0.9650.246700.0783j0.3502j1线路上功率：~*****S12U1[U1y10(U1U2)y12]1.0(1.01.00.2103)j100.0001j0.0367~***S21U2[(U2U1)y21]1.00.2103(1.00.21031.0)100.00003j0.0367~***S14U1[(U1U4)y14]1.0(1.00.9650.2467)100.3501j0.0416~***S41U4[(U4U1)y41]0.9650.2467(0.9650.24671.0)100.338j0.0386~***S23U2[(U2U3)y23]1.00.2103(1.00.21031.00.2941)100.00005j0.0146~***S32U3[(U3U2)y32]1.00.2941(1.00.29411.00.2103)100.00007j0.0146~***S32U3[(U3U2)y32]1.00.2941(1.00.29411.00.2103)100.00007j0.0146~***S24U2[(U2U4)y24]1.00.2103(1.00.21030.9650.2467)100.34998j0.00618~***S42U4[(U4U2)y42]0.9650.2467(0.9650.24671.00.2103)100.3377j0.0061~***S35U3[(U3U5)y35]1.00.2941(1.00.29410.9550.3579)100.4501j0.0107~***S53U5[(U5U3)y53]0.9550.3579(0.9550.35791.00.2941)100.4298j0.0107~***S45U4[(U4U5)y45]0.9650.2467(0.9650.24670.9550.3579)100.0965j0.0179~***S54U5[(U5U4)y54]0.9550.3579(0.9550.35790.9650.2467)100.0955j0.0179 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/第五章5-1解：（1）计算每台机组的单位调节功率和负荷的频率调节效应系数：1PGN1100水轮发电机100MW：KG180(MW/Hz)fN0.025501PGN17575MW：KG254.545(MW/Hz)fN0.0275501PGN1100汽轮发电机100MW：KG357.143(MW/Hz)fN0.035501PGN15050MW：KG425(MW/Hz)fN0.04501PGN11000较小容量发电机1000MW：KG5500(MW/Hz)fN0.0450PDN3000负荷3000MW：KLDKLD1.590(MW/Hz)fN50又PLD33003000300MWKG5KG15KG26KG320KG4KG52015.583(MW/Hz)PLD(KGKLD)f3000f0.142(Hz)2015.58390即系统新的稳定频率为：f500.14249.858(Hz)(2)∵全部机组都不参加一次调频，KG0KKLD90(MW/Hz)300f3.333(Hz)90即系统新的稳定频率为：f503.33346.667(Hz)(3)仅水轮机参加一次调整：KG5KG15KG2672.725(MW/Hz)300f0.393(Hz)672.72590即系统新的稳定频率为：f500.39349.607(Hz) 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/5-2解：(1)PGPLD300(MW)(2)∵要求系统频率不低于49.8Hz，f0.2PLDPG(KGKLD)fPG300(672.72590)0.2452.545(MW)∴发电机二次调整整发的功率不超过452.545MW5-3解：计算发电机的单位调节功率和负荷的调节效应系数：11001000MW：KG166.667(MW/Hz)0.03501200200MW：KG2100(MW/Hz)0.04501200负荷1200MW：KLD248(MW/Hz)50又KG5KG14KG2733.335(MW/Hz),PLD200(MW)PLD200f0.256(Hz)KGKLD733.33548即系统的频率变化量为0.256Hz110005-4解：KGA400(MW)0.055012400K1600(MW)GB0.0350P(10002400)20%680(MW)LDPLD680f0.34(Hz)KK4001600GAGBf500.3450.34(Hz)PKf4000.34136(MW)GAGAPKf16000.34544(MW)GBGBP1000-136864(MW)GAP2400-5441856(MW)GB∴负荷减少后系统频率为50.34Hz，A、B两机组分别出力864MW、1856MW 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/5-5解：计算系统的单位调节功率：PAN5000KGAKGA*202000(MW/Hz)fN50PAN5000KLDAKLDA*151500(MW/Hz)fN50PBN2000KGBKGB*251000(MW/Hz)fN50PBN2000KLDBKLDB*280(MW/Hz)fN50PCN8000KGCKGC*152400(MW/Hz)fN50PCN8000KLDCKLDC*2320(MW/Hz)fN50PLDAPABPGA(KGAKLDA)fAKAfAPLDBPBCPGBPAB(KGBKLDB)fBKBfBPLDCPDCPGC(KGCKLDC)fCKCfC又fAfBfCfPAPABKAfPBPDCPABKBfPCPDCKCfPAPBPCfKAKBKCKAPBKAPCKBPAKCPAPABKAKBKCKAPCKBPCKCPAKCPBPDCKAKBKC(1)PLDA300MW,PA300MW,PB0,PC0KA200015003500(MW/Hz)KB1000801080(MW/Hz)KC24003202720(MW/Hz)30000f0.041(Hz)3500108027200010803002720300PAB156.164(MW)3500108027200027203000PDC111.781(MW)350010802720 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/(2)∵实现无差调节，∴Δf=0PLDA300MW,PA300MW,PB0,PC300MW03500(300)10803002720300PAB300(MW)3500108027203500(300)1080(300)3002720PDC300(MW)350010802720(3)PLDB300MW,PB300MW,PA0,PC0300f0.0476(Hz)35008027203500300000PAB166.667(MW)3500108027200002720300PDC129.524(MW)350010802720PLDA5-6解：PLDA500MW,f14950Hz,KA500(MW/Hz)f1又PLDA500MW,PBA300MW,PA500MW500300PAPBAKAf,f0.4(Hz)5000300PBPBAKBf,KB750(MW/Hz)0.4即A、B两系统的单位调节功率分别为500MW/Hz,750MW/Hz5-7解：计算各系统的单位调节功率：PLDA10000KLDAKLDA*2400(MW/Hz)fN50PGNA12000KGAKGA*153600(MW/Hz)fN50PLDB1000KLDBKLDB*360(MW/Hz)fN50PGNB1200KGBKGB*8192(MW/Hz)fN50(1)PLDA500MW,PA500MW,PLDB0,PGB500MW,PB500MW 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/PA500fA0.125(Hz)KLDAKGA4003600PB500fB1.984(Hz)KLDBKGB60192(2)PB(KLDBKGB)fB,fB0.5PB(60192)0.5126(MW)PBPLDBPGBPLDBPBPGB374(MW)5-8解：UTmax8kV,UTmin4kV，采用逆调压U1maxUTmax2358U1tmaxU2N10.5243(kV)U2max10.5U1minUTmin2264U1tminU2N10.5241.5(kV)U2min10U1tmaxU1tmin取平均值：U1t242.25(kV)2选取最接近的分接头电压242kV，并进行校验U2N10.5U2maxs(U1maxUTmax)(2358)10.543(kV)10.5(kV)U1ts242U2N10.5U2mins(U1minUTmin)(2264)9.979(kV)10(kV)U1ts242∴所选分接头电压满足要求。PmaxRTQmaxXT242.4410405-9解：UTmax4.094(kV)U1max112PminRTQminXT102.44540UTmin1.951(kV)U1min115U1maxUTmax1124.094U1tmaxU2N6.3113.301(kV)U2max6U1minUTmin1151.951U1tminU2N6.3107.91(kV)U2min6.6U1tmaxU1tmin取平均值：U1t110.606(kV)2选取最接近的分接头电压110kV，并校验 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/U2N6.3U2maxs(U1maxUTmax)(1124.094)6.18(kV)6(kV)U1ts110U2N6.3U2mins(U1minUTmin)(1151.951)6.475(kV)6.6(kV)U1ts110∴所选分接头电压满足要求。