FACULTY OF ENGINEERING
LAB SHEET
EET 2036
POWER SYSTEMS ANALYSIS
TRIMESTER 2 (2010/2011)
EET 1 - Symmetrical Short Circuit Analysis and Determination of Circuit Breaker Rating
EET 2 - Economic Load Dispatch and Power Flow Analysis
*Note: On-the-spot evaluation will be carried out during or at the end of the experiment. Students are advised to read through this lab sheet before doing experiment. Your performance, teamwork effort, and learning attitude will count towards the marks.
Experiment: PS-1
SYMMETRICAL SHORT CIRCUIT ANALYSIS AND
DETERMINATION OF CIRCUIT BREAKER RATING
Objectives:
Ø To perform the symmetrical fault analysis and determine for circuit breaker rating.
Ø To calculate the current supplied by each generator and bus voltage magnitudes during fault
Introduction:
Fault in a circuit is any failure which interferes with the normal flow of current. Most faults on transmission lines of 115 kV 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 provides 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 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.
Most of the faults on the power system lead to a short-circuit condition. When such a condition occurs, a heavy current (called short-circuit current) flows through the equipment, causing considerable damage to the equipment and interruption of service to the consumers.
Fault Classifications
The types of fault commonly occurring in practice are as follows:
(a) Line to ground : Line to ground faults are caused by a transmission line touching the ground. Wind, ice loading, or an accident such as a tree limb falling on a line can cause a line to ground fault. This category accounts for about 70% of all transmission line short circuit faults.
(b) Line to line : These faults are normally caused by high winds blowing one line into another, or by a line breaking and falling on a line below it. These account for about 15% of transmission faults.
(c) Double line to ground : These category is caused by the same things that cause single line to ground faults, except two lines are involved instead of one. These account for about 10% of transmission line faults.
(d) Three-phase faults : If a line condition occurs in which all three phases are shorted together, either by something falling on the phase conductors, an equipment failure, or all three lines falling to the ground, it is called a three-phase fault. These are relatively rare, accounting for only about 5% of all transmission line faults.
The fault current that flows depends on the equivalent Thevenin voltage, and the equivalent impedance at the fault terminals and the fault impedance, as illustrated in Figure 1.1.
Figure 1-1 : Simple equivalent circuit for fault current calculation
Three-phase faults are called symmetrical faults which give rise to symmetrical currents (i.e. equal fault currents in the lines with 120 degree displacement). Other types of transmission-line faults (line to ground, line to line and double line to ground faults) cause an imbalance between the phases, and so they are called unsymmetrical faults. In this experiment, the symmetrical short-circuit analysis is performed and the ratings of CBs are determined.
Software to be used for Symmetrical Fault Calculations:
The software, ‘Power World Simulator’ is used in this experiment to study the symmetrical fault and determine CB ratings.
Experimental Procedure
Test System
A power system consisting of 3 generators is shown in Figure 1-2.
Figure 1-2: A Three-Generator Power System
(Note: Do not save any changes that you have made in the original file. Save the file in a new file. Record the results.)
1. Launch the program by double-clicking on the PowerWorld Simulator icon on the desktop. Click on the File Menu and Open File “Problem 7-20”.
2. Click on each symbol, and check out what they represent. All data are on a 1000MVA base.
3. Complete Tables 1-1, 1-2, and 1-3
Table 1-1: Generator data for Symmetrical Short Circuit Program
Bus / Generator Subtransient Reactance-X” (per unit) –positive sequence5
6
7
Table1-2: Line Data for Symmetrical Short Circuit Program
Bus to Bus / Equivalent Positive-Sequence Series Reactance (per unit)1-2
2-3
2-4
Table 1-3: Transformer Data for Symmetrical Short Circuit Program
Bus to Bus / Equivalent Positive-Sequence Series Reactance (per unit)1-5
7-3
4-6
4. Perform fault analysis and determine the current supplied by each generator and the per-unit bus voltage magnitudes at each bus for a three-phase fault at bus 2.
