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Lecture 13Newton-Raphson Power Flow
Professor Tom OverbyeDepartment of Electrical and
Computer Engineering
ECE 476
POWER SYSTEM ANALYSIS
2
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.
3
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
4
Two Bus Newton-Raphson Example
Line Z = 0.1j
One Two 1.000 pu 1.000 pu
200 MW 100 MVR
0 MW 0 MVR
For the two bus power system shown below, use the Newton-Raphson power flow to determine the voltage magnitude and angle at bus two. Assumethat bus one is the slack and SBase = 100 MVA.
2
2
10 10
10 10busj j
V j j
x Y
5
Two Bus Example, cont’d
i1
i1
2 1 2
22 1 2 2
General power balance equations
P ( cos sin )
Q ( sin cos )
Bus two power balance equations
(10sin ) 2.0 0
( 10cos ) (10) 1.0 0
n
i k ik ik ik ik Gi Dik
n
i k ik ik ik ik Gi Dik
V V G B P P
V V G B Q Q
V V
V V V
6
Two Bus Example, cont’d
2 2 2
22 2 2 2
2 2
2 2
2 2
2 2
2 2 2
2 2 2 2
P ( ) (10sin ) 2.0 0
( ) ( 10cos ) (10) 1.0 0
Now calculate the power flow Jacobian
P ( ) P ( )
( )Q ( ) Q ( )
10 cos 10sin
10 sin 10cos 20
V
Q V V
VJ
V
V
V V
x
x
x x
xx x
7
Two Bus Example, First Iteration
(0)
2 2(0)2
2 2 2
2 2 2(0)
2 2 2 2
(1)
0Set 0, guess
1
Calculate
(10sin ) 2.0 2.0f( )
1.0( 10cos ) (10) 1.0
10 cos 10sin 10 0( )
10 sin 10cos 20 0 10
0 10 0Solve
1 0 10
v
V
V V
V
V V
x
x
J x
x1 2.0 0.2
1.0 0.9
8
Two Bus Example, Next Iterations
(1)2
(1)
1(2)
0.9(10sin( 0.2)) 2.0 0.212f( )
0.2790.9( 10cos( 0.2)) 0.9 10 1.0
8.82 1.986( )
1.788 8.199
0.2 8.82 1.986 0.212 0.233
0.9 1.788 8.199 0.279 0.8586
f(
x
J x
x
(2) (3)
(3)2
0.0145 0.236)
0.0190 0.8554
0.0000906f( ) Done! V 0.8554 13.52
0.0001175
x x
x
9
Two Bus Solved Values
Line Z = 0.1j
One Two 1.000 pu 0.855 pu
200 MW 100 MVR
200.0 MW168.3 MVR
-13.522 Deg
200.0 MW 168.3 MVR
-200.0 MW-100.0 MVR
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
10
Two Bus Case Low Voltage Solution
(0)
2 2(0)2
2 2 2
This case actually has two solutions! The second
"low voltage" is found by using a low initial guess.
0Set 0, guess
0.25
Calculate
(10sin ) 2.0f( )
( 10cos ) (10) 1.0
v
V
V V
x
x
2 2 2(0)
2 2 2 2
2
0.875
10 cos 10sin 2.5 0( )
10 sin 10cos 20 0 5
V
V V
J x
11
Low Voltage Solution, cont'd
1(1)
(2) (2) (3)
0 2.5 0 2 0.8Solve
0.25 0 5 0.875 0.075
1.462 1.42 0.921( )
0.534 0.2336 0.220
x
f x x x
Line Z = 0.1j
One Two 1.000 pu 0.261 pu
200 MW 100 MVR
200.0 MW831.7 MVR
-49.914 Deg
200.0 MW 831.7 MVR
-200.0 MW-100.0 MVR
Low voltage solution
12
Two Bus Region of Convergence
Slide shows the region of convergence for different initialguesses of bus 2 angle (x-axis) and magnitude (y-axis)
Red regionconvergesto the highvoltage solution,while the yellow regionconvergesto the lowvoltage solution
13
Using the Power Flow: Example 1
slack
SLACK345
SLACK138
RAY345
RAY138
RAY69
FERNA69
A
MVA
DEMAR69
BLT69
BLT138
BOB138
BOB69
WOLEN69
SHI MKO69
ROGER69
UI UC69
PETE69
HI SKY69
TI M69
TI M138
TI M345
PAI 69
GROSS69
HANNAH69
AMANDA69
HOMER69
LAUF69
MORO138
LAUF138
HALE69
PATTEN69
WEBER69
BUCKY138
SAVOY69
SAVOY138
J O138 J O345
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
A
MVA
1.02 pu
1.01 pu
1.02 pu
1.03 pu
1.01 pu
1.00 pu1.00 pu
0.99 pu
1.02 pu
1.01 pu
1.00 pu
1.01 pu1.01 pu
1.01 pu
1.01 pu
1.02 pu
1.00 pu
1.00 pu
1.02 pu
0.997 pu
0.99 pu
1.00 pu
1.02 pu
1.00 pu1.01 pu
1.00 pu
1.00 pu 1.00 pu
1.01 pu
1.02 pu1.02 pu
1.02 pu1.03 pu
A
MVA
1.02 pu
A
MVA
A
MVA
LYNN138
A
MVA
1.02 pu
A
MVA
1.00 pu
A
MVA
218 MW 54 Mvar
21 MW 7 Mvar
45 MW 12 Mvar
140 MW 45 Mvar
37 MW
13 Mvar
12 MW 5 Mvar
150 MW 0 Mvar
56 MW
13 Mvar
15 MW 5 Mvar
14 MW
2 Mvar
42 MW 2 Mvar
45 MW 0 Mvar
58 MW 36 Mvar
36 MW 10 Mvar
0 MW 0 Mvar
22 MW 15 Mvar
60 MW 12 Mvar
20 MW 30 Mvar
23 MW 7 Mvar
33 MW 13 Mvar
16.0 Mvar 18 MW 5 Mvar
58 MW 40 Mvar 51 MW
15 Mvar
14.3 Mvar
33 MW 10 Mvar
15 MW 3 Mvar
23 MW 6 Mvar 14 MW
3 Mvar
4.8 Mvar
7.2 Mvar
12.8 Mvar
29.0 Mvar
7.4 Mvar
0.0 Mvar
106 MW 8 Mvar
20 MW 8 Mvar
150 MW 0 Mvar
17 MW 3 Mvar
0 MW 0 Mvar
14 MW 4 Mvar
Usingcasefrom Example6.13
14
Three Bus PV Case Example
Line Z = 0.1j
Line Z = 0.1j Line Z = 0.1j
One Two 1.000 pu 0.941 pu
200 MW 100 MVR
170.0 MW 68.2 MVR
-7.469 Deg
Three 1.000 pu
30 MW 63 MVR
2 2 2 2
3 3 3 3
2 2 2
For this three bus case we have
( )
( ) ( ) 0
V ( )
G D
G D
D
P P P
P P P
Q Q
x
x f x x
x
15
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
16
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
17
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
18
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
19
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
20
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
21
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
22
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
23
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
24
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
25
Honest N-R Region of Convergence
Maximum of 15
iterations
26
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.
27
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
28
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
29
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
30
Decoupled N-R Region of Convergence
31
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
32
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
33
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
34
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
35
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
36
FDPF Region of Convergence
37
“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
38
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
39
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
40
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%
41
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.
42
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
43
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
44
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