POWER SYSTEM ANALYSIS LAB MANUAL
POWER SYSTEM ANALYSIS
SUBMITTED TO:
ENGR.M.JUNAID
SUBMITTED BY:
ASAD NAEEM
2006-RCET-EE-22
DEPARTMENT OF ELECTRICAL ENGINEERING
(A CONSTITUENT COLLEGE: RACHNA COLLEGE OF ENGINEERING &
TECHNOLOGY GUJRANWALA)
UNIVERSITY OF ENGINEERING & TECHNOLOGY LAHORE, PAKISTAN
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
01
To plot the daily load curve for the given data using MATLAB
02
Introduction to basics of Electrical Transients Analyzer
Program (ETAP)
03
Evaluate the value of voltages for a 4-BUS system using node
equations in MATLAB
04
Modeling and Load flow analysis of RCET power distribution
network using ETAP
05
06
07
ASAD NAEEM
2006‐RCET‐EE‐22
Bus elimination of a 4-BUS system using MATLAB
To study the Concept of Modifications of an Existing BusImpedance Matrix & Implementing in MATLAB
Application of Gauss-Siedal and Newton-Raphson method for
load flow studies on a three bus system using MATLAB
POWER SYSTEM ANALYSIS LAB MANUAL
08
09
10
11
12
13
14
15
ASAD NAEEM
2006‐RCET‐EE‐22
Harmonic Load Modeling using built-in and user defined
models of ETAP
Impact of personal computer load on power distribution
network of RCET
Flow of triplen harmonics (zero-sequence harmonics) during 5
different schemes of connection for a 3-phase transformer with
presence of large non-linear load using ETAP
Three phase short circuit analysis (3-phase faults-device duty)
for a given power system using ETAP
Three phase short circuit analysis (3-phase faults-30 cycle
network) for a given power system using ETAP
Three phase short circuit analysis (LG, LL, LLG, & 3-Phase
Faults - ½ Cycle) for a given power system using ETAP
Three phase short circuit analysis (LG, LL, LLG, & 3-Phase
Faults - 1.5 to 4 Cycle) for a given power system using ETAP
Three phase short circuit analysis (LG, LL, LLG, & 3-Phase
Faults - 30 Cycle) for a given power system using ETAP
POWER SYSTEM ANALYSIS LAB MANUAL
EXPERIMENT#01
To plot the daily load curve for the given data using
MATLAB
Given data:
Interval from
12 A.M
2
6
9
12 P.M
2
4
6
8
10
11
To
2 A.M
6
9
12
2 P.M
4
6
8
10
11
12 A.M
Requirements:
1.
2.
3.
4.
Find average value of load
Find peak value of load
Find the load factor
Plot the load curve
ASAD NAEEM
2006‐RCET‐EE‐22
Load MW
6
5
10
15
12
14
16
18
16
12
6
POWER SYSTEM ANALYSIS LAB MANUAL
Theory
Loads:
Loads of power systems are divided into three main categories that are
given below.
1. Industrial Loads
2. Commercial Loads
3. Residential Loads
Very large industrial loads are served through the transmission lines.
Large industrial loads are served directly from the sub-transmission
level. And small industrial loads are served directly from the primary
distribution network. The industrial loads are composite loads and
induction motors from a high proportion of these loads. These
composite loads are functions of voltage and frequency and form a
major part of the system load. Commercial and residential load consist
largely of lighting, heating and cooling. These loads are independent of
frequency and consume negligibly small reactive power.
The real power of loads is expressed in terms of kilowatts or
megawatts. The magnitude of load varies throughout the day and
power must be available to the consumer on demand.
The daily load curve of a utility is a composite of demands made by
various classes of users. The greatest value of load during a twenty
four hours is called the peak or maximum demand. Smaller peaking
generators may be commissioned to meet the peak load that occurs
for only a few hours. In order to asses the usefulness of the generating
plant the load factor is defined.
The load factor is the ratio of average load over a designated period of
time to the peak load occurring in that period. Load factor may be
given for a day, a month or an year. Yearly or annual load factor is the
most useful since a year represents a full cycle of time. The daily load
factor is
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
Daily load factor = average load / peak load
Multiplying the numerator and denominator by a time period of 24 hr
we have
Daily load factor= average load*24 hr / (peak load*24 hrs)
= energy consumed during 24 hr/ (peak load*24 hr)
The annual load factor is
Annual load factor = total annual energy / (peak load*8760 hr)
Today’s typical system load factors are in range of 55-70%. In
Pakistan WAPDA standard for urban areas load factor is 60% and that
of rural areas is 65%.
Matlab code:
data=[0 2 6;
2 6 5;
6 9 10;
9 12 15;
12 14 12;
14 16 14;
16 18 16;
18 20 18;
20 22 16;
22 23 12;
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
23 24 6];
p=data(:,3);
Dt=data(:,2)-data(:,1);
w=p'*Dt;
pavg=w/sum(Dt)
peak=max(p)
LF=pavg/peak*100
L=length(data);
tt = [data(:,1) data(:,2)];
t = sort(reshape(tt, 1, 2*L));
for n = 1:L
pp(2*n-1)=p(n);
pp(2*n)=p(n);
end
plot(t,pp)
xlabel('TIME,Hr'),ylabel('P,MW')
Matlab results:
pavg =11.5417
peak =18
LF =64.1204
ASAD NAEEM
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POWER SYSTEM ANALYSIS LAB MANUAL
18
16
14
P,MW
12
10
8
6
4
0
5
10
15
20
25
TIME,Hr
COMMENTS:
In this experiment we learn how to find the daily load curve
for any power system using MATLAB. Load curve is very
important as we can achieve very important information
from it like:
• Peak load
• Average load
• Load factor
These quantities are very helpful for understanding any
power system.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
EXPERIMENT#02
Introduction to basics of Electrical Transients Analyzer
Program (ETAP)
What is ETAP?
ETAP is the most comprehensive analysis platform for the design,
simulation, operation, control, optimization, and automation of
generation, transmission, distribution, and industrial power systems.
Project Toolbar
The Project Toolbar contains icons that allow you to perform shortcuts of many
commonly used functions in PowerStation.
Create
Create a new project file
Open
Open an existing project file
Save
Save the project file
Print
Print the one‐line diagram or U/G raceway system
Cut
Cut the selected elements from the one‐line diagram or U/G raceway
system to the Dumpster
Copy
Copy the selected elements from the one‐line diagram or U/G raceway
system to the Dumpster
Paste
raceway
Paste elements from a Dumpster Cell to the one‐line diagram or U/G
system
Zoom In
Magnify the one‐line diagram or U/G raceway system
Zoom Out
Reduce the one‐line diagram or U/G raceway system
Zoom to Fit Page
Re‐size the one‐line diagram to fit the window
Check Continuity
Check the system continuity for non‐energized elements
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POWER SYSTEM ANALYSIS LAB MANUAL
Power Calculator
Help
Activate PowerStation Calculator that relates MW, MVAR, MVA,
kV, Amp, and PF together with either kVA or MVA units
Point to a specific area to learn more about PowerStation
Mode Toolbar
ETAP offers a suite of fully integrated software solutions including arc flash, load flow,
short circuit, transient stability, relay coordination, cable ampacity, optimal power flow,
and more. Its modular functionality can be customized to fit the needs of any company,
from small to large power systems.
