The University of Derby Faculty of Arts Design and Technology School of Technology BEng in Electrical and Electronic Engineering Electrical Power Utilisation ___________________________________________________________________ ______ Assignment No 2 An Investigation into Load Flow Studies Using MATLAB - Power Flow Analysis Tools Assessment Title: An Investigation into Load Flow Studies Using M ATLAB - Power Flow Analysis Tools Assessment No.: 2 of 2 Assessment Weighting: Estimate of Assessment Duration: At Least 25 hours
An Investigation into Load Flow Studies Using MATLAB - Power Flow Analysis Tools
Assessment Title: An Investigation into Load Flow Studies Using MATLAB - Power Flow Analysis Tools
Assessment No.: 2 of 2
Assessment Weighting: Estimate of Assessment Duration: At Least 25 hours
Learning Outcomes Assessed: 2
Common Skills OutcomesAssessed: None
Date Set: 8th of February 2011
Submission Date: 15/03/2011
Note: Work must be handed in at the Student Information Centre.
Please note there are changes in the undergraduate regulations, more information can be found in the module handbook.
Issue No 1 (01/11)
Work To Be Undertaken To Complete This Assignment
Introduction:
Load flow studies are used to determine the voltage, current and power, (both the active and
reactive components and their associate power factors), at all the node points in a specific power
system.
Load flow studies should indicate the likely behaviour of the system in terms of:
The flow of MW and MVAR in the various branches of the network
The precise bus-bar voltages
Effects of any partial rearrangement of the system inter-connections, or the consequences of
adding further networks
The outcome of any loss of overhead line or underground cable
The results of changes in transformer tap position
The optimum performance of the system with minimum losses
Justification for any recommended system improvements.
The complexity of obtaining a formal solution for load flow in a power system arises because of the
differences in the type of data specified for the various nodes, (bus bars), of the system.
At the SLACK BUS, (see lecture notes for definition), the voltage magnitude and angle will be
designated. At the remaining generator, (supply), nodes the usual information available is the active
power supplied and the voltage magnitude. Note, in this assignment there are no other generator
nodes except at the slack bus bar. This considerably simplifies the actual assignment work. Nodes
with output loading are usually conditioned by their active and reactive power demand. Although
the formulation of sufficient equations is not difficult, the closed form of solution is not practical.
Consequently, a digital solution of load flow problem is required. This will need to follow an iterative
process, initially requiring estimated values be assigned to the unknown bus voltages.
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The voltage at one of the bus bars is then recalculated from the estimated values at the other
nodes.A corrected value is obtained and this is then used in performing similar calculations to obtain
a corrected voltage at the next bus. Each bus voltage is corrected in turn around the system, the
process then being repeated until sufficient convergence is obtained.
2. Develop an Algorithm of the Load Flow Problem.
Various methods have been devised for solving this problem. In this assignment the student is
expected to adapt three separate well known iterative methods to solve the actual load flow
distribution in a given power supply network. Each of these iterative techniques will be performed
by writing suitable computer code in the powerful MATLAB program language or using power flow
simulation tools.
Iterative methods to be used are:
(1) The Jacobi Method
(2) The Gauss Seidal Method
(3) The Impedance Matrix Method.
3. Work to be undertaken:
(1) Carry out a series of computer simulations for power flow for the 5-Buse/Node - seven-line network - Power Network shown in figure 1-A and report it. Further information and necessary technical parameters are available on table 1 and support material available in the assignment and UDo.
(2) Analyse the results of each simulations and provide the necessary graphical analysis of the results
(3) Explain your own thoughts, observation for each simulations and comment on it
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Figure 1-A: 5-Buse/Node - seven-line network - Power Network
Table 1:From
BusbarTo Busbar R (p.u.) X (p.u.) B (p.u.)
1 2 0.05 0.11 0.02
1 3 0.05 0.11 0.02
1 5 0.03 0.08 0.02
2 4 0.04 0.09 0.02
2 5 0.04 0.09 0.02
3 4 0.06 0.13 0.03
4 5 0.04 0.09 0.02
4. In completing this assignment the student will have developed skills in:
a) Solving the load flow distribution problem in electrical power networks
b) Developing algorithms to describe the methodology for solving the electrical power load flow problem in power network.
c) Using the MATLAB / Power Flow Tools computer programming software.
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5. Assignment resources:
The University of Derby, Faculty of Arts Design and Technology, Electrical Power
Laboratory.
Assignment Support Material
Lecture notes available on the university black board
Reference text books available in the library
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Assessments
In order to pass the module THIS ASSESSED WORK MUST BE ATTEMPTED AND THE WORK SUBMITTED VIA THE STUDENT OFFICE. Assessment is based upon the achievement or non-achievement of learning outcomes and the grade achieved. In evaluating the student’s performance the assessor compares the product with the set of grade descriptors shown in the Assessment Grade table in the Assessment Regulations handbook or Rights, Responsibilities and Regulations handbook. The alpha grade, achievement or non-achievement of learning outcome(s) are reported on the Student Assessment Record and Receipt Form
Learning outcomes for the assessed work are specified within the module in the Programme Learning Outcomes, Module Learning Outcomes, Indicative Content, Teaching and Learning Strategies, Resource and Assessment Methods handbook.
Assessment Criteria:
A+ to B+: Outstanding, of exceptionally high standard to a very good standard. Completion of all parts of the work with correctly calculated results and extensive analysis as required. Your conclusions must be comprehensive, correct and thoroughly justified. The work presented must be of the highest standard, with appropriate diagrams and extensive numerical analysis. During the actual laboratory period you have demonstrated a leadership in organising and interpreting instructions, assembling the test circuits and proceeding with the test programme using safe and efficient practices.
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Suggested Contents
You may follow the following steps to write your assignment.
