nodal the voltages for a given load and generation schedule
known real (P) and reactive (Q) power injections
known real (P) power injection and the voltage magnitude (V)
known voltage magnitude (V) and voltage angle ()
must have one generator as the slack bus
takes up the power slack due to losses in the network
slack bus (swing bus)
generator bus
load bus
Types of network buses
The utility wants to know the voltage profile
Power Systems I
Power Flow Solution
n
j =1
*
ii
ji
Pi jQi
Ii =
Vi*
ji
n
n
Pi jQi
= Vi yij yijV j
*
Vi
j =1
j =0
Pi + jQi = V I
Power Law
j =0
= Vi yij yijV j
n
= ( yi 0 + yi1 + yi 2 + + yin )Vi yi1V1 yi 2V2 yinVn
I i = yi 0Vi + yi1 (Vi V1 ) + yi 2 (Vi V2 ) + + yin (Vi Vn )
KCL
Power Systems I
Power Flow Equations
find an iterative improvement of x[k], that is: x[k+1] = g( x[k] )
a solution is reached when the difference between two iterations is
less than a specified accuracy: x[k+1] - x[k]
make an an initial estimate of the variable x: x[0] = initial value
take a function and rearrange it into the form x = g(x)
{there are several possible arrangements}
x[k+1] = x[k] + ( g( x[k] ) - x[k] )
can improve the rate of convergence: > 1
modified step: the improvement is found as
acceleration factors
method of successive displacements
iterative steps:
A non-linear algebraic equation solver
Power Systems I
Gauss-Seidel Method
4
x = 1 x3 + 6 x2 + 9 = g ( x)
9
9
9 x = x3 + 6x2 + 4
Step 1. Cast the equation into the g(x) form.
Find the root of the equation: f(x) = x3 - 6x2 + 9x - 4 = 0
Power Systems I
Gauss-Seidel Example
x[8] = 4.0000
x[ 7 ] = 3.9988
x[ 6 ] = 3.9568
x[5] = 3.7398
x[ 4 ] = 3.3376
6
4
x[3] = g ( x[ 2 ] = 2.5173) = 1 (2.5173)3 + 9 (2.5173) 2 + 9 = 2.8966
9
6
4
x[ 2 ] = g ( x[1] = 2.2222) = 1 (2.2222)3 + 9 (2.2222) 2 + 9 = 2.5173
9
6
4
x[1] = g ( x[ 0 ] = 2) = 1 (2)3 + 9 (2) 2 + 9 = 2.2222
9
Step 2. Starting with an initial guess of x[0] = 2, several iterations
are performed.
Power Systems I
Gauss-Seidel Example
0
Power Systems I
0
0. 5
1
1. 5
2
2. 5
3
3. 5
4
4. 5
0. 5
1
Initial Value
1. 5
Iterations
3
2
x
2. 5
x
3
3. 5
x = g(x)
Solution Points
2
g(x) =-1/9x +6/9x +4/9
Matlab Results
Gauss-Seidel Example
4
4. 5
x[ 2 ] = 2.2778 + 1.25 [2.5902 2.2778] = 2.6683
6
4
g (2.2778) = 1 (2.2778) 3 + 9 (2.2778) 2 + 9 = 2.5902
9
x[1] = 2 + 1.25 [2.2222 2] = 2.2778
4
g (2) = 1 (2) 3 + 6 (2) 2 + 9 = 2.2222
9
9
x[ 0 ] = 2
Starting with an initial guess of x[0] = 2.
