Psa.docx

Experiment # 01: The Working of a single phase power system and its theoretical variation through per unit analysis Objective: To observe the working of a single phase power system and its theoretical variation through per unit analysis.

Requirement: MATLAB software Lap Top

Theory: The Per Unit System: Power-system quantities such as voltage, current, power, and impedance are often expressed in perunit or percent of specified base values. One advantage of the per-unit system is that by properly specifying base quantities, the transformer equivalent circuit can be simplified. The ideal transformer winding can be eliminated, such that voltages, currents, and external impedances and admittances expressed in per-unit do not change when they are referred from one side of a transformer to the other. This can be a significant advantage even in a power system of moderate size, where hundreds of transformers may be encountered. The per-unit system allows us to avoid the possibility of making serious calculation errors when referring quantities from one side of a transformer to the other. Another advantage of the per-unit system is that the per-unit impedances of electrical equipment of similar type usually lie within a narrow numerical range when the equipment ratings are used as base values. Because of this, per-unit impedance data can be checked rapidly for gross errors by someone familiar with per-unit quantities. In addition, manufacturers usually specify the impedances of machines and transformers in per-unit or percent of nameplate rating. Per-unit quantities are calculated as follows: 𝑎𝑐𝑡𝑢𝑎𝑙 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 = 𝑏𝑎𝑠𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 where actual quantity is the value of the quantity in the actual units. The base value has the same units as the actual quantity, thus making the per-unit quantity dimensionless. Also, the base value is always a real number. Therefore, the angle of the per-unit quantity is the same as the angle of the actual quantity. Two independent base values can be arbitrarily selected at one point in a power system. Usually the base voltage 𝑉𝑏𝑎𝑠𝑒𝐿𝑁 and base complex power 𝑆𝑏𝑎𝑠𝑒1∅ are selected for either a single-phase circuit or for one phase of a three phase circuit. Then, in order for electrical laws to be valid in the per-unit system, the following relations must be used for other base values:

In above equations the subscripts LN and 1∅ denote ‘‘line-to-neutral’’ and ‘‘per-phase,’’ respectively, for three-phase circuits. These equations are also valid for single-phase circuits, where subscripts can be omitted. By convention, we adopt the following two rules for base quantities: 1. The value of 𝑆𝑏𝑎𝑠𝑒1∅ is the same for the entire power system of concern.

2. The ratio of the voltage bases on either side of a transformer is selected to be the same as the ratio of the transformer voltage ratings. With these two rules, a per-unit impedance remains unchanged when referred from one side of a transformer to the other.

One Line Diagram with Parameters:

Figure 1:One line Diagram

Procedure:       

Start MATLAB and go to Start -> Simulink -> Sim Power Systems -> Block Library. Open a new file by clicking file ->new->model or pressing ctrl +n. Take Voltmeter and Multimeter from measurements section of block library. Take Transformer, RLC branch and ground node from Elements section. Take AC voltage Source from sources section. Also paste the ‘power gui’ from the block library Sim Power Systems in your file. You may also connect Scope with voltmeters and ammeter. Scope can be taken from Simulink -> Commonly Used Blocks.  Make the connections as shown in the figure 2.  Set the values of various blocks as instructed below.  Run the Simulation  Open Power gui window and observe values of voltages and currents in ‘Steady State Voltages and currents’ section.  Enter the values observed in step 2 in observations given below. Values of Different Parameters: Source: Voltage peak =√2 × 220 Vrms, Frequency = 50 Hz, Measurement: voltage T1: 30kVA, 240/480V, R=0 p.u., X=0.1p.u. Measurement: all voltages and currents T2: 20kVA, 460/115V, R=0 p.u., X=0.1p.u. Measurement: all voltages and currents Transmission Line = X = 2 Measurement: branch voltage and current Load (also made from series RLC branch) = 0.9+0.2j Measurement: branch voltage and current Power Gui Block: Check ‘Discretize Electrical model’ & set Sample time = 0.0002

Simulated Figure in MATLAB:

Figure 2: One line Figure in MATLAB

Figure 3: Steady State Value of System

Observation Table: Parameters V source

Values 220 V rms

T2 Primary Voltage V load

423.66 V rms 105.91 V rms

Parameters T1 Primary Voltage T 2 Secondary Voltage I load

Values 220 V rms

Values 439.98 V rms

105.91 V rms

Parameters T1 Secondary Voltage V line

114.85 A rms

I line

28.76 A rms

57.82 V rms

Table 1: Experiment No 01

Home Task: Solve the above system using per unit analysis, calculate and compare the values measured in observation sets.

Solution: Manually by hand

Experiment # 02: The working of a three phase power system and its theoretical variation through per unit analysis Objective: To observe the working of a three phase power system and its theoretical variation through per unit analysis

Requirement:  

MATLAB software Lap Top

Theory: The per unit system for three phase systems: The three phase problems involving balanced systems can be solved on a per unit basis. In that case, the equations that are developed for single phase system can be used for three phase system as long as per phase values are used consistently. Therefore, 𝑆𝑏𝑎𝑠𝑒 1 𝐼𝑏𝑎𝑠𝑒 1 = 𝑉𝑏𝑎𝑠𝑒𝐿𝑁 𝑍𝑏𝑎𝑠𝑒 =

(VbaseLN )2 Sbase1

Note that, for a balance system: 𝑉𝑏𝑎𝑠𝑒𝐿𝑁 =

VbaseL−L √3

𝑆𝑏𝑎𝑠𝑒 3 𝑆𝑏𝑎𝑠𝑒 1 = 3 However, it has been customary in three phase system analysis to use line to line voltage and three phase voltamperes as the base values. Therefore, Sbase3 𝐼𝑏𝑎𝑠𝑒 =

Zbase =

√3𝑉𝑏 𝐿−𝐿 (VbL−L )2 Sbase3

The per unit impedance of the transformer remains same without taking into account whether it is converted from physical impedance values that are that are found by referring to high voltage side or low voltage side of the transformer. This can be accomplished by selecting separate appropriate base values for each side of the transformer. In other words, the design per unit impedance values of transformer based on rating of the coil. Since the rating of the coil cannot alter by simile change in connection. The per-unit impedance remains the same regardless of the three phase connection.

