Load flow PPT

Power flow calculations

Dirk Van HertemHakan Ergun

Priyanko Guha Thakurta

Research group ElectaDepartment of electrical engineering (ESAT)

K.U.Leuven, Belgium

DVH, HE, PGT (KUL/ESAT/ELECTA) Power flow calculations September 19, 2011 1 / 33

Introduction

Outline

1 IntroductionExample

2 System representation

3 The load flow problem

4 Solving the problemGauss-SeidelNewton-Raphson power flowSimplified Newton-Raphson

5 Software

6 State estimation


Introduction

Introduction: load or power flow

What are “power flow calculations”

Calculating the power flow (active and reactive) through all thelines in the power system

Calculating the voltages (amplitudes and angles) at every node(substation)

Determination of the static state of a given system

Knowing only:

Grid configuration and parameters (R and X )Power outputs of generator unitsLoads (active and reactive)Some voltages

“Load flow” and “power flow” are synonyms


Introduction


Why is load flow important?

Assessing if the power system is:

Within operational limitsSafe (N-1)

Basis for other (e.g. dynamic) calculations

Checking whether future situations are valid


Introduction


When is it used?System planning

System operations

State estimation

Dynamic simulations (basis, first calculation)

. . .


Introduction


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Introduction Example

ExampleFile: case6 wh.m, from the book “computational methods for electricpower systems”, M. Crow.

Bus databus type Pd Qd Vm Va Vmax Vmin

1 3 25 10 1.05 0 1.05 1.052 2 15 5 1.05 0 1.05 1.053 2 27.5 11 1 0 1.05 0.954 1 0 0 1 0 1.05 0.955 1 15 9 1 0 1.05 0.956 1 25 15 1 0 1.05 0.95

Generator dataGen Pg Pq Pmax

1 0 0 2002 50 0 150

1#1

4 3

256

#2#3

#4

#5

#6

#7

G

G

Branch dataline from to R X B

1 1 4 0.020 0.185 0.0092 1 6 0.031 0.259 0.0103 2 3 0.006 0.025 0.0004 2 5 0.071 0.320 0.0155 4 6 0.024 0.204 0.0106 3 4 0.075 0.067 0.0007 5 6 0.025 0.150 0.017


Introduction Example

ExampleFile: case6 wh.m, from the book “computational methods for electricpower systems”, M. Crow.

Bus databus type Pd Qd Vm Va Vmax Vmin

1 3 25 10 1.05 0 1.05 1.052 2 15 5 1.05 0 1.05 1.053 2 27.5 11 1 0 1.05 0.954 1 0 0 1 0 1.05 0.955 1 15 9 1 0 1.05 0.956 1 25 15 1 0 1.05 0.95

Generator dataGen Pg Pq Pmax

1 0 0 2002 50 0 150

1#1

4 3

256

#2#3

#4

#5

#6

#7

G

G

Branch dataline from to R X B

1 1 4 0.020 0.185 0.0092 1 6 0.031 0.259 0.0103 2 3 0.006 0.025 0.0004 2 5 0.071 0.320 0.0155 4 6 0.024 0.204 0.0106 3 4 0.075 0.067 0.0007 5 6 0.025 0.150 0.017

How would you solve this simple example by hand?


System representation

Outline





5 Software

6 State estimation




Most power systems are three phase AC

Normal power flow uses one phase equivalents

⇒ We only focus on this one today

One phase power flow only valid for balanced systems

Systems are usually given in per unit values

Lines can be represented by a π-equivalent



Per-unit calculations

Normalized representation of the four basic properties: voltage,current, impedance and complex power

Of these, two can be chosen independently

Normally rated phase voltage and one phase rated power aretaken as basis

Upu = UUbase

and Spu = SSbase

Ibase = Sbase

Ubase

Zbase = Ubase

Ibaseor Zbase =

U2base

Sbase

Logical values: for a 11.8 kV , 60 MVA machine,Ubasis = 11.8√

3kV and Sbasis = 60

3 MVA

For a 400 kV line, with 100 MVA: Ubase = 400/√

3,

Sbase = 100/3 ⇒ Rbase =(

400√3

)2

· 3100 = 1600Ω

Why are voltage and complex power chosen as fixed values?



Example per-unit

Generator example of before

Ubasis = 11.8√3kV and Sbasis = 60

3 MVA

Basis for current: Ibase = Sbase

Ubase= 60·

√3

11.8·3 = 2.9357 kA

Basis for impedance: Zbase =U2

base

Sbase=

(11.8√

3

)2

603

= 2.3207 Ω

Line connecting load: 0, 5 + 1 Ω = 0.21546 + 0.43091 pu

Afterwards, calculate using per-unit instead of original values



Per-unit and transformers

Z1 Z2

n1 : n2

U2U1

Voltage at both sides of the transformer is different → differentbasisOne of the major advantages of per-unit calculations because ofsimplification

Z ′2 = Z2 ·(

n21

n22

)= Z2 ·

(U2

1

U22

)Zp = Z1 + Z ′2

Zs = Zp ·(

U22

U21

)and Zs(pu) = Zs

Zbase (sec)

The per-unit impedance is the same on both sides of thetransformer⇒ can be replaced by one series impedance!




Z1n1 : n2

Z ′2

U2U1


Z ′2 = Z2 ·(

n21

n22

)= Z2 ·

(U2

1

U22

)

Zp = Z1 + Z ′2

Zs = Zp ·(

U22

U21

)and Zs(pu) = Zs

Zbase (sec)





Zp

n1 : n2

U1 U2


Z ′2 = Z2 ·(

n21

n22

)= Z2 ·

(U2

1

U22

)Zp = Z1 + Z ′2

Zs = Zp ·(

U22

U21

)and Zs(pu) = Zs

Zbase (sec)





Zs

n1 : n2

U2U1


Z ′2 = Z2 ·(

n21

n22

)= Z2 ·

(U2

1

U22

)Zp = Z1 + Z ′2

Zs = Zp ·(

U22

U21

)and Zs(pu) = Zs

Zbase (sec)




Representation of a transmission line

B2

X

B2

G2

G2

R

π-equivalent

Valid for lines up to 240 km

All values are normally small

Other equivalents exist and are sometimes used in practice

Normally, G can be neglected

With overhead lines, B can be neglected as well, for cables this isnot the case (see chapters on lines and cables)


The load flow problem

Outline





5 Software

6 State estimation



A bit of algebra: incidence matrixDirected Graph

Incidence matrix (A0)

nodesline # 1 -1 1 0 0

# lines (branches) × # nodes





nodesline # 1 -1 1 0 0line # 2 -1 0 1 0






nodesline # 1 -1 1 0 0line # 2 -1 0 1 0line # 3 0 -1 1 0






nodesline # 1 -1 1 0 0line # 2 -1 0 1 0line # 3 0 -1 1 0line # 4 0 -1 0 1






nodesline # 1 -1 1 0 0line # 2 -1 0 1 0line # 3 0 -1 1 0line # 4 0 -1 0 1line # 5 0 0 -1 1






nodesline # 1 -1 1 0 0line # 2 -1 0 1 0line # 3 0 -1 1 0line # 4 0 -1 0 1line # 5 0 0 -1 1

# lines (branches) × # nodesthe columns are dependent



Incidence matrix

Incidence matrix is written as A0

Meaning of the incidence matrix

Describes the directed graph

Produces differences

−1 1 0 0−1 0 1 00 −1 1 0−1 0 0 10 −1 0 1

·U1

U2

U3

U4

=

U2 − U1

U3 − U1

U3 − U2

U4 − U1

U4 − U2

(1)

Some symbols

Ii : Current injected atnode i

Iij : Current from node ito node j

Ui : Potential of node i

Eij : Potential difference(voltage) betweennodes i and j

Cij : Conductance of theline between nodes iand j



Incidence matrix





−1 1 0 0−1 0 1 00 −1 1 0−1 0 0 10 −1 0 1

·U1

U2

U3

U4

=

U2 − U1

U3 − U1

U3 − U2

U4 − U1

U4 − U2

(1)

Setting U4 = 0

Resulting matrix is the incidence matrix: A

Some symbols








Incidence matrix





−1 1 0−1 0 10 −1 1−1 0 00 −1 0

·U1

U2

U3

=

U2 − U1

U3 − U1

U3 − U2

− U1

− U2

(1)

Setting U4 = 0

Resulting matrix is the incidence matrix: A

Some symbols








Meaning of the incidence matrix1 Incidence matrix is A0 with one node removed

(grounded, reference)2 Ii is a nodal current injection, Iij is a branch flowI1I2

I3

=

−1 1 0−1 0 10 −1 10 −1 00 0 −1

T

·

I12

I13

I23

I24

I34

3 Ui is a nodal voltage/potential, Eij represents a

potential drop over line ij4 The relation between the voltage difference (e)

and line flows (f): Ohms law (take Cij theconductance of i to j)

5 Link on youtube


http://video.google.com/videoplay?docid=-2098741567961991321&q=incidence+matrix&ei=UGOOSK3qHoeUjQKAwvU8



(grounded, reference)2 Ii is a nodal current injection, Iij is a branch flow3 Ui is a nodal voltage/potential, Eij represents a

potential drop over line ijE12

E13

E23

E24

E34

=

−1 1 0−1 0 10 −1 10 −1 00 0 −1

·U1

U2

U3

4 The relation between the voltage difference (e)

and line flows (f): Ohms law (take Cij theconductance of i to j)

5 Link on youtube





(grounded, reference)2 Ii is a nodal current injection, Iij is a branch flow3 Ui is a nodal voltage/potential, Eij represents a

potential drop over line ij4 The relation between the voltage difference (e)

and line flows (f): Ohms law (take Cij theconductance of i to j)I12

I13

I23

I24

I34

=

C12 0 0 0 00 C13 0 0 00 0 C23 0 00 0 0 C24 00 0 0 0 C34

·E12

E13

E23

E24

E34

5 Link on youtube




Putting it together

AT · C · A · U = I

Ybus · U = I

Ybus is the bus admittance matrix

Representation of the entirenetwork by an admittance matrix, avector of nodal voltages and avector of nodal current injections

Yij = −yij (admittance betweennode i and j)

Yii =∑n

j yij (sum of the rest of therow + yii , the impedance to thereference)



The power system represented

The power system consists of:

Generators: delivering P and Q

Loads: consuming P and Q

Lines or branches: connecting generation and load

Wanted: Power flow of P and Q through these lines

Nodes or busbars: connections points in the power system

Wanted: Voltage amplitude (U) and voltage angle (θ) at eachnode



Mathematical statement of the problem

Uc∠θcUa∠θaUb∠θb

Ia

IbIbc

Iac

Icybc

yabyacc©

b©

a©

Iab

Neutral




Uc∠θcUa∠θaUb∠θb

Ia

IbIbc

Iac

Icybc

yabyacc©

b©

a©

Iab

Neutral

Voltage of node i to neutral is Ui∠θiAdmittance between i and j is yij

Current from i to j is IijThe injected current at i is Ii

Ia = Iab + IacIa = (Ua − Ub) · yab + (Ua − Uc) · yacIa = Ua · (yab + yac)− Ub · yab − Uc · yac




Last equation repeated: Ia = Ua · (yab + yac)− Ub · yab − Uc · yacWe take Yaa = yab + yac = yaa +

∑ni 6=a yai

yaa = ya0 = the parallel branches to node a (in this example,yaa = 0)

We take Yai = −yai⇒ as with Ybus , the bus admittance matrix

Which results in:Ia = Yaa · Ua + Yab · Ub + Yac · Uc (2)

Or for the entire system: IaIbIc

=

Yaa Yab Yac

Yba Ybb Ybc

Yca Ycb Ycc

· Ua

Ub

Uc

(3)

or I = Ybus ·U (4)

and Yij = Yji in symmetrical systems

(e.g. not with phase shifting transformers)




Resulting equations for a general system with n nodes

Ii =n∑

j=1

Yij · Uj ∀i ∈ N ≤ n (5)

S∗i = U∗i · Ii (6)

Above equations form the basis of power flow

There are 4 basic quantities for each node in power flowcalculations:

Voltage amplitude |U|Voltage angle θ between the voltage vector and the voltagereferenceActive power injection, withdrawal at a nodeReactive power injection, withdrawal at a node




Where do shunt elements fit? IaIbIc

=

Yaa Yab Yac

Yba Ybb Ybc

Yca Ycb Ycc

· Ua

Ub

Uc

(7)

Yii =n∑

j=1

yij (8)

=n∑

j=1j 6=i

−Yij + yii (9)

yii is the term to the node that has been “grounded”

In practice: shunt elements



Types of nodes

Three distinct types of nodes (important)

PV bus: A generating source is connected to the bus; the nodalvoltage is controlled at a certain magnitude U by injectingor absorbing reactive energy. The generated power PG isset at a specified value. θ and QG are computed. Constantvoltage operation is only possible when the generator iswithin its reactive energy generation limits.

PQ bus: P and Q are the control variables. This is the case whenthere is only a load connected to the bus or the generator isoutside its reactive power limits.

Slack (swing) bus: one of the generator busses is chosen to be the slackbus where the nodal voltage magnitude, Uslack , and phaseangle θslack are specified. This bus is needed to provide a“compensation” for the electrical losses that are not knownin advance. The bus forms a reference for the voltage angle.



Types of nodes





What is the mathematical meaning of the slack bus?



Types of nodes





Ii =∑n

j=1 Yij · Uj ∀i ∈ N ≤ n and i 6= nslack


Solving the problem

Outline





5 Software

6 State estimation


Solving the problem

The problemKnown data:

Active power injections in my system at generator nodesVoltages at generator nodesActive and reactive withdrawals (load) at PQ nodesSlack node voltage and angleImpedances (Ybus)

Unknowns:

Rest of P (slack), Q (slack and PV), voltage amplitude (PQnodes) and voltage angle (all but slack)

Equations

I = Y · U (10)

S∗ = U∗ · I (11)

S∗ = U∗ · Y · U (12)

P − Q = U∗ · Y · U (13)

Pi − Qi = U∗i ·n∑

j=1

Yij · Uj (14)


Solving the problem Gauss-Seidel

Gauss-Seidel

Algorithm

Ii =∑n

j=1 Yij · Uj and S∗i = U∗i · Ii give:

Ui =1

Yii·

S∗iU∗i−

n∑j=1j 6=i

Yij · Uj

(15)

This is solved bus by bus, and solutions of previous calculationsare filled in directly

U(i+1)i =

1

Yii·

S∗i

U∗(i)i

−i∑

j=1j 6=i

Yij · U(i+1)j −

n∑j=i+1j 6=i

Yij · U(i)j

(16)



Gauss-Seidel

1 For busbar 21, calculate I2 =S∗2U∗2

2 Calculate∑n

j=1j 6=2

Y2j · Uj

3 Subtract solution 2 from solution 1 and divide the result by Y22

to obtain a new value for U2

4 For busbar 3, calculate I3 =S∗3U∗3

5 Using the new value of U2 of step 3, calculate∑n

j=1j 6=3

Y3j · Uj

6 Subtract solution 5 from solution 4 and divide the result by Y33

to obtain a new value for U3

7 Repeat for all busses

8 Compare latest set of voltages with previous and check tolerance:U(i+1) −U(i) < ε? If not, go to step 1.

1when 1 is the reference busDVH, HE, PGT (KUL/ESAT/ELECTA) Power flow calculations September 19, 2011 21 / 33


Gauss-Seidel

Convergence and acceleration

The Gauss-Seidel method converges linearly (slow) with systemsize

Each iteration itself requires limited processing power

Often, the method is corrected with an acceleration factorU

(new)i(acc) = α · U(new)

i − (α− 1) · U(old)i (17)

= α · U(new)i − α · U(old)

i + U(old)i (18)

= U(old)i + α · (U(new)

i − U(old)i ) (19)

1 < α < 2

For large systems, often a value of 1.6 is chosen



Gauss-Seidel

Gauss-Seidel properties

A starting vector must be chosen

Often, the starting voltages are set to 1∠0 pucalled “Flat start”

If the voltages are calculated in block (and not replaced after onehas calculated the former one), we call the method the Jacobimethod.

The Jacobi method has a slower convergence

The Gauss-Seidel method is not often used anymore


Solving the problem Newton-Raphson power flow

Newton-Raphson method: theory

Solution of equation y = f (x)

Taylor series expansion:

y = f [x(0)]+

df

dx

∣∣∣∣x=x(0)

1!·[x−x(0)]+

d2f

dx2

∣∣∣∣x=x(0)

2!·[x−x(0)]2+. . . (20)




Solution of equation y = f (x)

Taylor series expansion:

y = f [x(0)] +

df

dx

∣∣∣∣x=x(0)

1!· [x − x(0)] (20)

Solving this for x :

x = x(0) +1

df

dx

∣∣∣∣x=x(0)

· [y − f (x(0))] (21)




(U , θ)0

(U , θ)∗

f (U , θ)




f (U , θ)0

(U , θ)0

(U , θ)∗

f (U , θ)




derivative

f (U , θ)0

(U , θ)0

(U , θ)∗

f (U , θ)




(U , θ)1

f (U , θ)1

derivative

f (U , θ)0

(U , θ)0

(U , θ)∗

f (U , θ)




(U , θ)2

f (U , θ)2

(U , θ)1

f (U , θ)1

derivative

f (U , θ)0

(U , θ)0

(U , θ)∗

f (U , θ)




Solution for multivariable nonlinear equations

y1 = f1(x1, x2, . . . , xn)

y2 = f2(x1, x2, . . . , xn)

...

y3 = fn(x1, x2, . . . , xn)





y1

y2

...yn

=

f1(x1(0), x2(0), . . . , xn(0))f2(x1(0), x2(0), . . . , xn(0))

...fn(x1(0), x2(0), . . . , xn(0))

+

∂f1

∂x1

∂f1

∂x2· · · ∂f1

∂xn∂f2

∂x1

∂f2

∂x2· · · ∂f2

∂xn...

.... . .

...∂fn∂x1

∂fn∂x2

· · · ∂fn∂xn

·

x1 − x1(0)x2 − x2(0)

...xn − xn(0)

(22)





Summarized, we can write the following:

y = f [x(0)] + J(0) · [x− x(0)] (23)

or solving for x:

x = x(0) + J(0)−1 · [y − f (x(0))] (24)

or in its recursive form:

xi+1 = xi + J−1i · [y − f (xi )] (25)



Newton-Raphson and load flow

Remember

Ii =∑n

j=1 Yij · Uj

S∗i = U∗i · Ii

Equivalents

Si = f (Ui ) is equivalent to y = f (x)

Ui+1 = Ui + Ji · [S− f (Ui )] (26)

S is here the specified complex power at any busbar

f (Ui ) is here the specified complex power at any busbar

∆Si = Ji ·Ui+1 (27)

You normally know the active and reactive power injections ineach node (load and generation)

You want to know the complex voltages at the nodes




S∗i = U∗i∑n

j=1 Yij · Uj and i = 1, . . . , n

Newton-Raphson in rectangular coordinates

Pi = Ui

n∑j=1

Uj · (Gij · cos(θi − θj) + Bij · sin(θi − θj)) (28)

Qi = −Ui

n∑j=1

Uj · (Gij · sin(θi − θj)− Bij · cos(θi − θj)) (29)

Newton-Raphson in polar coordinates

Pi = Ui

n∑j=1

Uj · Yij · cos(θi − θj − φij) (30)

Qi = −Ui

n∑j=1

Uj · Yij · sin(θi − θj − φij) (31)

note: Yij = Gij + · Bij = |Yij |∠φij




∆Pi = Pi, scheduled − Pi, calc ∀PQ and PV (32)

∆Qi = Qi, scheduled − Qi, calc ∀PQ (33)

Pi, scheduled and Qi, scheduled are known from the input data

Pi, calc and Qi, calc are obtained from the calculation inrectangular or polar coordinates




Putting it in one equation

Writing the power flow equations (both rectangular and polar) inthe form of equation (25):

[∆P∆Q

](i)

= −

∂P

∂θ

∂P

∂UU

∂Q

∂θ

∂Q

∂UU

(i)

︸︷︷︸J((U,θ)(i−1))

·[

∆θ∆UU

](i)

(34)

Or written in a simplified form:[∆P∆Q

](i)

= −[

H NM L

](i)

︸︷︷︸J((U,θ)(i−1))

·[

∆θ∆UU

](i)

(35)




Equation explained

∆θi = θ(i+1)i − θ(i)

i

∆Ui = U(i+1)i − U

(i)i

Voltages and angles (i + 1) are updated after each iteration andused for the following step

J is the Jacobian, and forms the derivative (tangent, gradient) ofthe power flow equations

∂P

∂UU and

∂Q

∂UU simplify the equations and results in fewer

computations

There are n − 1 equations for ∆P

There are n −#pv − 1 equations for ∆Q

The Jacobian is a square matrix(2 · n −#pv − 2)× (2 · n −#pv − 2)

The Jacobian is a sparse matrix (Special techniques can be usedwhen numerical calculations are performed)




Equation explained

∆θi = θ(i+1)i − θ(i)

i

∆Ui = U(i+1)i − U

(i)i

Voltages and angles (i + 1) are updated after each iteration andused for the following step

J is the Jacobian, and forms the derivative (tangent, gradient) ofthe power flow equations

∂P

∂UU and

∂Q

∂UU simplify the equations and results in fewer

computations

There are n − 1 equations for ∆P ⇒ Why?

There are n −#pv − 1 equations for ∆Q ⇒ Why?

The Jacobian is a square matrix(2 · n −#pv − 2)× (2 · n −#pv − 2)

The Jacobian is a sparse matrix (Special techniques can be usedwhen numerical calculations are performed)




Full equations (polar form): off-diagonal∂∆Pi

∂θj= Hij = −Uj · Ui · Yij · sin(θi − θj − φij)

∂∆Pi

∂UjUj = Nij = −Uj · Ui · Yij · cos(θi − θj − φij)

∂∆Qi

∂θj= Mij = Uj · Ui · Yij · cos(θi − θj − φij)

∂∆Qi

∂UjUj = Lij = −Uj · Ui · Yij · sin(θi − θj − φij)




Full equations (polar form): diagonal∂∆Pi

∂θi= Hii = Ui ·

n∑j=1

UjYij sin(θi − θj − φij) + U2i · Yii · sin(φij)

∂∆Pi

∂UiUi = Nii = −Ui ·

n∑j=1

UjYij cos(θi − θj − φij)− U2i · Yii · cos(φij)

∂∆Qi

∂θi= Mii = −Ui ·

n∑j=1

UjYij cos(θi − θj − φij) + U2i · Yii · cos(φij)

∂∆Qi

∂UiUi = Lii = −Ui ·

n∑j=1

UjYij sin(θi − θj − φij) + U2i · Yii · sin(φij)

Notice the symmetry

Notice that the off-diagonal elements are also in the diagonalelements

M = −N for off-diagonal elements




Figure: Sparsity of the Jacobian matrix shown




Numerical aspects

Iterative process until mismatch is below threshold(max(∆P(i); ∆Q(i)) = ε < εlimit)

Quadratic convergence

Major computational effort is calculating the inverse of theJacobian

The Jacobian is sparse, so special techniques can be used (lessstorage)

Ordering schemes can increase speed

Convergence is not guaranteed

A good starting point is needed

Flat start?Previous outcomeDC load flow as starting point

Simplifications exist




derivative

f (U , θ)0

(U , θ)0

(U , θ)∗

f (U , θ)


