Transcript EE369 POWER SYSTEM ANALYSIS Lecture 6 Development of Transmission Line Models Tom Overbye and Ross Baldick.

EE369
POWER SYSTEM ANALYSIS
Lecture 6
Development of Transmission Line Models
Tom Overbye and Ross Baldick
1
Homework
• HW 5 is Problems 4.24, 4.25 (assume Cardinal
conductor and look up GMR in Table A.4),
4.26, 4.33, 4.36, 4.38, 4.49, 4.1, 4.3, 4.6; due
Thursday 10/8.
• HW 6 is problems 5.2, 5.4, 5.7, 5.9, 5.14, 5.16,
5.19, 5.26, 5.31, 5.32, 5.33, 5.36; case study
questions chapter 5 a, b, c, d, is due Thursday,
10/15.
2
Review of Electric Fields
To develop a model for transmission line capacitance
we first need to review some electric field concepts.
Gauss's law relating electric flux to enclosed charge):
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,
qe = total charge in coulombs enclosed
3
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
the wire.
D is perpendicular
A D da  D2 Rh  qe  qh
to ends of cylinder.
q
D 
ar where ar radially directed unit vector
4
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)
5
Voltage Difference
The voltage difference between any two
points P and P is defined as an integral
V

P
P
E dl ,
where the integral is along any path
from point P to point P .
6
Voltage Difference
In previous example, E

q
2 o R
ar , with ar radial.
Consider points P and P , located radial distance R and R
from the wire and collinear with the wire.
Define R to be the radial distance from the wire
q
on the path from points P to P , so E dl 
dR
2 o R
Voltage difference between P and P (assuming  =  o ) :
V
 
R
R
R
dR 
ln
2 o R
2 o R
q
q
7
Voltage Difference, cont’d
Repeating:
V
 
R
R
R
dR 
ln
2 o R
2 o R
q
q
So, if q is positive then those points closer to the
charge have a higher voltage.
The voltage between two points (in volts)
is equal to the amount of energy (in joules)
required to move a 1 coulomb charge
against the electric field between the two points.
Voltage is infinite if we pick one of the points to be
8
infinitely far away.
Multi-Conductor Case
Now assume we have n parallel conductors,
each with a charge density of qi coulombs/m.
The voltage difference between our two points,
P and P , is now determined by superposition
V
n
R i

qi ln

2 i 1
R i
1
where R i is the radial distance from point P
to conductor i, and R i the distance from P to i.
9
Multi-Conductor Case, cont’d
n
If we assume that
 qi  0 then rewriting
i =1
V
1
1 n

qi ln

qi ln R i


2 i 1
R i 2 i 1
1
n
n
We then subtract
 qi ln R 1  0
i 1
V
R i
1
1 n

qi ln

qi ln


2 i 1
R i 2 i 1
R 1
1
n
R i
As we move P to infinity, ln
0
R 1
10
Absolute Voltage Defined
Since the second term goes to zero as P goes to
infinity, we can now define the voltage of a
point w.r.t. a reference voltage at infinity:
V
1
n
1

qi ln

2 i 1
R i
This equation holds for any point as long as
it is not inside one of the wires!
Since charge will mostly be on the surface
of a conductor, the voltage inside will equal
the voltage at the surface of the wire.
11
Three Conductor Case
Assume we have three
infinitely long conductors,
A, B, & C, each with radius r
B
and distance D from the
other two conductors.
Assume charge densities such
that qa + qb + qc = 0
1 
1
1
1
Va 
q
ln

q
ln

q
ln
a
b
c
2 
r
D
D 
qa
D
Va 
ln
2 r
A
C
12
Line Capacitance
For a single capacitor, capacitance is defined as
qi  CiVi
But for a multiple conductor case we need to
use matrix relationships since the charge on
conductor i may be a function of V j
 q1 
 C11
   
 

 qn 
Cn1
q  CV
C1n  V1 
 
 
Cnn  Vn 
13
Line Capacitance, cont’d
We will not be considering the
cases with mutual capacitance. To eliminate
mutual capacitance we'll again assume we have
a uniformly transposed line, using similar arguments
to the case of inductance. For the previous
three conductor example:
qa
2
Since qa = C Va 
C 

