Power Systems Analysis

1. EE 369
POWER SYSTEM ANALYSIS
Lecture 10
Transformers, Load & Generator Models, YBus
Tom Overbye and Ross Baldick
1

2. Announcements
• Homework 7 is 5.8, 5.15, 5.17, 5.24, 5.27,
5.28, 5.29, 5.34, 5.37, 5.38, 5.43, 5.45; due
10/22.
• Homework 8 is 3.1, 3.3, 3.4, 3.7, 3.8, 3.9, 3.10,
3.12, 3.13, 3.14, 3.16, 3.18; due 10/29.
• Homework 9 is 3.20, 3.23, 3.25, 3.27, 3.28,
3.29, 3.35, 3.38, 3.39, 3.41, 3.44, 3.47; due
11/5.
• Start reading Chapter 6 for lectures 11 and 12.
2

3. Load Tap Changing Transformers
LTC transformers have tap ratios that can be
varied to regulate bus voltages.
The typical range of variation is ±10% from the
nominal values, usually in 33 discrete steps
(0.0625% per step).
Because tap changing is a mechanical process,
LTC transformers usually have a 30 second
deadband to avoid repeated changes to minimize
wear and tear.
Unbalanced tap positions can cause “circulating
VArs;” that is, reactive power flowing from one
winding to the next in a three phase transformer.3

4. Phase Shifting Transformers
Phase shifting transformers are used to
control the phase angle across the
transformer.
Since power flow through the transformer
depends upon phase angle, this allows the
transformer to regulate the power flow
through the transformer.
Phase shifters can be used to prevent
inadvertent "loop flow" and to prevent line
overloads by controlling power flow on lines. 4

5. Phase Shifting Transformer Picture
230 kV 800 MVA Phase Shifting
Transformer During factory testing
Source: Tom Ernst, Minnesota Power
Costs about $7 million,
weighs about 1.2
million pounds
5

6. Autotransformers
Autotransformers are transformers in which the
primary and secondary windings are coupled
magnetically and electrically.
This results in lower cost, and smaller size and
weight.
The key disadvantage is loss of electrical
isolation between the voltage levels. This can be
an important safety consideration when a is
large. For example in stepping down 7160/240
V we do not ever want 7160 on the low side!
6

7. Load Models
Ultimate goal is to supply loads with electricity
at constant frequency and voltage.
Electrical characteristics of individual loads
matter, but usually they can only be estimated
– actual loads are constantly changing, consisting of a
large number of individual devices,
– only limited network observability of load
characteristics
Aggregate models are typically used for analysis
Two common models
– constant power: Si = Pi+ jQi
– constant impedance: Si = |V|2
/ Zi
7

8. Generator Models
Engineering models depend on the application.
Generators are usually synchronous machines:
– important exception is case of wind generators,
For generators we will use two different
models:
– (in 369) a steady-state model, treating the generator
as a constant power source operating at a fixed
voltage; this model will be used for power flow and
economic analysis.
– (in 368L) a short term model treating the generator
as a constant voltage source behind a possibly time-8

9. 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. 9

10. Linear versus Nonlinear Systems
• A function H is linear if
H(α1µ1 + α2µ2) = α1H(µ1) + α2H(µ2)
• 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(x +x )2
≠ c(x )2
+ c(x )2 10

11. Linear Power System Elements
Resistors, inductors, capacitors, independent
voltage sources, and current sources are linear
circuit elements:
1
Such systems may be analyzed by superposition.
V R I V j L I V I
j C
ω
ω
= = =
11

12. Nonlinear Power System Elements
•Constant power loads and generator
injections are nonlinear and hence systems
with these elements cannot be analyzed
(exactly) by superposition.
Nonlinear problems can be very difficult to solve,
and usually require an iterative approach. 12

13. 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 to f(x) = 0
13

14. Multiple Solution Example 3
• The dc system shown below has two
solutions for a value of load resistance that
results in 18 W dissipation in the load:
That is, the 18 watt
load is an unknown
resistive load RLoad
2
2
Load Load
Load
Load
Load
The equation we're solving is:
9 volts
18 watts
1 +
One solution is 2
Other solution is 0.5
I R R
R
R
R
 
