Chapter 、對(duì)電力系統(tǒng)運(yùn)行的基本要求是什么？SUI 額定電壓的110%；既用于大城市或大工業(yè)企業(yè)網(wǎng)絡(luò)，也廣泛用于農(nóng)村網(wǎng)絡(luò)；10KV則是最常用的更低一級(jí)的配電電壓；只有負(fù)荷中高壓電的很大時(shí)，才考慮6KV配電的方案。的電能恰好等于全年耗電量W，即： PP

Tmax的大小反映了負(fù)荷隨時(shí)間變化波動(dòng)的幅度大小，Tmax大則表示負(fù)荷波動(dòng)小，Tmax年持續(xù)負(fù)荷曲線(xiàn)如圖所示，試求最大負(fù)荷利用小時(shí)。P(MW10630200

500 876 解：A年200010050002000)60(87605000)30492800(

A

492（h）331)假設(shè)C相接地，則C0。此時(shí)相量圖如圖一。由圖可知，故障相電壓變?yōu)?，非故障相得對(duì)地電壓變?yōu)榫€(xiàn)電壓，即為原來(lái)的3倍。32）則相量圖為圖二。由圖可知，∑Ic=33

時(shí)稱(chēng)為過(guò)補(bǔ)償，這是系統(tǒng)運(yùn)行中經(jīng)常使用的補(bǔ)償方式；當(dāng)ILICChapter2-1、電力線(xiàn)路由哪些部分組成？它們的作用如何？電力線(xiàn)路的桿塔由哪些形加而大為復(fù)雜，目前導(dǎo)線(xiàn)的根數(shù)一般不超過(guò)4根答：1)短線(xiàn)路，是指長(zhǎng)度不超過(guò)100km的線(xiàn)路，線(xiàn)路電壓為35KV及以下時(shí)，電納可這種線(xiàn)路的等值電路有T型和π型等值電路，如圖二。3）長(zhǎng)線(xiàn)路，是指長(zhǎng)度超過(guò)300km的線(xiàn)路和超過(guò)100km的電纜線(xiàn)路。這種線(xiàn)路的等PUU%UXT N

RT NS2NS2

U20U2N

I0%SNNN2）同——都有一串聯(lián)的阻抗Z=R+jX和并聯(lián)的導(dǎo)納Y=G+jB100%100%50%容量繞組空載時(shí)的有功損組，高低壓繞組間漏抗最大。繞組間漏抗不同，則XT就不同。2-10、發(fā)電機(jī)的等值電路有幾種形式？它們等效嗎？發(fā)電機(jī)電抗的百分?jǐn)?shù)XG%的含義是什XG%實(shí)際上是在XG上通過(guò)額定電流時(shí)產(chǎn)生的電壓降與額定電UGX% 3INXGUGN1是按變壓器實(shí)際變比計(jì)算，2是按平均額定電壓之KV解（1）K（2）

242(15%)K2(1K

=2202203K35(15%)3K35(15%)K5Ul1220(15%)231KVUl2110(15%)(3)Ul335(15%)Ul410(15%)10.5KVUl56(15%)6.3KVUl63(15%)2-17.某一回110kv電力線(xiàn)路,長(zhǎng)度為60km,導(dǎo)線(xiàn)型號(hào)LGJ-120,導(dǎo)線(xiàn)計(jì)算外徑為15.2mm,三相導(dǎo)線(xiàn)水平排列,兩相鄰導(dǎo)線(xiàn)三相間的距離為4m,試計(jì)算該電力線(xiàn)路的參數(shù),并作等效電解:r1

31.50.2625r15.27.6(mm)340003400040002X0.1445lg

0.01570.1445lg5039.70.01570.423(/DbD

6

6 61 r

lgg1R1r1l0.262560X1x1l0.42360則B1b1l

6601.614104B10.80710G1

2-18.220kv電力線(xiàn)路,LGJ-2×185的雙導(dǎo)線(xiàn),每一根導(dǎo)線(xiàn)的計(jì)算19mm,三相導(dǎo)線(xiàn)的不等邊三角形排列,D129mD238.5mD316.1m,導(dǎo)線(xiàn)的數(shù)n=2,間距為d=400mm,試計(jì)算該電力線(xiàn)路的參數(shù),并作其等值電路。

2

0.085(/22

223900850022390085001X0.1445lgDm0.01570,1445lg7756.460.01570.311(/1

b1

D

6

6 (s/ .

b11.80510g1

(s/2-19.SFL-40500/110,40500KVA,121/10.5KV,Pk234.4kw,Uk(%)11,P0PU 234.4103 解:`R 1.57102SN (40500103SNXT

UN 11(10.5103

2 1002 93.6U0 U0N

(10.5103

8.49104(SI0 2.315 BT 100U

100(10.5103)2 (S2-20.SFPSL-120000/220,,)PK(13)182.5KW PK(23)132.5KW,UK(12)(%)14.85,UK(13)(%)UK23(%)7.96,P0135KW,I0(%)0.663,求該變壓器的參數(shù),并作等值電路。解I:PK(12)601(KW)P'PK(1

