- This document summarizes a lecture on power flow analysis in power systems.
- Power flow analysis is used to determine the complex bus voltages and current injections in a power system given the complex power consumed by loads and injected by generators, as well as generator voltage magnitudes.
- The power flow problem is solved iteratively using methods like Gauss-Seidel iteration, as there is no closed-form solution to the power balance equations that relate bus voltages and power injections.
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Technical Specifications
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
Key Features
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface
• Compatible with MAFI CCR system
• Copatiable with IDM8000 CCR
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
Application
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
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1. 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
Special Guest Lecturer: TA Iyke Idehen
2. Announcements
• Please read Chapter 2.4, Chapter 6 up to 6.6
• HW 5 is 5.31, 5.43, 3.4, 3.10, 3.14, 3.19, 3.23, 3.60,
6.30 should be done before exam 1
• Exam 1 is Thursday Oct 6 in class
• Closed book, closed notes, but you may bring one 8.5 by 11
inch note sheet and standard calculators
• Last name A-M here, N to Z in ECEB 1013
1
3. Power Flow Analysis
• When analyzing power systems we know neither
the complex bus voltages nor the complex current
injections
• Rather, we know the complex power being
consumed by the load, and the power being
injected by the generators plus their voltage
magnitudes
• Therefore we can not directly use the Ybus
equations, but rather must use the power balance
equations
2
4. Power Balance Equations
1
bus
1
From KCL we know at each bus i in an n bus system
the current injection, , must be equal to the current
that flows into the network
Since = we also know
i
n
i Gi Di ik
k
n
i Gi Di ik k
k
I
I I I I
I I I Y V
I Y V
*
i
The network power injection is then S i i
V I
3
5. Power Balance Equations, cont’d
*
* * *
i
1 1
S
This is an equation with complex numbers.
Sometimes we would like an equivalent set of real
power equations. These can be derived by defining
n n
i i i ik k i ik k
k k
ik ik ik
i
V I V Y V V Y V
Y G jB
V
j
Recall e cos sin
i
j
i i i
ik i k
V e V
j
4
6. Real Power Balance Equations
* *
i
1 1
1
i
1
i
1
S ( )
(cos sin )( )
Resolving into the real and imaginary parts
P ( cos sin )
Q ( sin cos
ik
n n
j
i i i ik k i k ik ik
k k
n
i k ik ik ik ik
k
n
i k ik ik ik ik Gi Di
k
n
i k ik ik ik i
k
P jQ V Y V V V e G jB
V V j G jB
V V G B P P
V V G B
)
k Gi Di
Q Q
5
7. Power Flow Requires Iterative Solution
i
bus
*
* * *
i
1 1
In the power flow we assume we know S and the
. We would like to solve for the V's. The problem
is the below equation has no closed form solution:
S
Rath
n n
i i i ik k i ik k
k k
V I V Y V V Y V
Y
er, we must pursue an iterative approach.
6
8. Gauss Iteration
There are a number of different iterative methods
we can use. We'll consider two: Gauss and Newton.
With the Gauss method we need to rewrite our
equation in an implicit form: x = h(x)
To iterate we fir (0)
( +1) ( )
st make an initial guess of x, x ,
and then iteratively solve x ( ) until we
find a "fixed point", x, such that x (x).
ˆ ˆ ˆ
v v
h x
h
7
9. Gauss Iteration Example
( 1) ( )
(0)
( ) ( )
Example: Solve - 1 0
1
Let = 0 and arbitrarily guess x 1 and solve
0 1 5 2.61185
1 2 6 2.61612
2 2.41421 7 2.61744
3 2.55538 8 2.61785
4 2.59805 9 2.61798
v v
v v
x x
x x
v
v x v x
8
10. Stopping Criteria
( ) ( ) ( 1) ( )
A key problem to address is when to stop the
iteration. With the Guass iteration we stop when
with
If x is a scalar this is clear, but if x is a vector we
need to generalize t
v v v v
x x x x
( )
2
i
2
1
he absolute value by using a norm
Two common norms are the Euclidean & infinity
max x
v
j
n
i i
i
x
x
x x
9
11. Gauss Power Flow
*
* * *
i
1 1
* * * *
1 1
*
*
1 1,
*
*
1,
We first need to put the equation in the correct form
S
S
S
S
1
i i
i
i
n n
i i i ik k i ik k
k k
n n
i i i ik k ik k
k k
n n
i
ik k ii i ik k
k k k i
n
i
i ik k
ii k k i
V I V Y V V Y V
V I V Y V V Y V
Y V Y V Y V
V
V Y V
Y V
10
12. Gauss Two Bus Power Flow Example
•A 100 MW, 50 Mvar load is connected to a generator
•through a line with z = 0.02 + j0.06 p.u. and line
charging of 5 Mvar on each end (100 MVA base).
Also, there is a 25 Mvar capacitor at bus 2. If the
generator voltage is 1.0 p.u., what is V2?
SLoad = 1.0 + j0.5 p.u.
11
13. Gauss Two Bus Example, cont’d
2
2 bus
bus
22
The unknown is the complex load voltage, V .
To determine V we need to know the .
