Transcript Lecture 11

ECE 476
Power System Analysis
Lecture 11: Load and Generators, Bus
Admittance Matrix
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
[email protected]
Announcements
• Please read Chapter 2.4, Chapter 6 up to 6.6
• HW 5 is 5.31, 5.43, 3.4, 3.10, 3.14, 3.19, 3.23, 3.60,
6.30 should be done before exam 1
• Exam 1 is Thursday Oct 6 in class
•
•
Closed book, closed notes, but you may bring one 8.5 by 11
inch note sheet and standard calculators
Last name A-M here, N to Z in ECEB 1013
• Power Area Scholarships, Awards (Oct 1 deadlines,
except Nov 1 for Grainger)
•
•
http://energy.ece.illinois.edu/
Turn both into Prof Sauer (4046 or 4060)
1
UIUC Solar Farm Output: 9/7/16
Data from Keith Erickson, UIUC
Details on UIUC Solar Farm at icap.sustainability.illinois.edu/files/project/175/solar-farm-project-fact-sheet_withfaqs_121515%282%29.pdf
Autotransformers
• Autotransformers are transformers in which the
primary and secondary windings are coupled
magnetically and electrically.
• This results in lower cost, smaller size and weight,
lower leakage reactance
• The key disadvantage is loss of electrical isolation
between the voltage levels. Hence auto-transformers
are not used when turns ratio is large. For example in
stepping down 7160/240 V we do not ever want 7160
on the low side!
• Modeled as regular transformers in power flow
3
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
4
Scientific Modeling Quotes
• "All models are wrong but some are useful,“
–
–
George Box, Empirical Model-Building and Response
Surfaces, (1987, p. 424)
Box went on to say that the practical question is how wrong
they have to be to not be useful
• “Everything should be made as
simple as possible, but not simpler.”
–
Albert Einstein [maybe]
• “With four parameters I can fit an elephant, and
with five I can make him wiggle his trunk”.
–
John von Neumann
Drawing an elephant with four complex parameters” by Jurgen Mayer, Khaled Khairy, and Jonathon Howard, Am. J. Phys. 78,
648 (2010), DOI:10.1119/1.3254017
Generator Models
• Engineering models depend upon application
• Generators are usually synchronous machines
• For generators we will use two different models:
–
–
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
6
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
7
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
9
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
10
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:
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?
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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
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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
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
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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 )

 V1
Z
2
1
I1
 12  j16
0.03  j 0.04
What if
 I1 
12  j15.9 12  j16  V1 
we remove

I 
 12  j16 12  j15.9  V 
the shunts?

 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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