One Line Diagram Three Phase Circuit:

Figure 4:One line Diagram of three per unit system

Procedure:            

Start MATLAB and go to start Simulink Sim Power systems block library Open a new file by clicking file new model Take three phase VI measurement and Multimeter from measurement section of block library. Take three phase transformer, RLC branch and ground node from element section. Take three phase AC voltage source from source section. Also paste the “power gui” from the block library sim power system in file. Scope may also connect with voltmeter and ammeter. Scope can be taken from Simulink commonly used blocks. Make the connection Set the values of various blocks as instructed below. Run the simulation. Open the power gui window and observe values of voltage and current in steady state voltage and current section  Change values of different parameters and repeat the process Values of different parameters Source: Vrms = 13.2KV, Frequency = 50 Hz, Yg, Base voltage = 13.2KV T1: 50MVA, 13.2/132KV, R = 0pu, X= 0.1pu, Measurement: all voltages and currents T2: 45MVA, 132/11KV, R = 0pu, X= 0.1pu, Measurement: all voltages and currents Transmission line =X=20+30j, Measurement: branch voltage and current Load (also made from series RLC branch) = 3+4j, Measurement: branch voltage and current Power gui block: Check ‘discreet electrical model’ and set sample time =0.0002 .write the values

Simulated Figure in MATLAB:

Figure 5: Simulated three phase figure in MATLAB

Set 1:

Set 2: Now change the values as Transmission line=10+15j, load=1+3j

Observation Table: Set 1: Parameters V source

Values 7621.02 V rms

T2 Primary Voltage V load

59940.20 V rms 4994.33 V rms

Parameters T1 Primary Voltage T 2 Secondary Voltage I load

Values 6295.24 V rms 4994.33 V rms

Parameters T1 Secondary Voltage V line

Values 62944.95 V rms 3009.21 V rms

998.59 A rms

I line

83.43 A rms

Parameters

Values

Set 2: Now change the values as Transmission line=10+15j, load=1+3j Parameters Values Parameters Values

V source

7621.02 V rms

T2 Primary Voltage V load

53631.25 V rms 4468.12 V rms

T1 Primary Voltage T 2 Secondary Voltage I load

5569.73V rms 4468.12 V rms

T1 Secondary Voltage V line

55684.97 V rms 2125.50 V rms

1412.85 A rms

I line

117.91 A rms

Home Task: Solve the above system using per unit analysis, calculate and compare the values measured in observation sets. Solution: Manually by hand

Experiment # 03: Introduction to Power World Simulator Objective: Requirement:  

Power world Simulator software Lap Top

Theory: Overview PWS: Simulator seamlessly integrates two functions once commonly separated in power flow software. Graphical power system case editor. Power Flow package with many related analysis tools. Contingency Analysis, Time‐Step Simulation, Sensitivity Analysis, Loss Analysis, Fault Analysis, OPF, PVQV, ATC, SCOPF. Also, Transient Stability and Distributed Computing have recently become available Provide a better understanding of how to use Power World Simulator for power system analysis and visualization. Provide techniques for building good power system models and show how these techniques can be used to analyze system issues. Primary Goal of PWS make you aware of the capabilities of Simulator. We are frequently asked to add features to Simulator that are already available. We want you to make the most of our software.

Procedure: Starting power world simulator (PWS): Double click on the power world simulator icon present on the desktop; a blank window will be opened. Now click on the blue icon in the upper left corner and select ‘new case’. A white window will appear. Power world simulator is started. The next step is to implement a simple power system in PWS.

Making a new power system in PWS: Now we will make a power system in PWS. We will have to bring bus, generator, transmission line etc. in the blank window to do so. So each element is discussed here separately. First make sure that ‘Edit Mode’ is selected (present below the blue icon). Now go to the ‘Draw’ tab from the tabs present below the title bar. Now we can insert any element from the network menu. Bus: Select network->bus. Now in the white window click at any point where you want to place the bus. As soon as you click it a dialogue box as shown below will appear.

Keep the name and size etc. of the bus to the default value. Go to the ‘bus information’ tab and check the ‘system slack bus’ option. Click ‘ok’ to close the dialogue box. Now you can see the bus visible in the white window. Generator: Select network-> generator. Now in the white window click at bus1 so that generator will be attached to bus1. As soon as you click it a dialogue box as shown below will appear.

For the present case insert ‘500’ in the MW output box. Keep the remaining information to the default value and click ‘ok’. Now you can see the generator connected to bus1. Now select another bus (bus2) by following the same steps mentioned for bus1. But this time do not check the ‘system slack bus’ option. Load: Select network-> load. Now in the white window click at bus2 so that generator will be attached to bus2. As soon as you click it a dialogue box as shown below will appear.

Put ‘300’ and ‘100’ in ‘MW value’ and ‘MVAr value’ respectively under constant power option. You can change the orientation to ‘up’, ‘right’ etc as desired. Click ‘ok’ to close the dialogue box. Now load will be visible attached to bus2. Transmission Line: Transmission line is always connected between two busses. Select network-> Transmission Line. Now in the white window click at bus1 and drag your mouse to bus2 so that transmission line will be connected between bus1 and bus2. As soon as you do it a dialogue box as shown below will appear. Put 0.05, 0.1 and 0.15 for series resistance, series reactance and shunt charging respectively (note that these values are in per unit). Put 1000 in Limit A under MVA limits. Click ok to close the dialogue box.

