Power flow studies analyze steady-state conditions in power systems to determine voltage magnitudes and angles at each bus and the active and reactive power flows. The document introduces power flow analysis techniques including constructing an admittance matrix, defining the power flow equations, and using the Gauss-Seidel iterative method to solve the nonlinear power flow problem. Different types of buses - load, generator, and slack - are also defined based on the known and unknown power injections and voltages.

1. Introduction to Power Flow Studies
Introduction to Power Flow Studies
Prepared by,
KOK BOON CHING
2007@JEK/FKEE

2. Outlines
Outlines

 Introduction
Introduction

 Basic Techniques for Power Flow Studies
Basic Techniques for Power Flow Studies

 Basic Techniques for Power Flow Studies
Basic Techniques for Power Flow Studies

 The Bus Admittance Matrix
The Bus Admittance Matrix
P Fl E ti
P Fl E ti

 Power Flow Equations
Power Flow Equations

 Gauss
Gauss-
-Seidel Method
Seidel Method
G
G S id l P Fl S l i
S id l P Fl S l i

 Gauss
Gauss-
-Seidal Power Flow Solution
Seidal Power Flow Solution
BEE 3243 – Electric Power Systems – Module 5 2
2

3. Introduction
Introduction

 Power flow study is also known as
Power flow study is also known as load flow
load flow study.
study.

 It is an analysis during
It is an analysis during steady
steady-
-state
state conditions
conditions

 It is an analysis during
It is an analysis during steady
steady-
-state
state conditions.
conditions.

 It is used for
It is used for planning
planning and
and controlling
controlling a system.
a system.

 Assumptions:
Assumptions: balanced
balanced conditions and
conditions and single
single
Assumptions:
Assumptions: balanced
balanced conditions and
conditions and single
single
phase
phase analysis.
analysis.

 Problems:
Problems:
–
– determine the
determine the voltage magnitude
voltage magnitude and
and phase angle
phase angle at
at
each bus.
each bus.
–
– determine the active and reactive (
determine the active and reactive (P & Q
P & Q) power flow in
) power flow in
–
– determine the active and reactive (
determine the active and reactive (P & Q
P & Q) power flow in
) power flow in
each line
each line

 each bus has
each bus has four
four state variables:
state variables: voltage
voltage
BEE 3243 – Electric Power Systems – Module 5 3
3
magnitude, voltage phase angle, real power
magnitude, voltage phase angle, real power
injection, and reactive power injection
injection, and reactive power injection

4. Introduction
Introduction

 Each bus has
Each bus has two
two of the four state variables
of the four state variables
defined or given
defined or given
defined or given.
defined or given.

 Text book:
Text book: Hadi Saadat
Hadi Saadat –
– Power System Analysis
Power System Analysis
(
(Chapter 6
Chapter 6)
)
(
(Chapter 6
Chapter 6).
).
BEE 3243 – Electric Power Systems – Module 5 4
4

5. Types of Buses in Power Systems
Types of Buses in Power Systems

 Types of network buses:
Types of network buses:

 Load Bus
Load Bus or
or PQ Bus
PQ Bus

 Load Bus
Load Bus or
or PQ Bus
PQ Bus
–
– known real (P) and reactive (Q) power injections.
known real (P) and reactive (Q) power injections.
–
– No generator attach to load bus.
No generator attach to load bus.
No generator attach to load bus.
No generator attach to load bus.

 Generator Bus
Generator Bus or
or PV Bus
PV Bus
–
– known real (P) power injection and the voltage
known real (P) power injection and the voltage
magnitude (V).
magnitude (V).

 Slack Bus
Slack Bus or
or Swing Bus
Swing Bus
–
– known voltage magnitude (V) and voltage angle (
known voltage magnitude (V) and voltage angle (
),
),
often it is taken to be
often it is taken to be 1
1
0
0
 p.u.
p.u.
must have one generator as the slack bus
must have one generator as the slack bus
BEE 3243 – Electric Power Systems – Module 5 5
5
–
– must have one generator as the slack bus.
must have one generator as the slack bus.
–
– takes up the power slack due to losses in the network.
takes up the power slack due to losses in the network.

6. Basic Techniques for Power Flow Studies
Basic Techniques for Power Flow Studies

 Power flow analysis is an
Power flow analysis is an iterative
iterative problem.
problem.

