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1. Lecture 13
Newton-Raphson Power Flow
Professor Tom Overbye
Department of Electrical and
Computer Engineering
ECE 476
POWER SYSTEM ANALYSIS

2. 1
Announcements
 Homework 6 is 2.38, 6.8, 6.23, 6.28; you should
do it before the exam but need not turn it in.
Answers have been posted.
 First exam is 10/9 in class; closed book, closed
notes, one note sheet and calculators allowed. Last
year’s tests and solutions have been posted.
 Abbott power plant and substation field trip,
Tuesday 10/14 starting at 12:30pm. We’ll meet at
corner of Gregory and Oak streets.
 Be reading Chapter 6; exam covers up through
Section 6.4; we do not explicitly cover 6.1.

3. 2
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

4. 3
Two Bus Newton-Raphson Example
Line Z = 0.1j
One Two
1.000 pu 1.000 pu
200 MW
100 MVR
0 MW
0 MVR
For the two bus power system shown below, use the
Newton-Raphson power flow to determine the
voltage magnitude and angle at bus two. Assume
that bus one is the slack and SBase = 100 MVA.
2
2
10 10
10 10
bus
j j
V j j
 
   
 
   

 
 
x Y

5. 4
Two Bus Example, cont’d
i
1
i
1
2 1 2
2
2 1 2 2
General power balance equations
P ( cos sin )
Q ( sin cos )
Bus two power balance equations
(10sin ) 2.0 0
( 10cos ) (10) 1.0 0
n
i k ik ik ik ik Gi Di
k
n
i k ik ik ik ik Gi Di
k
V V G B P P
V V G B Q Q
V V
V V V
 
 




   
   
 
   



6. 5
Two Bus Example, cont’d
2 2 2
2
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 0
Now calculate the power flow Jacobian
P ( ) P ( )
( )
Q ( ) Q ( )
10 cos 10sin
10 sin 10cos 20
V
Q V V
V
J
V
V
V V




 
 
  
    
 
 
 
 
 

 
 
 
 
 
 
  
 
 
x
x
x x
x
x x

7. 6
Two Bus Example, First Iteration
(0)
2 2
(0)
2
2 2 2
2 2 2
(0)
2 2 2 2
(1)
0
Set 0, guess
1
Calculate
(10sin ) 2.0 2.0
f( )
1.0
( 10cos ) (10) 1.0
10 cos 10sin 10 0
( )
10 sin 10cos 20 0 10
0 10 0
Solve
1 0 10
v
V
V V
V
V V


 
 
 
   
 

   
 
   
    
 
 
   
 
   
   
 
  
 
 
  
x
x
J x
x
1
2.0 0.2
1.0 0.9


    

     
    

8. 7
Two Bus Example, Next Iterations
(1)
2
(1)
1
(2)
0.9(10sin( 0.2)) 2.0 0.212
f( )
0.279
0.9( 10cos( 0.2)) 0.9 10 1.0
8.82 1.986
( )
1.788 8.199
0.2 8.82 1.986 0.212 0.233
0.9 1.788 8.199 0.279 0.8586
f(

 
   
 
   
      
 

 
  

 
  
       
  
       

       
x
J x
x
(2) (3)
(3)
2
0.0145 0.236
)
0.0190 0.8554
0.0000906
f( ) Done! V 0.8554 13.52
0.0001175

   
 
   
   
 
    
 
 
x x
x

9. 8
Two Bus Solved Values
Line Z = 0.1j
One Two
1.000 pu 0.855 pu
200 MW
100 MVR
200.0 MW
168.3 MVR
-13.522 Deg
200.0 MW
168.3 MVR
-200.0 MW
-100.0 MVR
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

10. 9
Two Bus Case Low Voltage Solution
(0)
2 2
(0)
2
2 2 2
This case actually has two solutions! The second
"low voltage" is found by using a low initial guess.
0
Set 0, guess
0.25
Calculate
(10sin ) 2.0
f( )
( 10cos ) (10) 1.0
v
V
V V


 
   
 

 
  
  
 
 
x
x
2 2 2
(0)
2 2 2 2
2
0.875
10 cos 10sin 2.5 0
( )
10 sin 10cos 20 0 5
V
V V
 
 
 
