電力公司管理電力系統(tǒng)分析第三版于永源楊綺雯著中國電力出版

一、、:、電力系統(tǒng)的額定電壓是如何確定的？系統(tǒng)各元件的額定電壓是多少？什么叫電力線路答：各部分電壓等級之所以不同，是因三相功率S和線電壓U、線電流I之間的關(guān)系為3；；的或。答：500330則是最常用的更低一級、處？1）2）3）成。maxP8760W∫0PP、某一負(fù)荷的年持續(xù)負(fù)荷曲線如圖所示，試求最大負(fù)荷利用小時。W)10060300200050008760A年=?Mw.hA年P(guān)）、電力系統(tǒng)中性點的接地方式有哪些？它們各有什么特點？我國電力系統(tǒng)中性點接地情?3倍。、答假設(shè)CC00，非3倍。333倍。當(dāng)I=∑ICIL＞∑ICIL＜∑IC∑IC二2-1、架空電力線路由哪些部分組成？它們的作用如何？架空電力線路的桿塔由哪些形式？ρ2分裂數(shù)為根4根2TπKRTPU2KNS2NXUU2KN。T100SNI0B%S耗GN。PI00T2T2UNUN。異雙繞組變壓器的勵磁回路的等值導(dǎo)納并聯(lián)在變壓器繞組等值阻抗的首端；雙繞組變BT前是“BL前是“+”號。答：指兩個容量繞組中流過額定電流，另一個或容量繞組空載時的有功損耗。、對于升壓型和降壓型的變壓器，如果所給出的其他原始數(shù)據(jù)均相同，它們的參數(shù)相同數(shù)。、發(fā)電機的等值電路有幾種形式？它們等效嗎？發(fā)電機電抗的百分?jǐn)?shù)的含義是什么？XGXG%3INXGUN2-11、算。12122抽1K）110.5K2(1?2)=2(1?2)=220121K2(1?3)2205.7138.5K?311K?5%)410K53.15U

Ul1220231l2110Ul3355%)Ul41010.5KV

Ul56Ul633.15KV2-17.某一回架空電力線路,長度為導(dǎo)線型號導(dǎo)線計算外徑為三相導(dǎo)線水平排列,兩相鄰導(dǎo)線三相間的距離為4m,試計算該電力線路的參數(shù),并作等效電路。解:1Σs0.2625?km120r)2Dm34000400024000)D0.01570.1445lg0.0157?/km)

X0.1445lgm1r7.67.58?610?67.58611010Dlgmlgr7.6g10.R11l0.262560)X1l0.42360)6?41l2.6910601.61410(s)

則BB10.80710

(s)?4210.有一回架空電力線路,采用型號為的雙分裂導(dǎo)線,每一根導(dǎo)線的計算三相導(dǎo)線的不等邊三角形排列,線間距離129m,23,31,,解：r1Σns2185?/km)r219400)2D3900850061007756.46(mm)mX0.1445lg1D0.01570.0157?/km)mn61.64217.58Dlgmr?6104610lg61.64?63.6110?6(s/km).b11.805102g1?6(s/km)2-19.三相雙繞組升壓變壓器的型號為額定容量為40500KVA,額定電壓為kkw,Uk(%)0,I0(%)KUN2234.410310

解:`RP3)2?2(?)T2(40500103)2SNNUN2113)2XU?)T100S10040500103N3P1008.4910?4(S)

T103)2U2N3ISB0N2.315405001010?3(S)T100U2100103)2N三相三繞組降壓變壓器的型號為額定容量為K?2)601KW,PK?3)182.5KW,PK(2?3)5,UK?2)(%)UK?3)(%)UK(2?3)(%)0135KW,I0(%)解:I:PK?2)601(KW)P'4PK?K3)(1?3)'PK(2?3)4K(2?730(KW)530()3)P1(PP''P))K121P(PK(1?2)K?3)'P–P'K(2?3)))K221P(P'K(1?2)P'K(2?3)K(1?3)?P))K32K(1?3)K(2?3)K(1?2)R1PK1U2332400.510(22010)1.346(?)NT12SN2RPK2UN1T22SN2KURN1104103)2332200.510(22010)0.674(?)104103)2332329.510(22010)1.107(?)T3210S4103)2UK(1?2)14.85,UK(1?3)28.25,UK(2?3)7.96(%)1UU(%)?(%)U(%))17.57K1K?K?K(2?2U2)3)3)1(%)UU(%)U(%)?(%))?2.72K2K(1?K(2?3K?2U2))3)(%)1UU(%)U(%))10.68K32(%)?K?3)K(2?3)K?2)UUU217.57(2203)2)XK1N1T1100S10012104103N232?(22010)T2UK2UN1X?)100S100121043NU(%)U210.68(2203)2)XK3N1T3100S10012104103NP：G13510310