5. Repeat step 4, for a fault at bus 4.
6. Repeat step 4, for a fault midway between buses 1 and 2. Determining the values for line faults requires that the line be split, with a fictitious bus added at the point of the fault. The original line impedance is then allocated to the two new lines based on the fault location, 50% each for this case. Fault calculations are then the same as for a bus fault. This is done automatically in PowerWorld Simulator by first right-clicking on a line, and then selecting ‘Fault..’. The Fault dialog appears as before, except now the fault type is changed to ‘In-Line Fault’. Set the location percentage field to 50% to model a fault midway between buses 1 and 2.
7. For the test system shown in Figure 1-2;
a) draw the positive-sequence reactance diagram in per-unit on a 1000MVA base.
b) calculate the equivalent Thevenin reactance in per-unit at the fault terminals if the short-circuit occurs at bus 2.
c) if the prefault voltage is 1.05 p.u, calculate the subtransient fault current in per unit at the fault and verify the results with the fault current obtained in step 4.
8. Determine the rating of the circuit breaker CB1 which is placed between Bus 4 and the transformer if the nominal at Bus 4 is 345kV.
9. Disconnect the generator at bus 7, and determine the fault current for three-phase faults at each of the buses 1, 2, 3 and 4.
10. A large power system data is stored in file “Example 6.13”. Open the file and determine the per-unit current and actual current in amps supplied by each of the generators for a fault at the LAUF69 bus. During the fault, what percentage of the system buses has voltage magnitudes below 0.75 per-unit.
11. Repeat step 10, for the fault at the AMANS69 bus.
Exercise
1. What can you conclude from this experiment?
2. What software are you using? What is the number of buses of the system used in the first part of the experiment? How many generator buses are there?
3. Name the different types of faults.
4. Suggest methods to limit the fault current during fault.
5. How do you determine the rating of circuit breakers based on symmetrical fault study?
Lab Assessment:
· On-Spot evaluation (40%).
· Experiment results & Lab report (60%).
Experiment: PS-2
Part I: ECONOMIC LOAD DISPATCH
Objectives
Ø To obtain economic load dispatch for a general power system using the MATLAB program
Introduction
The simplest economic dispatch problem is the case when transmission line losses are neglected. That is, the problem model does not consider the system configuration and line impedances. In essence, the model assumes that the system is only one bus with all generation and loads connected to it as shown schematically in Figure 2-1
Figure 2-1: Plants connected to a common bus
Since transmission losses are neglected, the total demand PD is equal to the sum of all power generation. A cost function Ci is assumed to be known for each plant. The problem is to find the real power generation for each plant such that the objective function (i.e: total production cost) as defined by the equation
(2-1)
is minimum, subject to the constraint
(2-2)
where Ct is the total production cost, Ci is the production cost of ith plant, Pi is the generation of ith plant, PD is the total load demand, and ng is the total number of dispatchable generating plants.
A typical approach is to augment the constraints into objective function by using the Lagrange multiplier
(2-3)
The minimum of L is found by equating the partials derivatives of the function with respect to Pi and λ to zero.
(2-4)
(2-5)
Equation (2-4), results in
Since
and therefore the condition for optimum dispatch is
i=1,….,ng (2-6)
or
(2-7)
Second condition, given by (2-5), results in
(2-8)
Equation (2-8) is precisely the power flow equality constraint. In summary, when losses are neglected with no generator limits, for most economic operation, all plants must operate at equal increment production cost, while satisfying the equality constraint given by (2-8). In order to find the solution, (2-7) is solved for Pi as
i=1,….,ng (2.9)
The relations given by (2.9) are known as the coordination equations. They are functions of λ. An analytical solution can be obtained for λ by substituting (2.9) in (2-8), i.e
(2-10)
or
(2-11)
The value of λ found from (2-11) is substituted in (2-7) to obtain the optimal scheduling of generation. The solution for economic dispatch neglecting losses has been found analytically.
Here, an iterative procedure is introduced. In the iterative search technique, starting with two values, λ is obtained by extrapolation, and the process is continued until Δpi is within a specified accuracy. However, a rapid solution is obtained by the use of the gradient method.
To do this, (2-10) is written as
f(λ) = PD (2-12)
Expanding the left-hand side of the above equation in Taylor’s series about an operating point λ(k), and neglecting the higher order terms results in
(2-13)
or
(2-14)
or
(2-15)
and therefore,
(2-16)
where
(2-17)
The process is continued until ΔP(k) is less than a specified accuracy.
Example
The fuel-cost functions for three thermal plants in $/h are given by
where P1, P2 and P3 are in MW. The total load, PD is 800MW. Neglecting line losses and generator limits, find the optimal dispatch and the total cost in $/h
a) by analytical method using (2-11)
b) by graphical demonstration
a) From (2-11), λ is found to be
substituting for λ in the coordination equation given in (2-9), the optimal dispatch is
b) From equation (2-6), the necessary conditions for optimal dispatch are
subject to
P1+P2+P3=PD
To demonstrate the concept of equal increment cost for optimal dispatch, we can use MATLAB plot command to plot the incremental cost of each plant on the same graph as shown in Figure 2-2. To obtain a solution, various values of λ could be tried until one is found which produces. For each λ, if , we increase λ otherwise, if , we reduce λ. Therefore, the horizontal line shown in the graph is moved up or down until at the optimum point. For this example, with PD=800MW, the optimal dispatch is P1=400 MW, P2=250 MW and P3=150 MW at λ=8.5$/MWh
Figure 2-2: Ilustratrating the concept of equal increment cost production cost.
To demonstrate the above iterative method, the following simple program is written for this example, which can be accessed from “File 2-1”
% Iterative solution Using Newton method
alpha =[500; 400; 200];
beta = [5.3; 5.5; 5.8]; gama=[.004; .006; .009];
PD=800;
DelP = 10; % Error in DelP is set to a high value
lambda = input('Enter estimated value of Lambda = ');
fprintf('\n ')
disp([' Lambda P1 P2 P3 DP'...
' grad Delambda'])
iter = 0; % Iteration counter
while abs(DelP) >= 0.001 % Test for convergence
iter = iter + 1; % No. of iterations
P = (lambda - beta)./(2*gama);
DelP =PD - sum(P); % Residual
J = sum( ones(length(gama), 1)./(2*gama)); % Gradient sum
Delambda = DelP/J; % Change in variable
disp([lambda, P(1), P(2), P(3), DelP, J, Delambda])
lambda = lambda + Delambda; % Successive solution
end
totalcost = sum(alpha + beta.*P + gama.*P.^2)
%Graphical Demonstration of Example 7.4
axis([0 450 6.5 10.5]);
P1=250:10:450; P2 = 150:10:350; P3=100:10:250;
IC1= 5.3 + 0.008*P1;
IC2= 5.5 + 0.012*P2;
IC3= 5.8 + 0.018*P3;
Px = 0:100:400;
plot(P1, IC1, P2, IC2, P3, IC3, Px, lambda*ones(1, length(Px)),'-m'),
xlabel('P, MW'), ylabel(' $/MWh'), grid
When the program is run, the result is
Enter estimated value of Lambda =
Lambda P1 P2 P3 DP grad Delambda
1.0e+003 *
0.0020 -0.4125 -0.2917 -0.2111 1.7153 0.2639 0.0065
8.5000 400.0000 250.0000 150.0000 0 263.8889 0
totalcost =
6.6825e+003
Experimental Procedure
Do not save any change you have made in the original file. Save the changes in a new file. Record all the results and observations.
The fuel-cost functions in $/h for three thermal plants of a power system are given by
where P1, P2 and P3 are in MW. Neglect line losses and generator limits.
1. Using the “File-Ch7ex3”, edit the file accordingly in order to find the optimal dispatch and total cost in $/h for the above system when the total load is
i) PD=450MW
ii) PD= 745MW
iii) PD=1335MW
2. Using the analytical expression derived in (2-11) and (2.7), obtain the optimal dispatch and the total cost in $/h when the total load, PD=745MW. Show that these results are similar to the results obtained for Q1 (ii).
Exercise
1. State briefly what you have learned from this experiment.
2. Define incremental fuel cost?
3. What is the objective of studying economic operation of a power system?
4. What is the condition to be satisfied for the economic scheduling of generators?
5. What is Marginal Cost?
6. How do you modify the optimal load dispatch solution method if generator limits are specified?
Lab Assessment: