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Chapter 6: Power Flow Analysis

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Introduction to Power Flow Analysis

In a three phase ac power system, active and reactive power flows from the

generating stations to the loads through transmission line/network.

The flow of active and reactive powers on a transmission line of a power

system/network is called power flow or load flow.

Power flow analysis is very important in planning stages of power network to

add new transmission lines and new generation sides.

The results of power flow analysis is widely used by electrical power

engineers during the planning stages and operation of power systems.

In power flow studies the power system is represented by one-line (single-

line) diagram.

In power studies, the analyzed is assumed to be balanced and operating

under steady-state and normal conditions.

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To find bus voltage magnitude and bus voltage phase angle of each bus

in a given power system.

To find real and reactive power flows on each line of a given power

system.

To find real and reactive power losses of each line of a given power

system.

The solution of the power flow study is used to determine

under/overloaded transmission lines so that suitable line compensation

methods can be designed and applied.

For example to add a FACTS device in a power system, before studying

power flow analysis, it is impossible to define the type, parameters, and

the best location of the FACTS device which is considered for

compensation.

Objective of Power Flow Study

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Properties of Power Flow Study

The bus admittance matrix of the power system is required for power flow

study giving impedance/admittance information of each

line/generator/transformer.

Non-linear power flow equations for each bus should be described,

linearized, and solved by iterative methods.

Commonly used iterative methods are Gauss-Seidel, Newton-Raphson,

and fast decoupled methods.

For a power system having many buses and branches (lines) the

equations are generally put in matrix form for calculation efficiency.

For example in a 30-bus power system, there are 2*30=60 unknows to be

solved by power flow study.

Hand-made calculations are very hard to implement so computer software

is used for power flow (load flow) analysis.

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Some Examples to Power Flow Analysis Software

PSS/E (commercial)

DigSILENT (commercial)

EasyPower (commercial)

Eurostag (commercial)

ETAP (commercial)

PSASP (power system analysis software package)

Powerworld (commercial, demo is available)

Matpower (non-commercial, matlab based free program)

PSAT (non-commercial matlab based free program)

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Bus Admittance Matrix

Generator

Generator impedance

Bus (node)

Line reactancePU admittance

between

bus i & j

PU impedance

between

bus i & j

resistance

reactance

Impedance diagram of a simple power system

Bus-0 is ground node taken as

Reference (not shown)

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Admittance diagram of a simple power system

Transformations Made to Obtain Admittance Diagram: All line impedances are converted into line admittances

All voltage source+impedance are converted into a current source+parallel admittance

admittance

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Applying KCL to Buses 1-4:

Rearranging the above equations:

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We introduce the following admittances:

The node equations reduces to:

Y14=Y41=0 and Y24=Y42=0

Since there is no connection between Bus 1 and Bus 4

and there is no connection between Bus 2 and Bus 4

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For a general n-bus power system the following matrix equation can be written:

or

Injected

current vector

Positive if injected into bus

Negative if outgoing from bus

Bus admittance matrixBus voltage

vector

self-admittance

or driving-point admittancemutual or transfer admittance

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Since

The inverse of bus admittance matrix is called Bus Impedance Matrix (Zbus)

If injected current vector is known

voltage bus vector can be calculated

Notes on Bus Admittance Matrix (Ybus):

Zbus can be calculated only if Ybus is inversible (non-singular)

Ybus is a symmetrical matrix since Yij=Yji (there is only one connection configuration between buses i & j)

In real, many off-diagonal elements are zero, since each bus is connected to only a few nearby buses in a power

system

Ybus is a sparse matrix because of many zeros and there are efficient numerical methods to find its inverse

rather than calculating its inverse directly. Since for example for a 300-bus system to find Zbus we need to find the

inverse of a 300x300 matrix, which is not so efficient and slow even with a fast computer.

In real, Zbus which is required for short-circuit analysis can be obtained more efficiently and fast by applying

building algorithm without the need for matrix inversion.

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Example: Find the bus admittance matrix of the following simple power system with the given

admittance diagram

Impedance diagram of the power system

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Example:

E1 E2

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% Bus Admittance Matrix% Copyright (c) 1998-2010 by H. Saadat.

function[Ybus] = ybus1(zdata)nl=zdata(:,1); nr=zdata(:,2); R=zdata(:,3); X=zdata(:,4);nbr=length(zdata(:,1)); nbus = max(max(nl), max(nr));Z = R + j*X; %branch impedancey= ones(nbr,1)./Z; %branch admittanceYbus=zeros(nbus,nbus); % initialize Ybus to zerofor k = 1:nbr; % formation of the off diagonal elements

if nl(k) > 0 && nr(k) > 0Ybus(nl(k),nr(k)) = Ybus(nl(k),nr(k)) - y(k);Ybus(nr(k),nl(k)) = Ybus(nl(k),nr(k));end

endfor n = 1:nbus % formation of the diagonal elements

for k = 1:nbrif nl(k) == n || nr(k) == nYbus(n,n) = Ybus(n,n) + y(k);else, endend

end

The code of Matlab function «ybus1.m»

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Let’s execute the following commands in Matlab’s workplace

Put a semicolon ; after writing each X

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Notes on the Previous Solution Technique:

The solution of the equation Vbus = Zbus.Ibus by matrix inversion is very inefficient.

Actually, it is not necessary to obtain the inverse of Ybus. Instead, direct solution is obtained by optimally

ordered triangular factorization.

In Matlab, the solution of the linear equation AX=B can be obtained by using the matrix division operator

(X=A\B), which is based on the triangular factorization and Gaussian elimination.

This technique is superior in both execution time (two to three times faster) and numerical accuracy.

We can obtain the direct solution of the previous example by executing the following command in

Matlab.

Vbus = Y \ Ibus

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Example: Find the voltages of the buses of the following power system with the given impedance diagram

using Matlab function «ybus1.m»

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Solution:

First we write the following Matlab command to enter power system data:

>> z= [0 1 0 0.8;0 2 0 1.0;0 3 0 2.0;1 2 0 0.5; 1 3 0 0.2;2 3 0 0.4];

z =

0 1.0000 0 0.8000

0 2.0000 0 1.0000

0 3.0000 0 2.0000

1.0000 2.0000 0 0.5000

1.0000 3.0000 0 0.2000

2.0000 3.0000 0 0.4000

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>> Y = ybus1(z)

Y =

0.0000 - 8.2500i 0.0000 + 2.0000i 0.0000 + 5.0000i

0.0000 + 2.0000i 0.0000 - 5.5000i 0.0000 + 2.5000i

0.0000 + 5.0000i 0.0000 + 2.5000i 0.0000 - 8.0000i

Bus Admittance Matrix

of the power system

>> 1.28/(j*0.8)

ans =

0.0000 - 1.6000i

>> (1.2*cos(pi/6)+j*1.2*sin(pi/6))/j

ans =

0.6000 - 1.0392i

>> Ibus=[-j*1.6; 0.6-j*1.0392; 0]

Ibus =

0.0000 - 1.6000i

0.6000 - 1.0392i

0.0000 + 0.0000i

No injected current at Bus 3

(no Gen. is connected)

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>> Vbus=inv(Y)*Ibus

Vbus =

0.9791 + 0.1860i

0.9594 + 0.2676i

0.9118 + 0.1999i

or

>> Vbus = Y \ Ibus

Vbus =

0.9791 + 0.1860i

0.9594 + 0.2676i

0.9118 + 0.1999i

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Let’s extend the previous example to find other power system parameters:

P12

Q12

P12=((V1*V2)/X12)*sin(theta1-theta2)

Q12=(V1*V2*cos(theta1-theta2)-V2*V2)/X12

Real power flow from Bus 1 to Bus 2

Reactive power flow from Bus 1 to Bus 2

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>> Vbus1= 0.9791 + 0.1860i

Vbus1 =

0.9791 + 0.1860i

>> Vbus2= 0.9594 + 0.2676i

Vbus2 =

0.9594 + 0.2676i

>> Vbus1_mag=sqrt(0.9791^2+0.1860^2)

Vbus1_mag =

0.9966

>> Vbus1_ang=atan2(0.1860,0.9791)

Vbus1_ang =

0.1877

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>> Vbus2_mag=sqrt(0.9594^2+0.2676^2)

Vbus2_mag =

0.9960

>> Vbus2_ang=atan2(0.2676,0.9594)

Vbus2_ang =

0.2720

>> P12=((Vbus1_mag*Vbus2_mag)/(0.5))*sin(Vbus1_ang-Vbus2_ang)

P12 =

-0.1671

>> Q12=(Vbus1_mag*Vbus2_mag*cos(Vbus1_ang-Vbus2_ang)-Vbus2_mag^2)/0.5

Q12 =

-0.0059

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Solution of Non-Linear Algebraic Equations

The previous method is very practical if all voltage magnitudes of generators are known

Actually, only one bus voltage magnitude and phase angle is known to solve power flow

problem (slack bus)

In the previous method, the load information is not available, however should be

considered in an actual power flow problem.

