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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:
 To perform the symmetrical fault analysis and determine for circuit breaker rating.
 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.
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(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.
I FAULT 
VTH
Z TH  Z f
ZTH
ZFAULT
VTH
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.
3
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 sequence
5
6
7
4
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 perunit 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.
5
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%).
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Experiment: PS-2
Part I: ECONOMIC LOAD DISPATCH
Objectives
 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
Png
P2
PD
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
ng
Ct   Ci
i 1
(2-1)
n
   i   i Pi   i Pi
2
i 1
is minimum, subject to the constraint
ng
P  P
i 1
i
(2-2)
D
where Ct is the total production cost, C i is the production cost of ith plant, P i is the
generation of ith plant, P D is the total load demand, and n g is the total number of
dispatchable generating plants.
A typical approach is to augment the constraints into objective function by using the
Lagrange multiplier
ng



L  Ct   PD   Pi 


i 1


(2-3)
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The minimum of L is found by equating the partials derivatives of the function with
respect to Pi and λ to zero.
L
0
Pi
(2-4)
L
0

(2-5)
Equation (2-4), results in
Ct
  (0  1)  0
Pi
Since
Ct  C1  C2  .....  Cng
then
Ct Ci


Pi Pi
and therefore the condition for optimum dispatch is
i=1,….,ng
(2-6)
 i  2 i Pi  
(2-7)
or
Second condition, given by (2-5), results in
ng
P  P
i
i 1
(2-8)
D
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
Pi 
  i
2 i
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
ng

i 1
  i
 PD
2 i
(2-10)
or
8
ng

( PD  
i 1
i
)
2 i
(2-11)
ng
1

i 1 2 i
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 Δp i 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
f ( ) ( k )  (
df ( ) ( k ) ( k )
)   PD
d
(2-13)
or
( k ) 

P ( k )
df ( ) ( k )
(
)
d
(2-14)
P
 ( Pi )(k )
(k )
or
( k ) 
P ( k )
1
 2
i
(2-15)
and therefore,
( k 1)  ( k )  ( k )
(2-16)
where
P
ng
(k )
 PD   Pi
(k )
(2-17)
i 1
The process is continued until ΔP (k) is less than a specified accuracy.
9
Example
The fuel-cost functions for three thermal plants in $/h are given by
C1  500  5.3P1  0.004 P1
2
C2  400  5.5 P2  0.006 P2
C3  200  5.8 P3  0.009 P3
2
2
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 $/h
a) by analytical method using (2-11)
b) by graphical demonstration
a) From (2-11), λ is found to be
5.3
5.5
5.8


2 * 0.004 2 * 0.006 2 * 0.009

1
1
1


0.008 0.012 0.018
800  14443.0555

 8.5$ / MWh
263.8889
800 
substituting for λ in the coordination equation given in (2-9), the optimal dispatch is
8.5  5.3
 400 MW
2 * 0.004
8.5  5.5
P2 
 250 MW
2 * 0.006
8.5  5.8
P3 
 150 MW
2 * 0.009
P1 
b) From equation (2-6), the necessary conditions for optimal dispatch are
dC1
 5.3  0.004 P1  
dP1
dC2
 5.5  0.012 P2  
dP2
dC3
 5.3  0.018P3  
dP3
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
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shown in Figure 2-2. To obtain a solution, various values of λ could be tried until one is
found which produces
 P  P . For each λ, if  P  P
i
D
i
D
, we increase λ otherwise, if
 P  P , we reduce λ. Therefore, the horizontal line shown in the graph is moved up or
down until at the optimum point  ,  P  P . For this example, with P =800MW, the
i
D
i
D
D
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
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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
Delambda
1.0e+003 *
0.0020
-0.4125 -0.2917
-0.2111
8.5000
400.0000
250.0000 150.0000
DP
1.7153
0
grad
0.2639
0.0065
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
C1  350  7.2 P1  0.004 P1
2
C2  500  7.3P2  0.0025P2
C3  600  6.74 P3  0.003P3
2
2
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
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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:

On-Spot evaluation (40%).

