ECE 333 Renewable Energy Systems Lecture 14: Power Flow Prof. Tom Overbye Dept. of Electrical and...
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ECE 333 Renewable Energy Systems Lecture 14: Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana- Champaign [email protected]
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Transcript of ECE 333 Renewable Energy Systems Lecture 14: Power Flow Prof. Tom Overbye Dept. of Electrical and...
ECE 333 Renewable Energy Systems
Lecture 14: Power Flow
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
Announcements
• Power flow is only covered in lecture; not in the book• No quiz today• HW 6 is posted on the website and is due on Thursday
March 19. HW 6 must be turned in and will count the same as a quiz. There will be no quiz on March 19.
• Start Reading Chapter 4 (The Solar Resource)
2
Bus Admittance Matrix or Ybus
• First step in solving the power flow is to create what is known as the bus admittance matrix, often call the Ybus.
• The Ybus gives the relationships between all the bus current injections, I, and all the bus voltages, V,
I = Ybus V
• The Ybus is developed by applying KCL at each bus in the system to relate the bus current injections, the bus voltages, and the branch impedances and admittances
3
Ybus Example
Determine the bus admittance matrix for the network
shown below, assuming the current injection at each
bus i is Ii = IGi - IDi where IGi is the current injection into the bus from the generator and IDi is the current flowing into the load
4
Ybus Example, cont’d
1 1 1
1 2 1 31 12 13
1 1 2 1 3 j
1 2 3
2 21 23 24
1 2 3 4
By KCL at bus 1 we have
1( ) ( ) (with Y )
( )
Similarly
( )
G D
A B
A Bj
A B A B
A A C D C D
I I I
V V V VI I I
Z Z
I V V Y V V YZ
Y Y V Y V Y V
I I I I
Y V Y Y Y V Y V Y V
5
Ybus Example, cont’d
1 1
2 2
3 3
4 4
We can get similar relationships for buses 3 and 4
The results can then be expressed in matrix form
0
0
0 0
bus
A B A B
A A C D C D
B C B C
D D
I Y Y Y Y V
I Y Y Y Y Y Y V
I Y Y Y Y V
I Y Y V
I Y V
For a system with n buses, Ybus is an n by n
symmetric matrix (i.e., one where Aij = Aji)
6
Ybus General Form
•The diagonal terms, Yii, are the self admittance terms, equal to the sum of the admittances of all devices incident to bus i.
•The off-diagonal terms, Yij, are equal to the negative of the sum of the admittances joining the two buses.•With large systems Ybus is a sparse matrix (that is, most entries are zero)
7
Power Flow Analysis
• When analyzing power systems we know neither the complex bus voltages nor the complex current injections
• Rather, we know the complex power being consumed by the load, and the power being injected by the generators plus their voltage magnitudes
• Therefore we can not directly use the Ybus equations, but rather must use the power balance equations
8
Power Flow Slack Bus
• We can not arbitrarily specify S at all buses because total generation must equal total load + total losses
• We also need an angle reference bus.• To solve these problems we define one bus as the
"slack" bus. This bus has a fixed voltage magnitude and angle, and a varying real/reactive power injection.
• A “slack bus” does not exist in the real power system.
9
Power Balance Equations
1
bus
1
From KCL we know at each bus i in an n bus system
the current injection, , must be equal to the current
that flows into the network
Since = we also know
i
n
i Gi Di ikk
n
i Gi Di ik kk
I
I I I I
I I I Y V
I Y V
*iThe network power injection is then S i iV I
10
Power Balance Equations, cont’d
** * *
i1 1
S
This is an equation with complex numbers.
