電力系統(tǒng)分析理論課本習(xí)題MATLAB做

1、%例題3- 111=80rl=0.21xl =0.416b= 2 74/100 0 000vn= 1 1 0Sl=l 5dp0=40. 5dps=l 2 8v s =10. 5i0=3. 5sld b =30+1 2isldc=20+ 1 5i%(1)計(jì)算參數(shù)并作岀等值電路rl= 0 .5* 1 l*rlxl= 0.5* 1 l*xlb c=2*ll*bdqb=-0.5* b c*v n *vnrt=05*dps *vnA2 / 10 00 / S 1 八 2xt=O.5*vs*vnA2/l 0 0/S 1dq0=i0*Sl/100dpl=2*dp0/ 1 000d q l=2*d q 0 *

2、 is 0 = d pl+dq 1sb=s 1 db+s0+dqb*isc = sl d c%計(jì)算由母線A輸出的功率dst=(a b s (sc)/vn廠2*( r t + xt*i)%變壓器繞組中的功率損耗為s cl= s c +dsts 1 1 =s c 1 + s bdsl=( a b s (si 1) /vn ) A2*(rl+x 1 * i ) % 線路中的功率損耗為s 1 =sll+ds 1sa=sl+dq b* i%由母線A輸出的功率為%(3)計(jì)算各節(jié)點(diǎn)電壓pl= 5 2.54ql= 32.1v a =117%線路中電壓降落的縱分量和橫分量分別為：dvl=( p l*r I +q

3、l*xl)/v apvl=(p 1 *x 1 ql 次 r 1) / v avb=s q r t(va-dvl)八 2+( p vl廠 2)%b 點(diǎn)電壓為pc=20.18qc= 1 79vcc=ll%變壓器中電壓降落的縱分量和橫分量分別為：dvt=( pc*rt+q c *xt) / vbp v t=(pc * xt-qc*rt) / vbv c 1 =sqrt (vb-dvt)A 2 +(pv t )A2)%歸算到高壓側(cè)的c點(diǎn)電壓vc=v c 1 *vcc/vn%變電所低壓母線c的實(shí)際電壓1 1 =80rl =0.2100xl =0 4160b =2.7400e-0 0 6v n =110S

4、I =1 5dp 0 =40.50 0 0dps =12 8v s =10. 5 0 00i 0=3.50 0 0s 1 db = 3 0. 0 0 00+12. OOOOiside = 2 0.00 0 0 +15.0 0 00ir 1 =8.4 0 00xl=16.6 400be =4.3 8 40 e -004dqb =- 2 .6523rt = 3.4418x t =42.35 0 0dqO= 0.52 5 0dpi = 0.0810dq I =0 + 1.0 500isO =0. 0 81 0 + 1.0 500isb =30.0 8 10+10.397 7 isc =20.00 0

5、 0+ 15.OOOOidst =0. 1778 + 2.1 8 75isc 1 =2 0778+1 78 75isll =50 2 588 + 2 7.58 5 2 idsl =2.2818 + 4.520 1 is 1=52.5406 +32.10 5 3 is a =52.540 6 +29.4530ipl =52.540 0q I =32. 1000va=ll 7d vl = 8 337 4p vl =5.1677vb =10&78 5 4pc = 2 0.1800qc=17.1900vcc =11d v t =7.3305pvt =7.3122vc 1 =1 0 1.7180 vc=

6、10.1718 %例題32s2 = 0 3+02is 3 =0.5+ 0 .3is 4 =0.2+0. 1 5iZ12=1.2+2.4iz23=l.0+2.0 iz24= 1 .5 + 3.0 ivn =10% (1)求線路始端功率d s 23=( a bs (s3 ) A2/vn 2) *z23ds24= (abs(s 4 )A2/vnA 2 )*z 2 4s23=s3+ds23s24=s4+d s 24sll2=s 23+s2 4+s2dsl2=(abs(sll2)/vn)人2* z 1 2sl2=sll 2+ds 1 2% (2)求線路各點(diǎn)電壓dvl2=(r e al(sl2) *re