5-10解：计及变压器功率损耗，设UN110kV22~PmaxQmax1SZmax2(1.8j40.3)0.186j4.163MVAU2N22~PminQmin1SZmin2(1.8j40.3)0.046j1.041MVAU2N~"~~SmaxSmaxSZmax40.186j34.163MVA~"~~SminSminSZmin20.046j16.041MVA""1PmaxRTQmaxXT140.1861.834.16340.3UTmax6.9kV2U1max2105""1PminRTQminXT120.0461.816.04140.3UTmin3.145kV2U1min2108.5U1maxUTmax1056.9U1tmaxU2N10.598.1(kV)U2max10.5U1minUTmin108.53.145U1tminU2N10.5110.623(kV)U2min10U1tmaxU1tminU1t104.362kV2选取分接头电压为104.5kV，并校验U2N10.5U2maxs(U1maxUTmax)(1056.9)9.857(kV)10.5(kV)U1ts104.5U2N10.5U2mins(U1minUTmin)(108.53.145)10.586(kV)10(kV)U1ts104.5∴所选分接头满足要求。变比为104.5/10.55-11解：不计变压器功率损耗~~~S1maxS2maxS3max12j9MVA~~~S1minS2minS3min6j4MVA计及变压器各绕组的电压损耗及中低压母线电压的归算值 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/P1maxR1Q1maxX1123965UT1max5.545kVU1max112PminR1QminXT63465U1Tmin2.417kVU1min115U0max1125.545106.455kVU0min1152.417112.583kV645(1)UT2max0.178kV106.455443(1)UT2min0.115kV112.583"U2max106.4550.178106.277kV"U2min112.5830.115112.468kV65430UT3max1.409kV106.45525130UT3min0.355kV112.583"U3max106.4551.409105.046kV"U3min112.5830.355112.228kV(1)根据低压母线的调压要求，选择高压绕组分接头。低压母线要求在最大负荷时为6kV，在最小负荷时为6.5kV"U3N6.6U1tmaxU3max105.046115.551kVU3max6"U3N6.6U1tminU3min112.228113.955kVU3min6.5115.551113.955U1t114.753kV2选取分接头电压为115.5kV，并校验"U3N6.6U3maxU3max105.0466.003kV6kVU1t115.5"U3N6.6U3minU3min112.2286.413kV6.5kVU1t115.5∴所选分接头电压满足要求。(2)根据中压母线的调压要求，选择中压绕组的分接头电压中压母线的要求值：最大负荷时为35kV，最小负荷时为38kV"U2NU1tmax由U1tmaxU2max得U2N"U2maxU2maxU2max 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/115.5U2Nmax3538.037kV106.277115.5U2Nmin3839.024kV112.46838.03739.024U2t38.531kV2选取分接头电压为38.5kV"U2t38.5U2maxU2max106.27735.426kV35kVU1t115.5"U2t38.5U2minU2min112.46837.489kV38kVU1t115.5所选分接头电压满足要求。变压器变比为115.5/38.5/6.65-12解：设置补偿设备前变电所低压侧归算到高压侧的电压为："PmaxRABQmaxXAB810620UBmaxUA3630.444kVUA36"PminRABQminXAB410320UBminUA3633.222kVUA36装电容器，最小负荷时补偿设备全部退出运行，降压变压器分接头电压为："UBmaxU2N11Ut33.22235.828kVUBmin10.2即分接头电压选为35.875kV。按最大负荷时，调压要求确定补偿容量Qc"UBCmaxUBmax2QC(UBCmax)kXABk10.51135.8752(10.530.444)()6.507Mvar2035.87511选取补偿容量为7Mvar，并校验"UBCmax810(67)2011UBCmax(36)10.527kVk3635.875"UBCmin11UBCmin33.22210.187kVk35.87510.52710.5电压偏移量：最大负荷时：100%0.27%1010.18710.2最小负荷时：100%0.13%10均没有超过1.25%∴所选电容器满足调压要求。 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/5-13解：线路参数：R1r1l0.277018.9,X1x1l0.4167029.1222Us%UN10.5110变压器电抗：X240.333100SN10031.511∴总阻抗ZRjX(R1jX1)jX29.45j34.72722"UBmin112装电容器最小负荷时：UtU2N11104.186kVUBmin11(17.5%)即分接头电压为11022.5%为104.5kV按最大负荷时调压要求确定补偿容量Qc"UBCmaxUBmax2QC(UBCmax)kXk11(12.5%)111102[11(12.5%)100.5]()39.773Mvar34.72711011即补偿设备最小容量为40Mvar5-14解：不计功率损耗22212034U10.382kV110补偿后压降为：1106%6.6kV，电压为：110-6.6=103.4kV110(10.3826.6)XC20.80120222220线路通过的最大电流为：Imax0.156kA3110采用0.66kV，40kvar的单相电容器40其INC60.606A0.6630.6610∴单个电容器容抗为：XNC10.8960.6063Imax0.15610∴需要并联的支路数：m2.574INC60.606ImaxXNC15620.801需要串联的支路数：n4.917UNC660取m=3，n=5，则总的补偿容量为Qc3mnQNC335401800kvar 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/5XNC510.89对应的补偿容抗为：XC18.1533补偿后线路末端电压为：222120(3418.15)U2C110102.918kV110满足调压要求。第六章6-1解：最大负荷下的有功功率损耗：2222PLD1maxQLD1max2(2tan)PL1max2R1220.125MWUN102222PLD2maxQLD2max1(1tan)PL2max2R2210.015625MWUN102222PLD3maxQLD3max1(1tan)PL3max2R3210.015625MWUN10利用日负荷曲线进行电能损耗计算：AAL1AL2AL336512(PL1maxPL2maxPL3max)684.375MWh利用最大负荷损耗时间法计算：线路L1输送电能：AL136512PLD1max8760MWhAL1Tmax14380hPmax查表得：max13060h线路L2输送电能：AL236512PLD2max4380MWhAL2Tmax24380hPmax查表得：max23060h同理得：max33060h 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/AAL1AL2AL3PL1maxmax1PL2maxmax2PL3maxmax3487.125MWh22S2503626-2解：(1)Pmax2Ps()2203.4()241.283kWSN2406全年电能损耗：A2P0TmaxPmaxTmax1.29710kWh(2)单台变压器运行的临界功率值为：P041.5ScrSNn(n1)402125.552MVAPs203.46-3解：(1)PLD40MW,QLDPLDtan24.79MW~12162SYLjBUj22.8210100110j3.412MVA22~P0I0%38.50.8SYT2(jSN)2(j31.5)0.077j0.504MVA10001001000100~PsS/22Us%SNS/22SZT1002[()j()]1000SN100SN14840/2210.531.540/222[()j()]0.165j3.691MVA10000.8531.51000.8531.5~PsS/42Us%SNS/42SZT502[()j()]0.041j0.923MVA1000SN100SN~PsS/82Us%SNS/82SZT252[()j()]0.01j0.231MVA1000SN100SN~~~~SLD100PLDjQLDSZT100SYLSYL40j24.790.165j3.6910.077j0.504-j3.41240.242j25.573MVA~~~~SLD50PLDjQLDSZT50SYLSYL40.118j22.805MVA~~~~SLD25PLDjQLDSZT25SYLSYL40.087j22.113MVA2222PLD100QLD10040.24225.5731PLDmax1002R21000.171.597MVAU1102N22PLD50QLD50PLDmax502R1.496MVAUN22PLD25QLD25PLDmax252R1.472MVAUNALPLDmax1002000PLDmax504000PLDmax25276013240.72MVA 