Sometimes we would like an equivalent set of real
power equations. These can be derived by defining
n n
i i i ik k i ik kk k
ik ik ik
i
V I V Y V V Y V
Y G jB
V
jRecall e cos sin
iji i i
ik i k
V e V
j
11
Real Power Balance Equations
* *i
1 1
1
i1
i1
S ( )
(cos sin )( )
Resolving into the real and imaginary parts
P ( cos sin )
Q ( sin cos
ikn n
ji i i ik k i k ik ik
k k
n
i k ik ik ik ikk
n
i k ik ik ik ik Gi Dik
n
i k ik ik ik ik
P jQ V Y V V V e G jB
V V j G jB
V V G B P P
V V G B
)k Gi DiQ Q
12
Newton-Raphson Method (scalar)
( )
( ) ( )
( )( ) ( )
2 ( ) 2( )2
1. For each guess of , , define ˆ
-ˆ
2. Represent ( ) by a Taylor series about ( )ˆ
( )( ) ( )ˆ
1 ( )higher order terms
2
v
v v
vv v
vv
x x
x x x
f x f x
df xf x f x x
dx
d f xx
dx
General form of problem: Find an x such that
( ) 0ˆf x
13
Newton-Raphson Method, cont’d
( )( ) ( )
( )
1( )( ) ( )
3. Approximate ( ) by neglecting all terms ˆ
except the first two
( )( ) 0 ( )ˆ
4. Use this linear approximation to solve for
( )( )
5. Solve for a new estim
vv v
v
vv v
f x
df xf x f x x
dx
x
df xx f x
dx
( 1) ( ) ( )
ate of x̂v v vx x x
14
Newton-Raphson Example
2
1( )( ) ( )
( ) ( ) 2( )
( 1) ( ) ( )
( 1) ( ) ( ) 2( )
Use Newton-Raphson to solve ( ) - 2 0
The equation we must iteratively solve is
( )( )
1(( ) - 2)
2
1(( ) - 2)
2
vv v
v vv
v v v
v v vv
f x x
df xx f x
dx
x xx
x x x
x x xx
15
Newton-Raphson Example, cont’d
( 1) ( ) ( ) 2( )
(0)
( ) ( ) ( )
3 3
6
1(( ) - 2)
2
Guess x 1. Iteratively solving we get
v ( )
0 1 1 0.5
1 1.5 0.25 0.08333
2 1.41667 6.953 10 2.454 10
3 1.41422 6.024 10
v v vv
v v v
x x xx
x f x x
16
Newton-Raphson Comments
• When close to the solution the error decreases quite quickly -- method has quadratic convergence
• f(x(v)) is known as the mismatch, which we would like to drive to zero
• Stopping criteria is when f(x(v)) < • Results are dependent upon the initial guess. What
if we had guessed x(0) = 0, or x (0) = -1?• A solution’s region of attraction (ROA) is the set of
initial guesses that converge to the particular solution. The ROA is often hard to determine
17
Multi-Variable Newton-Raphson
1 1
2 2
Next we generalize to the case where is an n-
dimension vector, and ( ) is an n-dimension function
( )
( )( )
( )
Again define the solution so ( ) 0 andˆ ˆn n
x f
x f
x f
x
f x
x
xx f x
x
x f x
x
ˆ x x
18
Multi-Variable Case, cont’d
i
1 11 1 1 2
1 2
1
n nn n 1 2
1 2
n
The Taylor series expansion is written for each f ( )
f ( ) f ( )f ( ) f ( )ˆ
f ( )higher order terms
f ( ) f ( )f ( ) f ( )ˆ
f ( )higher order terms
nn
nn
x xx x
xx
x xx x
xx
x
x xx x
x
x xx x
x
19
Multi-Variable Case, cont’d
1 1 1
1 21 1
2 2 22 2
1 2
1 2
This can be written more compactly in matrix form
( ) ( ) ( )
( )( ) ( ) ( )
( )( )ˆ
( )( ) ( ) ( )
n
n
nn n n
n
f f fx x x
f xf f f
f xx x x
ff f fx x x
x x x
xx x x
xf x
xx x x
higher order terms
