电力系统导论实验报告

Experiment Introduction to powersystems学生姓名：学号：专业班级：实验名称：电力系统导论（双语）2014年 6 月 5 日CONTENTS1、EXPERIMENT 1 (1)B US A DMITTANCE M A TRIX ...................................................................................................................... 1-62、EXPERIMENT 2 (6)B US I MPEDANCE M A TRIX....................................................................................................................... 6-133、EXPERIMENT 3 (13)G AUSS-S EIDEL AND NEWTON METHOD................................................................................................. 13-164、PERSENAL SUMMARY (16)Experiment 1Bus Admittance Matrix1. ObjectiveTo write a simple program in MATLAB® for the algorithm of bus admittance matrix. 2. System RequirementComputer with MATLAB® 6 or above installed.3. Procedure1.0 Launch the MATLAB program.2.0 Go to FILE NEW M-file.3.0 Write a function Y = The_Node_Admittance_Matrix(TopoStructureAndBranchPara) for theformation of the bus admittance matrix.124.0 TopoStructureAndBranchPara is the transmission line, cable and transformer input data andcontains five columns parameters. The first two columns are the line bus numbers and the remaining columns contain the line resistance and reactance in per-unit and transformer tap ratio or capacitor of transmission line.5.0 The function should return the bus admittance matrix.4. ExercisesUse the written function, Y = The_Node_Admittance_Matrix (TopoStructureAndBranchPara ) to obtain the Y bus of the following power system network:Q1. You are required to write the Ybus topological structure and parameter into a text file. (Hint:use the matlab text compiler to write down the table 1 data, using the comma to separate the parameters, and save it use the name of 4_Power_System_Data.dbf)Q2. You are required to write out the program flow figure of forming a nodal admittance matrix. Hint. You are required to compile a program to form the Y bus Matrix, the following program is areference program to you.Figure : One-line diagram of power system For example ,from the textbook “power system analysis” No.2 edition 3 on page 61~62 NodalAdmittanceMatrix =1.0421 - 8.2429i -0.5882 +2.3529i 0 +3.6667i -0.4539 + 1.8911i-0.5882 + 2.3529i 1.0690 - 4.7274i 0 0Table 1：Transformer and transmissssion Line dataFrom Bus#To Bus# R(p.u) X(p.u) B(p.u)or ratio K Others 12 0.1 0.4 j0.01528 13 0 0.3 1.1 14 0.12 0.5 j0.01920 2 4 0.08 0.40 J0.014130 + 3.6667i 0 0 - 3.3333i 0-0.4539 + 1.8911i 0 0 0.9346 - 4.2616i5.The flow chartFigure : The flow chart of Forming Nodal Admittance MatrixThe program is:%function OutPut=The_Node_Admittance_Matrix(handles)%is a subroutine of PowerSystemCalculationfunction OutPut=The_Node_Admittance_Matrix(handles)%the following program is open a data file and get the Number of% Node and Branch data to form a nodal addmittance matrix%the following code is open a file and read the data of power system network[fname,pname] = uigetfile('*.dbf','Select the network parametre data-file'); TopoStructureAndBranchPara= csvread(fname);[NumberOfBranch,NumberOfPara]=size(TopoStructureAndBranchPara);Temporary1=max(TopoStructureAndBranchPara(:,1));Temporary2=max(TopoStructureAndBranchPara(:,2));if Temporary1 > Temporary2NumberOfNode=Temporary1;elseNumberOfNode=Temporary2;end%The following program is to form the Nodal Admittance Matrix% and the Topologic structure and Branch Parametres are arranged% I,J,R,X,C/K, and pay attention to the inpedence of transformer is in the% side of Node J and the ratio of transformer 1:K is in the side of Node Ifor CircleNumber1=1:NumberOfBranchfor CircleNumber2=1:NumberOfBranchNodalAdmittanceMatrix(CircleNumber1,CircleNumber2)=0;endendfor CircleNumber=1:NumberOfBranchif TopoStructureAndBranchPara(CircleNumber,5) > 0.85NodalAdmittanceMatrix(TopoStructureAndBranchPara(TopoStructureAndBranchPara(CircleNumber, 1),TopoStructureAndBranchPara(CircleNumber,1)))=...3Experiment2 Bus Impedance MatrixNodalAdmittanceMatrix(TopoStructureAndBranchPara(TopoStructureAndBranchPara(CircleNumber, 1),TopoStructureAndBranchPara(CircleNumber,1)))+...TopoStructureAndBranchPara(CircleNumber,5)^2/...