 Steps
Steps to be taken in power flow analysis:
to be taken in power flow analysis:

 Steps
Steps to be taken in power flow analysis:
to be taken in power flow analysis:
1)
1) One line diagram or
One line diagram or load flow data
load flow data of a power system
of a power system
2)
2) Construct Bus Admittance Matrix (
Construct Bus Admittance Matrix (Ybus
Ybus)
)
2)
2) Construct Bus Admittance Matrix (
Construct Bus Admittance Matrix (Ybus
Ybus)
)
3)
3) Calculate the
Calculate the power flow
power flow analysis equations
analysis equations

 Power flow is a
Power flow is a nonlinear
nonlinear problem and it is
problem and it is
Power flow is a
Power flow is a nonlinear
nonlinear problem and it is
problem and it is
commonly solved by the iterative solution of
commonly solved by the iterative solution of
nonlinear algebraic equations
nonlinear algebraic equations:
:
–
– Gauss
Gauss-
-Seidal
Seidal
–
– Newton
Newton-
-Raphson
Raphson
BEE 3243 – Electric Power Systems – Module 5 6
6
–
– Fast Decoupled
Fast Decoupled

7. Basic Techniques for Power Flow Studies
Basic Techniques for Power Flow Studies
Example of
Example of load flow input data
load flow input data:
:

 Bus data
Bus data
Bus no Bus
code
Voltage Load Generator
Magnitude
(p.u.)
Angle
(degree)
P
(MW)
Q
(Mvar)
P
(MW)
Q
(Mvar)
(p ) ( g ) ( ) ( ) ( ) ( )
1 1 1.06 0 0.0 0.0 0.0 0.0
2 2 1.043 0 21.70 12.7 40.0 0.0
3 0 1 0 0 2 4 1 2 0 0 0 0

 Line data
Line data
3 0 1.0 0 2.4 1.2 0.0 0.0
Line bus no
(From)
Line bus no
(To)
Line
resistance
R (p.u.)
Line
reactance
X (p.u.)
½ line
susceptance
½ B (p.u.)
Transfor-
mer tap
setting
BEE 3243 – Electric Power Systems – Module 5 7
7
(p ) (p ) (p )
1 2 0.0192 0.0575 0.02640 0.978

8. The Bus Admittance Matrix
The Bus Admittance Matrix

 The matrix equation for relating the
The matrix equation for relating the nodal voltages
nodal voltages
to the
to the currents
currents that flow into and out of a network
that flow into and out of a network
to the
to the currents
currents that flow into and out of a network
that flow into and out of a network
using the
using the admittance
admittance values of circuit branches.
values of circuit branches.
V 1
1
V1
Vi
Ii
yi1
yi2
…
V2
Vn
ij
ij
ij
ij
jx
r
z
y



1
1
yin
…
yi0
Vn
Iinj = YbusVnode
Iinj = YbusVnode





 V
Y
Y
Y
I
yi0


























n
n
V
V
Y
Y
Y
Y
Y
Y
I
I








2
1
2
22
21
1
12
11
2
1
BEE 3243 – Electric Power Systems – Module 5 8
8

















 n
nn
n
n
n V
Y
Y
Y
I







2
1

9. The Bus Admittance Matrix
The Bus Admittance Matrix
One line diagram of a power system
BEE 3243 – Electric Power Systems – Module 5 9
9

10. The Bus Admittance Matrix
The Bus Admittance Matrix
Impedance Diagram
BEE 3243 – Electric Power Systems – Module 5 10
10

11. The Bus Admittance Matrix
The Bus Admittance Matrix
Admittance Diagram
BEE 3243 – Electric Power Systems – Module 5 11
11

12. The Bus Admittance Matrix
The Bus Admittance Matrix
Kirchhoff’s c rrent la
Kirchhoff’s current law:
BEE 3243 – Electric Power Systems – Module 5 12
12

13. The Bus Admittance Matrix
The Bus Admittance Matrix
Rearranging the KCL Equations:
Rearranging the KCL Equations:
Matrix Formation of the Equations
Matrix Formation of the Equations:
:
BEE 3243 – Electric Power Systems – Module 5 13
13

14. The Bus Admittance Matrix
The Bus Admittance Matrix





 V
Y
Y
Y
I
Completed Matrix Equation:
Completed Matrix Equation:
































n
n
V
V
Y
Y
Y
Y
Y
Y
I
I








2
1
2
22
21
1
12
11
2
1
Co p e ed a qua o
Co p e ed a qua o











 n
nn
n
n
n V
Y
Y
Y
I 
2
1
BEE 3243 – Electric Power Systems – Module 5 14
14

15. Y
Y-
-Bus Matrix Building Rules
Bus Matrix Building Rules

 Square matrix
Square matrix with dimensions equal to the
with dimensions equal to the
number of buses
number of buses
number of buses.
number of buses.

 Convert all network impedances into
Convert all network impedances into admittances
admittances.
.

