BEKP 4773 : POWER SYSTEM ANALYSIS
BY
DR AZIAH KHAMIS
Introduction
To determine the steady state analysis of an interconnected power
system during normal operation
This system is assumed to be operating under balanced condition and is
represented by a single-phase network
The network contains hundreds of nodes and branches with impedances
specified in per-unit on a common MVA base
LOAD/POWER FLOW STUDIES
 Load-flow studies are performed to determine the steady-state operation of an electric power
system. It calculates the voltage drop on each feeder, the voltage at each bus, and the power flow
in all branch and feeder circuits.
 It also determine if system voltages remain within specified limits under various contingency
(unforeseen event) conditions, and whether the equipment such as transformers and conductors
are overloaded.
 Load-flow studies are often used to identify the need for additional generation, capacitive, or
inductive VAR support, or the placement of capacitors and/or reactors to maintain system
voltages within specified limits.
 Losses in each branch and total system power losses are also calculated.
 Necessary for planning, economic scheduling, and control of an existing system as well as
planning its future expansion
THE IMPORTANCE OF POWER FLOW ANALYSIS
➢ Back bone of power system
➢ Planning
➢ Operation
➢ Economic scheduling
➢ Exchange of power between utilities
➢ Transient stability
➢ Contingency studies
REQUIREMENT FOR SUCCESSFUL OPERATION OF POWER SYSTEM
▪ Generator supplies the demand & cater for the losses
▪ Bus voltages magnitudes → close to rated values
▪ Generators operates within specified real & reactive power
limit
▪ Transmission lines & transformers are not overloaded
Therefore, the utilities need some kind of computation tool or
program in order to successfully obtained these information
for operation purposes- it’s called power flow or load flow
program
SCOPE
▪ Model the power system using the nodal voltage
method.
▪ Present a careful formulation of the basic power flow
problem
▪ Investigate its solution by the:
▪ Gauss Seidel Method and Newton Raphson Method
and analyze the solution characteristics
THE IMPORTANT STEPS IN SOLVING THE POWER FLOW
STUDIES:
▪ Previously (recall your earlier power engineering subjects), you have
learned how to model the components of power system (Gen, Tx, T-line)
from single line diagram
▪ Now, you’re going to learn the steady-state analysis of an interconnected
power system during normal operation
▪ which have hundreds/thousands of nodes and branches
▪ impedances is converted into per units on a common MVA base.
▪ The way to analyze these are using “network equations” such as:
▪ If analyze Voltages & Current only and formulate the admittance matrix →”NodeVoltage Method” (if currents known and want to solve for voltages only , or vice versa)
▪ If analyze Power Flow → “Iteration Method” (if power are known, and will become nonlinear) thus, is called power flow/load flow equations. E.g. Gauss Seidel, Newton Raphson
▪ TO APPLY THE PER UNIT SYSTEM IN ORDER TO GENERATE THE
IMPEDANCE, REACTANCE AND ADMITTANCE DIAGRAM FROM
ONE LINE DIAGRAM.
▪ TO FURTHER DETERMINE THE ADMITTANCE MATRIX IN ORDER
TO USE FOR FURTHER PERFORMANCE IN THE POWER SYSTEM
ANALYSIS.
G1
T1
G2
T2
Line 1
Bus 1
Bus 2
Line 2
Line 3
Using per unit
system taught
earlier
Bus 3
Bus 4
1) single line diagram
Analysis using
Gauss
Seidel/NR/FD
to solve for P, Q
& V.
Using conversion
Y=1/Z
OR
Analysis using
Nodal Voltage
again to solve V
&I
2) reactance/impedance diagram
4) bus admittance
matrices
Using Nodalvoltage method
(I=YV)
NOTE : Figure 2,3 & 4 are in PER UNIT!
3) Admittance diagram
▪ The most common way to represent a power system network is
to use the node-voltage method:
“Given the voltages of generators at all generator nodes, and
knowing all impedances of machines and loads, one can solve for
all the currents in the typical node voltage analysis methods using
Kirchoff's current law (KCL)”.
▪ First the generators are replaced by equivalent current sources
and the node equations are written in the form:
[I]=[Y][V]
I = injected current vector/matrix,
Y = is the admittance vector/matrix
V = is the node voltage vector/matrices.
Bus admittance matrix is just a matrix of all interconnected admittances
between busses.
