Transcript EE 369 POWER SYSTEM ANALYSIS Lecture 13 Newton-Raphson Power Flow Tom Overbye and Ross Baldick.

EE 369
POWER SYSTEM ANALYSIS
Lecture 13
Newton-Raphson Power Flow
Tom Overbye and Ross Baldick
1
Announcements
• Homework 10 is: 3.49, 3.55, 3.57, 6.2, 6.9,
6.13, 6.14, 6.18, 6.19, 6.20; due 11/19. (Use
infinity norm and epsilon = 0.01 for any
problems where norm or stopping criterion
not specified.)
• Homework 11 is 6.24, 6.26, 6.28, 6.30 (see
figure 6.18 and table 6.9 for system), 6.38,
6.42, 6.43, 6.46, 6.52, 6.54; due Tuesday
11/24. Note that HW is due on Tuesday
because Thanksgiving is on Thursday.
2
Dishonest Newton-Raphson
Since most of the time in the Newton-Raphson
iteration is spent dealing with the Jacobian, one
way to speed up the iterations is to only calculate
(and factorize) the Jacobian occasionally:
– known as the “Dishonest” Newton-Raphson or
Shamanskii method,
– an extreme example is to only calculate the Jacobian
for the first iteration, which is called the chord method.
Honest:
x ( v 1)  x ( v ) - J ( x ( v ) )-1 f ( x ( v ) )
Dishonest: x ( v 1)  x ( v ) - J ( x (0) )-1 f ( x ( v ) )
Stopping criterion f ( x ( v ) )   used in both cases.
3
Dishonest Newton-Raphson
Example
Use the Dishonest Newton-Raphson (chord method)
to solve f ( x )  0, where:
f ( x)  x - 2
2
x ( v )
x ( v )
x ( v 1)
1
df (0) 

 
(x )
f ( x(v) )
 dx

1  (v) 2

 
(( x ) - 2)
(0)
 2 x 
(v)  1 
(v) 2
 x 
((
x
) - 2)
(0)
 2 x 
4
Dishonest N-R Example, cont’d
x
( v 1)
 x
(v)
1  (v) 2

  (0)  (( x ) - 2)
2x 
Guess x (0)  1. Iteratively solving we get

x ( v ) (honest)
x ( v ) (dishonest)
0
1
1
1.5
1
1.5
2
1.41667
1.375
3
1.41422
1.429
4
1.41422
1.408
We pay a price
in increased
iterations, but
with decreased
computation
per iteration
5
Two Bus Dishonest ROC
Region of convergence for different initial
guesses for the 2 bus case using the Dishonest N-R
Red region
converges
to the high
voltage
solution,
while the
yellow region
converges
to the low
voltage
solution
6
Honest N-R Region of Convergence
Maximum
of 15
iterations
7
Decoupled Power Flow
The “completely” Dishonest NewtonRaphson (chord), where we only
calculate the Jacobian once, is not
usually used for power flow analysis.
However several approximations of the
Jacobian matrix are used that result in a
similar approximation.
One common method is the decoupled
power flow. In this approach
approximations are used to decouple the
real and reactive power equations.
8
Coupled Newton-Raphson Update
Standard form of the Newton-Raphson update:
 P ( v )

θ


 Q ( v )

 θ
P ( v ) 
  θ ( v )  
V 
P( x ( v ) ) 
(v)


f
(
x
)




(v)
(v) 
(v)

Q
(
x
) 

V
Q

 

 V 
 P2 ( x ( v ) )  PD 2  PG 2 


(v)
where P( x )  
.
 P (x( v ) )  P  P 
 n
Dn
Gn 
Note that changes in angle and voltage magnitude
both affect (couple to) real and reactive power.
9
Decoupling Approximation
P ( v )
Q ( v )
Usually the off-diagonal matrices,
and
V
θ
are small. Therefore we approximate them as zero:
 P ( v )

0 
(v) 


 P( x ( v ) ) 

θ

θ
(v)

 


f
(
x
)



(v)
(
v
)
(
v
)

Q    V 
Q( x ) 


 0


V


Then the update can be decoupled into two separate updates:
θ
(v)
 P
 
 θ
( v ) 1

(v)

P
(
x
),


V
(v)
 Q
 
 V
( v ) 1

(v)

Q
(
x
).


10
Off-diagonal Jacobian Terms
So, angle and real power are coupled closely, and
voltage magnitude and reactive power are coupled closely.
Justification for Jacobian approximations:
1. Usually r
x, therefore Gij
Bij
2. Usually ij is small so sin ij  0
Therefore
Pi
 Vi  Gij cosij  Bij sin ij 
 Vj
Qi
θ j
 0
  Vi V j  Gij cosij  Bij sin ij   0
11
Decoupled N-R Region of
Convergence
12
Fast Decoupled Power Flow
By further approximating the Jacobian we obtain
a typically reasonable approximation that is
independent of the voltage magnitudes/angles.
This means the Jacobian need only be built and
factorized once.
This approach is known as the fast decoupled
power flow (FDPF)
FDPF uses the same mismatch equations as
standard power flow so it should have same
solution if it converges
The FDPF is widely used, particularly when we
only need an approximate solution.
13
FDPF Approximations
The FDPF makes the following approximations:
1.
Gij  0
2.
Vi
3.
sin ij  0
 1 (for some occurrences),
cos ij  1
Then: θ( v )  B 1diag{| V |( v ) }1 P( x ( v ) ),
V
(v)
 B 1diag{| V |( v ) }1 Q( x ( v ) )
Where B is just the imaginary part of the Ybus  G bus  jB bus ,
except the slack bus row/column are omitted. That is,
B is B bus , but with the slack bus row and column deleted.
Sometimes approximate diag{| V |( v ) } by identity.
14
FDPF Three Bus Example
Use the FDPF to solve the following three bus system
Line Z = j0.07
One
Two
Line Z = j0.05
Three
Line Z = j0.1
200 MW
100 MVR
1.000 pu
200 MW
100 MVR
Ybus
 34.3 14.3 20 
 j  14.3 24.3 10 


