電力系統解析課后答案電力系統解析課后答案電力系統解析課后答案2-17.某一回110kv架空電力線路,長度為60km,導線型號LGJ-120,導線計算外徑為,三相導線水平排列,兩相鄰導線三相間的距離為4m,試計算該電力線路的參數,并作等效電路。解:r131.50.2625kms12015.2r7.6(mm)2Dm340004000240005039.7(mm)X1Dm0.01570.1445lg5039.70.01570.423(/km)0.1445lg7.67.58r7.58b1662.69106Dm1010s5039.7lglgr7.6g10.R1r1l0.26256015.75()X1x1l0.4236025.38()64則B1b1l2.6910601.61410(s)B10.8071042(s)G10.b11.8051062(s/km)g10.2-19.三相雙繞組升壓變壓器的型號為SFL-40500/110,額定容量為40500KVA,額定電壓為121/,Pk234.4kw,Uk(%)11,P093.6kw,I0(%)2.315,求該變壓器的參數,并作等值電路。2234.4103(10.5103)2解:`RTPKUN2()SN2(40500103)21.5710XTUN(%)UN211(10.5103)20.3()100SN10040500103GTP093.61038.49104(S)UN2(10.5103)2BTI0(%)SN2.315405001038.5103(S)100UN2100(10.5103)22-20.三相三繞組降壓變壓器的型號為SFPSL-120000/220,額定容量為120000/120000/60000KVA,額定電壓為220/121/11KV,PK(12)601KW,PK(13)182.5KW,PK(23)132.5KW,UK(12)(%)14.85,UK(13)(%)28.25,UK(23)(%)7.96,P0135KW,I0(%)0.663,求該變壓器的參數,并作等值電路。解:I:PK(12)601(KW)P'3)4P3)730(KW)K(1K(1P'3)4P3)530(KW)K(2K(2PK11(PK(12)PK'(13)PK'(23))400.5(KW)2PK21(PK(12)PK'(23)PK'(13))200.5(KW)2PK31(PK'(13)PK'(23)PK(12))329.5(KW)2RT1PK1UN21400.5103(220103)2)SN2(12104103)21.346(RT2PK2UN21200.5103(220103)2)2(12410320.674(SN10)RT3PK3UN21329.5103(220103)2)SN2(12104103)21.107(II：UK(12)(%)14.85,UK(13)(%)28.25,UK(23)(%)7.96UK1(%)12)(%)UK(13)(%)UK(23)(%))17.57(UK(12UK2(%)1(UK(12)(%)UK(23)(%)UK(13)(%))2.722UK3(%)1(UK(13)(%)UK(23)(%)UK(12)(%))10.682XT1UK1(%)UN2117.57(220103)270.866()100SN10012104103XT2UK2(%)UN212.72(220103)210.971()100SN10012104103XT3UK3(%)UN2110.68(220103)243.076()100SN10012104103III：GTP01351032.79106SUN2(220103)2( )1IV：BI0(%)SN0.663121041031.644105(ST100UN2100(220103)2)12-22.雙水內冷汽輪同步發電機的型號為TQFS-125-2,PN125MW,cos

N

0.85,UN

13.8KV,Xd

1.867,Xd

0.257,Xd''

0.18,試計算該發電機的直軸同步電抗

Xd,暫態電抗

Xd',直軸次暫態電抗

Xd''