nx
20
Jacobian Matrix
1 1 1
1 2
2 2 2
1 2
1 2
The n by n matrix of partial derivatives is known
as the Jacobian matrix, ( )
( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( ) ( )
n
n
n n n
n
f f fx x x
f f fx x x
f f fx x x
J x
x x x
x x x
J x
x x x
21
Multi-Variable N-R Procedure
1
( 1) ( ) ( )
( 1) ( ) ( ) 1 ( )
( )
Derivation of N-R method is similar to the scalar case
( ) ( ) ( ) higher order termsˆ
( ) 0 ( ) ( )ˆ
( ) ( )
( ) ( )
Iterate until ( )
v v v
v v v v
v
f x f x J x x
f x f x J x x
x J x f x
x x x
x x J x f x
f x
22
Multi-Variable Example
1
2
2 21 1 2
2 22 1 2 1 2
1 1
1 2
2 2
1 2
xSolve for = such that ( ) 0 where
x
f ( ) 2 8 0
f ( ) 4 0
First symbolically determine the Jacobian
f ( ) f ( )
( ) =f ( ) f ( )
x x
x x x x
x x
x x
x f x
x
x
x x
J xx x
23
Multi-variable Example, cont’d
1 2
1 2 1 2
11 1 2 1
2 1 2 1 2 2
(0)
1(1)
4 2( ) =
2 2
Then
4 2 ( )
2 2 ( )
1Arbitrarily guess
1
1 4 2 5 2.1
1 3 1 3 1.3
x x
x x x x
x x x f
x x x x x f
J x
x
x
x
x
24
Multi-variable Example, cont’d
1(2)
(2)
2.1 8.40 2.60 2.51 1.8284
1.3 5.50 0.50 1.45 1.2122
Each iteration we check ( ) to see if it is below our
specified tolerance
0.1556( )
0.0900
If = 0.2 then we wou
x
f x
f x
ld be done. Otherwise we'd
continue iterating.
25
Newton-Raphson Power Flow
i1
In the Newton-Raphson power flow we use Newton's
method to determine the voltage magnitude and angle
at each bus in the power system.
We need to solve the power balance equations
P ( cosn
i k ik ikk
V V G
i1
sin )
Q ( sin cos )
ik ik Gi Di
n
i k ik ik ik ik Gi Dik
B P P
V V G B Q Q
Power Flow Variables
2 2 2
n
2
Assume the slack bus is the first bus (with a fixed
voltage angle/magnitude). We then need to determine
the voltage angle/magnitude at the other buses.
( )
( )
G
n
P P
V
V
x
x f x
2
2 2 2
( )
( )
( )
D
n Gn Dn
G D
n Gn Dn
P
P P P
Q Q Q
Q Q Q
x
x
x
N-R Power Flow Solution
( )
( )
( 1) ( ) ( ) 1 ( )
The power flow is solved using the same procedure
discussed last time:
Set 0; make an initial guess of ,
While ( ) Do
( ) ( )
1
End While
v
v
v v v v
v
v v
x x
f x
x x J x f x
Power Flow Jacobian Matrix
1 1 1
1 2
2 2 2
1 2
1 2
The most difficult part of the algorithm is determining
and inverting the n by n Jacobian matrix, ( )
( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( ) ( )
n
n
n n n
n
f f fx x x
f f fx x x
f f fx x x
J x
x x x
x x x
J x
x x x
Power Flow Jacobian Matrix, cont’d
i
i
i1
Jacobian elements are calculated by differentiating
each function, f ( ), with respect to each variable.
For example, if f ( ) is the bus i real power equation
f ( ) ( cos sin )n
i k ik ik ik ik Gik
x V V G B P P
x
x
i
1
i
f ( )( sin cos )
f ( )( sin cos ) ( )
Di
n
i k ik ik ik iki k
k i
i j ik ik ik ikj
xV V G B
xV V G B j i
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. Assume
that bus one is the slack and SBase = 100 MVA.
