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ECE4334
Dr. C.Y. Evrenosoglu
ECE4334
POWER FLOW ANALYSIS
Dr. E
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ECE4334Power Flow (Load Flow) Analysis A steady-state analysis tool
Applied to three-phase balanced power systems
One-line diagram is used
Input data is the bus data, transmission line data and transformer
data
The transmission system is modeled by a set of buses (nodes)
interconnected by transmission lines (links).
Generators and loads connected to the various buses of the systeminject and remove power from the transmission system.
The purpose is to determine the current state of the network by
determining the complex bus voltages and subsequently computing
the real and reactive transmission line flows.
The power flow function is an integral part of most studies in
system planning and operation and the most common of power
system computer calculations.Dr. C.Y. Evrenosoglu
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ECE4334Power Flow Analysis Each node can represent an individual bus; or a substation that
supplies power to a distribution network.
Aggregated nodes are used widely in power system analysis.
Power flow analysis can be used for transmission or distribution
networks. (Different layers of the system) We will concentrate onthe transmission network because it is the backbone of the overall
system.
There are systems as small as a few hundred buses and as large as
tens of thousands of buses. Effective and fast solution of the
network is always extremely important especially during real-time
operations.
Use of efficient data storage and programming techniques isextremely important.
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ECE4334Power Flow Analysis
Demand changes
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CAISO
PJM
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ECE4334Power Flow Analysis During steady-state operation the consumption varies which
subsequently has an impact on the power system state (i.e. busvoltages, bus angles, line flows etc). We want to know if there are
any limit violations in the system (Is the system secure?):
Are the voltage magnitudes within acceptable limits? Are there any lines which are (thermally) overloaded?
Is the steady-state stability margin for a transmission line too small
(i.e. thepower angle across the line is too big)?
Is a generatoroverloaded?
How does the network behave when there is a single (or multiple)
contingency outages (i.e. a line failure, a generator failure etc.)?
In system operation it is desirable to operate the system in such a
way that it is not overloaded in any way nor will it become so in the
event of a likely emergency.
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ECE4334Power Flow Analysis In system operation and planning it is also extremely important to
consider the economy of operation.
We wish to consider among all the possible allocations of generation
assignments what is optimal in the sense of minimum production
costs (i.e. the fuel cost per hour to generate all the power needed tosupply the loads.)
The objectives (i.e. economy of operation and secure operation)
frequently give conflicting operating requirements and compromises
are often required.
In power system operation to achieve the two objectives we need to
know the relationships between the generation, the demand and the
voltages. These relationships are derived from Kirchhoff's CurrentLaw and known aspower flow equations (power balance
equations).
Before we start with the formulation we will introduce bus
admittance matrix, Ybus.Dr. C.Y. Evrenosoglu
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ECE4334Thvenins & Nortons equivalent circuits
Dr. C.Y. Evrenosoglu
Equivalent circuits are widely used to reduce the models in
large interconnections.
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ECE4334Bus admittance matrix
It provides the relationship governing the behavior of the nodevoltages and currents.
Widely used in various power system analysis tools.
It finds widespread application in determining the network solution
and forms an integral part of most modern-day power systemanalysis
Bus impedance matrix, Zbus is the inverse of Ybus and mainly used in
fault analysis. In the bus admittance matrix representation, the injected currents at
nodes of the interconnected network are related to the voltages at the
nodes via an admittance representation.
Ybus of an interconnected power system is large and has a large
number of zero entities (sparse matrix). This is because each node
in the physical power system is connected to at most three or five
other buses. However the bus impedance matrix, Zbus is a full matrix.Dr. C.Y. Evrenosoglu
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ECE4334Per phase bus admittance matrix
In developing this representation the neutral is taken asthe reference node. The relationship between the injected
node currents and the node voltages is whereIis the vector of injected node currents (from generators
and to loads) and Vis the vector of node voltages.