Find the root of the equation: f(x) = x3 - 6x2 + 9x - 4 = 0 with
an acceleration factor of 1.25
Power Systems I
Gauss-Seidel Example
x[8] = 4.0005
x[ 7 ] = 3.9978
x[ 6 ] = 4.0084
x[5] = 3.7238
x[ 4 ] = 3.1831
x[3] = 3.0801
Additional iterations
Power Systems I
Gauss-Seidel Example
0
Power Systems I
0
0. 5
1
1. 5
2
2. 5
3
3. 5
4
4. 5
0. 5
1
1. 5
2
x
2. 5
x
3
3. 5
x = g(x)
Solution Points
2
with acceleration factor: 1.25
Initial Value
Iterations
3
g(x) = -1/9x +6/9x +4/9
Matlab Results
Gauss-Seidel Example
4
4. 5
xn = cn + g n ( x1 , x2 , , xn )
x2 = c2 + g 2 ( x1 , x2 , , xn )
x1 = c1 + g1 ( x1 , x2 , , xn )
Rearrange each equation for one of the variables
f n x1 ( , x2 , , xn ) = cn
f 2 ( x1 , x2 , , xn ) = c2
f1 ( x1 , x2 , , xn ) = c1
Consider a system of n equations
Power Systems I
Gauss-Seidel for a System of Equations
0
2
0
n
k +1]
[
[
, x2k +1] , , xnk +1]
)
in the Gauss-Seidel method, the updated values of the variables
calculated in the preceding equations are used immediately in
the solution of the subsequent equations
1
(x[
find the results in a new approximate solution
0
1
(x[ ], x[ ],, x[ ] )
assume an approximate solution for the independent variables
s t e ps
Power Systems I
Gauss-Seidel for a System of Equations
Power Systems I
j =0
y
n
ij
Pi jQi n
+ yijV j
*
Vi
j =1
j i Vi[ k +1] =
The Gauss-Siedel form
ji
Pi jQi
Ii =
Vi*
n
n
Pi jQi
= Vi yij yijV j
*
Vi
j =1
j =0
Pi + jQi = V I
*
ii
T h e e q u a t io n
Vi =
The Power Flow Equation
j =0
y
n
ij
Pi jQi n
+ yijV j[ k ]
Vi*[ k ]
j =1
ji
ji
ji
for generation the powers are positive
for loads the powers are negative
the scheduled power is the sum of the generation and load powers
the real and reactive powers are scheduled for the load buses
that is, they remain fixed
the currents and powers are expressed as going into the bus
n
*[k ] [k ] n
[k +1]
[k ]
Qi = Vi Vi yij yij V j
j =0
j =1
n
*[k ] [k ] n
[k +1]
[k ]
Pi
= Vi Vi yij yij V j
j =0
j =1
Rewriting the power equation to find P and Q
Power Systems I
Power Injections
j =0
y
n
ij
ji
n
*[k ] [k ] n
[k +1]
[k ]
Qi = Vi Vi yij yij V j
j =0
j =1
n
*[k ] [k ] n
[k +1]
[k ]
= Vi Vi yij yij V j
Pi
j =0
j =1
Vi [k +1] =
Pi [sch ] jQi[sch ] n
+ yij V j[k ]
Vi*[k ]
j =1
ji
ji
The complete set of equations become:
Power Systems I
Solution by Gauss-Seidel
n
*[k ] [k ]
[k +1]
[k ]
= Vi Vi Yii + Yij V j
Pi
j =1
j i
n
Qi[k +1] = Vi*[k ] Vi [k ]Yii + Yij V j[k ]
j =1
j i
n
Pi [sch ] jQi[sch ]
Yij V j[k ]
Vi*[k ]
j =1, j i
Vi [k +1] =
Yii
Rewriting the equations in terms of the Y-Bus
Power Systems I
Solution by Gauss-Seidel
the voltage magnitude and angle must be estimated
in per unit, the nominal voltage magnitude is 1 pu
the angles are generally close together, so an initial value of 0
degrees is appropriate
Since both components (V & ) are specified for the slack bus,
there are 2(n - 1) equations which must be solved iteratively
For the load buses, the real and reactive powers are known:
scheduled
System characteristics
Power Systems I
Solution by Gauss-Seidel
e
[ k +1]
i
= Vi
[ sch ] 2
(
fi
)
[ k +1] 2
Vi = ei + j f i
the real power is scheduled
the reactive power is computed based on the estimated voltage
values
the voltage is computed by Gauss-Seidel, only the imaginary part is
kept
the complex voltage is found from the magnitude and the iterative
imaginary part
For the generator buses, the real power and voltage magnitude
are known
Power Systems I
Solution by Gauss-Seidel
V1 = 1.050
0.01 + j0.03
0.0125 + j0.025
0.02 + j0.04
0.452 pu
1.386 pu
1.102 pu
2.566 pu
Using the Gauss-Seidel method, determine the phasor
values of the voltage at the load buses 2 and 3, accurate
to 2 decimal places
Power Systems I
Example