ECE 476
POWER SYSTEM ANALYSIS
Lecture 13
Newton-Raphson Power Flow
Professor Tom Overbye
Department of Electrical and
Computer Engineering
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.
1
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
2
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. Assume
that bus one is the slack and SBase = 100 MVA.
Line Z = 0.1j
One
1.000 pu
0 MW
0 MVR
2 
x   
 V2 
Ybus
Two
1.000 pu
200 MW
100 MVR
  j10 j10 
 

j
10

j
10


3
Two Bus Example, cont’d
General power balance equations
Pi 
Qi 
n
 Vi Vk (Gik cosik  Bik sin ik )  PGi  PDi
k 1
n
 Vi Vk (Gik sin ik  Bik cosik )  QGi  QDi
k 1
Bus two power balance equations
V2 V1 (10sin  2 )  2.0  0
V2 V1 ( 10cos  2 )  V2 (10)  1.0  0
2
4
Two Bus Example, cont’d
P2 (x)  V2 (10sin  2 )  2.0  0
Q2 (x)  V2 (10 cos 2 )  V2 (10)  1.0  0
2
Now calculate the power flow Jacobian
 P2 (x)
 
2

J ( x) 
 Q 2 (x)
 
2

P2 ( x) 
V 2 

Q 2 ( x) 
 V 2 
10 V2 cos 2
 
10 V2 sin  2
10sin  2

10 cos 2  20 V2 
5
Two Bus Example, First Iteration
Set v  0, guess x
(0)
0 
 
1 
Calculate
V2 (10sin  2 )  2.0


f(x )  
 
2
 V2 (10 cos 2 )  V2 (10)  1.0 
10sin  2
10 V2 cos 2

(0)
J (x )  


10 V2 sin  2 10 cos 2  20 V2 
(0)
1
Solve x
(1)
0  10 0   2.0 
 
 1.0 
1
0
10
  
  
 2.0 
1.0 
 
10 0 
 0 10 


 0.2 
 

0.9


6
Two Bus Example, Next Iterations
0.9 (10sin(0.2))  2.0

 0.212 
f(x )  


2
0.279

0.9(10 cos(0.2))  0.9  10  1.0  
 8.82 1.986 
(1)
J (x )  


1.788
8.199


(1)
1
 0.2   8.82 1.986  0.212 
 0.233
(2)
x 

 





0.9

1.788
8.199
0.279
0.8586

 
 



 0.0145
 0.236 
(2)
(3)
f(x )  
x
 


0.0190
0.8554




0.0000906 
f(x )  

 0.0001175
(3)
Done!
V2  0.8554  13.52
7
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
200.0 MW
168.3 MVR
One
-200.0 MW
-100.0 MVR
Line Z = 0.1j
1.000 pu
200.0 MW
168.3 MVR
Two
0.855 pu -13.522 Deg
200 MW
100 MVR
8
Two Bus Case Low Voltage Solution
This case actually has two solutions! The second
"low voltage" is found by using a low initial guess.
Set v  0, guess x
(0)
 0 


0.25


Calculate
V2 (10sin  2 )  2.0


f(x )  
 
2
 V2 (10cos 2 )  V2 (10)  1.0 
10sin  2
10 V2 cos 2

(0)
J (x )  


10 V2 sin  2 10cos 2  20 V2 
(0)
 2 
 0.875


 2.5 0 
 0 5


9
Low Voltage Solution, cont'd
1
 0   2.5 0   2 
 0.8 
Solve x  

 





0.25
0

5

0.875
0.075

 
 