Each component element of the interconnected network is
referred to as abranch (line and xfmr).
For the purpose of modeling we will represent a branch
by thebranch admittance, y which is also referred to as
primitive admittance. Sometimes we will also use the
branch impedance, z (primitive impedance).
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ECE4334Example Construct Ybus
Dr. C.Y. Evrenosoglu Thanks to Dr. Thomas Baldwin, FAMU/FSU for slide content
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ECE4334Example Construct Ybus
Dr. C.Y. Evrenosoglu Thanks to Dr. Thomas Baldwin, FAMU/FSU for slide content
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ECE4334Example Construct Ybus
Dr. C.Y. Evrenosoglu
Rearrange the KCL equations
Thanks to Dr. Thomas Baldwin, FAMU/FSU for slide content
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ECE4334Example Construct Ybus
Dr. C.Y. Evrenosoglu Thanks to Dr. Thomas Baldwin, FAMU/FSU for slide content
Pattern?
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ECE4334Construct Ybus by inspection
In terms of primitive admittances, the steps in developing thebus admittance matrix or Ybus by inspection is as follows
1. Convert all the network impedances into admittances.
2. Yii
= Y(i,i), the diagonal term is the self-admittance and it
is equal to the sum of the primitive admittances of all the
components connected to the ith node.
3. Yij = Y(i,j), the ijth element (off-diagonal term) of the
matrix, is equal to the negative of the equivalent
primitive admittance of the components connected
between nodes i andj. ( ij)
4. The Ybus is symmetric. VERY IMPORTANT: The bus impedance Matrix, Zbus can
NOT be written by inspection. Zbus = [Ybus]-1
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ECE4334Example Line
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ECE4334Example Regulating XFMR
Dr. C.Y. Evrenosoglu
Single-line representation of an off-nominal turns ratio xfmr. The xfmr turns ratio is normalized as a:1 and the non-unity
side is called the tap side.
In the representation the series primitive admittance
(reciprocal of series primitive impedance) of the xfmr isconnected to the unity side.
LTC a is real
PAR
a is complex
Ip a*
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ECE4334Example Regulating XFMR
Dr. C.Y. Evrenosoglu
Ip a*
When a is real (LTC xfmr)
?
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ECE4334Example LTC
Dr. C.Y. Evrenosoglu
Ip a
x2x1
y/a
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ECE4334Example
The parameters for the branch data is provided. The series
impedance and total line charging susceptance for each branch are
in pu on an appropriately chosen base. A generatorwith emf equal
to 0.9 pu and a reactance ofj1.25 pu is connected to bus 1 whilea motorwith internal voltage equal to 0.8 pu and areactance ofj1.25 pu is connected to bus 5. Write the nodalequations.
Dr. C.Y. Evrenosoglu
From To R X B
1 2 0.004 0.0533 0
2 3 0.02 0.25 0.22
3 4 0.02 0.25 0.22
2 4 0.01 0.15 0.11
4 5 0.006 0.08 0
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ECE4334Example
Dr. C.Y. Evrenosoglu
From To R X B
1 2 0.004 0.0533 0
2 3 0.02 0.25 0.22
3 4 0.02 0.25 0.22
2 4 0.01 0.15 0.11
4 5 0.006 0.08 0
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ECE4334Example
Dr. C.Y. Evrenosoglu
jB24/2
jB23/2 jB34/2
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ECE4334Power flow
When analyzing power systems we know neither thecomplex bus voltages nor the complex current
injections
Rather, we know the complex power being consumedby the load, and the power being injected by the
generators plus their voltage magnitudes
Therefore we can not directly use the Ybus equations,but rather must use thepower balance equations.
We want to know the voltage profile of the network
that is the nodal (bus) voltages for a given load and
generation schedule.