的有名值

.解:UN2UN2cosN1.295( )ZNPNSNXd1.867ZN1.8671.2952.418()Xd'0.257ZN0.2571.2950.333()Xd''0.18ZN0.181.2950.233( )2-23.電抗器型號為NKL-6-500-4,UN6KV,IN500A,電抗器電抗百分數XL(%)4.試計算該電抗器電抗的有名值.XL(%)UN46103解:XL10030.277( )1003IN500Chapter三3-1.電力系統阻抗中的功率耗費表達式是什么電力線路始.尾端的電容功率表達式是什么答:SLPLjQLPi2Qi2(RLjXL)Ui21SC1jU12BL221SC22BL2jU223-8．110kv雙回架空電力線路，長度為150km，導線型號為LGJ-120，導線計算外徑為,三相導線幾何平均距離為5m,已知電力線路尾端負荷為30+j15MVA,尾端電壓為106kv,求始端電壓、功率,并做出電壓相量圖。解:r131.50.2625(/)s120kmDm5m5000(mm),r15.227.6(mm)X10.1445lgDm0.01570.1445lg50000.423(/km)r0.01577.6b17.581067.581062.69106(s/km),g10lgDmlg5000r7.6RL1r1l10.262515019.687()22XL1x1l10.42315031.725()22BL2b1l22.691061508.07104(s)GL0..'.jU22BL)(j10628.07104S2S2((30j15))30j10.466(MVA)22.'2'222SLPLjQLP2Q2(RLjXL)3010.466(19.687j31.725)U2210621.769j2.85(MVA).'.'.S1S2SL(30j10.466)(1.769j2.85)31.769j13.316(MVA).''''P2RLQ2XLjP2XLQ2RLdU2U2jU2U2U23019.68710.46631.725j3031.7251061068.7j7.03(kv)...。U1U2dU21068.7j7.03114.7j7.03114.93.5（kv)..'2BL28.07104S1S1(jU)(31.769j13.316)(j114.9)12231.769j7.99(MVA)。.設U28.7(kv),U27.03(kv),則.(kv)1060（kv),U2單回架空電力線路,長度為200KM,導線型號LGJ—300,導線計算外徑為24.2mm,三相導線幾何平均距離為7.5m,已知其始端輸出的功率為120j50MVA,始端的電壓為240KV.求末端電壓及功率,并作出電壓向量圖.解:r131.50.105(/Km)s300Dm7.5m7500(mm)r24.2912.1(mm)2x1Dm0.01570.1445lgr0.1445lg75000.01570.419(/km)12.1b17.581067.581062.715106(S/km)Dm7500lglgr12.1RLr1l0.10520021( )XLx1l0.41920083.8( )Blb1l2.7151062005.43104(S)Gl02Bl)(120j50)(j24025.434S1S1(jU122)120j65.6384(MVA)SLP12Q12(RLjXL)120265.63842(21j83.8)U1224026.82j27.22(MVA)S2S1SL(120j65.6384)(6.82j27.22)113.18j38.418(MVA)dU1U1jU1P1RLQ1XLjP1XLQ1RLU1U11202165.638483.8j12024024033.419j36.157(MVA)U2U1dU124033.419j36.157206.58j36.157209.7210(KV)2BL25.43104S2S2(jU22)(113.18j38.418)(j209.722)113.18j50.36(MVA).∴S2113.18j50.36(MVA)；U2209.7210(KV)。單回架空電力線路,長度為80km,導線型號為LGJ-95,導線計算外徑為,三相導線幾何平均距離為5m,電力線路始端電壓為116kv,尾端負荷為15+j10MVA,求該電力線路尾端電壓及始端輸出的功率.解:r131.50.332(/)s95kmDm5m5000(mm),r13.76.85(mm)2x10.1445lg50000.01570.425(/km)6.85b17.581067.581062.647106(s/km)Dm5000lglgr6.85g10RLr1l0.3328026.56()XLx1l0.4298034.32()BLb1l2.647106802.12104(s)GL0.'.(-jUN2BL)j11022.12104S2S2(15j10)()15j8.717(MVA)22.'2'222P2Q2158.717SL(RLjXL)(26.56j34.32)10.66j0.854(MVA)UN21102.'.'.S1S2SL(15j8.717)(10.66j0.854)15.66j9.57(MVA)..'jU12BL)(j11622.12104S1S1((15.66j9.57))15.66j8.14(MVA)22dU1U1jU1P1'RLQ1'XLjP1'XLQ1'RL15.66U1U1116+j15.6634.329.5726.566.417j2.442(KV)116..U2U1dU11166.417j2.442109.583j2.442(KV)單回架空電力線路,長度為220km,電力線路每公里的參數為r10.108/km,x10.42/km,b12.66106s/km,線路空載運行,當線路尾端電壓為205kv時,求線路始端電壓.解:RLr1l0.10822023.76(KV)XLx1l0.4222092.4()BLb1l2.661062205.852104(s).因為線路空載運行,所以S20.'.2BL25.852104S2S2(jU)j12.3(MVA)22j2052.'2'22SLP2Q2(RLjXL)12.3(23.76j92.4)0.085j0.33(MVA)U222052.'..'1SLj12.3j11.97(MVA)SS2.''''dU2U2jU2P2RLQ2XLP2XLQ2RLU2j205jU22055.544j1.4256(KV)...U1U2dU22055.544j1.4256199.456j1.4256(KV)..2BL225.852104'S1S1(jU12)(0.085j11.97)[j(119.4561.4256)2]0.085j23.61(MVA)U1119.45621.42562119.46(kV)3-12.有一臺三繞組變壓器,其歸算至高壓側的等效電路如圖所~示,ZT12.47j65,ZT22.47j115,ZT32.47j37.8,S25j4MVA,~S38j6MVA,當變壓器變比為110/(1+5%)/時,試計算高、中壓側的實質電壓....解:I:由U3',S3求U1'.1106110U3'U3100(KV)6.66.6.2222P3Q386ST3(RT3jXT3)(2.47j37.8)0.0247j0.378(MvA)U3'21002...'S3T3(8j6)(0.0247j0.378)8.0247j6.378(MVA)S3S.P3RT3Q3XT3jP3XT3Q3RT3dU3U3jU3U3'U3'82.47637.8j837.862.472.4656j2.87589(KV)100100.U3'.U1'dU3100j2.4656j2.8757102.4656j2.8758(KV)...II.由U1',S2求U2.2222P2Q254ST2(RT2jXT2)(2.47j1.5)0.0084j0.005(MVA)UN21102...S2'S2ST2(5j4)(0.0084j0.005)5.0084j3.995(MVA).''''P2RT2Q2XT2jP2XT2Q2RT2dU2U2jU2U1'U1'5.00842.473.9951.5j5.0084(1.5)3.9952.470.062j0.17(KV)102.465622.87582102.465622.87582...U1'U1dU2102.4j3.04(KV)...S1''S2'S3'13.033j10.373(MVA)...III.由U1',S1''求U1.P''2Q''213.033210.3732ST111(RT1jXT1)(2.47j65)0.0652j1.716(MvA)U1'2102.465622.87582...S1'S1''ST1(13.033j10.373)(0.0652j1.716)13.0982j12.089(MVA).''''''''P1RT1Q1XT1jP1XT1Q1RT1dU1U1jU1U1'U1'13.0332.4710.37365j13.0336510.3732.476.89j8(KV)102.5102.5.U1'.U1dU1109.3556j10.8758(KV)U1109.9(KV)U2102.423.042102.445(KV)U2實U238.5(15%)102.44538.51.0537.65(KV)1101103-13.某電力線路導線為LGJ-185,長度為100km,導線計算外徑為19mm,線路尾端負荷為90+j20MVA,該電力線路由110kv升壓至220kv運行,假設升壓后導線截面和負荷大小保持不變,且不計電暈損失的增加,導線水平排列,升壓前后線間距離由4m增加到,試問升壓后,該電力線路的功率耗費減少了多少電壓耗費的百分數減少了多少131.50.17(/km)解:r1185sRLr1l0.1710017( )Dm1340004000240005.04103(mm)Dm2355005500255006.93103(mm)r199.5(mm)2x10.1445lgDm15.041030.01570.41(/km)r9.5x20.1445lgDm26.931030.01570.43(/km)r9.5XL1x1l0.4110041( )XL2x2l0.4310043( )b17.582.78106(s/km)lgDm1106rb27.582.65106(s/km)lgDm2106rBL1b1l2.781061002.78104(s)BL2b2l2.651061002.65104(s).BL1.41222.7810SC1j110j1.68(MVA)2jU1N22.41.2BL222.6510SC2j220j6.4(MVA)2jU2N22..1.S2'1S2SC1(90j20)j1.6890j18.32(MVA)2..1.S2'2S2SC2(90j20)j6.490j13.6(MVA)2.'2'222P21Q219018.32S1(RLjXL1)(17j41)11.85j28.58(MVA)U12N1102.'2Q'290213.62S1P2222(RLjXL2)(17j43)2.9j97.36(MVA)U22N2202...SS1S2(11.85j28.58)(2.9j7.36)8.95j21.22(MVA).''''P21RLQ21XL1jP21XL1Q21RLdU1U1jU1U1NU1N901718.3241j904118.321720.74j30.7(kv)110110.''''P22RLQ22XL2jP22XL2Q22RLdU2U2jU2U2NU2N901713.643j904313.6179.6j16.54(kv)220220.dU120.74230.72dU1%100%100%33.68%110110.dU29.6216.542dU2%100%100%8.7%110220...dU%dU1%dU2%33.68%8.7%25%3-14.對圖所示環式等值網絡進行潮流計算,圖中各線路的阻抗為Zl110j17.32,Zl220j34.6,Zl325j43.3,Zl410j17.3,各點的運~~~算負荷為S290j40MVA,S350j30MVA,S440j15MVA,且U1235KV.解:在1處把環網解開,可得以下列圖:Zl110j17.321017.3Zl220j34.6(1017.3)2Zl325j43.3(1017.3)2.5Zl410j17.3....******則1S2(Zl2Zl3Zl4)S3(Zl3Zl4)S4Zl4SZl*1Zl*2Zl*3Zl*4...S2(22.51)S3(2.51)S4=122.51(90j40)5.5(50j30)3.5(40j15)=6.5=109.2j52.3(MvA)..***.*.*S*S1'4(Zl3Zl2Zl1)S3(Zl2Zl1)S2Zl1Zl*1Zl*2Zl*3Zl*4....S4(2.521)S3(21)S2=122.51(40j15)5.5(50j30)3(90j40)=6.570.77j32.7(MvA)...S23S1S2109.2j52.3(90j40)19.2j12.3(MvA)...S43S1'S470.77j32.7(40j15)30.77j17.7(MvA)..S1S1'109.2j52.370.77j85(MVA)...S2S3S490j4050j3040j15180j85(MVA).....所以S1S1'S2S3S4又3點為功率分點,于是網絡從3點打開,變成兩個開式網..2222P23Q2319.212.3S2(Rl2jXl2)(20j34.6)0.2j0.37(MVA)UN22202...S2"S23S2'19.2j12.30.2j0.3719.4j12.67(MVA)...S2'S2''S290j4019.4j12.67109.4j52.67(MVA).'2'222S1P2Q2(Rl1jXl1)109.452.67(10j17.32)3j5.27(MVA)UN22202...S1S2'S1'109.4j52.673j5.27122.4j57.94(MVA).P1RL1Q1XL1jP1XL1Q1RL1dU1U1jU1U1U1112.41057.9417.32j57.94109.05j5.82(KV)235235...。U2U1dU1235(9.05j5.82)225.95j5.82(kv)2261.5（KV）.“”“”dU2U2jU2P2RL2Q2XL2jP2XL2Q2RL2U2U219.420126734.6j19.434.612.67203.656j1.85(KV)226226...222.4262。U3U2dU2225.95j5.82(3.656j1.85)222.294j7.67(kv).P43RL3Q43XL3P43XL3Q43RL3dU3jU3U330.772517.743.330.7743.317.725j4(kV)222.426j222.4266.9...U4U3dU3222.4266.9j4229.326j4(KV)3-15.某35KV變電全部二臺變壓器并聯運行，如圖，其參數是T1為SN8MVA,PK25KW,UK(%)7.5;T2為SN2MVA,PK24KVA,UK(%)6.5~二臺變壓器均忽略勵磁支路，變電所低壓側經過的總功率為S=+,試求:當變壓器的變比為KT1KT235/11KV時,每臺變壓器的總功率為多少當KT134.125/11KV,KT235/11KV時,每臺變壓器經過的功率是多少解RT1PKUN225103(35103)20.48()SN2(8106)2XT1UK(%)UN27.5(35103)211.48()100SN1008106RT2PKUN224103(35103)27.35()SN2(2106)2XT2UK(%)UN26.5(35103)239.8()100SN1002106ZT10.48j11.48(),ZT27.35j39.8( )在1點處把環網解開,則:..*(8.5j5.3)(7.35j39.8)I:ST1SZT2Z*(0.486.48j4.34(MVA)j11.48)(7.35j39.8)..*(8.5j5.3)(0.48j11.48)ST2SZT1Z*(0.482.00j0.96(MVA)j11.48)(7.35j39.8)II:如右圖,設U235KV則U11134.1253534.125(KV)3511UU2U10.875(KV)U*UN0.87535SC*(0.48j11.48)0.092j0.58(MVA)Z(7.35j39.8)ST1(6.48j4.34)(0.092j0.58)6.572j3.76(MVA)ST2(2.00j0.96)(0.092j0.58)1.91j1.54(MVA)3-16.電力網的電能耗費怎樣計算什么是最大負荷使用時間Tmax什么是最大負荷耗費時間max答:電能耗費=I2RtP2Q2Rt,為某個時段t內的電能耗費.U2Tmax:指一年中負荷花銷的電能除以一年中的最大負荷Pmax,即Tmax

WPmaxmax:指全年電能耗費除以最大負荷時的功率耗費,即max

WPmax3-17.什么叫自然功率分布什么叫經濟功率分布答:自然功率分布是指環網中功率分布未加任何調治手段,完好取決于自然情況進行的分布，也即按阻抗共軛值成反比分布的。經濟功率分布是使有功功率耗費最小的分布，是按線段的電阻分布。3-18.某一由發電廠A供電的220KV環式網絡,其運算網絡以下列圖,圖中Z116j120,Z233j89,Z348j120,Z4~60j152,S1~~170+j40MVA,S250j30MVA,S340j15MVA.計算:網絡的自然功率分布;網絡的經濟功率分布;實現經濟功率分布后,每年節約的電能解:在A點處把環網解開,如圖:....***(Z**)S*23I:SA1Z1*Z2*Z3*Z4*=168.86j62.39(MVA).***.*..**S1(Z2Z3Z4)S2(Z3Z4)S3Z4SAZ1*Z2*Z3*Z4*=91.2922.38(MVA).....SASA260.15j84.77(MVA),S1S2S260j85(MVA)'3...S12.S23.SA'.SAO

SAS1(168.86j62.39)(170j40)1.14j22.39(MVA)..(50j30)(1.14j22.39)51.14j7.61(MVA)S2S12..S23S3(51.14j7.61)(40j15)91.1422.61(MVA)...S1(R2R3R4)S2(R3R4)S3R4R1R2R3R4(170j40)(334860)(50j30)(4860)(40j15)6016334860202.26j62.23(MVA).'...S3(R1R2R3)S2(R1R2)SR1SAO1R1R2R3R4(40j15)(334816)(50j30)(1633)(170j40)161633486057.64j22.7(MVA)..SAOSAO'260j84.93(MVA)SAOS23OS3(17.74j7.77)(40j15)57.74j22.77(MVA)III:Pi1(SA)2R1(S12)2R2(S23)2R3(SA)2R4UNUNUNUN24.625(KW)Pi2(SAO)2R1(S12O)2R2(S23O)2R3(SAO)2R4UNUNUNUN14.81.0460.374.77620.992(KW)WPi1tPi2t(24.62520.992)87603.633876031825.08(J)3-19.有一長

20KM,35KV的雙回平行電力線路

,其負荷為

10MW,cos

0.8.電力線路導線型號為

LGJ-70,導線計算外徑為

11.4mm,三相導線的幾何平均距離為

3.5m,在用戶變電所內裝設有兩

臺

SFZ-7500/35

型變壓

器并聯運

行

,

SN

7500KVA

,UN

為35/11KV

,Pk

75KW

,UK

(%)

7.5,P0

9.6KW,I0(%)