1.462  (2)  1.42 
 0.921
(2)
(3)
f (x )  
x 
x 



0.534
0.2336
0.220






(1)
Low voltage solution
200.0 MW
831.7 MVR
One
-200.0 MW
-100.0 MVR
Line Z = 0.1j
1.000 pu
200.0 MW
831.7 MVR
Two
0.261 pu -49.914 Deg
200 MW
100 MVR
10
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
11
Using the Power Flow: Example 1
SLA C K3 4 5
A
MVA
A
MVA
1 .0 2 pu
2 1 8 MW
5 4 M var
RA Y 3 4 5
sla ck
1 .0 2 pu
T IM 3 4 5
A
A
MVA
MVA
A
SLA C K1 3 8
1 .0 1 pu
A
MVA
RA Y 1 3 8
A
3 3 MW
1 3 M var
A
MVA
A
1 6 .0 M var
1 .0 2 pu
MVA
1 8 MW
5 M var
MVA
MVA
1 .0 2 pu
RA Y 6 9
T IM 6 9
P A I6 9
1 .0 1 pu
MVA
A
2 3 MW
7 M var
1 .0 1 pu
3 7 MW
1 7 MW
3 M var
A
1 .0 2 pu
A
MVA
T IM 1 3 8
A
Using
case
from
Example
6.13
1 .0 3 pu
A
MVA
1 .0 0 pu
GRO SS6 9
A
1 3 M var
MVA
A
MVA
FERNA 6 9
MVA
A
MVA
M O RO 1 3 8
MVA
H ISKY 6 9
MVA
A
1 2 MW
5 M var
A
A
2 0 MW
8 M var
1 .0 0 pu
A
1 .0 0 pu
P ET E6 9
DEM A R6 9
MVA
H A NNA H 6 9
5 1 MW
1 5 M var
5 8 MW
4 0 M var
2 9 .0 M var
1 4 .3 M var
1 .0 0 pu
A
MVA
MVA
BO B6 9
1 4 0 MW
4 5 M var
A
MVA
5 6 MW
MVA
5 8 MW
3 6 M var
MVA
0 .9 9 pu
A
1 4 MW
4 M var
MVA
A
MVA
MVA
1 .0 1 pu
A
H A LE6 9
MVA
A
BLT 6 9
1 .0 1 pu
A
1 5 MW
3 M var
1 .0 0 pu
BLT 1 3 8
1 .0 0 pu
LY NN1 3 8
A
3 3 MW
1 0 M var
MVA
A
A
MVA
A M A NDA 6 9
1 3 M var
0 MW
0 M var
MVA
A
0 .9 9 7 pu
A
MVA
1 .0 2 pu
A
MVA
MVA
MVA
1 2 .8 M var
A
BO B1 3 8
A
4 5 MW
1 2 M var
0 .9 9 pu
UIUC 6 9
A
H O M ER6 9
WO LEN6 9
4 .8 M var
MVA
1 .0 0 pu
1 .0 1 pu
2 1 MW
7 M var
A
SH IM KO 6 9
7 .4 M var
1 .0 2 pu
A
MVA
MVA
MVA
1 0 6 MW
8 M var
A
1 5 MW
5 M var
A
MVA
3 6 MW
1 0 M var
A
A
6 0 MW
1 2 M var
MVA
MVA
A
MVA
A
1 .0 0 pu
0 .0 M var
1 .0 1 pu
MVA
A
7 .2 M var
1 .0 0 pu
P A T T EN6 9
MVA
MVA
A
MVA
1 .0 0 pu
LA UF6 9
1 .0 2 pu
2 0 MW
3 0 M var
1 .0 0 pu
A
A
MVA
MVA
2 3 MW
6 M var
2 2 MW
1 5 M var
0 MW
0 M var
LA UF1 3 8
1 .0 1 pu
4 5 MW
0 M var
WEBER6 9
1 .0 2 pu
BUC KY 1 3 8
RO GER6 9
2 M var
1 4 MW
3 M var
A
MVA
SA V O Y 6 9
1 .0 2 pu
A
4 2 MW
2 M var
JO 1 3 8
MVA
A
MVA
1 4 MW
1 .0 1 pu
A
MVA
SA V O Y 1 3 8
JO 3 4 5
A
1 5 0 MW
0 M var
MVA
A
MVA
1 5 0 MW
0 M var
A
MVA
1 .0 2 pu
A
MVA
1 .0 3 pu
12
Three Bus PV Case Example
For this three bus case we have
 2 
x   3 
 
 V2 
 P2 (x)  PG 2  PD 2 
f (x)   P3 (x)  PG 3  PD3   0


 Q2 (x)  QD 2 
Line Z = 0.1j
0.941 pu
One
170.0 MW
68.2 MVR
1.000 pu
Line Z = 0.1j
Three
Two
Line Z = 0.1j
-7.469 Deg
200 MW
100 MVR
1.000 pu
30 MW
63 MVR
13
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 load. This
is done by making PDi and Q Di a function of Vi :
n
 Vi
Vk (Gik cos ik  Bik sin  ik )  PGi  PDi ( Vi )  0
 Vi
Vk (Gik sin  ik  Bik cos ik )  QGi  QDi ( Vi )  0
k 1
n
k 1
14
Voltage Dependent Load Example
In previous two bus example now assume the load is
constant impedance, so
P2 (x)  V2 (10sin  2 )  2.0 V2
2
 0
Q2 (x)  V2 (10 cos 2 )  V2 (10)  1.0 V2  0
2
2
Now calculate the power flow Jacobian
10 V2 cos 2
J ( x)  
10 V2 sin  2
10sin  2  4.0 V2

