7/16/2014
1
Chapter 6: Power Flow Analysis
1
2
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.
7/16/2014
2
3
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
4
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.
7/16/2014
3
5
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)
6
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)
7/16/2014
4
7
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
8
Applying KCL to Buses 1-4:
Rearranging the above equations:
7/16/2014
5
9
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
10
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
7/16/2014
6
11
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.
12
Example: Find the bus admittance matrix of the following simple power system with the given
admittance diagram
Impedance diagram of the power system
7/16/2014
7
13
Example:
E1 E2
14
% 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»
7/16/2014
8
15
Let’s execute the following commands in Matlab’s workplace
Put a semicolon ; after writing each X
16
7/16/2014
9
17
18
7/16/2014
10
19
20
7/16/2014
11
21
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
22
Example: Find the voltages of the buses of the following power system with the given impedance diagram
using Matlab function «ybus1.m»
7/16/2014
12
23
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
24
>> 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)
7/16/2014
13
25
>> 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
26
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
7/16/2014
14
27
>> 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
28
>> 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
7/16/2014
15
29
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
30
Gauss-Seidel Method:
7/16/2014
16
31
Solution:
initial estimate of x
32
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
7/16/2014
17
33
Newton-Raphson Method:
34
7/16/2014
18
35
Equations of NR algorithm
k: iteration number
36
Solution:
7/16/2014
19
37
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
38
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
7/16/2014
20
39
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.
40
N-dimensional Newton-Raphson Method
7/16/2014
21
41
Jacobian Matrix consisting of partial derivatives
42
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
7/16/2014
22
43
-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
44
>> 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
7/16/2014
23
45
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
46
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
7/16/2014
24
47
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
48
This equation is the non-linear power flow equation for Bus i
that should be solved using numerical techniques
Power Flow Equation
7/16/2014
25
49
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
50
If Pi and Qi are solved from the above equation:
Gauss-Seidel Power Flow Solution
7/16/2014
26
51
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)
52
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
7/16/2014
27
53
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
54
For Slack Bus Use:
The voltage magnitude and the
phase angle of the slack bus are
known
Gauss-Seidel Power Flow Solution
7/16/2014
28
55
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)
56
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)
7/16/2014
29
57
58
7/16/2014
30
59
60
7/16/2014
31
61
62
7/16/2014
32
63
64
7/16/2014
33
65
66
7/16/2014
34
67
68
7/16/2014
35
69
70
End of Chapter 6
Recommended