Edit Mode
Edit mode enables you to build your one‐line diagram, change system connections, edit
engineering properties, save your project, and generate schedule reports in Crystal
Reports formats. The Edit Toolbars for both AC and DC elements will be displayed to the
right of the screen when this mode is active. This mode provides a wide variety of tasks
including:
∙
∙
∙
∙
∙
∙
∙
∙
∙
∙
∙
∙
∙
∙
Drag & Drop Elements
Connect Elements
Change IDs
Cut, Copy, & Paste Elements
Move from Dumpster
Insert OLE Objects
Cut, Copy & OLE Objects
Merge PowerStation Project
Hide/Show Groups of Protective Devices
Rotate Elements
Size Elements
Change Symbols
Edit Properties
Run Schedule Report Manager
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
Example implementation:
ASAD NAEEM
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POWER SYSTEM ANALYSIS LAB MANUAL
EXPERIMENT#03
Evaluate the value of voltages for a 4­BUS system using
node equations in MATLAB
GIVEN ONE LINE DIAGRAM
REACTANCE DIAGRAM
In the first step, we draw the reactance diagram of the given
one-line diagram as shown below:
ASAD NAEEM
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POWER SYSTEM ANALYSIS LAB MANUAL
SOURCE TRANSFORM
• After making the reactance diagram, we apply source
transformation on the given network by replacing the
voltage sources with current sources
• Replace all the reactance by admittances using the
relation:
• Y=1/X
• The resultant diagram now can be shown as:
NODE EQUATIONS
Now, using the above figure write the node equations of the
system:
• Applying KCL at node-1:
I1= (V1-0) y10 + (V1-V4) y14+ (V1-V3) y13
I1= (y10+y14+y13) V1 + 0V2 + (-y13) V3+ (-y14) V4
ASAD NAEEM
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POWER SYSTEM ANALYSIS LAB MANUAL
• Applying KCL at node-2:
I2= (V2-0) y20 + (V2-V3) y23+ (V2-V4) y24
I2= 0V1+ (y20+y23+y24) V2 + (-y23) V3+ (-y24) V4
• Applying KCL at node-3:
I3= (V3-0) y30 + (V3-V1) y31+ (V3-V4) y34 + (V3-V2) y32
I3= (-y31) V1+ (-y32) V2+ (y30+y31+y34) V3 + (-y34) V4
• Applying KCL at node-4:
0= (V4-V1) y14+ (V4-V3) y43 + (V4-V2) y42
0= (-y14) V1+ (-y42) V2 + (-y34) V3+ (y14+y43+y42) V4
Matrix form of the node equations is:
Where:
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CALCULATIONS
MATLAB CODE
YBUS=
[0-9.80i
0
0+4.00i
0+5.00i;
0
0-8.30i
0+2.50i
0+5.00i;
0+4.00i
0+2.50i
0-15.30i
0+8.00i;
0+5.00i
0+5.00i
0+8.00i
0-18.00i];
I= [0-1.20i; 0-0.7200-0.9600i; 0-1.2000i; 0];
ZBUS=inv (YBUS);
V=ZBUS*I
MATLAB RESULTS
V=
1.4111 - 0.2668i
1.3831 - 0.3508i
1.4059 - 0.2824i
1.4010 - 0.2971i
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
COMMENTS:
In this experiment we learn that using the bus impedance or
admittance matrix we can find the voltages and currents for
all buses of a given power system.
Moreover, we use MATLAB for the calculation of these
quantities by just entering the bus impedance matrix and
one given quantity (current or voltage) and MATLAB gives
the results of very complex networks within no time.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
EXPERIMENT#04
Modeling and Load flow analysis of RCET power
distribution network using ETAP
INTRODUCTION:
LOAD FLOW STUDIES
In power engineering, the power flow study (also known as
load-flow study) is an important tool involving numerical
analysis applied to a power system. Unlike traditional circuit
analysis, a power flow study usually uses simplified notation
such as a one-line diagram and per-unit system, and focuses
on various forms of AC power (i.e: reactive, real, and
apparent) rather than voltage and current. It analyses the
power systems in normal steady-state operation. There exist
a number of software implementations of power flow
studies.
The great importance of power flow or load-flow studies is in
the planning the future expansion of power systems as well
as in determining the best operation of existing systems.
The principal information obtained from the power flow
study is the magnitude and phase angle of the voltage at
each bus and the real and reactive power flowing in each
line.
LOAD FLOW STUDIES IN ETAP
ETAP load flow analysis software calculates bus voltages,
branch power factors, currents, and power flows throughout
the electrical system. ETAP allows for swing, voltage
regulated, and unregulated power sources with multiple
power grids and generator connections. It is capable of
performing analysis on both radial and loop systems. ETAP
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
allows you to select from several different methods in order
to achieve the best calculation efficiency and accuracy.
Run Load Flow Studies
Update Cable Load Currents
Load Flow display Option
Alert View
Report Manager
Halt current calculations
Net on line data
STEPS
¾
¾
¾
¾
Modeling of the main network
Modeling of composite networks
Running of load flow analysis
Complete report from ETAP load flow analyzer
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
MODELING OF BASIC RCET NETWORK
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POWER SYSTEM ANALYSIS LAB MANUAL
MODELING OF COMPOSITE NETWORKS
STAFF COLONY:
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POWER SYSTEM ANALYSIS LAB MANUAL
OLD BUILDING:
NEW BUILDING:
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POWER SYSTEM ANALYSIS LAB MANUAL
HOSTEL-A,B:
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
HOSTEL-E:
Complete ETAP load flow analysis report of the given
network is attached with this experiment.
COMMENTS:
In this experiment we learn how to:
•
•
•
•
Model a power system in ETAP
Model composite networks in a basic network
Assign properties of components added
Study the load flow analysis for that network
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
EXPERIMENT#05
Bus elimination of a 4­BUS system using MATLAB
REACTANCE DIAGRAM
It is given that the transformer and generator at bus-3 are
disconnected, so the reactance diagram now becomes:
SOURCE TRANSFORM
• After making the reactance diagram, we apply source
transformation on the given network by replacing the
voltage sources with current sources
• Replace all the reactance by admittances using the
relation:
• Y=1/X
• The resultant diagram now can be shown as:
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
Part‐1: Elimination of Bus‐3&4
MATRIX FORM
Where:
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POWER SYSTEM ANALYSIS LAB MANUAL
MATLAB CODE
>>YBUS= [0-9.80i 0
0+4.00i
0+5.00i;
0-8.30i
0+2.50i
0+5.00i;
0+4.00i
0+2.50i
0-14.5i
0+8.00i;
0+5.00i
0+5.00i
0+8.00i
0-18.00i];
0
>>K= [0-9.80i 0; 0 0-8.30i];
>>L= [0+4.00i 0+5.00i; 0+2.50i 0+5.00i];
>>M= [0-14.5i 0+8.00i; 0+8.00i 0-18.00i];
>>LT= [0+4.00i 0+2.50i; 0+5.00i 0+5.00i];
>>N=inv (M);
>>P=L*N*LT;
>>Ybus=K-P
MATLAB RESULTS
Ybus =
0 - 4.8736i
0 + 4.0736i
0 + 4.0736i
0 - 4.8736i
ASAD NAEEM
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POWER SYSTEM ANALYSIS LAB MANUAL
Part-2: Elimination Bus-4
MATLAB CODE:
>>Ybus=[-9.8i
0
4.0i
5i;
0
-8.3i
2.5i
5i;
4i
2.5i
-14.5i
8i;
5i
5i
8i
-18i];
>>K=[-9.8i 0 4i;0 -8.3i 2.5i;4i 2.5i -14.5i];
>>L=[5i;5i;8i];
>>M=[-18i];
>>P=L';
>>T=inv(M);
>>A=K-L*T*P
MATLAB RESULTS
A=
0 -11.1889i
0 - 1.3889i
0 + 1.7778i
0 - 1.3889i
0 - 9.6889i
0 + 0.2778i
0 + 1.7778i
0 + 0.2778i
0 -18.0556i
Part-3: Elimination Bus-3
MATLAB CODE:
>>P=[-11.1889i -1.3889i;-1.3889i -9.6889i];
>>Q=[1.7778i;0.2778i];
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POWER SYSTEM ANALYSIS LAB MANUAL
>>R=[-18.0556i];
>>S=Q';
>>T=inv(R);
>>B=P-Q*T*S
MATLAB RESULTS
B=
0 -11.3639i
0 - 1.4163i
0 - 1.4163i
0 - 9.6932i
COMMENTS:
Bus impedance matrix is a very important tool for the
calculation of voltages and currents at all the buses of a
given network. Suppose that any fault occurs in the power
system then we can get a task to modify the bus impedance
matrix by eliminating the faulty node which will reduce the
order of matrix by eliminating the faulty node.
