ECE 476
Power System Analysis
Lecture 12: Power Flow
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
overbye@illinois.edu
Announcements
• Read Chapter 6
• H6 is 6.19, 6.30, 6.31, 6.34, 6.38, 6.45. It does not
need to be turned in, but will be covered by an in-class
quiz on Oct 15.
1
Transmission Line
Corridors from the Air
Image Source: Jamie Padilla
Slack Bus
• In previous example we specified S2 and V1 and
then solved for S1 and V2.
• We can not arbitrarily specify S at all buses
because total generation must equal total load +
total losses
• We also need an angle reference bus.
• To solve these problems we define one bus as the
"slack" bus. This bus has a fixed voltage
magnitude and angle, and a varying real/reactive
power injection.
3
Stated Another Way
• From exam problem 4.c we had
Bus 2
Bus 1
j0.2
j0.1
j0.1
Bus 3
Ybus
10 
 15 5
 j  5 15 10 
 10 10 20 
• This Ybus is actually singular!
• So we cannot solve V   Ybus 1 I
• This means (as you might expect), we cannot
independently specify all the current injections I
4
Gauss with Many Bus Systems
With multiple bus systems we could calculate
new Vi ' s as follows:
Vi
( v 1)

n
1  S*i
(v )
 ( v )*   YikVk 


Yii  V
k

1,
k

i
 i

 hi (V1( v ) ,V2( v ) ,...,Vn( v ) )
But after we've determined Vi( v 1) we have a better
estimate of its voltage , so it makes sense to use this
new value. This approach is known as the
Gauss-Seidel iteration.
5
Gauss-Seidel Iteration
Immediately use the new voltage estimates:
V2( v 1)  h2 (V1,V2( v ) ,V3( v ) ,,Vn( v ) )
V3( v 1)  h2 (V1,V2( v 1) ,V3( v ) ,,Vn( v ) )
V4( v 1)  h2 (V1,V2( v 1) ,V3( v 1) ,V4( v ) ,Vn( v ) )
Vn( v 1)  h2 (V1,V2( v 1) ,V3( v 1) ,V4( v 1) ,Vn( v ) )
The Gauss-Seidel works better than the Gauss, and
is actually easier to implement. It is used instead
of Gauss.
6
Three Types of Power Flow Buses
• There are three main types of power flow buses
–
–
–
Load (PQ) at which P and Q are fixed; iteration solves for
voltage magnitude and angle.
Slack at which the voltage magnitude and angle are fixed;
iteration solves for P and Q injections
Generator (PV) at which P and |V| are fixed; iteration
solves for voltage angle and Q injection
• special coding is needed to include PV buses in the
Gauss-Seidel iteration (covered in book, but not in
slides since Gauss-Seidel is no longer commonly
used)
7
Accelerated G-S Convergence
Previously in the Gauss-Seidel method we were
calculating each value x as
x ( v 1)  h( x ( v ) )
To accelerate convergence we can rewrite this as
x ( v 1)  x ( v )  h( x ( v ) )  x ( v )
Now introduce acceleration parameter 
x ( v 1)  x ( v )   (h( x ( v ) )  x ( v ) )
With  = 1 this is identical to standard gauss-seidel.
Larger values of  may result in faster convergence.
8
Accelerated Convergence, cont’d
Consider the previous example: x - x  1  0
x ( v 1)  x ( v )   (1  x ( v )  x ( v ) )
Comparison of results with different values of 
k
 1
  1.2
  1.5   2
0
1
1
1
1
1
2
2.20
2.5
3
2
2.4142
2.5399
2.6217
2.464
3
2.5554
2.6045
2.6179
2.675
4
2.5981
2.6157
2.6180
2.596
5
2.6118
2.6176
2.6180
2.626
9
Gauss-Seidel
Advantages/Disadvantages
• Advantages
–
–
Each iteration is relatively fast (computational order is
proportional to number of branches + number of buses in the
system
Relatively easy to program
• Disadvantages
–
–
–
–
Tends to converge relatively slowly, although this can be
improved with acceleration
Has tendency to miss solutions, particularly on large systems
Tends to diverge on cases with negative branch reactances
(common with compensated lines)
Need to program using complex numbers
10
Newton-Raphson Algorithm
• The second major power flow solution method is
the Newton-Raphson algorithm
• Key idea behind Newton-Raphson is to use
sequential linearization
General form of problem: Find an x such that
f ( xˆ )  0
11
Newton-Raphson Method (scalar)
1. For each guess of xˆ , x ( v ) , define
x ( v )  xˆ - x ( v )
2. Represent f ( xˆ ) by a Taylor series about f ( x )
(v)
df
(
x
) (v)
(v)
f ( xˆ )  f ( x ) 
x 
dx
1 d 2 f ( x( v ) )
(v)