Similarly insert a bus3 of 13.8kV, connect transformers between bus2 and bus 3 and between bus1 and bus3 (transformer connection are similar to transmission line). Connect the other elements as shown in the following figure.

Note: save your case periodically (save option is available in the blue icon menu of power world in the upper left corner). Electricity fluctuations can make you lose your work if you do not save.

Adding Bus and Line Fields: Values of voltage, power angle and power flow can be made visible on the transmission lines and busses. To do so write click on any bus or transmission line, select the ‘add new fields around bus/line’ option; a dialogue box will appear like the one below.

Click on any desired position and choose the quantity you want to make visible. Click ‘ok’ to finalize the selection. Repeat the procedure and select other desired quantity. Do this practice for every element present in the network.

Running the Simulation: Having completed the power system, click on the run mode present just under the blue icon. Go to the ‘tools’ tab and click on the play button. Now you can see the power flow represented by the moving arrows. If you have selected voltage, angle, MW flow and MVA flow on each transformer/ Transmission line, you will see the actual values of these quantities on your network. Show your simulation to the instructor. Your first simulation on PWS is complete. Now try to explore different options available in PWS and try to increase your knowledge for this software.

Analysis: After adding bus and line field and running simulation final diagram is

Home Task: Questions: 1. What is the meaning of slack bus? State the other types of bus as well. Mention the type of each bus in the above network. Answer:

In the above network bus 1 is slack bus, bus 3 is voltage controlled and bus 1 is load bus. Slack Bus:

Slack bus also called reference bus or swing bus. Its angle should be standard. It is also used to balance the active and reactive power in the system and serve as an angular reference for all other busses in the system which is set to 0.The voltage magnitude is also assumed to be 1 p.u . Load Bus (PQ bus):

It is the where may or ont be load connected. PQ are known and V and angle are unknown. Real and reactive will be assume to be zero. This bus used to find out of voltage and angle. Power will be flowing out thus the real and reactive power will be both negative. Voltage Controlled Bus (PV bus):

PV are known and Q and angle are unknown. These buses have a constant power generation control through a prime mover and constant bus voltage. Mostly, controlled where generator connected. We can say slack bus also voltage controlled bus.

2. What is the purpose and meaning of the circles present on each transformer and Transmission line. How does they change when your click on play button? Answer: It is used as a breaker when the fault occurs at the transmission lines. They are used to disconnect the faulted area from the main system.

3. What is the meaning of red boxes present at the corners of each element? Try to click on them and observe the corresponding change. Write your findings here. Answer: It is used as a breaker when the fault occurs at the transmission lines. They are used to disconnect the faulted area from the main system.

Experiment # 04: To calculate the Line Flows and Line Losses Objective: Requirement:  

Power world Simulator software Lap Top

Theory: Losses in Power System: Losses in electrical system can be determined in different ways. Electric technical losses occur as current flows through resistive materials and the magnetizing energy in the lines transformers and motors. However, the losses incurred in resistance materials can be reduced by adopting the following means a. Reducing the current b. reducing the resistance and the impedance c. Minimizing voltages. Electrical power system losses can be computed using several formulae in consideration of pattern of generation and loads, by means of any of the following methods: 1. Computing transmission losses as I2R 2. By differential power loss method 3. By computing line flows and line losses. 4. Analysing system parameters 5. By using B-loss coefficient formula 6. Load flow simulation 𝑃𝑙𝑜𝑠𝑠 𝑖𝑛 𝑂𝑛𝑒 𝑝ℎ𝑎𝑠𝑒 = 𝐼 2 𝑅 𝑃𝑙𝑜𝑠𝑠 𝑖𝑛 𝑡ℎ𝑟𝑒𝑒 𝑝ℎ𝑎𝑠𝑒 = 3𝐼 2 𝑅

Power Flow through the Transmission Line: An important goal of power transmission systems is to reliably deliver power at constant voltages at the receiving end under varying load conditions. The changes in load conditions result in changing the current through the transmission line causing line drop to change. That result in fluctuations in the voltage at the receiving end, which is undesirable. In the medium and long transmission lines, the line reactance is much higher than the line resistance and, therefore, for all practical purposes the line resistance can safely ignored and these lines are considered reactive in nature. From the power flow perspective, a higher line current for a

given transmission line means more reactive power absorbed by the line, which the source has to supply. If the receiving end, somehow, provides some of the reactive power absorbed by the lines, it would reduce the burden on the source thus reducing the line current and the line drop.

One Line Diagram in PWS:

Figure 6: One-line diagram of power system with line flows and line losses

Procedure:  



       

Double-click on the power world simulator icon to start the program. Then go to File > New Case. Go to Draw > Network and select Bus. Click on the one-line background at desired location. Enter Bus name as one, nominal voltage as 138 KV, and orientation as Right. Go to tab Bus Information select as Slack Bus. Click OK. Go to Draw > Network and select Generator. Click on the Bus one to attach the generator. Enter Generator MW output as 413 MW. Go to tab Display information and select orientation down. To save the case go to File > Save Case. Insert bus two by repeating the same procedure done for inserting bus one. Insert Load at bus two of values 200 MW, 100 MVAR. Select orientation as Up. Repeat the steps to insert another bus three somewhere on the bottom side of Bus one. Enter nominal voltage= 69 kV. Go to Draw > Network and select Transformer. Click on Bus 2 and then draw a line to Bus 3. Enter the parameters Series Resistance =0.02, Series Reactance(X)=0.08, Shunt Charging=0.1 Repeat the above steps to insert transformer between Bus1 and Bus 3. Insert Generator at bus 3 and enter value 300 MW and select orientation down. Insert Load at bus 3 of value as 400 MW, 200 MVAR. Select orientation as Up. Insert switched shunt capacitor bank of 10 MVAR at bus 3 and select the orientation Up. Save the case

Inserting Additional Fields:   

Go to Draw > Field and select Transmission Line Field. Click on the place near bus one. A box appears with different field options as shown in the figure. Enter 1 in the Near Bus and 2 in the Far Bus. Enter 4 in the Total Digits in Field and 2 in the Digits to Right of Decimal. Select Field value as MW Flow



Repeat the above three steps to insert another transmission line field near bus one and select Mvar flow.