 Diagonal
Diagonal elements:
elements:

 Diagonal
Diagonal elements:
elements:

 Off
Off-
-diagonal
diagonal elements:
elements:
BEE 3243 – Electric Power Systems – Module 5 15
15

 Matrix is
Matrix is symmetrical
symmetrical along the leading diagonal.
along the leading diagonal.

16. Power Flow Equations
Power Flow Equations
V1
Vi
yi1
yi2
V2
Ii
yin
…
V3
yi0
KCL Equations:
KCL Equations:
BEE 3243 – Electric Power Systems – Module 5 16
16

17. Power Flow Equations
Power Flow Equations
Power flow equation:
Power flow equation:
BEE 3243 – Electric Power Systems – Module 5 17
17

18. Gauss
Gauss-
-Seidel Method
Seidel Method

 Gauss
Gauss-
-Seidel
Seidel is a
is a nonlinear
nonlinear algebraic equation
algebraic equation
solver. It is a method of successive displacements.
solver. It is a method of successive displacements.

 Its
Its iterative steps
iterative steps:
:
–
– take a function and rearrange it into the form
take a function and rearrange it into the form x = g(x)
x = g(x)
–
– make an initial estimate of the variable x:
make an initial estimate of the variable x: x
x[0]
[0] = initial value
= initial value
–
– find an iterative improvement of x
find an iterative improvement of x[k]
[k], that is:
, that is: x
x[k+1]
[k+1] = g(x
= g(x[k]
[k])
)
l ti i h d h th diff b t t
l ti i h d h th diff b t t
–
– a solution is reached when the difference between two
a solution is reached when the difference between two
iterations is less than a specified accuracy:
iterations is less than a specified accuracy: |x
|x[k+1]
[k+1] –
– x
x[k]
[k]| ≤
| ≤ ε
ε

 Acceleration
Acceleration factors (
factors (
)
):
:
Acceleration
Acceleration factors (
factors (
)
):
:
–
– can improve the rate of convergence:
can improve the rate of convergence: 
 > 1
> 1
–
– modified step: the improvement is found as
modified step: the improvement is found as
BEE 3243 – Electric Power Systems – Module 5 18
18
p p
p p
x
x[k+1]
[k+1] = x
= x[k]
[k] +
+ 
 [
[g( x
g( x[k]
[k] )
) –
– x
x[k]
[k]]
]

19. Gauss
Gauss-
-Seidel Method
Seidel Method

 Example of the Gauss
Example of the Gauss-
-Seidel method:
Seidel method:
Find x of the equation: f(x) = x
Find x of the equation: f(x) = x3
3-
-6x
6x2
2+9x
+9x-
-4 = 0.
4 = 0.
9
9 3
3+6
+6 2
2+4
+4
9x =
9x = -
-x
x3
3+6x
+6x2
2+4
+4
)
(
9
4
9
6
9
1 2
3
x
g
x
x
x 




Start with
Start with initial
initial guess x
guess x[0]
[0] = 2,
= 2,
9
9
9
4
6
1
2222
.
2
9
4
)
2
(
9
6
)
2
(
9
1
)
2
( 2
3
]
0
[
]
1
[






 x
g
x
BEE 3243 – Electric Power Systems – Module 5 19
19
5173
.
2
9
4
)
2222
.
2
(
9
6
)
2222
.
2
(
9
1
)
2222
.
2
( 2
3
]
1
[
]
2
[






 x
g
x

20. Gauss
Gauss-
-Seidel Method
Seidel Method
8966
.
2
4
)
5173
.
2
(
6
)
5173
.
2
(
1
)
5173
.
2
( 2
3
]
2
[
]
3
[






 x
g
x
3376
.
3
9
4
)
8966
.
2
(
9
6
)
8966
.
2
(
9
1
)
8966
.
2
(
9
)
(
9
)
(
9
)
(
2
3
]
3
[
]
4
[






 x
g
x
g
7398
.
3
9
4
)
3376
.
3
(
9
6
)
3376
.
3
(
9
1
)
3376
.
3
(
9
9
9
2
3
]
4
[
]
5
[






 x
g
x
9988
3
9568
.
3
9
4
)
7398
.
3
(
9
6
)
7398
.
3
(
9
1
)
7398
.
3
(
]
7
[
2
3
]
5
[
]
6
[






 x
g
x
0000
.
4
9988
.
3
]
8
[
]
7
[


x
x
BEE 3243 – Electric Power Systems – Module 5 20
20

21. Gauss
Gauss-
-Seidel Method
Seidel Method

 Matlab Results of all iterations:
Matlab Results of all iterations:
Iter g dx x
1 2.2222 0.2222 2.2222
2 2.5173 0.2951 2.5173
3 2 8966 0 3793 2 8966
3 2.8966 0.3793 2.8966
4 3.3376 0.4410 3.3376
5 3.7398 0.4022 3.7398
6 3.9568 0.2170 3.9568
7 3.9988 0.0420 3.9988
8 4.0000 0.0012 4.0000
BEE 3243 – Electric Power Systems – Module 5 21
21
9 4.0000 0.0000 4.0000

22. Gauss
Gauss-
-Seidel Method
Seidel Method

 Graphical illustration:
Graphical illustration:
Iterations
x = g(x)
Initial value
Solution points
BEE 3243 – Electric Power Systems – Module 5 22
22

23. Gauss
Gauss-
-Seidel Method with
Seidel Method with 


 Find the root of the equation: f(x) = x
Find the root of the equation: f(x) = x3
3 -
- 6x
6x2
2 + 9x
+ 9x -
- 4
4
= 0 with an
= 0 with an acceleration factor
acceleration factor of 1 25
of 1 25
= 0 with an
= 0 with an acceleration factor
acceleration factor of 1.25.
of 1.25.

 Starting with an
Starting with an initial
initial guess of x
guess of x[0]
[0] = 2.
= 2.
BEE 3243 – Electric Power Systems – Module 5 23
23

24. Gauss
Gauss-
-Seidel Method with
Seidel Method with 


 Matlab results of all iterations:
Matlab results of all iterations:
Iter g dx x
1 2.2222 0.2778 2.2778
2 2 5902 0 3905 2 6683
2 2.5902 0.3905 2.6683
3 3.0801 0.5148 3.1831
4 3.6157 0.5407 3.7238
5 3.9515 0.2846 4.0084
6 4.0000 -0.0106 3.9978
7 4.0000 0.0027 4.0005
BEE 3243 – Electric Power Systems – Module 5 24
24
8 4.0000 -0.0005 4.0000

25. Gauss
Gauss-
-Seidel Method with
Seidel Method with 


 Graphical illustration:
Graphical illustration:
BEE 3243 – Electric Power Systems – Module 5 25
25

26. Gauss
Gauss-
-Seidel Method with
Seidel Method with 


 Do not use a
Do not use a very large
very large number of
number of 
 as the larger
as the larger
step size
step size may result in an overshoot
may result in an overshoot
step size
step size may result in an overshoot.
may result in an overshoot.

 If we take the previous example with
If we take the previous example with 
 = 1.8
= 1.8, we
, we
will need
will need more iterations
more iterations to obtain the answer:
to obtain the answer:
will need
will need more iterations
more iterations to obtain the answer:
to obtain the answer:
Iter g dx x
1 2.2222 0.4000 2.4000
Iter g dx x
6 3.9940 0.1619 4.0659
2 2.7484 0.6272 3.0272
3 3.4714 0.7996 3.8268
7 3.9971 -0.1239 3.9420
8 3.9978 0.1004 4.0424
3 3.4714 0.7996 3.8268
4 3.9806 0.2768 4.1036
5 3 9927 0 1996 3 9040
8 3.9978 0.1004 4.0424
9 3.9988 -0.0785 3.9639
10 3 9991 0 0634 4 0273
BEE 3243 – Electric Power Systems – Module 5 26
26
5 3.9927 -0.1996 3.9040 10 3.9991 0.0634 4.0273

27. Gauss
Gauss-
-Seidel Method with
Seidel Method with 

Iter g dx x
11 3.9995 -0.0501 3.9772
Iter g dx x
19 4.0000 -0.0084 3.9963
12 3.9997 0.0404 4.0176
13 3 9998 -0 0320 3 9856
20 4.0000 0.0067 4.0030
21 4 0000 -0 0054 3 9976
13 3.9998 -0.0320 3.9856
14 3.9999 0.0257 4.0113
21 4.0000 -0.0054 3.9976
22 4.0000 0.0043 4.0019
15 3.9999 -0.0205 3.9908
16 3.9999 0.0165 4.0073
23 4.0000 -0.0034 3.9985
24 4.0000 0.0027 4.0012
17 4.0000 -0.0132 3.9941
18 4.0000 0.0106 4.0047
25 4.0000 -0.0022 3.9990
26 4.0000 0.0018 4.0008
BEE 3243 – Electric Power Systems – Module 5 27
27

28. Gauss
Gauss-
-Seidel Method with
Seidel Method with 


 Graphical illustration:
Graphical illustration:
OVERSHOOT
BEE 3243 – Electric Power Systems – Module 5 28
28