0.8.兩臺變壓器全年投入運行,其年負荷曲線如圖.試計算:①電力網中的有功功率耗費;②電力網的網損率;③電力網一年中的電能耗費;④當用戶分別接功率因數為和運行時(用戶的有功功率不變)重復前三項計算.解:r131.50,45(/KM)s7011.4r5.7(mm)2Dm3.5m3500(mm)x10.1445lgDm0.0157r0.1445lg35000.4186(/KM)5.70.0157b17.581067.581062.72106(s/km)lgDmlg3500r5.7RL1r1l10.45204.5( )22XL1x1l10.4186204.186()22BL2b1l12.72106201.088104(S)2RTPKUN275103(35103)21.63()SN2(7.5106)2XTUK(%)UN275(35103)212.25()100SN1007.5106GTPO9.61037.84106(S)UN2(35103)2BTI0(%)SN0.87.51064.9105(S)100UN2100(35103)2變壓器在額定負荷下的耗費為:PKTP09.6KW0.0096(MW)I0(%)SNQKT0.06(MVar)100100PZTPK0.075(MW)UK(%)SNQKT0.5625(MVar)100100等效電路以下:cos0.810arccos10a.Sarccos0.812.536.8710j7.5(MvA)cos0.8STSKTSZT2(0.0096j0.06)2[(12.5)20.075j(12.5)20.5625]2j0.9(MvA)SASSTj8.4(MVA)S1PA2QA2(R1jX1)10.123228.42(4.5j4.186)0.636j0.59(MVA)UN2352S1SAS110.1232j8.40.636j0.590.7592j1.49(MVA)SSTS1j1.49(MVA)0.5b.Sarccos0.86.2536.875j3.75(MVA)0.8STSKTSZT2(0.0096j0.06)2[(6.25)20.075j(6.25)20.5625]7.527.520.0192j0.120.026j0.195.0452j4.06(MVA)SASST5j3.750.0452j0.315.0452j4.06(MVA)S1PA2QA2(R1jX1)5.045224.062(4.5j4.186)0.154j0.143(MVA)UN2352S1SAS10.154j0.1435.1992j4.203(MVA)SSTS10.0452j0.310.154j0.1430.1992j0.453(MVA)C.S100.25100.25arccos0.82.5j1.875(MvA)cosarccos0.8STSKTSZT2(0.0096j0.06)2[(3.125)20.075j(3.125)20.5625]7.527.520.0192j0.120.0065j0.04880.0257j0.1688(MVA)SASSTj2.044(MVA)S1PA2QA2(R1jX1)2.525722.0442(4.5j4.186)0.039j0.036(MVA)UN2352S1SAS12.5257j2.0440.039j0.0362.5647j2.08(MVA)SSTS10.0257j0.16880.039j0.0360.0647j0.205(MVA)有功功率耗費:P10.7592(MVA),P20.1992(MVA),P30.039(MVA)網損率:P11000.75921007.06P10.75921P21000.19923.83P21005.1992P31000.0391.52P31002.5647WP1t1P2t2P3t32262640(KWh)cos0.7a.S10arccos0.710j10.2(MVA)0.7STSKTSZT2(0.0096j0.06)2[(14.28)20.075j(14.28)20.5625]7.527.52SASST10j1.1210.1552j11.32(MVA)S1PA2QA2(R1jX1)10.1552211.322(4.5j4.186)0.85j0.79(MVA)UN2352S1SAS1j12.11(MVA)SSTS10.1552j1.120.85j0.791j1.91(MVA)b.S100.5arccos0.77.1445.575j5.1(MVA)0.7STSKTSZT2(0.0096j0.06)2[(7.14)20.075j(7.14)20.5625]7.527.520.0192j0.120.034j0.250.0532j5.47(MVA)SASST5j0.375.0532j5.47(MVA)S1PA2QA2(R1jX1)5.053225.472(4.5j4.186)0.2j0.189(MVA)UN2352S1SAS1j5.659(MVA)SSTS10.0532j0.370.2j0.1890.2532j0.559(MVA)c.S100.253.5745.572.5j2.55(MVA)0.7arccos0.7STSKTSZT2(0.0096j0.06)2[(3.57)20.075j(3.57)20.5625]7.527.520.0192j0.120.0085j0.0640.0277j0.184(MVA)SASSTj2.734(MVA)S1PA2QA2(R1jX1)2.52722.7342(4.5j4.186)0.05j0.047(MVA)UN2352S1SAS1j2.78(MVA)SSTS1j0.23(MVA)有功功率耗費:P11(MW),P20.2532(MW),P30.0777(MW)網損率:P110019.1P110011P21000.25324.82P21005.2532P31000.07773P31002.577WP1t1P2t2P3t33051752(KWh)III:cos0.9a.S10arccos0.91125.849.9j4.79(MVA)0.9STSKTSZT2(0.0096j0.06)2[(11)20.075j(11)20.5625]7.527.520.01920.092j0.725SASSTj5.515(MVA)S1PA2QA2(R1jX1)1025.5152(4.5j4.186)0.48j0.44(MVA)UN2352S1SAS110j5.5150.48j0.4410.48j5.955(MVA)SSTS10.0992j0.7250.48j0.440.5792j1.165(MVA)b.S100.5arccos0.95.525.844.9j2.4(MVA)0.9STSKTSZT2(0.0096j0.06)2[(5.52)20.075j(5.5)20.5625]7.57.52j0.27(MVA)SASSTj2.67(MVA)S1PA2QA2(R1jX1)4.9422.672(4.5j4.186)0.116j0.108(MVA)UN2352S1SAS12.49j1.3580.029j0.02752.52j1.385(MVA)SSTS10.029j0.02750.0532j0.1853(MVA)c.S100.252.7525.842.47j1.2(MVA)0.9arccos0.9STSKTSZT2(0.0096j0.06)2[(2.75)20.075j(2.75)20.5625]7.527.520.0192j0.120.005j0.03780.0242j0.1578(MVA)SASST2.49j1.20.0242j0.15782.49j1.358(MVA)S1PA2QA2(R1jX1)2.4921.3582(4.5j4.186)0.029j0.0275(MVA)UN2352S1SAS1j1.385(MVA)SSTS1j0.1853(MVA)有功功率耗費:P10.5792(MW),P20.1552(MW),P30.0532(MW)網損率:P11000.57925.53P110010.48P21000.15523.07P21005.056P31000.05322.1P31002.52WP1t1P2t2P3t31824032(KWh)Chapter五5-7.何為發電機組的有功功率---頻率靜態特點發電機的單位調治功率是什么答:假設電壓為額定電壓且不變時，發電機發出的有功功率隨頻率變化的特點稱為發電機組的有功功率---頻率靜態特點。由特點曲線可知隨著頻率的下降，發電機有功功率的輸出將上升。我們把發電機有功功率輸出增加的量和對應的頻率下降的量的比值稱為發電機的單位調治功率或調治系數，即：KGPG(MWHz)，KG*PG*。ff*5-8.什么叫調差系數它與發電機單位調治功率的標幺值有什么關系答:所謂的機組調差系數，是以百分數表示的機組空載運行時f0與額定條件下fN的差值，f0fN100%即：%fN調差系數與發電機單位調治功率的標幺值是互為倒數的關系，即：KG*1100%。.%5-11.互聯電力系統怎樣調頻才為合理為什么答:如圖，設A、B兩系統發電機組均能參加一次調頻，且均有調頻廠即均可以參加二次調頻。以