Dr. C.Y. Evrenosoglu Thanks to Dr. Tom Overbye, University of Illinois for the content
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ECE4334Power flow bus types
There three main types of network buses1. Load bus (PQ bus)
The power injection is known real power, P and
reactive power, Q injections are known2. Generator bus (PV bus)
The real power (P) injection is known and the voltage
magnitude, |V| is known.3. Slack bus (swing bus)
Reference bus; voltage angle is set to 0
Takes up the losses in the network. We will assume that there is one slack bus.
In practice distributed slack and generator
participation factors are in use.Dr. C.Y. Evrenosoglu
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ECE4334
PDk QDk
Power flow bus typesThe power delivered (injected) to bus kis
If there is no generation at bus k, then
1) Load bus (PQ bus)
Input PDkand QDk
Output |Vk| and k
2) Generator bus (PV bus)
Input |Vk| and PGk(sometimes PDkand QDk)
The maximum and minimum limits of reactive generation
If the QGkhits the limit, its value is held at the limit and the bus type is
switched to a PQ bus.
Output QGkand k
Slack bus (Swing bus)
Input |Vk| and k
Output
PGkand QGk
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ECE4334Power flow equations
KCL says that the current injectionIi in an n-bus system must be
equal to the current flows from bus i into the network:
Since I = YbusV then;
We also know that the power injection is Si = ViIi*
Dr. C.Y. Evrenosoglu
(ith row and kth column of Ybus)
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ECE4334Power flow equations
OR Yikis in polar form
)
)
Dr. C.Y. Evrenosoglu
P fl i
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ECE4334Power flow equations
We would like to solve for the voltage magnitudes and angles
HOW?In some cases the power flow solution can be solved
analytically because only one of the multiple solutions is
reasonable from power system operation point of view.However usually it is impossible to solve the power flow
analytically in large networks. We use iterative methods:
I. Gauss-Seidel II.NewtonDr. C.Y. Evrenosoglu
ECE4334G i i
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ECE4334Gauss iteration
Assume that we are trying to find the solution of a functionf(x).
We have to rewrite the equation in an implicit form:x = h(x).
Then, we first make an initial guess ofx asx(0) and then iteratively
solvex(k+1) = h(x(k)) until we reach a value such that Example: Solve
For the first iteration, k=0 we guess thatx(0)=1 and we start iterating:
Dr. C.Y. Evrenosoglu
k x(k) k x(k)
0 1 5 2.61185
1 2 6 2.61612
2 2.41421 7 2.61744
3 2.55538 8 2.61785
4 2.59805 9 2.61798
Thanks to Dr. Tom Overbye, University of Illinois for the content
ECE4334G it ti
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ECE4334Gauss iteration
Dr. C.Y. Evrenosoglu
0 0.5 1 1.5 2 2.5 3 3.5 40
0.5
1
1.5
2
2.5
3
3.5
4
k x(v) k x(v)
0 1 5 2.61185
1 2 6 2.61612
2 2.41421 7 2.61744
3 2.55538 8 2.61785
4 2.59805 9 2.61798
x(0)
x(1)
x(2)
ECE4334G it ti
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ECE4334Gauss iteration
Stopping criteria is a key problemWhen to stop the
iteration?
With Gauss iteration we stop when
Ifx is a scalar; stopping criteria is clear; however, ifx is a
vector, that is the problem has multiple variables then we need
to use a norm.
Common norms are the two-norm (Euclid) and infinity norm;
Dr. C.Y. Evrenosoglu Thanks to Dr. Tom Overbye, University of Illinois for the content
ECE4334Power flow Gauss
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ECE4334Power flow Gauss
We have to put the equations in the proper form for Gauss;x=f(x)
Dr. C.Y. Evrenosoglu Thanks to Dr. Tom Overbye, University of Illinois for the content
ECE4334Two bus example
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ECE4334Two-bus example
A 100 MW, 50 MVAr load is connected to a generator through a
line withz = 0.02 +j0.06 pu and line charging of 5 MVAr on each
end. Also, there is a 25 MVAr capacitor at bus 2. If the generator
voltage is 1 pu, find V2.