= = ÷Ω 
= Ω
= Ω
A different
problem:
What is the
resistance to
achieve maximum
PLoad? 14

15. Bus Admittance Matrix or Ybus
First step in solving the power flow is to create
what is known as the bus admittance matrix,
often called the Ybus.
The Ybusgives the relationships between all the
bus current injections, I, and all the bus voltages,
V, I = Ybus V
The Ybusis 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.
15

16. 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 thecurrent injection
into the bus from the generator and IDi is the current
flowing into the load.
16

17. Ybus Example, cont’d
1 1 1
1 31 2
1 12 13
1 1 2 1 3
1 2 3
2 21 23 24
1 2 3 4
By KCL at bus 1 we have
1
( ) ( ) (with )
( )
Similarly
( )
G D
A B
A B j
j
A B A B
A A C D C D
I I I
V VV V
I I I
Z Z
I V V Y V V Y Y
Z
Y Y V Y V Y V
I I I I
Y V Y Y Y V Y V Y V
−
−−
= + = +
= − + − =
= + − −
= + +
= − + + + − −
@
17

18. Ybus Example, cont’d
bus
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
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 Ybuskl = Ybuslk).
From now on, we will mostly write Y for Ybus,
but be careful to distinguish Ykl from line admittances.18

19. Ybus General Form
•The diagonal terms, Ykk, are the “self admittance”
terms, equal to the sum of the admittances of all
devices incident to bus k.
•The off-diagonal terms, Ykl, are equal to the
negative of the admittance joining the two buses.
•With large systems Ybusis a sparse matrix (that is,
most entries are zero):
–sparsity is key to efficient numerical calculation.
•Shunt terms, such as in the equivalent π line
model, only affect the diagonal terms. 19

20. Modeling Shunts in the Ybus
from other lines
2 2
Since ( )
2
2
1 1
Note
kc
ij i j k i
kc
ii ii k
k k k k
k
k k k k k k k
Y
I V V Y V
Y
Y Y Y
R jX R jX
Y
Z R jX R jX R X
= − +
= + +
− −
= = =
+ − +
20

21. Two Bus System Example
1 2
1 1
1 1
2 2
( ) 1 1
, where 12 16.
2 0.03 0.04
12 15.9 12 16
12 16 12 15.9
cYV V
I V j
Z Z j
I Vj j
I Vj j
−
= + = = −
+
− − +    
=    − + −    
21

22. Using the Ybus
bus
1
bus bus
1
bus 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 impedan
−
−
=
= =
Y V I
Y I V Z I
Z Y ce matrix.
22

23. 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:
1.0 (5
j
j j j
j j j j
S V I
 
=  − 
− − + −    
=    − + − − − +    
= = ×
V
*
2 2 2
.60 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
+ = +
= = − × − − = − +
23

24. Solving for Bus Voltages
1
*
1 1 1
As another 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
(0.0738 0.
j j j
j j j
S V I j
−
 
=  − 
− − + −     
=     − + − − − −     
= = −
I
*
2 2 2
902) 5 0.37 4.51
( 0.0738 1.098) ( 4.8) 0.35 5.27
j
S V I j j
× = −
= = − − × − = +
24

25. 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 and their voltage
magnitudes.
Therefore we can not directly use the Ybus
equations, but rather must use the power
balance equations. 25

26. Power Balance Equations
1
bus
1
From KCL we know at each bus in an 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 ik
k
n
i Gi Di ik k
k
i n
I
I I I I
I I I Y V
=
=
= − =
= − =
∑
∑
I Y V
*
The network power injection is then i i iS V I=
26

27. Power Balance Equations, cont’d
*
* * *
1 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 n
i i i i ik k i ik k
k k
ik ik ik
i
S V I V Y V V Y V
Y G jB
V
= =
 
= = = ÷
 
+
∑ ∑
@
@
j
Recall e cos sin
ij
i i i
ik i k
V e V
j
θ
θ
θ
θ θ θ
θ θ
= ∠
−
= +
@
27