2
2
10 10
10 10busj j
V j j
x Y
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
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
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
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
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
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
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
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
40 Bus Power Flow Example
slack
Island Electric Company (IEC)
350 MW
505 MW
396 MW
268 MW 128 Mvar
115 MW 25 Mvar
150 MW 70 Mvar
110 MW 30 Mvar
150 MW 39 Mvar
60 MW 15 Mvar
55 MW 15 Mvar
130 MW 30 Mvar
175 MW 40 Mvar
140 MW 32 Mvar 176 MW
15 Mvar
165 MW 30 Mvar
132 MW
15 Mvar
160 MW
35 Mvar
112 MW 40 Mvar
200 MW 60 Mvar
95 MW
23 Mvar
75 MW 15 Mvar
198 MW 35 Mvar
87 MW 19 Mvar
161 MW
21 Mvar
135 MW 20 Mvar
140 MW 20 Mvar
88 MW 11 Mvar
130 MW 45 Mvar
128 MW
28 Mvar
45%
22%
33%
35%
4%
13%
23%
14%
13%
28%
35%
8%
17%
13%
57%
28%
6%
3%
6%
16%
30%
22%
10%
27%
12%
30%
44%
9%
5%
46%
78%
45%
53%
47%
19%
900 MW
70 Mvar
51 Mvar
78 Mvar
47%
3%
10%
19%
16%
22%
48%
RobinEagle
Rook
SparrowBluebird
Hawk
Parrot
Turkey
Mallard
Condor
Hen
Dove
Piper
Cardinal
24%
1120 MW
2%
2%
29%
8%
36%
7%
18%
7%
11%
24%
17%
38% 39%
30%
7%
78%
14%
83 Mvar
300 MWSystem Losses: 35.22 MW
0.000 pu
NewWind
0 MW
56 Mvar
0.996 pu
1.018 pu
1.018 pu
1.015 pu
1.019 pu
1.015 pu
1.002 pu
0.997 pu
0.996 pu
1.004 pu
0.988 pu
1.001 pu
1.010 pu
1.020 pu
0.999 pu
0.989 pu
0.993 pu
0.996 pu
1.014 pu0.993 pu
1.010 pu
1.015 pu
1.023 pu
1.015 pu
1.010 pu1.006 pu
0.999 pu
1.018 pu1.020 pu
0.993 pu
1.012 pu
1.013 pu
0.992 pu
0.991 pu
1.005 pu
1.014 pu
1.015 pu0.994 pu
1.025 pu
57 Mvar
Ostrich
Crow
Peacock
Lark Finch
Oriole
Heron
Finch
Owl
Woodpecker
Pheasant
Canary
Flamingo
Good Power System Operation
• Good power system operation requires that there be no reliability violations for either the current condition or in the event of statistically likely contingencies• Reliability requires as a minimum that there be no transmission
line/transformer limit violations and that bus voltages be within acceptable limits (perhaps 0.95 to 1.08)
• Example contingencies are the loss of any single device. This is known as n-1 reliability.
• North American Electric Reliability Corporation now has legal authority to enforce reliability standards (and there are now lots of them). See http://www.nerc.com for details (click on Standards)
Looking at the Impact of Line and Transformer Outages
slack
Island Electric Company (IEC)
350 MW
505 MW
398 MW
268 MW 128 Mvar