Dr. C.Y. Evrenosoglu
V1 V2
SD2
ZLine
SG1Ycap
ECE4334Two bus example solution
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ECE4334Two-bus example solution
Sbase=100 MVA
Ratings for capacitors or line charging are given/calculated at |V|=1pu
and Sbase.
at both ends 25-MVAr shunt capacitor
Unknown complex voltage at bus 2.
Slack bus is bus 1 Slack bus will pick up the losses. SG1 will be calculated at the end once V2 is determined.
Dr. C.Y. Evrenosoglu
V1 V2
SD2ZLine
SG1
Ycap
ECE4334Two-bus example solution
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ECE4334Two-bus example solution
Determine Ybus
S2 = SG2 SD2 = 0 SD2 = 1 j0.5
Dr. C.Y. Evrenosoglu
V1 V2
SD2=1+j0.5y12
SG1
Ycap=j0.25
y/2=j0.05 y/2=j0.05
ECE4334Two-bus example solution
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C 33Two bus example solution
Dr. C.Y. Evrenosoglu
V1 V2
SD2
y12
SG1
Ycap
ECE4334Two-bus example solution
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Two bus example solution
Dr. C.Y. Evrenosoglu
V1 V2
SD2
y12
SG1
Ycap
Once the voltages are known; all other quantities can be
calculated such as generation at slack bus; line flows etc.
ECE4334Power flow
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Power flow
Study Example 6.9 in 5th edition.
Dr. C.Y. Evrenosoglu
ECE4334Gauss acceleration factor
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Gauss acceleration factor
The procedure for finding a solution tox = h(x)can be accelerated by using
If >>1 or 0 <
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k = 0, 1, 2,
In carrying out the computation we process the equations
from top to bottom.
You can observe that once you obtain , you can usethis updated value when you calculate . Thismodification is called Gauss-Seidel iteration
Dr. C.Y. Evrenosoglu
ECE4334Gauss-Seidel Multivariable
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k = 0, 1, 2,
Easier to program.
Faster than Gauss.
Acceleration factor is used.
Dr. C.Y. Evrenosoglu
ECE4334Power flow Multi-bus with Gauss-Seidel
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Load Bus Generator Bus
Given Find
Dr. C.Y. Evrenosoglu
Load Bus
ECE4334Power flow Multi-bus with Gauss-Seidel
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Dr. C.Y. Evrenosoglu
Generation Bus
is specified. At each iteration replace
by
but
keep the new angle
In each iteration check if . If hits one ofthe limits. It is assigned to the limit; the bus type is changed to PQ andthe is not specified and kept constant anymore.
ECE4334Example
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Find S1, Q2 and 2 using Gauss iteration. Ignore the Q generation
limits at bus 2.
Bus 1 is slack.S
1 will be calculated in the end. Bus 2 is PV bus: |V2| is known and PG2 is known P2 is known
QG2 is not known; once it is calculated then you can calculate
Dr. C.Y. Evrenosoglu
V1=1 V2=1SD2 = 1 +j0.5
Zline =j0.5
SG1 SG2 = 0.25 +jQG2
SD1
ECE4334Example
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Dr. C.Y. Evrenosoglu
V1=1 V2=1
SD2 = 1 +j0.5
Zline =j0.5
SG1 SG2 = 0.25 +jQG2
SD1
Bus 2: PV (Generator) Bus
ECE4334Example
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Dr. C.Y. Evrenosoglu
V1=1 V2=1
SD2 = 1 +j0.5
Zline =j0.5
SG1 SG2 = 0.25 +jQG2
SD1
Bus 2: PV (Generator) Bus
and
ECE4334Example
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Dr. C.Y. Evrenosoglu
V1=1 V2=1SD2 = 1 +j0.5
Zline =j0.5
SG1 SG2 = 0.25 +jQG2
SD1
ECE4334Power flow
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Study Example 6.10 in 5th edition.
