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Lecture 3Three Phase, Power System Operation
Professor Tom OverbyeDepartment of Electrical and
Computer Engineering
ECE 476
POWER SYSTEM ANALYSIS
2
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
3
Three Phase Transmission Line
4
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
5
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.
6
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.
7
Per Phase Example
Assume a 3, Y-connected generator with Van = 10 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” = 10 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
8
Per Phase Example, cont’d
First convert the delta load and source to equivalent
Y values and draw just the "a" phase circuit
9
Per Phase Example, cont’d
' ' 'a a a
To solve the circuit, write the KCL equation at a'
1(V 1 0)( 10 ) V (3 ) (V j
3j j
10
Per Phase Example, cont’d
' ' 'a a a
'a
' 'a b
' 'c ab
To solve the circuit, write the KCL equation at a'
1(V 1 0)( 10 ) V (3 ) (V j
310
(10 60 ) V (10 3 10 )3
V 0.9 volts V 0.9 volts
V 0.9 volts V 1.56
j j
j j j j
volts
11
Per Phase Example, cont’d
*'*
ygen
*" '"
S 3 3 5.1 3.5 VA0.1
3 5.1 4.7 VA0.1
a aa a a
a agen a
V VV I V j
j
V VS V j
j
12
Example 2.14
13
Example 2.21
14
Example 2.29
15
Example 2.44
16
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
17
Primary Methods for Power Transfer
The most common methods for transfer of electric power are
1) Overhead ac
2) Underground ac
3) Overhead dc
4) Underground dc
5) other
18
Magnetics Review
Ampere’s circuital law:
e
F = mmf = magnetomtive force (amp-turns)
= magnetic field intensity (amp-turns/meter)
d = Vector differential path length (meters)
= Line integral about closed path (d is tangent to path)
I =
eF d I
H l
H
l
l
Algebraic sum of current linked by
19
Line Integrals
Line integrals are a generalization of traditional integration
Integration along thex-axis
Integration along ageneral path, whichmay be closed
Ampere’s law is most useful in cases of symmetry, such as with an infinitely long line
20
Magnetic Flux Density
Magnetic fields are usually measured in terms of flux density
0-7
0
= flux density (Tesla [T] or Gauss [G])(1T = 10,000G)
For a linear a linear magnetic material
= where is the called the permeability
=
= permeability of freespace = 4 10
= relative permea
r
r
H m
B
B H
bility 1 for air
21
Magnetic Flux
Total flux passing through a surface A is
=
= vector with direction normal to the surface
If flux density B is uniform and perpendicular to an area A then
=
Ad
d
BA
B a
a
22
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 givenby the “Right-hand” Rule
Easiest way to solve the problem is to take advantage of symmetry. For an integration path we’ll choose acircle with a radius of x.
23
Single Line Example, cont’d
0
5 00 4
70
5
4
22
25 5
ln 2 10 ln2 4 4
4.46 10 Wb
2 10 2B T Gauss
x
A
IxH I H
xB H
IH dA dx
xI
I
x
For reference, the earth’smagnetic field is about0.6 Gauss (Central US)
24
Flux linkages and Faraday’s law
N
i=1
Flux linkages are defined from Faraday's law
dV = where V = voltage, = flux linkages
The flux linkages tell how much flux is linking an
N turn coil:
=
If all flux links every coil then
i
dt
N
25
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
= L i
where L has units of Henrys (H)
26
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
27
Inductance Example, cont’d
0
0
20
2 (path length is 2 R)
H2
2
H2
e
r
r
r
I d
NI H R
NIB H H
RAB N LI
NINAB NA
R
N AL
R
H l
28
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.
29
Flux Linkages outside of the wire
R0A r
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
d length 2Idxx
B a
30
Flux Linkages outside, cont’d
R0A r
R 00r
d length 2
Since length = we'll deal with per unit length values,
assumed to be per meter.
ln2 2
Note, this quantity still goes to infinity as R
Idxx
I Rdx Imeter x r
B a
31
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
1the skin depth = where f is
f the frequency in Hz
and is the conductivity in mhos/meter.
0.066 mFor copper skin depth 0.33 inch at 60HZ.
fFor derivation we'll assume a uniform current density.
32
Flux linkages inside, cont’d
Wire cross section
x
r
2
2
2
Current enclosed within distance
x of center I
2 2
e
ex
xI
rI Ix
Hx r
2 30
inside 2 2 40 0
Flux only links part of current
2 82
r r rIx x Ixdx dx I
r r r
33
Line Total Flux & Inductance
0 0
0
0
(per meter) ln2 8
(per meter) ln2 4
L(per meter) ln2 4
Note, this value still goes to infinity as we integrate
R out to infinity
rTotal
rTotal
r
RI I
rR
Ir
Rr
34
Inductance Simplification
0 0 4
0 4
Inductance expression can be simplified using
two exponential identities:
aln(ab)=ln a + ln b ln ln ln ln( )
b
ln ln ln ln2 4 2
ln ln2
r
r
a
r
a b a e
RL R r e
r
L R re
0
4r
ln2 '
Where r' 0.78 for 1r
Rr
r e r
35
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.
R
Creates counter-
clockwise field
Creates a
clockwise field
To determine the
inductance of each
conductor we integrate
as before. However
now we get some
field cancellation
36
Two Conductor Case, cont’d
R R
Direction of integration
Rp
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.
0 0left
For distance Rp, greater than 2R, from left line
ln ln2 ' 2
Rp Rp RI I
r R
Left Current Right Current
37
Two Conductor Inductance
0left
0
0
0
0
Simplifying (with equal and opposite currents)
ln ln2 '
ln ln ' ln( ) ln2
ln ln2 '
ln as Rp2 '
ln H/m 2 'left
Rp Rp RI
r R
I Rp r Rp R R
R RpI
r Rp R
RI
r
RL
r