In this experiment we learn how to:
• Eliminate last two nodes together
• Eliminate only one last node
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
EXPERIMENT#06
To study the Concept of Modifications of an Existing Bus­
Impedance Matrix & Implementing in MATLAB
IMPEDANCE MATRIX
Impedance matrix is a very important tool in power system
analysis. Using this matrix we can find:
• Voltages at all buses when currents are given
• Currents at all buses when voltages are given
So it is very important that how to modify the bus
impedance matrix when any new impedance is add into the
original system.
Suppose a power system with n-buses having the
impedances matrix of order n*n:
There are four cases that can take place while adding a new
impedance Zb in the system:
•
•
•
•
Adding
Adding
Adding
Adding
ASAD NAEEM
2006‐RCET‐EE‐22
Zb
Zb
Zb
Zb
from a new bus-P to reference bus
from a new bus-P to an existing bus-K
from an existing bus-K to reference bus
between two existing buses
POWER SYSTEM ANALYSIS LAB MANUAL
MODIFICATION CASES
CASE‐1: ADDING Zb FROM A NEW BUS TO REFERENCE BUS
This condition is explained in the following diagram:
Clearly,
Vp-0=Ib*Zb
Vp=Ib*Zb
Hence the modified matrix will take the form as:
MATLAB CODE
function [Z]=Case1(Zorg,Zb)
Zorg=[1 2 3 4;2 5 6 7;3 6 8 9;4 7 9 10]
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POWER SYSTEM ANALYSIS LAB MANUAL
Zb=17;
l=length(Zorg);
for i=1:l+1
for j=1:l+1
if i<=l && j<=l
Znew(i,j)=Zorg(i,j);
elseif i==l+1 && j==l+1
Znew(i,j)=Zb;
else
Znew(i,j)=0;
end
end
end
Znew
MATLAB RESULTS
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CASE‐2: ADDING Zb FROM A NEW BUS‐P TO AN EXISTING BUS‐K
This condition is explained in the following diagram:
Clearly,
Vp-Vk,new=Ip*Zb
Vp=Vk,new+Ip*Zb
Where,
Vk,new=Vk,org+Ip*Zkk
Vp= Vk,org+Ip(Zkk+Zb)
Hence the modified matrix will take the form as:
ASAD NAEEM
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POWER SYSTEM ANALYSIS LAB MANUAL
MATLAB CODE:
function [Z]=CASE2(Zorg,Zb)
Zorg=[1 2 3 4;2 5 6 7;3 6 8 9;4 7 9 10]
Zb=5;
l=length(Zorg);
row =Zorg(l,:);
column =Zorg(:,l);
for i=1:l+1
for j=1:l+1
if i<=l && j<=l
Znew(i,j)=Zorg(i,j);
elseif i==l+1
for p=1:l
Znew(i,p)=row(p);
end
elseif j==l+1
for q=1:l
Znew(q,j)=column(q);
end
end
if i==l+1 && j==l+1
Znew(i,j)=Zb+Zorg(l,l);
end
end
end
Znew
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POWER SYSTEM ANALYSIS LAB MANUAL
MATLAB RESULTS
CASE‐3: ADDING Zb FROM AN EXISTING BUS‐K TO REFERENCE
BUS
This condition is explained in the following diagram:
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POWER SYSTEM ANALYSIS LAB MANUAL
Here we can apply the same case as in case-2 and then put
Vp=0. This task can be achieved by eliminating the last row
and column of the Znew matrix.
Now this matrix is of the order (n+1)*(n+1), we have to
achieve a matrix of order n*n using formula:
Zkj(new)=Zkj(org)-(Zk(n+1)Z(n+1)j/Zkk+Zb)
In this case,
K=n
MATLAB CODE:
function [Z]=CASE3(Zorg,Zb)
Zorg=[1 2 3 4;2 5 6 7;3 6 8 9;4 7 9 10]
Zb=5;
l=length(Zorg);
row =Zorg(l,:);
column =Zorg(:,l);
for i=1:l+1
for j=1:l+1
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POWER SYSTEM ANALYSIS LAB MANUAL
if i<=l && j<=l
Znew(i,j)=Zorg(i,j);
elseif i==l+1
for p=1:l
Znew(i,p)=row(p);
end
elseif j==l+1
for q=1:l
Znew(q,j)=column(q);
end
end
if i==l+1 && j==l+1
Znew(i,j)=Zb+Zorg(l,l);
end
end
end
Znew
for a=1:l
for b=1:l
K(a,b)=Znew(a,b);
end
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POWER SYSTEM ANALYSIS LAB MANUAL
end
for a=1:l
L(a,1)=Znew(a,5);
end
M=Znew(l+1,l+1);
P=L';
T=inv(M);
Zwithnewbusrefferenced=K-L*T*P
MATLAB RESULTS
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POWER SYSTEM ANALYSIS LAB MANUAL
CASE‐4: ADDING Zb BETWEEN TWO EXISTING BUSES
This condition is explained in the following diagram:
In this case,
Zbb=Zb+Zjj+Zkk-2Zjk
Hence the modified matrix will take the form as:
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POWER SYSTEM ANALYSIS LAB MANUAL
Here again we have to eliminate the last row and column to
achieve the final matrix.
MATLAB CODE:
function [Z]=CASE4(Zorg,Zb)
Zorg=[1 2 3 4;2 5 6 7;3 6 8 9;4 7 9 10]
Zb=5;
l=length(Zorg);
R1 =Zorg(l,:);
C1 =Zorg(:,l);
R2 =Zorg(l-1,:);
C2 =Zorg(:,l-1);
for i=1:l+1
for j=1:l+1
if i<=l && j<=l
Znew(i,j)=Zorg(i,j);
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POWER SYSTEM ANALYSIS LAB MANUAL
elseif i==l+1
for p=1:l
Znew(i,p)=R1(p)-R2(p);
End
elseif j==l+1
for q=1:l
Znew(q,j)=C1(q)-C2(q);
End
End
if i==l+1 && j==l+1
Znew(i,j)=Zb+Zorg(l,l)+Zorg(l-1,l-1)-(2*Zorg(l,l-1));
end
end
end
Znew
for a=1:l
for b=1:l
K(a,b)=Znew(a,b);
end
end
for a=1:l
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POWER SYSTEM ANALYSIS LAB MANUAL
L(a,1)=Znew(a,5);
end
M=Znew(l+1,l+1);
P=L';
T=inv(M);
Zfinal=K-L*T*P
MATLAB RESULTS
ASAD NAEEM
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POWER SYSTEM ANALYSIS LAB MANUAL
COMMENTS:
Bus impedance matrix is a very important tool for the
calculation of voltages and currents at all the buses of a
given network. Suppose that any improvement occurs in the
power system then we can get a task to modify the bus
impedance matrix by adding the new impedance in the
system. The new impedance can be added in four different
conditions:
• Addition of new
bus
• Addition of new
bus
• Addition of new
reference bus
• Addition of new
impedance from a new bus to reference
impedance from a new bus to existing
impedance from an existing bus to
impedance between two existing buses
In this experiment we learn how to modify the bus
impedance matrix for all four cases using MATLAB.