(TopoStructureAndBranchPara(CircleNumber,3)+...j*TopoStructureAndBranchPara(CircleNumber,4)) ;NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranchPa ra(CircleNumber,2))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranchPa ra(CircleNumber,2))+...1/((TopoStructureAndBranchPara(CircleNumber,3)+j*TopoStructureAndBranchPara(CircleNumber,4) ));NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa ra(CircleNumber,2))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa ra(CircleNumber,2))...-TopoStructureAndBranchPara(CircleNumber,5)/...((TopoStructureAndBranchPara(CircleNumber,3)+j*TopoStructureAndBranchPara(CircleNumber,4))) ;NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranchPa ra(CircleNumber,1))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa ra(CircleNumber,2));elseNodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa ra(CircleNumber,1))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa ra(CircleNumber,1))+...+1/(TopoStructureAndBranchPara(CircleNumber,3)+...j*TopoStructureAndBranchPara(CircleNumber,4))+j*TopoStructureAndBranchPara(CircleNumber,5);4NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranchPa ra(CircleNumber,2))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranchPa ra(CircleNumber,2))+...+1/(TopoStructureAndBranchPara(CircleNumber,3)+...j*TopoStructureAndBranchPara(CircleNumber,4))+j*TopoStructureAndBranchPara(CircleNumber,5)NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa ra( CircleNumber,2))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa ra( CircleNumber,2))...-1/(TopoStructureAndBranchPara(CircleNumber,3)+...j*TopoStructureAndBranchPara(CircleNumber,4));NodalAdmittanceMatrix(TopoStructureAndBranchPara( CircleNumber,2),TopoStructureAndBranchPa ra(CircleNumber,1))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa ra( CircleNumber,2));endendThe result is：NodalAdmittanceMatrix =1.0421 - 8.2429i -0.5882 +2.3529i 0 +3.6667i -0.4539 + 1.8911i-0.5882 + 2.3529i 1.0690 - 4.7274i 0 00 + 3.6667i 0 0 - 3.3333i 0-0.4539 + 1.8911i 0 0 0.9346 - 4.2616i5Experiment2 Bus Impedance Matrix Experiment 2Power GridBus Impedance Matrix1. Objective•To write a simple program in MATLAB® for the algorithm of bus impedance matrix. 2. System RequirementComputer with MATLAB® 6 or above installed.3. Procedure1.0 Launch the MATLAB program.2.0 Go to FILE NEW M-file.63.0 Write a function Z = znbus (z) for the formation of the bus impedance matrix.4.0 z is the line input and contains three columns. The first two columns are the line bus numbersand the remaining columns contain the line resistance in per-unit.5.0 The function should return the bus impedance matrix.4. ExercisesUse the written function, Z = znbus(z) to obtain the Ybus of the following power system network:Example 1Figure 3: One-line diagram of power systemFor example ,from the textbook “power system analysis” No.2 edition 3 on page 61~62Table 1：Transformer and transmissssion Line dataFrom Bus# To Bus# R(p.u) X(p.u) B(p.u)or ratio K Others1 2 0.1 0.4 j0.015281 3 0 0.3 1.11 4 0.12 0.5 j0.019202 4 0.08 0.40 J0.01413Q2. You are required to write the Z bus into a text file. (Hint: use the matlab text compiler) Example 2For the system shown, form Zbus matrix using the building algorithmSolutionA line listApply Kron reduction to eliminate the last rowHint. You are required to compile a program to form the Z bus Matrix.the following program is a reference program to you.The program is:%function OutPut=The_Node_impedance_Matrix(handles)%is a subroutine of PowerSystemCalculationfunction OutPut=The_Node_impedance_Matrix(handles)%the following program is open a data file and get the Number of% Node and Branch data to form a nodal impedance matrix%the following code is open a file and read the data of power system network[fname,pname] = uigetfile('*.dbf','Select the network parametre data-file');Topo_Structure_And_Branch_Para= csvread(fname);%get the electric power system the number of branch and the parametre of% elements[NumberOfBranch,NumberOfPara]=size(Topo_Structure_And_Branch_Para);%Temporary1---temporary variable 1%Temporary2---temporary variable 2Temporary1=max(Topo_Structure_And_Branch_Para(:,1));Temporary2=max(Topo_Structure_And_Branch_Para(:,2));if Temporary1 > Temporary2NumberOfNode=Temporary1;elseNumberOfNode=Temporary2;end% The following program is to form the Nodal impedance Matrix% and the Topologic structure and Branch Parametres are arranged% I,J,R,X,C/K, and pay attention to the inpedence of transformer is in the% side of Node J and the ratio of transformer 1:K is in the side of Node %% set the initial value of Nodal Admittance Matrix to zerofor CircleNumber1=1:NumberOfNodefor CircleNumber2=1:NumberOfNodeNodal_impedance_Matrix(CircleNumber1,CircleNumber2)=0;endendfor CircleNumber=1:NumberOfBranchif Topo_Structure_And_Branch_Para(CircleNumber,5) > 0.85Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(Topo_Structure_And_Branch_Para(Cir cleNumber,1),Topo_Structure_And_Branch_Para(CircleNumber,1)))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(Topo_Structure_And_Branch_Para(Cir cleNumber,1),Topo_Structure_And_Branch_Para(CircleNumber,1)))+Topo_Structure_And_Branch_P ara(CircleNumber,5)^2/(Topo_Structure_And_Branch_Para(CircleNumber,3)+...j*Topo_Structure_And_Branch_Para(CircleNumber,4)) ;Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_And _Branch_Para(CircleNumber,2))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_And _Branch_Para(CircleNumber,2))+...1/((Topo_Structure_And_Branch_Para(CircleNumber,3)+j*Topo_Structure_And_Branch_Para(Circle Number,4)));Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And _Branch_Para(CircleNumber,2))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And _Branch_Para(CircleNumber,2))...-Topo_Structure_And_Branch_Para(CircleNumber,5)/...((Topo_Structure_And_Branch_Para(CircleNumber,3)+j*Topo_Structure_And_Branch_Para(CircleN umber,4)));Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_And _Branch_Para(CircleNumber,1))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And _Branch_Para(CircleNumber,2));elseNodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And _Branch_Para(CircleNumber,1))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And _Branch_Para(CircleNumber,1))+...+1/(Topo_Structure_And_Branch_Para(CircleNumber,3)+...j*Topo_Structure_And_Branch_Para(CircleNumber,4))+j*Topo_Structure_And_Branch_Para(Circle Number,5);Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_And _Branch_Para(CircleNumber,2))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_And _Branch_Para(CircleNumber,2))+...+1/(Topo_Structure_And_Branch_Para(CircleNumber,3)+...j*Topo_Structure_And_Branch_Para(CircleNumber,4))+j*Topo_Structure_And_Branch_Para(Circle Number,5)Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And _Branch_Para( CircleNumber,2))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And _Branch_Para( CircleNumber,2))...-1/(Topo_Structure_And_Branch_Para(CircleNumber,3)+...j*Topo_Structure_And_Branch_Para(CircleNumber,4));Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para( CircleNumber,2),Topo_Structure_And _Branch_Para(CircleNumber,1))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And _Branch_Para( CircleNumber,2));endendformat shortNodal_impedance_Matrix*inv(Nodal_impedance_Matrix)运行结果：Nodal_impedance_Matrix =1.0421e+000 -8.2429e+000i -5.8824e-001 +2.3529e+000i 0 +3.6667e+000i 0-5.8824e-001 +2.3529e+000i 5.8824e-001 -2.3377e+000i 0 00 +3.6667e+000i 0 0 -3.3333e+000i 00 0 0 4.5386e-001 -1.8719e+000iNodal_impedance_Matrix =1.0421e+000 -8.2429e+000i -5.8824e-001 +2.3529e+000i 0 +3.6667e+000i -4.5386e-001 +1.8911e+000i-5.8824e-001 +2.3529e+000i 1.0690e+000 -4.7274e+000i 0 00 +3.6667e+000i 0 0 -3.3333e+000i 0-4.5386e-001 +1.8911e+000i 0 0 9.3463e-001 -4.2616e+000ians =1.