In order to find transmission line losses, insert transmission line fields on the midpoint of transmission line between bus 1 and 2. Enter 4 in the Total Digits in Field and 2 in the Digits to Right of Decimal. Select Field value as MW Losses.

 



Repeat the above step and select Mvar Losses. Similarly insert transmission line fields (MW Flow and Mvar Flow) near bus 2. Make sure that 2 is entered in the Near Bus and 1 is entered in the Far Bus. Enter 4 in the Total Digits in Field and 2 in the Digits to Right of Decimal. By following the same procedure find all the unknown quantities indicated in the fig. 6. And match all the results of line flows and line losses as given in Fig. 7.

Analysis:

Figure 7: Results of line flows and line losses of the power system

Experiment # 05: Admittance matrix (Ybus) formulation in PWS and MATLAB Objective: Requirement:  

Power world Simulator software Lap Top

Theory: Admittance Matrix (Ybus): Bus admittance matrix (YBus) for an n-bus power system is square matrix of size n × n. The diagonal elements represent the self or short circuit driving point admittances with respect to each bus. The off-diagonal elements are the short circuit transfer admittances (or) the admittances common between any two numbers of buses. In other words, the diagonal element yii of the YBus is the total admittance value with respect to the ith bus and yik is the value of the admittance that is present between ith and kth buses. In power engineering, nodal admittance matrix (or just admittance matrix) or Y Matrix or Ybus is an N x N matrix describing a power system with N buses. It represents the nodal admittance of the buses in a power system. In realistic systems which contain thousands of buses, the Y matrix is quite sparse. Each bus in a real power system is usually connected to only a few other buses through the transmission lines. The Y Matrix is also one of the data requirements needed to formulate a power flow study.

In general the format of the Ybus matrix for an n-bus power system is as follows

Where

It is to be noted that Ybus is a symmetric matrix in which the sum of all the elements of the kth column is Ykk.

One Line Diagram for Ybus:

Figure 8: Diagram for Ybus

Procedure Network Formation:  



Open the power world simulator, take new case. Implement the above circuit by inserting the impedance values as given.  The properties of different elements are as follows Bus1: 138kV, slack bus Bus2: 138kV Bus3: 13.8kV Generator at bus1: Enter any random value in the MW and MVAR boxes (as it is attached to slack bus, it will adjust its power according to the requirement). Now go to the ‘fault parameters’ tab in the properties, the tab will look like the following fig:

 

  

Now enter the values in different boxes as entered in the fig above. Generator at bus2: MW a shown in the fig.1; go to the ‘fault parameters’ and enter ‘1.0’ in the reactance value boxes under positive, negative, zero and neutral to ground impedance.  Transmission line b/w bus 1 and 2: reactance 0.5, MVA limit A: 50  Transformer b/w bus1 and 3: reactance 0.2 MVA limit A: 50  Transformer b/w bus2 and 3: reactance 0.4 MVA limit A: 50 Go to case information ->solution details -> Ybus. Or go to fault analysis option and see the positive sequence y bus. Save Y bus in MATLAB format from the file drop down menu->save y bus or Jacobean. Now compute manually Y bus for the system and match the results.

Observations: Ybus from PWS

Ybus from manual calculations

𝑌𝑏𝑢𝑠

Manually Solution: Y-bus 𝑌11 𝑌12 𝑌13 𝑌𝑏𝑢𝑠 = [𝑌21 𝑌22 𝑌23 ] 𝑌31 𝑌32 𝑌33 𝑦10 + 𝑦12 + 𝑦13 −𝑦12 −𝑦13 −𝑦21 𝑦20 + 𝑦21 + 𝑦23 −𝑦23 𝑌𝑏𝑢𝑠 = [ ] −𝑦31 −𝑦32 𝑦30 + 𝑦31 + 𝑦32 −𝑗1 − 𝑗2 − 𝑗5 −(−𝑗2) −(−𝑗5) −𝑗1.25 − 𝑗2 − 𝑗2.5 −(−𝑗2.5) ] 𝑌𝑏𝑢𝑠 = [ −(−𝑗2) −(−𝑗5) −(−𝑗2.5) −0𝑗 − 𝑗5 − 𝑗2.5 −𝑗8 𝑗2 𝑗5 𝑌𝑏𝑢𝑠 = [ 𝑗2 −𝑗5.75 𝑗2.5 ] 𝑗5 𝑗2.5 −𝑗7.5

Ybus as a Jacobian save in MATLAB: Jac = sparse(6); Jac(1,1)= 1.00000;

−𝑗8 𝑗2 𝑗5 = [ 𝑗2 −𝑗5.75 𝑗2.5 ] 𝑗5 𝑗2.5 −𝑗7.5

Jac(4,4)= Jac(2,3)= Jac(2,6)= Jac(2,2)= Jac(2,5)= Jac(5,5)= Jac(3,3)= Jac(6,6)= Jac(3,2)= Jac(3,5)= Jac(6,2)= Jac(6,5)=

1.00000; -2.34670; 0.04676; 4.34591; 0.10000; 1.00000; 7.04072; 6.58332; -2.34670; -0.04390; 0.04390; -2.34670;

Home Task: Use the same bus data and code it for Y-bus formulation in MATLAB and show it in next lab. To input bus data to the Ybus code, a separate m file will be made in the following format S E R X Where S is the starting bus, E is the ending bus, R is the value of resistance of branch while X is its reactance. e.g. to input the data of branch between bus 1 and bus 2 in the above figure, the data will be written as 1 2 0.5 0.25 Similarly the data of branch between bus 2 and bus3 will be written in the next line and so on .Using these values formulate the complete Ybus matrix. Your code should not be general and it should itself decide the total number of busses present in the network.