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/对变压器：S2402Pmax1002PS()2148()165.155kW2SN20.8531.50.5S2Pmax502PS()41.289kW2SN0.25S2Pmax252PS()10.322kW2SNAT28760P02000Pmax1004000Pmax502760Pmax251198.47MWhAALAT14439.19MWh(2)当负荷低于50%的最大负荷时，切除一台变压器运行：~~由(1)知：SYLj3.412MVA,SYT0.077j0.504MVA~~SZT1000.165j3.691MVA,SZT500.041j0.923MVA~PSS2US%SNS2SZT25()j()0.33j7.382MVA1000SN100SN~~~~SLD100PLDjQLDSZT100SYLSYL40.242j25.573MVA~~~~SLD50PLDjQLDSZT50SYLSYL40.118j22.805MVA~~~~SLD25PLDjQLDSZT25SYL25SYL40.369j29.012MVA2222PLD100QLD10040.24225.5731PLDmax1002R21000.171.597MVAU1102N22PLD50QLD50PLDmax502R1.496MVAUN22PLD25QLD25PLDmax252R1.736MVAUNALPLDmax1002000PLDmax504000PLDmax25276013969.36MVA对变压器：Pmax100165.155kW,Pmax5041.289kW0.25S20.25402Pmax25PS()148()20.644kWSN0.8531.5AT28760P02000Pmax1004000Pmax502760Pmax251120.7MWhAALAT15090.06MWh 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/6-4解：变压器的电能损耗：Tmax4500h,cos0.85,查表得：max3000hATP0TPmax100max0.07787600.16530009255.077MWh线路的电能损耗：25.573Tmax4500h,arctan(),得出cos0.844,查表得：max3000h40.242ALPLDmax100max1.59730004791MWhAALAT14046.077MWhP06-5解：ScrSNn(n1)PS19取SN16MVA时,Scr1162111.24MVA7722.5取SN20MVA时,Scr2202113.91MVA93∴当负荷小于Scr1时，一台型号为SPT-16000/35的变压器单独运行；当负荷介于Scr1与Scr2之间时，一台型号为SF-20000/35的变压器单独运行；当负荷大于Scr2时，两台变压器并联运行。6-6解：三个电厂的耗量特性为：10.10.0024PG120.130.0024PG230.160.0036PG3按等耗量微增率准则计算：123,PG1PG2PG3300PG1123.4375MW,PG2110.9375MW,PG365.625MW若平均分配负荷：即PG1PG2PG3100MWW[F1(100)F1(123.4375)F2(100)F2(110.9375)F3(100)F3(65.625)]800023437.5(t)即8000h内浪费23437.5吨煤。6-7解：耗量特性为：11.20.02PG1,21.50.02PG2 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/(1)按等耗量微增率准则计算：12,PG1PG2120PG167.5MW,PG252.5MW(2)∵只有一台发电机运行∴哪一台机组的耗量特性小，即运行最经济令PG1PG2,F1F2，得出PG1PG250MW当负荷大于50MW时，F1F2，当负荷小于50MW时，F1F2即一台发电机运行时，负荷在0~50MW内采用2号机组运行最经济。6-8解：由水、火电耗量特性可得协调方程式为：0.30.003PT(10.004PH)又PTPHPLD0.30.004PLDPT,PHPLDPT0.0030.004取0.9，可得:在下午18时至次日上午8时(PLD600MW)""PH181.818MW,PT418.182MW在上午8时到下午18时(PLD1000MW)""PH363.636MW,PT636.364MW水电厂的全日降水量为：""73KW143600W1036003.5510m可见，γ很小，以致于水电厂分担负荷过大。取1.0，可得:在下午18时至次日上午8时(PLD600MW)""PH157.143MW,PT442.857MW在上午8时到下午18时(PLD1000MW)""PH328.571MW,PT671.429MW 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/水电厂的全日降水量为：""73KW143600W1036003.04410m可见，γ很小，以致于水电厂分担负荷过大。取1.4，可得:在下午18时至次日上午8时(PLD600MW)""PH81.395MW,PT518.605MW在上午8时到下午18时(PLD1000MW)""PH220.930MW,PT779.070MW水电厂的全日降水量为：""73KW143600W1036001.66710m可见，γ很小，以致于水电厂分担负荷过大。取1.45，可得:在下午18时至次日上午8时(PLD600MW)""PH73.864MW,PT526.136MW在上午8时到下午18时(PLD1000MW)""PH210.227MW,PT789.773MW水电厂的全日降水量为：""73KW143600W1036001.54510m可见，γ很小，以致于水电厂分担负荷过大。取1.47，可得:在下午18时至次日上午8时(PLD600MW)""PH70.946MW,PT529.054MW在上午8时到下午18时(PLD1000MW)""PH206.081MW,PT793.919MW水电厂的全日降水量为：""73KW143600W1036001.49910m 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/计算结果如表：(负荷经济分配方案)γP"/MWP"/MWP"/MWP"/MW73THTHK/10m0.9418.182181.818636.364363.6363.551.0442.857157.143671.429328.5713.0441.4518.60581.395779.07220.931.6671.45526.13673.864789.773210.2271.5451.47529.05470.946793.919206.0811.499122226-9解：(1)P2[(P1Q1)0.1(P2Q2)0.04]U12222Q2[(P1Q1)0.4(P2Q2)0.08]UPP计算网损微增率：0.2Q1,0.08Q2Q1Q2按等网损微增率准则有：0.2Q10.08Q2又Q1Q2QLQ02222QQ0.70.4(0.6Q)0.08(0.6Q)01212"Q0.2678,Q3.621(舍去),Q0.6695112(2)计及无功功率网损修正系数P1P1由Q1QQ2Q11Q1Q2PPQQ0.2Q1,0.08Q2,0.8Q1,0.16Q2Q1Q2Q1Q20.2Q10.08Q2得出：1-0.8Q110.16Q2又Q1Q2QLQ02222Q1Q20.70.4(0.6Q1)0.08(0.6Q2)0Q10.248,Q20.6886-10解：1222(1)P2[(1QC1)19.95(1.2QC2)4.07(1QC11.2QC2)6.75]UN计算网损微增率： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/P12[19.952(1QC1)6.752(1QC11.2QC2)]QC135P12[4.072(1.2QC2)6.752(1QC11.2QC2)]QC235按等网损微增率准则和约束条件建立方程：PP,19.95QC14.07QC215.066QC1QC2QC1QC2QC不同补偿容量下的经济分配方案及对应的网损微增率见下表：PQCQC1QC2QCi1.40.8640.536-0.0131.50.8810.619-0.0161.60.8980.702-0.011.80.9320.868-0.0072.00.9661.034-0.0032.211.20P()KC(2)根据，可得出最优网损微增率QCimaxP()KC(0.150.15)2000.011MvarQCimax5400P12[19.952(1QC1)6.752(1QC11.2QC2)]0.011QC135P12[4.072(1.2QC2)6.752(1QC11.2QC2)]0.011QC235由此可得：QC10.8873Mvar,QC20.6476MvarQCQC1QC21.5349Mvar 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/第七章7-1解：(1)由派克变换式：Udcoscos(120)cos(120)Umsin(t0)U2sinsin(120)sin(120)Usin(t120)q3m0U111Usin(t120)0m0222cossin(t0)cos(120)sin(t0120)cos(120)sin(t0120)2Umsinsin(t0)sin(120)sin(t0120)sin(120)sin(t0120)31sin(t)1sin(t120)1sin(t120)000222t03sin[t0(t0)]Ud2sin(00)23UqUmcos[t0(t0)]Umcos(00)32U000(2)由派克变换式： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/Udcoscos(120)cos(120)UmU2sinsin(120)sin(120)Uq3mU01113Um222coscos(120)3cos(120)2Usinsin(120)3sin(120)m33223sin2Um23cos332t0Udsin(t0)43UqUmcos(t0)3U3047-2解：(1)由派克变换：icossin1Icos(30)am0ibcos(120)sin(120)1Imsin(300)icos(120)sin(120)10ccoscos(30)sinsin(30)00Imcos(120)cos(300)sin(120)sin(300)cos(120)cos(30)sin(120)sin(30)00cos(30)0Imcos(900)cos(150)0t0icos(t30)aibImcos(t90)icos(t150)c(2)由派克变换： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/icossin1Icos(t30)am0ibcos(120)sin(120)1Imsin(t030)icos(120)sin(120)10ccoscos(t30)sinsin(t30)00Imcos(120)cos(t030)sin(120)sin(t030)cos(120)cos(t30)sin(120)sin(t30)00cos[(t30)]0Imcos[(120)(t030)]cos[(120)(t30)]0t03icos30a2ibImcos(90)Im0icos1503c27-3解：当0时， 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/acos0sin01Ldcos0cos(120)cos120ia2cos(120)sin(120)1Lsin0sin(120)sin120ib3qbcos120sin1201L111ic0c222cos0sin01Ldcos0cos(120)cos120ia2cos(120)sin(120)1Lsin0sin(120)sin120a2i3qacos120sin1201L111ai0a222L0LdL0LdL0Ld22222ia2LdL0LdL03LdL032LLai322424q424qaLdL0LdL03LdL03aiaLL22424q424qa2L02LdL0LdL0L[(L)a()a()]adi322222ab2LdL02LdL03LdL032L[()a(L)a(L)]/abqqi322424424bc2LdL02LdL03LdL03L[()a(L)a(L)]/acqqi322424424c当90时，acos90sin901Ldcos90cos(90120)cos(2702cos(90120)sin(90120)1Lsin90sin(90120)sin270b3qcos(90120)sin(90120)1L111c0222cos90sin901Ldcos90cos(90120)cos(270)i2cos(90120)sin(90120)1Lsin90sin(90120)sin270a23qcos(90120)sin(90120)1L111ai0222L0LqL0LqL0Lq22222ia2LqL0LqL03LLqL03La2i322424d424daLqL0LqL03LqL03aiaLdLd22424424 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/a2L02LqL0LqL0L[(L)a()a()]aqi322222ab2LqL02LqL03LqL032L[()a(L)a(L)]/abddi322424424bc2LqL02LqL03LqL03L[()a(L)a(L)]/acddi322424424c若i0,iisint，上述结果不变。abc7-4解：acossin1Ldcoscos(120)cos(120)2cos(120)sin(120)1Lsinsin(120)sin(120)b3qcos(120)sin(120)1L111c0222ia2pppi3123aia22L0LcosLsindq2L0其中，pLcoscos(120)Lsinsin(120)1dq2L0Lcoscos(120)Lsinsin(120)dq2L0Lcoscos(120)Lsinsin(120)dq222L0pLcos(120)Lsin(120)2dq2L0Lcos(120)cos(120)Lsin(120)sin(120)dq2L0Lcoscos(120)Lsinsin(120)dq2L0pLcos(120)cos(120)Lsin(120)sin(120)3dq222L0Lcos(120)Lsin(120)dq2a23LLLLLa00bci32a 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/7-5解：EqUjIxd∵发电机额定满载运行，cosN0.8531.788,U1.00,I1.031.788NE1.00j1.831.7881.95j1.532.47938.118q38.118向量图如下：7-6解：∵发电机额定满载运行，cosN0.8531.788,U1.00,I1.031.788NEUjIx1.00j0.5631.7881.295j0.4761.379720.18Qq20.18I1.0sin(31.78820.18)0.788dI1.0cos(31.78820.18)0.616qEEI(xx)1.37970.788(0.950.56)1.687qQddqE1.68720.18q向量图见上。 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/第八章8-1解：(1)短路前变压器空载运行，且3022UmUmz0.020.40.4,Ipmz0.40.4arctan87.140.02tiIsin(t)Isin()eTaapmpmtUmsin(t57.14)2.1UeTam0.4(2)短路前空载运行，又A相非周期分量为0，ttiIsin(120)eTa2.165UeTabpbpmmttiIsin(120)eTa2.165UeTacpcpmm(3)cos0.85,31.800.4Um又arctan87.14,Im0Ipm0.020.4tUim[sin(31.8)sin(87.14)]eTaapa0.4SN62.58-2解：外接电抗标幺值：0.60.60.3422UN10.5*1稳态电流：I0.4671.80.34*SN62.5II0.4671.605kA3Uav310.5"*1次暂态电流：I2.1740.120.34""*SN62.5II2.1747.471kA3Uav310.5 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/8-3解：(1)cos0.8,36.87EUjIx10j1.0836.871.648j0.8641.8627.67Qq27.67Eq0UcosIsin()xdcos27.67sin(27.6736.87)1.11.878"Eq0UcosIsin()xdcos27.67sin(27.6736.87)0.231.093"Eq0UcosIsin()xdcos27.67sin(27.6736.87)0.120.994""Ed0UdIqxqUsinIcos()xq0.4EUjIx"10j0.2336.871.138j0.1841.1539.1850d22EEE1.0710d0q0(2)短路后瞬间：E1.093,E0.994,E0.4q0q0d0Eq01.093Id4.752xd0.23Eq00.994Id8.283xd0.12Ed00.4Iq2.667xq0.15EqEq01.878I1.707xdxd1.1Iid4.75222IIdIq8.702(3)EQUIxq8.7020.151.305Eq0EQ(xdxq)Id1.305(0.120.15)8.2831.0578-4解：短路前空载额定电压下，有E1q00,UUE1,U000q0q0d0EUixE1q0q0d0dq0EUixE1q0q0d0dq0Ed0Ud0iq0xq0 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/tttEEEEEEq0q0Tdq0q0Tdq0d0Tqia[()e()e]cos(t0)esin(t0)xdxdxdxdxdxdttUU011T011T()cos()ea()cos(2t)ea00002xdxq2xdxqtt11111[()e0.04()e0.97]cos(t30)00.1740.2290.2291.81.8tt111111()cos(30)e0.22()cos(2t30)e0.2220.1740.17220.1740.172tttt(1.38e0.043.81e0.970.56)cos(t30)5.01e0.220.03cos(2t30)e0.22tttti(1.38e0.043.81e0.970.56)cos(t90)5.01e0.220.03cos(2t90)e0.22btttti(1.38e0.043.81e0.970.56)cos(t150)5.01e0.220.03cos(2t150)e0.22c8-5解：计算参数：取SB100MVA,UBUavSB100G:xdxd0.240.204SN100/0.85US%SB10.5100T1:xT10.0875100SN1100120US%SB10.5100T2:xT20.167100SN210063SB100L1:xL1x1l120.410020.302Uav1115SB100L2:xL2x2l220.082024.031Uav26.3xR%UNSB66100R:xR1.309100210023INUav30.46.3简化电路为：xxdxT1xL1xT2xRxL26.1005*11100∴三相短路电流：I0.164，有名值：I0.1641.503kAx6.100536.3 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/冲击电流(KM1.8)：iM2KMI21.81.5033.826kA8-6解：计算参数：取SB100MVA,UBUavSB100G1:xd1xd10.140.0595SN1200/0.85SB100G2:xd2xd20.190.0538SN2300/0.85US%SB14100T1:xT10.0583100SN1100240US%SB14100T2:xT20.0389100SN2100360SB100L:xLx1l20.4188020.0632Uav230简化电路为：x1xd1xT10.118,x2xLxd2xT20.156*E1E21.051.05∴f点的三相短路电流：I15.629x1x20.1180.156100有名值为：I15.6293.923kA3230冲击电流为(KM1.8)：iM21.83.9239.986kA8-7解：(1)当线路L3断开时，简化电路如下：取SB100MVA 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/1x1xdxT10.4,x2xL1xT2xd0.4,x3x1//x2xL20.42E1∴f点起始次暂态电流：I2.5x30.4冲击电流为(KM1.8)：iM2KMI21.82.56.364短路容量为：StISB2.5100250MVA(2)当线路L3投入时，等效电路如下：0.5xL1xL2x1xdxT10.4,x2xT2xd0.3,x30.040.5xL1xL3xL20.5xL1xL3xL2xL3x40.04,x50.080.5xL1xL3xL20.5xL1xL3xL2x(x1x3)//(x2x4)x50.272 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/E1∴f点起始次暂态电流：I3.676x0.272冲击电流为(KM1.8)：iM2KMI21.83.6769.358短路容量为：StISB3.676100367.6MVA8-8解：计算参数：取SB100MVA,UBUavSB100G1,2:xdxd0.120.4SN30US%SB10.5100T1,T2:xT1xT20.333100SN10031.5US%SB10.5100T3:xT30.525100SN10020US%SB10.5100T4:xT41.4100SN1007.5SB100L1,L2:xL1xL2x1l120.48020.242Uav115SB100L3:xL3x3l320.42020.06Uav115100LD1:xLD10.351.94418100LD2:xLD20.355.836根据母线上的断开容量可以确定系统的等值电抗，f2处发生三相短路，则G1,G2及系统供给的短路电流都要通过QF，其中G1,G2相供给的短路电流决定于电抗X111X(xdxT1)xL1xL20.60852221.05短路瞬间，这两个电源供给的短路功率为：S100172.555MVA0.60850.80.8负荷侧也通过QF，SL10010032.402MVAxT3xLD10.5251.944此时QF允许系统供给的短路功率为：1130-172.555-32.402925.043MVA 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/925.0431.05IQ9.25043,xQ0.1141009.25043简化电路为：111x1(xdxT1)xL10.4875,x2xL20.121,x3xL3xT41.46222x2xQx2x4x4xT3xLD12.469,x5xQx20.241,x6x4x25.463x4xQ1111Y7.069x1x3x5x6x7x1x3Y5.031,x8x5x3Y2.487,x9x6x3Y56.382*EEELD1ELD2∴f点起始次暂态电流为：I0.782x7x8x9xLD2 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/100起始次暂态电流有名值为：I0.7824.3kA310.5冲击电流：1.051.050.80.8100iM2[1.8()1.0()]10.008kA5.0312.4875.83356.382310.58-9解：计算参数，取SB1000MVA,UBUavSB1000G1~G4:xdxd0.242.04SN100/0.85US%SB10.51000T1~T2:xT1xT20.875100SN100120111000T3:xT31(23148)1.2082100120111000xT32(23814)0.7082100120111000xT33(14823)0.0422100120US%SB10.51000T4:xT40.875100SN1001201000S1:xS11.00.520001000S2:xS20.50.51000f1点发生短路时，简化电路为： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/11x1(xdxT1)(2.040.875)1.457522xS2xT320.50.708x2xS2xT320.50.7081.329xT4xd0.8752.04xT32(xT4xd)0.708(0.8752.04)x3xT4xT32xd0.7080.8752.047.751xS20.5x1xT311.45751.208x4x1xT311.45751.2086.187xS10.5xS1xT31x5xS1xT312.122x111111Y22.295,x6x4xT33Y5.793x4x5x2x3xT33x7x5xT33Y1.987,x8x2xT33Y1.244,x9x3xT33Y7.258计算运算电抗：SN200/0.85xc12x65.7931.363SB1000SN100/0.85xc3xd2.040.24SB1000SN100/0.85xc4x97.2580.854SB1000查运算曲线得：I120.72,I34.6,I41.2,I12,0.20.66,I3,0.23.3,I4,0.21.09 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/1200011000200/0.85100/0.85100/0.85I0.724.61.2146.38kA1.987310.51.244310.5310.5310.5310.51200011000200/0.85100/0.85100/0.85I0.20.663.31.09136.48kA1.987310.51.244310.5310.5310.5310.5iM2KMI21.9146.38393.32kAf2点发生三相短路时，简化电路为：11x1(xdxT1)(2.040.875)1.457522xS2xT320.50.708x2xS2xT320.50.7081.329xT4xd0.8752.04xT32(xT4xd)0.708(0.8752.04)x3xT4xT32xd0.7080.8752.047.751xS20.5x4xdxT332.040.0421.99811111111Y2.21x2x3x4xT311.3297.7511.9981.208x5x2xT31Y3.548，x6x4xT31Y5.334，x7x3xT31Y20.693计算运算电抗：SN200/0.85xc12x11.45750.343SB1000SN100/0.85xc3x65.3340.628SB1000SN100/0.85xc4x720.6932.434SB1000查运算曲线得：I123.18，I31.69，I40.42，I12,0.22.5，I3,0.21.49，I4,0.20.38 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/1200011000200/0.85100/0.85100/0.85I3.181.690.4213.25kA0.532303.54832303230323032301200011000200/0.85100/0.85100/0.85I0.22.51.490.3812.777kA0.532303.5483230323032303230iM2KMI21.913.2535.603kAf3点发生三相短路时，简化电路为：由f1点发生短路时的简化过程可知：x12.122，x26.187x3xdxT332.040.0421.998x4xT4xd0.8752.042.91511111111Y2.546x1x2x3xT322.1226.1871.9980.708x5x1xT32Y3.825，x6x2xT32Y11.152，x7x3xT32Y3.602计算运算电抗：SN200/0.85xc12x611.1522.624SB1000SN100/0.85xc3x73.6020.424SB1000SN100/0.85xc4x42.9150.343SB1000查运算曲线得：I120.38，I32.5，I43.18，I12,0.20.36，I3,0.22.04，I4,0.22.51200011000200/0.85100/0.85100/0.85I0.382.53.1816.47kA3.82531150.531153115311531151200011000200/0.85100/0.85100/0.85I0.20.362.042.515.773kA3.82531150.53115311531153115iM2KMI21.916.4744.255kA8-10解：取 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/10001000SB1000MVA，UBUav，xS11，xS20.6131800017000100008368简化等效电路为：11x1xS1xT110.1681.0842211x2xS2xT20.613(0.011)0.60752211x1xT31.0840.3811212x3x1xT31.0840.3811.6142x220.607511x2xT30.60750.3811212x4x2xT30.60750.3810.9052x121.084118000110000I0.01348.8kA1.614336.750.905336.75即变电所35kV母线三相短路t=0.01s时的短路电流为348.8kA。