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POWER SYSTEM ANALYSIS LAB MANUAL
EXPERIMENT#07
Application of Gauss­Siedal and Newton­Raphson method for load
flow studies on a three bus system using MATLAB(Implimentation
of example#6.7,6.8 & 6.10 from POWER SYSTEM ANALYSIS by Hadi­
Saadat)
EXAMPLE 6.7
Given figure shows the one line diagram of a simple three
bus system with generation at bus-1. The magnitude of
voltage at bus-1 is adjusted to 1.05 per unit. The scheduled
loads at buses-2 and 3 are as marked on the diagram. Line
impedances are marked in per unit on a 100-MVA base and
the line charging susceptances are neglected.
(A) Using the Gauss-Siedal method, determine the
phasor values of the voltage at the load buses 2 and
3 (P&Q buses) accurate to four decimal places
(B) Find the slack bus real and reactive power
(C) Determine the line flows and line losses. Construct a
power flow diagram showing the direction of line flow
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SOLUTION
Line impedances are converted to admittances:
At the P-Q buses, the complex loads expressed in per unit
are:
S2sch=-(256.6+j110.2)/100= -2.566-j1.102 pu
S3sch=-(138.6+j45.2)/100= -1.386-j0.452 pu
Starting from an initial estimate of V2(0)=1.0+j0.0 and
V3(0)=1.0+j0.0
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GAUSS-SIEDEL FARMULA
SLACK1=conj(V1)*[V1*(y12+y13)-( y12*V2+y13*V3)]
Sij=Vi*conj(Iij)
Iij=yij*(Vi-Vj)
MATLAB CODE
y12=10-j*20;
y13=10-j*30;
y23=16-j*32;
V1=1.05+j*0;
%CODE FOR PART-A
iter=0;
s2=-2.566-j*1.102;
s3=-1.386-j*0.452;
V2=1+j*0.0;
V3=1+j*0.0;
for I=1:10;
iter=iter+1;
V2=(conj(s2)/conj(V2)+y12*V1+y23*V3)/(y12+y23);
V3=(conj(s3)/conj(V3)+y13*V1+y23*V2)/(y13+y23);
end
V2
V3
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%CODE FOR PART-B
Pslack=conj(V1)*[V1*(y12+y13)-(y12*V2+y13*V3)]
%CODE FOR PART-C
I12=y12*(V1-V2)
I21=-I12
I13=y13*(V1-V3)
I31=-I13
I23=y23*(V2-V3)
I32=-I23
s12=V1*conj(I12)
s21=V2*conj(I21)
s13=V1*conj(I13)
s31=V3*conj(I31)
s23=V2*conj(I23)
s32=V3*conj(I32)
SL12=s12+s21
SL13=s13+s31
SL23=s23+s32
MATLAB RESULTS
PART-A RESULTS
V2 =
0.9800 - 0.0600i
V3 =
1.0000 - 0.0500i
PART-B RESULTS
SLACK-BUS POWER
Pslack = 4.0949 - 1.8900i
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PART-C RESULTS
I12 =
1.9000 - 0.8000i
I21 = -1.9000 + 0.8000i
I13 =
2.0000 - 1.0000i
I31 = -2.0000 + 1.0000i
I23 = -0.6400 + 0.4800i
I32 =
0.6400 - 0.4800i
LINE FLOWS
s12 =
1.9950 + 0.8400i
s21 = -1.9100 - 0.6700i
s13 =
2.1000 + 1.0500i
s31 = -2.0500 - 0.9000i
s23 = -0.6560 - 0.4320i
s32 =
0.6640 + 0.4480i
LINE LOSSES
SL12 =
0.0850 + 0.1700i
SL13 =
0.0500 + 0.1500i
SL23 =
0.0080 + 0.0160i
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
EXAMPLE 6.8
Given figure shows the one line diagram of a simple three
bus system with generators at buses-1 and 3. The
magnitude of voltage at bus-1 is adjusted to 1.05pu. voltage
magnitude at bus-3 is fixed at 1.04 pu with a real power
generation of 200MW. A load consisting of 400MW and
250MVAR is taken from bus-2. Line impedances are marked
in per unit on a 100MVA base, and the line charging
susceptances are neglected. Obtain the power flow solution
by the Gauss-Siedal method including line flows and line
losses.
Where,impedances are replaced by admittances as:
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FARMULA’S
S3=conj(V3)*(y33*V3-y13*V1-y23*V2)
Q3=-imag(conj(V3)*(y33*V3-y13*V1-y23*V2))
MATLAB CODE
y12=10-j*20;
y13=10-j*30;
y23=16-j*32;
y33=y13+y23;
V1=1.05+j*0;
format long
iter=0;
s2=-4.0-j*2.5;
p3=2;
V2=1+j*0.0;
Vm3=1.04;
V3=1.04+j*0;
for I=1:10;
iter=iter+1;
E2=V2;
E3=V3;
V2=(conj(s2)/conj(V2)+y12*V1+y23*V3)/(y12+y23)
DV2=V2-E2;
Q3=-imag(conj(V3)*(y33*V3-y13*V1-y23*V2))
s3=p3+j*Q3;
Vc3=(conj(s3)/conj(V3)+y13*V1+y23*V2)/(y13+y23);
Vi3=imag(Vc3);
Vr3=sqrt(Vm3^2-Vi3^2);
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V3=Vr3+j*Vi3
DV3=V3-E3;
end
V2
V3
Q3
format short
I12=y12*(V1-V2);
I21=-I12;
I13=y13*(V1-V3);
I31=-I13;
I23=y23*(V2-V3);
I32=-I23;
s12=V1*conj(I12);
s21=V2*conj(I21);
s13=V1*conj(I13);
s31=V3*conj(I31);
s23=V2*conj(I23);
s32=V3*conj(I32);
I1221=[I12,I21];
I1331=[I13,I31];
I2332=[I23,I32];
SL12=s12+s21
SL13=s13+s31
SL23=s23+s32
S1=(s12+s13)
S2=(s23+s21)
S3=(s31+s32)
S12=s12
S21=s21
S13=s13
S31=s31
S23=s23
S32=s32
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MATLAB RESULTS
ITERATION RESULTS:
1ST ITERATION
V2 = 0.974615384615385 - 0.042307692307692i
Q3 = 1.160000000000002
V3 = 1.039987148574197 - 0.005170183798502i
2ND ITERATION
V2 = 0.971057059512953 - 0.043431876337850i
Q3 = 1.387957731052817
V3 = 1.039974378708180 - 0.007300111679686i
3RD ITERATION
V2 = 0.970733708554698 - 0.044791724463619i
Q3 = 1.429040300785471
V3 = 1.039966679445820 - 0.008325001047174i
4TH ITERATION
V2 = 0.970652437281433 - 0.045329920732880i
Q3 = 1.448333275594840
V3 = 1.039963173621928 - 0.008752000354604i
5TH ITERATION
V2 = 0.970623655331095 - 0.045554240372625i
Q3 = 1.456209166612119
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V3 = 1.039961668920058 - 0.008929007616053i
6TH ITERATION
V2 = 0.970612037114234 - 0.045646940090561i
Q3 = 1.459469889628077
V3 = 1.039961037734205 - 0.009002221658867i
7TH ITERATION
V2 = 0.970607253520093 - 0.045685276728252i
Q3 = 1.460818201396914
V3 = 1.039960775170297 - 0.009032502820155i
8TH ITERATION
V2 = 0.970605276281561 - 0.045701131870879i
Q3 = 1.461375872168914
V3 = 1.039960666313617 - 0.009045027392915i
9TH ITERATION
V2 = 0.970604458527297 - 0.045707689707255i
Q3 = 1.461606535170454
V3 = 1.039960621244008 - 0.009050207830587i
10TH ITERATION
V2 = 0.970604120282796 - 0.045710402176455i
Q3 = 1.461701943643423
V3 = 1.039960602594413 - 0.009052350604469i
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FINAL RESULTS:
V2 = 0.970604120282796 - 0.045710402176455i
V3 = 1.039960602594413 - 0.009052350604469i
Q3 = 1.461701943643423
SL12 = 0.0839 + 0.1679i
SL13 = 0.0018 + 0.0055i
SL23 = 0.0985 + 0.1969i
S1 = 2.1841 + 1.4085i
S2 = -3.9999 - 2.5000i
S3 = 2.0000 + 1.4618i
S12 = 1.7936 + 1.1874i
S21 = -1.7096 - 1.0195i
S13 = 0.3906 + 0.2212i
S31 = -0.3887 - 0.2157i
S23 = -2.2903 - 1.4805i
S32 = 2.3888 + 1.6775i
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EXAMPLE 6.10
Given figure shows the one line diagram of a simple three
bus system with generators at buses-1 and 3. The
magnitude of voltage at bus-1 is adjusted to 1.05pu. voltage
magnitude at bus-3 is fixed at 1.04 pu with a real power
generation of 200MW. A load consisting of 400MW and
250MVAR is taken from bus-2. Line impedances are marked
in per unit on a 100MVA base, and the line charging
susceptances are neglected. Obtain the power flow solution
by the Newton-Raphson method including line flows and line
losses.
