lecture 14 power flow professor tom overbye department of electrical and computer engineering ece...
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Lecture 14Power Flow
Professor Tom OverbyeDepartment of Electrical and
Computer Engineering
ECE 476
POWER SYSTEM ANALYSIS
2
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)
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 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
4
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
5
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
6
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 atbus 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
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( 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
8
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
9
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
10
Two Bus Dishonest ROC
Slide shows the region of convergence for different initialguesses for the 2 bus case using the Dishonest N-R
Red regionconvergesto the highvoltage solution,while the yellow regionconvergesto the lowvoltage 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
( ) ( )
( ) ( )( )
( )( ) ( ) ( )
( )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
14
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 V
15
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
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
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
19
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
20
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
21
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
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
1θ B P
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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
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DC Power Flow Example
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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.
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 isthat the changein generation isabsorbed by theslack 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 = j0.1Limit 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), resultsin 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 jij i j
ij ij
i j ijij
ij Gk
V VP
X X
PX 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
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
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
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
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
41
WE to TVA PTDFs
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 kP 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
NERCis theNorthAmericanElectricReliabilityCouncil
45
NERC Reliability Coordinators
Source: http://www.nerc.com/page.php?cid=5%7C67%7C206