29. Gauss
Gauss-
-Seidel Method with
Seidel Method with 


 Example 2
Example 2
BEE 3243 – Electric Power Systems – Module 5 29
29

30. Gauss
Gauss-
-Seidel Method with
Seidel Method with 


 Solution
Solution
BEE 3243 – Electric Power Systems – Module 5 30
30

31. Gauss
Gauss-
-Seidal for a System of n Equations
Seidal for a System of n Equations

 Consider a system of
Consider a system of n equations
n equations:
:
R h i f h f h i bl
R h i f h f h i bl

 Rearrange each equation for each of the variables:
Rearrange each equation for each of the variables:
BEE 3243 – Electric Power Systems – Module 5 31
31

32. Gauss
Gauss-
-Seidal for a System of n Equations
Seidal for a System of n Equations

 Steps:
Steps:

Assume an
Assume an approximate solution
approximate solution for the independent
for the independent

Assume an
Assume an approximate solution
approximate solution for the independent
for the independent
variables,
variables,

Find the results in a
Find the results in a new approximate
new approximate solution
solution

I th
I th G
G S id l
S id l th d th d t d l f th
th d th d t d l f th

In the
In the Gauss
Gauss-
-Seidel
Seidel method, the updated values of the
method, the updated values of the
variables calculated in the preceding equations are used
variables calculated in the preceding equations are used
immediately in the solution of the subsequent equations.
immediately in the solution of the subsequent equations.
BEE 3243 – Electric Power Systems – Module 5 32
32

The rate of
The rate of convergence
convergence can be increased by suitable
can be increased by suitable 
.
.

33. Gauss
Gauss-
-Seidal Power Flow Solution
Seidal Power Flow Solution

 Previously derived
Previously derived power flow equation
power flow equation,
,
G
G S id l
S id l f
f

 Gauss
Gauss-
-Seidal
Seidal form,
form,
BEE 3243 – Electric Power Systems – Module 5 33
33

34. Gauss
Gauss-
-Seidal Power Flow Solution
Seidal Power Flow Solution

 Rewriting the power equation to find
Rewriting the power equation to find P and Q
P and Q

the
the real and reactive powers
real and reactive powers are scheduled for the
are scheduled for the load
load

the
the real and reactive powers
real and reactive powers are scheduled for the
are scheduled for the load
load
buses
buses that is, they remain fixed
that is, they remain fixed

the currents and powers are expressed as going into the
the currents and powers are expressed as going into the
b
b
bus
bus

 for
for generation
generation the powers are
the powers are positive
positive

 for
for loads
loads the powers are
the powers are negative
negative
BEE 3243 – Electric Power Systems – Module 5 34
34

 the
the scheduled power
scheduled power is the
is the sum
sum of the generation and load
of the generation and load
powers
powers

35. Gauss
Gauss-
-Seidal Power Flow Solution
Seidal Power Flow Solution

 The
The complete set
complete set of equations become:
of equations become:
BEE 3243 – Electric Power Systems – Module 5 35
35

36. Gauss
Gauss-
-Seidal Power Flow Solution
Seidal Power Flow Solution

 Since the
Since the off
off-
-diagonal
diagonal elements of Y
elements of Ybus
bus, Y
, Yij
ij =
= -
-y
yij
ij
and the
and the diagonal
diagonal elements Y
elements Yii
ii =
= 
y
yij
ij
and the
and the diagonal
diagonal elements, Y
elements, Yii
ii =
= 
y
yij
ij,
,
n
k
j
ij
k
sch
i
sch
i
V
Y
jQ
P


 ]
[
]
[
*
ii
i
j
j
j
ij
k
i
k
i
Y
V
V





,
1
]
[
*
]
1
[

 





















 



n
i
j
j
k
j
ij
ii
k
i
k
i
k
i V
Y
Y
V
V
P
,
1
]
[
]
[
]
[
*
]
1
[




















 



n
i
j
j
k
j
ij
ii
k
i
k
i
k
i V
Y
Y
V
V
Q
,
1
]
[
]
[
]
[
*
]
1
[
BEE 3243 – Electric Power Systems – Module 5 36
36



 


 i
j
j ,
1

37. Gauss
Gauss-
-Seidal Power Flow Solution
Seidal Power Flow Solution

 System characteristics:
System characteristics:

Since both components (V &
Since both components (V & 
) are specified for the
) are specified for the

Since both components (V &
Since both components (V & 
) are specified for the
) are specified for the
slack bus
slack bus, there are 2(n
, there are 2(n -
- 1) equations which must be
1) equations which must be
solved iteratively
solved iteratively

