ece 476 power system analysis lecture 11: ybus, power flow prof. tom overbye dept. of electrical and...
TRANSCRIPT
![Page 1: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/1.jpg)
ECE 476 Power System Analysis
Lecture 11: Ybus, Power Flow
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
![Page 2: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/2.jpg)
Announcements
• Please read Chapter 2.4; start on Chapter 6• H5 is 3.4, 3.10, 3.14, 3.19, 3.23, 3.60, 2.38, 6.9
• It should be done before the first exam, but does not need to be turned in
• First exam is Tuesday Oct 6 during class• Closed book, closed notes, but you may bring one 8.5 by
11 inch note sheet and standard calculators. • Covers up to end of today's lecture• Last name starting with A to 0 in 3017; P to Z in 3013
• Won will give optional review on Thursday; no new material
2
![Page 3: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/3.jpg)
Power Flow Analysis
• We now have the necessary models to start to develop the power system analysis tools
• The most common power system analysis tool is the power flow (also known sometimes as the load flow)– power flow determines how the power flows in a network– also used to determine all bus voltages and all currents– because of constant power models, power flow is a
nonlinear analysis technique– power flow is a steady-state analysis tool
3
![Page 4: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/4.jpg)
Linear versus Nonlinear Systems
A function H is linear if
H(a1m1 + a2m2) = a1H(m1) + a2H(m2)
That is
1) the output is proportional to the input
2) the principle of superposition holds
Linear Example: y = H(x) = c x
y = c(x1+x2) = cx1 + c x2
Nonlinear Example: y = H(x) = c x2
y = c(x1+x2)2 ≠ (cx1)2 + (c x2)2
4
![Page 5: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/5.jpg)
Linear Power System Elements
Resistors, inductors, capacitors, independent
voltage sources and current sources are linear
circuit elements
1V = R I V = V =
Such systems may be analyzed by superposition
j L I Ij C
5
![Page 6: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/6.jpg)
Nonlinear Power System Elements
•Constant power loads and generator injections are nonlinear and hence systems with these elements can not be analyzed by superposition
Nonlinear problems can be very difficult to solve,and usually require an iterative approach
6
![Page 7: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/7.jpg)
Nonlinear Systems May Have Multiple Solutions or No Solution
Example 1: x2 - 2 = 0 has solutions x = 1.414…
Example 2: x2 + 2 = 0 has no real solution
f(x) = x2 - 2 f(x) = x2 + 2
two solutions where f(x) = 0 no solution f(x) = 0
7
![Page 8: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/8.jpg)
Multiple Solution Example 3
• The dc system shown below has two solutions:
where the 18 watt
load is a resistive
load
22
Load
Load
Load
The equation we're solving is
9 voltsI 18 watts
1 +R
One solution is R 2
Other solution is R 0.5
Load LoadR R
What is themaximumPLoad?
8
![Page 9: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/9.jpg)
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
9
![Page 10: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/10.jpg)
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
10
![Page 11: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/11.jpg)
Ybus Example, cont’d
1 1 1
1 31 21 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 VV 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
11
![Page 12: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/12.jpg)
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)
12
![Page 13: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/13.jpg)
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)
• Shunt terms, such as with the p line model, only affect the diagonal terms.
13
![Page 14: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/14.jpg)
Modeling Shunts in the Ybus
from other lines
2 2
Since ( )2
21 1
Note
kcij i j k i
kcii ii k
k k k kk
k k k k k k k
YI V V Y V
YY Y Y
R jX R jXY
Z R jX R jX R X
14
![Page 15: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/15.jpg)
Two Bus System Example
1 21 1
1 1
2 2
( ) 112 16
2 0.03 0.04
12 15.9 12 16
12 16 12 15.9
cYV VI V j
Z j
I Vj j
I Vj j
15
![Page 16: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/16.jpg)
Using the Ybus
1
bus
If the voltages are known then we can solve for
the current injections:
If the current injections are known then we can
solve for the voltages:
where is the bus impedance matr
bus
bus bus
Y V I
Y I V Z I
Z ix
16
![Page 17: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/17.jpg)
Solving for Bus Currents
*1 1 1
For example, in previous case assume
1.0
0.8 0.2
Then
12 15.9 12 16 1.0 5.60 0.70
12 16 12 15.9 0.8 0.2 5.58 0.88
Therefore the power injected at bus 1 is
S 1.0 (5.60
j
j j j
j j j j
V I
V
*2 2 2
0.70) 5.60 0.70
(0.8 0.2) ( 5.58 0.88) 4.64 0.41
j j
S V I j j j
17
![Page 18: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/18.jpg)
Solving for Bus Voltages
1
*1 1 1
For example, in previous case assume
5.0
4.8
Then
12 15.9 12 16 5.0 0.0738 0.902
12 16 12 15.9 4.8 0.0738 1.098
Therefore the power injected is
S (0.0738 0.902) 5 0
j j j
j j j
V I j
I
*2 2 2
.37 4.51
( 0.0738 1.098) ( 4.8) 0.35 5.27
j
S V I j j
18
![Page 19: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/19.jpg)
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
19
![Page 20: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/20.jpg)
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
20
![Page 21: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/21.jpg)
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
21
![Page 22: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/22.jpg)
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
22
![Page 23: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/23.jpg)
Power Flow Requires Iterative Solution
i
bus
** * *
i1 1
In the power flow we assume we know S and the
. We would like to solve for the V's. The problem
is the below equation has no closed form solution:
S
Rath
n n
i i i ik k i ik kk k
V I V Y V V Y V
Y
er, we must pursue an iterative approach.