第九章9-1解：f1点发生不对称短路故障：正序：x(1)xG(1)xT1xL1 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/负序：x(2)xG(2)xT1(2)xL1(2)(xT1(0)xL1(0))(xL2(0)xT2(0)xL3(0)xT3Ⅰ)零序：x(0)xT1(0)xL1(0)xL2(0)xT2(0)xL3(0)xT3Ⅰf2点发生不对称短路故障：正序：x(1)xG(1)xT1xL1xL2xT2负序：x(2)xG(2)xT1(2)xL1(2)xL2(2)xT2(2)(xT1(0)xL1(0)xL2(0)xT2(0))(xL3(0)xT3Ⅰ)零序：x(0)xT1(0)xL1(0)xL2(0)xT2(0)xL3(0)xT3Ⅰ 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/f3点发生不对称短路故障：正序：x(1)xG(1)xT1xL1xL2xT2xL3xT3负序：x(2)xG(2)xT1(2)xL1(2)xL2(2)xT2(2)xL3(2)xT3(2)零序：x(0)9-2解：(1)正序、负序、零序等效电路图如下： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/(2)正序、负序、零序等效电路图如下：9-3解：正序、负序、零序等效电路图如下： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/9-4解：在f处发生单相断线，其正序、负序、零序等效电路图如下：UE1.0ff0x(1)xGxT1xLxT2xLD(1)0.250.20.150.21.22x(2)xG(2)xT1(2)xL(2)xT2(2)xLD(2)0.250.20.150.20.351.15x(0)xT1(0)xL(0)xT2(0)0.20.50.20.9各序电流分量为：Uff01If(1)j0.399x(2)x(0)j1.15j9xj2(1)j1.15j9x(2)x(0)x(0)j0.9If(2)If(1)(j0.399)j0.175x(2)x(0)j1.15j9x(2)j1.15If(0)If(1)(j0.399)j0.224x(2)x(0)j1.15j9各序电压分量为：Uf(1)Uf(2)Uf(0)x(0)If(0)j0.9j0.2240.2016故障处的各相电压电流为：Ufa3Uf130.20160.6048，UfbUfc0 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/I0faIa2IaII(0.5j0.866)(j0.399)(0.5j0.866)j0.175j0.224fbf(1)f(2)f(0)0.497j0.3360.5999145.939IaIa2II(0.5j0.866)(j0.399)(0.5j0.866)j0.175j0.224fcf(1)f(2)f(0)0.497j0.3360.599934.061在f处发生单相接地故障：其正序、负序、零序等效电路图如下：x(1)(xGxT1)//(xLxT2xLD(1))(0.250.2)//(0.150.21.2)0.349x(2)(xG(2)xT1(2))//(xL(2)xT2(2)xLD(2))(0.250.2)//(0.150.20.35)0.274x(0)xT1(0)//(xL(0)xT2(0))0.2//(0.50.2)0.156各序电流分量为：Uf01If(1)If(2)If(0)j1.284x(1)x(2)x(0)j(0.3490.2740.156)各序电压分量为：Uf(1)If(1)(x(2)x(0))j1.284(j0.274j0.156)0.552Uf(2)If(1)x(2)j1.284j0.2740.352Uf(0)If(1)x(0)j1.284j0.1560.2故障处的各相电流电压为：Ifa3If13(j1.284)j3.852，IfbIfc0 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/U0faUa2UaUU(0.5j0.866)0.552(0.5j0.866)(-0.352)0.2fbf(1)f(2)f(0)0.3j0.7830.839110.964UaUa2UU(0.5j0.866)0.552(0.5j0.866)(-0.352)0.2fcf(1)f(2)f(0)0.3j0.7830.839110.9649-5解：(1)计算参数，取SB100MVA,UBUavSB100xd0.130.130.221SN50/0.85SB100x20.20.20.34SN50/0.85US%SB10.5100xT10.175100SN10060SB100xLx1l20.420020.151Uav230US%SB10.5100xT20.35100SN10030x03xL30.1510.45311正序：x(1)(xdxT1xL)//(xLxT2)0.2242211负序：x(2)(x2xT1xL)//(xLxT2)0.2472211零序：x(0)(xT1x0)//(x0xT2)0.23722 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/建立正序增广网络：zΔx(2)x(0)0.4841x1xdxT1xL0.471521x2xLxT20.42552x1zΔ0.47150.484x3x1zΔ0.47150.4841.492x20.4255x2zΔ0.42550.484x4x2zΔ0.42550.4841.346x10.471550/0.85计算运算电抗：xc1.4920.878100I1.18，I0.21.0850/0.85130I1.180.23kA32301.346323050/0.85130I0.21.080.215kA32301.3463230故障相电流：I3a2I3(0.5j0.866)0.230.345j0.598kAfbI3a2I3(0.5j0.866)0.2150.323j0.559kAfb0.20.2(2)a2IaIIa2I0.23120f(1)f(2)f(0) 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/I0.23，I0.23120，I0.23-120f(1)f(2)f(0)1xLxT2I2I0.42550.230.109kA(1)f(1)xdxT1xLxT20.8971xLxT2I2I0.42550.231200.109120kA(2)f(2)xdxT1xLxT20.897∴机端电流为：I1110.109ej300.094j0.1640.18960.18Ga2-j30IGbaa10.109120e0.094j0.1640.189119.82Iaa21000Gc电流向量图：9-6解：正序：10.3x(1)(xdG1xT1)//(xLxT2xdG2)(0.30.12)//(0.10.25)0.22822负序：10.3x(2)(x2G1xT1)//(xLxT2x2G2)(0.20.12)//(0.10.15)0.1782210.7零序：x(0)xT1//(x0xT23xn)0.12//(0.130.35)0.11722(1)在f点发生b,c两相短路： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/Uf01If(1)If(2)j2.463x(1)x(2)j0.228j0.178Uf(1)Uf(2)If(2)x(2)j2.463j0.1780.438故障处的电流电压为：I0faIa2IaIj3I4.266fbf(1)f(2)f(1)IaIa2Ij3I4.266fcf(1)f(2)f(1)Ufa2Uf(1)20.4380.876UU(a2a)UU0.438fbfcf(1)f(1)故障处电流电压向量图：(2)在f点发生b,c两相短路接地：Uf01If(1)j3.349x(2)x(0)j0.178j0.117xj0.228(1)j0.178j0.117x(2)x(0)x(0)j0.117If(2)If(1)j3.349j1.328x(2)x(0)j0.178j0.117x(2)j0.178If(0)If(1)j3.349j2.021x(2)x(0)j0.178j0.117x(2)x(0)j0.178j0.117Uf(1)Uf(2)Uf(0)If(1)j3.3490.236x(2)x(0)j0.178j0.117故障处的电流电压为： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/Ifa0，UfbUfc0Ia2IaII4.05j3.035.058143.2fbf(1)f(2)f(0)IaIa2II4.05j3.035.05836.8fcf(1)f(2)f(0)Ufa3Uf(1)30.2360.708故障处电流电压向量图：9-7解：由题可得，令机端正序、负序、零序分别为I(1)，I(2)，I(0)I1aa230130(1)12I(2)1aa01303I11131800(0)取变压器星形侧正序、负序为I(1)，I(2)IIej30130(1)(1)I-j30(2)I(2)e1-30I10，I10II(1)(2)(1)(2)由此判断发生a相单相接地短路故障。