For the
For the load buses
load buses, the real and reactive powers are
, the real and reactive powers are
known/ scheduled:
known/ scheduled:

 the voltage magnitude and angle must be estimated
the voltage magnitude and angle must be estimated

 the voltage magnitude and angle must be estimated
the voltage magnitude and angle must be estimated

 in per unit, the nominal voltage magnitude is 1 pu
in per unit, the nominal voltage magnitude is 1 pu

 the angles are generally close together, so an initial value of 0
the angles are generally close together, so an initial value of 0
degrees is appropriate
degrees is appropriate
degrees is appropriate
degrees is appropriate
BEE 3243 – Electric Power Systems – Module 5 37
37

38. Gauss
Gauss-
-Seidal Power Flow Solution
Seidal Power Flow Solution

For the
For the generator buses
generator buses, the real power and voltage
, the real power and voltage
magnitude are known,
magnitude are known,
g ,
g ,

 the real power is scheduled
the real power is scheduled

 the reactive power is computed based on the estimated voltage
the reactive power is computed based on the estimated voltage
values
values
values
values

 the voltage is computed by Gauss
the voltage is computed by Gauss-
-Seidel, only the imaginary
Seidel, only the imaginary
part is kept
part is kept

 the complex voltage is found from the magnitude and the
the complex voltage is found from the magnitude and the

 the complex voltage is found from the magnitude and the
the complex voltage is found from the magnitude and the
iterative imaginary part
iterative imaginary part
BEE 3243 – Electric Power Systems – Module 5 38
38

39. Line Flows and Losses
Line Flows and Losses

 After solving for bus voltages and angles,
After solving for bus voltages and angles, power
power
flows
flows and
and losses
losses on the network branches are
on the network branches are
flows
flows and
and losses
losses on the network branches are
on the network branches are
calculated.
calculated.

Transmission lines and transformers are network
Transmission lines and transformers are network
branches
branches

The direction of positive current flow are defined as
The direction of positive current flow are defined as
f ll f b h l t (d t t d
f ll f b h l t (d t t d
follows for a branch element (demonstrated on a
follows for a branch element (demonstrated on a
medium length line)
medium length line)

Power flow is defined for each end of the branch
Power flow is defined for each end of the branch
BEE 3243 – Electric Power Systems – Module 5 39
39

40. Line Flows and Losses
Line Flows and Losses

 Example
Example: the power leaving
: the power leaving bus i
bus i and flowing to
and flowing to
bus j
bus j:
:
bus j
bus j:
:
BEE 3243 – Electric Power Systems – Module 5 40
40

41. Line Flows and Losses
Line Flows and Losses

 Current and power flows
Current and power flows:
:

 Power losses
Power losses
BEE 3243 – Electric Power Systems – Module 5 41
41

42. Example
Example
1.
1. Figure below show the OLD of a simple three
Figure below show the OLD of a simple three-
-bus power
bus power
system with generation at bus 1. The magnitude of
system with generation at bus 1. The magnitude of
voltage at bus 1 is adjusted to 1 05 per unit The schedule
voltage at bus 1 is adjusted to 1 05 per unit The schedule
voltage at bus 1 is adjusted to 1.05 per unit. The schedule
voltage at bus 1 is adjusted to 1.05 per unit. The schedule
loads at buses 2 and 3 are as marked on the diagram.
loads at buses 2 and 3 are as marked on the diagram.
Line impedances are marked in per unit on a 100MVA
Line impedances are marked in per unit on a 100MVA
base and the line charging
base and the line charging susceptances
susceptances are neglected.
are neglected.
BEE 3243 – Electric Power Systems – Module 5 42
42

43. Example
Example
a)
a) Using the
Using the Gauss
Gauss-
-Seidel method
Seidel method, determine the
, determine the phasor
phasor
values of the
values of the voltage
voltage at the load buses 2 and 3, accurate
at the load buses 2 and 3, accurate
to 4 decimal places
to 4 decimal places
to 4 decimal places.
to 4 decimal places.
b)
b) Find the slack bus
Find the slack bus P
P and
and Q
Q.
.
c)
c) Determine the
Determine the line flows
line flows and
and line losses
line losses. Construct a
. Construct a
c)
c) Determine the
Determine the line flows
line flows and
and line losses
line losses. Construct a
. Construct a
power flow diagram
power flow diagram showing the direction of the line flow.
showing the direction of the line flow.
BEE 3243 – Electric Power Systems – Module 5 43
43