?6(S)0T2(220103)2UN1TI0(%)SNIV：B100U2N1430.663121010100(220103)21.64410?5(S)．三相自耦三繞組降壓變壓器的型號，額定容量為，額定電壓，PK?2)417KW,PK(1?3)318.5KW,PK(2?3)314KW;UK(1?2)8.98,UK(1?3)(%)UK(2?3)(%)057.7,I0(%),.解:I:PK?2))P'K?3)4PK(1?3))P'K(2?3)PK(2?3))P1(PP''P))K121PK221P(P'K(1?KK(2?3)(1?3)))K?3)?P))'P2)(PP'–P'K(1?K(2?2)3)K32K?3)K(2?3)K(1?2)2332R1RT2RT3UN1217.510(22010)?)S2104103)2N2332PK2UN1199.510(22010))S104103)22N2332PKUN110(22010)?)S104103)22SII:UK(1?2)(%)U'(%)U(%)K(1?3)KU'(%)U(%)K(2?3)K(%)1UU(%)U(%)?'K1K?K?22)3)1(%)UU(%)(%)?'K2K(1?K(2?22)3)1U(%)U'(%)U'(%)?UK?K(2?3K323))'K(2?3)'K(1?3)K?2)(%))(%))?1.31(%)23.01UU23)2N1)XT1K1S104103NK2N13)2X)TU

U22100S43NUU23)21)XT3K3S43NPIII:G57.71031.1910

?6(S)0T2(220103)2UN10(%)SN0.7121210410

IV:BI31.765?5(S)T100U2100(220103)2N1N,cosΣN0.85,UN13.8KV,Xd1.867,Xd0.257,X''0.18,試計算該發(fā)電機的直d'''XdXdXd.U2解:ZNNSNU2ΣNN)PNXdZN1.8671.295)X'0.257Z0.2571.2950.333(?)dNX''0.18Z0.181.2950.233(?)dN2-23.電抗器型號為UN6KV,IN500A,電抗器電抗百分?jǐn)?shù)XL(%).3解:XL4610)1003500三?答:?SLPLQLP2Q2ii(RLU2ijXL)1SB?jU2L212C12B1S?jUL2C222?答:UPΣUPiRiiX?iRiXUiiUi3U和U前面33,將U和U前的3變成333與UUUUU3母UU。?...(U1?U2)或dU.1?U2U1?U2100.U%=UN1?UN或U2?U1),以百分?jǐn)?shù)表示1。N。N%=U1?UNUN2。N。N%=U2?UNUN100.U20?U2).以百分?jǐn)?shù)表示,電壓調(diào)整ΔU?UU220U20:21,,P2P1算??答::答：S?ST?SOKjUKSNP100OjIO%SNP100KPOj(UK%SN100IO%SN100)n.*.∑SmZm答：Sa1*ZΣ110kv30+j15MVA,末端電壓為106kv,求始Σ?/解:1

km)120sDmmm),r)m2D50000.0157?/km)X0.1445lgm0.01570.1445lg1r17.58?610D10500010?6(s/km),10?6lgmlgr7.6R1rl10.2625150)L212X1xl10.423150)L212Bbl22.6910?61508.0710?4(s)L1GL0.2BL)(30(?28.0710)30).'.(?jU?S2S2222.?SLPL?QL'2Q'PL)3010.4662(RLjX2222U210622(19.687j31.725)1.769).'.'.S1S2?SLj31.769).dU2U2U2P'RQ'P'X–Q'L2LL2L2j2UU223019.68710.46631.725j30?10.46619.6871061068.7j)...。U1U2dU21068.7jj5kv)..'B?4L)(31.769(?28.0710)

