PQ分解法潮流计算

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1、//PQ分解法解潮流程序#include#include#defineN5//节点数#definen_PQ3//PQ节点数#definen_PV1//PV节点数#definePI3.1415floatG[N][N]={{0.4347,-0.6,-0.2747,0,0},{-0.6,0.418,-0.258,0,0},{-0.2747,-0.258,0.5327,0,0},{0,0,0,0,0},{0,0,0,0,0}};//导纳矩阵的实部floatB[N,N]={{-1.7115,1,0

2、.9615,0,0},{1,-64.9577,0.9677,63.49,0},{0.9615,0.9677,-33.4252,0,31.746},{0,-63.49,0,63.49,0},{0,0,31.746,0,-31.746}};//导纳矩阵的虚部floatU_amp[N]={1.0,1.0,1.0,1.05,1.05,};//电压初值floatU_ang[N]=[0,0,0,0,0]//电压相角初值floatPs[N-1]={1.6,2,3.7,5};//有功初值floatQs[N-2]={0.8,1,1.3};

3、//无功初值floatdU_amp[n_PQ],dU_ang[N-1];//电压幅值和相角修正值floatP1[N],Q1[N],dP[N],dQ[n_PQ],H[N-1,N-1],L[n_PQ-1,n_PQ-1];floatJ[N+n_PQ-1,N+n_PQ-1]，M[N+n_PQ-1,1]，N[N+n_PQ-1,N+n_PQ-1];//J[N+n_PQ-1-1,N+n_PQ-1]雅可比矩阵,N[N+n_PQ-1-1,N+n_PQ-1]雅可比矩阵逆矩阵floatdS[N+n_PQ-1,1],T,U[N],SS;floa

4、tprecision=1.0;//precision计算精度intkk=0;//迭代次数inti,j,k;//主程序===================================================================while(precision>0.0001){for(i=1;i

5、;j++){P1[i]=P1[i]+U_amp[i]*(G[i][j]*U_amp[j]*cos(U_ang[j])+B[i][j]*U_amp[j]*sin(U_ang[j]));Q1[i]=Q1[i]+U_amp[i]*(G[i][j]*U_amp[j]*sin(U_ang[j])-B[i][j]*U_amp[j]*cos(U_ang[j]));}dP[i]=Ps[i]-P1[i];if(i

6、>N-1)M[i,1]=dQ[i];}for(i=1;i

7、])-B[i][j]*cos(U_ang[j]);}}if(i=j){H[i,i]=H[i,i]-U_amp[i]*U_amp[j]*(G[i][j]*sin(U_ang[j])-B[i][j]*cos(U_ang[j]);If(i

8、--------------------------------------------------------------------------------------//形成雅可比矩阵----------------------------------------------------------------