x
2
2 dx


2
 higher order terms
12
Newton-Raphson Method, cont’d
3. Approximate f ( xˆ ) by neglecting all terms
except the first two
(v )
df
(
x
) (v)
(v )
f ( xˆ )  0  f ( x ) 
x
dx
4. Use this linear approximation to solve for x ( v )
1
 df ( x ) 
(v)
x
 
f
(
x
)

 dx 
5. Solve for a new estimate of x̂
(v)
(v)
x ( v 1)  x ( v )  x ( v )
13
Newton-Raphson Example
Use Newton-Raphson to solve f ( x)  x 2 - 2  0
The equation we must iteratively solve is
x
(v)
x ( v )
x
x
1
 df ( x ) 
(v)
 
f
(
x
)

 dx 
1  (v) 2

   ( v )  (( x ) - 2)
2x 
(v )
( v 1)
 x
(v)
( v 1)
 x
(v)
 x
(v)
1  (v) 2

  ( v )  (( x ) - 2)
2x 
14
Newton-Raphson Example, cont’d
x
( v 1)
 x
(v)
1  (v) 2

  ( v )  (( x ) - 2)
2x 
Guess x (0)  1. Iteratively solving we get
v
x(v)
f ( x(v) )
x ( v )
0
1
1
0.5
1
1.5
0.25
0.08333
2
1.41667
6.953  103
2.454  103
3
1.41422
6.024  106
15
Sequential Linear Approximations
Function is f(x) = x2 - 2 = 0.
Solutions are points where
f(x) intersects f(x) = 0 axis
At each
iteration the
N-R method
uses a linear
approximation
to determine
the next value
for x
16
Newton-Raphson Comments
• When close to the solution the error decreases quite
quickly -- method has quadratic convergence
• f(x(v)) is known as the mismatch, which we would
like to drive to zero
• Stopping criteria is when f(x(v))  < 
• Results are dependent upon the initial guess. What
if we had guessed x(0) = 0, or x (0) = -1?
• A solution’s region of attraction (ROA) is the set of
initial guesses that converge to the particular
solution. The ROA is often hard to determine
17
Multi-Variable Newton-Raphson
Next we generalize to the case where x is an ndimension vector, and f (x) is an n-dimension function
 x1 
x 
x   2
 
x 
 n
 f1 (x) 
 f ( x) 
f ( x)   2 


 f ( x) 
 n 
Again define the solution xˆ so f (xˆ )  0 and
x  xˆ  x
18
Multi-Variable Case, cont’d
The Taylor series expansion is written for each fi (x)
f1 (x)
f1 ( x)
f1 ( xˆ )  f1 ( x) 
x1 
x2 
x1
x2
f1 (x)
xn  higher order terms
xn
f n (x)
f n ( x)
f n (xˆ )  f n ( x) 
x1 
x2 
x1
x2
f n (x)
xn  higher order terms
xn
19
Multi-Variable Case, cont’d
This can be written more compactly in matrix form
 f1 (x)
 x
1

 f1 (x) 
 f (x)   f 2 (x)
f (xˆ )   2    x1

 
 f ( x)  
 n   f (x)
 n
 x1
f1 (x)
x2
f 2 (x)
x2
f n (x)
x2
f1 (x) 
xn 
  x1 
f 2 ( x)  


x
xn   2 





x
f n (x)   n 

xn 
 higher order terms
20
Jacobian Matrix
The n by n matrix of partial derivatives is known
as the Jacobian matrix, J (x)
 f1 (x)
 x
1

 f 2 (x)
J (x)   x1


 f (x)
 n
 x1
f1 ( x)
x2
f 2 ( x)
x2
f n ( x)
x2
f1 ( x) 
xn 

f 2 ( x) 
xn 


f n ( x) 

xn 
21
Multi-Variable Example
 x1 
Solve for x =   such that f ( x)  0 where
x2 
f1 (x)  2 x12  x22  8  0
f 2 (x)  x12  x22  x1 x2  4  0
First symbolically determine the Jacobian
 f1 (x)
 x
1

J (x) =
 f 2 (x)
 x1
f1 ( x) 
x2 

f 2 ( x) 
x2 
22
Multi-variable Example, cont’d
 4 x1
J (x) = 
 2 x1  x2
Then
 x1 
 4 x1
 x     2 x  x
 2
 1 2
Arbitrarily guess x(0)

x1  2 x2 
2 x2
1
  f1 ( x) 
x1  2 x2   f 2 ( x) 
1
 
1
2 x2
1
x
(1)
1  4 2   5
 2.1
  
  



1  3 1  3
1.3 
23
Multi-variable Example, cont’d
1
 2.1 8.40 2.60   2.51
1.8284
x
  
 




1.3
5.50

0.50
1.45
1.2122
  
 



Each iteration we check f (x) to see if it is below our
(2)
specified tolerance 
0.1556 
f (x )  

0.0900


If  = 0.2 then we would be done. Otherwise we'd
(2)
continue iterating.
24
NR Application to Power Flow
We first need to rewrite complex power equations
as equations with real coefficients
*
n