S1S1(?jU1222)...設(shè)U2106（),U28.7(),U2),則U5()單回架空電力線路,長度為導(dǎo)線型號24.2mm,三相導(dǎo)線幾何平均距離為7.5m,已知其始端輸出的功率為120,始端的電壓為240KV.解:1Σs?/Km)300mm)r24.29)2x10.1445lgDmr0.01570.1445lg75000.0157?/km)12.117.58?6?6?6102.71510(S/)D10lg7500mlgrR1l0.10520021(?)LrXLl0.419200)Bbl2.71510?62005.4310?4(S)l1l0?′??(?jU2l5.43?4)j?(?j2402)11122120j)P′QS′(RjX)120222265.6384L2LL2U1240(21jj27.22(MVA)??S′′?Lj?(6.82j27.22)113.18)?1?U1jΣU1RL′XLU1jXL?′RLU11202165.6384j120?638421240240)?2U1??1240?33.419?j36.157?j36.157Σ?()??′?(?jU5.432BL)j?(?j210?4)22222113.18)∴S2j)；?.2Σ?()。單回架空電力線路,長度為導(dǎo)線型號為導(dǎo)線計算外徑為三116kv,15+j10MVA,.解:1Σs?/km)95Dmm),r)2x1445lg50000.0157?/km)17.58?6?6?61102.64710(s/)D10lg5000mlgrg10RLl0.33280)XLl0.42980)Bbl2.64710?6802.1210?4(s)L1GL0.'.B2.1210?4L)(?2S2S2(-jU2N22)15).?SLQ'2(RLjXL)158.7172(26.56j10.66j)P'2222U21102N.'.'.S1S2?SLjj15.66)..'2BLS1S1(?jU21)(15.6622.12104?j9.57)(?)15.66j8.14(MVA)21?U1U1P'RQ'X1L1U1Lj1P'X–Q'15.6626.5634.32RL1LU1116+j15.6634.32?9.576.417j)116..U2U1?dU1116?6.417?j2.442583?j)單回架空電力線路,長度為電力線路每公里的參數(shù)為10.108?/km,1?/,12.6610?s/km,線路空載運行,當(dāng)線路末端電壓為.解:RLl0.108220)XLl0.42220)Bbl2.6610?62205.85210?4(s)L1.,所以S202BL)?j20525.85210?).'.(?jU?S2S2222'2'22.L)?SP(RLjX2(23.76j0.085j)2Q2052LU22.'..S1S'?SL?0.085j0.330.085?)2.P'RQ'XP'X–Q'?12.323.76RjdU2U2U2

L2L

Lj2L2U205U2205225.544)...U1U2dU2205?5.544j1.4256199.456)..B5.852S'(?jU2L)(0.085?[?j45621.42562)S11122?4]0.085?j)1kV)有一臺三繞組變壓器,其歸算至高壓側(cè)的等效電路如圖所~示,ZT1j65?,ZT2?,ZT3,S25,~S38,..'..':I:由U3,S3求U1.'110U3U36.66110)6.6.P2Q222?ST333(RjX)86UT31002'2T33(2.47j0.0247j0.378(A)...'S3?STS33j6)(0.0247j8.0247j6.378().dUjUU33Σ3?PRQXjPX?QR3T33T3'3T33T3'U3U38637.8j8?62.4656j)100100..'U'dU3100j2.4656j2.8757102.4656j)U13...由U',S2求U21.P2Q222?ST222(RjX)542T2T21102UN(2.47?0.0084?j)...'S2?ST2j4)(0.0084?j5.0084j)S2.P'RQ'XP'X–Q'RdU2U2U22T22T22T2UU''115.00842.47?3.9951.5102.465622.8758222.87582j5.0084(?1.5)?3.9952.470.062?102.465622.8758222.87582j0.17(KV)...UU'?dU2102.411j3.04(KV)...'''S'S13.033)123...由U',S''求U11122.P''Q''?11ST12U'12(2.47j60.0652A)2(R)T1T1102.465622.87582...S'S''?ST1(0.065213.0982)11.P''RQ''XP''X–Q''RdU1?U111T111T11T1T1'U'U1j110.37365j13.03365?3732.476.89j8(KV)..UU'dU1109.3556)11U1109.9(KV)U2)102.4451.05)

UU2實21101103-13.某電力線路導(dǎo)線為100km,導(dǎo)線計算外徑為線路末端負(fù)荷為90+j20MVA,,變,且不計電暈損失的增加,導(dǎo)線水平排列,升壓前后線間距離由增加到試問升壓后,?1Σ解:r1s?/km)185RL1l0.1710017()Dm140004000240005.041033(mm)Dm2355005500255006.93103)r19)210.01570.1445lg5.04100.0157?/km)

x0.1445lg31r9.5Dm2x1445lg2r0.01570.1445lg6.931030.0157?/km)X11l0.41100)XL2x2l0.43100)b1lgD110?6r2.7810?6(s/km)b2lg7.58Dm2?6r2.65?6(s/)1bl2.78?61002.7810?4(s)1BL2bl10?61002.6510?4(s)21.?S1?jU221NB.L12.7810?4?j110222?)1.?SC2222?jUNB.L210?4?j220222?)..S'2121.?S12(90j?1.6890)..S'2221.?SC22j?j9013.6()2Q'222.P'?S12121LjX)90(R18.32U21N2110j11.85j)2Q'222.P'13.6?S12222(R2j4j)jX)90U2220LL22N...?S?S1??S2j?(2.9jj21.22().'''?'21X121LLQ21XL1jP

