Transcript ECE 310 - University of Illinois at Urbana–Champaign

ECE 476
POWER SYSTEM ANALYSIS
Lecture 3
Three Phase, Power System Operation
Professor Tom Overbye
Department of Electrical and
Computer Engineering
Reading and Homework
•
•
For lecture 3 please be reading Chapters 1 and 2
For lectures 4 through 6 please be reading Chapter 4
–
•
we will not be covering sections 4.7, 4.11, and 4.12 in
detail though you should still at least skim those sections.
HW 1 is 2.9, 22, 28, 32, 48; due Thursday 9/8
•
For Problem 2.32 you need to use the PowerWorld
Software. You can download the software and cases at
the below link; get version 15.
http://www.powerworld.com/gloversarma.asp
1
Three Phase Transmission Line
2
Per Phase Analysis


Per phase analysis allows analysis of balanced 3
systems with the same effort as for a single phase
system
Balanced 3 Theorem: For a balanced 3 system
with
–
–
All loads and sources Y connected
No mutual Inductance between phases
3
Per Phase Analysis, cont’d

Then
–
–
–
All neutrals are at the same potential
All phases are COMPLETELY decoupled
All system values are the same sequence as sources. The
sequence order we’ve been using (phase b lags phase a
and phase c lags phase a) is known as “positive”
sequence; later in the course we’ll discuss negative and
zero sequence systems.
4
Per Phase Analysis Procedure
To do per phase analysis
1. Convert all  load/sources to equivalent Y’s
2. Solve phase “a” independent of the other phases
3. Total system power S = 3 Va Ia*
4. If desired, phase “b” and “c” values can be
determined by inspection (i.e., ±120° degree phase
shifts)
5. If necessary, go back to original circuit to determine
line-line values or internal  values.
5
Per Phase Example
Assume a 3, Y-connected generator with Van = 10
volts supplies a -connected load with Z = -j
through a transmission line with impedance of j0.1
per phase. The load is also connected to a
-connected generator with Va”b” = 10 through a
second transmission line which also has an impedance
of j0.1 per phase.
Find
1. The load voltage Va’b’
2. The total power supplied by each generator, SY and
S
6
Per Phase Example, cont’d
First convert the delta load and source to equivalent
Y values and draw just the "a" phase circuit
7
Per Phase Example, cont’d
To solve the circuit, write the KCL equation at a'
(Va'
 10)(10 j )  Va' (3 j )  (Va'
1

   j  
3
8
Per Phase Example, cont’d
To solve the circuit, write the KCL equation at a'
(Va'
 10)(10 j )  Va' (3 j )  (Va'
1

   j  
3
10
(10 j 
60)  Va' (10 j  3 j  10 j )
3
Va'  0.9  volts
Vb'  0.9  volts
Vc'  0.9 volts
'
Vab
 1.56 volts
9
Per Phase Example, cont’d
*
'
 Va  Va 
Sygen  3Va I a*  3Va 
  5.1  j3.5 VA
 j 0.1 
"

V
Sgen  3Va"  a
' *
 Va
  5.1  j 4.7 VA
 j 0.1 
10
Example 2.14
11
Example 2.21
12
Example 2.29
13
Example 2.44
14
Development of Line Models
Goals of this section are
1)
develop a simple model for transmission lines
2)
gain an intuitive feel for how the geometry of the
transmission line affects the model parameters
15
Primary Methods for Power Transfer

1)
2)
3)
4)
5)
The most common methods for transfer of electric
power are
Overhead ac
Underground ac
Overhead dc
Underground dc
other
16
Magnetics Review
Ampere’s circuital law:
F
  H dl  I e
F = mmf = magnetomtive force (amp-turns)
H = magnetic field intensity (amp-turns/meter)
dl = Vector differential path length (meters)