7、a 1 ( z 1 2 )+ima g (sl2)*imag (z 1 2) )/(1.0 5*vn)v2 = 1 .05*vn-dvl2d v 24=(re a l(s 2 4)* real ( z 24)+ i mag( s 2 4)*ima g (z24)/v2dv23 = %例題33vn=101 1 =101 2=413 = 314=2v a = 1 0.5vb=l 0 4z 1=0.17+0.3 8*iz 2=0. 45+0.4 *i%的線路等值阻抗zl 1 =10* z 1Z12 = 4 *zlz 13=3 * z 1zl4= 2 *z2%求（2和D點(diǎn)的運(yùn)算負(fù)荷，為s c 1 =2

8、 6 0 0+1600引s e =3 0 0+160* isd 1=600+2 0 0引s d2= 1 60 0 + 1 0 0 0*i d sce=(a b s (s e / 1000) / v n)A2* z 14*1000sc=s c 1+ s e +dscesd=s d l+sd2%循環(huán)功率z 1 1 =0. 1 7 一 0 38*is cc=100 0 *(va-v b ) * v n /( 1 7 * z 1 1)s a c= (real( s c)*7+ i mag ( s c ) * 7 * i +real ( s d )*3+imag(sd)*3 * i)/1 7+sc c

9、sb d =(r e al (sc)*10+imag(sc)* 1 0*i+r e al (sd)* 1 44- i ma g (sd)* 1 4 * i ) /17-scc s 1 = s ac+ s bds2=s c + s ds d c=s b d sd%C點(diǎn)位功率分點(diǎn)，可推測出E點(diǎn)電壓最低點(diǎn)，進(jìn)一步可求得E點(diǎn)電壓dsac= (abs( sac/ 1000) / vn)A2* z 11*1000sla c =sac+ d sacdv a c=( r e al (sla c/100 0 )*re a 1( z 1 1 )+ i mag(s 1 a c / 1 0 0 0) *im a g(

10、 z ll)/va vc=va d v a csce=se+dsc ed v c e =( r eal(sce / 1000) *real(zl4)+ i mag( s ce / 10 0 0) *im a g (z I 4) /vc ve=v c dvcevn =1011=101 2=41 3 =314=2va=10.5 0 0 0vb =10.40 0 0zl =0. 1 700 +0 3 8 00iz2 =0.4 500+ 0 4000izll =1.70 0 0 + 3 .80 0 0 iZ12 =0.680 0 + 1.520 OiZ13 =0.510 0+1 14 0 0iz 1

11、4 = 0 900 0 + 0. 8 0 OOiscl =2.6 000C+003 +1.600 0 e+00 3 ise =3.0000e+ 0 02+1.6 000e+002isd 1 =6. 0 000e + 0 02 +2.0 0 00e+002 is d2 = 1 .60 0 0 e+0 0 3+1.0 0 0 0e+ 0 03 idsce = 1 .0404 + 0.9248 isc = 2.901 Oe+OO3 +1.7 6 09e+ 0 03isd=2 2 0 00e+0 0 3 +1. 200 0 e+00 3 izll =0. 1 70 0 0. 3 800is c c =

12、5.77O3e+O 0 1 + 1.2 8 98e+002isac= 1 640 5 e+003 +1.065 8 e+0 0 3isbd =3.4 6 06e+0 0 3 +L 8 9 51e+ 003isi =5. 1010 e+003+2.96 0 9c+0 0 3 is2 =5.1010e+0 0 3 +2. 9 6 0 9e+003 isdc = 1 2 606e+003 +6.9509e+002idsac =6.5062 e +001 + 1.4543e+002is lac =1. 7 0 55e+ 0 03 +1.211 3e+00 3i dvac = 0 .7145vc =9.

13、7855see =3.0 10 4e+00 2 +1. 60 9 2 e +002idvee =0.04 0 8ve =9. 7 4 47 %例題34rl=O 27xl=0.4 2 3bl =2.69 / 1 0 0000 0r 2=rlx2=xlb2 =blr 3=0.45x 3 =0.44b3=2 58/1 0 0 0 00011=6012=5013=40vn=ll 0snb=20d sOb= 0 05+0.6引rtb=4. 8 4xtb=63.5snc= 1 0d sO c =0. 0 3+ 0 .3 5 * irt c =11 4x t c=127sldb=24+ 1 8*is 1 d