PLA

、

PLB

表示

A、B兩系統的負荷變化，以

PGA、

PGB表示

A,B兩系統的發電機組二次調頻增發用心，以KA、KB表示

A,B兩系統的單位調治功率，

聯系線功率為

Pab，且設其正方向是由

A系統流向

B系統。對A系統

Pab可看作一負荷

:

PLA

Pab

PGA

KA

f對B系統

Pab可看作一電源

:

PLB

(

Pab

PGB)

KB

f.(PLAPGA)(PLBPGB)fKBKAPabKA(PLBPGB)KB(PLAPGA)，KAKB在進行聯合系統調頻時需注意的是不但是要保持整個系統的一個有功功率的平衡，還要考慮聯系線上功率Pab不高出贊同的范圍，在調頻收效相同的前提下應盡量使每個系統的功率缺額由該系統的發電機補償，從而使聯系線上功率Pab盡可能小。、B兩系統由聯系線相連以下列圖,已知A系統KGA800MW/HZKLA50MWHZPLA100MWB系統KGB700MWHZKLB40MW/HZ,PLB50MW.求在以下情況下頻率的變化量f和聯系線功率的變化量Pab;I.當兩系統機組都參加一次調頻時;II.當A系統機組參加一次調頻,而B系統機組不參加一次調頻時;III.當兩系統機組都不參加一次調頻時.解:I.PGAPGB0,PLA100,PLB50；KGA800,KGB700,KLA50,KLB40.則PAPLAPGB100,PBPLBPGB50KAKLAKGA850,KBKLBKGB740.PAPB10050fKB8500.0943(HZ)所以KA740KAPBKBPA85050740100PabKB85019.8(MW)KA740II.PGAPGB0,PLA100,PLB50;KGA800,KGB0,KLA50,KLB40.則PAPLAPGA100,PBPLBPGB50.KAKLAKGA850,KBKLBKGB40.所以PAPB100500.168.fKB85040KAKAPBKBPA8505040100PabKB85043.26(MW)KA40III.PGAPGB0,PLA100,PLB50;KGAKGB0,KLA50,KLB40.則PAPLAPGA100,PBPLBPGB50KAKLAKGA50,KBKLBKGB40.所以PAPB10050fKB501.67(HZ)KA40KAPBKBPA505040100PabKAKB5016.67(MW).405-13.仍按題5-12中已知條件,試計算以下情況下的頻率變化量f和聯系線上流過的功率Pab。,B兩系統機組都參加一、二次調頻,A,B兩系統機組都增發50MW；,B兩系統機組都參加一次調頻,A系統并有機組參加二次調頻,增發60MW；,B兩系統都參加一次調頻,B系統并有機組參加二次調頻,增發60MW。解:I.PGAPGB50,PLA100,PLB50;KAKLAKGA850,KBKLBKGB740.PAPLAPGA50,PBPLBPGB0.所以PAPB5000.0314(HZ)fKB850740KAKAPBKBPA850074050PabKB85023.27(MW)KA740II.PGA60,PGB0,PLA100,PLB50;KA=KLAKGA850,KBKLBKGB740.PAPLAPGA40,PBPLBPGB50.所以PAPB40500.0566(HZ)fKB850740KAKAPBKBPA8505074040PabKB8507408.11(MW)KAIII.PGA0,PGB60,PLA100,PLB50;KAKLAKGB850,KBKLBKGB740.PAPLAPGA100,PBPLBPGB10.所以PAPB100100.0566(HZ)fKB850740KAKAPBKBPA850(10)740100PabKB85051.89(MW)KA7405-14.某電力系統負荷的頻率調治效應KL2,主調頻電廠額定容量為系統負荷的20%,當系統運行于負荷PL1,fN50HZ時,主調頻電廠用心為其額定值的50%,若是負荷增加,而主調頻電廠的頻率調整不動作,系統的頻率就下降,此時測得PL1.1(發電機組仍不滿載),現在頻率調整器動作,使頻率上升,問二次調頻作用增加的功率是多少解:如圖:KL2,調頻廠SN0.2PLSBPL,則PL1,SN0.2PL0.2,S用心50%SN0.1不滿載表示發電機將自動參加一次調頻，調治系數為KG，PL'PG'1.1發電機在一次調頻的作用下用心增加了PG'0.1KGff0.350PL''KGfKLf0.10.30.1PGPL250由KGf0.1得KG0.10.15050f0.33二次調頻投入后頻率上升,f'0.35050由PLf'PGOKLKG0.112PGO2500.118.673解得：PGO0.07470.1調頻廠可調容量大于二次調頻所需容量。Chapter六6-13.有一降壓變壓器歸算至高壓側的阻抗為+j40,變壓器的額定電壓為1102*%/,在最大負荷時,變壓器高壓側經過功率為28j14MVA,高壓母線電壓為113KV,低壓母線要求電壓為6KV;在最小負荷時,變壓器高壓側經過功率為10+j6MVA,高壓母線電壓為115KV,低壓母線要求電壓為,試選擇該變壓器的分接頭.