正序、负序等效电路：x(1)x(2)0.160.210.050.42零序： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/150xL(0)3xL0.15,xL(0)30.20.45200x(0)(0.150.21)//(0.450.21)0.2311I(1)jj0.935j(x(1)x(2)x(0))1.07Ifa3I(1)-j2.805，IfbIfc09-8解：正序：1(0.210.18)x(1)(xdxT1)//[xL1(1)//(xL2(1)xL3(1))]//(0.2//0.3)0.0722负序：x(2)x(1)0.07零序：10.18x(0)xT1//xL1(0)//[xL2(0)(xL3(0)//xT3)]//0.6//[0.6(0.3//0.12)]0.0722 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/故障处各序电流电压：Uf01If(1)j9.524x(2)x(0)j0.07j0.07xj0.07(1)j0.07j0.07x(2)x(0)x(0)j0.07If(2)If(1)j9.524j4.762x(2)x(0)j0.07j0.07x(2)j0.07If(0)If(1)j9.524j4.762x(2)x(0)j0.07j0.07x(2)x(0)j0.07j0.07Uf(1)Uf(2)Uf(0)If(1)j9.5240.333x(2)x(0)j0.07j0.07故障处的电流电压为：Ifa0，UfbUfc0Ia2IaII12.372j2.71412.666167.627fbf(1)f(2)f(0)IaIa2II12.372j2.71412.66612.373fcf(1)f(2)f(0)Ufa3Uf(1)30.3330.9999-9解：计算参数，取SB100MVA,UBUavSB100G:xdx20.10.10.222SN45US%SB10.5100T1:xT10.333100SN10031.510.5100SB100T2,T3:xT2xT30.525，xn2022020.0410020U230avSB100L1,L2:xL1xL20.45020.45020.04Uav230xL1(0)xL2(0)3xL130.040.12正序、负序电路： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/x(1)x(2)xdxT1xL1xT20.2220.3330.040.5251.12零序：x(0)xT2xL1(0)xT1//(xL2(0)xT33xn)0.5250.120.333//(0.120.52530.04)0.877各序电流电压分量为：Uf01.1If(1)If(2)If(0)j0.353x(1)x(2)x(0)j(1.121.120.877)Uf(1)If(1)(x(2)x(0))j0.353(j1.12j0.877)0.705Uf(2)If(1)x(2)j0.353j1.120.395Uf(0)If(1)x(0)j0.353j0.8770.31机端电压为：UG(1)1.1(j0.353)j0.2221.022UG(1)(j0.353)j0.2220.078U1111.022ej300.818j0.550.98633.92Ga2-j30UGbaa10.078ej1.11.190Uaa2100.818j0.550.986146.08Gc9-10解：计算参数，取SB1000MVA,UBUav 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/SB10001000S1:x1(1)x1(2)0.50.50.03，x1(0)0.750.04SN1800018000SB10001000S2:x2(1)x2(2)0.20.20.02，x2(0)0.650.065SN1000010000US1%SB12.61000T1:xT10.056100SN1003750US2%SBT2:xT20100SNUS3%SB28.551000T3:xT30.127100SN1003750xT1xT2xT3正序、负序电路：x(1)x(2)(x1(1))//(x2(1))0.08222xT1xT2xT3零序电路：x(0)(x1(0))//(x2(0))0.097222 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/故障处各序电流电压分量为：Uf01If(1)If(2)If(0)j3.891x(1)x(2)x(0)j(0.080.080.097)Uf(1)If(1)(x(2)x(0))j3.891(j0.08j0.097)0.689Uf(2)If(1)x(2)j3.891j0.080.311Uf(0)If(1)x(0)j3.891j0.0970.377故障处的不对称电流为：Ifa3If(1)3(j3.891)-j11.673IfbIfc0第十章10-1解：系统的综合电抗为：1xdxdxT1xLxT21.8121xqxqxT1xLxT21.4121xdxdxT1xLxT20.9121xexT1xLxT20.612运行参数：U1,P01,0arccos0.925.84,Q0P0tan00.484Q0xe2P0xe2U(U)()1.432G0UUQ0xq2P0xq2E(U)()2.195Q0UU1.41arctan39.965010.4841.4122P0Q0I1.1110UId0I0sin(00)1.013Eq0EQ0Id0(xdxq)2.6002Eq0EQ0Id0(xqxd)1.6885Q0xd2P0xd2E(U)()1.7040UU(1)Eq保持不变的功率特性，有 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/Eq0UU2xxdqPEsinsin21.437sin0.078sin2qx2xxddqdPEq由0得，1.437cos0.0782cos20d解得，cos0.106,cos4.712(舍去)，83.921EqmaxPE1.437sin83.920.078sin(283.92)1.445qmaxPEP0qmaxKP100%44.5%P0(2)Eq保持不变的功率特性，有Eq0UU2xxqdPEsinsin21.855sin0.195sin2qx2xxddqdPEq由0得，1.855cos0.1952cos20d解得，cos0.194,cos2.572(舍去)，101.1861EqmaxPE1.855sin101.1860.195sin(2101.186)1.894qmaxPEP0qmaxKP100%89.4%P0(3)E保持不变的功率特性，有EU0PEsin1.873sinxdUxdarcsin[(1)sin]arcsin(0.208sin)E0xqdPE由0得，arcsin(0.208sin)90d解得，sin0.979,sin10.979(舍去)，E101.76,90maxPEmaxP0PEmax1.873sin901.873,KP100%87.3%P0(4)UG保持不变的功率特性，有 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/UUG0PUGsinG2.348sinGxeUxeGarcsin[(1)sin]arcsin(0.396sin)UG0xqdPUG由0得，arcsin(0.396sin)90d解得，sin0.9298,115.6UGmaxPUGmaxP0PUGmax2.348sin902.348,KP100%134.8%P010-2解：(1)取SB250MVA,UB115kV,U1200P0*0.8,0arccos0.9811.478,Q0*P0*tan00.162250综合电抗：xxxxx0.8,xxxx0.7dGT1LT2eT1LT2QxPx0d20d2E(U)()1.2980UU0.80.8arctan29.53010.1620.8E保持不变的功率特性，有EU0PEsin1.623sin,PE1.623xmaxdPEmaxP01.6230.8K100%100%102.88%PP0.80即正常运行功角为29.53，K为102.88%0P(2)P*1,011.478,Q*P*tan00.203Q*xe2P*xe2U(U)()1.34G0UUQ*xd2P*xd2E(U)()1.4110UU即发电机的最小电动势为1.3410-3解：(a)有公式可得： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/222222rx2j[r(x1x2)x1x2]0.20.5j[0.2(0.50.5)0.50.5]Y110.4878j1.609822222222x1x2r(x1x2)0.50.50.2(0.50.5)rx2j[r2(xx)x2x]222112120.20.5j[0.2(0.50.5)0.50.5]Y220.4878j1.609822222222x1x2r(x1x2)0.50.50.2(0.50.5)22rx1x2jr(x1x2)0.20.50.5j0.2(0.50.5)Y120.4878j0.390222222222x1x2r(x1x2)0.50.50.2(0.50.5)x1x20.50.512arctanarctan51.34r(x1x2)0.2(0.50.5)1202222P1EqG11EqUcY12sin(12)1.20.48781.210.48780.3902sin(51.34)0.70240.7496sin(51.34)(b)由公式可得：x2x30.50.5Y11jjj1.333xxxxxx2221213230.50.50.5x1x30.50.5Y22jjj1.333xxxxxx2221213230.50.50.5x30.5Y12jjj0.667xxxxxx2221213230.50.50.5120EqUc1.21P1sinsin0.8sinx1x20.50.5x1x20.50.5x30.5EqU1.81.010-4解：(1)PEqsinsin1.2sinxd1.5 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/dPEqSEq01.2cos01.2cos540.705ddS0250314dtdSEqD0.7051.50.07050.15dtTjTj10100314(2)由(1)知，A0.07050.152又由EA0，特征方程为：0.1522.1370解得：10.075j4.705,20.075j4.705∵特征根为负实部的共轭根，∴,作衰减震荡，系统具有小干扰稳定性。