# EE 369 POWER SYSTEM ANALYSIS

Document Sample

```					          EE 369
POWER SYSTEM ANALYSIS

Lecture 13
Newton-Raphson Power Flow
Tom Overbye and Ross Baldick

1
Announcements
Read Chapter 6 and (fourth edition) Chapter
11, concentrating on sections 11.4 and 11.5;
(fifth edition) Chapter 12, concentrating on
sections 12.4 and 12.5.
Homework 10 is fourth edition: 6.27, 6.35,
6.36, 6.44, 6.46, 6.57, 11.24, 11.29; fifth
edition 6.28, 6.36, 6.37, 6.45, 6.51, 6.57,
12.24, 12.28; due November 3. Turn in to in
box next to my office door.
2
Using the Power Flow: Example 1
A
SLA C K3 4 5
MVA
A

MVA

2 1 8 MW
1 .0 2 pu                                   RA Y 3 4 5
sla ck        5 4 M var
A                      A                                                                   A

1 .0 2 pu                                                                                                                                                              MVA                      MVA                                        SLA C K1 3 8              MVA
T IM 3 4 5
1 .0 1 pu                                           RA Y 1 3 8

1 .0 0 pu
A

MVA

T IM 1 3 8
3 3 MW
A

MVA
A
A

MVA
1 .0 3 pu                                                                     Using
1 .0 2 pu
1 3 M var

case
MVA
A
A
1 6 .0 M var                        1 8 MW
1 .0 2 pu                              RA Y 6 9
MVA
MVA                                                                                               5 M var                                                                                                                                                    3 7 MW
A
1 7 MW                            A

1 .0 2 pu                            T IM 6 9                                                          P A I6 9                                                                                                                                                                                       1 3 M var
1 .0 1 pu                                                                                                                                    3 M var

from
MVA                                                                                MVA
A

2 3 MW                                                                                                                      1 .0 1 pu                GRO SS6 9                                 A
A                                                              MVA
7 M var                                                                                                                                                                                      MVA
FERNA 6 9
MVA                                                                                                                        A
1 .0 1 pu                     WO LEN6 9
2 1 MW
M O RO 1 3 8
Example
A
MVA

MVA
H ISKY 6 9                                                                                                                                              7 M var
A
A
4 .8 M var
1 2 MW            MVA
A                             MVA

5 M var                                2 0 MW                                                                                                                                                         1 .0 0 pu

1 .0 0 pu
H A NNA H 6 9
8 M var                                    A

MVA
P ET E6 9
5 8 MW
DEM A R6 9                       A
1 .0 0 pu
A
BO B1 3 8
MVA
A
MVA

6.13
MVA                                                       MVA
5 1 MW                                                                              4 0 M var
4 5 MW
1 5 M var                                                                                                                                                               A
1 .0 2 pu                                              BO B6 9
1 2 M var
2 9 .0 M var                                                                                                                            MVA
UIUC 6 9                0 .9 9 pu
1 4 .3 M var
1 .0 0 pu                                                                                 1 4 0 MW                                                                   5 6 MW
1 2 .8 M var                        A

4 5 M var
A
MVA                                                                                    1 3 M var                                         LY NN1 3 8
A
0 MW
MVA
A                                                                                                                                                                                     0 M var
MVA
A
MVA                                                    5 8 MW                                            A
1 4 MW
0 .9 9 7 pu                                                                                                                                                                                 BLT 1 3 8
3 6 M var                                      MVA                                1 .0 0 pu                                                   MVA                                                         4 M var
0 .9 9 pu                                  A M A NDA 6 9                                                  A
A

A
SH IM KO 6 9                                        1 .0 2 pu
H O M ER6 9                                                       3 3 MW
MVA
MVA
7 .4 M var
A
MVA
1 0 M var                                                 1 .0 1 pu                                                                                                                         A

BLT 6 9                                                                                                             MVA
A                                                1 .0 1 pu                                                                            MVA

1 5 MW                                       A                                                                          MVA
1 5 MW
