Transcript Slides

ECE 476
POWER SYSTEM ANALYSIS
Lecture 10
Transformers, Load & Generator Models, YBus
Professor Tom Overbye
Department of Electrical and
Computer Engineering
Announcements
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Homework #4 4.34, 4.35, 5.14, 5.26 is due now
Homework #5 is 3.12, 3.14, 3.19, 60 due Oct 2nd
(Thursday)
First exam is 10/9 in class; closed book, closed
notes, one note sheet and calculators allowed
Start reading Chapter 6 for lectures 11 and 12
1
In the News: Solar Energy
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On 9/23/08 US Senate passed a bill that would
greatly expand the tax credits available for solar
energy installations.
The professor Chapman experience is available at
http://www.patrickchapman.com/solar.htm.
But there are environmental
objections to largescale solar projects
Picture source: NYTimes, 9/23/08
2
ECE PowerLunches
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ECE Department (through ECE Student Advisory
Committee) sponsors “PowerLunches” to encourage
undergraduates to go out to lunch with a professor
of their choice
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The lunch, which takes place in the Illini Ballroom, is
free for the students (no more than three) and the
professor
Details can be found at
http://sac.ece.uiuc.edu/newpage/pwrlunch.php
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Load Tap Changing Transformers
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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.
Unbalanced tap positions can cause "circulating
vars"
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Phase Shifting Transformers
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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.
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Phase Shifting Transformer Picture
Costs about $7 million,
weighs about 1.2
million pounds
230 kV 800 MVA Phase Shifting
Transformer During factory testing
Source: Tom Ernst, Minnesota Power
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ComED Control Center
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ComED Phase Shifter Display
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Autotransformers
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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!
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Load Models
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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
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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
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constant power: Si = Pi + jQi
constant impedance: Si = |V|2 / Zi
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Generator Models
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Engineering models depend upon application
Generators are usually synchronous machines
For generators we will use two different models:
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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
a short term model treating the generator as a constant
voltage source behind a possibly time-varying reactance
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Power Flow Analysis
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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)
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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
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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
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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
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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
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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
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Multiple Solution Example 3
The dc system shown below has two solutions:
where the 18 watt
load is a resistive
load
The equation we're solving is
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 9 volts 
I RLoad  
RLoad  18 watts

 1 +R Load 
What is the
One solution is R Load  2
maximum
Other solution is R Load  0.5
PLoad?
2
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Bus Admittance Matrix or Ybus
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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
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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
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Ybus Example, cont’d
By KCL at bus 1 we have
I1
IG1  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
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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)
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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.
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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
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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
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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
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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
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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
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Power Flow Analysis
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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
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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*
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Power Balance Equations, cont’d
*
n


* *
Si 
 Vi   YikVk   Vi  YikVk
 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
*
Vi I i
n
Yik
Gik  jBik
Vi
Vi e ji  Vi  i
 ik
i   k
Recall e j  cos  j sin 
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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
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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   YikVk   Vi  Yik*Vk*
 k 1

k 1
Rather, we must pursue an iterative approach.
Vi I i*
n
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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
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Gauss Iteration Example
Example: Solve x - x  1  0
x ( v 1)  1  x ( v )
Let k = 0 and arbitrarily guess x (0)  1 and solve
k
x
0
1
2
3
4
(v)
(v)
k
x
1
2
2.41421
5
6
7
2.61185
2.61612
2.61744
2.55538
2.59805
8
9
2.61785
2.61798
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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
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Gauss Power Flow
We first need to put the equation in the correct form
n


 Vi   YikVk   Vi  Yik*Vk*
 k 1

k 1
Si  Vi I i*
*
Si

S*i

*
Vi
*
n
 Vi
*
Vi I i
*
n
 YikVk
k 1
n
 YikVk
k 1
 YiiVi 
 Vi
*
n
 YikVk
k 1
n

k 1,k i
YikVk
n

1  S*i
Vi 
 *   YikVk 

Yii  V
k 1,k i

i
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