4

K

730(KWP'PK

4

K(2

530(KW 1 P P )400.5(KW K(1 K(1 K(2 1 P )200.5(KWK K(12 K(2 K 1 )329.5(KWK K(1 K(23 K(1PU 400.5103(220103)RT1

K NS2SN

(12104103)

1.346(PU 200.5103(220103)RT2

K NS2SN

(12104103)

0.674(PU 329.5103(220103)RT3

K NS2SN

(12104103)

1.107(II：UK(12)(%)14.85,UK(13)(%)28.25,UK(23)(%)

(%)1 K

(%)

K(1

(%)

K(2

(%))UKUK

(%)1(%)1(U2

K(1K(1

(%)(%)

K(23K(2

(%)(%)

K(1K

(%))(%)) (%)U 17.57(220103XT1 N1

10012104

XT

UK2(%)U

2.72(22010310012104

N

10.68(220103XT3 K N1

10012104

P0

U (220103)2NI0(%)SN0.66312104N

(SIV：

100U

100(220103

(S2-21OSSPSL-120000/220 PK(12)417KW,PK(13)318.5KW,PK(23)314KW UK(12)(%)UK(13(%)16.65,UK23(%)10.85;P057.7KW,I0(%)0.712求該變壓器的參數(shù)解IPK(12)417(KWP'PK(1P'PK(2

4PK(2

1274(KW1256(KW 1 P P )217.5(KW K(1 K K(2 1 P )199.5(KWK K(1 K(2 K 1 P )1056.5(KWK K K(2 K(1PU 217.5103(220103RT1 N1S

(12104103

3K

(22010S2RT2 NS2N

(12104103

PU

1056.5103(220103)2(12104103RT3K N1 SII:UK(12)(%)U'UK

(%)

K

(%)U'UK(2

(%)

K

(%) (%)

1

(%)U

(%)U

(%))KUK

(%)

K1 K(1

K(1(%)(%)K(2

K(2(%)(%)K

(%)) (%)1(U (%)U (%) (%)K K(1 K(23 K(12 XT1 K N1 %U 1.31(220103)XT2 K

5.284()10012104103 %U XT3 P0 III:GTU (220103)21.19I0(%)SN0.71212104

(SNIV:BTN

100U

100(220103 1.765

(SdcosN0.85,UN13.8KVXd1.867,Xd0.257,X''0.18,dX,XX的有名值 U U 解: 1.295( Xd1.867ZN1.8671.295 X'0.257 0.2571.295 X'' 0.181.295 1003解:XLXL(%)UN1003解:XL1003I

6KVIN500AXLChapterPi2QU2答:SLPLjQL i(RLjXLU2i1 jU2 1 jU2 C 2答

PiRQiX

PiXQiRUi 都要乘3倍。而線(xiàn)電壓是相電壓的3倍,將ΔU和δU前的3變成3 ΔU、δU相乘則使相電壓的ΔU、δU變成了線(xiàn)電壓的ΔU、δU3則和等式右側(cè)分母上的相電壓U相乘得到線(xiàn)電壓U。答:電壓降落:線(xiàn)路始.末兩端電壓的相量差(U1U2或dU電壓損耗:線(xiàn)路始末兩端電壓的數(shù)值差(U1U2從百分?jǐn)?shù)表示,電壓損耗ΔUU1U2U電壓偏移:線(xiàn)路始.末端電壓與線(xiàn)路額定電壓的數(shù)值差(U1UN)(U2U1),始端電壓偏移ΔU1

N%=U1UNU末端電壓偏移ΔU2

N%=U2UNU電壓調(diào)整:線(xiàn)路末端空載與負(fù)載時(shí)電壓的數(shù)值差(U20U2).以百分?jǐn)?shù)表示,電壓調(diào)整0%=U20U2U輸電效率η%=P2100.變電所的運(yùn)算負(fù)荷=低壓側(cè)負(fù)荷+變壓器的繞組損耗+變壓器的勵(lì)磁損耗+所有與其高壓母線(xiàn)答：SS PjUK%SN

jIO

P

j(UK%SNIO%SN

SmZSam18110解:r1

31.50.2625(/km) 5m5000(mm),r15.2 X0.1445lgDm0.01570.1445lg50000.01570.423(/ b

106

1062.69106(s/km),g lgDm

lg1R1rl10.26251501 2 1xl10.423150 2 2bl22.691061508.07104 GL 2 S2S2(jU2 )(30j15)(j106 )30 SLPL

P'2Q'U2jQL 2(RLU22

jXL)

30210.466106

(19.687j1.769 S1S2SL(30j10.466)(1.769j2.85)31.769j13.316(MVA)P'RQ'X P'XQ'RdU2U2jU22 U

j 2U3019.68710.46631.725j3031.7258.7

U1UdU1068.7j7.03114.7j7.03114.93.5（ 2 S1S1(jU1 )(31.769j13.316)(j114.9 31.769 設(shè)U1060（.kvU8.7(kv),U7.03(kv則U114.9 3-9.220KV單回電力線(xiàn)路,長(zhǎng)度為200KM,導(dǎo)線(xiàn)型號(hào)LGJ—300,導(dǎo)線(xiàn)計(jì)算外徑為24.2mm,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r1