MATLAB Code: Code Ybus: % Line Data for Y-Bus Formation. % | From | To | R | X | B/2 | % | Bus | Bus | | | linedata = [3 1 0 0.25 0; 1 4 0 0.4 0; 4 2 0 0.2 0; 1 2 0 0.125 0; 2 3 0 0.25 0;]; fb = linedata(:,1); % From bus number... tb = linedata(:,2); % To bus number... r = linedata(:,3); % Resistance, R... x = linedata(:,4); % Reactance, X... b = linedata(:,5); % Ground Admittance, B/2... z = r + i*x; % Z matrix...

y = 1./z; % To get inverse of each element... b = i*b; % Make B imaginary... nbus = max(max(fb),max(tb)); % no. of buses... nbranch = length(fb); % no. of branches... Y = zeros(nbus,nbus); % Initialise YBus... % Formation of the Off Diagonal Elements... for k=1:nbranch Y(fb(k),tb(k)) = Y(fb(k),tb(k))-y(k); Y(tb(k),fb(k)) = Y(fb(k),tb(k)); end % Formation of Diagonal Elements.... for m =1:nbus for n =1:nbranch if fb(n) == m Y(m,m) = Y(m,m) + y(n) + b(n); elseif tb(n) == m Y(m,m) = Y(m,m) + y(n) + b(n); end end end Y % Bus Admittance Matrix... Z = inv(Y)

Result: Y= 0.0000 -14.5000i 0.0000 + 8.0000i 0.0000 + 4.0000i 0.0000 + 2.5000i Z= 1.0e+14 * 0.0000 - 5.6295i 0.0000 - 5.6295i 0.0000 - 5.6295i 0.0000 - 5.6295i

0.0000 + 8.0000i 0.0000 -17.0000i 0.0000 + 4.0000i 0.0000 + 5.0000i

0.0000 - 5.6295i 0.0000 - 5.6295i 0.0000 - 5.6295i 0.0000 - 5.6295i

0.0000 + 4.0000i 0.0000 + 4.0000i 0.0000 - 8.0000i 0.0000 + 0.0000i

0.0000 - 5.6295i 0.0000 - 5.6295i 0.0000 - 5.6295i 0.0000 - 5.6295i

0.0000 + 2.5000i 0.0000 + 5.0000i 0.0000 + 0.0000i 0.0000 - 7.5000i

0.0000 - 5.6295i 0.0000 - 5.6295i 0.0000 - 5.6295i 0.0000 - 5.6295i

Experiment # 06: Verification of Kron Reduction of 4 bus power system through Power World Simulator Objective: Requirement:  

Power world Simulator software Lap Top

Theory: Nodal Elimination by Kron Reduction: Gaussian elimination removes the need for matrix inversion when solving the nodal equations of a large-scale power system. At the same time it is also shown that elimination of variables is identical to network reduction since it leads to a sequence of reduced-order network equivalents by node elimination at each step. This may be important in analyzing a large interconnected power system if there is special interest in the voltages at only s0me of the buses of t h e overall system. The Kron Reduction is a relatively simple technique for eliminating nodes from a network when the voltage or current at that node is zero. Consider an equation of the form Ax= b where A is an ( n X n ) real or complex valued matrix, x and b are vectors in either Rn or Cn . Assume that the b vector has a zero element in the n th row such that Ax= b is given as

We can then eliminate the kth row and kth column to obtain a reduced ( n - 1) number of equations of the form

The elimination is performed using the following elementary operations 𝑎𝑗𝑘 𝑛𝑒𝑤 = 𝑎𝑗𝑘 𝑜𝑙𝑑 −

𝑎𝑗𝑝 𝑎𝑝𝑘 𝑎𝑝𝑝

Original Figure:

One Line Diagram in PWS:

Procedure Network Formation:  



Open the power world simulator and take new case. Implement the above circuit figure system by inserting the impedance values as given.  The properties of different elements are as follows Bus1: 13.8kV, Slack bus Bus3: 138kV Bus4: 138 kV Bus 1&2: 138 kV Generator at bus 3: Enter any random value in the MW and MVAR boxes (as it is attached to slack bus, it will adjust its power according to the requirement). Now go to the ‘fault parameters’ tab in the properties, the tab will look like the following fig :  X= 1.25 positive, x=1 negative and x=1 zero

  

Now go to “power and voltage control” tab and enter the set point voltage 1.25. Now enter the values in different boxes as entered in the fig above. Generator at bus 4: MW a shown in the fig.1; go to the ‘fault parameters’ and enter X= 1.25 positive, x=1 negative and x=1 zero  Now go to “power and voltage control” tab and enter the set point voltage 0.85  Transmission line b/w bus 1 and 2: x=0.125 MVA limit A: 200  Transmission line b/w bus3 and 2: x= 0.25, MVA limit A: 200  Transmission line b/w bus 1 and 3: x=0.25, MVA limit A: 200  Transmission line b/w bus 4 and 2: x= 0.2, MVA limit A: 200  Transmission line b/w bus 1 and 4: x= 0.4, MVA limit A: 200  Go to case information ->solution details -> Ybus. Or go to fault analysis option and see the positive sequence y bus Verification Krown Reduction:  Go to “tools” tab and click on the “Equivalence” option as shown in Fig below.

 

The bus we want to remove we click on it select the “External “option like Bus 2 as shown above fig. After that we click on “Create the Equivalent” there will open a window like this as shown fig 5 below.







Then we check the following options as shown above fig.  Remove external objects from online  Delete empty Areas/Zones /Substation that occur from Equivalencing Then click on “Build Equivalent System” there will come a system in which 2nd bus is removed as shown in fig below.

Then again go to case information ->solution details -> Ybus. Or go to “fault analysis” option and see the positive sequence y bus

Observations: Ybus from PWS

Ybus from manual calculations

𝑌𝑏𝑢𝑠

−10.735𝑗 5.88𝑗 −7.85𝑗 = [ 5.88𝑗 4.85𝑗 1.176𝑗

4.85𝑗 1.176𝑗 ] −7.464𝑗

Home Task: Solve the System which we get after Kron reduction by hand compare the results with Ybus from PWS and attach the with manual submit in the lab. Solution: manually

Experiment # 07: Gauss Seidel Method MATLAB Programming Objective: Requirement:  

MATLAB software Lap Top

Theory: Consider a nonlinear equation of f(x)=0 with x being the variable. It can be rearranged to unknown value like x = g(x) (𝑘) If 𝑥 is a initial estimate of the variable x the following iterative sequence is formed. 𝑥 (𝑘+1) =𝑔(𝑥 (𝑘) ) Then at the end of the each iteration tolerance will be checked. |𝑥 (𝑘+1) -𝑥 (𝑘) | ≤ 𝜖

dx=1; x=input('Enter initial estimate'); iter=0; disp('Iter g dx x'); while abs(dx)>=0.001 && iter <50 iter=iter+1; g=-1/9*x^3+6/9*x^2+4/9; dx=g-x; x=x+dx; fprintf('%g',iter) disp([g,dx,x]) end

Example # 01: Now for multiple equations 𝑥1 + 𝑥1 𝑥2 = 10 𝑥1 + 𝑥2 = 6 Solution: These can be transformed like 10 1 + 𝑥2 𝑥2 = 6 − 𝑥1 𝑥1 = (0)

(0)

With initial estimate 𝑥1 = 1 and 𝑥2 = 1, the iterative sequence becomes 10 (1) 𝑥1 = =5 1+1

(1)

𝑥2 = 6 − 5 = 1 10 (2) 𝑥1 = =5 1+1 (2) 𝑥2 = 6 − 5 = 1

Via MATLAB Code: x=input('Enter initial estimates [x1; x2] -> '); dx=[1;1] xc=x; while max(abs(dx)) >=.001 xc(1)=10/(1+x(2)); xc(2)=6-xc(1); dx=x-xc x=xc end disp('Graphical display of of x1 and x2') x2=0:.1:10; x1=10*ones(1,length(x2))./(1+x2); x3 =-x2+6; plot(x2,x1,x2,x3),grid

Result: Enter initial estimates [x1; x2] -> [1;1] dx = 1 1 dx = -4 0 x= 5 1 dx = 0 0 x= 5 1

Graphical display of x1 and x2:

Home Task: The Gauss Seidel Method is an iterative algorithm for solving set of non-linear load flow equations. Nonlinear load flow equation is given by

The reactive power of system is given by

Write the program of Gauss seidal and solve for following circuit:

MATLAB Code: % IMPLEMENTATION OF GAUSS SEIDEL METHOD IN MATLAB % DESIGNED BY: % SHAHID NADEEM 2015-EE-731 % CREATED: 12- May-2018 % SEMESTER # 6, % SUBJECT: POWER SYSTEM ANALYSIS AND DESIGN % MNS UNIVERSITY OF ENGINEERING AND TECHNOLOGY, MULTAN format short g disp (' TABLE 6.1 & 6.4 PAGE # 327 LINE DATA FOR EXAMPLE 6.9 ') linedata=[1 2 0.01008, 0.05040, 3.815629, -19.078144, 10.25, 0.05125; 1 3 0.00744, 0.03720, 5.169561, -25.847809, 7.75, 0.03875; 2 4 0.00744, 0.03720, 5.169561, -25.847809, 7.75, 0.03875; 3 4 0.01272, 0.06360, 3.023705, -15.118528, 12.75, 0.06375]; disp (' TABLE 9.3 PAGE # 338 BUS DATA FOR EXAMPLE 9.2 ') busdata=[1 0, 0, 50, 30.99, 1.00, 0 1; 2 0, 0, 170, 105.35, 1.00, 0 2; 3 0, 0, 200, 123.94, 1.00, 0 2; 4 318, 0 , 80, 49.58, 1.02, 0 3]; % Bus Type: 1.Slack Bus 2.PQ Bus 3.PV Bus ss=i*linedata(:,8); y=linedata(:,5)+i*linedata(:,6); totalbuses = max(max(linedata(:,1)),max(linedata(:,2))); % total buses totalbranches = length(linedata(:,1)); % no. of branches ybus = zeros(totalbuses,totalbuses); for b=1:totalbranches ybus((linedata(b,1)),(linedata(b,2)))=-y(b); ybus((linedata(b,2)),(linedata(b,1))) =ybus((linedata(b,1)),(linedata(b,2))); end for c=1:totalbuses for d=1:totalbranches if linedata(d,1) == c || linedata(d,2) == c ybus(c,c) = ybus(c,c) + y(d) + ss(d); end end end disp('TABLE 9.3 PAGE # 338 BUS ADMITTANCE MATRIX FOR EXAMPLE 9.2') ybus; z=zeros(totalbuses,4); busnumber=busdata(:,1);

