電力系統(tǒng)潮流短路-計(jì)算課程設(shè)計(jì)

1、一、 設(shè)計(jì)題目:1 潮流計(jì)算用圖21（5W8。5kW，uk=10.5,I0=0.9；。76kW,uk=10。5，I0=1。1；變壓器 T42SFL116000/110(11022.5）/10.5Pk=62kW，P0=11.6kW，uk=10.5，I0=1.1。電網(wǎng)潮流計(jì)算(1)計(jì)算各元件參數(shù)，畫出等值電路;（2）進(jìn)行網(wǎng)絡(luò)潮流計(jì)算；(3）不滿足供電要求,進(jìn)行調(diào)壓計(jì)算。二、 題目分析：這是一道潮流計(jì)算題，依據(jù)一般潮流計(jì)算的步驟將元件轉(zhuǎn)換為等值參數(shù)，這里我們進(jìn)行真實(shí)值的直接計(jì)算，并用近似計(jì)算計(jì)算 .由于負(fù)載給出，線路長(zhǎng)度已知，我們可以將如圖閉環(huán)的潮流計(jì)算4的必要.三、 潮流計(jì)算過程：(一) 各元件參

2、數(shù)計(jì)算：120Km1R= r l= 0.21 120 = 25.21011X= x l= 0.4120 = 48101B= b l10= 2.8 106 120S = 3.36 104S100Km2R= r l= 0.21 100 = 212022X= x l= 0.4 100 = 402022B= b l= 2.8 106 100S = 2.8 104S20270Km3R= r l= 0.21 70 = 14.73033X= x l= 0.4 70 = 283033B= b l= 2.8 106 70S = 1.96 104S303T,T12P U2110 1212RT1 = RT2 =kN

3、103 = 103 = 6.188S2160002NXT1 = XT2= UK(%)UN2100SN103 =10.5 1212100 16000 103 = 96.082S= P+ jI0%S= (18.5 + j0.9 16000KVA10100N100)= (0.0185+ j0.144)MVAT3RT3 =P U2kN52 1102 103 = 103 = 9.831SNUK(%)U8000210.5 1102XT3 =N100S103 =100 8000 103 = 158.813NS= P + jI0%S= (9.76 + j1.1 8000KVA30100N100)= (0.00

4、976+ j0.0088)MVAT4RT4 =P U2kN62 1102 103 = 103 = 2.93SNUK(%)U16000210.5 1102XT4 =N100S103 =100 16000 103 = 79.406NS= P+ jI0%S= (11.6 + j1.1 16000KVA40100N100)= (0.0116+ j0.176)MVA(二) 繪制等效電路：(三) 功率分布計(jì)算：1、各元件功率損耗： 兩臺(tái)T4變壓器并聯(lián)損耗：So= 2S0= 0.232+ j0.352SS2(1R+1X) =10.22 + 6.22 (1.465+ j34.705)T4UN222T41102

5、= (0.0173+ j0.468)MVAT3S=S2(R+X)=5.62 + 4.42 (9.831 + j158.813)T3UN2T31102= (0.0412+ j0.666)MVA100Km70Km4 末端功率損耗S= S4+SLD4+ So= (0.0173+ 10.2+ 0.0232)+ j(0.468+ 6.2 + 0.352)MVA= (10.2405+ 7.02)MVA120Km100Km3S= S+S+S3T3LD30= (0.0412+ 0.00976+ 5.6)+ j(0.666+ 0.088+ 4.4)MVA= (5.651+ j5.154)MVA1。4100KmS

6、LD1474= 17S4=7 (10.2405+ j7.02) = (4.217+ j2.891)MVA17SLD14 = SLD14SjB2 U2N2= (4.217+ j1.197)MVABS ( ) = S+LD14(R+X)j2 U21 1LD14UN2222N= (4.217+ j1.197)+ (0.0333 j1.6305)MVA= (4.2503 j0.433)MVA1。3120KmSLD13= 10S22= 10(5.651+ j5.154) = (2.569 + j2.343)MVA22SLD13 = SLD13SLD132BNj1 UN2= (2.569+j0.3102)