解:在最大負荷時,28*2.4414*405.56(kv)Umax113低壓側歸算至高壓側的電壓為,U低maxU高maxUmax1135.56107.44(kv)UtmaxU低maxU低額107.44*6.3112.8(kv)‘U低max6在最小負荷時,Umin10*2.446*402.3(kv)115低壓側歸算至高壓側的電壓為,U低minU高minUmin1152.3112.7(kv)UtminU低minU低額112.7*6.3107.58(kv)‘U低min6.6則公共抽頭UtI1Utmin)110.19(kv)(Utmax2選擇110kv分接頭,U低'U低額107.44*6.36.156maxU低maxUtI'110U低'U低minU低額112.7*6.36.45(KV)6.6min‘UtI110U%max6.156*100%2.5%6U%min6.456.6*100%2.27%6.6滿足調壓要求,所以分接頭電壓110KV6-14.有一降壓變壓器,SN20MVA,UN11022.5%/11KV,PK163KVUK(%)10.5,變壓器低壓母線最大負荷為18MVA,cos0.8,最小負荷為7MVA,cos0.7,已知變壓器高壓母線在任何方式下均保持電壓為,若是變壓器低壓側要求順調壓,試選擇該變壓器的分接頭.解RTPKUN216311021034.93()SN2202XTUK(%)UN210.5110263.525()100SN10020最大、最小負荷為:Smax18arccos0.814.4j10.8(MVA)Sminj5(MVA)14.4210.82STmax(4.93j63.525)0.13j1.7(MVA)11024.9252STmin2(4.93j63.525)0.02j0.26(MVA)110S高maxSmaxSTmax14.53j12.5(MVA)S高minSminSTmin4.92j5.26(MVA)最大、最小負荷時變壓器電壓耗費為:Umax14.534.9312.563.525107.58.05(kv)Umax4.924.935.2663.5253.33(kv)107.5U低maxU高maxUmax107.58.0599.45(kv)U低minU高minUmin107.53.33104.17(kv)UtmaxU低maxU低N99.4511106.73(kv)U‘低max10102.5%U低N11UtminU低minU‘低min104.1710107.5%107.1(kv)則公共抽頭UtI1(UtmaxUtmin)106.92(kv)2選擇分接頭'107.25KV%,UtI'U低N99.4511U低maxU低max‘10.2(kv)10.25KVUtI107.25不滿足調壓要求,則重新選擇.選擇11022.5%分接頭,UtI'104.5kv'U低minU低N104.71111(kv)10.75KVU低min‘104.5UtI不滿足調壓要求,則選擇有載調壓變壓器.最大負荷時選擇11022.5%分接頭'104.5,,UtImaxkv'U低maxU低N99.451110.5(kv)10.25KVU低max‘104.5UtImax最小負荷時,選擇11022.5%分接頭,Ut'Imin107.25kv'U低minU低N104.1711U低min‘10.68(kv)10.75KVUtImin107.25滿足調壓要求.6-15.某變電全部一臺降壓變壓器,變壓器額定電壓為11022.5%/11kv變壓器已運行于抽頭電壓,在最大、最小負荷時,變壓器低壓側電壓的偏移為-7%和-2%,若是變壓器低壓側要求順調壓,那么調壓器的抽頭電壓,能否滿足調壓的要求,應該怎樣辦理解:Ui'max1.025UN10.25Ui'min1.075UN10.75Ui'max(17%)109.310.25Ui'min(12%)109.810.75則UtI115.5不滿足調壓要求.kv,最大負荷時,UimaxUi'maxUtIU低N1197.65(kv)UtImaxUiminU低N97.6511104.8(kv)U‘imax1.02510最小負荷時,UiminUi'minUtI102.9(kv)U低N11U低N11UtIminUiminU‘imin102.91.07510105.3(kv)UtI1(UtImaxUtImin)105.05(kv)2選擇一臺湊近的標準抽頭UtI'104.5kvU'U低N97.6511imaxUimax‘10.28(kv)10.25KV(合格)UtI104.5U'UiminU低N102.91110.7510.83( )(不合格)imin‘kvKVUtI104.5則重新選擇UtI'107.25kvU'imax97.651110.01(kv)10.25KV107.25U'imin102.91110.55(kv)10.75KV107.25所以選擇有載調壓變壓器U'104.5(kv)U'tImin107.25(kv)tImax

