lecture4 power flow analysis

8/13/2019 Lecture4 Power Flow Analysis

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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


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


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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Rearrange the KCL equations

Thanks to Dr. Thomas Baldwin, FAMU/FSU for slide content


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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


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


Ip a*

When a is real (LTC xfmr)

?


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ECE4334Example LTC


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.


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


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


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


27/93

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*


(ith row and kth column of Ybus)


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OR Yikis in polar form

)

)


P fl i


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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:


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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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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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;


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)


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.


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.


V1 V2

SD2ZLine

SG1

Ycap



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Determine Ybus

S2 = SG2 SD2 = 0 SD2 = 1 j0.5


V1 V2

SD2=1+j0.5y12

SG1

Ycap=j0.25

y/2=j0.05 y/2=j0.05



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C 33Two bus example solution


V1 V2

SD2

y12

SG1

Ycap



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Two bus example solution


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.


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


ECE4334Gauss-Seidel Multivariable


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k = 0, 1, 2,

Easier to program.

Faster than Gauss.

Acceleration factor is used.


ECE4334Power flow Multi-bus with Gauss-Seidel


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Load Bus Generator Bus

Given Find


Load Bus

ECE4334Power flow Multi-bus with Gauss-Seidel


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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


V1=1 V2=1SD2 = 1 +j0.5

Zline =j0.5

SG1 SG2 = 0.25 +jQG2

SD1

ECE4334Example


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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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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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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.


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


ECE4334NR scalar


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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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,

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


56/93

Remember

,

,

In DC all the quantities are real numbers


Ybus

ECE4334Example


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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


59/93


ECE4334Example


60/93


ECE4334Example


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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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=

=

=

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


N


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( ) ( ) ( )[ ]

( ) ( ) ( )[ ]

( )

( )

( )( )

( )

( )

( )( )

( )

=

=

=

+=

=

=

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


( )[ ] ( )+ kkkk 11


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( )[ ] ( )

( )[ ] ( )( )[ ] ( )

[ ] ( )( )

( )( )

=

=

=

==

=

+

+

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.


( ) kkkk PJJ ( ) ( ) ( )[ ]kiikkiikkN

iisinBcosGVVxP +=


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( )( )

=

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)


( ) kkkk xPJJ ( ) ( ) ( )[ ]kiikkiikkN

iisinBcosGVVxP +=


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( )( )

( ) ( )

( )( )

( ) ( )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( ) ( ) ( )[ ]


1k

ii

kiikkiikk

1k

ii

cosBsinGVVxQ

sinBcosGVVxP

=

+

=

=



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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.



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( ) ( ) ( )[ ]

( ) ( ) ( )[ ] ( )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.


Bus 1: Slack Bus 2: PV bus, Bus 3: PQ bus,

ECE4334Example


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unknowns knowns

ECE4334Example


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( ) ( ) ( )[ ]


k

ii

kiikkiikk

N

k

ii

BGVVxQ

BGVVxP

=

+=

=

=

cossin

sincos

1

1

ECE4334Example


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( ) ( ) ( )[ ]


k

ii

kiikkiikk

N

k

ii

BGVVxQ

BGVVxP

=

+=

=

=

cossin

sincos

1

1

ECE4334Example


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( ) ( ) ( )[ ]


k

ii

kiikkiikk

N

k

ii

BGVVxQ

BGVVxP

=

+=

=

=

cossin

sincos

1

1

ECE4334Example


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( ) ( ) ( )[ ]


kii

kiikkiikk

N

k

ii

BGVVxQ

BGVVxP

=

+=

=

=

cossin

sincos

1

1

ECE4334Example


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( ) ( ) ( )[ ]


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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( ) ( ) ( )[ ]


kii

kiikkiikk

N

k

ii

BGVVxQ

BGVVxP

=

+=

=

=

cossin

sincos

1

1

Big mismatch!


ECE4334Example


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( ) ( ) ( )[ ]


kii

kiikkiikk

N

k

ii

BGVVxQ

BGVVxP

=

+=

=

=

cossin

sincos

1

1


After 1 iteration:

Better mismatch!

ECE4334Example


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( ) ( ) ( )[ ]


kii

kiikkiikk

N

k

ii

BGVVxQ

BGVVxP

=

+=

=

=

cossin

sincos

1

1


After 2 iterations:

Sufficient mismatch!

STOP

ECE4334Example


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( ) ( ) ( )[ ]


kii

kiikkiikk

N

k

ii

BGVVxQ

BGVVxP

=

+=

=

=

cossin

sincos

1

1


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


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


ECE4334Real-sized Power Flow Cases

Real power flow studies are usually done with


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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



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ECE4334Power flow simulation before

One way to determine the impact of a generator change


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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%


ECE4334Power flow simulation after

Increasing the generation at bus 3 by 95 MW (and hence


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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.


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


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


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.

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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.


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.


V1= V2SD2= 0.3 +j1.0

Zline=j0.4

SG1jQG2 = j1.1

SD1


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).


1 2

3


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.


lecture4 power flow analysis

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