Experiment results & Lab report (60%).
13
Part II: Power-Flow Analysis
Objectives:
*
*
*
*
*
*
To understand with the components of a power system
To incorporate any alteration/modification in the power system
To edit any generation and/or load data
To calculate the load-flow study
To investigate the overloading of any equipment and voltage limit at any busbar
To perform corrective action for removing overloading and voltage violation.
Introduction:
An electric power system consists of three principal divisions: the generating stations, the
transmission lines, and the distribution systems. Generating stations generate electrical
power and deliver it to the customers through transmission lines. Transmission lines are
the connecting links between the generating stations and the distribution systems and lead
to other power systems over interconnections. A distribution system connects all the
individual loads to the transmission lines at substations, which perform voltage
transformation and switching functions.
Under normal conditions, electrical systems operate in their steady-state mode and the
basic calculation required determining the characteristic of this state is termed load-flow
or power-flow study. The objective of load-flow calculations is to determine the steadystate operating characteristics of the power generation/transmission system for a given set
of loads. The solution is expected to provide information of voltage magnitudes and angles
at all buses, active and reactive power flows in the individual transmission lines,
transmission line losses, and the reactive power generated or absorbed at voltagecontrolled buses. The great importance of load-flow studies is in planning the future
expansion of power systems as well as in determining the best operation of the existing
system without overloading any equipment and to keep bus voltage magnitudes within
specified limits.
Operating conditions must always be selected for each study. At each bus except one, the
net real power injected into the network must be specified. The power drawn by a load is
the negative power input to the system. The power from generator is the positive power
input. In addition, at these buses either the net flow of reactive power into the network or
the magnitude of the voltage must be specified; that is, at each bus a decision is required
whether the voltage magnitude or the reactive-power flow is to be maintained constant.
The usual case is to specify reactive power at load buses and voltage magnitude at
generator buses. In digital computer programs provision is made for the calculation to
consider voltage to be maintained constant at a bus only so long as the reactive-power
generation remains within designated limits.
The bus, at which real-power flow is not specified, called the swing bus, is usually a bus to
which a generator is connected. Obviously, the net power flow to the system cannot be
fixed in advance at every bus because the loss in the system is not known until the study is
complete. The generators at the swing bus supply the difference between the specified real
power into the system at the other buses and the total system output plus losses. Both voltage
14
magnitude and angle are specified at the swing bus. The computer as part of the solution determines
real and reactive powers at this bus.
Power/load-flow study of a power system deals with four variables at any bus. They are P, Q, V,
and . Depending on the type of bus two of them are specified and the remaining two are
determined by the load-flow study.
Software to be used for load flow results:
The software, ‘Power World Simulator’ will be used to study the power flow in the system.
Test System:
395 MW
114 MVR
One
Five
1.000 pu
0.000 Deg
Four
1.000 pu
0.000 Deg
1.000 pu
0.000 Deg
1.000 pu
0.000 Deg
Three
1.050 pu
0.000 Deg
520 MW
337 MVR
80 MW
40 MVR
Two
800 MW
280 MVR
0.0 MVR
Figure 2.1- Example 6-14
395 MW
0 MVR
One
Five
1.000 pu
0.000 Deg
Four
1.000 pu
0.000 Deg
1.000 pu
0.000 Deg
1.000 pu
0.000 Deg
Three
1.000 pu
0.000 Deg
520 MW
0 MVR
80 MW
40 MVR
Two
800 MW
280 MVR
Figure 2.2- Example 6-9