▪
Step 1: Number all the nodes of the system from 0 to n. Node 0 is
the reference node (or ground node).
▪
Step 2: Replace all generators by equivalent current sources in
parallel with an admittance.
▪
Step 3: Replace all lines, transformers, loads to equivalent
admittances whenever possible. The rule for this is simple:
where and y, z are generally complex numbers.
▪
Step 4: The bus admittance matrix Y is then formed by inspection as
follows :
▪
The current vector/matrix is next found from the sources connected to
nodes 0 to n. If no source is connected, the injected current would be
0.
▪
The equations which result are called the node-voltage equations and
are given the "bus" subscript in power studies thus:
or
I = YV
DERIVATION OF NODE VOLTAGE EQUATION
In order to obtain the
node voltage equations :
• Impedances are expressed in
per-unit on a common MVA
base
• Impedances are converted to
admittance
• Nodal solutions is based upon
kirchoff’s current law
DERIVATION OF NODE VOLTAGE EQUATION
DERIVATION OF NODE VOLTAGE EQUATION (USING KCL) :
I1 = y10V1 + y12 (V1 − V2 ) + y13 (V1 − V3 )
I 2 = y20V2 + y12 (V2 − V1 ) + y23 (V2 − V3 )
0 = y23 (V3 − V2 ) + y13 (V3 − V1 ) + y34 (V3 − V4 )
0 = y34 (V4 − V3 )
Rearrange,
I1 = ( y10 + y12 + y13 )V1 − y12V2 − y13V3
I 2 = − y12V1 + ( y20 + y y12 + y23 )V2 − y23V3
0 = − y13V1 − y23V2 + ( y13 + y23 + y34 )V3 − y34V4
0 = − y34V3 + y34V4
Y11 = y10 + y12 + y13
Node equation reduces to:
Y22 = y 20 + y y12 + y23
Y33 = y13 + y23 + y34
Y44 = y34
Y12 = Y21 = − y12
Y13 = Y31 = − y y13
Y23 = Y32 = − y23
Y34 = Y43 = − y34
Y14 = Y41 = 0 , Y24 = Y42 = 0
BUS ADMITTANCE MATRIX
Ibus = Ybus Vbus
Vector of
the injected
bus currents
Vector of
bus
voltages
measured
from the
reference
node


Bus Admittance Matrix
Diagonal element ➔ self-admittance / driving point admittance
Off-diagonal element ➔ mutual admittance / transfer admittance
➔ equal to the negative of the admittance between the node
EXAMPLE (TO FIND THE Y BUS)
Ibus = Ybus Vbus
Bus admittance matrix for the network in previous figure is:
EXAMPLE 1:
1
2
3
4
▪ Figure above shows the one line diagram of a simple four-bus system. Table below gives
the line impedances identified by the buses on which these terminate. The shunt
admittance at all the buses is assumed negligible.
(a) Find the Ybus assuming that the line shown dotted is not connected.
(b) What modifications need to be carried out in Ybus if the line shown dotted is connected.
Line,
Bus to bus
1-2
1-3
2-3
2-4
3-4
R, pu
X, pu
0.05
0.10
0.15
0.10
0.05
0.15
0.30
0.45
0.30
0.15
SOLUTION:
▪ (a) Ybus without dotted line.
Line,
Bus to bus
1-2
1-3
2-3
2-4
3-4
Ybus = 1 – j3
G, pu
B, pu
2.0
1.0
0.666
1.0
2.0
- 6.0
- 3.0
- 2.0
- 3.0
- 6.0
0
-1 + j3
0
0
1.666 – j5 -0.666 + j2 -1 + j3
-1 + j3 -0.666 + j2 3.666 – j11 -2 + j6
0
-1 + j3
-2 + j6
3 – j9
SOLUTION
▪ (b) Ybus with dotted line.
Line,
Bus to bus
1-2
1-3
2-3
2-4
3-4
Ybus = 3 – j9
G, pu
B, pu
2.0
1.0
0.666
1.0
2.0
- 6.0
- 3.0
- 2.0
- 3.0
- 6.0
-2 + j6
-1 + j3
0
-2 + j6 3.666 – j11 -0.666 + j2 -1 + j3
-1 + j3 -0.666 + j2 3.666 – j11 -2 + j6
0
-1 + j3
-2 + j6
3 – j9
EXAMPLE 2
SOLUTION:
Solution: