Electronics and communication engineering_Lect14.ppt

Lecture 14
Power Flow
Professor Tom Overbye
Department of Electrical and
Computer Engineering
ECE 476
POWER SYSTEM ANALYSIS

1
Announcements
 Be reading Chapter 6, also Chapter 2.4 (Network Equations).
 HW 5 is 2.38, 6.9, 6.18, 6.30, 6.34, 6.38; do by October 6
but does not need to be turned in.
 First exam is October 11 during class. Closed book, closed
notes, one note sheet and calculators allowed. Exam covers
thru the end of lecture 13 (today)
 An example previous exam (2008) is posted. Note this is
exam was given earlier in the semester in 2008 so it did not
include power flow, but the 2011 exam will (at least
partially)

2
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
 
 


   
   



3
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
 
 
 
  
 

4
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

 
 

5
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

6
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

7
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


 
 
   
 
 
   
 
 
   
 

8
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

9
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

10
Honest N-R Region of Convergence
Maximum
of 15
iterations

11
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.

12
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

13
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

14
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

15
Decoupled N-R Region of Convergence

16
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

17
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

18
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

19
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



 
 
  
         
  
         
  
       
 

20
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

22
“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

23
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

26
DC Power Flow 5 Bus Example
slack
One
Two
Three
Four
Five
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
1.000 pu 1.000 pu
1.000 pu
1.000 pu
1.000 pu
0.000 Deg -4.125 Deg
-18.695 Deg
-1.997 Deg
0.524 Deg
360 MW
0 Mvar
520 MW
0 Mvar
800 MW
0 Mvar
80 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
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%

29
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.

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
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
 
 
  

  
   
 


31
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

 
 

 
 
 
 


32
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




  
   

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

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
 Ideally the ACE should always be zero.
 Because the load is constantly changing, each
utility must constantly change its generation
to “chase” the ACE.
int sched
ace 10
P P f

   

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
PTDFs
 Power transfer distribution factors (PTDFs) show
the linear impact of a transfer of power.
 PTDFs calculated using the fast decoupled power
flow B matrix
1
( )
Once we know we can derive the change in
the transmission line flows
Except now we modify several elements in ( ),
in portion to how the specified generators would
participate in the pow

  


θ B P x
θ
P x
er transfer

38
Nine Bus PTDF Example
10%
60%
55%
64%
57%
11%
74%
24%
32%
A
G
B
C
D
E
I
F
H
300.0 MW
400.0 MW 300.0 MW
250.0 MW
250.0 MW
200.0 MW
250.0 MW
150.0 MW
150.0 MW
44%
71%
0.00 deg
71.1 MW
92%
Figure shows initial flows for a nine bus power system

39
Nine Bus PTDF Example, cont'd
43%
57%
13%
35%
20%
10%
2%
34%
34%
32%
A
G
B
C
D
E
I
F
H
300.0 MW
400.0 MW 300.0 MW
250.0 MW
250.0 MW
200.0 MW
250.0 MW
150.0 MW
150.0 MW
34%
30%
0.00 deg
71.1 MW
Figure now shows percentage PTDF flows from A to I

40
Nine Bus PTDF Example, cont'd
6%
6%
12%
61%
12%
6%
19%
21%
21%
A
G
B
C
D
E
I
F
H
300.0 MW
400.0 MW 300.0 MW
250.0 MW
250.0 MW
200.0 MW
250.0 MW
150.0 MW
150.0 MW
20%
18%
0.00 deg
71.1 MW
Figure now shows percentage PTDF flows from G to F

42
Line Outage Distribution Factors (LODFS)
 LODFs are used to approximate the change in the
flow on one line caused by the outage of a second
line
– typically they are only used to determine the change in
the MW flow
– LODFs are used extensively in real-time operations
– LODFs are state-independent but do dependent on the
assumed network topology
,
l l k k
P LODF P
 

43
Flowgates
 The real-time loading of the power grid is accessed
via “flowgates”
 A flowgate “flow” is the real power flow on one or
more transmission element for either base case
conditions or a single contingency
– contingent flows are determined using LODFs
 Flowgates are used as proxies for other types of
limits, such as voltage or stability limits
 Flowgates are calculated using a spreadsheet

44
NERC Regional Reliability Councils
NERC
is the
North
American
Electric
Reliability
Council

45
NERC Reliability Coordinators
Source: http://www.nerc.com/page.php?cid=5%7C67%7C206

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