7、MVAB1S (2) =1S+LD13UN2(R1 + X1)j1 U22N= (2.596+ j0.3102)+(0.0139+j0.0266) j2.0328MVA= (2.61 j1.696)MVA2.470KmS= 10S= 10(10.2405+ j7.02) = (6.0238+ j4.1294)MVALD24S174 = SjB4 U2 = (6.0238+j2.9436)MVALD24LD242NS= S + SLD242 (R+ X ) jB4 U22(1)LD24UN2442N= (6.0238+ j2.9436)+ (0.0546+ j0.104) j1.1856MVA=

8、 (6.0784+ j1.862)MVA2。3100KmSLD24= 12S22= 12(5.651 + j5.154) = (3.082+ j2.811)MVA22SLD23 = SLD23jB3 U22N= (3.082+ j1.1172)MVAS= S + SLD232 (R+X ) jB3 U22(2)LD23UN2332N= (3.082+ j1.1172)+ (0.0187 + j0.0355) j1.694MVA= (3.101j0.5413)MVA1S= S ( ) +S () = (4.2503 j0.433)+ (2.61 j1.696)11 11 2= (6.8603j2

9、.129)MVA2( )( )S= S+ S= (6.0784+ j1.862)+ ( )( )22 12 2= (9.1794+ j1.3207)MVA(四) 調(diào)壓計(jì)算1.1。4S = S 2 = SS2+LD14+X )1(1)1(1)2NLD14UN222= (4.2503+ j1.261)MVA4G1U41= UU1P= U1(1)R2 + Q1U1(1)X2= 121 4.2503 21+ 1.261 40121= 121 1.155= 119.84KV119.84110（閱歷3）121KV110T (1212 2.5%)/6.3 121 (15%) =1114.95KvT (110

10、2 2.5%)/10.5 110 (15%) =4115.5Kv調(diào)壓后：= UU1P= U1(1)R2 + Q1U1(1)X2= 114.95 4.250321+ 1.261 40114.95= 113.73KV 115.5= 114.95 1.155所以滿足調(diào)壓關(guān)系。閱歷證其他三路均滿足，調(diào)壓成功。結(jié)論：將變壓器 T，T 置于(121 22.5%)和12將變壓器 T，T 置于(110 + 2 2.5%)調(diào)壓檔可滿足條件。34一、 設(shè)計(jì)題目：(1）繪出等值網(wǎng)絡(luò)并化簡(jiǎn)；（2）計(jì)算各元件的標(biāo)幺值，求出計(jì)算電抗；（30s、0。2s、0。5s二、 題目分析：電流的思路,考慮到有兩個(gè)短路點(diǎn)同時(shí)作用，需進(jìn)

11、行分解分析,F1 發(fā).F2于無窮大系統(tǒng)距離短路點(diǎn)較遠(yuǎn)，所以產(chǎn)生的影響在下面計(jì)算忽視不計(jì).二、短路計(jì)算過程：取Sd = 100MVA Ud = Uav(一) 各元件電抗的標(biāo)幺值：1G ，G2:X11 = X2= 0.14110080= 0.1763G ，G4:X33 = X4= 0.1410060= 0.2335G ，G6:X55 = X6= 0.142100100= 0.1421T ，T2:X17 = X8= 10.5 10010090= 0.1173T ，T4:X39 = X10= 10.5 10010063= 0.1675T ，T6:X511 = X12= 10.5 100100120=

12、0.0875LX1= 0.441601001152= 0.532L2 X14= 0.44 80 1001152= 0.266L3 X15= 0.44 1101001152= 0.366X16= XL(%) 100UN3INSdUd2= 0.07103 1.51006.32= 0.679(二) 繪制等效電路圖(三) 化簡(jiǎn)等效電路圖X17= 1(X2+X8) +X14= 1(0.176+ 0.117) + 0.266 = 0.4132X18= 1(X2+X11)+X15= 1(0.0875 + 0.142) + 0.366 = 0.4812X19 = (X9 + X3) = (0.167+ 0.2