Where,impedances are replaced by admittances as:
The bus impedance matrix can be constructed as:
YBUS=[20-j50
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-10+j20
-10+j30
POWER SYSTEM ANALYSIS LAB MANUAL
-10+j20
26-j52
-10+j30
-16+j32
-16+j32
26-j62];
FARMULA’S
P1=V1^2*Y11*cos(Ѳ11)+V1*V2*Y12*cos(Ѳ12-d1+d2)+...
V1*V3*Y13*cos(Ѳ13-d1+d3)
Q1=-V1^2*Y11*sin(Ѳ11)-V1*V2*Y12*sin(Ѳ12-d1+d2)-...
V1*V3*Y13*sin(Ѳ13-d1+d3)
Q3=-V3*V1*Y31*sin(Ѳ31)-d3+d1)-V3*V2*Y32*...
sin(Ѳ32-d3+d2)-V3^2*Y33*sinѲ33
MATLAB CODE
V=[1.05;1.0;1.04];
d=[0;0;0];
Ps=[-4;2.0];
Qs=-2.5;
YB=[20-j*50 -10+j*20 -10+j*30
-10+j*20 26-j*52 -16+j*32
-10+j*30 -16+j*32 26-j*62];
Y=abs(YB);
t=angle(YB);
iter=0;
pwracur=0.00025;
%power accuracy
DC=10;
%set the maximun power residue to a high value
while max(abs(DC))>pwracur
iter=iter+1
P=[V(2)*V(1)*Y(2,1)*cos(t(2,1)d(2)+d(1))+V(2)^2*Y(2,2)*cos(t(2,2))+...
V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3));
V(3)*V(1)*Y(3,1)*cos(t(3,1)d(3)+d(1))+V(3)^2*Y(3,3)*cos(t(3,3))+...
V(3)*V(2)*Y(3,2)*cos(t(3,2)-d(3)+d(2))];
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Q=-V(2)*V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))V(2)^2*Y(2,2)*sin(t(2,2))-...
V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3));
J(1,1)=V(2)*V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))+...
V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3));
J(1,2)=-V(2)*V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3));
J(1,3)=V(1)*Y(2,1)*cos(t(2,1)d(2)+d(1))+2*V(2)*Y(2,2)*cos(t(2,2))+...
V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3));
J(2,1)=-V(3)*V(2)*Y(3,2)*sin(t(3,2)-d(3)+d(2));
J(2,2)=V(3)*V(1)*Y(3,1)*sin(t(3,1)-d(3)+d(1))+...
V(3)*V(2)*Y(3,2)*sin(t(3,2)-d(3)+d(2));
J(2,3)=V(3)*Y(2,3)*cos(t(3,2)-d(3)+d(2));
J(3,1)=V(2)*V(1)*Y(2,1)*cos(t(2,1)-d(2)+d(1))+...
V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3));
J(3,2)=-V(2)*V(3)*Y(2,3)*cos(t(2,3)-d(2)+d(3));
J(3,3)=-V(1)*Y(2,1)*sin(t(2,1)-d(2)+d(1))-2*V(2)*Y(2,2)
*sin(t(2,2)) ...
V(3)*Y(2,3)*sin(t(2,3)-d(2)+d(3));
DP=Ps-P;
DQ=Qs-Q;
DC=[DP;DQ]
J
DX=J\DC
d(2)=d(2)+DX(1);
d(3)=d(3)+DX(2);
V(2)=V(2)+DX(3);
V, d, delta=180/pi*d;
end
P1=V(1)^2*Y(1,1)*cos(t(1,1))+V(1)*V(2)*Y(1,2)*cos(t(1,2)d(1)+d(2))+...
V(1)*V(3)*Y(1,3)*cos(t(1,3)-d(1)+d(3))
Q1=-V(1)^2*Y(1,1)*sin(t(1,1))-V(1)*V(2)*Y(1,2)*sin(t(1,2)d(1)+d(2))-...
V(1)*V(3)*Y(1,3)*sin(t(1,3)-d(1)+d(3))
Q3=-V(3)*V(1)*Y(3,1)*sin(t(3,1)-d(3)+d(1))-V(3)*V(2)*Y(3,2)*...
sin(t(3,2)-d(3)+d(2))-V(3)^2*Y(3,3)*sin(t(3,3))
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MATLAB RESULTS
1ST ITERATION
DC =
-2.8600
1.4384
-0.2200
J=
54.2800 -33.2800
24.8600
-33.2800
66.0400 -16.6400
-27.1400
16.6400
DX =
-0.0453
-0.0077
-0.0265
V=
1.0500
0.9735
1.0400
d=
0
-0.0453
-0.0077
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2ND ITERATION
DC =
-0.0992
0.0217
-0.0509
J=
51.7247 -31.7656
21.3026
-32.9816
65.6564 -15.3791
-28.5386
17.4028
DX =
-0.0018
-0.0010
-0.0018
V=
1.0500
0.9717
1.0400
d=
0
-0.0471
-0.0087
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3RD ITERATION
DC =
1.0e-003 *
-0.2166
0.0382
-0.1430
J=
51.5967 -31.6939
21.1474
-32.9339
65.5976 -15.3516
-28.5482
17.3969
DX =
1.0e-005 *
-0.3856
-0.2386
-0.4412
V=
1.0500
0.9717
1.0400
d=
0
-0.0471
-0.0087
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FINAL RESULTS:
P1 =
2.1842
Q1 =
1.4085
Q3 =
1.4618
COMMENTS:
Power system calculations are mostly very complex for large
power systems. To analyze such power systems, there are
two very important iterative methods:
• Gauss Siedel Method
• Newton Raphson method
In this experiment we learn how to apply these two methods
using MATLAB.