Solving the problem Simplified Newton-Raphson

Decoupled load flow

In a power system with mostly inductive lines, the power flowequations can be decoupled. (φij ≈ 90)

Active power is related to the angle between nodesReactive power is related to the voltage[

∆P∆Q

](i)

= −[

H NM L

](i)

︸︷︷︸J((U,θ)(i−1))

·[

∆θ∆UU

](i)

(36)

Advantages and disadvantages

+ Two small inverses instead of one big

+ Faster as only 2 · n3 calculations are needed, and not(2 · n)3 = 8 · n3

- The two subsystems may converge differently

- Convergence rate is slightly reduced

- Not often used nowadays



Decoupled load flow

In a power system with mostly inductive lines, the power flowequations can be decoupled. (φij ≈ 90)

Active power is related to the angle between nodesReactive power is related to the voltage[

∆P∆Q

](i)

= −[

H 00 L

](i)

︸︷︷︸J((U,θ)(i−1))

·[

∆θ∆UU

](i)

(37)

Advantages and disadvantages

+ Two small inverses instead of one big

+ Faster as only 2 · n3 calculations are needed, and not(2 · n)3 = 8 · n3

- The two subsystems may converge differently

- Convergence rate is slightly reduced

- Not often used nowadays



Decoupled load flow

Approximation

derivative

f (U , θ)0

(U , θ)0

(U , θ)∗

f (U , θ)



Fast decoupled load flow

In decoupled load flow, a new reduced Jacobian is determinedduring each iteration

Of each new Jacobian, the inverse needs to be calculated

Fast decoupled does not calculate a new Jacobian for eachiteration [

∆P(i)]

= [B′] ·[∆θ(i+1)

](38)[

∆Q(i)

U

]= [B′′] ·

[∆Ui+1

](39)

B′ and B′′ are real, sparse and constant matrices

Only series elements are involved (no shunts)

If the system has high R/X -ratio, large voltage angle deviationsor voltages which seriously differ from 1 pu, convergenceproblems can arise

Slower convergence (more iterations) but each iteration is muchfaster



Fast decoupled load flow

Approximation

derivative

f (U , θ)0

(U , θ)0

(U , θ)∗

f (U , θ)



DC load flow

If we consider the system to be lossless (Y = B)

And voltages to be around 1.0 pu (∆U = 0)

Voltage angles between busses are small (sin(θi − θj) ≈ (θi − θj)and cos(θi − θj) ≈ 1)

One equation of Newton-Raphson:

∆P = Ui

n∑j=1

Uj · (Gij · cos(θi − θj) + Bij · sin(θi − θj)) (40)

[∆P] = [B′] · [∆δ] (41)

B′ is real

Linear system

One calculation, no iterations

Easy for optimizations

Not correct (approximation)


Software

Outline





5 Software

6 State estimation


Software

Power flow software

Demo’s + try at home

Free available (open source)

Matpower (matlab based):http://www.pserc.cornell.edu/matpower/

PSAT (matlab based):http://www.power.uwaterloo.ca/~fmilano/psat.htm

InterPSS (Java based): http://www.interpss.org/

Professional software

PSS/EEurostagDigSilentPowerworld (demo athttp://www.powerworld.com/downloads.asp)


http://www.pserc.cornell.edu/matpower/

http://www.power.uwaterloo.ca/~fmilano/psat.htm

http://www.interpss.org/

http://www.powerworld.com/downloads.asp

State estimation

Outline





5 Software

6 State estimation


State estimation

State estimation

Known and unknown variables in the real power system

Lines, cables, transformers, location of generation and load ⇒ allknown and constant in time

Voltages, currents, actual generation and load (at that moment),position of power switches, tap-changer settings,. . .⇒ mostlyunknown or variable

Measurements:

P, Q: Generation and load, some linesVoltage: |U| every substation. θ only with PMU (phasormeasurement unit)Tap-changer settingsIncompleteMeasurement errors


