Lecture 6

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
A
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):
d = (integrate over closed surface)eq∫ D ag
2
where
= electric flux density, coulombs/m
d = differential area da, with normal to surface
A = total closed surface,
= total charge in coulombs enclosedeq
D
a
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
to ends of cylinder.
A
d 2
where radially directed unit vector
2
eD Rh q qh
q
R
π
π
= = =
=
∫
r r
D a
D a a
g
4

Electric Fields
•The electric field, E, is related to the electric flux
density, D, by
• D = ε E
•where
• E = electric field (volts/m)
• ε = permittivity in farads/m (F/m)
• ε = εoεr
• εo = permittivity of free space (8.854×10-12
F/m)
• εr = relative permittivity or the dielectric
constant
(≈1 for dry air, 2 to 6 for most dielectrics)
5

Voltage Difference
P
P
The voltage difference between any two
points P and P is defined as an integral
V ,
where the integral is along any path
from point P to point P .
dβ
α
α β
βα
α β
−∫ E l@ g
6

Voltage Difference
In previous example, , with radial.
2
Consider points P and P , located radial distance and
from the wire and collinear with the wire.
Define to be the radial distance from the wir
o
q
R
R R
R
α β α β
πε
= r rE a a
e
on the path from points P to P , so
2
Voltage difference between P and P (assuming = ) :
V ln
2 2
o
o
R
R
o o
q
d dR
R
Rq q
dR
R R
β
α
α β
α β
α
βα
β
πε
ε ε
πε πε
=
= − =∫
E lg g
g
7

Voltage Difference, cont’d
V ln
2 2
So, if 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 ene
Repeating:
rg
R
R
o o
Rq q
dR
R R
q
β
α
α
βα
βπε πε
= − =∫ g
y (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
infinitely far away. 8

Multi-Conductor Case
1
Now assume we have parallel conductors,
each with a charge density of coulombs/m.
The voltage difference between our two points,
P and P , is now determined by superposition
1
V ln
2
i
n
i
i
ii
n
q
R
q
R
α β
α
βα
βπε =
=
where is the radial distance from point P
to conductor , and the distance from P to .
i
i
R
i R i
α α
β β
∑
9

Multi-Conductor Case, cont’d
=1
1 1
1
1
11 1
1
If we assume that 0 then rewriting
1 1 1
V ln ln
2 2
We then subtract ln 0
1 1 1
V ln ln
2 2
As we move P to infinity, ln 0
n
i
i
n n
i i i
ii i
n
i
i
n n
i
i i
ii i
i
q
q q R
R
q R
R
q q
R R
R
R
βα α
β
α
α
βα
β α
α
α
α
πε πε
πε πε
= =
=
= =
=
= +
=
= +
→
∑
∑ ∑
∑
∑ ∑
10

Absolute Voltage Defined
1
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:
1 1
V ln
2
This equation holds for any point as long a
n
i
ii
q
R
α
β
βπε =
= ∑
s
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
A
BC
Assume we have three
infinitely long conductors,
A, B, & C, each with radius r
and distance D from the
other two conductors.
Assume charge densities such
that qa + qb + qc = 0
1 1 1 1
ln ln ln
2
ln
2
a a b c
a
a
V q q q
r D D
q D
V
r
πε
πε
 = + +  
=
12

Line Capacitance
1 11
For a single capacitor, capacitance is defined as
But for a multiple conductor case we need to
use matrix relationships since the charge on
conductor may be a function of
i i i
j
n
q CV
i V
q C
q
=
 
  =
 
  
L
M
1 1
1
n
n nn n
C V
C C V
  
  
  
    
=q C V
M L M M
L
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:
2
Since = C
ln
a
a a
a
q
q V C
DV
r
πε
⇒ = =
14

Bundled Conductor Capacitance
1
1
12
Similar to the case for determining line
inductance when there are bundled conductors,
we use the original capacitance equation just
substituting an equiva
Note for the ca
lent radius
( )
p
n
c n
b
n
R rd d= L
acitance equation we use rather
than ' which was used for in the inductance
equation
b
r
r R
15

Line Capacitance, cont’d
[ ]
1
1
3
1
12
-12
o
For the case of uniformly transposed lines we
use the same GMR, , as before.
2
ln
where
( ) (note NOT ')
ε in air 8.854 10 F/m
n
m
m
c
b
m ab ac bc
c n
b
D
C
D
R
D d d d
R rd d r r
πε
ε
=
 
 ÷
 
=
=
= = ×
L
16

Line Capacitance Example
•Calculate the per phase capacitance and susceptance
of a balanced 3φ, 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
1
3
1
3
12
11
11
8
(0.01 0.3 0.3) 0.0963 m
(10 10 20) 12.6 m
2 8.854 10
1.141 10 F/m
12.6
ln
0.0963
1 1
2 60 1.141 10 F/m
2.33 10 -m (not / m)
c
b
m
c
R
D
C
X
C
π
ω π
−
−
−
= × × =
= × × =
× ×
= = ×
= =
× ×
= × Ω Ω
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 π × 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
-8
-8
Line resistance per unit length is given by
= where is the resistivity
A
Resistivity of Copper = 1.68 10 Ω-m
Resistivity of Aluminum = 2.65 10 Ω-m
Example: What is the resistance in Ω / mile of a
R
ρ
ρ
×
×
-8
2 2
1" diameter solid aluminum wire (at dc)?
2.65 10 Ω-m m
1609 0.084
mile mile(0.0127) m
R
π
× Ω
= =
×
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)
Inductance and Capacitance
assume a geometric mean
distance Dm of 1 ft.
GMR is equivalent to
effective radius r’
23

ACSR Data, cont’d
7
3
3 3
2 4 10 ln 1609 /mile
1
2.02 10 ln ln
1
2.02 10 ln 2.02 10 ln
m
L
m
m
D
X f L f
GMR
f D
GMR
f f D
GMR
π π −
−
− −
= = × × Ω
 = × +  
= × + ×
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.
0
6
To use the phase to neutral capacitance from table
21
-m where
2 ln
1
1.779 10 ln -mile (table is in M -mile)
1 1 1
1.779 ln 1.779 ln M -mile
C
m
m
m
X C
Df C
r
D
f r
D
f r f
πε
π
= Ω =
= × × Ω Ω
= × × + × × Ω
Term from table,
depending on conductor type,
but assuming a one foot spacing
Term independent
of conductor, but
with spacing Dm in feet.25

Dove Example
7
0.0313 feet
Outside Diameter = 0.07725 feet (radius = 0.03863)
Assuming a one foot spacing at 60 Hz
1
2 60 2 10 1609 ln Ω/mile
0.0313
0.420 Ω/mile, which matches the table
For the capacitance
a
a
C
GMR
X
X
X
π −
=
= × × × ×
=
6 41 1
1.779 10 ln 9.65 10 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

Lecture 6

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Lecture 6