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潮流計(jì)算--電力系統(tǒng)大作業(yè)(c++編寫)-預(yù)覽頁

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【正文】 edouble K,double P0,double void PreOrder(line_part *amp。t)。} else if(vallget_val()){insertline_part(tlchild, vall,u_end,u_begin,r,x,b,Pe,Xe,K,P0,Q0)。t){if(tget_val()0amp。ofstream outfile1。// 計(jì)算并存放各個(gè)節(jié)點(diǎn)的電壓相角(始端為零)U=atan2(tU[1],(tget_Ubegin()tU[0]))/*180。的有功損耗P_cost=tget_Pcost()?!眊et_Pbegin()get_Qbegin()//計(jì)算并存放各個(gè)支“get_Pend()get_Qend()//計(jì)算并存放各個(gè)支//計(jì)算并存放各個(gè)支路 “//計(jì)算并存放各個(gè)支路 ”}。tget_val()get_lchild()get_val()0amp。t{tUend()。}else{tUend()。amp。}else{} }PreOrder(tlchild)。amp。amp。trchildSbegin()。pass_w2(t,tlchild)。t{trchildSbegin()。void main(){ofstream outfile。支路數(shù)據(jù)數(shù)組while(l}}//輸出數(shù)據(jù)流定義 (“e:”)。(“e:”)。//支路節(jié)點(diǎn)數(shù)常//輸入流輸入各infileA[l]。(“e:”,ios::ate)。BinTree elec。()。double aa =eeget_Pbegin()。bb =aap_cost_all。p_cost_all=。同時(shí)基于MATLAB的計(jì)算機(jī)算法以雙精度類型進(jìn)行數(shù)據(jù)的存儲(chǔ)和運(yùn)算, 數(shù)據(jù)精確度高,能進(jìn)行潮流計(jì)算中的各種矩陣運(yùn)算,使得傳統(tǒng)潮流計(jì)算方法更加優(yōu)化。二 牛頓拉夫遜法潮流計(jì)算 1 基本原理牛頓法是取近似解x(k)之后,在這個(gè)基礎(chǔ)上,找到比x(k)更接近的方程的根,一步步地迭代,找到盡可能接近方程根的近似根。2 基本步驟和設(shè)計(jì)流程圖形成了雅克比矩陣并建立了修正方程式,運(yùn)用牛頓拉夫遜法計(jì)算潮流的核心問題已經(jīng)解決,已有可能列出基本計(jì)算步驟并編制流程圖。y(4,5)=0。y(2,3)=1/(+)。endendY=0。B=imag(Y)。S(4)=。% 賦初值,U為節(jié)點(diǎn)電壓的幅值,a為節(jié)點(diǎn)電壓的相位角U=ones(1,5)。x2=ones(8,1)。M(i,j)=0。endend% 求有功、無功功率不平衡量for i=1:4for j=1:5oP(i)=oP(i)U(i)*U(j)*(G(i,j)*cos(a(i)a(j))+B(i,j)*sin(a(i)a(j)))。endx2=[oP,oQ]39。M(i,j)=N(i,j)。L(i,i)=L(i,i)U(i)*U(j)*(G(i,j)*sin(a(i)a(j))B(i,j)*cos(a(i)a(j)))endendN(i,i)=N(i,i)2*(U(i))^2*G(i,i)。% x1為所求△x的列向量% 求節(jié)點(diǎn)電壓新值,準(zhǔn)備下一次迭代for i=1:4oa(i)=x1(i)。endk=k+1。endS(5)=P(5)+Q(5)*sqrt(1)。利用x(*)=x(k+1)+DX(k+1)進(jìn)行多次迭代,通過迭代判據(jù)得到所需要的精度值即準(zhǔn)確值x(*)。ph=input(‘n請(qǐng)輸入平衡母線的節(jié)點(diǎn)號(hào):ph=’); B1=input(‘n請(qǐng)輸入支路信號(hào):B1=’)。f=zeros(1,n);no=2*ph=1; Y(q,q)=Y(q,q)+1./B1(i3)+B1(i4)/2;End for i=1:nG=real(Y)。End f(i)=imag(B2(i3))。p=B1(i2)。J=jacci(Y,G,B,P,Q,e,f,V,C,D,B2,n,ph,no)。max(abs(Dfi)ipt=t+1。for i=1:n hh(i)=conj(Y(ph,i)*v(i))。