31.50.105(/Km)Dm7.5mr24.2912.1(mm)x0.1445lgDm 0.1445lg75000.01570.419(/km)b r

106 lg

1062.715106(S/RLr1l0.105200XLx1l0.419200Blb1l2.7151062005.43104(SGl 2 S1S1(jU1 )(120j50)(j240 120SL

P12U2U1

(RLjXL)

120265.63842(216.82

1202165.638483.8j12083.865.638433.419

2U1dU124033.419j36.157206.58j36.157209.7210(KV 2 S2S2(jU2 )(113.18j38.418)(j209.72 113.18S2113.18j50.36(MVAU2209.7210(KV3-10.110kv單回電力線(xiàn)路,長(zhǎng)度為80km,導(dǎo)線(xiàn)型號(hào)為L(zhǎng)GJ-95,導(dǎo)線(xiàn)計(jì)算外徑為13.7mm,三解r1

31.50.332(/km)Dm5mr13.76.85(mm)x0.1445lg50000.01570.425(/ 1b7.58106 1

1062.647106(s/lgr

lgg1RLr1l0.33280XLx1l0.42980BLb1l2.647106802.12104GL 2 S2S2(-jUN )(15j10)(j110

)152P'2Q 152SL 2(RLjXL)

(26.56j34.32)10.66UN UN S1S2SL(15j8.717)(10.66j0.854)15.66'S1S1(jU

2.12L)(15.66j9.57)(j116 )15.66 P'RQ' P' Q' 15.6626.569.57dU1U1jU11 Lj 1L +j15.6634.329.5726.566.417j2.442(KV)U2U1dU11166.417j2.442109.583j2.442(KV3-11.220kv單回電力線(xiàn)路,長(zhǎng)度為220km,電力線(xiàn)路每公里的參數(shù)為r10.108km,x10.42kmb12.66106skm,線(xiàn)路空載運(yùn)行,當(dāng)線(xiàn)路末端電壓為205kv時(shí),求線(xiàn)路始端電壓.解:RLr1l0.10822023.76(KVXLx1l0.42220BLb1l2.661062205.852104 所以S2 2S2S2(jU2

L)

2

U2SL (RLjXL)U22

(23.76j92.4)0.085'2S1S'SLj12.30.085j0.330.0852P'RQ' P' Q' 12.3 12.3dU2U2jU2 2 U5.544j1.4256(KV

Lj 2LU

U1U2dU22055.544j1.4256199.456j1.4256(KV 2 5.852S1S1(jU1 )(0.085j11.97)[2

1.4256) 0.085U13-12.,~示ZT12.47j65ZT22.47j115ZT32.47j37.8S25~解:I:由US求U U'U1106110100(KV P2Q TST3 TU'U3

jXT3)

82

(2.47j37.8)0.02473S'S3ST3(8j6)(0.0247j0.378)8.02473dU3U

jU

P3RT3Q3XTUU

P3XT3Q3RTUjUj82.47637.8j837.862.472.4656j2.87589(KV U'U'dU3100j2.4656j2.8757102.4656j2.8758(KV 1II.由U'S2求U1 P2

52T2 U

2(RT2jXT2)2

(2.47j1.5)0.00842S'S2ST2(5j4)(0.0084j0.005)5.00842P Q' P' Q'U'dU2U2jU2 2T T2jU'1

T2 2T2U1U5.00842.473.9951.5102.465622.8758

j5.0084(1.5)3.9952.470.062102.465622.8758

j0.17(KV1U1U'dU2102.4j3.04(KV1S''S'S'13.033 III.由U'S''求 P''2Q'' 13.0332ST1 1(RT1jXT1)

(2.47j65)0.0652U1' 102.46562U1S'S''ST1(13.033j10.373)(0.0652j1.716)13.0982 P'' Q'' P'' QdU1U1jU1 T1j T 1U U 13.0332.4710.37365j13.0336510.3732.476.89j8(KV 1U1U'dU1109.3556j10.8758(KV1U 109.9(KV102.42U102.42

102.445(KV38.5(15%)102.44538.51.0537.65(KV2 3-13.某電力線(xiàn)路導(dǎo)線(xiàn)為L(zhǎng)GJ-185,長(zhǎng)度為100km,導(dǎo)線(xiàn)計(jì)算外徑為19mm線(xiàn)路末端負(fù)荷為90+j20MVA,110kv220kv運(yùn)行,假設(shè)升壓后導(dǎo)線(xiàn)截面和負(fù)荷大小保持不變,且不計(jì)電暈損失的增加,導(dǎo)線(xiàn)水平排列,升壓前后線(xiàn)間距離由4m增加到5.5m,試問(wèn)升壓后,解

s

31.50.17(/km)RLr1l0.171001Dm340004000240005.041031Dm2355005500255006.93103r199.5(mm)1x0.1445lgDm10.01570.1445lg1