Si  Vi I i*  Vi   YikVk   Vi  Yik*Vk*
 k 1

k 1
These can be derived by defining
n
Yik
Gik  jBik
ji
Vi
Vi e
 ik
i   k
 Vi  i
Recall e j  cos  j sin 
25
Real Power Balance Equations
n
Si  Pi  jQi  Vi  Yik*Vk* 
k 1

n
 Vi Vk
k 1
n
j ik
V
V
e
(Gik  jBik )
 i k
k 1
(cos ik  j sin  ik )(Gik  jBik )
Resolving into the real and imaginary parts
Pi 
Qi 
n
 Vi Vk (Gik cos ik  Bik sin  ik )  PGi  PDi
k 1
n
 Vi Vk (Gik sin ik  Bik cos ik )  QGi  QDi
k 1
26
Newton-Raphson Power Flow
In the Newton-Raphson power flow we use Newton's
method to determine the voltage magnitude and angle
at each bus in the power system.
We need to solve the power balance equations
Pi 
Qi 
n
 Vi
Vk (Gik cos  ik  Bik sin  ik )  PGi  PDi
 Vi
Vk (Gik sin  ik  Bik cos  ik )  QGi  QDi
k 1
n
k 1
27
Power Flow Variables
Assume the slack bus is the first bus (with a fixed
voltage angle/magnitude). We then need to determine
the voltage angle/magnitude at the other buses.
 2 




 n 
x  

V
 2




 Vn 
 P2 (x)  PG 2  PD 2 




 Pn (x)  PGn  PDn 
f ( x)  
Q2 (x)  QG 2  QD 2 






 Qn (x)  QGn  QDn 
28
N-R Power Flow Solution
The power flow is solved using the same procedure
discussed last time:
Set v  0; make an initial guess of x, x( v )
While f (x( v ) )   Do
x( v 1)  x( v )  J (x( v ) ) 1 f (x( v ) )
v
 v 1
End While
29
Power Flow Jacobian Matrix
The most difficult part of the algorithm is determining
and inverting the n by n Jacobian matrix, J (x)
 f1 (x)
 x
1

 f 2 (x)
J (x)   x1


 f (x)
 n
 x1
f1 ( x)
x2
f 2 (x)
x2
f n ( x)
x2
f1 ( x) 
xn 

f 2 ( x) 
xn 


f n ( x) 

xn 
30
Power Flow Jacobian Matrix, cont’d
Jacobian elements are calculated by differentiating
each function, fi ( x), with respect to each variable.
For example, if fi (x) is the bus i real power equation
fi ( x) 
fi ( x)

i
n
 Vi Vk (Gik cosik  Bik sin ik )  PGi  PDi
k 1
n
 Vi Vk (Gik sin ik  Bik cosik )
k 1
k i
fi ( x)
 Vi V j (Gij sin ij  Bij cosij ) ( j  i )
 j
31
Two Bus Newton-Raphson Example
For the two bus power system shown below, use the
Newton-Raphson power flow to determine the
voltage magnitude and angle at bus two. Assume
that bus one is the slack and SBase = 100 MVA.
Line Z = 0.1j
One
1.000 pu
Two
0 MW
0 MVR
2 
x   
 V2 
1.000 pu
200 MW
100 MVR
Ybus
  j10 j10 
 

j
10

j
10


32
Two Bus Example, cont’d
General power balance equations
Pi 
Qi 
n
 Vi Vk (Gik cosik  Bik sin ik )  PGi  PDi
k 1
n
 Vi Vk (Gik sin ik  Bik cos ik )  QGi  QDi
k 1
Bus two power balance equations
V2 V1 (10 sin  2 )  2.0  0
V2 V1 ( 10 cos  2 )  V2 (10)  1.0  0
2
33
Two Bus Example, cont’d
P2 (x)  V2 (10sin  2 )  2.0  0
Q2 (x)  V2 ( 10 cos  2 )  V2 (10)  1.0  0
2
Now calculate the power flow Jacobian
 P2 (x)
 
2

J ( x) 
 Q 2 (x)
 
2

P2 ( x) 
V 2 

Q 2 ( x) 
 V 2 
10 V2 cos 2
 
10 V2 sin  2
10sin  2

10 cos  2  20 V2 
34
Two Bus Example, First Iteration
Set v  0, guess x
(0)
0 
 
1 
Calculate
V2 (10sin  2 )  2.0


f(x )  
 
2
 V2 (10 cos 2 )  V2 (10)  1.0 
10sin  2
10 V2 cos 2

(0)
J (x )  


10 V2 sin  2 10 cos 2  20 V2 
(0)
1
Solve x
(1)
0  10 0   2.0 
 
 1.0 
1
0
10
  
  
 2.0 
1.0 
 
10 0 
 0 10 


 0.2 
 

0.9


35