This document discusses the Newton-Raphson power flow method. It begins with announcements about homework assignments. It then provides an overview of the dishonest or shamanskii Newton-Raphson method, which calculates the Jacobian less frequently than the honest method to reduce computation time. An example is shown comparing the two methods. The document also discusses decoupled and fast decoupled power flow methods, which make additional approximations to further reduce computation time. It concludes with a brief discussion of power system control and indirect methods of controlling transmission line flows.

1. EE 369
POWER SYSTEM ANALYSIS
Lecture 13
Newton-Raphson Power Flow
Tom Overbye and Ross Baldick
1

2. 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 (note in Ybus in problem 6.34 should have
Y32 = Y23 = j5, not j2 as stated), 6.43, 6.46, 6.49,
6.50; due Tuesday 11/24. Note that HW is
due on Tuesday because Thanksgiving is on 2

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. ( 1) ( ) ( ) -1 ( )
( 1) ( ) (0) -1 ( )
( )
Honest: - ( ) ( )
Dishonest: - ( ) ( )
Stopping criterion ( ) used in both cases.
v v v v
v v v
v
ε
+
+
=
=
<
x x J x f x
x x J x f x
f x 3

4. Dishonest Newton-Raphson
Example
2
1
( ) (0) ( )
( ) ( ) 2
(0)
( 1) ( ) ( ) 2
(0)
Use the Dishonest Newton-Raphson (chord method)
to solve ( ) 0, where:
( ) - 2
( ) ( )
1
(( ) - 2)
2
1
(( ) - 2)
2
v v
v v
v v v
f x
f x x
df
x x f x
dx
x x
x
x x x
x
−
+
=
=
 ∆ = −
  
 ∆ = −
  
 = −
  
4

5. Dishonest N-R Example, cont’d
( 1) ( ) ( ) 2
(0)
(0)
( ) ( )
1
(( ) - 2)
2
Guess 1. Iteratively solving we get
(honest) (dishonest)
0 1 1
1 1.5 1.5
2 1.41667 1.375
3 1.41422 1.429
4 1.41422 1.408
v v v
v v
x x x
x
x
x xν
+  = −   
=
We pay a price
in increased
iterations, but
with decreased
computation
per iteration
5

6. 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

7. Honest N-R Region of Convergence
Maximum
of 15
iterations
7

8. Decoupled Power Flow
The “completely” Dishonest Newton-
Raphson (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

9. Coupled Newton-Raphson Update
( ) ( )
( ) ( )
( )
( )( ) ( ) ( )
( )
2 2 2
( )
( )
Standard form of the Newton-Raphson update:
( )
( )
( )
( )
where ( ) .
( )
Note
v v
v v
v
vv v v
v
D G
v
v
n Dn Gn
P P P
P P P
 ∂ ∂
    ∆∂ ∂ ∆ − = =   
  ∆∆  ∂ ∂      
∂ ∂  
 + −
 
∆ =  
 + − 
P P
θθ V P x
f x
Q xVQ Q
θ V
x
P x
x
M
that changes in angle and voltage magnitude
both affect (couple to) real and reactive power. 9

10. Decoupling Approximation
( ) ( )
( )
( ) ( )
( )
( ) ( ) ( )
Usually the off-diagonal matrices, and
are small. Therefore we approximate them as zero:
( )
( )
( )
Then the update c
v v
v
v v
v
v v v
∂ ∂
∂ ∂
 ∂
    ∆ ∆∂ − = =   
 ∂ ∆∆      
 ∂ 
P Q
Vθ
P
0
θ P xθ
f x
Q Q xV0
V
1 1( ) ( )
( )( ) ( ) ( )
an be decoupled into two separate updates:
( ), ( ).
v v
vv v v
− −
   ∂ ∂
∆ = − ∆ ∆ = − ∆   
∂ ∂   
P Q
θ P x V Q x
θ V
10

11. Off-diagonal Jacobian Terms
So, angle and real power are coupled closely, and
voltage magnitude and reactive power are coupled cl
Justification for Jacobian approximations:
1. Usually , therefore
2. U
os
su
ely.
ally is s
ij ij
ij
r x G B
θ
= =
( )
( )
mall so sin 0
Therefore
cos sin 0
cos sin 0
ij
i
i ij ij ij ij
j
i
i j ij ij ij ij
j
V G B
V V G B
θ
θ θ
θ θ
≈
∂
= + ≈
∂
∂
= − + ≈
∂
P
V
Q
θ 11

12. Decoupled N-R Region of
Convergence
12

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. 13

14. FDPF Approximations
( ) 1 ( ) 1 ( )
( ) 1 ( ) 1 ( )
The FDPF makes the following approximations:
1. 0
2. 1 (for some occurrences),
3. sin 0 cos 1
Then: {| | } ( ),
{| | } ( )
Where is just the imaginary pa
ij
i
ij ij
v v v
v v v
G
V
diag
diag
θ θ
− −
− −
=
=
= =
∆ = ∆
∆ = ∆
θ B V P x
V B V Q x
B bus bus bus
bus
( )
rt of the ,
except the slack bus row/column are omitted. That is,
is , but with the slack bus row and column deleted.
Sometimes approximate {| | } by identity.v
j
diag
= +Y G B
B B
V 14