0.01570.41(/ x20.1445lgr

0.01570.1445lg

0.01570.43(/XL1x1l0.41100XL2x2l0.43100b 2.78106(s/ lgDm1rb2

m2r

2.65106(s/BL1b1l2.781061002.78104BL2b2l2.651061002.651041SC1jU2

2

21S

jU BL

2

2

S S21SC1(90j20)j1.68902 S S21SC22

(90j20)j6.490P'2Q' S

21(R )9018.32(17j41)11.85 2 U 2P'2Q'

U2S1 22U22

L2)

9013.6(17j43)2.9SS1S2(11.85j28.58)(2.9j7.36)8.95P'RQ' P' Q'dU1U1jU1 21L

L1j 21 901718.3241j904118.321720.74 P'RQ' P' Q'dU2U2jU222L

L2j L 222 U2901713.643j904313.6179.6 dU

%dU1100%

20.74220.742

100%dU2%

dU2100%

9.62

100%dU%dU1%dU2%33.68%8.7%3-14.,Zl110j17.32Zl220j34.6Zl325j43.3Zl410j17.3 S290j40MVA,S350j30MVA,S440j15MVA且U1235KVZl110j17.3210Zl220j34.6(1017.3)Zl325j43.3(1017.3)Zl410S2(Z*Z*Z*)S3(Z*Z*)S4Z則S1 l l l l l lZ*Z*Z*Z l l lS2(22.51)S3(2.51)S 122.5=(90j40)5.5(50j30)3.5(40j15)=109.2 S4(Z*Z*Z*)S3(Z*Z*)S2ZZS1 l lZ

*l1

Z

l2Z

Z l l lZS4(2.521)S3(21)S 122.5=(40j15)5.5(50j30)3(90j40)=70.7711S23S1S2109.2j52.3(90j40)19.2j12.3(MvA)S43S'S470.77j32.7(40j15)30.77j17.7(MvA)S1S'109.2j52.370.77j32.7179.97j85(MVA)11S2S3S490j4050j3040j151801S1S'S2S3S1P2Q 19.22S2 23(Rl2jXl2)

(20j34.6)0.2UN UNS" S'19.2j12.30.2j0.3719.4 S'S2S''90j4019.4j12.67109.4 P'2Q' U2S1 2U2N

)109.452.67(10j17.32)3S1S'S'109.4j52.673j5.27122.42dU1

jUP1RL1Q1XL1jP1XL1Q1

112.41057.9417.32j112.417.3257.94109.05j5.82(KV U2U1dU1235(9.05j5.82)225.95j5.82(kv)2261.5.P“ Q” P“ Q”dU2U2jU2 U

j 2LU19.4 34.6j19.434.612.67203.656j1.85(KV U3U2dU2225.95j5.82(3.656j1.85)222.294j7.67(kv)222.426dU3P43RL3Q43XL3jP43XL3Q43U U30.772517.743.3j30.7743.317.7256.9j4(kV U4U3dU3222.4266.9j4229.326j4(KVSN8MVAPK25KW,UK7.5;T2SN2MVAPK24KVA,UK~的變比為KT1KT235/ 時(shí),每臺(tái)變壓器的總功率為多少?KT134.125/11KVKT235/11KV時(shí),每臺(tái)變壓器通過(guò)的功率是多少解PU 25103(35103 SN (8106SNU(%)U 7.5(35103XT1

100RPU 24103(35103RT2 NS (2106NU(%)U 6.5(35103XT2

1002ZT10.48j11.48(),ZT27.35SZZ*I: Z*

(8.5j5.3)(7.35(0.48j11.48)(7.35

6.48SZZ*ST2 T1Z*

(8.5j5.3)(0.48(0.48j11.48)(7.35

2.00II:如右圖,設(shè)U2則U351134.12534.125(KV UU2U10.875(KVSU*USZ Z

0.875 (0.48j11.48)(7.35

T1(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(wǎng)的電能損耗如何計(jì)算?什么是最大負(fù)荷使用時(shí)間Tmax?什么是最大負(fù)荷損耗時(shí)max

2Rt

P2QU

Rt,為某個(gè)時(shí)段t內(nèi)的電能損耗Tmax:指一年中負(fù)荷消費(fèi)的電能除以一年中的最大負(fù)荷Pmax,Tmax

max:指全年電能損耗除以最大負(fù)荷時(shí)的功率損耗,即

3-18.A220KV,,~Z116j120,Z233j89,Z348j120,Z460j152,S1 170+j40MVA,S250j30MVAS340j15MVA計(jì)算:網(wǎng)絡(luò)的自然功率分布;網(wǎng)絡(luò)的解:在A點(diǎn)處把環(huán)網(wǎng)解開(kāi),如圖S1(Z*Z*Z*)S2(Z*Z*)S3ZI:SA 3Z*4Z*Z*Z* =168.86S1(Z*Z*Z*)S2(Z*Z*)S3ZSA 3Z*4Z*Z*Z* =91.29ASAS'260.15j84.77(MVA),S1S2S3260AS12SAS1(168.86j62.39)(170j40)1.14j22.39(MVA)S23S2S12(50j30)(1.14j22.39)51.14j7.61(MVA)AS'S23S3(51.14j7.61)(40j15)91.14AII:SAOS1(R2R3R4)S2(R3R4)S3R1R2R3(170j40)(334860)(50j30)(4860)(40j15)163348202.26' S3(R1R2R3)S2(R1R2)S1' j40)R1R2R3j40)(40j15)(334816)(50j30)(1633)16334857.64SAOS 260 AO III:P(SA)2R(S12)2R(S23)2R(SA)2UUUU UUUU 10.7050.342.6510.9324.625(KW