PG=busdata(:,2); QG=busdata(:,3); PL=busdata(:,4); QL=busdata(:,5); V=busdata(:,6); VV=V; ANG=busdata(:,7); type = busdata(:,8); P = (PG-PL)./100; % per unit active power at buses Q = (QG-QL)./100; % per unit reactive power at buses tol=1; iter=0; kk=input('Enter the tolerance for iteration '); %alfa=input('Enter the value of ALPHA '); alfa=1.6; while tol > kk for i = 2:totalbuses YV = 0; for k = 1:totalbuses if i~=k YV = YV + ybus(i,k)* V(k); % multiplying admittance & voltage end YV; end if busdata(i,8) == 3 %Calculating Qi for PV bus %Q(i) = -imag(conj(V(i))*(YV + ybus(i,i)*V(i))); Q(i) = -imag(conj(V(i))*(YV + ybus(i,i)*V(i))); busdata(i,3)=Q(i); end % end V(i) = (1/ybus(i,i))*((P(i)-1i*Q(i))/conj(V(i)) - YV); % Compute Bus Voltages. % Calculating Corrected Voltage for PV bus if busdata(i,8) == 3 vc(i)=abs(VV(i))*(V(i)/abs(V(i))); busdata(i,6)=vc(i); V(i)=vc(i); end % Calculating Accelerated Voltage for PQ bus if busdata(i,8) == 2 VACC(i)= VV(i)+alfa*(V(i)-VV(i)); busdata(i,6)=VACC(i); V(i)=VACC(i); end %V(i)=V; end iter = iter + 1; % Increment iteration count. tol = max(abs(abs(V) - abs(VV))); % Calculate tolerance. VV = V; end Q; iter; YV;

V; %real(VACC') z(1:totalbuses,1)=busdata(:,1); z(1:totalbuses,2)=busdata(:,8); z(1:totalbuses,3)=abs(busdata(:,6)); z(1:totalbuses,4)=radtodeg(angle(V)); disp(' Bus No. Bus Type Voltage z;

Angle

');

Conclusion:

Experiment # 08: Load Flow Analysis by Gauss Seidel Method Objective: 

To study load flow analysis of the given power system network by Gauss Seidel method using Power World Simulator.

Requirement:  

Power world Simulator software Lap Top

Theory: Power-Flow Solution by Gauss–Seidal: The Gauss Seidel Method is an iterative algorithm for solving set of non-linear load flow equations. Nonlinear load flow equation is given by

The reactive power of system is given by

Ik can be calculated from

One Line Diagram:

Bus Input Data for System:

Input Data and Unknowns:

Procedure:    

Open Power World Simulator and draw required circuit diagram bus is given Set value of generator(keep MW& MVAR Limit high), transmission lines, and buses according to requirement Select Run mode > Tools > Solve > Gauss Seidel Power Flow Check different iteration of Gauss Seidel Method. For which you need to go to tools >simulator options then click “Do only one iteration” shown below. [By default it will be unchecked and you will get just the final solution].



Now Set the MVAR limit to 200MVAR of Generator at bus 3 and go to tools->simulator options then click “check immediately” and perform same experiment.

Analysis: First Iteration

2nd Iteration:

Final Iteration:

Home Task: DO your calculations by manually for 2 iterations of above diagram and compare the results.

Solution:

Manually by hand

Experiment # 09: Load Flow Analysis by Newton Raphson Method Objective: To study load flow analysis of the given power system network by Newton Raphson method using Power World Simulator.

Requirement:  

Power world Simulator software Lap Top

Theory: Power Flow by Newton Raphson: The Newton Raphson method of load flow analysis is an iterative method which approximates the set of nonlinear simultaneous equations to a set of linear simultaneous equations using Taylor’s series expansion and the terms are limited to first order approximation. The load flow equations for Newton Raphson method are given

Where the Jacobian matrix

And

Jacobian matrix elements When

Use Gauss elimination and back substitution to solve

After solving above system of equation you need to update the values

Circuit Diagram:

Procedure:    



Open Power World Simulator and draw required circuit diagram. Set values of generator, transmission lines, and buses according to requirement. Select Run mode > Tools > Solve > Newton Raphson. Check different iteration of Gauss Seidel Method. For which you need to go to tools->simulator options then click “Do only one iteration” shown below. [By default it will be unchecked and you will get just the final solution]. Show each iteration of PWS and calculation for only 2 iteration

Analysis: First iteration:

2nd Iterartion:

Final Result:

Home Task: DO your calculations by manually for 2 iterations and compare the results.

The Slack bus voltage is 𝑉1 = 1.05 0 p.u, and the bus voltage magnitude is |𝑉3 = 1.04|p.u. Starting with (0) (0) (0) an initial estimate of |𝑉2 |=1.0, 𝛿2 = 0.0, and 𝛿3 = 0.0. Solution: Manually by hand

Experiment # 10: Three phase symmetrical faults Objective: Requirement:  