14、c=12+9*i% (Dil-算網(wǎng)絡(luò)參數(shù)及制左等值電路z 1 =1 1 * ( r 1+xl* i )%線路 1B 1 =1 1 *bldq b 1=-B I *vnA2/2Z2=12 * ( r 2+x2*i )%線路 2B2=12*b2dqb2 =-B2* vnA2/2Z3=13*(r3+x3*i)%線路 3B3=13*b 3d qb3=-B3*vn 2/2z t b=(rt b +xtb*i)/2%變電所 bd S 0b=2*ds0bztc=( r tc+xtc*i)/2 %變電所 cdS0c=2* d sOc弘計(jì)算節(jié)點(diǎn)b和c得運(yùn)算負(fù)荷d s t b=(a b s(s 1 db)/vn)

15、A2* z tbs b =sld b +ds t b+ d SOb+dq b 1 * i + d qb 3 * id stc=(a b s (sldc)/vn) * 2* z t csc=s 1 dc+ds t c + d SOc+dqb3*i+dq b 2* i%(3)計(jì)算閉式網(wǎng)絡(luò)的功率分布s l=(sb*( c o nj (Z2)+ c on j (Z3)+sc*conj(Z2) /( c on j (Zl) +conj (Z2)+co nj (Z3)s 2=( s c*(c o nj(Z 1 ) +conj( Z 3 )+sb*conj( Z 1 ) / (conj(Zl)+con j

16、(Z2) +c o nj(Z3)s 1 2=sl+s2sbc=sb+scs3=s b -si%(4)計(jì)算電壓損耗dsll=(ab s (si)/ vn) * 2 *Z 1sal=sl+dsl 1va=117d v 1 = ( r eal(sal)*real( Z l)+imag (sal)*im a g (Zl) /vav b =vadvlrl =0.27 0 0xl =0.4230bl =2.6 9 00e-006r2 = 0 .270 0x 2 =0.42 3 0b 2=2.6900 e-0 06r3 =0. 45 0 0x3 =0.4400b3 =2. 5 8 0 0 e -0061 1

17、 =601 2 =5013 =40vn =110s n b =20dsOb =0.05 00+ 0 .600 0 irtb =4. 8400xtb =63. 5 0 0 0s nc = 1 0dsOc = 0 030 0 + 0 35 0 0 ir tc =11. 4 00 0xtc =127sldb = 2 4. 0 000 +18.0000is 1 de =12.00 0 0 + 9.0 0 OOiZ1 =16.2000 +25.3 800 iBl = 1 6 14 0 e-0 0 4dqb 1 =- 0 .9765Z2 = 1 3.50 0 0+21.15001B2 =1. 3450 e

18、 - 004dqb 2 =-0.8 137Z3 = l&OOOO +17.6 000 iB3 = 1 .0320e-004dqb3 =-0.6244ztb =2.4 200 + 3 1.7 50 OidSO b =0. 1 0 00 + 1 .200 0 iztc = 5 .7000 +6 3 .5000idSOc =0.0600 +0 7000ids t b =0.1800 + 2 36 16isb =2 4.2 8 00+19 9607idstc = 0.1 060+18 0 8isc=l 26 60+ 9. 4 4 2 7isi = 1 8. 64 1 0 +15.7 9 72is2 =

19、1 7.8 05 0 +13.60 6 2 i si 2 =36.4 4 60 +29 4 0 34is be =36. 4 46 0+29 4034is3 =5.63 9 0 +4.163 5 id s 1 1 =0. 7 9 9 3 + 1.2 523isa 1 =19.4404+17.0 4 95iv a = 1 17dvl =6. 3 902v b = 1 10.6 0 9 8 %例題35kl= 1 10/11k2= 1 15.5 / 11z t 1 = 1 *izt2= 1 *ivb=10% (1)計(jì)算變壓器的功率分布sld= 1 6+12*is 1 I d=sld/2s2 1d=s

20、 lid%(2)求循環(huán)功率dE=vb * (k2/k 1 -1)%故循環(huán)功率為sc=(v b * d E)/( c on j ( z t 1 )4-conj(z t 2)%(3)計(jì)算倆臺(tái)變壓器的實(shí)際功率分布st 1 =slld+s cst2=s 2 Id- s c%計(jì)算髙壓側(cè)電壓vat 1 =(vb+(imag(st 1 ) *imag( z 11 ) )/vb)* k 1%按變壓器T 1 計(jì)算vat2=(vb+(ima g (st2)*im a g (z t 2) / v b ) * k2%按變壓器 T-2 計(jì)算%計(jì)及電壓降落的橫分量，按T-1和T-2計(jì)算可分別得va t 1 = 10&7