(SAO)2RUN

(S12O)2RUN

(S23O)2RUN

UN123414.81.0460.374.77620.992(KW1234WPi1tPi2t(24.62520.992)3.63387603-19.20KM,35KV的雙回平行電力線(xiàn)路,10MWcos0.8.電力線(xiàn)路導(dǎo)線(xiàn)裝設(shè)有兩臺(tái)SFZ-7500/35SN7500KVA,UN3511KV,Pk75KWUK7.5,P09.6KW,I00.8運(yùn)行,其年負(fù)荷曲線(xiàn)如圖.7和9r1

31.50,45(/KM)r11.45.7(mm)Dm3.5mx0.1445lgDm 0.1445lg35000.01570.4186(/KM)b

106

2.72106(s/ lgDm

lgR1rl10.4520 2 1xl10.418620 2 2bl12.72106201.088104(S RT

PKUNS2NSN

75103(35103(7.5106

UK(%)U N

75(35103

N 100NGT

PONU N

7.84106(SBI0(%)SN0.8

NN 100U 100(35103PKTP09.6KW0.0096(MW I0(%)SN0.8

PZTPK0.075(MW UK(%)SN7.5

I:cosS

arccos10arccos0.8

10S

2(0.0096j0.06)2[( 7.5

)20.075j(7.5

)20.0192j0.120.104j0.780.1232SASST10j7.50.1232j0.910.1232P2Q

10.12322 2A(R1jX1)

(4.5j4.186)0.636U1SAS110.1232j8.40.636j0.590.7592 T 10.1232j0.90.636j0.590.7592S100.5arccos0.86.2536.875j3.75(MVA)

KT

2(0.0096j0.06)2[( 7.5

)20.075j(7.5

)20.0192j0.120.026j0.195.0452SASST5j3.750.0452j0.315.0452P2Q

5.04522 2A(R1jX1)

(4.5j4.186)0.154U1SA

15.0452j4.060.154j0.1435.1992 T 10.0452j0.310.154j0.1430.1992 100.25arccos100.25arccos0.83.12536.872.5j1.875(MvA) 2(0.0096j0.06)2[(3.125)20.075j(3.125)2 7.5 7.50.0192j0.120.0065j0.04880.0257SASST2.5j1.8750.0257j0.16882.5257P2Q

2.52572 A2A(R1jX1)

(4.5j4.186)0.039U1SAS12.5257j2.0440.039j0.0362.5647 T 10.0257j0.16880.039j0.0360.0647P1100

0.7592100P21000.1992100 P3100

100WP1t1P2t2P3t30.759220000.199230000.0391518.4597.6146.64II.cos 10arccos0.714.2845.57

10 2(0.0096j0.06)2[(14.28)20.075j(14.28)2 7.5 SASST10j10.20.1552j1.1210.1552P2

10.15522 2A(R1jX1)

(4.5j4.186)0.85U1SAS110.1552j11.320.85j0.7911 T 10.1552j1.120.85j0.791 100.5arccos0.77.1445.57

5S

2(0.0096j0.06)2[( 7.5

)20.075j(

)20.0192j0.120.034j0.250.0532SASST5j5.10.0532j0.375.0532P2Q 5.053225.47S1

2A(R1UNU

jX1)

35

(4.5j4.186)0.21SAS15.0532j5.470.2j0.1895.2532 T 10.0532j0.370.2j0.1890.2532 100.25arccos0.73.5745.572.5j2.55(MVA)S

2(0.0096j0.06)2[( 7.5

)20.075j(

)20.0192j0.120.0085j0.0640.0277SASST2.5j2.550.0277j0.1842.527jP2Q 2.52722.734S1

2A(R1UNU

jX1)

35

(4.5j4.186)0.05j1SAS12.527j2.7340.05j0.0472.577 T 10.0277j0.1840.05j0.0470.0777P11(MW),P20.2532(MW),P30.0777(MWP1100

1100P21000.2532100 P31000.0777100 WP1t1P2t2P3t3120000.253230000.07772000759.6292.152III:cos 10arccos0.91125.849.9j4.79(MVA)S

2(0.0096j0.06) 7.5

)20.075

)20.0192j0.120.08j0.6250.092SASST9.9j4.790.0992j0.72510P2Q 102S1

2A(R1UNU

jX1)

35

(4.5j4.186)0.481SAS110j5.5150.48j0.4410.48 T 10.0992j0.7250.48j0.440.5792 100.5arccos0.95.525.844.9j2.4(MVA)S

2(0.0096j0.06)2[( 7.5

)20.075

)20.0192j0.120.02j0.150.0392SASST4.9j2.40.0392j0.274.94P2Q 4.9422.67S1

2A(R1UNU

jX1)