Gauss-Seidel iteration has two advantages:
•
Errors do not accumulate during the calculation. If the
procedure converges, it approaches the correct answer
without rounding errors such as can occur during
inversion of large matrices.
•
The method can be used for nonlinear sets of
equations.
While Newton Raphson method is readily applied to nonlinear equations, and can use finite-difference estimates
of the derivatives to evaluate the gradients.
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EXPERIMENT#08
Harmonic Load Modeling using built­in and user defined
models of ETAP
HARMONIC ANALYSIS
Because of the wide and ever increasing applications of power
electronic devices, such as variable speed drives, uninterruptible
power supplies (UPS), static power converters, etc., power system
voltage and current quality has been severely affected in some
areas. In these areas components other than that of fundamental
frequency can be found to exist in the distorted voltage and
current waveforms. These components usually are the integer
multipliers of the fundamental frequency, called harmonics. In
addition to electronic devices, some other non‐linear loads, or
devices including saturated transformers, arc furnaces,
fluorescent lights, and cycloconverters are also responsible for the
deterioration in power system quality.
HARMONIC SOURCES
The following components can be modeled as a harmonic voltage
source in PowerStation:
•
•
•
•
Power Grid
Synchronous Generator
Inverter
Charger/Converter
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• Static Load
IMPORTANT DEFINITIONS
Transients
The term transient has long been used in the analysis of power
system variations to denote an event that is undesirable and
momentary in nature. Transient is “that part of the change in a
variable that disappears during transition from one steady state
operating condition to another.”
Impulsive transient
An impulsive transient is a sudden; non–power frequency change
in the steady‐state condition of voltage, current, or both that is
unidirectional in polarity (primarily either positive or negative).
Oscillatory transient
An oscillatory transient is a sudden, non–power frequency change
in the steady‐state condition of voltage, current, or both, that
includes both positive and negative polarity values.
Long-Duration Voltage Variations
Long‐duration variations encompass root‐mean‐square (rms)
deviations at power frequencies for longer than 1 minute.
Overvoltage
An overvoltage is an increase in the rms ac voltage greater than
110 percent at the power frequency for duration longer than 1
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min. Over voltages are usually the results of load switching (e.g.,
switching off a large load or energizing a capacitor bank).
Under voltage
An under voltage is a decrease in the rms ac voltage to less than
90 percent at the power frequency for a duration longer than 1
min. Under voltages are the results of switching events that are
the opposite of the events that cause over voltages.
Short-Duration Voltage Variations
This category encompasses the IEC category of voltage dips and
short interruptions. Each type of variation can be designated as
instantaneous, momentary, or temporary, depending on its
duration.
Short‐duration voltage variations are caused by fault conditions,
the energization of large loads which require high starting
currents, or intermittent loose connections in power wiring.
Interruption
An interruption occurs when the supply voltage or load current
decreases to less than 0.1 pu for a period of time not exceeding 1
min.
Sags (dips)
Sag is a decrease to between 0.1 and 0.9 pu in rms voltage or
current at the power frequency for durations from 0.5 cycle to 1
min.
Swells
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A swell is defined as an increase to between 1.1 and 1.8 pu in rms
voltage or current at the power frequency for durations from 0.5
cycle to 1 min.
Voltage Imbalance
Voltage imbalance (also called voltage unbalance) is sometimes
defined as the maximum deviation from the average of the three‐
phase voltages or currents, divided by the average of the three‐
phase voltages or currents, expressed in percent.
Waveform Distortion
Waveform distortion is defined as a steady‐state deviation from
an ideal sine wave of power frequency principally characterized
by the spectral content of the deviation.
Harmonics
Harmonics are sinusoidal voltages or currents having frequencies
that are integer multiples of the frequency at which the supply
system is designed to operate (termed the fundamental
frequency usually 50 or 60 Hz).
Interharmonics
Voltages or currents having frequency components that are not
integer multiples of the frequency at which the supply system is
designed to operate (e.g., 50 or 60 Hz) are called Interharmonics.
Odd harmonics
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Voltages or currents having frequency components that are odd
integer multiples of the frequency at which the supply system is
designed to operate (e.g., 50 or 60 Hz) are called odd harmonics.
Even harmonics
Voltages or currents having frequency components that are even
integer multiples of the frequency at which the supply system is
designed to operate (e.g., 50 or 60 Hz) are called even harmonics.
Voltage Fluctuation
Voltage fluctuations are systematic variations of the voltage
envelope or a series of random voltage changes, the magnitude of
which does not normally exceed the voltage ranges specified by
ANSI C84.1 of 0.9 to 1.1 pu.
Power Frequency Variations
Power frequency variations are defined as the deviation of the
power system fundamental frequency from it specified nominal
value (e.g., 50 or 60 Hz).
Power factor, displacement
The power factor of the fundamental frequency components of
the voltage and current waveforms
Power factor (true)
The ratio of active power (watts) to apparent power (volt
amperes)
Total harmonic distortion (THD)
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The ratio of the root mean square of the harmonic content to the
rms value of the fundamental quantity, expressed as a percent of
the fundamental.
Triplen harmonics
A term frequently used to refer to the odd multiples of the third
harmonic, which deserve special attention because of their
natural tendency to be zero sequence.
ONE LINE DIAGRAM
MODELING OF HARMONIC LOAD
• Double click on the charger
• Select the harmonics section
• Select the type of harmonics from the given library of
harmonics
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BUILT-IN MODELS OF ETAP
ROCKWELL (12-Pulse-VFD)
ROCKWELL (18-Pulse-VFD)
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ROCKWELL (6-Pulse-VFD)
TOSHIBA (PWM-ASD)
TYPICAL (LCI)
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TYPICAL-IEEE (12-Pulse1)
TYPICAL-IEEE (12-Pulse2)
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TYPICAL-IEEE (18-Pulse-CT)
TYPICAL-IEEE (18-Pulse-VT)
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POWER SYSTEM ANALYSIS LAB MANUAL
TYPICAL-IEEE (6-Pulse1)
TYPICAL-IEEE (6-Pulse2)
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
TYPICAL-IEEE (Fluorescent)
TYPICAL-IEEE (Large-ASD)
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
TYPICAL-IEEE (SPC)
TYPICAL-IEEE (XFMR-Magnet)
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
USER DEFINED MODELING
•
•
•
•
•
•
Go to the library tab given on the main window of ETAP
Select the harmonic section
Click on add tab
Enter the name of new harmonic model
Click on edit tab
Enter the spectrum components in the table
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
COMMENTS
In this experiment, we learnt:
• How to model a harmonic load using built-in models
• How to built a user-defined harmonic model
• How to model a harmonic load using user-defined
models
So, ETAP is a very powerful tool for harmonic analysis of any
power distribution network.
EXPERIMENT#09
Impact of personal computer load on power distribution
network of RCET
Harmonics
Harmonics are sinusoidal voltages or currents having frequencies
that are integer multiples of the frequency at which the supply
system is designed to operate (termed the fundamental
frequency usually 50 or 60 Hz).
