Transcript EE369 POWER SYSTEM ANALYSIS

EE369
POWER SYSTEM ANALYSIS
Lecture 6
Development of Transmission Line Models
Tom Overbye and Ross Baldick
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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.
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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
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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  D 2 Rh  qe  qh
to ends of cylinder.
q
D 
ar where ar radially directed unit vector
2 R
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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)
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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 .
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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
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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
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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.
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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
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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.
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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
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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 
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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
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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
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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
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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.
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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)
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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.
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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.
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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 Ω-m
m

R 
1609
 0.084
2 2
mile
mile
  (0.0127) m
-8
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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.
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ACSR Table Data (Similar to Table A.4)
GMR is equivalent to
effective radius r’
Inductance and Capacitance
assume a geometric mean
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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
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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
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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!
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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).
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