44. Example
Example
a)

 Line admittances:
Line admittances:
1
a)
1
20
10
04
.
0
02
.
0
1
12 j
j
y 



32
16
1
30
10
03
.
0
01
.
0
1
13
j
j
j
y 



32
16
025
.
0
0125
.
0
23 j
j
y 




 At
At P
P-
-Q buses
Q buses, the complex loads in p.u.:
, the complex loads in p.u.:
p.u.
102
.
1
566
.
2
2 j
Ssch



BEE 3243 – Electric Power Systems – Module 5 44
44
p.u.
452
.
0
386
.
1
3 j
Ssch




45. Example
Example
BEE 3243 – Electric Power Systems – Module 5 45
45

46. Example
Example

 Initial estimate,
Initial estimate,
0
0
0
1
0
0
0
1 )
0
(
)
0
(
j
V
j
V 




 Compute V
Compute V2
2 and V
and V3
3 using:
using:
0
.
0
0
.
1
,
0
.
0
0
.
1 )
(
3
)
(
2 j
V
j
V 



jQ
P sch
sch
23
12
)
0
(
3
23
1
12
)
0
(
*
2
2
2
)
1
(
2
y
y
V
y
V
y
V
jQ
P
V
sch
sch





)
52
26
(
)
0
0
.
1
)(
32
16
(
)
0
05
.
1
)(
20
10
(
0
0
.
1
102
.
1
566
.
2
23
12
j
j
j
j
j
j
j
y
y










031
.
0
9825
.
0
)
52
26
(
j
j



BEE 3243 – Electric Power Systems – Module 5 46
46

47. Example
Example
)
1
(
2
23
1
13
)
0
(
*
3
3
3
)
1
(
3
V
y
V
y
V
jQ
P
V
sch
sch




)
031
.
0
9825
.
0
)(
32
16
(
)
0
05
.
1
)(
30
10
(
0
0
.
1
452
.
0
386
.
1
23
13
3
j
j
j
j
j
j
y
y










0353
.
0
0011
.
1
)
62
26
(
j
j
j





 For 2
For 2nd
nd iteration,
iteration,
)
0353
.
0
0011
.
1
)(
32
16
(
)
0
05
.
1
)(
20
10
(
102
.
1
566
.
2
j
j
j
j
j








052
.
0
9816
.
0
)
52
26
(
)
0353
.
0
0011
.
1
)(
32
16
(
)
0
05
.
1
)(
20
10
(
031
.
0
9825
.
0
)
2
(
2
j
j
j
j
j
j
j
V










BEE 3243 – Electric Power Systems – Module 5 47
47
052
.
0
9816
.
0 j

48. Example
Example
)
62
26
(
)
052
.
0
9816
.
0
)(
32
16
(
)
0
05
.
1
)(
30
10
(
0353
.
0
0011
.
1
452
.
0
386
.
1
)
1
(
3
j
j
j
j
j
j
j
V











 The remaining iterations until the solution is
The remaining iterations until the solution is
0459
.
0
0008
.
1
)
62
26
(
j
j




 The remaining iterations until the solution is
The remaining iterations until the solution is
converged with an accuracy of 5 x 10
converged with an accuracy of 5 x 10-
-5
5 p.u.:
p.u.:
0488
0
0004
1
0578
0
9808
0 )
3
(
)
3
(
j
V
j
V 

0499
0
0001
1
0598
0
9801
0
0497
.
0
0002
.
1
0594
.
0
9803
.
0
0488
.
0
0004
.
1
0578
.
0
9808
.
0
)
5
(
3
)
5
(
2
)
4
(
3
)
4
(
2
)
(
3
)
(
2
j
V
j
V
j
V
j
V
j
V
j
V












0500
.
0
0000
.
1
0600
.
0
9800
.
0
0500
.
0
0000
.
1
0599
.
0
9801
.
0
0499
.
0
0001
.
1
0598
.
0
9801
.
0
)
7
(
3
)
7
(
2
)
6
(
3
)
6
(
2
3
2
j
V
j
V
j
V
j
V
j
V
j
V










Final solution
BEE 3243 – Electric Power Systems – Module 5 48
48
0500
.
0
0000
.
1
0600
.
0
9800
.
0 3
2 j
V
j
V Final solution

49. Example
Example
b)

 The slack bus power:
The slack bus power:
b)
)]
(
)
(
[
*
V
y
V
y
y
y
V
V
jQ
P 




)]
05
.
0
0
.
1
)(
30
10
(
)
06
.
0
98
.
0
)(
20
10
(
)
50
20
(
05
.
1
[
05
.
1
)]
(
)
(
[ 3
13
2
12
13
12
1
1
1
1
j
j
j
j
j
V
y
V
y
y
y
V
V
jQ
P