dU1?UjΣU11UNU1N901718.3241j9041?18.321720.74j30.7(kv)110110.'''?'22XL2L22L22XL2jP

dU2UjU22U2NU2N901713.643j9043?17)220220.dU1%dU230.72120.74110110.dU2%dU229.6216.54110220...dU%dU1%?dU2%?對圖所示環(huán)式等值網(wǎng)絡(luò)進行潮流計算,圖中各線路的阻抗為Zl110,Zl220j,Zl325j,Zl410,~~~S290j40MVA,S350,S440,且U1235KV.解:在1:Zl110≈10Zl220j34.62Zl325j43.32.5Zl410.則S1.S(Z*2l2..3Z4)S3(Z3Z4)S4Z4Z****l1l2l3l4=...S2(22.5S3S4121=jj30)3.5(406.5=).******...S'S4(Zl3Zl2Zl1)S3(Zl2Zl1)S2Zl11Z*Z**ZZ*l1l2l3l4=....S4(2.52S3(2S2122.51=(40j15)j30)3j40)6.5=70.77)...S23S1?S2109.2j52.3?(90j40))...S43S'?S470.771j32.7?(4030.77j17.7(MvA)..S1S'109.2j52.370.77j32.7)1...S2S3S490j4050j3040180).所以S1.S'1..S2S.3S4又33.22S2P23(RjX)12.32(20j0.2).Q232UN2202l2l22...S"SS'j0.37)2232...'S2S''90j40j)S22.P'Q'2222?S1(R22jXl1)52.67U2l1N22023)SS'S109.4j52.673j5.27)...121.dUjU?1PRQX111U1PX?QRj11L111U1112.41057.9417.32j17.32?57.9410j5.82(KV)235235...U2U1?dU1235?(9.05j5.82)?)Σ?.）.P“RQ”XP“X–Q”RdU2U2UL222L22L22L2UU22

2j201267j34.6?12.67203.656)226226...U3U2?dU2225.95?j5.82?(3.656?j7.67(kv)Σ?..dU3P43RL3QXL3j43XL3?Q43U3U3RL330.7725j30.77?25j4(kV)222.426222.426...U4U3dU3222.426j4229.326j4(KV)某1為SN,K25KW,UK(%)T2為SN,K24KVA,UK(%)6.5~S=8.5+j5.3MVA,:的變比為KT1KT235/11KV時,每臺變壓器的總功率為多少?當(dāng)KT134.125/11,KT235/11?解2332RUN251010))

?T12S6)2NT1UK(%)UNX100SN27.5103)21008106)2332RUN2410(3510))

?T22S(2106)2NT2UK(%)UNX100SN2103)21002106)ZT10.48ZT27.35)在1:..*SZI:ST1T2Z*?j(0.48?(7.35?j6.48)..*SZST2T1Z*j5.3)(0.48?(0.48?(7.35?j32.00)II:設(shè)U2則U351134.125)13511UU2?U1)?CUUNZ*0.87535(0.48?(7.35?0.092?j)?T1(6.48j4.34)(0.092?j6.572j)?T2(2.00j0.96)?(0.092?j1.54()??Σ?=I2PQ22t.U2T,即TWPΣ即Σ?WP?某一由發(fā)電廠A供電的環(huán)式網(wǎng)絡(luò),其運算網(wǎng)絡(luò)如圖所示,圖中Z116j120?,Z233j89?,Z348,Z116j120?,Z233j89?,Z348,Z4~60,S1~~S250,S340.計算:網(wǎng)絡(luò)的自然功率分布;網(wǎng)絡(luò)的?解:在A:.S(Z*Z*Z*)(Z*Z*).Z*..S1I:SA23423434Z*Z*Z*Z*1234=).S(Z*Z*Z*)(Z*Z*)SZ*...S123434SA234Z*Z*Z*Z*1234=91.29).....SAS'260.15j84.77(S1S2S3260j85(MVA)A...S12SA?S1j?j?j)...S23S2?S12j?(j51.14j)...'S23S3j(4091.14)SA.S.S1(R2R3R4.)S23R4.)S3R4R1234j40)(3348j30)(48(4060

16334860).'S.S3R1R2R3.)S21R2.)S1R11234(4048j3j16

1633486057.64j)..SS'260)?′??j(4057.74j)O3SA22S232SA2III:1()2()3()41()UUUUNNNN10.70510.93)SΣO2ΣS23Σ)2R(S′ΣO)2?P)2i2R4R2(U1NUNUN1.0464.776)Wit?i2t(24.625?87603.6338760J)的雙回平行電力線路,其負(fù)荷為10MW,cosΣm裝設(shè)有兩臺型變壓器并聯(lián)運行,SN,UN為k,UK(%),0,I0(%)0.8.兩臺變壓器全年投入