115 MW 25 Mvar
150 MW 70 Mvar
110 MW 30 Mvar
150 MW 39 Mvar
60 MW 15 Mvar
55 MW 15 Mvar
130 MW 30 Mvar
175 MW 40 Mvar
140 MW 32 Mvar 176 MW
15 Mvar
165 MW 30 Mvar
132 MW
15 Mvar
160 MW
35 Mvar
112 MW 40 Mvar
200 MW 60 Mvar
95 MW
23 Mvar
75 MW 15 Mvar
198 MW 35 Mvar
87 MW 19 Mvar
161 MW
21 Mvar
135 MW 20 Mvar
140 MW 20 Mvar
88 MW 11 Mvar
130 MW 45 Mvar
128 MW
28 Mvar
49%
23%
37%
37%
8%
15%
25%
16%
13%
30%
37%
9%
19%
13%
62%
31%
7%
9%
3%
17%
32%
32%
16%
32%
13%
30%
43%
9%
5%
45%
46%
52%
42%
19%
900 MW
68 Mvar
51 Mvar
78 Mvar
40%
4%
12%
20%
9%
20%
41%
RobinEagle
Rook
SparrowBluebird
Hawk
Parrot
Turkey
Mallard
Condor
Hen
Dove
Piper
Cardinal
23%
1120 MW
6%
3%
29%
9%
41%
8%
23%
4%
2%
24%
16%
39% 39%
33%
11%
15%
83 Mvar
300 MWSystem Losses: 37.50 MW
0.000 pu
NewWind
0 MW
56 Mvar
0.988 pu
1.014 pu
1.014 pu
1.015 pu
1.016 pu
1.011 pu
0.996 pu
0.987 pu
0.982 pu
0.990 pu
0.974 pu
0.988 pu
0.994 pu
1.020 pu
0.986 pu
0.983 pu
0.990 pu
0.993 pu
1.011 pu0.991 pu
1.009 pu
1.013 pu
1.022 pu
1.015 pu
1.010 pu1.006 pu
0.999 pu
1.018 pu1.020 pu
0.992 pu
1.011 pu
1.013 pu
0.991 pu
0.990 pu
1.003 pu
1.011 pu
1.015 pu0.991 pu
1.025 pu
57 Mvar
Ostrich
Crow
Peacock
Lark Finch
Oriole
Heron
Finch
Owl
Woodpecker
Pheasant
Canary
Flamingo
139%
40 Bus Example: New Wind Generation
43
slack
Island Electric Company (IEC)
350 MW
505 MW
-110 MW
268 MW 128 Mvar
115 MW 25 Mvar
150 MW 70 Mvar
110 MW 30 Mvar
150 MW 39 Mvar
59 MW 15 Mvar
55 MW 15 Mvar
130 MW 30 Mvar
175 MW 40 Mvar
140 MW 32 Mvar 176 MW
15 Mvar
165 MW 30 Mvar
132 MW
15 Mvar
160 MW
35 Mvar
112 MW 40 Mvar
200 MW 60 Mvar
95 MW
23 Mvar
75 MW 15 Mvar
198 MW 35 Mvar
87 MW 19 Mvar
161 MW
21 Mvar
135 MW 20 Mvar
140 MW 20 Mvar
88 MW 11 Mvar
130 MW 45 Mvar
128 MW
28 Mvar
30%
15%
54%
36%
11%
5%
13%
14%
13%
45%
15%
11%
8%
16%
47%
28%
52%
30%
27%
26%
28%
17%
8%
24%
9%
29%
45%
12%
4%
53%
71%
44%
54%
40%
10%
900 MW
62 Mvar
49 Mvar
76 Mvar
42%
5%
30%
10%
70%
23%
43%
RobinEagle
Rook
SparrowBluebird
Hawk
Parrot
Turkey
Mallard
Condor
Hen
Dove
Piper
Cardinal
35%
1120 MW
9%
1%
27%
6%
35%
11%
11%
19%
35%
31%
36% 36%
33%
13%
71%
14%
83 Mvar
300 MWSystem Losses: 129.92 MW
0.657 pu
NewWind
600 MW
55 Mvar
0.653 pu
0.850 pu
0.907 pu
1.012 pu
0.953 pu
0.947 pu
0.885 pu
0.843 pu
0.933 pu
0.941 pu
0.925 pu
0.940 pu
0.983 pu
1.020 pu
0.942 pu
0.944 pu
0.952 pu
0.955 pu
0.973 pu0.973 pu
1.004 pu
0.998 pu
1.009 pu
1.012 pu
1.010 pu1.005 pu
0.997 pu
1.017 pu1.020 pu
0.987 pu
1.004 pu
1.012 pu
0.987 pu
0.985 pu
0.997 pu
0.989 pu
1.015 pu0.956 pu
1.025 pu
40 Mvar
Ostrich
Crow
Peacock
Lark Finch
Oriole
Heron
Finch
Owl
Woodpecker
Pheasant
Canary
Flamingo
131%