  

 
   
 
   
  
 
 
J x

11. 10
Low Voltage Solution, cont'd
1
(1)
(2) (2) (3)
0 2.5 0 2 0.8
Solve
0.25 0 5 0.875 0.075
1.462 1.42 0.921
( )
0.534 0.2336 0.220


       
  
       
 
       
 
     
  
     
     
x
f x x x
Line Z = 0.1j
One Two
1.000 pu 0.261 pu
200 MW
100 MVR
200.0 MW
831.7 MVR
-49.914 Deg
200.0 MW
831.7 MVR
-200.0 MW
-100.0 MVR
Low voltage solution

12. 11
Two Bus Region of Convergence
Slide shows the region of convergence for different initial
guesses of bus 2 angle (x-axis) and magnitude (y-axis)
Red region
converges
to the high
voltage
solution,
while the
yellow region
converges
to the low
voltage
solution

13. 12
Using the Power Flow: Example 1
slack
SLA CK345
SLA CK138
RA Y345
RA Y138
RA Y69
FERNA 69
A
MVA
DEMA R69
BLT 69
BLT 138
BOB138
BOB69
WOLEN69
SHIMKO69
ROGER69
UIUC69
PET E69
HISKY69
T IM69
T IM138
T IM345
PA I69
GROSS69
HA NNA H69
A MA NDA 69
HOMER69
LA UF69
MORO138
LA UF138
HA LE69
PA T T EN69
WEBER69
BUCKY138
SA VOY69
SA VOY138
JO138 JO345
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
1.02 pu
1.01 pu
1.02 pu
1.03 pu
1.01 pu
1.00 pu
1.00 pu
0.99 pu
1.02 pu
1.01 pu
1.00 pu
1.01 pu
1.01 pu
1.01 pu
1.01 pu
1.02 pu
1.00 pu
1.00 pu
1.02 pu
0.997 pu
0.99 pu
1.00 pu
1.02 pu
1.00 pu
1.01 pu
1.00 pu
1.00 pu 1.00 pu
1.01 pu
1.02 pu
1.02 pu
1.02 pu
1.03 pu
A
MVA
1.02 pu
A
MVA
A
MVA
LYNN138
A
MVA
1.02 pu
A
MVA
1.00 pu
A
MVA
218 MW
54 Mvar
21 MW
7 Mvar
45 MW
12 Mvar
140 MW
45 Mvar
37 MW
13 Mvar
12 MW
5 Mvar
150 MW
0 Mvar
56 MW
13 Mvar
15 MW
5 Mvar
14 MW
2 Mvar
42 MW
2 Mvar
45 MW
0 Mvar
58 MW
36 Mvar
36 MW
10 Mvar
0 MW
0 Mvar
22 MW
15 Mvar
60 MW
12 Mvar
20 MW
30 Mvar
23 MW
7 Mvar
33 MW
13 Mvar
16.0 Mvar 18 MW
5 Mvar
58 MW
40 Mvar
51 MW
15 Mvar
14.3 Mvar
33 MW
10 Mvar
15 MW
3 Mvar
23 MW
6 Mvar 14 MW
3 Mvar
4.8 Mvar
7.2 Mvar
12.8 Mvar
29.0 Mvar
7.4 Mvar
0.0 Mvar
106 MW
8 Mvar
20 MW
8 Mvar
150 MW
0 Mvar
17 MW
3 Mvar
0 MW
0 Mvar
14 MW
4 Mvar
Using
case
from
Example
6.13

14. 13
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
( )
( ) ( ) 0
V ( )
G D
G D
D
P P P
P P P
Q Q


 
   
   
    
   

   
 
 
x
x f x x
x

15. 14
Modeling Voltage Dependent Load
So far we've assumed that the load is independent of
the bus voltage (i.e., constant power). However, the
power flow can be easily extended to include voltage
depedence with both the real and reactive l
Di Di
1
1
oad. This
is done by making P and Q a function of :
( cos sin ) ( ) 0
( sin cos ) ( ) 0
i
n
i k ik ik ik ik Gi Di i
k
n
i k ik ik ik ik Gi Di i
k
V
V V G B P P V
V V G B Q Q V
 