35/11KV,P運行,其年負(fù)荷曲線如圖.試計算①電力網(wǎng)中的有功功率損耗②電力網(wǎng)的網(wǎng)損率③電力網(wǎng)和重.解:1Σs?/)70r5.7(mm)2m5m)10.1445lgDmr0.01570.1445lg35000.0157?/)17.58?67.5810D35002.7210?6(s/)?6lgmlgrR1rl10.4520)L212X1xl10.418620)L212B2bl12.7210?6201.088?4(S)L122332RUN751010))T26)2SNU(%)U275(353)2)XKNT100S100106NPG9.61037.8410

?6(S)OT2103)2UNBI0SNT100U2N7.5106100103)210?5(S):KTP0)IQ0(%)SN1007.51000.06(MVar)?PK0.075()UQK(%)SN1007.5100MVar):I:cosΣa.S10cosΣΣarccos?10Σarccos0.8?0.810jA)T??KTZT2(0.0096j)27.520.075j()220.0192j0.120.104j0.780.1232)??AT10j7.50.1232j0.910.1232)??P2Q22AA(R10.1232jX)8.42j0.636)AA(R10.12321U211352???1SA?S110.12320.636j0.590.7592)??T10.1232j0.636j0.590.7592)10.1232j0.636j0.590.7592)S100.5Σ?5j)?SZT2(0.0096jTKTTKT6.25)27.520.075j(625)220.0192j0.120.026j0.195.0452j)??A??T5j3.750.0452j0.315.0452)??P2Q2AA(RjX25.0452)4.062j0.154j)1211U2N35???1SAS15.0452j4.060.154j0.1435.1992j)??T10.0452j0.310.154j0.1430.1992j0.453()10.0452j0.310.154j0.1430.1992j0.453()C.?100.25ΣarccosΣ10Σ3.12536.87

?2.5)Σ0.8T??KTZT2(0.0096j3.125)27.520.075j(3.125)7.5220.0192j0.120.0065j0.04880.0257j)??ATj1.8750.0257j0.16882.5257j)??2222P(RjXAQA2.044)

(4.52.5

j2570.039j0.0

)1U211N352???S1SA?S12.5257j2.0440.039j0.0362.5647j)??T10.0257j0.16880.039j0.0360.0647j)10.0257j0.16880.039j0.0360.0647j):P1),?2),3):P1100P10.759210010.7592P21000.19921003.8325.1992310030.0392.56471001.52W1t12t23t30.759220000.199230000.0393760146.64)II.cosΣa.?10Σarccos0.7?0.710)T??KTKTZT2(0.0096j0.06)14.28)220.075j(14.28)22??AT100.155210.1552)2222AA10.1552(4.5j0.85j)

?P?jX)(R1211UN2???1SAS110.1552j0.7911)??T10.15521.120.85j1)10.15521.120.85j1)?100.5Σarccos0.745.57?5)T??KTZT2(0.0096jZT2(0.0096j714)220.075j(714)220.0192j0.120.034j0.250.0532)??AT5j5.10.0532j0.375.0532)??P2Q12(R1jX1)5.053225.4722(4.5j4.186)0.2j0.189(MVA)2AAUN35???1SAS15.0532j0.2j0.1895.2532j)??T10.0532j0.372j0.1890.2532j)c.?100.25Σarccos0.745.57

?j)T??KTZT2(0.0096j3.57)220.075j(3.57)220.0192j0.120.0085j0.0640.0277j)?A???2.5j2.550.0277j0.1842.527j2.734(MVA)T??P2Q12(R1jX1)2.52722.73422(4.5j4.186)0.05j0.047(MVA)2AAUN35???1SA?S12.527j2.734j0.0472.577j)??T10.0277j0.1840.05j0.0470.0777j)10.0277j0.1840.05j0.0470.0777j):1),?2),30.0777(MW):?1100111009.111P0.25321004.82

210025.2532P0.07771003

310032.577W1t12t23t3120000.253230000.077737602000292.152)III:cosΣ0.9a.?10Σarccos0.91125.84

?9.9j)0.9T??KTKTZT2(0.0096j0.06)11)220.075j(11)22??A??Tj4.790.0992j0.72510j5.515()??P2Q12(R1jX1)1025.51522(4.5j4.186)0.48j0.44(MVA)2AAUN35???1SAS110j5.5150.48j10.48j)??T10.0992j0.7250.48j0.5792)?100.5Σ25.84?j2.4()T??KTZT2(0.0096jZT2(0.0096j5)220.075j(5)220.0192j0.120.02j0.150.0392)??ATj2.40.0392j0.274.94j2.67(MVA)??P2Q12(R1jX1)4.9422.672(4.5j4.186)0.116j0.108(MVA)2AA2UN35???S1SA?S12.49j1.3580.029j0.0275)??T10.024j0.15780.029j0.02750.0532j)c.?100.25Σarccos0.925.84?2.47)T??KTZT2(0.0096j275)220.075j(2.75)220.0192j0.120.005j0.03780.0242j)??AT0.0242j0.1578)2222AQA1.358j0.029j)