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
PC LOAD THD’S
Harmonic No.
rd
%THD
3
91.63
5th
86.61
7th
69.87
th
9
44.76
11th
54.81
13th
46.44
15th
46.44
th
17
33.05
19th
24.70
23rd
11.74
th
25
7.900
29th
5.120
%THD
178.97
ONE LINE DIAGRAM
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
LOAD FLOW DIAGRAM
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
LOAD FLOW ANALYSIS REPORT
HARMONIC ANALYSIS RESULTS
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
VOLTAGE WAVEFORMS & SPECTRAS
BUS-5:
BUS-11:
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POWER SYSTEM ANALYSIS LAB MANUAL
BUS-27:
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BUS-33:
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CURRENT WAVEFORMS & SPECTRAS
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CABLE-2:
CABLE-7:
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POWER SYSTEM ANALYSIS LAB MANUAL
CABLE-11:
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POWER SYSTEM ANALYSIS LAB MANUAL
CABLE-18:
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POWER SYSTEM ANALYSIS LAB MANUAL
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
HARMONIC ANALYSIS REPORT
COMMENTS
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
In this experiment, we learnt:
• How to apply harmonic analysis to a power distribution
network
• How to perform load flow analysis on a power
distribution network
• How to perform harmonic analysis on a power
distribution network
• How to get harmonic spectra at required locations
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
EXPERIMENT#10
Flow of triplen harmonics (zero­sequence harmonics)
during 5 different schemes of connection for a 3­phase
transformer with presence of large non­linear load using
ETAP
ONE LINE DIAGRAM
HARMONIC MODEL USED
PC load is used as a source of harmonics in this experiment
that has the following range of THD’S.
Harmonic No.
3rd
ASAD NAEEM
2006‐RCET‐EE‐22
%THD
91.63
POWER SYSTEM ANALYSIS LAB MANUAL
5th
86.61
7th
69.87
9th
44.76
11th
54.81
th
13
46.44
15th
46.44
17th
33.05
th
19
24.70
23rd
11.74
25th
7.900
th
5.120
29
%THD
178.97
5 DIFFERENT SCHEMES OF TRANSFORMER
WINDING
PRIMARY SIDE
SECONDARY SIDE
Y-Grounded
Y-Grounded
Y-Grounded
Y-Ungrounded
Delta
Y-Ungrounded
Y-Grounded
Delta
Delta
Delta
1
2
3
4
5
HARMONIC ANALYSIS RESULTS
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CASE-1:
Transformer primary side Y-Grounded and
secondary Y-Ungrounded
VOLTAGE SPECTRA ON LT-SIDE
VOLTAGE SPECTRA ON HT-SIDE
VOLTAGE WAVEFORM ON LT-SIDE
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
VOLTAGE WAVEFORM ON HT-SIDE
CURRENT SPECTRA ON LT-SIDE
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CURRENT SPECTRA ON HT-SIDE
CURRENT WAVEFORM ON LT-SIDE
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CURRENT WAVEFORM ON HT-SIDE
OBSERVATIONS:
Harmonic source is connected on the LT side of transformer,
so the triplen harmonics are blocked due to ungrounded Yconnection. The magnitude of remaining harmonic
components is reduced on the HT side of transformer.
CASE-2:
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POWER SYSTEM ANALYSIS LAB MANUAL
Transformer primary side Y-Grounded and
secondary Y-Grounded
VOLTAGE SPECTRA ON LT-SIDE
VOLTAGE SPECTRA ON HT-SIDE
VOLTAGE WAVEFORM ON LT-SIDE
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
VOLTAGE WAVEFORM ON HT-SIDE
CURRENT SPECTRA ON LT-SIDE
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CURRENT SPECTRA ON HT-SIDE
CURRENT WAVEFORM ON LT-SIDE
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CURRENT WAVEFORM ON HT-SIDE
OBSERVATIONS:
Harmonic source is connected on the LT side of transformer,
so the triplen harmonics are not blocked due to grounded Yconnection. The triplen harmonics are also present on the HT
side of transformer as that is also Y-grounded.
CASE-3:
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
Transformer primary side Y-Grounded and
secondary Delta
VOLTAGE SPECTRA ON LT-SIDE
VOLTAGE SPECTRA ON HT-SIDE
VOLTAGE WAVEFORMS ON LT-SIDE
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
VOLTAGE WAVEFORM ON HT-SIDE
CURRENT SPECTRA ON LT-SIDE
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CURRENT SPECTRA ON HT-SIDE
CURRENT WAVEFORM ON LT-SIDE
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CURRENT WAVEFORM ON HT-SIDE
OBSERVATIONS:
Harmonic source is connected on the LT side of transformer,
so the triplen harmonics are blocked due to deltaconnection. The triplen harmonics are also blocked on the
HT side of transformer as there are no triplen harmonics on
secondary side of transformer.
CASE-4:
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
Transformer primary side Y-Ungrounded and
secondary Delta
VOLTAGE SPECTRA ON LT-SIDE
VOLTAGE SPECTRA ON HT-SIDE
VOLTAGE WAVEFORM ON LT-SIDE
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
VOLTAGE WAVEFORM ON HT-SIDE
CURRENT SPECTRA ON LT-SIDE
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CURRENT SPECTRA ON HT-SIDE
CURRENT WAVEFORM ON LT-SIDE
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CURRENT WAVEFORM ON HT-SIDE
OBSERVATIONS:
Harmonic source is connected on the LT side of transformer,
so the triplen harmonics are blocked due to deltaconnection. The triplen harmonics are also blocked on the
HT side of transformer as there are no triplen harmonics on
secondary side of transformer.
CASE-5:
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POWER SYSTEM ANALYSIS LAB MANUAL
Transformer primary side Delta and secondary
Delta
VOLTAGE SPECTRA ON LT-SIDE
VOLTAGE SPECTRA ON HT-SIDE
VOLTAGE WAVEFORM ON LT-SIDE
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
VOLTAGE WAVEFORM ON HT-SIDE
CURRENT SPECTRA ON LT-SIDE
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CURRENT SPECTRA ON HT-SIDE
CURRENT WAVEFORM ON LT-SIDE
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
CURRENT WAVEFORM ON HT-SIDE
OBSERVATIONS:
Harmonic source is connected on the LT side of transformer,
so the triplen harmonics are blocked due to deltaconnection. The triplen harmonics are also blocked on the
HT side of transformer as there are no triplen harmonics on
secondary side of transformer. Moreover, primary side is
also delta-connected.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
EXPERIMENT#11
Three phase short circuit analysis (3­phase faults­device
duty) for a given power system using ETAP
SHORT CIRCUIT ANALYSIS
The power station short circuit analysis program analyze the
effect of three phase, line to ground, line to line, and line to
line to ground faults on the electrical distribution networks.
The program calculates the total short circuit currents as
well as the contributions of individual motors, generators,
and utility ties in the system. Fault duties are in compliance
with the latest editions of the ANSI/IEEE standards and IEC
standards.
The ANSI/IEEE Short-Circuit Toolbar and IEC Short-Circuit
Toolbar sections explain how you can launch a short-circuit
calculation, open and view an output report, or select display
options. The Short-Circuit Study Case Editor section
explains how you can create a new study case, what
parameters are required to specify a study case, and how to
set them. The Display Options section explains what options
are available for displaying some key system parameters
and the output results on the one-line diagram, and how to
set them.
Short-Circuit Toolbar
This toolbar is active when you are in Short-Circuit mode and
the standard is set to ANSI in the Short-Circuit Study Case
Editor.
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
3-Phase Faults - Device Duty
Click on this button to perform a three-phase fault study per
ANSI C37 Standard. This study calculates momentary
symmetrical and asymmetrical rms, momentary
asymmetrical crest, interrupting symmetrical rms, and
interrupting adjusted symmetrical rms short-circuit currents
at faulted buses. The program checks the protective device
rated close and latching, and adjusted interrupting capacities
against the fault currents, and flags inadequate devices.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
Generators and motors are modeled by their positive
sequence sub-transient reactance.