 


   
   



16. 15
Voltage Dependent Load Example
2
2 2 2 2
2 2
2 2 2 2 2
2 2 2 2
In previous two bus example now assume the load is
constant impedance, so
P ( ) (10sin ) 2.0 0
( ) ( 10cos ) (10) 1.0 0
Now calculate the power flow Jacobian
10 cos 10sin 4.0
( )
10
V V
Q V V V
V V
J


 
  
    


x
x
x
2 2 2 2 2
sin 10cos 20 2.0
V V V
 
 
 
  
 

17. 16
Voltage Dependent Load, cont'd
(0)
2
2 2 2
(0)
2 2
2 2 2 2
(0)
1
(1)
0
Again set 0, guess
1
Calculate
(10sin ) 2.0 2.0
f( )
1.0
( 10cos ) (10) 1.0
10 4
( )
0 12
0 10 4 2.0 0.1667
Solve
1 0 12 1.0 0.9167
v
V V
V V V



 
   
 
 
  
 
   
 
 
  
 
 
  
 

      
  
     
     
x
x
J x
x

 
 

18. 17
Voltage Dependent Load, cont'd
Line Z = 0.1j
One Two
1.000 pu
0.894 pu
160 MW
80 MVR
160.0 MW
120.0 MVR
-10.304 Deg
160.0 MW
120.0 MVR
-160.0 MW
-80.0 MVR
With constant impedance load the MW/Mvar load at
bus 2 varies with the square of the bus 2 voltage
magnitude. This if the voltage level is less than 1.0,
the load is lower than 200/100 MW/Mvar

19. 18
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

20. 19
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

21. 20
Dishonest Newton-Raphson
 Since most of the time in the Newton-Raphson
iteration is spend calculating the inverse of the
Jacobian, one way to speed up the iterations is to
only calculate/inverse the Jacobian occasionally
– known as the “Dishonest” Newton-Raphson
– an extreme example is to only calculate the Jacobian for
the first iteration
( 1) ( ) ( ) -1 ( )
( 1) ( ) (0) -1 ( )
( )
Honest: - ( ) ( )
Dishonest: - ( ) ( )
Both require ( ) for a solution
v v v v
v v v
v






x x J x f x
x x J x f x
f x

22. 21
Dishonest Newton-Raphson Example
2
1
(0)
( ) ( )
( ) ( ) 2
(0)
( 1) ( ) ( ) 2
(0)
Use the Dishonest Newton-Raphson to solve
( ) - 2 0
( )
( )
1
(( ) - 2)
2
1
(( ) - 2)
2
v v
v v
v v v
f x x
df x
x f x
dx
x x
x
x x x
x


 
 
   
 
 
   
 
 
   
 

23. 22
Dishonest N-R Example, cont’d
( 1) ( ) ( ) 2
(0)
(0)
( ) ( )
1
(( ) - 2)
2
Guess x 1. Iteratively solving we get
v (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
  
   
 

We pay a price
in increased
iterations, but
with decreased
computation
per iteration

24. 23
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

25. 24
Honest N-R Region of Convergence
Maximum
of 15
iterations

26. 25
Decoupled Power Flow
 The completely Dishonest Newton-Raphson is not
used for power flow analysis. However several
approximations of the Jacobian matrix are used.
 One common method is the decoupled power flow.
In this approach approximations are used to
decouple the real and reactive power equations.

27. 26
Decoupled Power Flow Formulation
( ) ( )
( ) ( )
( )
( )
( ) ( ) ( )
( )
2 2 2
( )
( )
General form of the power flow problem
( )
( )
( )
where
( )
( )
( )
v v
v v
v
v
v v v
v
D G
v
v
n Dn Gn
P P P
P P P
 
 
     

  
 
  
   
  
  
     
 
 
 
 
 
 
 
 
   
 
 
 
P P
θ
θ V P x
f x
Q x
V
Q Q
θ V
x
P x
x

28. 27
Decoupling Approximation
( ) ( )
( )
( ) ( )
( )
( ) ( ) ( )
Usually the off-diagonal matrices, and
are small. Therefore we approximate them as zero:
( )
( )
( )
Then the problem
v v
v
v v
v
v v v
 