13、33) = 0.4X20 = X3 + X16 = 0.912X21 = X4 = 0.233(四)計(jì)算各電源電抗:(2)G:X1.2= 0.413 802c1100= 0.661G X3c2 = 0.460100= 0.24G5.6 Xc3= 0.481 1002 = 0.961100(1)G X3G X4c4 = 0.912c5 = 0.2336010060100= 0.547= 0.140(五)查計(jì)算曲線數(shù)字表，求電流周期重量的標(biāo)幺值短路計(jì)算時(shí)間電流值供應(yīng)短路電流的機(jī)組短路點(diǎn)總電流2短路計(jì)算時(shí)間電流值供應(yīng)短路電流的機(jī)組短路點(diǎn)總電流2G1.2(0.661)G3(0.24)G5.6(0.96

14、7)標(biāo)幺值1.6334。5261.105出名值1.3121.3631.331標(biāo)幺值1。5303。2861.090出名值1。2290。9901。313標(biāo)幺值1.6322。8161.145出名值1.3110.8481.380KA04.0060.23.532KA04.0060.23.5320.53.539歸算至短路點(diǎn)電壓級(jí)的各等值電源的額定電流和基準(zhǔn)電流IN(12)=IN1=2 803 115= 0.833KAIN(3)=603 115= 0.3012KAIN(56)=IN5+I6=2 1203 115= 1.2049KA基準(zhǔn)電流:I3115= 0.3012KA1 點(diǎn)短路計(jì)算結(jié)果：1G (0.547

15、)G4(0.14)標(biāo)幺值1.9137.718出名值0。5762。325標(biāo)幺值1.6654.878出名值0.5011。4691。5810.4763。8291。153間間電流值供應(yīng)短路電流的機(jī)組短路點(diǎn)總電流302.9010。21.9100.51。629IN(3)= IN(3)=603 115= 0.3012KA四、 結(jié)論(期間工作).與同學(xué)探討做題思路、方法，初步得出幾輪.Word輸入公式，計(jì)算結(jié)果，應(yīng)用AutoCadWord文檔。增加文字?jǐn)⑹觯虐嫘?duì)，打印出稿。參考資料1.,2008.22、水利電力部西北電力設(shè)計(jì)院.電力工程電氣設(shè)計(jì)手冊(cè)。中國電力出版社，1989。123、韓學(xué)山、張文編。電力系

16、統(tǒng)工程基礎(chǔ).機(jī)械工業(yè)出版社,2003。2五、 內(nèi)容擴(kuò)展系統(tǒng)的進(jìn)展過程輸電技術(shù)全部進(jìn)展史的特征便是不斷地提高線路電壓。,必需增加導(dǎo)線截面積和提高線電壓,增加導(dǎo)線截面積勢(shì)必增加導(dǎo)線材料及成本，而提高線路電壓相對(duì)較為經(jīng)濟(jì)成為輸電進(jìn)展的制約因素 面不夠大時(shí)會(huì)猛烈增加電暈損耗，從而導(dǎo)致鋼芯鋁線的應(yīng)用。150kV我國電壓等級(jí)的標(biāo)準(zhǔn)1000kV 特高壓電壓等級(jí)500kV800kV,正在爭(zhēng)辯中的有直流 1000kV全部連接于電力網(wǎng)中的用電設(shè)備均用溝通電源直流輸電我國負(fù)荷與能源分布的沖突造成了電能需要遠(yuǎn)距離傳送500kV 300km 時(shí)，則屬于超高壓長(zhǎng)距離輸電網(wǎng)超高壓長(zhǎng)距離輸電在穩(wěn)定把握、參數(shù)補(bǔ)償及快速繼電愛護(hù)方面都提出了新的要求超高壓溝通輸電的特點(diǎn)保持同步發(fā)電機(jī)并列運(yùn)行的穩(wěn)定性成為突出問題.300km 以上線路受制于靜態(tài)穩(wěn)定的限制。500kV線路故障時(shí)，先是電壓下降，而后電壓崩潰，然后兩側(cè)電勢(shì)間功角增大失去穩(wěn)定。壓太高，無功補(bǔ)償對(duì)輸送容量起重要作用內(nèi)部過電壓的影