jd=angle(v)*180/p。c(i)=v(B1(i2))*a(i)j*B1(i4)*v(B1(i2))^2/2。D(i)=0。Q1=C(i)*f(i)D(i)*e(i)。p=p+1。Df(p)=Q(i)Q1。DX1=DX。For i=(no+2):(x1+2)DX(i)=DX1(i2)。For i=1:2:x K=k+1。X2=G(i,j)*e(i)+B(i,j)*f(i)。continue q=2*j1。J=ph J(m,q)=X3。J(p,q)=X2。X1=D(i)+G(i,j)*f(i)B(i,j)*e(i)。X3=0。P=2*i1。J(p,q)=X1。J(m,q)=X3。J(p,q)=X2。end q=2*j1。if ph=n q=q+1。J(m,q)=X4。cess dyn鋤ic Voltage stabil時(shí)of hybrid AC/DC poWer tmnsmission System舶m Yu衄an proVince to G啪gdong province【J】.Electric Power,2008,4l(10): l4.[4] 朱新立,湯涌,等.大電網(wǎng)安全分析的全過程動(dòng)態(tài)仿 真技術(shù)[J】.電網(wǎng)技術(shù),2008,32(22):23—28. SONG Xin—Ii,TANG Yof唱,et a1. Full dyn鋤ic simulation for the stabilhy a眥lysis of large power system【J】.Power System融IlrIolo影,2008,32(22): 23.28.[5]Roytelm鋤I,Shallidehpour S M.A prehsivc long teml dynaIIlic simulation for powcr system recoVery【J】. IEEE Transactions 0n Power Systems,1994,9(3). [6] 石雩梅,汪志宏,等.發(fā)電機(jī)勵(lì)磁系統(tǒng)數(shù)學(xué)模型及參 數(shù)對(duì)電網(wǎng)動(dòng)態(tài)穩(wěn)定性分析結(jié)果影響的研究[J】.繼電 器,2007,35(21):2227.SHI Xue.mei,WANG Zllihon舀et a1.Iksearch on the innuence of g鋤e翰to璐baScd ∞de詛iled excitation system models柚d parameterS t0 power鏟id dyn鋤ic stabil時(shí)【J】.Relay,2007,35(2 1):2227.[7] 方思立,朱方.快速勵(lì)磁系統(tǒng)對(duì)系統(tǒng)穩(wěn)定的影響[J】.中 國電機(jī)工程學(xué)報(bào),1986,6(1):20.28.FANG Si.1i,ZHU Fang.The effbct of f弧t.respon∞excitation system on the stability of power netwofk【J】. Proceedings ofthe CSEE,1986,6(1):2028.[8] 劉?。娏ο到y(tǒng)穩(wěn)定性及發(fā)電機(jī)勵(lì)磁控制[M】.北京: 中國電力出版社,2007.LIU Qu.Power system S詛bility鋤d generator excitation control【M】.BeUing:ChiIla Electric Powef Press,2007. [9] Dallachy J L,Anderson T.EXperience with rcplacing ro詛ting exciters wim static exciters【J】.1k InStitution of Electrical Engineers,1 996.[10] 陳利芳,陳天祿.淺談自并勵(lì)勵(lì)磁系統(tǒng)在大容量機(jī)組 中的應(yīng)用【J】.繼電器,2007,35(1):8l培4. CHEN Lif抽島CHEN Tian—lIL Application of 辯l仁exci組tion mode in large capacity髫memtor unit【J】. ReIay39。int _tmain(int argc, _TCHAR* argv[]){int i。amp。//輸入支路參數(shù)Line::PrintfLineData(line,L)。bus[i].=0。X=new Complex *[N]。//輸出節(jié)點(diǎn)導(dǎo)納矩陣int NN=(N1)*2。idouble *x。//計(jì)算節(jié)點(diǎn)注入電流Bus::JisuanNodeScal(X,bus,N)。