?P?jX)(R1211UN2???1SA?S12.49j1.3580.029j0.0275)??T10.0242j0.15780.029j0.02750.0532j)10.0242j0.15780.029j0.02750.0532j):10.5792(),20.1552(),3):P11000.5792100110.4821000.15521003.07

25.05631000.05321002.1

32.52W1t12t2?3t30.579220000.155230000.05323760032)五?:I:,:II:低頻率運行將增加汽輪機葉片所受的應(yīng)力,引起葉片的共振,縮短葉片的壽命,甚至使葉低頻率運轉(zhuǎn)時,發(fā)電機的通風(fēng)量將減少,而為了維持正常的電壓,又要求增加勵磁電流,以致使發(fā)電機定子和轉(zhuǎn)子的溫升都得增加,為了不超越溫升限額,不得不降低發(fā)電機所發(fā)功.低頻率運行時,由于磁通密度的增大,變壓器的鐵芯損耗和勵磁電流都得增大,也為了不..總之,由于所有設(shè)備都是按系統(tǒng)額定頻率設(shè)計的,系統(tǒng)頻率質(zhì)量的下降將影響各行各業(yè),而.?答:電力系統(tǒng)有功功率負(fù)荷的變化可分解為三種,第一種變動幅度很小,周期又很短,這種負(fù)荷變動有很大的偶然性.第二種變動幅度較大,周期也較長,屬于這一種的主要有電爐、壓延機械、電氣機車等帶有沖擊性的負(fù)荷變動.第三種變動幅度最大,周期也最大,這一種是由.,次頻率指由發(fā)電機組的調(diào)速器進行的、對第一種負(fù)荷變動引起的頻率偏移作調(diào)整；二次調(diào)整或頻率的二次調(diào)整指由發(fā)電機的調(diào)頻器進行的、對第二種負(fù)荷變動引起的頻率偏移的調(diào)整.三次調(diào)整是指按最優(yōu)化準(zhǔn)則分配等三種有規(guī)律變動的負(fù)荷,即責(zé)成各發(fā)電廠按事.?答:I:∑P∑PL∑S∑P∑P∑L∑P∑PSII:用；：IIIIII間。。II：III：充分合理地利用水力資源，盡量避免棄水。由于防洪、灌溉、航運、供水等原因必須：原子能電廠的可調(diào)容量雖大，但因原子能電廠的一次投資大，運行費用小，建成后應(yīng)：執(zhí)行國家的燃料政策，減少燒油電廠的發(fā)電量，增加燒劣質(zhì)煤和當(dāng)?shù)禺a(chǎn)煤電廠的發(fā)則。KLPL()，KL()，KfL*P*。f*?答:出增加的量和對應(yīng)的頻率下降的量的比值稱為發(fā)電機的單位調(diào)節(jié)功率或調(diào)節(jié)系數(shù)，即：K?PG(Hz)G(Hz)，GfKG*?PG*。G*。?f*?f0fN即：Σ%f0?fNfNK1。.*Σ%??功O點LGf1L,負(fù)荷由PL變?yōu)長?LGL,即負(fù)荷大于出力,轉(zhuǎn)速減小,頻率也減小.在系統(tǒng)頻率下,,O'.KSKGΣKLKSKGΣKL。KLKSKΣKG也即調(diào)差系數(shù)不能過小，否則就要出現(xiàn)負(fù)荷變化量在各發(fā)電機組之間的分配無法固定,即將KGΣKS5-10.電力系統(tǒng)頻率的二次調(diào)整(二次調(diào)頻)的基本原理是什么?如何才能做到頻率的無差調(diào)節(jié)?答::,f,電機組發(fā)出的功率,使頻率特性向上移動.設(shè)發(fā)電機組增發(fā)?P,則運行點又將從點O'轉(zhuǎn)移到點O".LOf0.?答:BLA、LB表示AB、?表示KA、KB表示abAB對A系統(tǒng)?ab:LAab?KAf對B系統(tǒng)Pab:LB?(ab)KBf.?f(?P