(4.5j4.186)0.1161SAS12.49j1.3580.029j0.02752.52 T 10.024j0.15780.029j0.02750.0532 100.25arccos0.92.7525.84

2.47S

2(0.0096j0.06)2[( 7.5

)20.075 )2.752.750.0192j0.120.005j0.03780.0242SASST2.49j1.20.0242j0.15782.49P2

2.492 2A(R1jX1)

(4.5j4.186)0.029U1SAS12.49j1.3580.029j0.02752.52 T 10.0242j0.15780.029j0.02750.0532P10.5792(MW),P20.1552(MW),P30.0532(MWP11000.5792100 P21000.1552100 P31000.0532100 WP1t1P2t2P3t30.579220000.155230000.05321158.4465.6200.032Chapter五I:火力發(fā)電廠的主要廠用機(jī)械—通風(fēng)機(jī)和給水泵,在頻率降低時(shí),所能供應(yīng)的風(fēng)量和水量將II:低頻率運(yùn)行將增加汽輪機(jī)葉片所受的應(yīng)力,引起葉片的,縮短葉片的,甚至使葉致使發(fā)電機(jī)定子和轉(zhuǎn)子的溫升都得增加,為了不溫升限額,不得不降低發(fā)電機(jī)所發(fā)功頻率過(guò)低時(shí)甚至?xí)拐麄€(gè)系統(tǒng),造成大面積停電.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(dòng)幅度很小,周期又很短,這種負(fù)荷變動(dòng)有很大的偶然性.第二種變動(dòng)幅度較大,周期也較長(zhǎng),機(jī)械、電氣機(jī)車(chē)等帶有沖擊性的負(fù)荷變動(dòng).第三種變動(dòng)幅度最大,周期也最大,.,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(dòng)的負(fù)荷,即責(zé)成各發(fā)電廠按事.都是和系統(tǒng)的總負(fù)荷相平衡的，系統(tǒng)的總負(fù)荷包括所有用戶(hù)的有功負(fù)荷PL、所有發(fā)PGiPLPPS（備用容量 額定負(fù)荷的10%-15%。

PL(MWHz)，

PL**:出增加的量和對(duì)應(yīng)的頻率下降的量的比值稱(chēng)為發(fā)電機(jī)的單位調(diào)節(jié)功率或調(diào)節(jié)系數(shù)，即：KPG(MW

答:所謂的機(jī)組調(diào)差系數(shù)，是以百分?jǐn)?shù)表示的機(jī)組空載運(yùn)行時(shí)f0fN的差值，即：%f0fN100%

1100。率特性曲線(xiàn)如圖,相交于點(diǎn)O,則在OPLPG,穩(wěn)態(tài)運(yùn)行,運(yùn)行頻率為f1.負(fù)荷出現(xiàn)增量PLPLPLPLPGPL,即負(fù)荷大于出力,轉(zhuǎn)速減小,頻率也減小.調(diào)節(jié)功率（負(fù)荷的調(diào)節(jié)效應(yīng)）KLKSKGKL。對(duì)一個(gè)系統(tǒng)而言KL不可調(diào)，調(diào)節(jié)KSKG著手。對(duì)每臺(tái)發(fā)電機(jī)而言KG不能過(guò)大也即調(diào)差系數(shù)不能過(guò)小，否則就要出現(xiàn)負(fù)荷變化量在各發(fā)電機(jī)組之間的分配無(wú)法固定,即將調(diào)節(jié)功率之和KG不可能過(guò)大，KS自然不可能過(guò)大。5-10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(wú)差調(diào)電機(jī)組發(fā)出的功率,使頻率特性向上移動(dòng).設(shè)發(fā)電機(jī)組增發(fā)P ,則運(yùn)行點(diǎn)又將從點(diǎn)O'轉(zhuǎn)移到點(diǎn)O",由于進(jìn)行了二次調(diào)整,系統(tǒng)的運(yùn)行質(zhì)量有了改善.無(wú)差調(diào)節(jié):如負(fù)荷的變化量等于發(fā)電機(jī)出力的變化量PLOPGO，則

0,答:如圖，設(shè)A、B兩系統(tǒng)發(fā)電機(jī)組均能參加一次調(diào)頻，且均有調(diào)頻廠即均可以參加二次調(diào) 表示A、B兩系統(tǒng)的負(fù)荷變化，以PGA、PGB表示A,B兩系統(tǒng)的KAKB表示A,BPab，且設(shè)其正方向是由A系統(tǒng)流向B系統(tǒng)對(duì)A系統(tǒng)Pab可看成一負(fù)荷PLAPabPGAKA對(duì)B系統(tǒng)Pab可看成一電源PLB(PabPGBKBff(PLAPGA)(PLBPGBKAK KA(PLBPGB)KB(PLAPGA)KKA