Va
ln D
r
14
Bundled Conductor Capacitance
Similar to the case for determining line
inductance when there are n bundled conductors,
we use the original capacitance equation just
substituting an equivalent radius
Rbc
 (rd12
d1n )
1
n
Note for the capacitance equation we use r rather
than r ' which was used for Rb in the inductance
equation
15
Line Capacitance, cont’d
For the case of uniformly transposed lines we
use the same GMR, Dm , as before.
2
C 
 Dm 
ln 
c
R

b
where
Dm
c
Rb

d abd ac d bc 
 ( rd12
d 1n )
1
n
1
3
(note r NOT r ')
ε in air   o  8.854  10-12 F/m
16
Line Capacitance Example
•Calculate the per phase capacitance and susceptance
of a balanced 3, 60 Hz, transmission 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 a 1cm radius.
17
Line Capacitance Example, cont’d
Rbc
Dm
C
Xc
 (0.01  0.3  0.3)
 (10  10  20)
1
3
1
3
 0.0963 m
 12.6 m
2  8.854  1012
11

 1.141  10 F/m
12.6
ln
0.0963
1
1


C
2 60  1.141  1011 F/m
 2.33  108 -m (not  / m)
18
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.
19
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, and so has area   0.00052 square inches
 Example: what is the area of a solid, 1” diameter
circular wire?
Answer: 1000 kcmil (kilo circular mils)
 Because conductors are stranded, the inductance and
resistance are not exactly given by using the actual
diameter of the conductor.
 For calculations of inductance, the effective radius
must is provided by the manufacturer. In tables this
value is known as the GMR and is usually expressed
in feet.
20
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 Ω-m
Example: What is the resistance in Ω / mile of a
-8
1" diameter solid aluminum wire (at dc)?
2.65  10-8 Ω-m
m

R 
1609
 0.084
2 2
mile
mile
  (0.0127) m
21
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.
22
ACSR Table Data (Similar to Table A.4)
GMR is equivalent to
effective radius r’
Inductance and Capacitance
assume a geometric mean
23
distance Dm of 1 ft.
ACSR Data, cont’d
Dm
X L  2 f L  4 f  10 ln
 1609 /mile
GMR
1

3 
 2.02  10 f ln
 ln Dm 
 GMR

1
3
 2.02  10 f ln
 2.02  103 f ln Dm
GMR
7
Term from table,
depending on conductor type,
but assuming a one foot spacing
Term independent
of conductor, but
with spacing Dm in feet
24 .
ACSR Data, Cont.
To use the phase to neutral capacitance from table
2 0
1
XC 
-m where C 
Dm
2 f C
ln
r
Dm
1
6

 1.779  10 ln
-mile (table is in M-mile)
f
r
1
1 1

 1.779  ln   1.779  ln Dm M-mile
f
r f
Term from table,
Term independent
depending on conductor type,
of conductor, but
but assuming a one foot spacing with spacing Dm in feet
25 .
Dove Example
GMR  0.0313 feet
Outside Diameter = 0.07725 feet (radius = 0.03863)
Assuming a one foot spacing at 60 Hz
1
7
X a  2 60  2  10  1609  ln
Ω/mile
0.0313
X a  0.420 Ω/mile, which matches the table
For the capacitance
1
1
6
X C   1.779  10 ln  9.65  104 Ω-mile
f
r
26
Additional Transmission Topics
 Multi-circuit lines: Multiple lines often share a
common transmission right-of-way. This DOES cause
mutual inductance and capacitance, but is often
ignored in system analysis.
 Cables: There are about 3000 miles of underground ac
cables in U.S. Cables are primarily used in urban areas.
In a cable the conductors are tightly spaced, (< 1ft)
with oil impregnated paper commonly used to provide
insulation
– inductance is lower
– capacitance is higher, limiting cable length
27
Additional Transmission topics
 Ground wires: Transmission lines are usually
protected from lightning strikes with a ground
wire. This topmost wire (or wires) helps to
attenuate the transient voltages/currents that
arise during a lighting strike. The ground wire is
typically grounded at each pole.
 Corona discharge: Due to high electric fields
around lines, the air molecules become ionized.
This causes a crackling sound and may cause the
line to glow!
28
Additional Transmission topics
 Shunt conductance: Usually ignored. A small
current may flow through contaminants on
insulators.
 DC Transmission: Because of the large fixed
cost necessary to convert ac to dc and then back
to ac, dc transmission is only practical for
several specialized applications
– long distance overhead power transfer (> 400 miles)
– long cable power transfer such as underwater
– providing an asynchronous means of joining
different power systems (such as the Eastern and
ERCOT grids).
29