Transcript Lecture 6

ECE 476
Power System Analysis
Lecture 6: Transmission Line Parameters
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
[email protected]
Special Guest Lecturer: TA Iyke Idehen
Announcements
• Please read Chapters 4 and 5
• Quiz today on HW 2
• HW 3 is 4.9 (use lecture results for 4.8
comparison), 4.12, 4.19 (just compare 4.19a to
4.19b), 4.25
•
•
•
It does not need to be turned in, but will be covered by an
in-class quiz on Thursday Sept 15
Positive sequence is same as per phase; it will be covered
in Chapter 8
Use Table A.4 values to determine the Geometric Mean
Radius of the wires (i.e., the ninth column).
1
Many-Conductor Case, cont’d
0 
R1
R2
1 

i1 ln '  i2 ln
2 
d12
r1
Rn 
 in ln

d1n 
0 
1
1
1 
i1 ln '  i2 ln


2 
d12
r1
1 
 in ln


d1n 
0
i1 ln R1  i2 ln R2 
2
 in ln Rn 
As R1 goes to infinity R1  R2  Rn so the second
0  n 
term from above can be written =   i j  ln R1
2  j 1 
2
Many-Conductor Case, cont’d
n
Therefore if
 i j  0, which is true in a balanced
j 1
three phase system, then the second term is zero and
0 
1
1
1 
i1 ln '  i2 ln


2 
d12
r1
1  L11i1  L12i2
1 
 in ln
d1n 
 L1nin
System has self and mutual inductance. However
the mutual inductance can be canceled for
balanced 3 systems with symmetry.
3
Symmetric Line Spacing – 69 kV
4
Birds Do Not Sit on the Conductors
5
Line Inductance Example
Calculate the reactance for a balanced 3, 60Hz
transmission line with a conductor geometry of an
equilateral triangle with D = 5m, r = 1.24cm (Rook
conductor) and a length of 5 miles.
Since system is assumed
balanced
ia  ib  ic
0 
1
1
1 
a 
ia ln( )  ib ln( )  ic ln( ) 

2 
r'
D
D 
6
Line Inductance Example, cont’d
Substituting
ia  ib  ic
Hence
0 
1
1 


a 
ia ln    ia ln   

2 
 r '
 D 
0
D


ia ln  
2
 r'
7
0  D  4  10
5


La 
ln   
ln 
3 
2  r ' 
2
 9.67  10 
 1.25  106 H/m
7
Conductor Bundling
To increase the capacity of high voltage transmission
lines it is very common to use a number of
conductors per phase. This is known as conductor
bundling. Typical values are two conductors for
345 kV lines, three for 500 kV and four for 765 kV.
8
Bundled Conductor Pictures
The AEP Wyoming-Jackson
Ferry 765 kV line uses
6-bundle conductors.
Conductors in a bundle are
at the same voltage!
Photo Source: BPA and American Electric Power
9
Bundled Conductor Flux Linkages
For the line shown on the left,
define dij as the distance between conductors i and j. We
can then determine  for each
i  1


1
1
1
 ln
 ln
 
 a  ln  ln

d12
d13
d14 
4  r'

 


0  ib
1
1
1
1
1 
 ln
 ln
 ln
 ln
  
2  4  d15
d16
d17
d 18  


 ic  1
1
1
1 
 ln
 ln
 ln

  ln
d1,10
d1,11
d1,12  
 4  d19
10
Bundled Conductors, cont’d
Simplifying




1
i ln 



1
a 


4
 (r ' d12 d13d14 ) 






0 
1



1 
i
ln

b

2
 ( d d d d ) 14  

 15 16 17 18
 






1


ic ln

1

4 

(
d
d
d
d
)
 19 1,10 1,11 1,12

11
Bundled Conductors, cont’d
Rb
geometric mean radius (GMR) of bundle
 (r ' d12 d13d14 )
 (r ' d12
D1b
d1b
1
4
1
) b
for our example
in general
geometric mean distance (GMD) of
conductor 1 to phase b.
 (d15d16 d17 d18 )
D1c
1
4
 D2b  D3b  D4b  Dab
 (d19 d1,10 d1,11d1,12 )
1
4
 D2c  D3c  D4c  Dac
12
Inductance of Bundle
If D ab  Dac  Dbc  D and i a  ib  ic
Then
 1 
0 
1 

1 
ia ln    ia ln   

2 
 D 
 Rb 
D
0

I a ln  
2
 Rb 
D
0

4 I1 ln  
2
 Rb 
D
0
L1 
 4  ln  
2
 Rb 
13
Inductance of Bundle, cont’d
But remember each bundle has b conductors
in parallel (4 in this example). So
0  D 
La  L1 / b 
ln  
2  Rb 
14
Bundle Inductance Example
Consider the previous example of the three phases
symmetrically spaced 5 meters apart using wire
with a radius of r = 1.24 cm. Except now assume
each phase has 4 conductors in a square bundle,
spaced 0.25 meters apart. What is the new inductance
per meter?
r  1.24  102 m r '  9.67  103 m