10 cos 2  20 V2  2.0 V2 
15
Voltage Dependent Load, cont'd
Again set v  0, guess x
(0)
0 
 
1 
Calculate
2


V
(10sin

)

2.0
V
 2.0 
2
2
2
(0)
f(x )  
  
2
2
1.0 
 V2 (10 cos 2 )  V2 (10)  1.0 V2 
10 4 
(0)
J (x )  

0
12


1
Solve x
(1)
0  10 4   2.0 
 
 1.0 
1
0
12
  
  
 0.1667 
 

0.9167


16
Voltage Dependent Load, cont'd
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
160.0 MW
120.0 MVR
One
-160.0 MW
-80.0 MVR
Line Z = 0.1j
1.000 pu
160.0 MW
120.0 MVR
Two
0.894 pu
-10.304 Deg
160 MW
80 MVR
17
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
18
Newton-Raphson Power Flow

Advantages
–
–

Disadvantages
–
–

fast convergence as long as initial guess is close to
solution
large region of convergence
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
19
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
Honest:
x(v 1)  x( v ) - J (x( v ) )-1f (x( v ) )
Dishonest: x(v 1)  x( v ) - J (x(0) )-1 f (x( v ) )
Both require f (x(v ) )   for a solution
20
Dishonest Newton-Raphson Example
Use the Dishonest Newton-Raphson to solve
f ( x)  x 2 - 2  0
x
(v )
x ( v )
x ( v 1)
1
 df ( x ) 
(v)
 
f
(
x
)

 dx 
1  (v) 2

   (0)  (( x ) - 2)
2x 
(v)  1 
 x   (0)  (( x ( v ) )2 - 2)
2x 
(0)
21
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
v
0
1
2
3
4
x (v ) (honest)
1
1.5
1.41667
1.41422
1.41422
x ( v ) (dishonest)
1
1.5
1.375
1.429
1.408
We pay a price
in increased
iterations, but
with decreased
computation
per iteration
22
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
23
Honest N-R Region of Convergence
Maximum
of 15
iterations
24
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.
25
Decoupled Power Flow Formulation
General form of the power flow problem
 P ( v )

θ


 Q ( v )

 θ
P
V
(v) 

(v) 
(v) 


θ

P
(
x
)
(v )



f
(
x
)



(
v
)
Q ( v )    V   Q(x( v ) ) 

 V 
where
 P2 (x( v ) )  PD 2  PG 2 


(v)
P (x )  

 P (x(v ) )  P  P 
 n
Dn
Gn 
26
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 ) 

(v) 



θ

P
(
x
)

θ
(v)

 


f
(
x
)



(
v
)
(
v
)
(v )

Q    V 

Q
(
x
) 


0



V


Then the problem can be decoupled
θ
(v)
 P
 
 θ
( v )  1
 P (x

(v)
) V
(v)
 Q
 
 V
( v )  1
(v)

Q
(
x
)


27
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 
 Vj
Qi
θ j
 0
  Vi V j  Gij cos ij  Bij sin  ij   0
28
Decoupled N-R Region of Convergence
29
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
30
FDPF Approximations
The FDPF makes the following approximations:
1.
G ij  0
2.
Vi
3.
sin  ij  0
 1
cos ij  1
Then
(v )
(v )

P
(
x
)

Q
(
x
)
(v)
(v )
1
1
θ  B
 V B
(v)
(v)
V
V
Where B is just the imaginary part of the Ybus  G  jB,
except the slack bus row/column are omitted
31
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
20 
 34.3 14.3
 j  14.3 24.3 10 


10
30 
 20
32
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


20
10

30


 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
  
  

33
FDPF Three Bus Example, cont’d
V 2 
V 
 3
(1)

1  0.0477 0.0159  1  0.9364
1   0.0159 0.0389  1   0.9455
 
  

PDi  PGi
Pi (x ) n
  Vk (Gik cos ik  Bik sin  ik ) 
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


34
FDPF Region of Convergence
35
“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
36
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
37
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
38
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
39
Power Flow Simulation - After
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.
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
40
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
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
PGk
41
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.
P
0
 
 
  1  Bus k
0
 
 
42
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
Then P3 to 1 
 0.667 pu
0.1
P3 to 2  0.333 pu
P 2 to 1  0.333 pu
43