ONE LINE DIAGRAM
FAULTY POINT
• BUS-15
There is a short circuit fault on bus-15.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
LOAD FLOW DIAGRAM
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
SHORT CIRCUIT ANALYSIS DIAGRAM
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
COMMENTS:
In this experiment, we use three phase fault-device duty
analysis to analyze the effect of fault on the system.
Following results are obtained in this experiment:
At bus-15:
Current
Power flow
Before fault
568A
345KW
After fault
4.8KA
3.6KW
We observe that the current flowing through bus-15 is
increased up to many times as compared to the current
before fault.
We observe that the power flowing through bus-15 is
decreased up to many times as compared to the power
before fault due to the short circuit at bus-15 as the load
connected to that bus is now shorted and no power is
flowing into that load.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
EXPERIMENT#12
Three phase short circuit analysis (3­phase faults­30 cycle
network) for a given power system using ETAP
3-Phase Faults – 30-Cycle Network
Click on this button to perform a three-phase fault study per
ANSI standards. This study calculates short-circuit currents
in their rms values after 30 cycles at faulted buses.
Generators are modeled by their positive sequence transient
reactance’s, and short-circuit current contributions from
motors are ignored.
ONE LINE DIAGRAM
FAULTY POINT
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2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
• BUS-15
There is a short circuit fault on bus-15.
LOAD FLOW DIAGRAM
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
SHORT CIRCUIT ANALYSIS DIAGRAM
COMMENTS:
In this experiment, we use three phase fault-device duty
analysis to analyze the effect of fault on the system.
Following results are obtained in this experiment:
At bus-15:
Current
Power flow
ASAD NAEEM
2006‐RCET‐EE‐22
Before fault
568A
345KW
After fault
3.6KA
3.6KW
POWER SYSTEM ANALYSIS LAB MANUAL
We observe that the current flowing through bus-15 is
increased up to many times as compared to the current
before fault.
We observe that the power flowing through bus-15 is
decreased up to many times as compared to the power
before fault due to the short circuit at bus-15 as the load
connected to that bus is now shorted and no power is
flowing into that load.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
EXPERIMENT#13
Three phase short circuit analysis (LG, LL, LLG, & 3­Phase
Faults ­ ½ Cycle) for a given power system using ETAP
LG, LL, LLG, & 3-Phase Faults - ½ Cycle
Click on this button to perform line-to-ground, line-to-line,
line-to-line-to-ground, and three-phase fault studies per
ANSI standards. This study calculates short-circuit currents
in their rms values at ½ cycles at faulted buses.
Generators and motors are modeled by their positive,
negative, and zero sequence sub-transient reactance.
In all the unbalanced fault calculations (½ cycle, 1.5-4 cycle
and 30 cycles), it is assumed that the negative sequence
impedance of a machine is equal to its positive sequence
impedance. Generator, motor, and transformer grounding
types and winding connections are taken into consideration
when constructing system positive, negative, and zero
sequence networks.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ONE LINE DIAGRAM
FAULTY POINT
• BUS-15
There is a short circuit fault on bus-15.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
LOAD FLOW DIAGRAM
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
SHORT CIRCUIT ANALYSIS DIAGRAM
COMMENTS:
In this experiment, we use three phase fault-device duty
analysis to analyze the effect of fault on the system.
Following results are obtained in this experiment:
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
At bus-15:
Current
Power flow
Before fault
568A
345KW
After fault
6.2KA
5.1KW
We observe that the current flowing through bus-15 is
increased up to many times as compared to the current
before fault.
We observe that the power flowing through bus-15 is
decreased up to many times as compared to the power
before fault due to the short circuit at bus-15 as the load
connected to that bus is now shorted and no power is
flowing into that load.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
EXPERIMENT#14
Three phase short circuit analysis (LG, LL, LLG, & 3­Phase
Faults ­ 1.5 to 4 Cycle) for a given power system using
ETAP
LG, LL, LLG, & 3-Phase Faults - 1.5 to 4 Cycle
Click on this button to perform three-phase, line-to-ground,
line-to-line, line-to-line-to-ground, and three-phase fault
studies per ANSI standards. This study calculates shortcircuit currents in their rms values between 1.5 to 4 cycles
at faulted buses.
Generators are modeled by their positive, negative, and zero
sequence sub-transient reactance, and motors are modeled
by their positive, negative and zero sequence transient
reactance. Generator, motor and transformer grounding
types and winding connections are taken into considerations
when constructing system positive, negative, and zero
sequential networks.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ONE LINE DIAGRAM
FAULTY POINT
• BUS-15
There is a short circuit fault on bus-15.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
LOAD FLOW DIAGRAM
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
SHORT CIRCUIT ANALYSIS DIAGRAM
COMMENTS:
In this experiment, we use three phase fault-device duty
analysis to analyze the effect of fault on the system.
Following results are obtained in this experiment:
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
At bus-15:
Current
Power flow
Before fault
568A
345KW
After fault
5.7KA
5KW
We observe that the current flowing through bus-15 is
increased up to many times as compared to the current
before fault.
We observe that the power flowing through bus-15 is
decreased up to many times as compared to the power
before fault due to the short circuit at bus-15 as the load
connected to that bus is now shorted and no power is
flowing into that load.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
EXPERIMENT#15
Three phase short circuit analysis (LG, LL, LLG, & 3­Phase
Faults ­ 30 Cycle) for a given power system using ETAP
LG, LL, LLG, & 3-Phase Faults - 30 Cycle
Click on this button to perform three-phase, line-to-ground,
line-to-line, line-to-line-to-ground, and three-phase fault
studies per ANSI standards. This study calculates shortcircuit currents in their rms values at 30-cycles at faulted
buses.
Generators are modeled by their positive, negative, and zero
sequence reactance, and short-circuit current contributions
from motors are ignored. Generator, motor, and
transformer grounding types and winding connections are
taken into consideration when constructing system positive,
negative, and zero sequence networks.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
ONE LINE DIAGRAM
FAULTY POINT
• BUS-15
There is a short circuit fault on bus-15.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
LOAD FLOW DIAGRAM
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
SHORT CIRCUIT ANALYSIS DIAGRAM
COMMENTS:
In this experiment, we use three phase fault-device duty
analysis to analyze the effect of fault on the system.
Following results are obtained in this experiment:
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
At bus-15:
Current
Power flow
Before fault
568A
345KW
After fault
4.8KA
4.8KW
We observe that the current flowing through bus-15 is
increased up to many times as compared to the current
before fault.
We observe that the power flowing through bus-15 is
decreased up to many times as compared to the power
before fault due to the short circuit at bus-15 as the load
connected to that bus is now shorted and no power is
flowing into that load.
ASAD NAEEM
2006‐RCET‐EE‐22
POWER SYSTEM ANALYSIS LAB MANUAL
COMPARISON OF SHORT CIRCUIT ANALYSIS
3‐phase faults‐
device duty
3‐phase faults‐30
cycle network
LG, LL, LLG, & 3‐
Phase Faults ‐ ½
Cycle
LG, LL, LLG, & 3‐
Phase Faults ‐ 1.5 to
4 Cycle
LG, LL, LLG, & 3‐
Phase Faults ‐ 30
Cycle
ASAD NAEEM
2006‐RCET‐EE‐22
CURRENT
POWER
4.8
3.6
3.6
3.6
6.2
5.1
5.7
5
4.8
4.8