Ie
= Line integral about closed path 
(dl is tangent to path)
= Algebraic sum of current linked by 
17
Line Integrals
Line integrals are a generalization of traditional
integration
Integration along the
x-axis
Integration along a
general path, which
may be closed
Ampere’s law is most useful in cases of symmetry,
such as with an infinitely long line
18
Magnetic Flux Density
Magnetic fields are usually measured in terms of flux
density
B = flux density (Tesla [T] or Gauss [G])
(1T = 10,000G)
For a linear a linear magnetic material
B =  H where  is the called the permeability
 = 0  r
0 = permeability of freespace = 4  10-7 H m
 r = relative permeability  1 for air
19
Magnetic Flux
Total flux passing through a surface A is
 =
A B da
da = vector with direction normal to the surface
If flux density B is uniform and perpendicular to an
area A then
 = BA
20
Magnetic Fields from Single Wire
Assume we have an infinitely long wire with current
of 1000A. How much magnetic flux passes through a
1 meter square, located between 4 and 5 meters from
the wire?
Direction of H is given
by the “Right-hand” Rule
Easiest way to solve the problem is to take advantage
of symmetry. For an integration path we’ll choose a
circle with a radius of x.
21
Single Line Example, cont’d
2 xH  I  H 
B  0 H
I
2 x
0 I
  A 0 H dA  4
dx
2 x
5
I
5
  0 ln
2 4
5
 2  10 I ln
4
7
  4.46  105 Wb
For reference, the earth’s
2  104
2
B 
T 
Gauss magnetic field is about
x
x
0.6 Gauss (Central US)
22
Flux linkages and Faraday’s law
Flux linkages are defined from Faraday's law
d
V =
where V = voltage,  = flux linkages
dt
The flux linkages tell how much flux is linking an
N turn coil:
 =
N
i
i=1
If all flux links every coil then   N
23
Inductance
For a linear magnetic system, that is one where
B =H
we can define the inductance, L, to be
the constant relating the current and the flux
linkage
 =Li
where L has units of Henrys (H)
24
Inductance Example
Calculate the inductance of an N turn coil wound
tightly on a torodial iron core that has a radius of R
and a cross-sectional area of A. Assume
1) all flux is within the coil
2) all flux links each turn
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Inductance Example, cont’d
Ie 
 H dl
NI  H 2 R (path length is 2 R)
NI
B   H  r 0 H
H 
2 R
  N  LI
  AB
NI
  NAB  NA r  0
2 R
N 2 A r  0
H
L 
2 R
26
Inductance of a Single Wire
To development models of transmission lines, we first
need to determine the inductance of a single, infinitely
long wire. To do this we need to determine the wire’s
total flux linkage, including
1. flux linkages outside of the wire
2. flux linkages within the wire
We’ll assume that the current density within the wire is
uniform and that the wire has a radius of r.
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Flux Linkages outside of the wire
We'll think of the wire as a single loop closed at
infinity. Therefore  = since N = 1. The flux linking
the wire out to a distance of R from the wire center is
 
A B da
 length
R
r
0
I
2 x
dx
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Flux Linkages outside, cont’d
   
A B da
 length
R
r
0
I
dx
2 x
Since length =  we'll deal with per unit length values,
assumed to be per meter.


R
r
0
I
dx
0
R

I ln
2
r
2 x
Note, this quantity still goes to infinity as R  
meter
29
Flux linkages inside of wire
Current inside conductor tends to travel on the outside
of the conductor due to the skin effect. The pentration
of the current into the conductor is approximated using
the skin depth =
1
where f is the frequency in Hz
 f
and  is the conductivity in mhos/meter.
0.066 m
For copper skin depth 
 0.33 inch at 60HZ.
f
For derivation we'll assume a uniform current density.
30
Flux linkages inside, cont’d
Wire cross section Current enclosed within distance
x2
x of center  Ie  2 I
r
Ie
Ix
Hx 

2 x 2 r 2
x
r
Flux only links part of current
inside
Ix x 2

 
dx 
2
2
0
2
2 r r
r
0  r
Ix3
0 r 4 dx  8 I
r
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Line Total Flux & Inductance
0
R 0  r
Total (per meter) 
I ln 
I
2
r
8
0  R  r 
Total (per meter) 
I  ln  
2  r 4 
0  R  r 
L(per meter) 
 ln  
2  r 4 
Note, this value still goes to infinity as we integrate
R out to infinity
32
Inductance Simplification
Inductance expression can be simplified using
two exponential identities:
a
ln(ab)=ln a + ln b
ln  ln a  ln b a  ln(e a )
b
 r  

0  R  r  0 
4
L
ln
R

ln
r

ln
e
 ln   



2  r 4  2 


  r 4   0 R
0 
L
ln
 ln R  ln  re


2 

  2 r '
Where r'
re
 r
4
 0.78r for r  1
33
Two Conductor Line Inductance
Key problem with the previous derivation is we
assumed no return path for the current. Now consider
the case of two wires, each carrying the same current
I, but in opposite directions; assume the wires are
separated by distance R.
To determine the
R
inductance of each
conductor we integrate
as before. However
Creates counter- Creates a
now we get some
clockwise field clockwise fieldfield cancellation
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Two Conductor Case, cont’d
R
R
Rp
Direction of integration
Key Point: As we integrate for the left line, at distance 2R from
the left line the net flux linked due to the Right line is zero!
Use superposition to get total flux linkage.
For distance Rp, greater than 2R, from left line
left
0
Rp 0
Rp  R 


I ln

I ln 

2
r ' 2
 R 
Left Current
Right Current
35
Two Conductor Inductance
Simplifying (with equal and opposite currents)
left
Lleft
0  Rp
Rp  R  


I  ln
 ln 

2  r '
R


0

I  ln Rp  ln r ' ln( Rp  R)  ln R 
2
0  R
Rp 

I  ln  ln
2  r '
Rp  R 
0  R 

I  ln  as Rp  
2  r ' 
0  R 

 ln  H/m
2  r ' 
36