Dr. C.Y. Evrenosoglu
ECE4334Newton-Raphson
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Quadratic convergence
Mathematically superior to Gauss-Seidel method
More efficient for large networks
Number of iterations required for solution is independent of
system size
The Newton-Raphson (NR) equations are cast in natural
power system form
Solving for voltage magnitude and angle, given real and reactive
power injections.
General form of the problem is to find anx such that f() = 0 Use 1st order Taylor expansion around a point;x0
Dr. C.Y. Evrenosoglu Thanks to Dr. Thomas Baldwin, FAMU/FSU for slide content
ECE4334NR scalar
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Dr. C.Y. Evrenosoglu
NR iterations f(xk) is known as the mismatch and we are trying to
drive it to zero.The stopping criteria is |f(xk)| <
ECE4334NR non-linear algebraic equations
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Dr. C.Y. Evrenosoglu
,
H.o.t. (higher order terms) are neglected.
NR iterations
ECE4334NR non-linear algebraic equations
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In power flow analysis flat start is used as initial guess; however,
to improve the initial guess a few steps of Gauss-Seidel iterationmay be used.
A disadvantage of NR is the need to update the Jacobian (J) every
iteration. Sometimes we can update less often and still get goodresults.
In practice we do not evaluate the inverse matrix. Taking inverses is
computationally expensive and not really needed. Instead we use;
(Gauss elimination or LU factorization can be used in the solution of A x = b.)
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ECE4334Example
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Remember
,
,
In DC all the quantities are real numbers
Dr. C.Y. Evrenosoglu
Ybus
ECE4334Example
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Dr. C.Y. Evrenosoglu
Bus 1 is slack busWe will concentrate on P2 and P3.
For NR formulation we need the equations in the form of
f(x)=0. The easiest way is to subtract the left sides from the
right sides and use the slack bus information as V1 = 1.
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ECE4334Example
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ECE4334Example
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ECE4334Example
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Dr. C.Y. Evrenosoglu
The accuracy is acceptable. Two iterations are enough! Our
objective to find and x for which f(x)=0 is met.
Now, find P1:
P1=1.511800 .Losses in the transmission system is
ECE4334NR Application to Power Flow
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Dr. C.Y. Evrenosoglu
=
=
=
Vx
V
V
V
NN
...,...22
( )
( )
( ) ( )
( ) ( )( )
( )
=
=
=
=
=
=
=
xQ
xPxf
xQQxQ
xPPxP
xQQ
xPP
QQQPPP
i
i
iii
iii
ii
ii
DiGii
DiGii
)(
Power system
state vector
ECE4334NR Application to Power Flow
N
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Dr. C.Y. Evrenosoglu
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]
( )
( )
( )( )
( )
( )
( )( )
( )
=
=
=
+=
=
=
xQQ
xQQ
xPP
xPP
xQ
xQ
xP
xP
xf
BGVVxQ
BGVVxP
NN
NN
N
N
kiikkiikk
N
k
ii
kiikkiikk
N
kii
...
...
...
...
cossin
sincos
22
22
2
2
1
1
ECE4334NR Application to Power Flow
( )[ ] ( )+ kkkk 11
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Dr. C.Y. Evrenosoglu
( )[ ] ( )
( )[ ] ( )( )[ ] ( )
[ ] ( )( )
( )( )
=
=
=
==
=
+
+
k
ii
kii
k
k
k
k
kkk
kkkkk
kkkk
xQQ
xPP
xQ
xP
VJ
xfxxJ
xfxJxxx
xfxJxx
11
11
The right side represent the mismatch between the
specified values ofP and Q and the corresponding values
obtained with the trial value . As the iteration proceeds,we expect these mismatched terms go to zero.
We solve for and find the new . Wecan update the mismatch vector and the Jacobian matrix andcontinue iterating.
ECE4334NR Application to Power Flow
( ) kkkk PJJ ( ) ( ) ( )[ ]kiikkiikkN
iisinBcosGVVxP +=
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Dr. C.Y. Evrenosoglu
( )( )
=
k
k
kkk
kk
xQ
xP
VJJ
JJ
2221
1211
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]ijijijijij
iij22
ijijijijji
j
iij21
ijijijiji
j
iij12
ijijijijji
j
iij11
cosBsinGVV
xQJ
sinBcosGVVxQ
J
sinBcosGVV
xPJ
cosBsinGVVxPJ
=
=
+=
=
+=
=
=
=
i j
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]kiikkiikkN
1k
ii
kiikkiikk
1k
ii
cosBsinGVVxQ
sinBcosGVVxP
=
+
=
=
Partition theJacobian into block sub-matrices.
ith andjth bus(ith row andjth column)
ECE4334NR Application to Power Flow
( ) kkkk xPJJ ( ) ( ) ( )[ ]kiikkiikkN
iisinBcosGVVxP +=
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Dr. C.Y. Evrenosoglu
( )( )
( ) ( )
( )( )
( ) ( )iii
i
i
i
iii22
2
iiii
i
iii21
iii
i
i
i
iii12
2
iiii
i
iii11
VBV
xQ
V
xQJ
VGxPxQ
J
VGVxP
VxPJ
VBxQxP
J
=
=
=
=
+=
=
=
=
i = j
( )( )
=
kkkk
xQ
xP
VJJ
JJ
2221
1211( ) ( ) ( )[ ]
( ) ( ) ( )[ ]kiikkiikkN
1k
ii
kiikkiikk
1k
ii
cosBsinGVVxQ
sinBcosGVVxP
=
+
=
=
ECE4334NR Application to Power Flow
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Dr. C.Y. Evrenosoglu
PVbuses As long as the reactive power generated at the bus is
within the reactive power limit specified at the bus, we do
not solve for the |Vi| since |Vi| is specified.
This reduces the dimensions of the problem. We exclude the voltage magnitudes of the PV buses from
the state vector, corresponding entries in theJacobian
matrix and the mismatch vector. If during an iteration a PV bus violates the reactive power
limit at the bus, then the reactive power is held at the
limit, and the bus is treated as PQ bus. The voltage
magnitude has to be reintroduced to the state vector andthe corresponding entries have to be reintroduced to theJ
and the mismatch vector.
ECE4334NR Application to Power Flow
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Dr. C.Y. Evrenosoglu
( ) ( ) ( )[ ]
( ) ( ) ( )[ ] ( )ijshuntijijiijjiijjiij
iijjiijjiijjiij
bBVBGVVxQ
VGBGVVxP
,
2
2
cossin
sincos
+=
+=
Once the voltage magnitudes and angles at each bus isknown, the slack bus power injections and the line flows
can be calculated. For line flows the following equations
can be used:
Line losses can be easily calculated as follows:
ECE4334Example
Find 2, |V2|, 3, SG1 and QG2. All the transmission line impedances
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Find 2, |V2|, 3, SG1 and QG2. All the transmission line impedances
are same andj0.1 and the shunt admittances are same andj0.01.Ignore the generator reactive limits at bus 2.
Dr. C.Y. Evrenosoglu
Bus 1: Slack Bus 2: PV bus, Bus 3: PQ bus,
ECE4334Example
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Dr. C.Y. Evrenosoglu
unknowns knowns
ECE4334Example
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Dr. C.Y. Evrenosoglu
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]kiikkiikkN
k
ii
kiikkiikk
N
k
ii
BGVVxQ
BGVVxP
=
+=
=
=
cossin
sincos
1
1
ECE4334Example
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Dr. C.Y. Evrenosoglu
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]kiikkiikkN
k
ii
kiikkiikk
N
k
ii
BGVVxQ
BGVVxP
=
+=
=
=
cossin
sincos
1
1
ECE4334Example
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Dr. C.Y. Evrenosoglu
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]kiikkiikkN
k
ii
kiikkiikk
N
k
ii
BGVVxQ
BGVVxP
=
+=
=
=
cossin
sincos
1
1
ECE4334Example
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Dr. C.Y. Evrenosoglu
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]kiikkiikkN
kii
kiikkiikk
N
k
ii
BGVVxQ
BGVVxP
=
+=
=
=
cossin
sincos
1
1
ECE4334Example
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Dr. C.Y. Evrenosoglu
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]kiikkiikkN
kii
kiikkiikk
N
k
ii
BGVVxQ
BGVVxP
=
+=
=
=
cossin
sincos
1
1
We are ready to start iterating.
Initial guess
Flat start For angles: For voltage magnitude:
ECE4334Example
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( ) ( ) ( )[ ]
( ) ( ) ( )[ ]kiikkiikkN
kii
kiikkiikk
N
k
ii
BGVVxQ
BGVVxP
=
+=
=
=
cossin
sincos
1
1
Big mismatch!
Dr. C.Y. Evrenosoglu
ECE4334Example
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( ) ( ) ( )[ ]
( ) ( ) ( )[ ]kiikkiikkN
kii
kiikkiikk
N
k
ii
BGVVxQ
BGVVxP
=
+=
=
=
cossin
sincos
1
1
Dr. C.Y. Evrenosoglu
After 1 iteration:
Better mismatch!
ECE4334Example
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( ) ( ) ( )[ ]
( ) ( ) ( )[ ]kiikkiikkN
kii
kiikkiikk
N
k
ii
BGVVxQ
BGVVxP
=
+=
=
=
cossin
sincos
1
1
Dr. C.Y. Evrenosoglu
After 2 iterations:
Sufficient mismatch!
STOP
ECE4334Example
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( ) ( ) ( )[ ]
( ) ( ) ( )[ ]kiikkiikkN
kii
kiikkiikk
N
k
ii
BGVVxQ
BGVVxP
=
+=
=
=
cossin
sincos
1
1
Dr. C.Y. Evrenosoglu
Now; calculate SG1 and QG2
ECE4334NR in power flow analysis
Advantages
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fast convergence as long as initial guess is close to
solution
large region of convergence
Disadvantages
each iteration takes much longer than a Gauss-Seidel
iteration more complicated to code, particularly when
implementing sparse matrix algorithms
Newton-Raphson algorithm is very common inpower flow analysis
Dr. C.Y. Evrenosoglu Thanks to Dr. Tom Overbye, University of Illinois for the content
ECE4334Power flow control
A major problem with power system operation is
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the limited capacity of the transmission system
lines/transformers have limits (usually thermal)
no direct way of controlling flow down atransmission line (e.g., there are no valves to close to
limit flow)
open transmission system access associated withindustry restructuring is stressing the system in new
ways
We need to indirectly control transmission lineflow by changing the generator outputs
Dr. C.Y. Evrenosoglu Thanks to Dr. Tom Overbye, University of Illinois for the content
ECE4334Real-sized Power Flow Cases
Real power flow studies are usually done with
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Dr. C.Y. Evrenosoglu
p y
cases with many thousands of buses
Buses are usually group in to various balancing
authority areas, with each area doing its owninterchange control
Cases also model a variety of different automatic
control devices, such as generator reactive powerlimits, load tap changing transformers, phase
shifting transformers, switched capacitors,
HVDC transmission lines, and (potentially)FACTS devices
Thanks to Dr. Tom Overbye, University of Illinois for the content
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ECE4334Power flow simulation before
One way to determine the impact of a generator change
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Dr. C.Y. Evrenosoglu
is to compare a before/after power flow.
For example below is a three bus case with an overload
Z for all lines = j0.1
One Two
200 MW
100 MVR200.0 MW
71.0 MVR
Three 1.000 pu
0 MW
64 MVR
131.9 MW
68.1 MW 68.1 MW
124%
Thanks to Dr. Tom Overbye, University of Illinois for the content
ECE4334Power flow simulation after
Increasing the generation at bus 3 by 95 MW (and hence
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Dr. C.Y. Evrenosoglu
Z for all lines = j0.1Limit for all lines = 150 MVA
One Two
200 MW
100 MVR105.0 MW
64.3 MVR
Three1.000 pu
95 MW
64 MVR
101.6 MW
3.4 MW 98.4 MW
92%
100%
decreasing it at bus 1 by a corresponding amount), resultsin a 31.3 drop in the MW flow on the line from bus 1 to 2.
Thanks to Dr. Tom Overbye, University of Illinois for the content
ECE4334More
There are different types of power flow under certain
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assumptions based on NR: Decoupled Power Flow
Fast Decoupled Power Flow
DC Power Plow Sensitivities are calculated by using DC power flow
Contingency analysis
Dr. C.Y. Evrenosoglu
ECE4334Homework 6.1 (due Thursday, 10/27)
Find one of the roots of the equation
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by using Gauss iteration. You can assume the initial guess
forx(0) = 2. Observe the number of iterations.
Upload your file to the assignment area.
Name your file as ECE4334_HW6_1_lastname.m
Dr. C.Y. Evrenosoglu
ECE4334Homework 6.2 (due Tuesday, 11/1)
A. Repeat Homework 6.1 by using an acceleration factor of
1 25
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1.25.
B. Repeat 6.1 by using an acceleration factor of 2.
C. Repeat 6.1 by using an acceleration factor of 0.1.
Upload your file(s) to the assignment area.
Name your file as ECE4334_HW6_2_lastname.m
Dr. C.Y. Evrenosoglu
ECE4334Homework 6.3 (due Tuesday, 11/1)
Solve the class (Slide 46) example using Gauss-Seidel and an
l ti f t
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acceleration factor.Find S1, Q2 and 2. Ignore the Q generation limits at bus 2.
Dr. C.Y. Evrenosoglu
V1=1 V2=1SD2 = 1 +j0.5
Zline =j0.5
SG1 SG2 = 0.25 +jQG2
SD1
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ECE4334Homework 7.2 (due Tuesday 11/8)
In the following network;
a) Find V2 exactly (take the larger of two possible values) Hint: Use
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a) Find V2 exactly (take the larger of two possible values)Hint: Usepower circle
b) Find V2 by Gauss iteration and use flat start for initial condition. If
you use hand calculation; stop after one iteration. If you use
MATLAB; please provide your code (print-out).
c) Find S1.
Dr. C.Y. Evrenosoglu
V1= V2SD2= 0.3 +j1.0
Zline=j0.4
SG1jQG2 = j1.1
SD1
ECE4334Homework 7.3 (due Tuesday 11/8)
In the following network SD1=1.0, SD2=1.0 j0.8 and SD3 = 1.0+j0.6.
Zline = j0 4 for all lines and line charging susceptances are neglected
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Zline =j0.4 for all lines and line charging susceptances are neglected. PG2 = 0.8 and |V2| = 1.0
Bus 1 is slack.
Use Gauss-Seidel to find V2 and V3. Use flat start for iterations. If you do hand calculations do one iteration only. If you use
MATLAB please provide your code (print-out).
Dr. C.Y. Evrenosoglu
1 2
3
ECE4334Homework 7.4 (due Tuesday 11/8)
For the system on Slide 69, assume that PG2 = 0.3, |V2|=0.95 and
SD3=0 5+j0 2 Use flat start for the iterations Carry out one NR
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SD3=0.5+j0.2. Use flat start for the iterations. Carry out one NRiteration and find , and |V3|1. Evaluate the mismatch vectorafter single iteration.
Dr. C.Y. Evrenosoglu