395 MW
114 MVR
One
1.000 pu
0.000 Deg
Total Real Power Losses:
Five
1.0000 tap
Four
1.000 pu
0.000 Deg
1.000 pu
0.000 Deg
34.80 MW
1.000 pu
0.000 Deg
Two
800 MW
280 MVR
Figure 2.3- Example 6-40
15
Three
1.050 pu
0.000 Deg
520 MW
337 MVR
80 MW
40 MVR
Experimental Procedure:
Note: Do not save any change you have made in the original file. Save the changes in a new file.
Record all the results and observation.
1. Launch the program by double clicking on the POWER WORLD SIMULATOR icon on the
desktop.
2. Load PowerWorld Simulator case Example 6_14 (see Figure 2.1). Right click on each symbol,
and check out what they represent.
i)
Using the existing information in the file, run the simulation, by selecting Simulation,
Gauss-Seidel Power Flow. Note down the voltage at bus 2, V2 and the total real and
reactive power losses (total real and reactive losses are shown on the Case Information,
Case Summary dialog).
ii) Determine the MVAr rating of the shunt capacitor bank so that V 2 can be increased to
1.00 per unit (to vary the capacitor’s nominal MVAr rating, right click on the capacitor
symbol to view the Switched Shunt dialog and then change the Nominal MVAr field).
Run the simulation using the Gauss-Seidel Power Flow method again.
iii) Determine the effect of this capacitor bank on line loadings and the total real power
losses.
3. Load PowerWorld Simulator case Example 6_9 (see Figure 2.2).
i)
Click on the Edit Mode. Right click on the transmission line between buses 2 and 4, and click
on the Information Dialog. Note down the parameters of the transmission line 2-4. Run the
simulation using Gauss-Seidel Power Flow, and note down your results in the column ‘Before
new line’ in Table 2.1. To obtain the percentage loading of the branch between buses, right
click the transmission line, go to Line Information Dialog.
ii) Insert a second line between buses 2 and 4. Give the new line a circuit identifier of “2” to
distinguish it from the existing line (go to Insert, Transmission Line). The line parameters of
the added line should be identical to those of the existing line 2 - 4. Run the simulation using
Gauss-Seidel Power Flow, and note down your results in the column ‘After new line’ in Table
2.1.
Table 2.1
Before new line
Bus voltage V2 (p.u)
Total real power losses (MW)
Line flow in branch between buses 1 and 5 at bus 1 (% loading)
Line flow in branch between buses 2 and 4 at bus 2 (% loading)
Line flow in branch between buses 2 and 5 at bus 2 (% loading)
Line flow in branch between buses 3 and 4 at bus 3 (% loading)
Line flow in branch between buses 4 and 5 at bus 4 (% loading)
16
After new line
iii) Using the original file of Example 6_9, change the voltage set point of generator 3
between 1.00 and 1.08 per unit in 0.2 per-unit steps. Using Gauss-Seidel Power Flow,
complete Table 2.2.
Table 2.2
V3
Reactive power output of generator 3
V2
Total real power losses
4. Load Example 6_40 (see Figure 2.3). This case is similar to Example 6-9, except that the
transformer between buses 1 and 5 is now a tap-changing transformer. Vary the tap setting
between 0.975 and 1.1 with a tap size of 0.025, and complete Table 2.3 after performing the
Power Flow study.
5.
Table 2.3
Tap Setting
Reactive output power of
Generator 1
V5
V2
Total real power
losses
Exercise:
1. State briefly what you have learned from this experiment.
2. List the specified variables for each bus of all the systems studied.
3. What is the information obtained from the load-flow solution? How can this information
be used for system planning or system expansion?
4. Refer to the system given below.
17
a) Using a MATLAB program or otherwise obtain the bus admittance matrix, Ybus.
b) Determine the power injections S2 and S3 at buses 2 and 3. Fill up the MATLAB
program given below. Using the program determine V2 and V3.
iter =0
S2=______________
S3=_________
V2=1+0i
V3=__________
for k=1:10
iter=iter+1;
V2 = (conj(S2)/conj(V2)+y12*V1+y23*V3)/(y12+y23);
V3 =________________________
disp([iter, V2, V3])
end
Lab Assessment:

On-Spot evaluation (40%).

Experiment results & Lab report (60%).
18