1. Executive Summary
I expect that each student is to conduct his own survey about the assignment topics and report his
own findings. The survey is to cover the assignment technology, applications, and needs. You need
also to make yourself familiar with the latest Industrial and Academic publications available related
to you assignment applications. This is to be followed by the structure of your report and main topics
that have been covered. Then explain the findings of your assignments and skills that you have
learned from the work carried out from Analysis, design, development and practical point of view.
2. Assignment Aims and Objectives
You need to specify clearly your assignment main aims and objectives.
3. Algorithm 1 & 2 and 3
Explain the basic principles of each algorithm
4. Flowchart of each Algorithm
You need to provide a flowchart for each Algorithm and CD contains your reporting and
programmes.
5. Computer Programmes and Simulations
Provide your programme using MATLAB and Simulation using Simulink and or Power Flow Tools
6. Results, graphs and Discussions
You need to explain the numerical and the graphical results of your programming. Then write your
own comments.
7. Conclusions
8. References
9. Appendices
10. Risk Assessment
a) You need to read and understand the risk Assessment for Electrical Power Laboratory MS123.
b) You need to read and understand the safety and operation of the machines test rigs available of
the Laboratory
c) You need to include a risk assessment for your assignment. A copy of the form can be obtained
from room MS123.
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Supporting Materials
Load Flow Studies Using MATLAB.In order to provide a common basis for this series of load flow studies a 5-node network has been configured. This will allow the relative performance of each of the load flow solution methods used to be compared objectively. The actual network contains only a limited range of system elements and there is only one source of generated power, which must therefore be connected to the designated slack busbar and referred to as node (1).
The five-busbar, seven-line network is shown in figure 1_B.
Figure 1_B: five-busbar, seven-line network
At the slack busbar the voltage is made reference at 1 p.u. The remaining busbars all have connected loads for which the values of P and Q are specified. Table 1 gives the series impedance of the interconnecting lines. Each line is associated with capacitive shunt admittance. This is to be BOTH ignored and included at the various stages of the analysis. Hence, the effect of the shunt capacitance in relation to the resulting node voltage distribution of the network will be determined.
The network base quantities are MVA base = 100 MVA and V base = 132 kV.
Table 1:
From Busbar To Busbar R (p.u.) X (p.u.) B (p.u.)
1 2 0.05 0.11 0.02
1 3 0.05 0.11 0.02
1 5 0.03 0.08 0.02
2 3 0.04 0.09 0.02
2 5 0.04 0.09 0.02
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3 4 0.06 0.13 0.03
4 5 0.04 0.09 0.02
The method of formulation of the network admittance matrix is covered in the ‘Introductory Lecture Notes to Load Flow Studies’. On this basis the line admittances can be calculated, see Table 2.
Table 2:
Line Admittance Conductance G. (S) Susceptance B. (S)
Y(1.2) 3.424658 - j 7.534247
Y(1.3) 3.424658 - j 7.534247
Y(1.5) 4.109589 - j 10.958904
Y(2.3) 4.123711 - j 9.278351
Y(2.5) 4.123711 - j 9.278351
Y(3.4) 2.926829 - j 6.341463
Y(4.5) 4.123711 - j 9.278351
The complete system admittance matrix can now be determined. In Table 3a: the effects of shunt line admittance have not been included.
Table 3a:
Admittance Matrix- shunt
capacitance
10.958905- j 26.027398
- 3.424658+ j 7.534247
- 3.424658+ j 7.534247
0- 4.109589
+ j 10.958904
- 3.424658+ j 7.534247
11.672080- j 26.090949
- 4.123711+ j 9.278351
0- 4.123711
+ j 9.278351
- 3.424658+ j 7.534247
- 4.123711+ j 9.278351
10.475198- j 23.154061
- 2.926829+ j 6.341463
0
0 0- 2.926829
+ j 6.3414637.050541
- j 15.619814- 4.123711
+j 9.278351
- 4.109589+ j 10.958904
- 4.123711+ j 9.278351
0- 4.123711
+j 9.27835112.357012
- j 29.515606
The effects of the shunt admittance in each line must now be included. The shunt admittance of each line which connects to a particular node must now be added to the diagonal element of that node. Since the shunt elements are capacitive, the result of incorporating them into the admittance matrix will reduce the magnitude of the complex element in the diagonal terms.
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Table 3b:
Admittance Matrix
+ shunt capacitance
10.958905- j 25.997397
- 3.424658+ j 7.534247
- 3.424658+ j 7.534247
0- 4.109589
+ j 10.958904
- 3.424658+ j 7.534247
11.672080- j 26.060948
- 4.123711+ j 9.278351
0- 4.123711
+ j 9.278351
- 3.424658+ j 7.534247
- 4.123711+ j 9.278351
10.475198- j 23.119061
- 2.926829+ j 6.341463
0
0 0- 2.926829
+ j 6.3414637.050541
- j 15.594814- 4.123711
+j 9.278351
- 4.109589+ j 10.958904
- 4.123711+ j 9.278351
0- 4.123711
+j 9.27835112.357012
- j 29.485605
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Method No1 - Jacobi Method:
The Jacobi method is the simplest of the iterative load-flow techniques to program. It is a straightforward process whereby successive sets of busbar voltages are calculated using the previously obtained values. The complete range of busbar voltages are calculated before they are then used in the next iteration.From the introductory load-flow notes it was shown that the basis of the iterative solution is based on the matrix form of the network’s nodal equation:
; Where k = 2, 3… n
; k = 2, 3, …. N (1)
Substituting for I k
or where k = 2, 3, …. N
(i) is the previous iteration, (i + 1) is the current
Calculation:The solution strategy is based on all system admittances and busbar conditions being available and an initial voltage assigned to each busbar. A flow diagram of the iterative process is indicated below.The number of nodes n = 5
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And since the voltage at node (1) is specified The actual range of k = 2, 3, 4, 5.
The step by step process is now demonstrated by numerically solving the first set of iterations for the network illustrated in Fig1.
Procedure (shunt capacitance included).
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The MATLAB program code for the JACOBI SOLUTION of the network load-flow problem can now be written.
% USING THE JACOBI METHOD % YEAR 2010/11 % STUDENT ........... % NETWORK (see Fig 1) % Input (1) % Insert the Network Admittance Matrix % (see page 2 of supplied assignment paperwork ignoring shunt capacitive admittance) format shortY(1,1) = 10.958905 - 26.027398i;Y(1,2) = -3.424658 + 7.534247i;Y(1,3) = -3.424658 + 7.534247i;Y(1,4) = 0.00;Y(1,5) = -4.109589 + 10.958904; Y(2,1) = -3.424658 + 7.534247i;Y(2,2) = 11.672080 - 26.090949i;Y(2,3) = -4.123711 + 9.278351i;Y(2,4) = 0.00;Y(2,5) = -4.123711 + 9.278351i; Y(3,1) = -3.424658 + 7.534247i;Y(3,2) = -4.123711 + 9.278351i;Y(3,3) = 10.457198 - 23.154061i;Y(3,4) = -2.926829 + 6.341463i;Y(3,5) = 0.00; Y(4,1) = 0.00;Y(4,2) = 0.00;Y(4,3) = -2.926829 + 6.341463i;Y(4,4) = 7.050541 - 15.619814i;Y(4,5) = -4.123711 + 9.278351i; Y(5,1) = -4.109589 + 10.958904i;Y(5,2) = -4.123711 + 9.278351i;Y(5,3) = 0.00;Y(5,4) = -4.123711 + 9.278351i;Y(5,5) = 12.357012 - 29.515606i; % Input (2)% Input the given node loadings .....see diagram on page 1 .... % .... of the supplied assignment paperwork% A positive (+ sign) is for generated power,% The negative (- sign) indicates load taken from the network.% Per unit loading is used with a base loading taken at 100MVA P2 = - 0.4; % p.u. active power loadingP3 = - 0.25; % p.u. active power loadingP4 = - 0.4; % p.u. active power loadingP5 = - 0.5; % p.u. active power loading
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Q2 = - 0.2i; % p.u. reactive power loadingQ3 = - 0.15i; % p.u. reactive power loadingQ4 = - 0.2i; % p.u. reactive power loadingQ5 = - 0.2i; % p.u. reactive power loading % The voltage at the generation busbar, node (1) is 1.0 p.u. % This voltage is fixed and therefore determines the reference ....% ... or SLACK BUSBAR must be Node(1) % JACOBI SOLUTION% The number of iteration will need to be set% Set the value of 'l' the chosen number of iterations% Typically l=40 should be sufficient for this problem. l = 40; % sets the number of iterationsm = l + 1; % allocates the space used to store each iteration vector = [1:m]; % assigns a row with the appropriate spaces row = ones(size(vector)); Vnode1 = row; % each node voltage has the required storage spacesVnode2 = row;Vnode3 = row;Vnode4 = row;Vnode5 = row; for n = 1:m Vnode1(n) = 1.0 + 0.00i; % inserts the assumed initial node voltagesVnode2(n) = 1.0 + 0.00i;Vnode3(n) = 1.0 + 0.00i;Vnode4(n) = 1.0 + 0.00i;Vnode5(n) = 1.0 + 0.00i; end S2star = P2 - Q2; % congugates of the specified node loadings S3star = P3 - Q3;S4star = P4 - Q4;S5star = P5 - Q5; % Now Proceed with the JACOBI SOLUTION% Following the same method as in the worked example % Page 5 of the assignment notes for n = 1:l % NOW UPDATE THE NODE VOLTAGES (l=40) TIMES % SIMILAR TO THE CALCULATION PROCEDURES % AS DETAILED ON PAGE 9 OF THE ASSIGNMENT NOTES V2star(n) = conj(Vnode2(n)); % conjugate of V2 V3star(n) = conj(Vnode3(n)); % conjugate of V3 V4star(n) = conj(Vnode4(n)); % conjugate of V4 V5star(n) = conj(Vnode5(n)); % conjugate of V5 % (Step 1) Solving for the updated values of the currents I2(n) = S2star/V2star(n); I3(n) = S3star/V3star(n); I4(n) = S4star/V4star(n);
The Matlab output program code for the JACOBI SOLUTION of the network load-flow problem. Command Window Output:: Electrical Power Utilisation 6EJ022 : Jacobi Solution of Load Flow Assignment (2) 2010/11: Successive Iteration Values : Vnode2 Vnode3 Vnode4 Vnode5 :
Jacobi Method convergence after 30 iterations for the given network
Method No2 - Gauss-Seidal Method:
Whereas the Jacobi method calculates new values for all the bus bar voltages before replacing the old values of voltage with those newly calculated, the Gauss-Seidal method updates the individual bus bar voltage immediately after it is calculated. This leads to a faster convergence to the final solution.
As in the previous analysis,
where k = 2, 3, …. n
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k = 2, 3, …. n
Combining the above two equations and rearranging gives:
Since the Gauss-Seidal approach updates each node voltage immediately an improved value is calculated, the actual solution strategy will need to follow steps indicated below.
(1) For node (2)
(2) Also since this summation can be calculated using the estimated values of
node voltage.(3) Subtract the solution of the summation from (2) from the value of obtained from (1) and
dividing the result by will give a new value for this then allows the node(2) voltage to be updated.
(4) The voltage at node(3) is then re-calculated using and but
with the updated value of obtained from Step (3) then the node(3) voltage can be updated.
(5) Subsequently, repeat the process, using the most recently calculated values of node voltages to update the node(4) voltage.
(6) And again, repeat the process, using the most recently calculated values of node voltages to update the node(5) voltage.
(7) Compare the latest set of voltages with the previous set. If the two sets agree within defined limits, then the process is complete, otherwise return to Step 1.
See program flow diagram
Gauss-Seidal Program Flow Diagram:
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This step by step process can now demonstrated by numerically solving the first set of iterations for the network illustrated in Fig1.
Procedure (shunt capacitance included).
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The Matlab program code for the GAUSS SEIDAL SOLUTION of the network load-flow problem can now be written. Note the many copy and paste opportunities that exist between the existing Jacobi codes when subsequently producing the Gauss Seidal code.
Matlab Program.
% BSc(Hons) Electrical Electronic Engineering % Module 6EJ998 % Assignment No2 % MODELLING LOAD FLOW STUDIES USING MATLAB % USING THE GAUSS SEIDAL METHOD
% Input (2)% Input the given node loadings .....see diagram on page 1 .... % .... of the supplied assignment paperwork% A positive (+ sign) is for generated power,% The negative (- sign) indicates load taken from the network.% Per unit loading is used with a base loading taken at 100MVA P2 = - 0.4; % p.u. active power loadingP3 = - 0.25; % p.u. active power loadingP4 = - 0.4; % p.u. active power loadingP5 = - 0.5; % p.u. active power loading Q2 = - 0.2i; % p.u. reactive power loadingQ3 = - 0.15i; % p.u. reactive power loadingQ4 = - 0.2i; % p.u. reactive power loadingQ5 = - 0.2i; % p.u. reactive power loading % The voltage at the generation busbar, node (1) is 1.0 p.u. % This voltage is fixed and therefore determines the reference ....% ... or SLACK BUSBAR must be Node(1) % GAUSS SEIDAL SOLUTION% The number of iteration will need to be set
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% Set the value of 'l' the chosen number of iterations% l=40 will be used since this is the number used in the JACOBI solution. l = 40; % sets the number of iterationsm = l + 1; % allocates the space used to store each iteration vector = [1:m]; % assigns a row with the appropriate spaces row = ones(size(vector)); Vnode1 = row; % each node voltage has the required storage spacesVnode2 = row;Vnode3 = row;Vnode4 = row;Vnode5 = row; for n = 1:m Vnode1(n) = 1.0 + 0.00i; % inserts the assumed initial node voltagesVnode2(n) = 1.0 + 0.00i;Vnode3(n) = 1.0 + 0.00i;Vnode4(n) = 1.0 + 0.00i;Vnode5(n) = 1.0 + 0.00i; end S2star = P2 - Q2; % conjugates of the specified node loadings S3star = P3 - Q3;S4star = P4 - Q4;S5star = P5 - Q5; % Now Proceed with the GAUSS SEIDAL SOLUTION% Following the same method as in the worked example % Page 16 of the assignment notes for n = 1:l % NOW UPDATE THE NODE VOLTAGES TWENTY FIVE TIMES % SIMILAR TO THE CALCULATION PROCEDURES % AS DETAILED ON PAGE 5 OF THE ASSIGNMENT NOTES V2star(n) = conj(Vnode2(n)); % conjugate of V2 V3star(n) = conj(Vnode3(n)); % conjugate of V3 V4star(n) = conj(Vnode4(n)); % conjugate of V4 V5star(n) = conj(Vnode5(n)); % conjugate of V5 % (Step 1) Solving for the updated values of the currents I2(n) = S2star/V2star(n); % (Step 2) Solving the summation of Y(kj)V(j) for all busbars Sum2(n) = (Y(2,1) * Vnode1(n)) + (Y(2,3) * Vnode3(n)) + (Y(2,5) * Vnode5(n)); % (Step 3) Solving for the new value of node voltage V2(n) = ( I2(n) - Sum2(n) ) / Y(2,2); Vnode2(n) = V2(n); % Immediately updates the current value of Vnode2 % (Step 4) % Now determine the new value of Vnode3 % This calculation uses the new updated value of Vnode2
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I3(n) = S3star/V3star(n); Sum3(n) = (Y(3,1) * Vnode1(n)) + (Y(3,2) * Vnode2(n)) + (Y(3,4) * Vnode4(n)); V3(n) = ( I3(n) - Sum3(n) ) / Y(3,3); Vnode3(n) = V3(n); % Immediately updates the current value of Vnode3 % (Step 5) % Now determine the new value of Vnode4 % This calculation uses the new updated value3 of Vnode2 and Vnode3 I4(n) = S4star/V4star(n); Sum4(n) = (Y(4,3) * Vnode3(n)) + (Y(4,5) * Vnode5(n)); V4(n) = ( I4(n) - Sum4(n) ) / Y(4,4); Vnode4(n) = V4(n); % Immediately updates the current value of Vnode4 % (Step 5) % Now determine the new value of Vnode5 % This calculation uses the new updated value3 of Vnode2 Vnode3 & Vnode4 I5(n) = S5star/V5star(n); Sum5(n) = (Y(5,1) * Vnode1(n)) + (Y(5,2) * Vnode2(n)) + (Y(5,4) * Vnode4(n)); V5(n) = ( I5(n) - Sum5(n) ) / Y(5,5); Vnode5(n) = V5(n); % Immediately updates the current value of Vnode4
The Matlab output program code for the GAUSS SEIDAL SOLUTION of the network load-flow problem. Command Window Output:: Electrical Power Utilisation 6EJ022
Gauss Seidal Method: Note the faster convergence: Sixteen Iterations
Method 3 Acceleration Factors with the Gauss Seidal method of solution:
To reduce the number of iterations required to reach a final solution, attempts have been made to speed up the convergence process. A simple yet effective approach is to project forward each calculated voltage in the direction of the way the iteration values are moving. If the trend indicate indicates the successive values are reducing a factor is applied to accelerate this progression. Conversely, if there is a rising trend, then the accelerating factor needs to project the voltage further upwards.
The acceleration factor needs to modify the voltages according to the following considerations:
Vk(acc) = Vk(new) ( 1) Vk(old)
= Vk(old) + (Vk(new) Vk(old)
Clearly, as convergence is approach the acceleration becomes less pronounced. The modification is applied immediately each busbar voltage is calculated. For large scale networks an acceleration factor of 1.6 is often used. For the limited network being analysed, an acceleration factor of 1.25 will be more effective.
Students could experiment with different values to predict an optimum value.