21、9v a 0=109%(5)計(jì)算從高壓母線輸入變壓器T一 1和T-2的功率st 1 1= s tl+ (abs (st 1 )/ v b)2 * zt 1stl2=st 2 +(ab s ( s t2)/v b ) 2 * z t 2%輸入髙壓母線的總功率為s=stll+s t 12kl =10k2 =10. 50 0 0z t 1 =0 + 1 OOOOizt2 =0 +1 .OOOOivb =10sld =1 6.0 000+12.0 000 islid =& 0000 + 6. 0 000 is21d =&0 0 00 + 6.00 0 Oid E =0.500 0sc = 0 + 2

22、5 0 0 0 is tl =&0000+ 8 .5 0 OOist2 =&0000 + 3.5 0 OOivatl =1 0 8.5000vat2 =1 0 8. 67 5 0va t 1 =10 8.79 0 0vat2 = 10 9stll = 8 .0000 + 9 .8625 is t 12 = 8.0000 + 4 2625is= 1 6.0000+14.1 2 50i%例題36 s=40+30*i c o sa=O 8Tm a x= 4 5 0 0r=0.17x= 0 .409b=2. 8 2/1 0 0 0000d p 0=86 d p s=200 10=2.7 Vs=10.5

23、 s n=3 1 .5%最大負(fù)荷時(shí)變壓器的繞組功率損耗為d s t = 2 *(dp s +10 0 0*Vs * s n * i /100) *(real(s) / 0.8/ 2 /sn) A2 dsO=2 *(dpO+l 0 0 0 *10* sn*i/100)%變壓器的鐵芯功率損耗為1=1 00 v=110qb 2 =-2*b* 1 * v A2/2%線路末端充電功率%等值電路中流過線路等值阻抗的功率為si = s +dst/1000+ d s 0/1000+ qb2 rt= r *1/2%線路上的有功功率損耗%變壓器全年的電能損耗%線路全年的電能損耗%輸電系統(tǒng)全年的總電能損耗d p l

24、=(a b s(sl) /v廠 2*r tT=8760t= 3 1 5 0dwt= 2 *dpO * T+re a l(dst)* td w 1= 1 OOO*dp 1 * td w =dw t +dwl s =40.0000 +30. 0 000ic o s a = 0 8 000T max = 4 500r = 0.1 7 00x =0.4090b =2.8200 e-0 06d p 0 = 8 6dps =20010 =2.7 0 0 0Vs =10.5000sn =31.5000ds t =2.5 1 95e+00 2+4.16 67e+00 3 idsO =1.72 0 0e+00

25、2 +1. 7 0 1 0 e+003i1 = 1 00v =1 1 0qb2 =3 4122si = 4 0.4 240 +3 2 .4555ir t = 8 . 500dpi =1.8879T =8760t =3150dw t =2. 3 0 0 4e+006dwl =5. 9 4 6 8 e + 0 06dw = 8 .2472e+ 0 06 %例題5-2zl=4.32+ 1 0.5*ivl=35v a =36vb=10Smax= 8 4-5* iSmin= 4 +3*izt= 0 .69+7. 84* i%變壓器阻抗與線路阻抗合并得等值阻抗z=zl+zt%線路首端輸送功率為Samax=

26、 S max+(ab s (Smax) / vl)A2* zSamin=Smi n +(abs(Smin)/v 1 )A2* z%B點(diǎn)折算到高壓側(cè)電壓為V lbmax=va (r eal(S a max) *real ( z )+ i ma g (Samax)* i mag (z)/v a Vlbm in=va(real(Samin)*real(z)+imag (Samin) *imag(z) / v a%最大和最小負(fù)荷時(shí)對應(yīng)的分接頭電壓V b max=0.95 * vbVbm i n=l. 05汝vbV2n=10. 5V tin a x=Vlbmax*V2n/V b maxVtm i n=V

27、lbmi n *V2n/V b min%取平均值Vt=(Vtmax+ V tm i n) / 2%選擇變壓器最接近的分接頭a=(Vt/v M) * 1 00%所以取25%的分接頭，即Vtl=(l-o. 0 25)*vl%按所選分接頭校驗(yàn)1 0KV母線的實(shí)際電壓Vbmax 1 = V lbmax* V 2 n / V t 1dvmax=(V b ma x I 10)/1 0Vbminl = V lbmi n*V2n/Vt 1 dvmax= (Vbmin 1-10)/10zl =4.3 2 0 0 + 1 0.5 0 OOivl =35v a =36vb = 1 0Sma x = 8 .0 0 0