Power world Simulator software Lap Top

Theory: Introduction: A fault in a circuit is any failure which interferes with the normal flow of current. Most faults on transmission lines of 115 k V and higher are caused by lightning, which results in the flashover of insulators. The high voltage between a conductor and the grounded supporting tower causes ionization, which pro-vides a path to ground for the charge induced by the lightning stroke. Once the ionized path to ground is established, the resultant low impedance to ground allows the flow of power current from the conductor to ground and through the ground to the grounded neutral of a transformer or generator, thus completing the circuit. Line-to -line faults not involving ground are less common. Opening circuit breakers to isolate the faulted portion of the line from the rest of the system interrupts the flow of current in the ionized path and allows deionization to take place. After an interval of about 20 cycles to allow deionization, breakers can usually be reclosed without reestablishing the arc. Experience in the operation of transmission lines has shown that ultra-high-speed reclosing breakers successfully rec1ose after most faults. Of those cases where reclosing is not successful, many are caused by permanent fault s where reclosing would be impossible regardless of the interval between opening and reclosing. Perm anent faults are caused by lines being on the ground, by insulator strings breaking because of ice loads, by permanent damage to towers, and by surge arrester failures. Experience has shown that between 70 and 80% of transmission-line faults are single line-to-ground faults, which arise from the flashover of only one line to the tower and ground. Roughly 5 % of all faults involve all three phases. These are the so- called symmetrical three- phase faults. Other types of transmission-line faults are li ne-to-line faults, which do not involve ground, and double line-la-ground faults. The currents which flow in different parts of a power system immediately after the occurrence of a fault differ from those flowing a few cycles later just before circuit breakers are called upon to open the line on both sides of the fault.

Transients in RL series circuit: The selection of a circuit breaker for a power system depends not only on the current the breaker is to carry under normal opera ting conditions, but also on the maxi mum current it may have to carry momentarily and the current it may have to interrupt at the voltage of the line in which it is placed. In order to approach the problem of calculating the initial current when a system is short-circuited, consider what happens when an AC voltage is applied to a circuit containing cons tan t values of resistance and inductance. Let the applied voltage be Vmax sin ( w t + a), where t is zero at the time of applying the voltage. Then, a determines the magnitude of the voltage when the circuit is closed. If the instantaneous voltage is zero and increasing in a positive direction when it is applied by closing a switch, a is zero. If the voltage is at its positive maximum instantaneous value, a is pi/2. The differential equation is

The solution of this equation is

Fault calculations using Z bus: Z bus is very important in fault calculations for a power system containing any number of buses. In order to understand, consider a network show in figure and fault is occur at bus 2 and pre-fault voltage at bus 2 is Vf. Z bus is very important in fault calculations for a power system containing any number of buses. In order to understand, consider a network show in figure and fault is occur at bus 2 and pre-fault voltage at bus 2 is Vf.

Reactance diagram a

power system with three phase fault at bus 2

a power system with three phase fault at bus 2 simulated by Vf and –Vf in series

The prefix is chosen to indicate the changes in the voltages at the buses due to the current - If injected into bus 2 by the fault. The Z bus building algorithm, or some other means such as Y bus triangularization and inversion, can be used to evaluate the bus impedance matrix. The changes in the bus voltages due to -If are given by

The second row of this equation shows that

We recognize Z22 as the diagonal element of Z bus representing the the-venin impedance of the network at bus 2 . Substituting the expression for If

Thus, the voltages at all buses of the network can be calculated using the pre-fault voltage Vf of the fault bus and the elements in the column of Z bus corresponding to the fault bus. The calculated values of the bus voltages will yield the sub transient currents in the branches of the ne t work if the system Z bus has been formed with sub transient values for the machine reactances. In more general terms, when the three-phase fault occurs on bus 2 of a large-scale network, we have

and neglecting pre-fault load currents, we can then write for the voltage at any bus j during the fault

we can calculate the sub transient current Iij “ from bus i to bus j in the line of impedance Zb connecting those two buses,

One Line Diagram:

Procedure Network Formation:  

Open the power world simulator and implement the above circuit. The properties of different elements are as follows Bus1: 13.8kV, slack bus Bus2: 13.8kV Bus3: 138kV



Generator at bus1: enter any random value in the MW and MVAR boxes (as it is attached to slack bus, it will adjust its power according to the requirement).  Now go to the ‘fault parameters’ tab in the properties.  Now enter the values in different boxes.  Generator at bus2: MW a shown in the fig.1; go to the ‘fault parameters’ and enter ‘1.0’ in the reactance value boxes under positive, negative, zero and neutral to ground impedance.  Transmission line b/w bus 1 and 2: reactance 0.5, MVA limit A: 500  Transformer b/w bus1 and 3: reactance 0.2 MVA limit A: 500  Transformer b/w bus2 and 3: reactance 0.4 MVA limit A: 500  Running the fault analysis: Having completely implemented the system, switch to the run mode, go the ‘tool’ ribbon and click on the fault analysis button. This button is shown in the following fig:

On clicking this button a window will appear, similar to the following:

Vbus1 = 0.61926 I12 = 1.83184 Ig1= 3.63744 Y bus matrix is

Vbus2 = 0.38991 I23 = 1.54815 Ig2 = 3.33230

Y bus after manual calculation is

Analysis: Fault at 3rd Bus:

Vbus3 = 0.0000 I13 = 1.94951 I load = 0.0000

Fault at 2nd Bus:

Task: Manual Calculation: For manual calculations we will use the Z bus matrix method to find different voltages and currents. For this purpose form the Z bus matrix first. Z bus is calculated by taking invers of Y bus matrix using MATLAB. Z bus matrix is

Now note the pre-fault voltage of each bus. For this run the above system on PWS from the play button under ‘tools’ ribbon and see the voltages at each bus. Pre- fault voltages: V1 = 1pu

V2 = 1pu

V3 = 0.90pu

Now having found these voltages and Z bus matrix, calculate all the currents and voltages under fault conditions Vbus1 = 0.6091

Vbus2 = 0.3338

Vbus3 = 0.0000

I12 = 1.7324 Manually Calculation by Hand:

I23 = 1.6236

I13 = 1.8754