ele b7 power systems engineering newton-raphson raphson...
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ELE B7 Power Systems Engineering ELE B7 Power Systems Engineering
NewtonNewton--RaphsonRaphson MethodMethod
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Slide # 2
Newton-Raphson Algorithm
• The second major power flow solution method is the Newton-Raphson algorithm
• Key idea behind Newton-Raphson is to use sequential linearizationGeneral form of problem: Find an x such that
( ) 0ˆf x
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Slide # 3
Newton-Raphson Method (scalar)
( )
( ) ( )
( )( ) ( )
2 ( ) 2( )2
1. For each guess of , , define ˆ
-ˆ2. Represent ( ) by a Taylor series about ( )ˆ
( )( ) ( )ˆ
( ) higher order terms
v
v v
vv v
vv
x x
x x xf x f x
df xf x f x xdx
d f x xdx
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Slide # 4
Newton-Raphson Method, cont’d
( )( ) ( )
( )
1( )( ) ( )
3. Approximate ( ) by neglecting all termsˆexcept 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 xdx
x
df xx f xdx
( 1) ( ) ( )
ate of x̂v v vx x x
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Slide # 5
Newton-Raphson Example
2
1( )( ) ( )
( ) ( ) 2( )
( 1) ( ) ( )
( 1) ( ) ( ) 2( )
Use Newton-Raphson to solve ( ) - 2 0The 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 xdx
x xx
x x x
x x xx
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Slide # 6
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 0.51 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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Slide # 7
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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Slide # 8
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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Slide # 9
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 ˆ
n n
x fx f
x f
xf x
xx
x f x
xx f x
x
ˆ x x
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Slide # 10
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
xx xx x
x
x xx x
x
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Slide # 11
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 f
x x x
x x x
xx x x
xf x
xx x x
higher order terms
nx
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Slide # 12
Jacobian Matrix
1 1 1
1 2
2 2 2
1 2
1 2
The n by n matrix of partial derivatives is knownas 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 xx x x
x x xJ x
x x x
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Slide # 13
Multi-Variable N-R Procedure
1
( 1) ( ) ( )
( 1) ( ) ( ) 1 ( )
( )
Derivation of N-R method is similar to the scalar case( ) ( ) ( ) higher order termsˆ( ) 0 ( ) ( )ˆ
( ) ( )
( ) ( )
Iterate until ( )
v v v
v v v v
v
f x f x J x xf x f x J x x
x J x f x
x x x
x x J x f x
f x
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Slide # 14
Multi-Variable Example
1
22 2
1 1 22 2
2 1 2 1 2
1 1
1 2
2 2
1 2
xSolve for = such that ( ) 0 where
x
f ( ) 2 8 0
f ( ) 4 0First symbolically determine the Jacobian
f ( ) f ( )
( ) =f ( ) f ( )
x x
x x x x
x x
x x
x f x
x
x
x x
J xx x
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Slide # 15
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 2Then
4 2 ( )2 2 ( )
1Arbitrarily guess
1
1 4 2 5 2.11 3 1 3 1.3
x xx x x x
x x x fx x x x x f
J x
xx
x
x
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Slide # 16
Multi-variable Example, cont’d
1(2)
(2)
2.1 8.40 2.60 2.51 1.82841.3 5.50 0.50 1.45 1.2122
Each iteration we check ( ) to see if it is below our specified tolerance
0.1556( )
0.0900If = 0.2 then we wou
x
f x
f x
ld be done. Otherwise we'd continue iterating.
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Slide # 17
NR Application to Power Flow
** * *
i1 1
We first need to rewrite complex power equationsas equations with real coefficients
S
These can be derived by defining
n n
i i i ik k i ik kk k
V I V Y V V Y V
ik ik ik
i
Y G jB
V
jRecall e cos sin
iji i i
ik i k
V e V
j
=
=
=
ik ik ik
i
Y G jB
V
jRecall e cos sin
iji i i
ik i k
V e V
j
=
=
=
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Slide # 18
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 kn
i k ik ik ik ikk
n
i k ik ik ik ik Gi Dikn
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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Slide # 19
Newton-Raphson Power Flow
i1
In the Newton-Raphson power flow we use Newton'smethod to determine the voltage magnitude and angleat each bus in the power system. We need to solve the power balance equations
P ( cosn
i k ik ikk
V V G
i1
sin )
Q ( sin cos )
ik ik Gi Di
n
i k ik ik ik ik Gi Dik
B P P
V V G B Q Q
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Slide # 20
Power Flow Variables
Assume the slack bus is the first bus (with a fixedvoltage angle/magnitude). We then need to determine the voltage angle/magnitude at the other buses.
DnGnn
DG
DnGnn
DG
n
n
QQQ
QQQPPP
PPP
V
V
(x)
(x)(x)
(x)
ΔQ(x)ΔP(x)
f(x)X
222
222
2
2
,
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Slide # 21
N-R Power Flow Solution
( )
( )
( 1) ( ) ( ) 1 ( )
The power flow is solved using the same procedurediscussed last time:
Set 0; make an initial guess of ,
While ( ) Do
( ) ( )1
End While
v
v
v v v v
v
v v
x x
f x
x x J x f x
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Slide # 22
Power Flow Jacobian Matrix
1 1 1
1 2
2 2 2
1 2
1 2
The most difficult part of the algorithm is determiningand inverting the n by n Jacobian matrix, ( )
( ) ( ) ( )
( ) ( ) ( )( )
( ) ( ) ( )
n
n
n n n
n
f f fx x x
f f fx x x
f f fx x x
J xx x x
x x xJ x
x x x
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Slide # 23
Power Flow Jacobian Matrix, cont’d
i
i
i1
Jacobian elements are calculated by differentiating each function, f ( ), with respect to each variable.For example, if f ( ) is the bus i real power equation
f ( ) ( cos sin )n
i k ik ik ik ik Gik
x V V G B P P
xx
i
1
i
f ( ) ( sin cos )
f ( ) ( sin cos ) ( )
Di
n
i k ik ik ik iki k
k i
i j ik ik ik ikj
x V V G B
x V V G B j i
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Slide # 24
Line Flows and Losses
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Slide # 25
Line Flows and Losses
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Slide # 26
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. Assumethat bus one is the slack and SBase = 100 MVA.