 
 

     
 

 
  
   
 
 
  
   
 
 

 
P Q
V θ
P
0
θ P x
θ
f x
Q Q x
V
0
V
1 1
( ) ( )
( )
( ) ( ) ( )
can be decoupled
( ) ( )
v v
v
v v v
 
   
 
       
   
 
   
P Q
θ P x V Q x
θ V

29. 28
Off-diagonal Jacobian Terms
 
 
Justification for Jacobian approximations:
1. Usually r x, therefore
2. Usually is small so sin 0
Therefore
cos sin 0
cos sin 0
ij ij
ij ij
i
i ij ij ij ij
j
i
i j ij ij ij ij
j
G B
V G B
V V G B
 
 
 


  


   

P
V
Q
θ

30. 29
Decoupled N-R Region of Convergence

31. 30
Fast Decoupled Power Flow
 By continuing with our Jacobian approximations we
can actually obtain a reasonable approximation that
is independent of the voltage magnitudes/angles.
 This means the Jacobian need only be built/inverted
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
 The FDPF is widely used, particularly when we
only need an approximate solution

32. 31
FDPF Approximations
ij
( ) ( )
( )
( ) 1 1
( ) ( )
bus
The FDPF makes the following approximations:
1. G 0
2. 1
3. sin 0 cos 1
Then
( ) ( )
Where is just the imaginary part of the ,
except the slack bus row/co
i
ij ij
v v
v
v
v v
V
j
 
 


 
 
   
 
P x Q x
θ B V B
V V
B Y G B
lumn are omitted

33. 32
FDPF Three Bus Example
Line Z = j0.07
Line Z = j0.05 Line Z = j0.1
One Two
200 MW
100 MVR
Three 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

34. 33
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




 

 
 
     
  
 

 
 
 
 
  
 
 
 
    
 
 
   
 
   
Y B
B
(1)
2
3
0 0.0477 0.0159 2 0.1272
0 0.0159 0.0389 2 0.1091



 
 
  
         
  
         
  
       
 

35. 34
FDPF Three Bus Example, cont’d
(1)
2
3
i
i i
1
(2)
2
3
1 0.0477 0.0159 1 0.9364
1 0.0159 0.0389 1 0.9455
P ( )
( cos sin )
V V
0.1272 0.0477 0.0159
0.1091 0.0159 0.0389
n
Di Gi
k ik ik ik ik
k
V
V
P P
V G B
 



 
         
  
         
 
       
 


  
  
     
 
     
  
   
 

x
(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

36. 35
FDPF Region of Convergence

37. 36
“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
1


θ B P

38. 37
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 valves to close to limit flow)
– 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

39. 38
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

40. 39
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 = j0.1
One Two
200 MW
100 MVR
200.0 MW
71.0 MVR
Three 1.000 pu
0 MW
64 MVR
131.9 MW
68.1 MW 68.1 MW
124%

41. 40
Power Flow Simulation - After
Z for all lines = j0.1
Limit for all lines = 150 MVA
One Two
200 MW
100 MVR
105.0 MW
64.3 MVR
Three
1.000 pu
95 MW
64 MVR
101.6 MW
3.4 MW 98.4 MW
92%
100%
Increasing the generation at bus 3 by 95 MW (and hence
decreasing it at bus 1 by a corresponding amount), results
in a 31.3 drop in the MW flow on the line from bus 1 to 2.

42. 41
Analytic Calculation of Sensitivities
 Calculating control sensitivities by repeat 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
 
 
  

  
   
 


43. 42
Analytic Sensitivities
1
From the fast decoupled power flow we know
( )
So to get the change in due to a change of
generation at bus k, just set ( ) equal to
all zeros except a minus one at position k.
0
1
0

  



  

θ B P x
θ
P x
P Bus k

 
 

 
 
 
 


44. 43
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





 

 
 
     
  
 

 
 
 
      
 
      
  
    
 
Y B
3 to 1
3 to 2 2 to 1
0.0667 0
Then P 0.667 pu
0.1
P 0.333 pu P 0.333 pu




  
   