15. FDPF Three Bus Example
Line Z = j 0.07
Line Z = j 0.05 Line Z = j 0.1
On e Tw o
200 MW
100 MVR
Th r ee 1.000 pu
200 MW
100 MVR
Use the FDPF to solve the following three bus system
34.3 14.3 20
14.3 24.3 10
20 10 30
bus j
− 
 = −
 
−  
Y
15

16. FDPF Three Bus Example, cont’d
1
(0)(0)
2 2
3 3
34.3 14.3 20
24.3 10
14.3 24.3 10
10 30
20 10 30
0.0477 0.0159
0.0159 0.0389
Iteratively solve, starting with an initial voltage guess
0 1
0 1
bus j
V
V
θ
θ
−
− 
−  = − ⇒ =    − 
−  
− − 
=  − − 
     
= =    
    
Y B
B
(1)
2
3
0 0.0477 0.0159 2 0.1272
0 0.0159 0.0389 2 0.1091
θ
θ

 
 
− − −         
= + =         − − −        
16

17. FDPF Three Bus Example, cont’d
(1)
2
3
1
(2)
2
3
1 0.0477 0.0159 1 0.9364
1 0.0159 0.0389 1 0.9455
( )
( cos sin )
0.1272 0.0477 0.0159
0.1091 0.0159 0.0389
n
i Di Gi
k ik ik ik ik
i ik
V
V
P x P P
V G B
V V
θ θ
θ
θ
=
− −        
= + =        − −       
∆ −
= + +
− − −     
= +     − − −    
∑
(2)
2
3
0.151 0.1361
0.107 0.1156
0.924
0.936
0.1384 0.9224
Actual solution:
0.1171 0.9338
V
V
−   
=   −   
   
=   
  
−   
= =   −   
θ V
17

18. FDPF Region of Convergence
18

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:
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.
1−
= −θ B P
19

20. DC Power Flow Example
20

21. DC Power Flow 5 Bus Example
sla ck
One
Tw o
Thr eeFou rFive
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
1.0 00 p u 1.0 00 p u
1.00 0 pu
1.000 pu
1 .000 pu
0.000 Deg -4.1 25 Deg
-1 8.69 5 Deg
-1.997 Deg
0.524 Deg
360 M W
0 M var
52 0 M W
0 M v ar
8 00 M W
0 M v ar
8 0 M W
0 M v ar
Notice with the dc power flow all of the voltage magnitudes are
1 per unit.
21

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 line22

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
23

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
overload.
Z for all lines = j 0.1
On e Tw o
200 MW
100 MVR
200.0 MW
71.0 MVR
Th r ee 1.000 pu
0 MW
64 MVR
131.9 M W
68.1 M W 68.1 M W
1 2 4 %
24

25. Power Flow Simulation - After
Z for all lines = j 0.1
Lim it for all lines = 150 M VA
On e Tw o
200 MW
100 MVR
105.0 MW
64.3 MVR
Th r ee
1.000 pu
95 MW
64 MVR
101.6 M W
3.4 M W 98.4 M W
92%
1 0 0 %
•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.
25

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
sin( )
So We just need to get
i j i j
ij i j
ij ij
i j ij
ij
ij Gk
V V
P
X X
P
X P
θ θ
θ θ
θ θ θ
−
≈ − ≈
∆ − ∆ ∆
∆ ≈
∆ 26

27. Analytic Sensitivities
1
From the fast decoupled power flow we know: ( ).
Sign convention in definition of ( ) is that entry in ( )
is negative if change in net injection (generation) is positive.
So to get the chan
−
∆ = ∆
∆ ∆
θ B P x
P x P x
ge in due to a change of generation at
bus , just set ( ) equal to all zeros except a minus one
at position :
0
For 1MW increase in generation at bus1
0
k
k
k
∆
∆
 
 
 
∆ = ¬− 
 
 
  
θ
P x
P
M
M 27

28. Three Bus Sensitivity Example
line
bus
1
2
3
For the previous three bus case with Z 0.1
20 10 10
20 10
10 20 10
10 20
10 10 20
Hence for a change of generation at bus 3
20 10 0 0.0333
10 20 1 0.0667
j
j
θ
θ
−
=
− 
−  = − ⇒ =    − 
−  
∆ −      
= =      ∆ − −     
Y B
3 to 1
3 to 2 2 to 1
0.0667 0
Changes in line flows are: 0.667 pu
0.1
0.333 pu 0.333 pu
P
P P



−
∆ = =
∆ = ∆ = 28