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The Matlab program code for the GAUSS SEIDAL SOLUTION of the network load-flow problem can now be modified to include a suitable ACCELERATION FACTOR.
Matlab Program.
% BSc(Hons) Elelectrical Electronic Engineering % Module 6EJ998 % Assignment No2 % MODELLING LOAD FLOW STUDIES USING MATLAB % USING THE GAUSS SEIDAL METHOD PLUS AN ACCELERATION FACTOR
% YEAR 2010/11 % STUDENT ........... % NETWORK (see Fig 1) % Input (1) % Insert the Network Admittance Matrix % (see page 2 of supplied assignment paperwork ignoring shunt capacitive admittance) format shortY(1,1) = 10.958905 - 26.027398i;Y(1,2) = -3.424658 + 7.534247i;Y(1,3) = -3.424658 + 7.534247i;
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Y(1,4) = 0.00;Y(1,5) = -4.109589 + 10.958904; Y(2,1) = -3.424658 + 7.534247i;Y(2,2) = 11.672080 - 26.090949i;Y(2,3) = -4.123711 + 9.278351i;Y(2,4) = 0.00;Y(2,5) = -4.123711 + 9.278351i; Y(3,1) = -3.424658 + 7.534247i;Y(3,2) = -4.123711 + 9.278351i;Y(3,3) = 10.457198 - 23.154061i;Y(3,4) = -2.926829 + 6.341463i;Y(3,5) = 0.00; Y(4,1) = 0.00;Y(4,2) = 0.00;Y(4,3) = -2.926829 + 6.341463i;Y(4,4) = 7.050541 - 15.619814i;Y(4,5) = -4.123711 + 9.278351i; Y(5,1) = -4.109589 + 10.958904i;Y(5,2) = -4.123711 + 9.278351i;Y(5,3) = 0.00;Y(5,4) = -4.123711 + 9.278351i;Y(5,5) = 12.357012 - 29.515606i; % Input (2)% Input the given node loadings .....see diagram on page 1 .... % .... of the supplied assignment paperwork% A positive (+ sign) is for generated power,% The negative (- sign) indicates load taken from the network.% Per unit loading is used with a base loading taken at 100MVA P2 = - 0.4; % p.u. active power loadingP3 = - 0.25; % p.u. active power loadingP4 = - 0.4; % p.u. active power loadingP5 = - 0.5; % p.u. active power loading Q2 = - 0.2i; % p.u. reactive power loadingQ3 = - 0.15i; % p.u. reactive power loadingQ4 = - 0.2i; % p.u. reactive power loadingQ5 = - 0.2i; % p.u. reactive power loading % The voltage at the generation busbar, node (1) is 1.0 p.u. % This voltage is fixed and therefore determines the reference ....% ... or SLACK BUSBAR must be Node(1) % GAUSS SEIDAL PLUS ACCELERATION FACTOR SOLUTION% The number of iteration will need to be set% Set the value of 'l' the chosen number of iterations% l=40 will be used since this is the number in the JACOBI solution. l = 40; % sets the number of iterationsm = l + 1; % allocates the space used to store each iteration vector = [1:m]; % assigns a row with the appropriate spaces row = ones(size(vector)); Vnode1 = row; % each node voltage has the required storage spacesVnode2 = row;Vnode3 = row;
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Vnode4 = row;Vnode5 = row; for n = 1:m Vnode1(n) = 1.0 + 0.00i; % inserts the assumed initial node voltagesVnode2(n) = 1.0 + 0.00i;Vnode3(n) = 1.0 + 0.00i;Vnode4(n) = 1.0 + 0.00i;Vnode5(n) = 1.0 + 0.00i; end S2star = P2 - Q2; % conjugates of the specified node loadings S3star = P3 - Q3;S4star = P4 - Q4;S5star = P5 - Q5; % Now Proceed with the GAUSS SEIDAL SOLUTION% Following the same method as in the worked example % Page 16 of the assignment notes for n = 1:l % NOW UPDATE THE NODE VOLTAGES TWENTY FIVE TIMES % SIMILAR TO THE CALCULATION PROCEDURES % AS DETAILED ON PAGE 5 OF THE ASSIGNMENT NOTES V2star(n) = conj(Vnode2(n)); % conjugate of V2 V3star(n) = conj(Vnode3(n)); % conjugate of V3 V4star(n) = conj(Vnode4(n)); % conjugate of V4 V5star(n) = conj(Vnode5(n)); % conjugate of V5 % (Step 1) Solving for the updated values of the currents I2(n) = S2star/V2star(n); % (Step 2) Solving the summation of Y(kj)V(j) for all busbars Sum2(n) = (Y(2,1) * Vnode1(n)) + (Y(2,3) * Vnode3(n)) + (Y(2,5) * Vnode5(n)); % (Step 3) Solving for the new value of node voltage V2(n) = ( I2(n) - Sum2(n) ) / Y(2,2); Vnode2(n+1) = V2(n); % Immediately updates the current value of Vnode2 % (Step 4) % Now determine the new value of Vnode3 % This calculation uses the new updated value of Vnode2 I3(n) = S3star/V3star(n); Sum3(n) = (Y(3,1) * Vnode1(n)) + (Y(3,2) * Vnode2(n+1)) + (Y(3,4) * Vnode4(n)); V3(n) = ( I3(n) - Sum3(n) ) / Y(3,3); Vnode3(n+1) = V3(n); % Immediately updates the current value of Vnode3 % (Step 5) % Now determine the new value of Vnode4 % This calculation uses the new updated value3 of Vnode2 and Vnode3
