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ECE 476
POWER SYSTEM ANALYSIS
Lecture7
Development of Transmission Line Models
Professor Tom Overbye
Department of Electrical and
Computer Engineering
Announcements
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
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For next two lectures read Chapter 5.
HW 2 is 4.10 (positive sequence is the same here as per
phase), 4.18, 4.19, 4.23. Use Table A.4 values to determine
the Geometric Mean Radius of the wires (i.e., the ninth
column). Due September 15 in class.
“Energy Tour” opportunity on Oct 1 from 9am to 9pm. Visit
a coal power plant, a coal mine, a wind farm and a bio-diesel
processing plant. Sponsored by Students for Environmental
Concerns. Cost isn’t finalized, but should be between $10
and $20. Contact Rebecca Marcotte at
[email protected] for more information or to sign up.
1
SDGE Transmission Grid (From CALISO
2009 Transmission Plan)
2
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.
3
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.
4
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
5
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.
6
Variation in Line Resistance Example
7
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
8
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.
A D da
D 
 D 2 Rh  q e  qh
q
2 R
Since D comes
radially out integrate over the
cylinder bounding
the wire
ar where ar radially directed unit vector
9
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)
10
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
11
Voltage Difference, cont’d
With
V
 
R
R
R
dR 
ln
2 o R
2 o R
q
q
if q is positive then those points closer in have
a higher voltage. Voltage is defined as the energy
(in Joules) required to move a 1 coulomb charge
against an electric field (Joules/Coulomb). Voltage
is infinite if we pick infinity as the reference point
12
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.
13
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 more P to infinity, ln
0
R 1
14
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!
15
Three Conductor Case
Assume we have three
infinitely long conductors,
A, B, & C, each with radius r
C
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
16
Line Capacitance
For a single line 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 Vj
 q1 
 C11
   
 

 qn 
Cn1
q  CV
C1n  V1 
 
 
Cnn  Vn 
17
Line Capacitance, cont’d
In ECE 476 we will not be considering theses
cases with mutual capacitance. To eliminate
mutual capacitance we'll again assume we have
a uniformly transposed line. For the previous
three conductor example:
Va  V
Since q a = C Va

qa
2
C 

Va
ln D r
18
Bundled Conductor Capacitance
Similar to what we did for determining line
inductance when there are n bundled conductors,
we use the original capacitance equation just
substituting an equivalent radius
R cb
 (rd12
1
d1n )
n
Note for the capacitance equation we use r rather
than r' which was used for R b in the inductance
equation
19
Line Capacitance, cont’d
For the case of uniformly transposed lines we
use the same GMR, D m , as before.
ln
2
Dm

 d ab d ac dbc 
C 
c
Rb
where
Dm
c
Rb
 (rd12
1
d1n )
n
1
3
(note r NOT r')
ε in air   o  8.854  10-12 F/m
20
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.
21
Line Capacitance Example, cont’d
Rbc
Dm
C
Xc

1
(0.01  0.3  0.3) 3

1
(10  10  20) 3
 0.0963 m
 12.6 m
2  8.854  1012

 1.141  1011 F/m
12.6
ln
0.0963
1
1


11
C
2 60  1.141  10 F/m
 2.33  108 -m (not  / m)
22
ACSR Table Data (Similar to Table A.4)
GMR is equivalent to r’
Inductance and Capacitance
assume a Dm of 1 ft.
23
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 assuming
a one foot spacing
Term independent
of conductor with
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 independent
Term from table assuming
of conductor with
a one foot 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
X a  2 60  2  10  1609  ln
Ω/mile
0.0313
X a  0.420 Ω/mile, which matches the table
7
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
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
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 Western grids).
29
Tree Trimming: Before
30
Tree Trimming: After
31
Transmission Line Models


Previous lectures have covered how to calculate the
distributed inductance, capacitance and resistance of
transmission lines.
In this section we will use these distributed
parameters to develop the transmission line models
used in power system analysis.
32
Transmission Line Equivalent Circuit
Our current model of a transmission line is shown
below
Units on
z and y are
per unit
length!
For operation at frequency  , let z = r + jL
and y = g +jC (with g usually equal 0)
33
Derivation of V, I Relationships
We can then derive the following relationships:
dV  I z dx
dI
 (V  dV ) y dx  V y dx
dV ( x)
dI ( x)
 zI
 yV
dx
dx
34
Setting up a Second Order Equation
dV ( x)
dI ( x)
 zI
 yV
dx
dx
We can rewrite these two, first order differential
equations as a single second order equation
d 2V ( x)
dI ( x)
z
 zyV
2
dx
dx
d 2V ( x)
 zyV  0
2
dx
35
V, I Relationships, cont’d
Define the propagation constant  as
  yz    j 
where
  the attenuation constant
  the phase constant
Use the Laplace Transform to solve. System
has a characteristic equation
( s 2   2 )  ( s   )( s   )  0
36
Equation for Voltage
The general equation for V is
V ( x)  k1e x  k2e  x
Which can be rewritten as
e x  e  x
e x  e  x
V ( x)  (k1  k2 )(
)  (k1  k2 )(
)
2
2
Let K1  k1  k2 and K 2  k1  k2 . Then
x
 x
x
e
e e
V ( x)  K1 (
)  K2 (
2
2
 K1 cosh( x)  K 2 sinh( x)
e
 x
)
37
Real Hyperbolic Functions
For real x the cosh and sinh functions have the
following form:
d cosh( x)
  sinh( x)
dx
d sinh( x)
  cosh( x)
dx
38
Complex Hyperbolic Functions
For x =  + j the cosh and sinh functions have the
following form
cosh x
 cosh  cos   j sinh  sin 
sinh x
 sinh  cos   j cosh  sin 
39