(不合格)(合格)6-16.水電廠經過SFL-40000/110型升壓變壓器與系統連接,變壓器歸算至高壓側阻抗為2.1j38.5,額定電壓為12122.5%/10.5kv,系統在最大、最小負荷時高壓母線電壓分別為,和,低壓側要求的電壓在系統最大負荷時不低于10KV,在系統最小負荷時不高于11KV,當在系統最大、最小負荷時水電廠輸出功率均為28+j21MVA時,試選該變壓器的分接頭.解:最大負荷時,Umax282.12138.57.74( )112.09kv低壓側歸算至高壓側的電壓為,U低maxU高maxUmax112.097.74119.83(kv)U低額119.8310.5125.8(kv)UtmaxU低max‘U低max10最小負荷時,,Umin282.12138.57.48(kv)115.92低壓側歸算至高壓側的電壓為,U低maxU高minUmin123.4(kv)UtminU低minU低額123.410.5117.8(kv)‘U低min111)12.18()則公共抽頭(UTIUTImaxUTIminKV2選擇121KV為標準抽頭,UTI121KVU低maxU低額119.8310.510.410U低maxUTI121U低minU低額123.410.510.711U低min121UTIU%max10.410100%4%10U%min10.711100%2.7%11滿足調壓要求,所以分接頭電壓為121KV6-17.有一個降壓變電所由兩回110KV,長70KM的電力線路供電,導線為LGJ-120型,導線計算外徑為15.2mm,三相導線幾何平均距離為5m.變電所裝有2臺變壓器并列運行,其型號為SFL31500/110型,SN31.5MVA,UN110/11KV,UK(%)10.5.最大負荷時變電所低壓側歸算至高壓側的電壓為,最小負荷時為.變電所二次母線贊同電壓偏移在最大,最小負荷時為二次網絡額定電壓的2.5%7.5%.試依照調壓的要求接并聯電容器和同期調相機兩種措施,確定變電所10KV母線上所需補償設備的最小容量.解:r131.50.2625(/)s120kmDm0.0157x10.1445lgr0.1445lg51030.01570.423(/km)15.2/2RL1r1L10.2625709.19()22XL1x1L10.4237014.8()22XT1UK%UN210.511022100SN210020.167( )31.5U2max100.5(KV),U2min112(KV)U2max10(12.5%)10.25(KV)U2min10(17.5%)10.75(KV)①并聯電容器U2N11UTIU2minU2min11210.75114.6(KV)選擇UT2U115.5KV,則KU

TI2N

115.510.511QCK2U2max(U2CmaxU2max)XK10.5210.25(10.25100.5)21.9(Mvar)14.820.16710.5②并聯調相機U2maxU2max2U2minU2minK2max2U2min2U210.252210.752

10.227UTIKU2N10.22711112.5(KV)選擇UTI112.75KVKUtI'112.7510.25U2N11QCK2U2C'max(U2'CmaxU2max)10.25210.25(10.25100.5XK20.167)13.7(MVar)14.810.25選擇QC20MVar的調相器.效驗:I:Umax21.920.1674(kv)110U2max100.54104.5(kv)'U2maxU2N104.511U2maxUtI'9.95(KV)115.59.9510U%max100%0.5%10U2min'U2minUU

N'tI

1111210.67(KV)115.510.6710U%min100