电力系统分析中国电力出版社苏小林闫晓霞版.docx
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1、第一章T 220x110/1001 I:,10x105/100242T6 110 _1103:10xl.l -1T,T. 35 二 354:6xl.l 6I-I 解:发电机G额定电压:IoX = 10.5KV100变压器额定变比:T220 r(110 ) :(35 X ) = 220:121:38.521001001-2解:(1):发电机G: 10. 5KV变压器高低压侧额定电压:T 121 T 110HO T 35351,:, 1=, 1 9=10.5351.1 38.53 10l.l 11(2)变压器实际变比:T 121x(1 + 2.5%) 124 TlIO T 35(1-5%) 33.
2、2511:=, , 1=10.510.52 38.53 101.1111-3 解:日用电量:W = 702 + 504 + 802 + 1004 + 802 + 904 + 1204 + 702 = 2040(MWh)W 2040日平均负荷:Pqv = HL =上= 85(MW) av 2424P 25负荷率:/ = - = = 0.708PM 120P 50最小负荷率:/=皿=0.417PmaX 12()W 11-4 解:Tmax = -(100 2000 + 60 4000 + 30 2760) = 5228(h)max 匕I(X)第二章2-1解:线路长度为65km,属短线路,不计分布性参
3、数= K = W = O 0 km),S 300D111 = Dab Dac Dhc = V6000 600012000 = 7559.5(mm)x1 =0.1445 Ig+ 0.0157 = 0.4215( / km)7 58b =106 =2.7106 (SZkm)ig4r.-./? = r1 = 0.10565 = 6.825()X xll = 0.421565 = 27.3975()8 = Z = 2.7 X10-6 * 65 = 1.755 X104 (S)等效电路:6.825+ j27.3975二二 j0.877510-4S2-2解:(1)不计分布性参数31.54x400= 0.0
4、197(km)D111 = Dab Dac Dbc = Vl 20001200024000 = 15119.1(mm)Vr-J12-J13 614 = V13.624004004002 = 187.38(mm)X1 = 0.1445Ig= 0.1445Ig% n187.384= 0.279(km)7.58IgDm106 =3.975 106 (SZkm).? = = 0.0197120 = 2.364()X =x1 = 0.279120 = 33.48()B = bJ = 3.975XlO-6 120 = 4.77X10-4(S)等效电路:2.364+ j33.48I1 j2.385 104S
5、(2)不计分布性参数r. = = - 315- = 0.0197(/km),S 4400Dlll = Dab Dac Dhc = V14000 1400014000 = 14000(mm)req = rd2 .九d, = N13.62 X 450 X 450X 450后=204.68(mm)a 1,z1c1 Dm 0.0157 八f/口 140000.0157 八“八小八、X1 =0.1445 Ig i2- += 0.1445 Ig+ - = 0.269( / km)7 58b、=-106 =4.131106 (SZkm)的Yeq:. = = 0.0197120 = 2.364()X =xl
6、= 0.269120 = 32.28()B = i = 4.131106120 = 4.9572 104 (S) 2-3 解: = = 0.1200 = 20()X = Lsl = 250203 200= 125.66()B = 2 / = 2) X 50 X 0.01X10-6 * 200 = 6.28 X104(S)I27002/: =l-x1-= 1-25021032-500.01106- = 0.974 r 1 1 33k=-(x1Z?, -= 0.987, E=I+ 再4 L = LOO7X1 612Z, = krR-vkxX = 0.974 20 + j.987 X125.66 =
7、 19.48 + jl 24.026()Y, = JkbB = jl .007 6.28 104 = j6.324 W4(S)20 + jl25.66oEZZJOO短线路等效电路1=1= j3.1410-4S =C20 + jl25.66中等长度线路等效电路= j3.16210-4S =19.48 + jl24.026长线路等效电路2-4 解:XT1 UkooUli _ 1 10.5 IlO22 100 SN 2 100GT = 2 - = 2IO(W ;10181000x1 IO263.525()2.976 10-6(S)= 2x* = l-型等效电路:O-jl .488 105 S2.97
8、610 6S3.63 j63.525I 12-5 解:%=g(%-2+”3-2.3)= 93(kW)=-3-3) = 52(kW)&=(+%-3 - -)= 65(kW)RTl =PsU1i 931212Rf3IOOOS 1000 202/12U = 5212121000S 1000 202生 U; 65l2f=3.404()1.903()IOOOS; 1000 202=2.379()% % = ; (USf % + UST % -42.3 %)= 1 1 42 % = ; (UST % 0S2.3 % - - %) = -0.5S3% = -(Us3T % + US2.3 % - US- %
9、)= 7XTlX2 =IOO SNUs2%U;IOO ST3 一IOO SNBr 二IOO% 0%SN11 1212=XIOO 20-0.5 1212=80.526()IOO201212=-3.66()=XIOO 2043.31000 12123.46x20100(7;1001212= 51.244()2.957 106 (S)=4.726 IO5(S)等效电路图:1.903-j3.662-6 解:SPs2-3 = (l)2 %一3 = 4啜-3 = 1080(kW), GT= 4GL= 1380(kW) 3N& =1-3-2-3) = 312.5(kW)& = (+ k - -)= lZ5(