28、0 + 5. OOOOiS m i n =4.0 0 0 0 + 3. 0 0 00 izt = O 6900 + 7 8400iz = 5. 0 100 4- 1 8 .3400iSama x =8.3640 + 6.33 2 5iS amin =4. 1 0 22 + 3 37 4 3iVI b max =31.6 1 0 0V 1 bmin =33. 7 10 1vbm a x = 9.50 0 0Vbmin =1 0 .5 0 0 0V2n = 1 0 .5 0 00V tmax = 3 4.9373Vtmi n =33.710 1V t =34. 32 3 7a =1. 9 322V

29、tl =34.125 0Vbmaxl =9.7 2 61d vmax =0.027 4V bm i n 1 =10.3 723dvm a x = 0 0 372 %例題5-3Sm a x=25+ 18iSmin=14+10 iZ t =3+30iV 1 max=120V 1 mi n =114Pm a x=real(Sm a x )Qm a x=imag( S max)Pmi n =rea 1 (Smin)Qm i n=ima g (Smin)R=r e a l(Zt)X=imag ( Z t)%最大負(fù)荷時(shí)變圧器的電壓降為d V max= (Pma x * R +Qmax*X)/V lmax%

30、規(guī)算至高電壓側(cè)的低電壓為V2ma x =V 1 max+dVm a x%最小負(fù)荷時(shí)變壓器的電壓降為d Vmin=( P m i n*R+Qmin*X)/V 1 m i n%規(guī)算至髙電壓側(cè)的低電壓為V2min=Vlmin+ d Vm i n%假立最大負(fù)荷時(shí)發(fā)電機(jī)電壓為6. 6 kv,最小負(fù)荷時(shí)電壓為6kvV 1 tma x =V2ma x *63/6.6Vltm i n=V 2 min* 6 .3/ 6Vlt=(Vltmax+ V 1 tmi n )/2%選擇最近的分接頭12 1 kv%校驗(yàn)：最大負(fù)荷時(shí)發(fā)電機(jī)端實(shí)際電壓為V2m a x *6.3/121%最大負(fù)荷時(shí)發(fā)電機(jī)端實(shí)際電壓為V2min*6

31、.3/1 21S max =2 5 .0 0 00 +1 8.00 0 OiSmin =14.00 0 0 +1 0 .00 OOiZt = 3 .0000 +30. OOOOiVI max = 1 20V 1 mi n =114Pmax =25Q max = 1 8Pm i n =14Q mi n =10R =3X =30d Vmax =5. 1 250V2ma x =125 . 1250dVmin =3V2min =1 1 7VI t ma x =11 9 .43 7Vltmin=l 2 2.8 5 00Vlt=121. 143 8ans =6. 5148ans = 6 .0 9 1 7

32、%例題5-4V 1 n =110V 2 n =35V3n=6Vn3=6.6Pl=12. 8Q 1 =9.6R 1 =2. 9 4Xl=65P2=6.4Q2=4. 8R2=4.4 2X2=-1.5P3=6. 4Q 3 =4.8R 3 =4. 4 2X3= 3 7.7V 1 m a x= 1 12Vlmin = ll5% (1)求最大、最小負(fù)荷時(shí)各繞組的電壓損耗dV lmax=(P 1 *R1+Q1 * X 1)/Vlmax%最大負(fù)荷時(shí)dV2ma x = ( P2* R2+Q 2*X 2)/(V lmax- d Vlma x)dV 3 ma x =(P3 *R 3 +Q3*X3)/(VImax-d

33、V 1 max)dVlmi (O.5*P1*R1+O.5*Q1*X1)/V I min %最小負(fù)荷時(shí)dV2mi n=(0 5*P2*R2+0. 5*Q2* X2)/ ( Vlmi ndVlmin)dV3m i n =(0. 5 *P3*R3+O 5*Q3* X3) / (V 1 min-dVlm i n)%(2)求最大、最小負(fù)荷時(shí)個(gè)母線電壓V 1 max= 1 12%最大負(fù)荷時(shí)V2maxl=Vlma x -dV 1 m a x d V 2 ma xV3maxl=V 1 max-dV1 m a x d V3max Vim i n=l 1 5%最小負(fù)荷時(shí)V2minl = V 1 min-dV 1