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 - 0.0000i 0.0000 - 0.0000i-0.0000 - 0.0000i 1.0000 - 0.0000i -0.0000 + 0.0000i -0.0000 - 0.0000i -0.0000 - 0.0000i -0.0000 - 0.0000i 1.0000 - 0.0000i -0.0000 0 - 0.0000i 0 + 0.0000i 0.0000 - 0.0000i 1.0000 + 0.0000i以上就是对阻抗矩阵的验证，其和其逆相乘为单位对角矩阵Experiment 3Gauss-Seidel Method 1. Objective• To write a simple program in MATLAB® for the algorithm to solution of nonlinear algebraic equations;• Known as the method of successive displacements. 2. DiscussionThe most common methods for solving nonlinear algebraic equations are Gauss-Seidel,Newtow-Rahpson, and quasi-Newton-Raphson methods. We start with one dimensional equations and then generalize to n-dimensional equations. 3. Mathmatics modelConsider the nonlinear equation 0)(=x f .The equation is broken into two parts thus:)(x g x =. We assume )0(xis an initial "guess" of the solution, then "refine" the solution using:)()0()1(x g x =This process is repeated thus)()1()2(x g x =and on the thn iteration we have: )()1()(-=n n x g x. If this process is convergent, then the successive solutions approach a value which is declared as the solution. Thus if at some step 1+k we have:ε≤-+)()1(k k x xwhere e ε is the desired "accuracy", then we claim the solution has been found to the accuracy specified.4. System RequirementComputer with MATLAB® 6 or above installed. 5. Procedure1.0 Launch the MATLAB program.2.0 Go to FILE NEW M-file.3.0 Write a function program of Gauss Seidel Method. 6. ExercisesExample: Using the Gauss-Seidel method to obtain the roots of the equation:0496)(23=-+-=x x x x fFirst the equation is expressed in a different form thus())(469123x g x x x =---=And the iteration can proceed. Take a good look at the shape of the iterations! Below is the program showing the process graphically (later showing how to do it iteratively).7.The flow chart of Gauss Seidel method (Omitted)8.Reference Program and result.程序是：clear allclcx0=0.5;n=0;while (abs(x0^3-6*x0^2+9*x0-4)>0.00001)x0=-(x0^3-6*x0^2-4)/9;y=x0;n=n+1;end结果是：n=1627 y=x0=0.99818clear allclcx0=2.5;n=0;while (abs(x0^3-6*x0^2+9*x0-4)>0.00001)x0=-(x0^3-6*x0^2-4)/9;y=x0;n=n+1;end结果是：n=7 y=x0=4仿照高斯--赛德尔法，我们可以写出简单的牛顿法的程序,如下：牛顿法解方程x0=0.5;n=0;while (abs((x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9))>0.00001)dx0=-(x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9);x0=x0+dx0;n=n+1;end结果是：dx0=1.7684e-005 n=15 x0=0.99998 y= -0.875x0=0.5;n=0;while (abs(x0^3-6*x0^2+9*x0-4)>0.00001)dx0=-(x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9);x0=x0+dx0;n=n+1;end结果是: dx0=0.0011305 n=9 x0=0.99887 y= -0.875x0=3.5;n=0;while (abs(x0^3-6*x0^2+9*x0-4)>0.00001)dx0=-(x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9);x0=x0+dx0;n=n+1;end结果是：dx0= -2.5283e-006 n=5 x0=4 y= -0.875Personal Summary：The experiment of bilingual class is over,here is my personal summary.In my opinion,first and foremost,I had to acknowledge that I have elementary know the base using of MATLAB,during approximately ten hours’hard working.Although I have spent ten hours or less on learning this software,I merely grasp the knowledge which is just like the tip of the iceberg.In terms with the application of this software,we just do some simple steps.For instance,the node admittance matrix of node impedance matrix in date input software,the experimental program input last run results,the corresponding node admittance matrix of node impedance matrix is obtained.Besides,I would say I haven’t master the method of how to edit a program.This is a pity.Nevertheless,I think it is just a program of time .I’ll pay more time on fulfill a deeper study.Last but not least,I am really appreciate for teacher’s patient teaching and conducting.Thank you very much!Because the experimental time coincides with exam review time, so I didn't understand a lot of knowledges in the experimental thoroughly, but I still hope the teacher can forgive me.。