Actually a power flow problem is a non-linear multi-variable problem that should be solved

by computer software iteratively.

The followings are the common methods used for iterative solution of non-linear power

flow equations:

Gauss-Seidel Method

Newton-Raphson Method (powerful and popular)

Quasi-Newton Method

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Gauss-Seidel Method:

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Solution:

initial estimate of x

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The Limitations of Gauss-Seidel Method:

This method needs many iterations to achieve the desired accuracy.

There is no guarantee for the convergence.

The solution is highly dependent on initial estimate of x. For example if initial estimate of x were

chosen as 6, the process would diverge.

For a power system having many buses, convergence test is difficult to apply and there is no general

method for convergence test.

The root of the polynomial after 9 iterations

0 1 2 3 4 50

0.5

1

1.5

2

2.5

3

3.5

4

4.5

x

g(x) =-1/9x3+6/9x

2+4/9

x

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Newton-Raphson Method:

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Equations of NR algorithm

k: iteration number

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Solution:

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The root of the polynomial after 6 iterations

0 1 2 3 4 5 6-10

0

10

20

30

40

50

x

f(x) = x3-6x

2+9x-4

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Type the following command in Matlab Workspace to examine the same example with different

initial estimates

>> chp6ex4

Enter the initial estimate -> 9

iter Dc J dx x

1 -320.0000 144.0000 -2.2222 6.7778

2 -92.7298 65.4815 -1.4161 5.3617

3 -25.9042 30.9022 -0.8383 4.5234

4 -6.4975 16.1025 -0.4035 4.1199

5 -1.1669 10.4817 -0.1113 4.0086

6 -0.0774 9.1029 -0.0085 4.0000

7 -0.0004 9.0006 -0.0000 4.0000

>> chp6ex4

Enter the initial estimate -> 21

iter Dc J dx x

1 1.0e+03 *

-6.8000 1.0800 -0.0063 0.0147

2 1.0e+03 *

-2.0101 0.4812 -0.0042 0.0105

3 -592.2208 215.0830 -2.7535 7.7726

4 -173.0465 96.9703 -1.7845 5.9881

5 -49.4669 44.7152 -1.1063 4.8818

6 -13.2884 21.9152 -0.6064 4.2755

7 -2.9557 12.5336 -0.2358 4.0397

8 -0.3665 9.4808 -0.0387 4.0010

9 -0.0091 9.0121 -0.0010 4.0000

10 -0.0000 9.0000 -0.0000 4.0000

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The Comments on NR Solution:

Newton-Raphson (NR) method converges considerably more rapidly than the Gauss-Seidel method.

Although the initial estimate of x=6 makes Gauss-Seidel method diverges, the same initial estimate

of x makes NR method converges.

NR method is generally more tolerant when choosing the initial estimate of x, even initial estimate

is not close enough to the root of the equation.

The limitation of NR method is that it requires the first order derivative of the function whose root is

under consideration.

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N-dimensional Newton-Raphson Method

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Jacobian Matrix consisting of partial derivatives

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The Comments on N-Dimensional NR Algorithm:

Elements of Jacobian matrix are the partial derivatives evaluated at (k)

It is assumed that Jacobian matrix has an inverse during each iteration.

N-dimensional NR algorithm reduces the non-linear equations to a N-dimensional linear equations

to be solved by iteration.

Actually the inversion of Jacobian matrix is inefficient and not necessary. Instead, a direct solution is

obtained by optimally ordered triangular factorization.

In Matlab, the solution of linear equations ΔC=JΔX is obtained by using the matrix division operator

which is based on the triangular factorization and Gaussian elimination.

For example,

ΔX = J \ ΔC

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-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

x

x2+y

2=4

ex+y=1

Solution:

(Jacobian matrix )

solution1

solution2

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>> chp6ex5

Enter initial estimates, col. vector [x1; x2] -> [1;1]

Iter DC Jacobian matrix Dx x

1 2.0000 2.0000 2.0000 -2.1640 -1.1640

-2.7183 2.7183 1.0000 3.1640 4.1640

2 -14.6933 -2.3279 8.3279 -2.8927 -4.0567

-3.4762 0.3122 1.0000 -2.5730 1.5910

3 -14.9879 -8.1134 3.1820 1.5979 -2.4588

-0.6083 0.0173 1.0000 -0.6360 0.9550

4 -2.9577 -4.9176 1.9101 0.5669 -1.8919

-0.0406 0.0855 1.0000 -0.0891 0.8660

5 -0.3293 -3.7838 1.7319 0.0742 -1.8177

-0.0168 0.1508 1.0000 -0.0279 0.8380

6 -0.0063 -3.6354 1.6761 0.0014 -1.8163

-0.0004 0.1624 1.0000 -0.0007 0.8374

7 -0.0000 -3.6325 1.6747 0.0000 -1.8163

-0.0000 0.1626 1.0000 -0.0000 0.8374

>> chp6ex5

Enter initial estimates, col. vector [x1; x2] -> [0.5;-1]