34、mindV2min V3minl=V 1 mi n dVlmi n d V 3 mi n %(3)選擇髙壓繞組分接頭V3max=V3n*( 1 + 0 )V3min=V3n*( 1 + 0.0 75)Vtl max =V3max 1 *V n 3/V 3 ma xVt 1 mi n =V3m i n 1 * V n 3/V3m i nV t 1 =(Vtlmax+V t lmin) / 2Vtl = ll 5.5%選用110+5%的分接頭%(4)校驗(yàn)低壓母線電壓V3ma x =V3maxl*Vn 3 /VtlV3min=V3min 1 *V n 3/Vtl%最大負(fù)荷時(shí)%最小負(fù)荷時(shí)掘低壓母線電壓

35、偏移d V 3max=(V3m a x-V3n)/V3n %最大負(fù)荷時(shí)dV3min=(V3min-V3n) / V3n%最小負(fù)荷時(shí)%(5)根據(jù)中壓母線的調(diào)壓要求，又高、中壓倆側(cè)，選擇中壓繞組的分接頭 %最大、最小負(fù)荷時(shí)中壓母線調(diào)壓要求電壓為V2ma x =V2n*( 1 + 0 )V2min =V2 n *(1+0.075)%最大、最小負(fù)荷時(shí)中壓繞組的分接頭電壓為V t 2max=V 2 max* V t 1 / V2max 1V t2m i n=V2min* V tl/V2minlV t 2 =(V t 2max+Vt 2 m i n) /2%于是就選電壓為3& 5 kV的主抽頭Vt2=3&

36、 5%(6)校驗(yàn)中壓側(cè)母線電壓V2max=V2max PVtWtl%最大負(fù)荷時(shí)V2min=V2min 1 *Vt2/Vtl% 最小負(fù)荷時(shí)%中壓母線電壓偏移：dV2m a x=( V 2m a x V2 n )/V2ndV2min=(V2min-V 2 n ) /V2 nVln=110V2n =3 5V3n =6Vn3 =6 60 0 0Pl = 1 2.8000QI =9.60 0 0R 1 =2 .9400XI =65P 2=6.4000Q2 =4.800 0R2 =4.4200X2 =- 1 5000P3 =6.4000Q3 =4. 8000R3 =4.4 2 00X3 =3 7.7000

37、Vim a x =112V 1 min = 1 15d V lmax =5.9 0 7 4dV 2 max =098 8 dV3m a x =1.9723 d V lmin =2. 8767 dV2m i n = 0 0 940 dV3min=0 9 331VI max = 1 12V 2maxl =105. 8 9 38V3max 1=104. 120 3V lmi n =115V2m i n 1 =1 1 2.0 2 93V3m i n 1 =111.1 9 02V 3 ma x =6V 3m i n =6.450 0Vtlmax =114. 5 3 2 3 Vt 1 min= 1 13.

38、7760V t 1 =114.1542Vtl = 1 15. 5 0 0 0V 3max = 5 .9497V 3 m i n = 6.3 537 dV3max =-0.0 0 84d V3mi n = 0 . 0 590 V2m a x =35V2min =37.6 2 5V t2max =38 1750Vt 2 m i n =38. 7 90 6Vt2 =38. 4 828V t 2 =38. 5 000V2m ax =35.2 979V 2 min =37.3431dV2max =0.00 8 5 dV2min =0.0 6 69%例題7-3Sgl = l 0 0,Xlld=0 183,

39、cosal 二 0 85Sg 2 =50.X22 d =0.14 1 , cosa 2 =0.8S t 1 =120,Vsl= 1 4 . 2St2=6 3 , V s 2 = 1 4.511 = 170 , x 1=0.4 271 2 = 1 20, x 2 = 0 .43213=100, x 3=0. 4 32SId=l 6 0 ,v n 1 =230 % (1)各電抗標(biāo)幺值SB=1 0 0, VB=230,x 1 =0.35, E3 = 0. 8)%發(fā)電機(jī)G1%發(fā)電機(jī)G-2%變壓器TJ%變壓器T-2%線路L-l%線路2%線路L-3X 1 =Xlld*SB/(Sg 1/co s a 1 X

