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
overbye@illinois.edu
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
1
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
2
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
3
Linear Power System Elements
Resistors, inductors, capacitors, independent
voltage sources and current sources are linear
circuit elements
1
V = R I V = j L I V =
I
j C
Such systems may be analyzed by superposition
4
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
5
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
two solutions where f(x) = 0
f(x) = x2 + 2
no solution f(x) = 0
6
Multiple Solution Example 3
• The dc system shown below has two solutions:
The equation we're solving is
where the 18 watt
load is a resistive
load
2
 9 volts 
I RLoad  
RLoad  18 watts

 1 +R Load 
One solution is R Load  2
2
Other solution is R Load  0.5
What is the
maximum
PLoad?
7
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
8
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
9
Ybus Example, cont’d
By KCL at bus 1 we have
I1
I G1  I D1
I1  I12  I13
V1  V2 V1  V3


ZA
ZB
I1  (V1  V2 )YA  (V1  V3 )YB
1
(with Yj  )
Zj
 (YA  YB )V1  YA V2  YB V3
Similarly
I 2  I 21  I 23  I 24
 YA V1  (YA  YC  YD )V2  YC V3  YD V4
10
Ybus Example, cont’d
We can get similar relationships for buses 3 and 4
The results can then be expressed in matrix form
I  Ybus V
YA
YB
 I1  YA  YB
 I   Y
YA  YC  YD
YC
2
A
  
YC
YB  YC
 I 3   YB
I   0
YD
0
 4 
0  V1 
YD  V2 
 
0  V3 
YD  V4 
For a system with n buses, Ybus is an n by n
symmetric matrix (i.e., one where Aij = Aji)
11
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.
12
Modeling Shunts in the Ybus
Ykc
Since I ij  (Vi  V j )Yk  Vi
2
Ykc
Yii 
 Yk 
2
Rk  jX k Rk  jX k
1
1
Note Yk 

 2
Z k Rk  jX k Rk  jX k Rk  X k2
Yiifrom other lines
13
Two Bus System Example
Yc
(V1  V2 )
I1 
 V1
Z
2
1
 12  j16
0.03  j 0.04
 I1 
12  j15.9 12  j16  V1 
 I    12  j16 12  j15.9  V 

 2
 2
14
Using the Ybus
If the voltages are known then we can solve for
the current injections:
Ybus V  I
If the current injections are known then we can
solve for the voltages:
1
Ybus
I  V  Zbus I
where Z bus is the bus impedance matrix
15
Solving for Bus Currents
For example, in previous case assume
 1.0 
V

0.8

j
0.2


Then
12  j15.9 12  j16   1.0   5.60  j 0.70 
 12  j16 12  j15.9  0.8  j 0.2    5.58  j 0.88


 

Therefore the power injected at bus 1 is
S1  V1I1*  1.0  (5.60  j 0.70)  5.60  j 0.70
S2  V2 I 2*  (0.8  j 0.2)  (5.58  j 0.88)  4.64  j 0.41
16
Solving for Bus Voltages
For example, in previous case assume
 5.0 
I


4.8


Then
1
12  j15.9 12  j16   5.0   0.0738  j 0.902 
 12  j16 12  j15.9   4.8   0.0738  j1.098

 
 

Therefore the power injected is
S1  V1I1*  (0.0738  j 0.902)  5  0.37  j 4.51
S2  V2 I 2*  (0.0738  j1.098)  (4.8)  0.35  j 5.27
17
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
18
Power Balance Equations
From KCL we know at each bus i in an n bus system
the current injection, I i , must be equal to the current
that flows into the network
I i  I Gi  I Di 
n
 Iik
k 1
Since I = Ybus V we also know
I i  I Gi  I Di 
n
 YikVk
k 1
The network power injection is then Si  Vi I i*
19
Power Balance Equations, cont’d
*
n


Si  Vi I i*  Vi   YikVk   Vi  Yik*Vk*
 k 1

k 1
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
Yik
Gik  jBik
Vi
Vi e ji  Vi  i
 ik
i   k
Recall e j  cos  j sin 
20
Real Power Balance Equations
n
Si  Pi  jQi  Vi  Yik*Vk* 
k 1

n
 Vi Vk
k 1
n
j ik
V
V
e
(Gik  jBik )
 i k
k 1
(cos ik  j sin  ik )(Gik  jBik )
Resolving into the real and imaginary parts
Pi 
Qi 
n
 Vi Vk (Gik cosik  Bik sinik )  PGi  PDi
k 1
n
 Vi Vk (Gik sinik  Bik cosik )  QGi  QDi
k 1
21
Power Flow Requires Iterative Solution
In the power flow we assume we know Si and the
Ybus . We would like to solve for the V's. The problem
is the below equation has no closed form solution:
*
n


Si  Vi I i*  Vi   YikVk   Vi  Yik*Vk*
 k 1

k 1
Rather, we must pursue an iterative approach.
n
22
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 first make an initial guess of x, x (0) ,
and then iteratively solve x (v +1)  h( x ( v ) ) until we
find a "fixed point", x,
ˆ such that xˆ  h(x).
ˆ
23
Gauss Iteration Example
Example: Solve x - x  1  0
x ( v 1)  1  x ( v )
Let v = 0 and arbitrarily guess x (0)  1 and solve
v
0
1
2
3
4
x(v )
1
2
2.41421
2.55538
2.59805
v
5
6
7
8
9
x(v)
2.61185
2.61612
2.61744
2.61785
2.61798
24
Stopping Criteria
A key problem to address is when to stop the
iteration. With the Guass iteration we stop when
x ( v )  
with x ( v )
x ( v 1)  x ( v )
If x is a scalar this is clear, but if x is a vector we
need to generalize the absolute value by using a norm
x
(v)
j

Two common norms are the Euclidean & infinity
x 2 
n
2

x
 i
i 1
x   max i x i
25
Gauss Power Flow
We first need to put the equation in the correct form
*
n


 Vi   YikVk   Vi  Yik*Vk*
 k 1

k 1
n
Si  Vi I i*
n
n
k 1
k 1
S*i  Vi* I i  Vi*  YikVk  Vi*  YikVk
S*i

*
Vi
n
 YikVk
k 1
 YiiVi 
n

k 1,k i
n

1  S*i
Vi 
 *   YikVk 

Yii  V
k 1,k i

i
YikVk
26
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.
27
Gauss Two Bus Example, cont’d
The unknown is the complex load voltage, V2 .
To determine V2 we need to know the Ybus .
1
 5  j15
0.02  j 0.06
5  j14.95 5  j15 
Hence Ybus  


5

j
15
5

j
14.70


( Note B22  - j15  j 0.05  j 0.25)
28
Gauss Two Bus Example, cont’d
n

1  S*2
V2 
 *   YikVk 
Y22  V2 k 1,k i

 -1  j 0.5

1
V2 
 (5  j15)(1.00) 

*
5  j14.70  V2

Guess V2(0)  1.00 (this is known as a flat start)
v
0
1
2
V2( v )
1.000  j 0.000
0.9671  j 0.0568
0.9624  j 0.0553
v
3
4
V2( v )
0.9622  j 0.0556
0.9622  j 0.0556
29
Gauss Two Bus Example, cont’d
V2  0.9622  j 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
S1*  V1* (Y11V1  Y12V2 )  1.023  j 0.239
In actual units P1  102.3 MW, Q1  23.9 Mvar
2
The capacitor is supplying V2 25  23.2 Mvar
30
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.
31