Iter DC Jacobian matrix Dx x

1 2.7500 1.0000 -2.0000 0.8034 1.3034

0.3513 1.6487 1.0000 -0.9733 -1.9733

2 -1.5928 2.6068 -3.9466 -0.2561 1.0473

-0.7085 3.6818 1.0000 0.2344 -1.7389

3 -0.1205 2.0946 -3.4778 -0.0422 1.0051

-0.1111 2.8499 1.0000 0.0092 -1.7296

4 -0.0019 2.0102 -3.4593 -0.0009 1.0042

-0.0025 2.7321 1.0000 0.0000 -1.7296

5 -0.0000 2.0083 -3.4593 -0.0000 1.0042

-0.0000 2.7296 1.0000 -0.0000 -1.7296

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Solution:

(Jacobian matrix )

iter =

6

DC =

1.0e-04 *

-0.3669

-0.1784

0.0957

J =

4.0000 -6.0000 8.0000

3.0000 8.0000 -3.0000

-3.0000 4.0000 1.0000

Dx =

1.0e-05 *

-0.5577

-0.1130

-0.2645

x =

2.0000

3.0000

4.0000

only the

results of

6. iteration

are shown

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Power Flow Solution

In a power flow study the power system buses are classified into three-types:

or known as PQ Bus

P,Q are known

V and angle of the bus

are calculated

or known as PV Bus

P, V are known

Q and angle of the bus

are calculated

V, angle of the bus are

known

P, Q are calculated

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Power Flow Equation

A typical bus of a power system

Transmission lines are represented by their equivalent pi models

All the quantities are converted into the pu values

Appliying KCL at Bus i

(Total sum of entering currents = Total sum of outgoing currents)

or

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This equation is the non-linear power flow equation for Bus i

that should be solved using numerical techniques

Power Flow Equation

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Gauss-Seidel Power Flow Solution

yij is the pu admittance of the branch (or line) ij

Pisch is the net injected real power at Bus i expressed in pu

Qisch is the net injected reactive power at Bus i expressed in pu

For a generator Bus, Pisch and Qisch are positive

For a load Bus, Pisch and Qisch are negative

k is the iteration number

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If Pi and Qi are solved from the above equation:

Gauss-Seidel Power Flow Solution

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Gauss-Seidel Power Flow Solution

Power flow equation is usually expressed in terms of the elements of bus admittance matrix Ybus.

Off-diagonal elements of Ybus are Yij = -yij

Diagonal elements of Ybus are Yii = sum(yij)

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Gauss-Seidel Power Flow Solution

Since voltage magnitude and phase angle of slack bus are known, there are 2(n-1) equations which must be

solved iteratively. Example if bus number is 4, there are 2(4-1) = 6 equations to be solved iteratively.

Under normal operating conditions, the voltage magnitude of the buses are in the neighbourhood of 1.0 pu.

Voltage magnitudes of the load buses are generally lower than the slack bus value.

Voltage magnitıdes of the generator buses are generally higher than the slack bus value.

Phase angle of the load buses are generally below the reference angle of the slack bus.

Phase angle of the generator buses are generally higher than the reference angle of the slack bus.

Thus for Gauss-Seidel method 1.0 + j0.0 is a satisfactory initial voltage estimate for the unknown bus voltage

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Gauss-Seidel Power Flow Solution

For PQ Buses Use:Pisch and Qisch are

known

For PV Buses

Use first:Pisch and magnitude

of bus voltage are

known

Since bus voltage magnitude is known to find the phase angle the following equation is used:

second

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For Slack Bus Use:

The voltage magnitude and the

phase angle of the slack bus are

known

Gauss-Seidel Power Flow Solution

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Gauss-Seidel Power Flow Solution

The rate of convergence speed can be increased with the following equation:

(α is the accelaration factor, generally chosen between 1.3 and 1.7)

Gauss-Seidel iteration continues until the following conditions are satisfied:

(ε is generally chosen between 1E-6 and 1E-5)

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Line Flows and Line Losses

After finding all bus voltages with the Gauss-Seidel method, line flows and line losses can be computed:

(Complex power flow from Bus i to j )

(Complex power flow from Bus j to i )

(The power loss in the line ij)

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End of Chapter 6