State estimation

State estimation

State estimation: what?Monitoring or supplementing data for load flow

Many measurements in the system

Determining measurement errors, estimate and (statistically)analyze

If needed, certain measurements should be rejected

Least Squares approach

Another Youtube video: least squares

Has to be solved iteratively for power systems


http://www.youtube.com/watch?v=HG43z9xopU8

State estimation

State estimation

Weighted least-square method

(measurements z , with errors e, h(x) is the true model of the state x)

z =

z1

z2

z3

z4

=

h1(x1, x2, . . . , xn)h2(x1, x2, . . . , xn)h3(x1, x2, . . . , xn)h4(x1, x2, . . . , xn)

+

e1

e2

e3

e4

= h(x) + e (42)

With errors having a zero average, and each independent we get acovariance matrix R:

R =

σ2

1 0 · · · 00 σ2

2 · · · 0...

.... . .

...0 0 · · · σ2

n

(43)

R is the inverse of what we could call the weighting matrixR = inv(W)


State estimation

State estimation

Solving the state estimation

The expected values are:

x =

x1

x2

. . .xn

=

HT ·W ·H︸︷︷︸G

−1

·HT ·W · z = G−1 ·HT ·W · z

(44)

x = G−1 ·HT ·W · (H · x + e) (45)

x = G−1 · (HT ·W ·H)︸︷︷︸G

·x + G−1 ·HT ·W · e (46)

z = H · x (47)


State estimation

State estimation: simple example

Figure: Example network


State estimation


We want to know x1 and x2, which are voltages U1 and U2

Two amperemeters measuring z1 = 9.01 A and z2 = 3.02 A

U1 = 16.0233 V and U2 = 8.0367 V

Two voltmeters measuring z3 = 6.98 V and z4 = 5.01 V

U1 = 15.93 V and U2 = 8.05 V

The system equations can be written as:

z1

z2

z3

z4︸︷︷︸measurements

====

58 · x1 − 1

8 · x2

− 18 · x1 + 5

8 · x238 · x1 + 1

8 · x218 · x1 + 3

8 · x2︸︷︷︸true values from model

++++

e1

e2

e3

e4︸︷︷︸errors

(48)


State estimation


We want to know x1 and x2, which are voltages U1 and U2

Two amperemeters measuring z1 = 9.01 A and z2 = 3.02 A

U1 = 16.0233 V and U2 = 8.0367 V ⇒ Conflict

Two voltmeters measuring z3 = 6.98 V and z4 = 5.01 V

U1 = 15.93 V and U2 = 8.05 V ⇒ Conflict

The system equations can be written as:

z1

z2

z3

z4︸︷︷︸measurements

====

58 · x1 − 1

8 · x2

− 18 · x1 + 5

8 · x238 · x1 + 1

8 · x218 · x1 + 3

8 · x2︸︷︷︸true values from model

++++

e1

e2

e3

e4︸︷︷︸errors

(48)


State estimation


Calculating the expected values x

We take the following weighting matrix (1/sigma):W = diag([100, 100, 50, 50])

The most probable values for U1 and U2 are 16.00719 and8.02614 resp.

The expected error will be: e =

0.008770.00456−0.02596−0.00070


State estimation


Measurement 4 changes

z4 = 4.4 instead of z4 = 5.01

The best estimate for the voltages: U1 = 15.86807 andU2 = 7.75860

In that case, the expected error will be: e =

0.062280.154380.05964−0.49298

When the expected error is too high, measurements can/shouldbe disregarded

Statistical test are needed to determine when errors are “high”

The weight matrix also has a serious influence on the results


State estimation

State estimation for power flow calculations

State estimator calculates voltage magnitudes and relative phaseangles of the system buses

Redundancy in input data

With errors on all measurement data

Non-DC circuit ⇒ non-linear equations: h = h(x)

Iterative solutions (as in the Newton-Raphson method) areneeded

The principle is the same


State estimation

References

Power System Analysis; Grainger, John J. and Stevenson,William D., Jr.

Computational Mehods for Electric Power Systems; Crow,Mariesa

Power System Load Flow Analysis; Powell, Lynn


Load flow PPT

Documents

Transcript of Load flow PPT