ece 476 power system analysis lecture 14: power flow prof. tom overbye dept. of electrical and...
TRANSCRIPT
ECE 476 Power System Analysis
Lecture 14: Power Flow
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
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
2
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
3
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
4
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 ik
n
i k ik ik ik ik Gi Di ik
V
V V G B P P V
V V G B Q Q V
5
Voltage Dependent Load Example
22 2 2 2
2 22 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 VJ
x
x
x2 2 2 2 2sin 10cos 20 2.0V V V
6
Voltage Dependent Load, cont'd
(0)
22 2 2(0)
2 22 2 2 2
(0)
1(1)
0Again set 0, guess
1
Calculate
(10sin ) 2.0 2.0f( )
1.0( 10cos ) (10) 1.0
10 4( )
0 12
0 10 4 2.0 0.1667Solve
1 0 12 1.0 0.9167
v
V V
V V V
x
x
J x
x
7
Voltage Dependent Load, cont'd
Line Z = 0.1j
One Two 1.000 pu 0.894 pu
160 MW 80 MVR
160.0 MW120.0 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
8
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
9
Dishonest Newton-Raphson Example
2
1(0)( ) ( )
( ) ( ) 2(0)
( 1) ( ) ( ) 2(0)
Use the Dishonest Newton-Raphson to solve
( ) - 2 0
( )( )
1(( ) - 2)
21
(( ) - 2)2
v v
v v
v v v
f x x
df xx f x
dx
x xx
x x xx
10
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 xx
x x
We pay a pricein increased iterations, butwith decreased computationper iteration
11
Two Bus Dishonest ROC
Slide shows the region of convergence for different initial
guesses for the 2 bus case using the Dishonest N-RRed regionconvergesto the highvoltage solution,while the yellow regionconvergesto the lowvoltage solution
12
Honest N-R Region of Convergence
Maximum of 15
iterations
13
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.
14
Decoupled Power Flow Formulation
( ) ( )
( ) ( )( )
( )( ) ( ) ( )
( )2 2 2
( )
( )
General form of the power flow problem
( )( )
( )
where
( )
( )
( )
v v
v vv
vv v v
vD G
v
vn Dn Gn
P P P
P P P
P Pθθ V P x
f xQ xVQ Q
θ V
x
P x
x
15
Decoupling Approximation
( ) ( )
( )
( ) ( )( )
( ) ( ) ( )
Usually the off-diagonal matrices, and
are small. Therefore we approximate them as zero:
( )( )
( )
Then the problem
v v
v
v vv
v v v
P QV θ
P0
θ P xθf x
Q Q xV0V
1 1( ) ( )( )( ) ( ) ( )
can be decoupled
( ) ( )v v
vv v v
P Qθ P x V Q x
θ V16
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
ii ij ij ij ij
j
ii j ij ij ij ij
j
G B
V G B
V V G B
P
V
Qθ
17
Decoupled N-R Region of Convergence
18
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
19
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 vvv
v v
V
j
P x Q xθ B V B
V VB Y G B
lumn are omitted
20
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 30bus j
Y
21
FDPF Three Bus Example, cont’d
1
(0)(0)2 2
3 3
34.3 14.3 2024.3 10
14.3 24.3 1010 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
22
FDPF Three Bus Example, cont’d
(1)2
3
i
i i1
(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
nDi Gi
k ik ik ik ikk
V
V
P PV G B
x
(2)2
3
0.151 0.1361
0.107 0.1156
0.924
0.936
0.1384 0.9224Actual solution:
0.1171 0.9338
V
V
θ V
23
FDPF Region of Convergence
24
“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
1θ B P
25
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 Example
27
DC Power Flow 5 Bus Example
slack
One
Two
ThreeFourFiveA
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.
28
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 isthat the changein generation isabsorbed by theslack bus
29
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%
30
Power Flow Simulation - After
Z for all lines = j0.1Limit for all lines = 150 MVA
One Two
200 MW 100 MVR
105.0 MW 64.3 MVR
Three1.000 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.
31
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 jij i j
ij ij
i j ijij
ij Gk
V VP
X X
PX P
32
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
33
Three Bus Sensitivity Example
line
bus
12
3
For the previous three bus case with Z 0.1
20 10 1020 10
10 20 1010 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 0Then P 0.667 pu
0.1P 0.333 pu P 0.333 pu
34
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 35
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 schedace 10P P f
36
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.
37
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
38