聯(lián)絡(luò)線(xiàn)上功率Pab不超過(guò)允許的范圍，在調(diào)頻效果相同的前提下應(yīng)盡量使每個(gè)系統(tǒng)的功缺額由該系統(tǒng)的發(fā)電機(jī)補(bǔ)償，從而使聯(lián)絡(luò)線(xiàn)上功率Pab盡可能小5-12.A、 兩系統(tǒng)由聯(lián)絡(luò)線(xiàn)相連如圖所示,已知 系KGA800MWHZKLA50MWHZPLA100MWB系統(tǒng)KGB700MWKLB40MWHZPLB50MW求在下列情況下頻率的變化量f化量Pab;I.當(dāng)兩系統(tǒng)機(jī)組都參加一次調(diào)頻時(shí);II.當(dāng)A系統(tǒng)機(jī)組參加一次調(diào)頻,而B(niǎo)系統(tǒng)機(jī)組不解:I.PGAPGB0,PLA100,PLBKGA800,KGB700,KLA50,KLB則PAPLAPGB100,PBPLBPGBKAKLAKGA850,KBKLBKGBfPAPB100

0.0943(HZ所 KA

850PKAPBKBPA85050740100 A

850II.PGAPGB0,PLA100,PLBKGA800,KGB0,KLA50,KLB則PAPLAPGA100,PBPLBPGBKAKLAKGA850,KBKLBKGBPA

10050所以 KA 850PKAPBKBPA850504010043.26(KKA

850III.PGAPGB0,PLA100,PLB50KGAKGB0,KLA50,KLB則PAPLAPGA100PBPLBPGBKAKLAKGA50,KBKLBKGB所以fPAPB100501.67(HZKA 50PKAPBKBPA505040100 K

505-13.5-12中已知條件,試計(jì)算下列情況下的頻率變化量f機(jī)組都參加一次調(diào)頻,A系統(tǒng)并有機(jī)組參加二次調(diào)頻,60MW；III.A,B兩系統(tǒng)都參加一次調(diào)頻,B系統(tǒng)并有機(jī)組參加二次調(diào)頻,增發(fā)60MW。解:I.PGAPGB50PLA100,PLBKAKLAKGA850,KBKLBKGBPAPLAPGA50,PBPLBPGB所以fPAPB 50KAK 850

0.0314(HZPKAPBKBPA85007405023.27(KA

850PGA60,PGB0,PLA100,PLBKA=KLAKGA850,KBKLBKGBPAPLAPGA40,PBPLBPGB所以fPAPB40500.0566(HZKAK

850PKAPBKBPA85050740408.11( K

850III.PGA0,PGB60,PLA100,PLBKAKLAKGB850,KBKLBKGBPAPLAPGA100,PBPLBPGB所以fPAPB100

0.0566(HZKA

850PKAPBKBPA850(10)74010051.89(KKA

850PL1,fN50HZ時(shí),主調(diào)頻電廠出力為其額定值的50%,如果負(fù)荷增加,調(diào)頻電廠的頻率調(diào)整不動(dòng)作,0.3HZ,PL1.1(發(fā)電機(jī)組仍不滿(mǎn)載),現(xiàn)在頻率調(diào)整器動(dòng)作,使頻率上升0.2HZ,問(wèn)二次調(diào)頻作用增加的功率是多少?解:如圖KL2,調(diào)頻廠SN0.2SBPLPL1,SN0.2PL0.2S出力50%SN不滿(mǎn)載表示發(fā)電機(jī)將自動(dòng)參加一次調(diào)頻，調(diào)節(jié)系數(shù)為KG P'P' 0.1

P'P'

fK

0.120.30.10.012

由 f0.1得 0.10.150 0.2HZ,f'0.30.2 PLPGOf

0.112PGO250 Chapter步電可發(fā)出一部分無(wú)功功率外,大多數(shù)都要消耗無(wú)功功率.Ub(K1UG

PbRiQbK1

)/KU---發(fā)電機(jī)出口電壓U'---發(fā)電機(jī)出口電壓歸算至高壓側(cè) U---用電設(shè)備端電壓U---用戶(hù)側(cè)電壓歸算至高壓側(cè) 通過(guò)改變發(fā)電機(jī)機(jī)端電壓UG,K1K2,改變無(wú)功功率分布或串聯(lián)電容補(bǔ)償來(lái)改變用電設(shè)備端電壓Ub.不可以.,如不能在正常電壓水平下保證無(wú)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(wú)功功率不足,則無(wú)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,雖可采用某些措施,,供應(yīng)的無(wú)功功率,則系統(tǒng)的電壓質(zhì)量總不能獲得全面改善.事實(shí)上,系統(tǒng)中無(wú)功功率電源不足時(shí)的無(wú)功功率平衡是由于系統(tǒng)電壓水平的下降,無(wú)功功率負(fù)荷(包括損耗)本身的具有正.UI---高壓側(cè)電壓Ui---低壓側(cè)歸算至高壓側(cè)電壓UUUiUU

