lecture 11 power flow professor tom overbye special guest appearance by professor sauer! department...
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Lecture 11Power Flow
Professor Tom Overbye
Special Guest Appearance by Professor Sauer!Department of Electrical and
Computer Engineering
ECE 476
POWER SYSTEM ANALYSIS
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Announcements
Homework #5 is 3.12, 3.14, 3.19, 3.60 due Oct 2nd (Thursday)
First exam is 10/9 in class; closed book, closed notes, one note sheet and calculators allowed
Start reading Chapter 6 for lectures 11 and 12
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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
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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 ikk
n
i Gi Di ik kk
I
I I I I
I I I Y V
I Y V
*iThe network power injection is then S i iV I
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Power Balance Equations, cont’d
** * *
i1 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 kk k
ik ik ik
i
V I V Y V V Y V
Y G jB
V
jRecall e cos sin
iji i i
ik i k
V e V
j
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Real Power Balance Equations
* *i
1 1
1
i1
i1
S ( )
(cos sin )( )
Resolving into the real and imaginary parts
P ( cos sin )
Q ( sin cos
ikn n
ji i i ik k i k ik ik
k k
n
i k ik ik ik ikk
n
i k ik ik ik ik Gi Dik
n
i k ik ik ik ik
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 DiQ Q
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Power Flow Requires Iterative Solution
i
bus
** * *
i1 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 kk k
V I V Y V V Y V
Y
er, we must pursue an iterative approach.
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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 vh x
h
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Gauss Iteration Example
( 1) ( )
(0)
( ) ( )
Example: Solve - 1 0
1
Let k = 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
k x k x
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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 vx x x x
( )
2i2
1
he absolute value by using a norm
Two common norms are the Euclidean & infinity
max x
v
j
n
i ii
x
x
x x
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Gauss Power Flow
** * *
i1 1
* * * *
1 1
*
*1 1,
*
*1,
We first need to put the equation in the correct form
S
S
S
S1
i i
i
i
n n
i i i ik k i ik kk k
n n
i i i ik k ik kk k
n ni
ik k ii i ik kk k k i
ni
i ik kii k k i
V I V Y V V Y V
V I V Y V V Y V
Y V Y V Y VV
V Y VY V
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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.
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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 .
15 15
0.02 0.06
5 14.95 5 15Hence
5 15 5 14.70
( Note - 15 0.05 0.25)
jj
j j
j j
B j j j
Y
Y
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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 kk k i
v v
V Y VY V
jV j
j V
V
v V v V
j j
j j
.9624 0.0553j
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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
22
, Q 23.9 Mvar
The capacitor is supplying V 25 23.2 Mvar
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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.
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Gauss with Many Bus Systems
*( )( 1)
( )*1,
( ) ( ) ( )1 2
( 1)
With multiple bus systems we could calculate
new V ' as follows:
S1
( , ,..., )
But after we've determined we have a better
estimate of
i
i
nvv i
i ik kvii k k i
v v vi n
vi
s
V Y VY 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.
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Gauss-Seidel Iteration
( 1) ( ) ( ) ( )2 12 2 3
( 1) ( 1) ( ) ( )2 13 2 3
( 1) ( 1) ( 1) ( ) ( )2 14 2 3 4
( 1) ( 1) ( 1)( 1) ( )2 1 2 3 4
Immediately use the new voltage estimates:
( , , , , )
( , , , , )
( , , , , )
( , , , ,
v v v vn
v v v vn
v v v v vn
v v vv vn 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.
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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 injectionspecial coding is needed to include PV buses in the
Gauss-Seidel iteration
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Inclusion of PV Buses in G-S
i
i
* *
1
( )( ) ( )*
1
( ) ( )
To solve for V at a PV bus we must first make a
guess of Q :
Hence Im
In the iteration we use
k
n
i i ik k i ik
nvv v
i i ikk
v vi i i
S V Y V P jQ
Q V Y V
S P jQ
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Inclusion of PV Buses, cont'd
( 1)
( )*( )( 1)
( )*1,
( 1)i i
Tentatively solve for
S1
But since V is specified, replace by V
i
vi
v nvv i
i ik kvii k k i
vi
V
V Y VY V
V
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Two Bus PV Example
Bus 1
(slack bus)
Bus 2V1 = 1.0 V2 = 1.05
P2 = 0 MW
z = 0.02 + j 0.06
Consider the same two bus system from the previousexample, except the load is replaced by a generator
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Two Bus PV Example, cont'd
*2
2 21 1*22 2
* *2 21 1 2 22 2 2
2
( ) ( 1) ( 1)2 2 2
1
Im[ ]
Guess V 1.05 0
0 0 0.457 1.045 0.83 1.050 0.83
1 0 0.535 1.049 0.93 1.050 0.93
2 0 0.545 1.050 0.96 1.050 0.96
v v v
SV Y V
Y V
Q Y VV Y V V
v S V V
j
j
j
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Generator Reactive Power Limits
The reactive power output of generators varies to maintain the terminal voltage; on a real generator this is done by the exciter
To maintain higher voltages requires more reactive power
Generators have reactive power limits, which are dependent upon the generator's MW output
These limits must be considered during the power flow solution.