KAKBLA?)(?LB?)PKabA(LB??)?KB(LA?)KAKB，?ab不超過允許的范圍，在調(diào)頻效果相同的前提下應(yīng)盡量使每個系統(tǒng)的功率ab、B兩系統(tǒng)由聯(lián)絡(luò)線相連如圖所示,已知A系統(tǒng)KGB700MWGA/KLAB系統(tǒng)KKLB/,50MW.求在下列情況下頻率的變化量f和聯(lián)絡(luò)線功率的變化量?ab;II.當(dāng)A,而B不.解LA?LB50；KKKLA5KLB則ALA?BLB?50KAKLAKGAKBKLBKGBfA?PB所以KAKB10050850740)PKAB?KBA??19.8(MW)

abKAKBII.?LALB5KKKLA5KLB則ALA?BLB??KAKLAKKBKLBK所以f?PAPBKAKB1005085040PKAB?KBA?)

abKAKBIII.PLA?LB50;KGAKGBKLAKLB則ALA?B?LB?50KAKLAKGAKBKLBKGB所以fPA?PBKAKB1005050401.67(HZ)PK?AB?KBPAabKAKB中已知條件,試計算下列情況下的頻率變化量f和聯(lián)絡(luò)線上流過的功率ab,增發(fā)。解LALB5KAKLAKKBKLBKPALA?BLB?所以?f?P?BAKAKB5008507400.0314()PKAB?KBA8500?74023.27()abKAKBII.LA?LB50;KA=KLAKKBKLBK740.ALA?BLB?5所以?f?A?BKAKB40508507400.0566()PKAB?KB?MW)PAabKAKBIII.LA?LBKAKLAKGBKBKLBKGBPALA?GABLB?GB10.所以fPA?PBKAKB100?10850740)PKAB?KBA850(??)

abKAKBKLΣ2ΣfN50HZΣ1.1(發(fā)電機組仍不滿?:KLΣ調(diào)頻廠SNPLSBL,則ΣSNΣ0.2Σ0.2,S出力ΣSNΣKG，P'P'1.1LG'?GΣ0.1KGΣΣfΣPLΣ''GLKGΣ?fΣKLΣ?fΣ2由KGΣ?fΣ0.1得KΣf3Σ'?fΣ由ΣP?PLf'ΣΣKKLΣGΣ?Σ23?GOΣ0.07470.1六?:.....而..?.;.?.?.;.?,Ub(K1UG?PbRibXiKU1G)/K2U,U'.GGU'b,Ub.UG,改變變壓器變比K1,K2Ub.II.不可以.因為進行無功功率平衡計算的前提是電壓水平正常,證無功功率的平衡,系統(tǒng)的電壓質(zhì)量不能保證,如系統(tǒng)電源所能供應(yīng)的無功功率不足,則無功功率雖也能平衡,平衡條件所決定的電壓將為低于正常的電壓.這種情況下,雖可采用某,如改變某臺變壓器的變比以提高局部地區(qū)的電壓水平,供應(yīng)的無功功率,則系統(tǒng)的電壓質(zhì)量總不能獲得全面改善.事實上,系統(tǒng)中無功功率電源不足時的無功功率平衡是由于系統(tǒng)電壓水平的下降,無功功率負(fù)荷(包括損耗)本身的具有正.?答.II..,?,UI.Ui.U'.iUiUI–UUI–UIUU'U低額UiiUi高抽U低額UtIΣUtIUiUU低額ΣU低額'tImaxi'UUiiU,U低額tImaximin'Uimin:異降壓變壓器由高壓母線推算低壓母線電壓時應(yīng)將高壓母線電壓和變壓器中電壓損耗相減;升壓變壓器,由高壓母線推算低壓母線電壓時,應(yīng)將高壓母線電壓和變壓器中電壓損耗..?答::;.對無功功率供應(yīng)較充裕的系統(tǒng)采用各種類型有載調(diào)壓變壓器調(diào)壓顯得靈活而有效.尤其是系統(tǒng)中個別負(fù)荷的變比規(guī)律以及它們距電源遠(yuǎn)近相差懸殊時,不采用有載調(diào)壓變壓器幾乎無法滿足所有負(fù)荷對電壓質(zhì)量的要求.有載調(diào)壓變壓器的特殊功能還能體現(xiàn)在系統(tǒng)間聯(lián)絡(luò).?UtIminU低Nin'Uin選擇,選一最接近的UtI標(biāo)準(zhǔn)抽頭UtI,則KU低N,QC'axXi(KU'?U)axax.II.,,,所以設(shè)調(diào)相機欠激運行時的容量為過激運行時額定容量的二分之一,并在最大負(fù)荷時過激,,',CaxXi'?U),,axax–1Q'Imax('?)IminUIminCX2i''UUUUUtIΣKIminIminIminax'2'2ΣUtIUImaxU低NIminU低。U'U',則KtI,將K代入QQtICCU低N.?.:.其端電壓的平方成正比當(dāng)系統(tǒng)電壓水平過于低下迫切需要補償器增加其感性無功功率輸.:)型)飽?答并聯(lián)電容器可以增加無功功率電源進而改善電壓質(zhì)量還可以降低網(wǎng)絡(luò)中的功率和能量損耗;串聯(lián)電容器雖可以增加無功功率電源,但并不能解決改善電壓質(zhì)量的問題.因個別串聯(lián),雖能調(diào)整網(wǎng)絡(luò)中無功功率的流通從而使局部地區(qū)的電壓有所提高,但其它地.在需要附加設(shè)備的調(diào)壓措施中,對無功功率不足的系統(tǒng),首要問題是增加無功功率電源,因此以采用并聯(lián)電容器為宜;作為調(diào)壓措施,串聯(lián)補償電容器由于設(shè)計、運行等方面的原因,.有一降壓變壓器歸算至高壓側(cè)的阻抗為?,變壓器的額定電壓為