40、 2 =X22 d *SB / (Sg2/ c osa2) X3=x 1 *SB / S IdX4=V s 1* SB/lOO/Stl X5=Vs2* S B/ 1 0 0 / St2 X6=xl* 1 l*SB/vnlA2X7= x 2*12 *SB/vnlA2X 8 =x3* I 3*SB/vnl A2E 1 = 1.08E 2= 1 .08%簡化網(wǎng)絡(luò)%發(fā)電機(jī)G-l%發(fā)電機(jī)G-2%負(fù)荷LD%變壓器T-l%變壓器T2%線路L- 1%線路L-2%線路L-3%取發(fā)電機(jī)的次暫態(tài)電勢X9=X 1 +X4X10=X2+X5%將乂6 X7 X 8構(gòu)成的三角形化為星型XI 1=X6*X7/ (X6+X7+

41、X8)X 12=X6* X 8/(X 6+X 7+X8)X13=X7*X8/ (X6+X7+X8)%將己1、E2倆條有源支路并聯(lián)X 14=(X9+XU)* (X10+X 1 2)/(X9+Xll + X10+X12) +X13 E12=1.08%(3)計(jì)算起始暫態(tài)電流%有發(fā)電機(jī)提供的起始次暫態(tài)電流為Ib=E12/X14%由負(fù)荷LD提供的起始次暫態(tài)電流為ILDb=E 3/X3%短路點(diǎn)總的起始次暫態(tài)電流為If b =Ib+ILDb%基準(zhǔn)電流IB=SB/sq rt(3)/VB%有起始次暫態(tài)電流有名值為Ifb =863*IB%( 4 )計(jì)算沖擊電流%有發(fā)電機(jī)沖擊系數(shù)kim= 1 .8,綜合負(fù)荷LD沖擊

42、系數(shù)kimLD=l,短路點(diǎn)的沖擊電流為 k i m= 1 &k i mLD=li im=(kim*Ib *sqrt(2)+kimLD*ILD b *sqrt( 2 )*IBSgl =100X 1 1 d =0. 1 830cosal =0.8 5 00Sg 2=50X2 2d =0. 1 410cosa2 =0.8000St 1 =120Vs 1 =14.20 0 0St 2 =6 3Vs 2 = 1 4. 500 011 =170xl=0 427 01 2 =120x2 =0. 432013=10 0x 3 =0.4320SId=l 60v n 1 =23 0SB =100VB =2 30x

43、 I =0.350 0E3 =0. 8 0 00XI =0.155 5X2 =0.22 5 6X 3 =0.21 8 8X4=0183X 5 =0.23 0 2X6=0.1372X7 =0.0980X8 =0.0 8 1 7E 1 = 1 .080 0E2 =1.0800X9 =0. 2739X10= 0 .455 8Xll=0. 04 2 4 X12 =0.035 4X13 =0.0253X14=0. 2 1 77E12 =1.08 0 0lb =4.9620I LDb = 3 .657 1Ifb = 8.6 1 9 1I B =0.2 5 1 0Ifb =2. 1663 kim = 1 .

44、8000 kimLD =1i im = 4 .4690 %例題8- 1SG=50,COSg=0 8,Xdb=0 1 5,X2=0.18,El=1.08ST1=6O.VS= 1 0. 5 ,xn=22SLD = 1 5,X 1 D=1.2,X2D=0 3 5L=50 xl=0. 4 , x 0 =3*xl%(1)各元件參數(shù)標(biāo)幺值計(jì)算%發(fā)電機(jī)G%變圧器TJ、T-2%負(fù)荷LD%輸電線路LSB = 100XGl=Xdb*SB/(SG/C OSg),XG 2 =X2*SB/(SG/C0Sg) %發(fā)電機(jī)X T 1=(VS /1OO)*(SB/ST1)XT2=XT1v = 37%變壓器TJ變壓器T2Xn=x

45、n* SB/ vA2XLD1=X 1 D*SB/SLD.XLD2=X2D * S B/SLD,XLl=L*xI*100/vA2 , XL0=3*XL 1%(2)制泄各序網(wǎng)絡(luò)(見課本222頁)%(3)網(wǎng)絡(luò)化簡，求組合電勢和各序組合電抗E = E1*X LD1/(XG 1 +XLD1)Xa=XGl*XL DI /(XG1+XLD 1 )+XT 1 +X L 1Xb=X G2*XLD2/(XG 2+XL D 2)+ XT1+XL1%中性點(diǎn)接地電阻 %負(fù)荷LD%輸電線路L%正序電抗%負(fù)序電抗X c =(XT1 + 3*xn+XLO)*XT2/(XTl+3*xn+ XL0+XT2) %零序電抗SG =5