U

PR

U'UU低額

U

U低額

U U UU

高

iUiU Ui

tIm

i i

tIm

i UiU減;升壓變壓器,由高壓母線(xiàn)推算低壓母線(xiàn)電壓時(shí),應(yīng)將高壓母線(xiàn)電壓和變壓器中電壓損耗答:I選用電容器,因電容器只能發(fā)出感性無(wú)功功率,提高電壓;不能吸取感性無(wú)功功率,降低電小負(fù)荷時(shí)電容器全部推出運(yùn)行的條件考慮,即按

t

U低UIm UIm

選擇,標(biāo)準(zhǔn)抽頭

K

UU

KUImaxX

低'Im

Im

)KU功電源QCImax(KU'U在最小負(fù)荷時(shí)作為無(wú)功負(fù)荷XImXi

Im'1QCKUImax(KU

Im Im2U U K IminIm IminImax UtIIm

Im

U低

低N選擇一最靠近的標(biāo)準(zhǔn)抽頭U',K

Ut'I,將K代入Q式即可解出Q,這樣,U U低 飽和電抗器型(SR型).答,耗;串聯(lián)電容器雖可以增加無(wú)功功率電源,但并不能解決改善電壓質(zhì)量的問(wèn)題.因個(gè)別串聯(lián)電容器的采用,,.在需要附加設(shè)備的調(diào)壓措施中,對(duì)無(wú)功功率不足的系統(tǒng),首要問(wèn)題是增加無(wú)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(bǔ)償電容器由于設(shè)計(jì)、運(yùn)行等方面的原因,.6-13.有一降壓變壓器歸算至高壓側(cè)的阻抗為2.44+j40,變壓器的額定電壓為1102*2.5%/6.3KV,在最大負(fù)荷時(shí),28j14MVA,高壓母線(xiàn)電壓高壓母線(xiàn)電壓為115KV,低壓母線(xiàn)要求電壓為6.6KV,試選擇該變壓器的分接頭.

28*2.4414*40U低maxU高maxUmax1135.56Ut

U低max低

107.44*6.3112.8(kv)10*2.446*在最小負(fù)荷時(shí)Umin

U低minU高minUmin1152.3Ut

U低低

112.7*6.3

1 t

Ut

)選擇110kv分接頭U'U低U

U低

UU'UU

107.44*6.36.15112.7*6.36.45(KV)低 低

U

6.156*100%66.45U%min *100%6.456-14.有一降壓變壓器SN20MVA,UN11022.5%/11KVPKUK(%)10.5,變壓器低壓母線(xiàn)最大負(fù)荷為18MVA cos0.8,最小負(fù)荷7MVA,cos0.7107.5KV,如果變壓器RT

SP NSP2N

103XT

UK(%)UNN

10.5110210020

Smax18arccos0.814.4j10.8(MVA)Smin7arccos0.74.9STmax

14.42 STmin

4.92 高maxSmax Tmax14.53 Smin

T

4.92UU

14.534.9312.563.5254.924.935.2663.525U低maxU高maxUmax107.58.05U低minU高minUmin107.53.33Ut

U低低max低U

99.45

低

t 低 低

則公共抽頭

1 t

Ut

)110-2.5%分接頭U'U'U低

U低

U低‘

99.45 選擇11022.5%分接頭U'U'U低

U低

U低‘

11(kv)最大負(fù)荷時(shí),選擇11022.5分接頭

UtImU

U'U低

U低低maxtIm

99.4511 10.5(kv)10.25KV最小負(fù)荷時(shí),選擇11022.5分接頭

UtImU

U'U低

U低低 tIm

104.17 解解i

1.025U

UiU

U'Ui

(17%)109.3

'UiU

(12%)109.8則UtI115.5kv,不滿(mǎn)足調(diào)壓要求U Uimax i tI

UtIm

U低Uimini

U Uimin imintI

U低 UtIm

U低imini

102.9

1tIm

UtIm

)選擇一臺(tái)靠近的標(biāo)準(zhǔn)抽頭U'U'Ui

Ui

97.6511 10.28(kv)10.25KV

Ui

Ui

102.911 10.83(kv)10.75KV

U則重新選擇'107UU'i

U 102.91110.55(kv)i

UtIm

U

tIm6-16.水電廠通過(guò)SFL-40000/110tImj38.5額定電壓為12122.5%10.5kv,282.121解:最大負(fù)荷時(shí),Umax U低maxU高maxUmax112.097.74Ut

U低max低

119.8310.5125.8(kv)282.121最小負(fù)荷時(shí),,Umin U低maxU高minUmin115.927.48Ut

U低低

123.410.5

1 TIm

UTIm

)12.18(KVUU U

低

UUU

119.8310.510.4123.410.510.7低 低min

UU

10.410100%10.7 100%10.76-17.有一個(gè)降壓變電所由兩回110KV,長(zhǎng)70KM的電力線(xiàn)路供電,導(dǎo)線(xiàn)為L(zhǎng)GJ-120型,導(dǎo)線(xiàn)計(jì)算外徑為15.2mm,5m.2臺(tái)變壓器并列運(yùn)行,其型號(hào)為SFL31500110型SN31.5MVAUN110/11KV,UK10.5.電壓偏移在最大,最小負(fù)荷時(shí)為二次網(wǎng)絡(luò)額定電壓的2.5%7.5%.試根據(jù)調(diào)壓的要求接并聯(lián)電容器和同期調(diào)相機(jī)兩種措施,確定變電所10KV母線(xiàn)上所需補(bǔ)償設(shè)備的最小容量.解r1

31.50.2625(/km)x0.1445lgDm 0.1445lg15.2/

0.01570.423(/R