Mvar
189
MW
5
.
409
p.u.
890
.
1
095
.
4
j
j




BEE 3243 – Electric Power Systems – Module 5 49
49

50. Example
Example
c)

 Line currents:
Line currents:
c)
8
0
9
1
)]
06
0
98
0
(
)
0
05
1
)[(
20
10
(
)
( j
j
j
j
V
V
I 
0
.
1
0
.
2
)]
05
.
0
0
.
1
(
)
0
05
.
1
)[(
30
10
(
)
(
8
.
0
9
.
1
8
.
0
9
.
1
)]
06
.
0
98
.
0
(
)
0
05
.
1
)[(
20
10
(
)
(
3
1
13
13
12
21
2
1
12
12
j
j
j
j
V
V
y
I
j
I
I
j
j
j
j
V
V
y
I























48
.
0
64
.
0
)]
05
.
0
0
.
1
(
)
06
.
0
98
.
0
)[(
32
16
(
)
(
0
.
1
0
.
2
)]
(
)
)[(
(
)
(
3
2
23
23
13
31
3
1
13
13
j
j
j
j
V
V
y
I
j
I
I
j
j
j
j
y















48
.
0
64
.
0
23
32 j
I
I 



BEE 3243 – Electric Power Systems – Module 5 50
50

51. Example
Example
Th li fl
Th li fl

 The line flows are:
The line flows are:
Mvar
0
84
MW
5
199
p.u.
84
.
0
995
.
1
)
8
.
0
9
.
1
)(
0
.
0
05
.
1
(
*
12
1
12
j
j
j
j
I
V
S








Mvar
0
.
67
MW
0
.
191
p.u.
67
.
0
91
.
1
)
8
.
0
9
.
1
)(
06
.
0
98
.
0
(
Mvar
0
.
84
MW
5
.
199
*
21
2
21
j
j
j
j
I
V
S
j













Mvar
0
.
105
MW
0
.
210
p.u.
05
.
1
1
.
2
)
0
.
1
0
.
2
)(
0
.
0
05
.
1
(
*
13
1
13
j
j
j
j
I
V
S
j








Mvar
0
.
90
MW
0
.
205
p.u.
90
.
0
05
.
2
)
0
.
1
0
.
2
)(
05
.
0
0
.
1
(
*
*
31
3
31
j
j
j
j
I
V
S











448
0
664
0
)
48
0
64
0
)(
05
0
0
1
(
Mvar
2
43
MW
6
.
65
p.u.
432
.
0
656
.
0
)
48
.
0
656
.
0
)(
06
.
0
98
.
0
(
*
*
23
2
23
j
j
j
I
V
S
.
j
j
j
j
I
V
S











BEE 3243 – Electric Power Systems – Module 5 51
51
Mvar
8
.
44
MW
4
.
66
p.u.
448
.
0
664
.
0
)
48
.
0
64
.
0
)(
05
.
0
0
.
1
(
32
3
32
j
j
j
j
I
V
S









52. Example
Example

 Line losses are:
Line losses are:
M
0
17
MW
5
8 j
S
S
S
Mvar
60
1
MW
8
0
Mvar
0
.
15
MW
0
.
5
Mvar
0
.
17
MW
5
.
8
32
23
31
13
13
21
12
12
j
S
S
S
j
S
S
S
j
S
S
S
L
L
L












Mvar
60
.
1
MW
8
.
0
32
23
23 j
S
S
SL 



BEE 3243 – Electric Power Systems – Module 5 52
52

53. Example
Example
Th fl di
Th fl di
191
199 5

 The power flow diagram:
The power flow diagram:
409 5
191
65 6
67.0
199.5
84.0 256.6
110 2
(8.5)
(17.0)
409.5 65.6
43.2
138 6
110.2
(0.8)
66.4
44.8
189
138.6
45.2
(1.6)
205
90.0
210.0
105.0
(5)
(15)
BEE 3243 – Electric Power Systems – Module 5 53
53
( )

54. BEE 3243 – Electric Power Systems – Module 5 54
54

55. BEE 3243 – Electric Power Systems – Module 5 55
55

56. BEE 3243 – Electric Power Systems – Module 5 56
56

57. BEE 3243 – Electric Power Systems – Module 5 57
57

58. BEE 3243 – Electric Power Systems – Module 5 58
58

59. BEE 3243 – Electric Power Systems – Module 5 59
59

60. BEE 3243 – Electric Power Systems – Module 5 60
60

61. BEE 3243 – Electric Power Systems – Module 5 61
61

62. BEE 3243 – Electric Power Systems – Module 5 62
62