10、kW)% = g(%-3 + %.3 - .2)= l67.5(kW)312.52202IOoOSt1000902= 1.867()2 = 票=E)Rt 3=6.379()1067.5x2202IOOOS- - IOoOX 9()2S么2 3% = (,% = 2U 3% = 24.2,么3 I % = 2U1 I % = 37.2 d3N% % = g (US2 % + USX % -42_3 %)=11542 % = : (US2 % Us2.3 % - %) = -l5Us3% = (Us3 % + u* 3%一4 2%) = 25.7 Iox J x ,A 一11.5 2202 = 6
11、1.844()100 SN10090一.S2%u;A n -1.5 2202 =X=-8.067()100 SN10090_ -S3% U; T425.7 2202 =X= 138.209()100 SN10090GT庶 104100 叫 1000 2202= 2.149x10-6(S)B 二-%SN IOo0.65901002202= 1.209 IO5(S)等效电路图:0.075 -j8.0676379+jl3S2091.867 +j61.844C0(2.149-jl2.09) IO-6S2-7 解:(1)以22OkV电压作为基本级RTPSUtIOOOS. 620x242211000 X
12、2402X = 4% /二 14/422 IOOSN 100 240G 二一二 185, IoOO 1000 2422B _ _ %SN 0.5x240 _TL 10(Wn - 100 2422 34.162()= 3.159106(S)2.049 105 (S)Ps2-3= ()2-3= = 592(kW), k = 4% = 880(kW)% = %匕.2 + %一3 一 之一3)= 4835(kW)& T+ -匕.3)= 1955(kW)% = ;(&-3 G2.3 一 匕-2)= 3965(kW)RrI =Rt2483.5x22021000S 10001802195.5x22()2 I
13、OOOS; IOoOXI 8()2 396.5x22()2=0.722()=0.292()IOOOS;10001802=0.592()Us% = JUSl一2% + Us3T% 一 42_3%)= 14.95Us2% = ; (USJ2% + 0S2-3% - Us3T%) = -0.95XTl100 SN14.95 2202、= 61.844()100180X2Us2%U1-0.95 2202S/。qqc100 SNX乙 QD4HSZJ100180X39.05 2202 C 、=X= 24.334()100 180100 SNGt22- 4 1 AOylrr6qIOo-*+. 11V ( D
14、)1000 2202Us3 = - (Us3 % + us2-3 % - US12 %)= 9.05B20%5n _ 0.8180 100(72 -ioo22O2= 2.975 105 (S)Ll: R1 = rll = 0.08 60 = 4.8()X1 二戈=0.417x60= 25.02(。)B1 = / = 2.91 106 60 = 1.746 X104 (S)22()L2: R2 =0.13250()2 =21.818()X2 = 0.405 50 ()2 = 66.942()220B2 =2.7610650(-)2 =4.1745 IO5 (S)等效电路:0.63 + j34.l
15、621(4.8 + j25.O2)r-=)-L jI.746104S :U1s(2)选取SB=I0MVA,4= 22OkV, Zb =- = 48.4(), Yb =- = 0.02066(S) SBUgT1: R;=殳= 0.013 X;=丘= 0.706,= il = 1.529 104, = = 9.918104ZBZBYq%:杵=殳 = 0.015 g=限= 0.006 =% = 0.012ZBZBZBX=&L = 0.831, X = 0.13250 = 0.499Uii1152X; =x2- = 0.40550 = 1.531 UlH52U21 1 C2B* = b4 = 2.76
16、 106 50 = 1.825 103-2 SB1000等效电路为:E。S+e0TO-XJ-Te0.011 + j.583- (0.091 + j.473)To-Xa_迩一) InTI-C)T j9.236IO3 -2-8解:(1)计算线路参数:7: RTPSUk _ 4101212XT =GTlBTl =10005; Us%U100 SN凡100(Wn,0%SN _IOoU;-,=0.417() 10001202=X = 12.811()100 12099 4,:7 = 6.789 10-6 (S) 100012120.61201001212PsW4.91810 5(S)300l IO2IO
17、OOS;2000X 6320.457()4I % = ; (UST % US3T % -42.3 %)= 1 42 % = J (US2 % + US2.3 % - dm % ) = 0.543% = (S3-I % US2.3% - %) = 6XTl 二100 SNX2IOO SX3Us3%UIOO SN10 HO2 X 100 630.5 HO2 =100 636 HO2 =X100 63= 19.206()=0.96()li.524()84.7Br2IOO喏 /0 %SN 100(721000 IlO27 106 (S)急爱= 6.248XmlvL2:lv叫=1 =0.422x80 =
18、 33.76(Q)X1 = x11 =0.42980 = 34.32()B1 =11 = 2.66 i 06 80 = 2.128 104 (S)R2 = r2l2 = 0.422 50 = 21.1()X2 = x2l2 =0.42950 = 21.45()B2 =b2l2 =2.6610-6 50 = 1.33l04(S)R3 =*=0.42260 = 25.32()X3 = x3l3 = 0.42960 = 25.74()B3 = b3l3 = 2.66 XIO-6 X 60 = 1.596 X10-4 (S)RT =PsU1 148i IO2XtGt3 =B3 =IOO SN4Io(W
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