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Generator Reactive Limits, cont'd
During power flow once a solution is obtained check to make generator reactive power output is within its limits
If the reactive power is outside of the limits, fix Q at the max or min value, and resolve treating the generator as a PQ bus
– this is know as "type-switching"– also need to check if a PQ generator can again regulate
Rule of thumb: to raise system voltage we need to supply more vars
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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 vx h x x
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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
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Gauss-Seidel Advantages
Each iteration is relatively fast (computational order is proportional to number of branches + number of buses in the system
Relatively easy to program
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Gauss-Seidel 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
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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
( ) 0ˆf x
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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
vv v
vv
x x
x x x
f x f x
df xf x f x x
dx
d f xx
dx
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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
vv v
v
vv v
f x
df xf x f x x
dx
x
df xx f x
dx
( 1) ( ) ( )
ate of x̂v v vx x x
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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
vv v
v vv
v v v
v v vv
f x x
df xx f x
dx
x xx
x x x
x x xx
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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 vv
v v v
x x xx
x f x x
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Sequential Linear Approximations
Function is f(x) = x2 - 2 = 0.Solutions are points wheref(x) intersects f(x) = 0 axis
At each iteration theN-R methoduses a linearapproximationto determine the next valuefor x
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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
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Multi-Variable Newton-Raphson
1 1
2 2
Next we generalize to the case where is an n-
dimension vector, and ( ) is an n-dimension function
( )
( )( )
( )
Again define the solution so ( ) 0 andˆ ˆn n
x f
x f
x f
x
f x
x
xx f x
x
x f x
x
ˆ x x
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38
Multi-Variable Case, cont’d
i
1 11 1 1 2
1 2
1
n nn n 1 2
1 2
n
The Taylor series expansion is written for each f ( )
f ( ) f ( )f ( ) f ( )ˆ
f ( )higher order terms
f ( ) f ( )f ( ) f ( )ˆ
f ( )higher order terms
nn
nn
x xx x
xx
x xx x
xx
x
x xx x
x
x xx x
x
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Multi-Variable Case, cont’d
1 1 1
1 21 1
2 2 22 2
1 2
1 2
This can be written more compactly in matrix form
( ) ( ) ( )
( )( ) ( ) ( )
( )( )ˆ
( )( ) ( ) ( )
n
n
nn n n
n
f f fx x x
f xf f f
f xx x x
ff f fx x x
x x x
xx x x
xf x
xx x x
higher order terms
nx
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Jacobian Matrix
1 1 1
1 2
2 2 2
1 2
1 2
The n by n matrix of partial derivatives is known
as the Jacobian matrix, ( )
( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( ) ( )
n
n
n n n
n
f f fx x x
f f fx x x
f f fx x x
J x
x x x
x x x
J x
x x x
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41
Multi-variable Example, cont’d
1 2
1 2 1 2
11 1 2 1
2 1 2 1 2 2
(0)
1(1)
4 2( ) =
2 2
Then
4 2 ( )
2 2 ( )
1Arbitrarily guess
1
1 4 2 5 2.1
1 3 1 3 1.3
x x
x x x x
x x x f
x x x x x f
J x
x
x
x
x
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42
Multi-variable Example, cont’d
1(2)
(2)
2.1 8.40 2.60 2.51 1.8284
1.3 5.50 0.50 1.45 1.2122
Each iteration we check ( ) to see if it is below our
specified tolerance
0.1556( )
0.0900
If = 0.2 then we wou
x
f x
f x
ld be done. Otherwise we'd
continue iterating.