//輸出節(jié)點(diǎn)功率差值int icon=wehcon1(bus,N)。//解方程組求出電壓差值Bus::ReviseNodeV(bus,x,N)。}else{for(i=0。=line[i].B。bb=Complex::subComplex(Complex::getconj(bus[statemp1].V), Complex::getconj(bus[endtemp1].V))。aa=Complex::productComplex(Complex::getconj(bus[endtemp1].V), B)。line[i].etos=Complex::productComplex(bus[endtemp1].V,dd)。//發(fā)電機(jī)功率=注入功率+負(fù)荷功率bus[i].= bus[i].+ bus[i].。coutt(bus[i].)。statemp=line[i].start。cout} }return 0。//輸入節(jié)點(diǎn)個(gè)數(shù)int L=Line::ScanflineNo()。} //如果找不到兩個(gè)文件中的任意一個(gè),退出Line *line=new Line[L]。//動(dòng)態(tài)分配結(jié)點(diǎn)結(jié)構(gòu)體for(int i=0。//輸入節(jié)點(diǎn)參數(shù)Bus::PrintfBusData(bus,N)。iBus::JisuanNodeDnz(X,line,bus,L,N)。JacAug=new double *[NN]。int count=1。//計(jì)算節(jié)點(diǎn)功率Bus::JisuanNodeSdelta(bus,N)。//計(jì)算雅可比增廣矩陣// Bus::PrintfJacAug(JacAug,N)。count++。Complex aa,bb,cc,dd,B。endtemp=line[i].end。dd=Complex::CaddC(aa,cc)。cc=Complex::productComplex(bb , Complex::getconj(line[i].Y))。//計(jì)算節(jié)點(diǎn)功率for(i=0。}coutfor(i=0。cout}coutfor(i=0。coutComplex::PrintfComplex(Complex::ComDivRea(line[i].stoe,))。class Complex//定義復(fù)數(shù)類 { public: double real。//求兩個(gè)復(fù)數(shù)和static Complex subComplex(Complex c1,Complex c2)。//除數(shù)static Complex getconj(Complex c1)。//求一個(gè)復(fù)數(shù)虛部static void PrintfComplex(Complex c1)。//取極坐標(biāo)Complex(){RecPolar=0。=atan2(,)*180/。=+。=。=**。=(*+*)/(pow(,2)+pow(,2))。=(r1)。=。=1。}double Complex::getComplexReal(Complex c1)//取實(shí)部 {return 。(ios::right)。(ios::left)。}class gauss { public: static void gauss_slove(double **a,double *x,int NN)。double eps,pivot,sum,aik,al。for(i=0。pivcol[k][1]=k。pivrow[k]=i。}if(pivrow[k]!=k)//行變換{for(j=k。}}if(pivcol[k][1]!=k)//列變換{for(i=0。}}if(k!=(n1))//將矩陣化為上三角形式{for(i=(k+1)。}}} } x[n1]=a[n1][n]/a[n1][n1]。for(j=(i+1)。k=0。}coutcout}====節(jié)點(diǎn)功率計(jì)算值==== icon=1進(jìn)行第2次迭代 節(jié)點(diǎn)電壓修正量================= ====節(jié)點(diǎn)功率計(jì)算值==== icon=1進(jìn)行第3次迭代 節(jié)點(diǎn)電壓修正量================= ====節(jié)點(diǎn)功率計(jì)算值==== icon=0,迭代結(jié)束。0節(jié)點(diǎn)25 。節(jié) 點(diǎn)27 。0節(jié)點(diǎn)29 。
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