10-5解：(1)xdxdxTxL0.30.150.61.05036,EUUGUS1.00.995PEsinsinsin0.948sin0.948sinxdxd1.05dPESE00.948cos00.948cos360.767ddS0250314dtdSED0.76750.1280.833dtTjTj660314(2)由(1)可知：A0.1280.8332又由EA0，特征方程为：0.83340.1920解得：10.417j6.326,20.417j6.326101314无阻尼自然频率：fSEq0.7671.008Hz2Tj26(3)∵特征根为负实部的共轭根，∴,作衰减震荡，系统具有小干扰稳定性。10-6解：该系统正常运行(a)，发生故障(c)(d)，切除故障(b)三阶段电路图： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/20070取S100MVA，P2，P0.7BⅠⅡ10010011(xxx)x111dTLL44xxxx，xxxxx，xxxxⅠdTLⅡdTLLⅢdTL2441xxL8EUEUEUP，P，PⅠⅡⅢxxxⅠⅡⅢ31综合上式解得：xx，x，P175MWdTLⅢ7710-7解：计算参数：SB100MVA22015.75UB(110)115kV,UB(220)115209kV,UB(15)20913.6kV121242 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/22UNSB15.75100G:xd0.250.250.142S2200/0.852NUB(15)13.6215.75100x00.080.046200/0.85213.6SN200/0.85Tj(B)Tj614.118(s)SB10022US%UNSB14242100T1:xT1220.078100SNUB(220)10024020922US%UNSB14220100T2:xT2220.078100SNUB(220)100200209SB100L:xⅠxⅡx1l20.410020.092UB(220)209x04xⅠ0.368115100U1,P01,0arccos0.9811.478,Q0P0tan00.2031151001x1xdxT1xⅠxT21.0462Q0x12P0x1222E(U)()(10.2031.046)1.0461.601UU1.0460arctan40.78710.2031.4061故障后，负序：x2(xdxT1)//(xⅡxT2)0.07921零序：x0(xT1)//(x0xT2)0.062∵在故障出发生的是单相接地短路，xx2x00.139∴增广网络： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/11(xdxT1)(xⅠxT2)2xⅡxdxT1xⅠxT20.542xEU1.6011∴故障时发电机的最大功率为：PⅡmax2.965xⅡ0.54故障切除后，xⅢxdxT1xⅠxT20.39EU1.6011P4.105Ⅲmaxx0.39ⅢPM1arcsin180arcsin165.9hP4.105Ⅲmax极限切除角为：P()PcosPcosMh0ⅢmaxhⅡmax0arccoscmPPⅢmaxⅡmax165.9-40.78714.105cos165.92.965cos40.787180arccos4.1052.965arccos(-3.546)10-8解：在线路中点发生单相接地故障：xxxxx0.2950.1380.2440.1220.799ddT1LT2Qxd2Pxd20.20.79922E(U)()(1)0.7991.408UU10.799arctan34.56010.20.799负序，零序等效电路： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/110.2440.244x(xxx)//(xx)(0.4320.138)//(0.122)0.1822T1LLT222221130.24430.244x(xx)//(xx)(0.138)//(0.122)0.2480T1L(0)L(0)T22222xxx0.180.2480.42820建立正序增广网络：11(xxx)(xx)11dT1LLT222xxxxxx1.115ⅡdT1LLT222xEU1.4081P1.263Ⅱmaxx1.115Ⅱ切除故障后，P0ⅢmaxPMarcsin180arcsin90hPⅢmaxP()PcosPcosMh0ⅢmaxhⅡmax0arccoscmPPⅢmaxⅡmax9034.56101.263cos34.56180arccos01.263arccos(0.058)86.7发生单相断线故障：负序、零序等效电路： 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/110.2440.244xxxxxx0.4320.1380.1220.93622T1LLT222221130.24430.244xxxxx0.1380.1220.9920T1L(0)L(0)T22222x2x00.9360.992x0.482xx0.9360.9922011(xxx)(xx)11dT1LLT222xxxxxx1.08ⅡdT1LLT222xEU1.4081P1.304Ⅱmaxx1.08Ⅱ切除故障后，P0ⅢmaxPMarcsin180arcsin90hPⅢmaxP()PcosPcosMh0ⅢmaxhⅡmax0arccoscmPPⅢmaxⅡmax9034.56101.304cos34.56180arccos01.304arccos(0.082)85.310-9解：由题意知：P1，Q0.4，U1.000xxxxx0.20.10.40.10.8ddT1LT2QxPx0d20d222E(U)()(10.40.8)0.81.544UUEU1.5441P1.93Ⅰmaxx0.8d77P0，PP，PP1.351ⅡmaxⅢⅠⅢmaxⅠmax10100.8arctan31.22010.40.8PmP01arcsinarcsin180arcsin132.25hPP1.351ⅢmaxⅢmaxP()PcosPcosmh0ⅢmaxhⅡmax0arccoscmPPⅢmaxⅡmax1(132.2531.22)1.351cos132.250180arccos50.731.3510 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/10-10解：(1)PP1.5，E1.2，U1.0m0Sxxx//(xx)0.20.4//(0.20.2)0.4dd123EUS1.21.0P3Ⅰmaxx0.4dPm1.5arcsinarcsin300P3Ⅰmax(2)xxx0.20.40.6Ⅲd1EUS1.21.0P2Ⅲmaxx0.6Ⅲxxxxx0.20.40.20.21.0Ⅱd123EUS1.21.0P1.2Ⅱmaxx1.0ⅡPmP01.5arcsinarcsin180arcsin131.41hPP2ⅢmaxⅢmaxP()PcosPcosmh0ⅢmaxhⅡmax0arccoscmPPⅢmaxⅡmax1.5(131.4130)2cos131.411.2cos30180arccos68.5321.2(3)若QF1、QF2不跳开，xxx//(xx)0.20.2//(0.40.2)0.35Ⅱd213EUS1.21.0P3.429Ⅱmaxx0.35Ⅱxxx0.20.40.6Ⅲd1EUS1.21.0P2Ⅲmaxx0.6ⅢPmP01.5arcsinarcsin180arcsin131.41hPP2ⅢmaxⅢmaxP()PcosPcosmh0ⅢmaxhⅡmax0coscmPPⅢmaxⅡmax1.5(131.4130)2cos131.413.429cos3018021.21.1461∴可知系统不稳定。若QF1、QF2可靠跳开， 欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/xxxxxx121323x0.1，x0.1，x0.05xxxxxxxxx123123123(xdx)x0.30.1xxxx0.20.10.11.0Ⅱdx0.05EUS1.21.0P1.2Ⅱmaxx1Ⅱxxx0.20.40.6Ⅲd1EU1.21.0SP2Ⅲmaxx0.6ⅢPmP01.5arcsinarcsin180arcsin131.41hPP2ⅢmaxⅢmaxP()PcosPcosmh0ⅢmaxhⅡmax0coscmPPⅢmaxⅡmax1.5(131.4130)2cos131.411.2cos3018021.20.366arccos0.36668.53cm'