1
5 15
0.02 0.06
5 14.95 5 15
Hence
5 15 5 14.70
( Note - 15 0.05 0.25)
j
j
j j
j j
B j j j
Y
Y
12
14. Gauss Two Bus Example, cont’d
*
2
2 *
22 1,
2
2 *
2
(0)
2
( ) ( )
2 2
1 S
1 -1 0.5
( 5 15)(1.0 0)
5 14.70
Guess 1.0 0 (this is known as a flat start)
0 1.000 0.000 3 0.9622 0.0556
1 0.9671 0.0568 4 0.9622 0.0556
2 0
n
ik k
k k i
v v
V Y V
Y V
j
V j
j V
V
v V v V
j j
j j
.9624 0.0553
j
13
15. Gauss Two Bus Example, cont’d
2
* *
1 1 11 1 12 2
1
0.9622 0.0556 0.9638 3.3
Once the voltages are known all other values can
be determined, such as the generator powers and the
line flows
S ( ) 1.023 0.239
In actual units P 102.3 MW
V j
V Y V Y V j
1
2
2
, Q 23.9 Mvar
The capacitor is supplying V 25 23.2 Mvar
14
16. 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.
15
17. Stated Another Way
• From exam problem 4.c we had
• This Ybus is actually singular!
• So we cannot solve
• This means (as you might expect), we cannot
independently specify all the current injections I
Bus 2 Bus 1
Bus 3
j0.2
j0.1 j0.1
15 5 10
5 15 10
10 10 20
j
bus
Y
1
bus
V Y I
16
18. Gauss with Many Bus Systems
*
( )
( 1)
( )*
1,
( ) ( ) ( )
1 2
( 1)
With multiple bus systems we could calculate
new V ' as follows:
S
1
( , ,..., )
But after we've determined we have a better
estimate of
i
i
n
v
v i
i ik k
v
ii k k i
v v v
i n
v
i
s
V Y V
Y V
h V V V
V
its voltage , so it makes sense to use this
new value. This approach is known as the
Gauss-Seidel iteration.
17
19. Gauss-Seidel Iteration
( 1) ( ) ( ) ( )
2 1
2 2 3
( 1) ( 1) ( ) ( )
2 1
3 2 3
( 1) ( 1) ( 1) ( ) ( )
2 1
4 2 3 4
( 1) ( 1) ( 1) ( 1) ( )
2 1 2 3 4
Immediately use the new voltage estimates:
( , , , , )
( , , , , )
( , , , , )
( , , , ,
v v v v
n
v v v v
n
v v v v v
n
v v v v v
n n
V h V V V V
V h V V V V
V h V V V V V
V h V V V V V
)
The Gauss-Seidel works better than the Gauss, and
is actually easier to implement. It is used instead
of Gauss.
18
20. Three Types of Power Flow Buses
• There are three main types of power flow buses
– Load (PQ) at which P/Q are fixed; iteration solves for
voltage magnitude and angle.
– Slack at which the voltage magnitude and angle are fixed;
iteration solves for P/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)
19
21. Accelerated G-S Convergence
( 1) ( )
( 1) ( ) ( ) ( )
(
Previously in the Gauss-Seidel method we were
calculating each value x as
( )
To accelerate convergence we can rewrite this as
( )
Now introduce acceleration parameter
v v
v v v v
v
x h x
x x h x x
x
1) ( ) ( ) ( )
( ( ) )
With = 1 this is identical to standard gauss-seidel.
Larger values of may result in faster convergence.
v v v
x h x x
20
22. Accelerated Convergence, cont’d
( 1) ( ) ( ) ( )
Consider the previous example: - 1 0
(1 )
Comparison of results with different values of
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 2.59
v v v v
x x
x x x x
k
81 2.6157 2.6180 2.596
5 2.6118 2.6176 2.6180 2.626 21
23. 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
22
24. 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
( ) 0
ˆ
f x
23
25. Newton-Raphson Method (scalar)
( )
( ) ( )
( )
( ) ( )
2 ( ) 2
( )
2
1. For each guess of , , define
ˆ
-
ˆ
2. Represent ( ) by a Taylor series about ( )
ˆ
( )
( ) ( )
ˆ
1 ( )
higher order terms
2
v
v v
v
v v
v
v
x x
x x x
f x f x
df x
f x f x x
dx
d f x
x
dx
24
26. Newton-Raphson Method, cont’d
( )
( ) ( )
( )
1
( )
( ) ( )
3. Approximate ( ) by neglecting all terms
ˆ
except the first two
( )
( ) 0 ( )
ˆ
4. Use this linear approximation to solve for
( )
( )
5. Solve for a new estim
v
v v
v
v
v v
f x
df x
f x f x x
dx
x
df x
x f x
dx
( 1) ( ) ( )
ate of x̂
v v v
x x x
25
27. Newton-Raphson Example
2
1
( )
( ) ( )
( ) ( ) 2
( )
( 1) ( ) ( )
( 1) ( ) ( ) 2
( )
Use Newton-Raphson to solve ( ) - 2 0
The equation we must iteratively solve is
( )
( )
1
(( ) - 2)
2
1
(( ) - 2)
2
v
v v
v v
v
v v v
v v v
v
f x x
df x
x f x
dx
x x
x
x x x
x x x
x
26
28. Newton-Raphson Example, cont’d
( 1) ( ) ( ) 2
( )
(0)
( ) ( ) ( )
3 3
6
1
(( ) - 2)
2
Guess x 1. Iteratively solving we get
v ( )
0 1 1 0.5
1 1.5 0.25 0.08333
2 1.41667 6.953 10 2.454 10
3 1.41422 6.024 10
v v v
v
v v v
x x x
x
x f x x
27
29. 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
28
30. 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
29