3 M var                                                                                             H A LE6 9                                  A                                                                       1 0 6 MW                                                                         5 M var
MVA
1 .0 0 pu                                                                                                                                                                                                         8 M var                  A

MVA

3 6 MW
MVA
A
A                                                                                                     A
1 .0 1 pu
6 0 MW                        MVA                                                       1 0 M var          7 .2 M var              MVA
A
A
MVA
1 2 M var
1 .0 0 pu                                                                 1 .0 0 pu                        P A T T EN6 9                                                                                                            MVA

0 .0 M var                                                                              A
MVA

4 5 MW                  1 4 MW                        RO GER6 9
MVA
1 .0 0 pu                       WEBER6 9                                         0 M var
LA UF6 9                                                                                                                                                      2 M var
1 .0 2 pu
2 3 MW
2 2 MW                            0 MW
A                    A
6 M var                                                                                        1 4 MW                                          A

2 0 MW                                                                                                         1 5 M var                         0 M var
MVA                  MVA                                                                                                                             3 M var                                    MVA
3 0 M var
1 .0 2 pu                                           JO 1 3 8          JO 3 4 5
LA UF1 3 8                                                                                                                       1 .0 2 pu             SA V O Y 6 9                                  4 2 MW
1 .0 0 pu
2 M var
1 .0 1 pu            BUC KY 1 3 8                                                                                                  A

A                                                                                                                                             MVA                                                                                            A

1 5 0 MW
MVA                                                                                                          1 .0 1 pu                          SA V O Y 1 3 8                                                                                MVA
A                                                                                                                         A
0 M var
MVA                                                                                                                     MVA

1 5 0 MW
A
0 M var
MVA
1 .0 3 pu
1 .0 2 pu   A

MVA
3
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 ) )-1f (x( v ) )
Dishonest: x(v1)  x(v ) - J (x(0) )-1f (x( v ) )
Both require f (x(v ) )   for a solution                4
Dishonest Newton-Raphson
Example
Use the Dishonest Newton-Raphson (chord method)
to solve f ( x )  0, where:
f ( x)  x - 2   2

1
x   (v)
   df ( x (0) )  f ( x ( v ) )
 dx
              

x ( v )       1  (( x ( v ) )2 - 2)
 2 x (0) 
         
(v)  1 
x ( v 1)    x                   (( x ( v ) )2 - 2)
 2 x (0) 
         
5
Dishonest N-R Example, cont’d
x   ( v 1)
 x   (v)      1  (( x ( v ) )2 - 2)
  (0) 
2x 
Guess x (0)  1. Iteratively solving we get
             x ( v ) (honest)       x ( v ) (dishonest)
0                   1                      1             We pay a price
1                1.5                       1.5           in increased
2                1.41667                   1.375         iterations, but
with decreased
3                1.41422                   1.429         computation
4                1.41422                   1.408         per iteration
6
Two Bus Dishonest ROC
Slide shows the 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

7
Honest N-R Region of Convergence

Maximum
of 15
iterations

8
Decoupled Power Flow
The “completely” Dishonest Newton-
Raphson (chord) 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.        9
Decoupled Power Flow
Formulation
General form of the power flow problem
 P ( v )   P   (v) 
                   
 θ         V        θ( v )   P (x( v ) ) 
                                                  f (x( v ) )
 Q ( v )   Q (v)         (v)
  V   Q(x ) 
(v)
                                             
 θ
            V    
where
 P2 (x( v ) )  PD 2  PG 2 
                            
P(x )  
(v)

 P (x(v ) )  P  P 
 n               Dn     Gn                             10
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 
 θ ( v )    P( x ( v ) ) 
θ
                                               f (x( v ) )
          Q    V 
(v)

(v)
   Q( x ) 

(v)

   0
          V    
Then the problem can be decoupled
( v ) 1                                   ( v ) 1
 P                             (v)       Q      
θ   (v)
           P( x ( v ) ),   V                     Q( x ( v ) )
 θ                                       V             11
Off-diagonal Jacobian Terms
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       0
 Vj
Qi
  Vi V j  Gij cos ij  Bij sin  ij   0
θ j

12
Decoupled N-R Region of
Convergence

13
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.           14
FDPF Approximations
The FDPF makes the following approximations:
1.      Gij  0
2.      Vi           1 (for some occurrences),
3.      sin ij  0          cos ij  1
Then: θ( v )  B 1diag{| V |( v ) }1 P( x ( v ) ),
 B 1diag{| V |( v ) }1 Q( x ( v ) )
(v)
V
Where B is just the imaginary part of the Ybus  G  jB,
except the slack bus row/column are omitted.