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I4(n) = S4star/V4star(n); Sum4(n) = (Y(4,3) * Vnode3(n+1)) + (Y(4,5) * Vnode5(n)); V4(n) = ( I4(n) - Sum4(n) ) / Y(4,4); Vnode4(n+1) = V4(n); % Immediately updates the current value of Vnode4 % (Step 5) % Now determine the new value of Vnode5 % This calculation uses the new updated value3 of Vnode2 Vnode3 & Vnode4 I5(n) = S5star/V5star(n); Sum5(n) = (Y(5,1) * Vnode1(n)) + (Y(5,2) * Vnode2(n+1)) + (Y(5,4) * Vnode4(n+1)); V5(n) = ( I5(n) - Sum5(n) ) / Y(5,5); Vnode5(n+1) = V5(n); % Immediately updates the current value of Vnode5 % Insert Accelerating Factor (alpha) alpha = 1.25; Vacc2(1) = 1; Vacc2(n+1) = Vnode2(n) + alpha * (Vnode2(n+1) - Vnode2(n)); Vacc3(n+1) = Vnode3(n) + alpha * (Vnode3(n+1) - Vnode3(n)); Vacc4(n+1) = Vnode4(n) + alpha * (Vnode4(n+1) - Vnode4(n)); Vacc5(n+1) = Vnode5(n) + alpha * (Vnode5(n+1) - Vnode5(n)); Vnode2(n+1) = Vacc2(n+1); Vnode3(n+1) = Vacc3(n+1); Vnode4(n+1) = Vacc4(n+1); Vnode5(n+1) = Vacc5(n+1);end disp (':')disp (' Electrical Power Utilisation 6EJ022 ')disp (':')disp (' Gauss Seidal Plus Acceleration Factor Solution of Load Flow Assignment(2) 2010/11')disp (':')disp (' Successive Iteration Values ')disp (':')disp (' Vnode2 Vnode3 Vnode4 Vnode5 ')disp(':') d = 0;for c = 1:m fprintf(' %1.4f %1.4fi\t, %1.4f %1.4fi\t, %1.4f %1.4fi\t, %1.4f %1.4fi\n',... real(Vnode2(1+d)), imag(Vnode2(1+d)), real(Vnode3(1+d)),... imag(Vnode3(1+d)), real(Vnode4(1+d)), imag(Vnode4(1+d)), real(Vnode5(1+d)), imag(Vnode5(1+d)));d=d+1;end
The Matlab program output code for the GAUSS SEIDAL with ACCELERATION FACTOR solution of the network load-flow problem. Command Window Output:: Electrical Power Utilisation 6EJ022 : Gauss Seidal Solution of Load Flow Assignment (2) 2010/11 : Successive Iteration Values :
Gauss Seidal Plus Accelerating Factor Note even faster convergence after Twelve Iterations
Method 4: The Z-Matrix Method. Using a block substitution procedure:
The Y-matrix methods used with the Jacobi and the Gauss Seidal techniques generally have a poor convergence and in some cases the success iteration values may actually diverge. Much work has been undertaken in an attempt to overcome the problem, but generally with inconclusive results.
Another approach to the solution of this problem is to use the Z-matrix method.In the Y-matrix methods the network relationships are expressed in the form of an admittance matrix.
I = Y VBy inverting the Y-matrix equation we can derive an equation of the form
V = Z I ….. where Z = Y 1
The Y-matrix is generally a sparse matrix, i.e. contains a significant number of zeros. On the other hand the Z-matrix is always full and therefore provides an improved mathematical linkage between V and I.It would appear that the system node voltages could be calculated in a more direct manner from the set of Z-matrix equations. Unfortunately, no direct solution is possible. Once more an iterative strategy must be
adopted which will again produce a convergence to the required set of system voltages.
Since busbar 1 is taken as the slack busbar, for which the voltage conditions are fully specified, the remaining nodal admittance equations will have the form:
I 2 = Y 2 1 V 1 + Y 2 2 V 2 + Y 2 3 V 3 + Y 2 4 V 4 + Y 2 5 V 5 I 3 = Y 3 1 V 1 + Y 3 2 V 2 + Y 3 3 V 3 + Y 3 4 V 4 + Y 3 5 V 5 I 4 = Y 4 1 V 1 + Y 4 2 V 2 + Y 4 3 V 3 + Y 4 4 V 4 + Y 4 5 V 5 I 5 = Y 5 1 V 1 + Y 5 2 V 2 + Y 5 3 V 3 + Y 5 4 V 4 + Y 5 5 V 5
In this form the Y-matrix is not square and therefore cannot be inverted. Rearranging:
I 2 Y 2 1 V 1 = Y 2 2 V 2 + Y 2 3 V 3 + Y 2 4 V 4 + Y 2 5 V 5 I 3 Y 3 1 V 1 = Y 3 2 V 2 + Y 3 3 V 3 + Y 3 4 V 4 + Y 3 5 V 5
I 4 Y 4 1 V 1 = Y 4 2 V 2 + Y 4 3 V 3 + Y 4 4 V 4 + Y 4 5 V 5 I 5 Y 5 1 V 1 = Y 5 2 V 2 + Y 5 3 V 3 + Y 5 4 V 4 + Y 5 5 V 5
Which can be written as I = Y V where Y ' is square and non-singular and can be inverted.
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After inversion the following equation is obtained:
V = Z I where for the k th busbar I k = I k Y k 1 V 1
V 2 Z 2 2 ( I 2 Y 2 1 V 1 ) + Z 2 3 ( I 3 Y 3 1 V 1 ) + Z 2 4 ( I 4 Y 4 1 V 1 ) + Z 2 5 ( I 5 Y 5 1 V 1 )V 3 Z 3 2 ( I 2 Y 2 1 V 1 ) + Z 3 3 ( I 3 Y 3 1 V 1 ) + Z 3 4 ( I 4 Y 4 1 V 1 ) + Z 3 5 ( I 5 Y 5 1 V 1 )V 4 Z 4 2 ( I 2 Y 2 1 V 1 ) + Z 4 3 ( I 3 Y 3 1 V 1 ) + Z 4 4 ( I 4 Y 4 1 V 1 ) + Z 4 5 ( I 5 Y 5 1 V 1 )V 5 Z 5 2 ( I 2 Y 2 1 V 1 ) + Z 5 3 ( I 3 Y 3 1 V 1 ) + Z 5 4 ( I 4 Y 4 1 V 1 ) + Z 5 5 ( I 5 Y 5 1 V 1 )
where the second summation is constant since V 1 is fixed
and
The required computational strategy is one that uses these last two equations iteratively until a converged voltage solution is obtained. Any Matlab code must be careful to designate the correct suffixes to the Z
which results from the inversion of Y matrix. The form of the computer program is illustrated below.
This program uses a block substitution technique similar to procedure followed in the Jacobi solution.
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The step by step process of the Z-matrix method with block substation illustrated above can now demonstrated by numerically solving the first set of iterations for the network shown in Fig1 on page 5.
Admittance Matrix but with the first row and column removed.
Admittance Matrix with the first row and column removed
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The Matlab program code for the Z-MATRIX WITH BLOCK SUBSTITUTION method of solution for the given network load-flow problem can now be written.
Y(4,3) = -2.926829 + 6.341463i;Y(4,4) = 7.050541 - 15.619814i;Y(4,5) = -4.123711 + 9.278351i; Y(5,1) = -4.109589 + 10.958904i;Y(5,2) = -4.123711 + 9.278351i;Y(5,3) = 0.00;Y(5,4) = -4.123711 + 9.278351i;Y(5,5) = 12.357012 - 29.515606i; % Input (2)% Input the given node loadings .....see diagram on page 1 .... % .... of the supplied assignment paperwork% A positive (+ sign) is for generated power,% The negative (- sign) indicates load taken from the network.% Per unit loading is used with a base loading taken at 100MVA P2 = - 0.4; % p.u. active power loadingP3 = - 0.25; % p.u. active power loadingP4 = - 0.4; % p.u. active power loadingP5 = - 0.5; % p.u. active power loading Q2 = - 0.2i; % p.u. reactive power loadingQ3 = - 0.15i; % p.u. reactive power loadingQ4 = - 0.2i; % p.u. reactive power loadingQ5 = - 0.2i; % p.u. reactive power loading % The voltage at the generation busbar, node (1) is 1.0 p.u. % This voltage is fixed and therefore determines the reference ....% ... or SLACK BUSBAR must be Node(1) % Z-MATRX SOLUTION% The sub-matrix of the actual 5x5 network admittances (Y) ....% but with the first row and column removed now needs to be obtained% This sub-matrix is called Ysub and is easily derived from ....% the network admittance matrix by using the following command Ysub = Y(2:5,2:5); % The inverse of this sub-matrix gives the required Z-matrix Z = inv(Ysub); % Some care needs to taken in respect of the Z-matrix suffixes% This can now be rectified by adding a further ....% row(1) and column(1) of zeros to the Z-matrix. % Note the pime (') after the matrix is a transpose command% Rows are changed columns and columns become rows A = [0;0;0;0]; Z = [A Z]; B = [0;0;0;0;0];Zt = [B Z']; % Zt is the transpose of the Z-matrix with an added set of zeros % for the first row and column% A further transposition brings back the modified Z-matrix Z = Zt'; % The number of iteration will need to be set% Set the value of 'l' the chosen number of iterations
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% l=40 will be used since this is the number in the JACOBI solution. l = 40; % sets the number of iterationsm = l + 1; % allocates the space used to store each iteration vector = [1:m]; % assigns a row with the appropriate spaces row = ones(size(vector)); Vnode1 = row; % each node voltage has the required storage spacesVnode2 = row;Vnode3 = row;Vnode4 = row;Vnode5 = row; for n = 1:m Vnode1(n) = 1.0 + 0.00i; % inserts the assumed initial node voltagesVnode2(n) = 1.0 + 0.00i;Vnode3(n) = 1.0 + 0.00i;Vnode4(n) = 1.0 + 0.00i;Vnode5(n) = 1.0 + 0.00i; end S2star = P2 - Q2; % conjugates of the specified node loadings S3star = P3 - Q3;S4star = P4 - Q4;S5star = P5 - Q5; % Now Proceed with the Z-MATRIX SOLUTION% Following the same method as in the worked example % Pages 28 and 29 of the assignment notes % STEP 1.% Obtain the values of the constant terms .... % from the summation of Z(kj)Y(j1)V(1) for n = 1 : mConst2 = Z(2,2) * Y(2,1) * Vnode1(n) + Z(2,3) * Y(3,1) * Vnode1(n) + Z(2,4) * Y(4,1) * Vnode1(n) + Z(2,5) * Y(5,1) * Vnode1(n);Const3 = Z(3,2) * Y(2,1) * Vnode1(n) + Z(3,3) * Y(3,1) * Vnode1(n) + Z(3,4) * Y(4,1) * Vnode1(n) + Z(3,5) * Y(5,1) * Vnode1(n);Const4 = Z(4,2) * Y(2,1) * Vnode1(n) + Z(4,3) * Y(3,1) * Vnode1(n) + Z(4,4) * Y(4,1) * Vnode1(n) + Z(4,5) * Y(5,1) * Vnode1(n);Const5 = Z(5,2) * Y(2,1) * Vnode1(n) + Z(5,3) * Y(3,1) * Vnode1(n) + Z(5,4) * Y(4,1) * Vnode1(n) + Z(5,5) * Y(5,1) * Vnode1(n); % STEP 2.% Obtain the values of the node current input terms .... % from Skstar/Vkstar V2star(n) = conj(Vnode2(n));V3star(n) = conj(Vnode3(n));V4star(n) = conj(Vnode4(n));V5star(n) = conj(Vnode5(n)); I2(n) = S2star / V2star(n); I3(n) = S3star / V3star(n);I4(n) = S4star / V4star(n);I5(n) = S5star / V5star(n); % STEP 3% Calculate the new (update) value of node Voltage
Z-MATRIX with BLOCK SUBSTITUTION Note even faster convergence after Four Iterations
Method 5 - The Z-Matrix Method: Using a forward substitution procedure
The block substitution procedure used in Method 4 is similar to some extent with the method used in the Jacobi solution.
Possibly, by using a forward substation technique similar to that adopted in the Gauss Seidal method the number of iterations in the Z-Matrix method can be reduced still further.
Using this approach, each value of node voltage must be immediately revised after each new calculated update.
The Matlab program code for the Z-MATRIX SOLUTION WITH FORWARD SUBSTITUTION can easily be produce by simple modification of the block substitution method, see program code below:
Y(3,5) = 0.00; Y(4,1) = 0.00;Y(4,2) = 0.00;Y(4,3) = -2.926829 + 6.341463i;Y(4,4) = 7.050541 - 15.619814i;Y(4,5) = -4.123711 + 9.278351i; Y(5,1) = -4.109589 + 10.958904i;Y(5,2) = -4.123711 + 9.278351i;Y(5,3) = 0.00;Y(5,4) = -4.123711 + 9.278351i;Y(5,5) = 12.357012 - 29.515606i; % Input (2)% Input the given node loadings .....see diagram on page 1 .... % .... of the supplied assignment paperwork% A positive (+ sign) is for generated power,% The negative (- sign) indicates load taken from the network.% Per unit loading is used with a base loading taken at 100MVA P2 = - 0.4; % p.u. active power loadingP3 = - 0.25; % p.u. active power loadingP4 = - 0.4; % p.u. active power loadingP5 = - 0.5; % p.u. active power loading Q2 = - 0.2i; % p.u. reactive power loadingQ3 = - 0.15i; % p.u. reactive power loadingQ4 = - 0.2i; % p.u. reactive power loadingQ5 = - 0.2i; % p.u. reactive power loading % The voltage at the generation busbar, node (1) is 1.0 p.u. % This voltage is fixed and therefore determines the reference ....% ... or SLACK BUSBAR must be Node(1) % Z-MATRX SOLUTION% The sub-matrix of the actual 5x5 network admittances (Y) ....% but with the first row and column removed now needs to be obtained% This sub-matrix is called Ysub and is easily derived from ....% the network admittance matrix by using the following command Ysub = Y(2:5,2:5); % The inverse of this sub-matrix gives the required Z-matrix Z = inv(Ysub); % Some care needs to taken in respect of the Z-matix suffixes% This can now be rectified by adding a further ....% row(1) and column(1) of zeros to the Z-matrix. % Note the pime (') after the matrix is a transpose command% Rows are changed columns and columns become rows A = [0;0;0;0]; Z = [A Z]; B = [0;0;0;0;0];Zt = [B Z']; % Zt is the transpose of the Z-matrix with an added set of zeros % for the first row and column% A further transposition brings back the modified Z-matrix
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Z = Zt'; % The number of iteration will need to be set% Set the value of 'l' the chosen number of iterations% l=40 will be used since this is the number in the JACOBI solution. l = 40; % sets the number of iterationsm = l + 1; % allocates the space used to store each iteration vector = [1:m]; % assigns a row with the appropriate spaces row = ones(size(vector)); Vnode1 = row; % each node voltage has the required storage spacesVnode2 = row;Vnode3 = row;Vnode4 = row;Vnode5 = row; for n = 1:m Vnode1(n) = 1.0 + 0.00i; % inserts the assumed initial node voltagesVnode2(n) = 1.0 + 0.00i;Vnode3(n) = 1.0 + 0.00i;Vnode4(n) = 1.0 + 0.00i;Vnode5(n) = 1.0 + 0.00i; end S2star = P2 - Q2; % conjugates of the specified node loadings S3star = P3 - Q3;S4star = P4 - Q4;S5star = P5 - Q5; % Now Proceed with the Z-MATRIX SOLUTION% Following the same method as in the worked example % Pages 28 and 29 of the assignment notes % STEP 1.% Obtain the values of the constant terms .... % from the summation of Z(kj)Y(j1)V(1) for n = 1 : mConst2 = Z(2,2) * Y(2,1) * Vnode1(n) + Z(2,3) * Y(3,1) * Vnode1(n) + Z(2,4) * Y(4,1) * Vnode1(n) + Z(2,5) * Y(5,1) * Vnode1(n);V2star(n) = conj(Vnode2(n));I2(n) = S2star / V2star(n);V2(n) = Z(2,2) * I2(n) + Z(2,3) * I3(n) + Z(2,4) * I4(n) + Z(2,5) * I5(n) - Const2; % Now update V2star I2 and V2 V2star(n) = conj(V2(n));I2(n) = S2star / V2star(n);V2(n) = Z(2,2) * I2(n) + Z(2,3) * I3(n) + Z(2,4) * I4(n) + Z(2,5) * I5(n) - Const2; Const3 = Z(3,2) * Y(2,1) * Vnode1(n) + Z(3,3) * Y(3,1) * Vnode1(n) + Z(3,4) * Y(4,1) * Vnode1(n) + Z(3,5) * Y(5,1) * Vnode1(n);V3star(n) = conj(Vnode3(n));I3(n) = S3star / V3star(n);V3(n) = Z(3,2) * I2(n) + Z(3,3) * I3(n) + Z(3,4) * I4(n) + Z(3,5) * I5(n) - Const3;
Z-MATRIX with FORWARD SUBSTITUTION Now even faster convergence only Two Iterations
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The solution of the network load flow problem network illustrated on page 5 has been solved using the Matlab computer code supplied, (see pages 6 to 41).
Assignment network connections:For the 2010/11 second assignment the network connections have now been modified, see figure 1A below. The network base quantities are still MVA base = 100 MVA and V base = 132 kV. The node loadings are shown on the modified diagram and the inter-connecting line impedances are given in Table 1A. By modifying the given admittance data to suit the new conditions, solve the new Load Flow Problem by the same methods as used in the procedures (pages 6-41) of the above notes. This gives the values of the voltages at each node. Using the node voltage results calculate the current flow in each of the interlinking cables in the network of FIG 1A. A computer programme solution of this problem would be preferred.