0.25 M
0.25 M
3
R b  9.67  10  0.25  0.25  2  0.25
0.25 M

1
4
 0.12 m (ten times bigger!)
0
5
La 
ln
 7.46  107 H/m
2 0.12
15
Transmission Tower Configurations
•The problem with the line analysis we’ve done so far
is we have assumed a symmetrical tower configuration.
Such a tower figuration is seldom practical.
Therefore in
general Dab 
Dac  Dbc
Typical Transmission Tower
Configuration
Unless something
was done this would
result in unbalanced
phases
16
Transposition
• To keep system balanced, over the length of a
transmission line the conductors are rotated so each
phase occupies each position on tower for an equal
distance. This is known as transposition.
Aerial or side view of conductor positions over the length
of the transmission line.
17
Line Transposition Example
18
Line Transposition Example
19
Transposition Impact on Flux Linkages
For a uniformly transposed line we can
calculate the flux linkage for phase "a"
1 0
a 
3 2

1
1
1 
 I a ln r '  I b ln d  I c ln d  

12
13 
1 0 
1
1
1 
I a ln  I b ln
 I c ln



3 2 
r'
d13
d 23 
1 0
3 2

1
1
1 
 I a ln r '  I b ln d  I c ln d 

23
12 
“a” phase in
position “1”
“a” phase in
position “3”
“a” phase in
position “2”
20
Transposition Impact, cont’d
Recognizing that
1
1
(ln a  ln b  ln c)  ln(abc) 3
3
We can simplify so
1
 I ln 1  I ln
1
 a r' b
 d12 d13d 23  3
0 
a 
2 
1
 I c ln
1

 d12 d13d 23  3
 




21
Inductance of Transposed Line
Define the geometric mean distance (GMD)
Dm
 d12 d13d 23 
1
3
Then for a balanced 3 system ( I a  - I b - I c )
0 
Dm
1
1  0
a 
I a ln  I a ln

I a ln


2 
r'
Dm  2
r'
Hence
0 Dm
Dm
7
La 
ln
 2  10 ln
H/m
2
r'
r'
22
Inductance with Bundling
If the line is bundled with a geometric mean
radius, R b , then
0
Dm
a 
I a ln
2
Rb
0 Dm
Dm
7
La 
ln
 2  10 ln
H/m
2 Rb
Rb
Inductance Example
• Calculate the per phase inductance and reactance of
a balanced 3, 60 Hz, line with horizontal phase
spacing of 10m using three conductor bundling
with a spacing between conductors in the bundle of
0.3m. Assume the line is uniformly transposed and
the conductors have a 1cm radius.
Answer: Dm = 12.6 m, Rb= 0.0889 m
Inductance = 9.9 x 10-7 H/m, Reactance = 0.6 /Mile
Line Conductors
• Typical transmission lines use multi-strand
conductors
• ACSR (aluminum conductor steel reinforced)
conductors are most common. A typical Al. to St.
ratio is about 4 to 1.
25
Line Conductors, cont’d
• Total conductor area is given in circular mils. One
circular mil is the area of a circle with a diameter
of 0.001 =   0.00052 square inches
• Example: what is the the area of a solid, 1”
diameter circular wire?
Answer: 1000 kcmil (kilo circular mils)
• Because conductors are stranded, the equivalent
radius must be provided by the manufacturer. In
tables this value is known as the GMR and is
usually expressed in feet.
26
Line Resistance
Line resistance per unit length is given by
R =

where  is the resistivity
A
Resistivity of Copper = 1.68  10-8 Ω-m
Resistivity of Aluminum = 2.65  10-8 Ω-m
Example: What is the resistance in Ω / mile of a
1" diameter solid aluminum wire (at dc)?
2.65  10-8 Ω-m
m

R 
1609
 0.084
2
mile
mile
  0.0127m
27
Line Resistance, cont’d
• Because ac current tends to flow towards the
surface of a conductor, the resistance of a line at 60
Hz is slightly higher than at dc.
• Resistivity and hence line resistance increase as
conductor temperature increases (changes is about
8% between 25C and 50C)
• Because ACSR conductors are stranded, actual
resistance, inductance and capacitance needs to be
determined from tables.
28
Review of Electric Fields
To develop a model for line capacitance we
first need to review some electric field concepts.
Gauss's law:
A D da
= qe
(integrate over closed surface)
where
D = electric flux density, coulombs/m 2
da = differential area da, with normal to surface
A = total closed surface area, m 2
q e = total charge in coulombs enclosed
Gauss’s Law Example
•Similar to Ampere’s Circuital law, Gauss’s Law is
most useful for cases with symmetry.
•Example: Calculate D about an infinitely long wire
that has a charge density of q coulombs/meter.
Since D comes
radially out integrate over the
cylinder bounding
D
d
a

D
2

Rh

q

qh
e
A
the wire
q
D 
ar where ar radially directed unit vector
2 R
Electric Fields
The electric field, E, is related to the electric flux
density, D, by
D = E
where
E = electric field (volts/m)
 = permittivity in farads/m (F/m)
 = o r
o = permittivity of free space (8.85410-12 F/m)
r = relative permittivity or the dielectric constant
(1 for dry air, 2 to 6 for most dielectrics)
Voltage Difference
The voltage difference between any two
points P and P is defined as an integral
V

P
P
E dl
In previous example the voltage difference between
points P and P , located radial distance R and R 
from the wire is (assuming  =  o )
V
 
R
R
R
dR 
ln
2 o R
2 o R
q
q