110在最大負(fù)荷時,變壓器高壓側(cè)通過功率為28MVA,高壓母線電壓為113KV,在最小負(fù)荷時,變壓器高壓側(cè)通過功率為MVA,115KV,.28*2.4414*40)

,U113,U低maxU高max??U113?5.56107.44(kv)UtU低maxU低額‘U低max*)6,Umin10*2.446*40)115,U低minU高min?Umin115?)UtminUmin低U低額‘U低min*)6.6UtI1U2tUtmin))選擇,U'低maxU低maxU低額U'tI107.44*6.156110U'低minUmin低U低額‘UtI*6.3)6.6110U%6.15?66U%min6.45?6.6,SN1102/11,KNUUK10.5,變壓器低壓母線最大負(fù)荷為,?0.8,最小負(fù)荷為7MVA,cos?.PU2解RKN163110102?3)T2202SNUU2XKNT100SN10.5110210020):?max18Σ0.814.410.8(MVA)?minΣj)??10.82(4.93j0.13)2T1102??225Tmin1102(4.93j0.02)?S高max?T14.53)???S高minSmin?STmin4.92):U14.534.9363.525kv)U4.924.935.2663.525)U低maxU高max??U107.5?8.0599.45(kv)U低minU高min??Umin107.5?3.33104.17(kv)UtU低maxU低N‘U低max99.451110)UUtmin低minU低N‘U低min1110)tI1UU2tUtmin))選擇,U'107.25KVtIU'低maxU低maxU低N‘UtI99.4511kv).選擇110?2,U'kvtIU'低minUmin低U低N‘UtI104.711).?22.5%,U'104.5kvtaxU'低maxU低maxU低N‘UtImax99.4511104.5)?22.5%,U'107.25kvtIminU低'minUmin低11)U低‘UtImin.22.5%/和-2%,,115.5KV,,?解:Ui≥UN10.25U''imin≤UN10.75Ui?109.310.25Uimin?2%)109.810.75''則UtI115.5kv.,UiU'UitI低N9.3)11UtImaxUiminU低N‘Ui97.65111.02510),UiminU'UimintI低N115.5)11UtIminUiminU低N‘Uimin111.07510)tI1UU2tImaxUtImin))U'104.5kvtIU'iUi97.6511)25)U低NU‘tIU'iminUimin102.911))U低NU‘tI'107.25kv

UtIU'i97.6511)25)U'11imin))U'tax)U'tImin)水電廠通過型升壓變壓器與系統(tǒng)連接,變壓器歸算至高壓側(cè)阻抗為,12122.5%/10.5kv和V,V,11KV,時,.282.1)

解:,U,U低maxU高maxU112.09)UtU低maxU低額‘U低max)10min282.121)

,,U,U低maxU高min?Umin115.927.48123.4(kv)UtminUmin低U低額‘U低min123.410.5)11UTI1U2TaxUTImin))選擇,UI121U′U低maxU低額U′I119.8310.410121U′minUmin低U低額U′I11121U%?1010U%min?1111100%,121110KV,長外徑為15.2mm2SFL?31500/110型,SNMVA,UN110,UK(%).最大負(fù)荷時變電所低壓側(cè)歸算至高壓側(cè)的電壓為最小負(fù)荷時為變電所二次母線允許電壓偏移在最大,最小負(fù)荷時為二次網(wǎng)絡(luò)額定電壓的?.試根據(jù)調(diào)壓的要求接

.解:1Σs?/)120x10.1445lgDmr0.01570.1445lg5100.0157?/km)3/2R1rL10.262570)L212X1xL10.42370)L2121UU2XKN1