ECE 476
Power System Analysis
Lecture 14: Power Flow
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
overbye@illinois.edu
Announcements
• Read Chapter 6, Chapter 12.4 and 12.5
• Quiz today on HW 6
• HW 7 is 6.50, 6.52, 6.59, 12.20, 12.26; due October
22 in class (no quiz)
• Power and Energy scholarships will be decided on
Monday; application on website; apply to Prof. Sauer;
Grainger Awards due on Nov 1; application on
website; apply to Prof. Sauer
•
energy.ece.illinois.edu
1
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
2
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
3
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
4
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 
5
Voltage Dependent Load, cont'd
Again set v  0, guess x
(0)
0 
 
1 
Calculate
2


V2 (10sin  2 )  2.0 V2
 2.0 
(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


6
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
7
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)
( v ) -1
(v)
- J (x ) f (x )
Dishonest: x(v 1)  x( v ) - J (x(0) )-1 f (x( v ) )
Both require f (x(v ) )   for a solution
8
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)
9
Dishonest N-R Example, cont’d
x
( v 1)
 x
(v)
1  (v) 2

  (0)  (( x ) - 2)
2x 
We pay a price
Guess x  1. Iteratively solving we get in increased
iterations, but
v
x (v ) (honest)
x ( v ) (dishonest)
with
decreased
0
1
1
computation
1
1.5
1.5
per iteration
2
1.41667
1.375
(0)
3
4
1.41422
1.41422
1.429
1.408
10
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
11
Honest N-R Region of Convergence
Maximum
of 15
iterations
12
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.
13
Decoupled Power Flow Formulation
General form of the power flow problem
 P ( v )

θ


 Q ( v )

 θ
P ( v ) 

(v) 
 P (x( v ) ) 
 V  θ
(v )


f
(
x
)




(v)
(v) 
(v)

Q
(
x
) 

V

Q

 

 V 
where
 P2 (x( v ) )  PD 2  PG 2 


(v)
P (x )  

 P (x(v ) )  P  P 
 n
Dn
Gn 
14
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
)


15
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
16
Decoupled N-R Region of Convergence
17
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
18
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
19
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
200 MW
100 MVR
Line Z = j0.1
1.000 pu
200 MW
100 MVR
Ybus
20 
 34.3 14.3
 j  14.3 24.3 10 


10
30 
 20
20
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
 2 
 
 3
(0)
0
 
0 
V 2 
V 
 3
(0)

1
1

(1)

 2
0   0.0477 0.0159   2   0.1272 
   0    0.0159 0.0389   2    0.1091
  
  

 3
21
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 
 
 3
(2)
V 2 
V 
 3
 0.1272   0.0477 0.0159   0.151  0.1361









0.1091

0.0159

0.0389
0.107

0.1156

 

 

(2)
0.924 
 

0.936 
 0.1384 
Actual solution: θ  

 0.1171
0.9224 
V

 0.9338
22
FDPF Region of Convergence
23
“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
24
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
25
DC Power Flow Example
26
DC Power Flow 5 Bus Example
One
360 MW
0 Mvar
A
Five
Four
MVA
A
Three
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.
27
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
28
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
29
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
30
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
31
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
 
 
32
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
33
Balancing Authority Areas
• An balancing authority area (use to be called
operating areas) has traditionally represented
the portion of the interconnected electric grid
operated by a single utility
• Transmission lines that join two areas are
known as tie-lines.
• The net power out of an area is the sum of
the flow on its tie-lines.
• The flow out of an area is equal to
total gen - total load - total losses = tie-flow
34
Area Control Error (ACE)
• The area control error (ace) is the difference
between the actual flow out of an area and
the scheduled flow, plus a frequency
component
ace
 Pint  Psched  10 f
• Ideally the ACE should always be zero.
• Because the load is constantly changing,
each utility must constantly change its
generation to “chase” the ACE.
35
Automatic Generation Control
• Most utilities use automatic generation
control (AGC) to automatically change their
generation to keep their ACE close to zero.
• Usually the utility control center calculates
ACE based upon tie-line flows; then the
AGC module sends control signals out to the
generators every couple seconds.
36
Power Transactions
• Power transactions are contracts between
generators and loads to do power
transactions.
• Contracts can be for any amount of time at
any price for any amount of power.
• Scheduled power transactions are
implemented by modifying the value of
Psched used in the ACE calculation
37