23
![Page 24: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/24.jpg)
Gauss Iteration
There are a number of different iterative methods
we can use. We'll consider two: Gauss and Newton.
With the Gauss method we need to rewrite our
equation in an implicit form: x = h(x)
To iterate we fir (0)
( +1) ( )
st make an initial guess of x, x ,
and then iteratively solve x ( ) until we
find a "fixed point", x, such that x (x).ˆ ˆ ˆ
v vh x
h
24
![Page 25: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/25.jpg)
Gauss Iteration Example
( 1) ( )
(0)
( ) ( )
Example: Solve - 1 0
1
Let = 0 and arbitrarily guess x 1 and solve
0 1 5 2.61185
1 2 6 2.61612
2 2.41421 7 2.61744
3 2.55538 8 2.61785
4 2.59805 9 2.61798
v v
v v
x x
x x
v
v x v x
25
![Page 26: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/26.jpg)
Stopping Criteria
( ) ( ) ( 1) ( )
A key problem to address is when to stop the
iteration. With the Guass iteration we stop when
with
If x is a scalar this is clear, but if x is a vector we
need to generalize t
v v v vx x x x
( )
2i2
1
he absolute value by using a norm
Two common norms are the Euclidean & infinity
max x
v
j
n
i ii
x
x
x x
26
![Page 27: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/27.jpg)
Gauss Power Flow
** * *
i1 1
* * * *
1 1
*
*1 1,
*
*1,
We first need to put the equation in the correct form
S
S
S
S1
i i
i
i
n n
i i i ik k i ik kk k
n n
i i i ik k ik kk k
n ni
ik k ii i ik kk k k i
ni
i ik kii k k i
V I V Y V V Y V
V I V Y V V Y V
Y V Y V Y VV
V Y VY V
27
![Page 28: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/28.jpg)
Gauss Two Bus Power Flow Example
•A 100 MW, 50 Mvar load is connected to a generator •through a line with z = 0.02 + j0.06 p.u. and line charging of 5 Mvar on each end (100 MVA base). Also, there is a 25 Mvar capacitor at bus 2. If the generator voltage is 1.0 p.u., what is V2?
SLoad = 1.0 + j0.5 p.u.
28
![Page 29: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/29.jpg)
Gauss Two Bus Example, cont’d
2
2 bus
bus
22
The unknown is the complex load voltage, V .
To determine V we need to know the .
15 15
0.02 0.06
5 14.95 5 15Hence
5 15 5 14.70
( Note - 15 0.05 0.25)
jj
j j
j j
B j j j
Y
Y
29
![Page 30: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/30.jpg)
Gauss Two Bus Example, cont’d
*2
2 *22 1,2
2 *2
(0)2
( ) ( )2 2
1 S
1 -1 0.5( 5 15)(1.0 0)
5 14.70
Guess 1.0 0 (this is known as a flat start)
0 1.000 0.000 3 0.9622 0.0556
1 0.9671 0.0568 4 0.9622 0.0556
2 0
n
ik kk k i
v v
V Y VY V
jV j
j V
V
v V v V
j j
j j
.9624 0.0553j30
![Page 31: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/31.jpg)
Gauss Two Bus Example, cont’d
2
* *1 1 11 1 12 2
1
0.9622 0.0556 0.9638 3.3
Once the voltages are known all other values can
be determined, such as the generator powers and the
line flows
S ( ) 1.023 0.239
In actual units P 102.3 MW
V j
V Y V Y V j
1
22
, Q 23.9 Mvar
The capacitor is supplying V 25 23.2 Mvar
31
![Page 32: ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader034.vdocuments.net/reader034/viewer/2022051316/56649f415503460f94c609bf/html5/thumbnails/32.jpg)
Slack Bus
• In previous example we specified S2 and V1 and then solved for S1 and V2.
• 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.
32