Sometimes approximate diag{| V |( v ) } by identity.             15
FDPF Three Bus Example
Use the FDPF to solve the following three bus system
Line Z = j0.07

One                                    Two

200 MW
100 MVR
Line Z = j0.05      Line Z = j0.1

Three                  1.000 pu

200 MW
100 MVR               34.3 14.3 20 
Ybus    j  14.3 24.3 10 
                
 20
        10  30 

16
FDPF Three Bus Example, cont’d
 34.3 14.3        20 
Ybus  j  14.3 24.3 10   B   24.3 10 
                                    10     30 
            
 20
           10     30 

1        0.0477 0.0159 
B        
 0.0159 0.0389   
Iteratively solve, starting with an initial voltage guess
(0)
 2 
(0)
0           V 2              1
                        V               1
 3            0            3               
 2 
(1)
0   0.0477 0.0159   2   0.1272 
                                  2    0.1091
 3             0   0.0159 0.0389                
17
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      k 1                                 Vi

 2 (2)  0.1272   0.0477 0.0159  0.151  0.1361
    0.1091   0.0159 0.0389  0.107    0.1156
 3                                                 
(2)
V 2            0.924 
V                   
 3              0.936
 0.1384      0.9224 
Actual solution: θ               V
 0.1171      0.9338

18
FDPF Region of Convergence

19
“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

20
DC Power Flow Example

  P

21
DC Power Flow 5 Bus Example

One                Five                                     Four             Three
A                                                          A

360 MW                                                                                                       520 MW
MVA                                                        MVA
A

0 Mvar
MVA

slack
0 Mvar

1.000 pu           1.000 pu      A            A
1.000 pu                      80 MW
0.000 Deg         -4.125 Deg     MVA          MVA
-1.997 Deg                     0 Mvar
1.000 pu
0.524 Deg

1.000 pu                   Two
-18.695 Deg

800 MW
0 Mvar

Notice with the dc power flow all of the voltage magnitudes are
1 per unit.
22
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.               23
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

24
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
131.9 MW

124%

One                                   Two

200.0 MW                                              200 MW
68.1 MW         68.1 MW
71.0 MVR                                              100 MVR

Z for all lines = j0.1
Three           1.000 pu

0 MW
64 MVR                25
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                                     Two

105.0 MW                                               200 MW
3.4 MW             98.4 MW
64.3 MVR                                               100 MVR

92%
Z for all lines = j0.1
Limit for all lines = 150 MVA
1.000 pu
Three
95 MW
64 MVR
26
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
Vi V j                        i   j
Pij              sin( i   j ) 
X ij                           X ij
 i   j                                   ij
So Pij                           We just need to get
X ij                                 PGk
27
Analytic Sensitivities
From the fast decoupled power flow we know
θ  B 1P (x)
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 .
0
 
 
P       1  Bus k
0
 
 
                                   28
Three Bus Sensitivity Example
For the previous three bus case with Zline  j 0.1
 20 10 10 
Ybus  j  10 20 10   B   20 10 
                        10 20 
        
 10 10 20 
               
Hence for a change of generation at bus 3
1
  2       20 10   0   0.0333
                    1   0.0667 
 3          10 20                  
0.0667  0
Then P3 to 1               0.667 pu
0.1
P3 to 2  0.333 pu     P 2 to 1  0.333 pu         29

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 16 posted: 11/23/2011 language: English pages: 29