28. Real Power Balance Equations
* *
1 1
1
1
1
( )
(cos sin )( )
Resolving into the real and imaginary parts
( cos sin )
( sin cos
ik
n n
j
i i i i ik k i k ik ik
k k
n
i k ik ik ik ik
k
n
i i k ik ik ik ik Gi Di
k
n
i i k ik ik ik i
k
S P jQ V Y V V V e G jB
V V j G jB
P V V G B P P
Q V V G B
θ
θ θ
θ θ
θ θ
= =
=
=
=
= + = = −
= + −
= + = −
= −
∑ ∑
∑
∑
∑ )k Gi DiQ Q= −
28

29. Power Flow Requires Iterative
Solution
bus
In the power flow we assume we know and the
. We would like to solve for the values .
The difficulty is that the following nonlinear
equation (solve for the values given )
has no closed
i
i
i i
S
V
V S
Y
*
* * *
1 1
form solution:
Rather, we must pursue an iterative approach.
n n
i i i i ik k i ik k
k k
S V I V Y V V Y V
= =
 
= = = ÷
 
∑ ∑
29

30. Gauss (or Jacobi) 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: ( ).
Our goal is to f
x h x=
(0)
( 1)
ind that satisfies this equation.
To seek a solution we first make an initial guess of ,
which we call ,and then iteratively plug into the right-
hand side to evaluate an updated guess (v
x
x
x
x h x+
= ( )
),
until we are close to a "fixed point," , such that ( ).ˆ ˆ ˆ
v
x x h x= 30

31. ( 1) ( )
( ) ( )
( 1)
Gauss Example: To solve 1 0, rearrange in the form
( ), where ( ) 1 . Iteration is: 1 .
That is, plug current iterate into:1 ; the answer is
the next iterate ; repea
v v
v v
v
x x
x h x h x x x x
x x
x
+
+
− − =
= = + = +
+
(0)
( ) ( )
t. Matlab code: x=x0; x=1+sqrt(x).
Start at = 0, arbitrarily guess 1 and iterate:
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
x
x x
ν
ν ν
=
31

32. Stopping Criteria
( ) ( ) ( 1) ( )
A key problem to address is when to stop the
iteration. With the Gauss iteration we stop when
with
If is a scalar this is clear, but if is a vector we
need to generalize t
v v v v
x x x x
x x
ε +
∆ < ∆ −@
( )
2
2
1
he absolute value by using a norm
Two common norms are the Euclidean & infinity
( ) max
v
n
i i i
i
x
x x
ε
∞
=
∆ <
∆ = ∆ ∆ = ∆∑x x
32

33. Gauss Power Flow
*
* * *
1 1
* * *
1
*
*
1 1,
We first need to put the equation in the appropriate form,
with power flow expressed in the form ( ) :
i
i
n n
i i i i ik k i ik k
k k
n
i i i ik k
k
n
i
ik k ii i ik
k k
V h V
S V I V Y V V Y V
S V I V Y V
S
Y V Y V Y
V
= =
=
= =
=
 
= = = ÷
 
= =
= = +
∑ ∑
∑
∑
*
*
1,
S1
.
i
n
k
k i
n
i
i ik k
ii k k i
V
V Y V
Y V
≠
= ≠
 
= − ÷
 ÷
 
∑
∑
33

34. Gauss Power Flow
*
*
1,
S1
We define ( ) by: ( ) ,
Collect the entries ( ) together to form the vector ( ).
Then we have expressed the power flow equations in
the form: ( ). (There are o
i
n
i
i i ik k
ii k k i
i
h V h V Y V
Y V
h V h V
V h V
= ≠
 
= − ÷
 ÷
 
=
∑
(0)
(1) (0)
(2) (1)
(
ther ways we can express
the power flow equations in this form.)
Start with an initial guess and then update according to:
( ),
( ),...
Continue until stopping criterion satisfied:
V
V h V
V h V
V
=
=
1) ( )
.Vν ν
ε+
− <
34