2
2
10 1010 10busj j
V j j
x Y
222 VV
1002002 jS
01 01V
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Slide # 27
Two Bus Example, cont’d
i1
i1
2 2 1 22
2 2 1 2 2
General power balance equations
P ( cos sin )
Q ( sin cos )
Bus two power balance equationsP (10sin ) 2.0 0
( 10cos ) (10) 1.0 0
n
i k ik ik ik ik Gi Dikn
i k ik ik ik ik Gi Dik
V V G B P P
V V G B Q Q
V V
Q V V V
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Slide # 28
Two Bus Example, cont’d
2 2 22
2 2 2 2
2 2
2 2
2 2
2 2
2 2 2
2 2 2 2
P ( ) (10sin ) 2.0 0
( ) ( 10cos ) (10) 1.0 0Now calculate the power flow Jacobian
P ( ) P ( )
( )Q ( ) Q ( )
10 cos 10sin10 sin 10cos 20
V
Q V V
VJ
V
VV V
x
x
x x
xx x
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Slide # 29
Two Bus Example, First Iteration
(0)
2 2(0)2
2 2 2
2 2 2(0)
2 2 2 2
(1)
0Set 0, guess
1Calculate
(10sin ) 2.0 2.0f( )
1.0( 10cos ) (10) 1.0
10 cos 10sin 10 0( )
10 sin 10cos 20 0 10
0 10 0Solve
1 0 10
v
V
V V
VV V
x
x
J x
x1 2.0 0.2
1.0 0.9
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Slide # 30
Two Bus Example, Next Iterations
(1)2
(1)
1(2)
0.9(10sin( 0.2)) 2.0 0.212f( )
0.2790.9( 10cos( 0.2)) 0.9 10 1.08.82 1.986
( )1.788 8.199
0.2 8.82 1.986 0.212 0.2330.9 1.788 8.199 0.279 0.8586
f(
x
J x
x
(2) (3)
(3)2
0.0145 0.236)
0.0190 0.85540.0000906
f( ) Done! V 0.8554 13.520.0001175
x x
x
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Slide # 31
Two Bus Solved Values
Once the voltage angle and magnitude at bus 2 are known we can calculate all the other system values,such as the line flows and the generator reactive power output
02 52.13855.0 V
1002002 jS
3.16820012 jS
01 01V
10020021 jS
3.680211212 jSSSloss
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Slide # 32
PV Buses
• Since the voltage magnitude at PV buses is fixed there is no need to explicitly include these voltages in x
or write the reactive power balance equations
– the reactive power output of the generator varies to maintain the fixed terminal voltage (within limits)
– optionally these variations/equations can be included by just writing the explicit voltage constraint for the generator bus
|Vi | – Vi setpoint = 0
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Slide # 33
Three Bus PV Case Example
Line Z = 0.1j
Line Z = 0.1j Line Z = 0.1j
One Two 1.000 pu 0.941 pu
200 MW 100 MVR
170.0 MW 68.2 MVR
-7.469 Deg
Three 1.000 pu
30 MW 63 MVR
2 2 2 2
3 3 3 3
2 2 2
For this three bus case we have( )
( ) ( ) 0V ( )
G D
G D
D
P P PP P P
Q Q
xx f x x
x
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Slide # 34
Solving Large Power Systems
• The most difficult computational task is inverting the Jacobian matrix– inverting a full matrix is an order n3 operation, meaning
the amount of computation increases with the cube of the size size
– this amount of computation can be decreased substantially by recognizing that since the Ybus is a sparse matrix, the Jacobian is also a sparse matrix
– using sparse matrix methods results in a computational order of about n1.5.
– this is a substantial savings when solving systems with tens of thousands of buses
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Slide # 35
Newton-Raphson Power Flow
• Advantages– fast convergence as long as initial guess is close to
solution– large region of convergence
• Disadvantages– each iteration takes much longer than a Gauss-Seidel
iteration– more complicated to code, particularly when
implementing sparse matrix algorithms• Newton-Raphson algorithm is very common in power flow
analysis
ELE B7 Power Systems Engineering � � Newton-Raphson MethodNewton-Raphson AlgorithmNewton-Raphson Method (scalar)Newton-Raphson Method, cont’dNewton-Raphson ExampleNewton-Raphson Example, cont’dSequential Linear ApproximationsNewton-Raphson CommentsMulti-Variable Newton-RaphsonMulti-Variable Case, cont’dMulti-Variable Case, cont’dJacobian MatrixMulti-Variable N-R ProcedureMulti-Variable ExampleMulti-variable Example, cont’dMulti-variable Example, cont’dNR Application to Power FlowReal Power Balance EquationsNewton-Raphson Power FlowPower Flow VariablesN-R Power Flow Solution Power Flow Jacobian MatrixPower Flow Jacobian Matrix, cont’dLine Flows and LossesLine Flows and LossesTwo Bus Newton-Raphson ExampleTwo Bus Example, cont’dTwo Bus Example, cont’dTwo Bus Example, First IterationTwo Bus Example, Next IterationsTwo Bus Solved ValuesPV BusesThree Bus PV Case ExampleSolving Large Power SystemsNewton-Raphson Power Flow
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