46、0COS g = 0 800 0X db =0.1 5 00X2 =080 0El =1. 0 80 0STI =6 0VS =10.5 000 xn = 2 2SLD =1 5X1D = 1.2 000X2D =0.35 0 0L=5 0x 1 =0. 400 0 xO =1.2000SB =1 00XG1 =0.2 4 00XG2 =0. 2 880XT1 =0.1 7 5 0X T2 =O 1750v =37Xn =1.6070XLD1 =8XLD2 = 2.33 33XL1 = 4609XLO =4. 382 8E =1.0485Xa=l. 86 8 9X b =1.8 9 23Xc

47、=O 1746 %例題8 -2Sgl=l 0 0c o s 1 = 0 .8 5Xd 1 =0.18 3X12=O2 23S g 2= 5 0 c o s2=0. 8Xd2=0.141Xx2 2=0.172St 1 = 1 2 0Vs 1 = 14.2St 2 =63Vs2=14.5Ll = l 20X1=0 4 32X0=5*X 1vb=230Sb=l 00X g 1 1 =Xdl*Sb/(S g 1/c o s 1 )X t ll=Vsl*Sb/ (100* St 1 ) Xlll=0.5*Ll*Xl*Sb/Vb 八 2Xt2 l=Vs2*Sb /(100*St2) Xg21=Xd2*Sb

48、/(Sg2/cos2) Xgl2=X 1 2* Sb/(S g 1/cosl)X t 12=Vs 1 * Sb/( 1 0O*S t 1 ) X112=0.5*Ll*Xl*Sb/Vb2Xt22=Vs2*Sb/(100*St2)Xg22=X t 22 * Sb / (Sg2 / cos 2 )X10=5 * X 1 11Xt 1 0=Vsl*Sb/ ( 1 0001)Xt2 0 = Vs2*Sb/(100*S t 2)% (1)制定各序等值電路Xl= ( Xg 1 1+Xt 1 1+X1 1 1 )*(Xt2 1+Xg21) /(Xgl 1 +X11I+X I 1 1) +(Xt21+Xg2 D

49、)X2=(Xgl2+X t 12+X112)*(X t 22+Xg2 2)/( ( Xgl2+Xtl2+Xl 1 2)+(X t 22+Xg 2 2) X0= (XIO+Xt 1 0)*Xt2 0 /(X10+X t 10) +Xt20)%( 2 )計(jì)算各種不對稱短路時(shí)的短路電流Xd 1 = X 2 +X0%單相接地短路m 1=3Ed=lIal l=Ed/ (Xl+Xdl)%基準(zhǔn)電流%倆相短路%:倆相短路接地I b =Sb/ (sqrt (3)*Vb)Ifl=ml*Ial 1 *IbX d 2=X 2m2=sqrt (3)I a I 2=Ed/ (X 1 +Xd2If2=m2*Ial2*IbX

50、dll=X 2 *X0/(X2+X 0 ) m 1 l=sqr t (3)*sqrt ( 1 -(X 2 *X0/(X2+X0)A2)I al 1 l = Ed/ (Xl+Xdi 1 )Ifl 1 =mll*Ia 1 1 1 * lbS g 1 =1 00 c o s 1 = 0 8500Xdl =0. 1 830XI 2 = 0 .22 3 0Sg2 = 50c os 2 = 0.8000X d 2 =0. 1 410Xx22 = 0.1720S t 1 =120Vsl =1 4 .2000St2 = 6 3V s 2 =14. 5 0 0 0LI =120XI =0. 4 32 0X0=2.16 0 0Vb =230Sb =100Xgll =0. 1 555X t 11 =0.1 1 83Xlll=0 .0490Xt21 =0.2 3 02Xg2 1 =0.2 2 56Xgl2 =0.18 9 6Xtl2 =0. 1 1 83X 1 1 2 =0.049 0Xt22 =0.2302Xg2 2 =0.3 683XI 0 =0.24 5 0XtlO=O .1183Xt20 =0 230 2Xl = 089 0X2 =0. 2236X0=04 0 9Xd 1=0.3