10
30 
 20
15
FDPF Three Bus Example, cont’d
Ybus
B 1
20 
 34.3 14.3
 24.3 10 


 j 14.3 24.3 10  B  



10

30


10
30 
 20
 0.0477 0.0159 
 


0.0159

0.0389


Iteratively solve, starting with an initial voltage guess
(0)

 2
0 
   0 
 
 3
 2 
 
 3
(1)
V 2 
V 
 3
(0)

1
1

0   0.0477 0.0159   2   0.1272 
  





0

0.0159

0.0389
2

0.1091
  
  

16
FDPF Three Bus Example, cont’d
(1)
V 2 
1  0.0477 0.0159  1  0.9364 
 V   1   0.0159 0.0389  1   0.9455
  
  

 3
Pi ( x ) n
P P
  Vk (Gik cosik  Bik sin ik )  Di Gi
Vi
Vi
k 1
(2)

 2
 0.1272   0.0477 0.0159   0.151  0.1361
    0.1091   0.0159 0.0389  0.107    0.1156

 

 

 3
V 2 
V 
 3
(2)
0.924 
 

0.936


 0.1384 
Actual solution: θ  


0.1171


0.9224 
V

0.9338


17
FDPF Region of Convergence
18
“DC” Power Flow
The “DC” power flow makes the most severe
approximations:
– completely ignore reactive power, assume all the
voltages are always 1.0 per unit, ignore line
conductance
This makes the power flow a linear set of
equations, which can be solved directly:
θ  B 1 P
where B is the imaginary part of the bus
admittance matrix with the row and column
corresponding to the slack bus deleted, and,
similarly, Θ and P omit the slack bus.
19
DC Power Flow Example
20
DC Power Flow 5 Bus Example
One
360 MW
0 Mvar
A
Five
Four
MVA
Three
A
520 MW
MVA
A
0 Mvar
MVA
slack
1.000 pu
0.000 Deg
1.000 pu
-4.125 Deg
A
A
MVA
MVA
1.000 pu
-18.695 Deg
1.000 pu
-1.997 Deg
80 MW
0 Mvar
1.000 pu
0.524 Deg
Two
800 MW
0 Mvar
Notice with the dc power flow all of the voltage magnitudes are
1 per unit.
21
Power System Control
A major problem with power system operation is
the limited capacity of the transmission system
– lines/transformers have limits (usually thermal)
– no direct way of controlling flow down a
transmission line (e.g., there are no low cost valves
to close to limit flow, except “on” and “off”)
– open transmission system access associated with
industry restructuring is stressing the system in new
ways
We need to indirectly control transmission line
flow by changing the generator outputs.
22
Indirect Transmission Line Control
What we would like to determine is how a change in
generation at bus k affects the power flow on a line
from bus i to bus j.
The assumption is
that the change
in generation is
absorbed by the
slack bus
23
Power Flow Simulation - Before
•One way to determine the impact of a generator
change is to compare a before/after power flow.
•For example below is a three bus case with an
overload.
131.9 MW
124%
One
200.0 MW
71.0 MVR
Two
68.1 MW
68.1 MW
200 MW
100 MVR
Z for all lines = j0.1
Three
1.000 pu
0 MW
64 MVR
24
Power Flow Simulation - After
•Increasing the generation at bus 3 by 95 MW
(and hence decreasing generation at the slack
bus 1 by a corresponding amount), results in a
31.3 MW drop in the MW flow on the line from
bus 1 to 2.
101.6 MW
100%
One
105.0 MW
64.3 MVR
Two
3.4 MW
Z for all lines = j0.1
Limit for all lines = 150 MVA
Three
98.4 MW
200 MW
100 MVR
92%
1.000 pu
95 MW
64 MVR
25
Analytic Calculation of Sensitivities
Calculating control sensitivities by repeated
power flow solutions is tedious and would
require many power flow solutions.
An alternative approach is to analytically
calculate these values
The power flow from bus i to bus j is
Pij 
Vi V j
So Pij 
X ij
sin( i   j ) 
 i   j
X ij
i   j
X ij
We just need to get
 ij
PGk26
Analytic Sensitivities
1
From the fast decoupled power flow we know: θ  B P(x ).
Sign convention in definition of P( x ) is that entry in P( x )
is negative if change in net injection (generation) is positive.
So to get the change in θ due to a change of generation at
bus k , just set P( x ) equal to all zeros except a minus one
at position k:
P
0
 
 
  1  For 1MW increase in generation at bus k
0
 
 
27
Three Bus Sensitivity Example
For the previous three bus case with Zline  j 0.1
 20 10 10 
 20 10 


Ybus  j 10 20 10  B  



10

20


 10 10 20
Hence for a change of generation at bus 3
  2 
  
 3
1
 20 10   0   0.0333






 10 20  1 0.0667 
0.0667  0
Changes in line flows are: P3 to 1 
 0.667 pu
0.1
P3 to 2  0.333 pu
P 2 to 1  0.333 pu
28