Transmission Lines
and
Transmission Line Parameters
Robert R. Krchnavek
Rowan University
Glassboro, New Jersey
Objectives
• Develop models/expressions for
transmission lines considering the distributed
nature of the impedances.
• View the transmission line as a two-port
network and develop ABCD parameters.
• Understand concepts of surge impedance
loading and voltage regulation.
• Understand concepts of line loading.
Transmission Line Model
Recall Engineering Electromagnetics
I(z) +
R z
V (z)
L z
+ I(z +
C z V (z +
G z
Note: R, G, L and
C are per meter
values.
z
V (z)
I(z) [R z + |!L z]
dV
=
dz
V (z +
[R + |!L] I(z)
z) = 0
z)
z)
Transmission Line Model
Recall Engineering Electromagnetics
I(z) +
R z
V (z)
L z
+ I(z +
G z
Note: R, G, L and
C are per meter
values.
C z V (z +
z)
z
I(z) + I(z +
z) + V (z +
dI
=
dz
z) [G z + |!C z] = 0
[G + |!C] V
z)
Transmission Line Model
Recall Engineering Electromagnetics
dV
=
dz
dI
=
dz
[R + |!L] I(z)
[G + |!C] V
These two equations are the coupled, time-harmonic, transmission-line equations.
d2 V
= (R + |!L) (G + |!C)V =
2
dz
V (z) = A1 e
z
+ A2 e
z
2
V
Design Considerations
• Conductors
• Insulators
• Support Structures
• Shield Wires
• Electrical Factors
• Mechanical Factors
• Environmental Factors
• Economic Factors
Conductors
• Aluminum, not copper – lower cost, lighter
weight, abundant supply, but higher loss.
• No insulating layer.
• Aluminum conductor, steel reinforced – ACSR.
Al around a steel core.
• All-aluminum conductor (AAC), All-aluminum
alloy conductor (AAAC), Aluminum conductor
alloy reinforced (ACAR), Aluminum clad steel
conductor (Alumoweld) and others.
• Goal: low loss, light weight, high strength.
http://en.wikipedia.org/wiki/Overhead_power_line
Insulators
380 kV
275 kV
cap and pin insulator
400 kV
http://en.wikipedia.org/wiki/Insulator_(electricity)
http://en.wikipedia.org/wiki/Corona_ring
Support Structures
Dead-End Tower
Max. distance between deadend towers is 5 km.
http://en.wikipedia.org/wiki/Suspension_tower
http://en.wikipedia.org/wiki/Dead-end_tower
Shield Wires
•
Shield wires are
grounded through the
tower and often called
ground wires.
•
Used to minimize direct
lightning strikes to the
phase conductors.
Electrical Factors
•
•
•
•
•
•
Type, size, and number of bundle conductors per phase.
•
Series impedance and shunt admittance requirements.
Thermal capacity and short-circuit current ratings.
Lowering E-field to eliminate corona.
Phase-to-Phase, and Phase-to-Ground/Tower clearance.
Line insulators.
Shield wires to intercept lightning strikes. Counterpoise
may be required.
Mechanical Factors
• Strength of conductors and insulator
strings.
• Must consider ice and wind-loading.
• Span length affects strength requirements.
• Dead-end towers approximately every
mile.
Environmental Factors
• Land usage.
• Visual impact.
• Public reaction.
• Biological effects of long-term exposure to
low-frequency (60 Hz) electric and
magnetic fields.
Economic Factors
• Cost to install.
• Cost to run – line losses.
Maintenance?
http://www.youtube.com/watch?v=LIjC7DjoVe8
•
•
•
•
Resistance
Temperature
Frequency
Spiraling
Current magnitude for
magnetic conductors
J. Duncan Glover, M. S. Sarma, T. J. Overbye,
Power System - Analysis and Design,
Cengage Learning
A
l
⇢T l
Rdc, T =
⌦
A
✓
◆
T2 + T
⇢T 2 = ⇢T 1
T1 + T
Conductance
• Real power loss between phase conductors
or between phase conductors and ground.
• Usually due to leakage currents on insulators
(dirt, moisture, salt) and corona (current
discharging into ionized air.)
•
2
I
Usually small compared to R losses in the
phase conductors.
• Often ignored for high-tension wires.
Inductance
• Need to consider multiple conductors.
• Geometry.
• Phase spacing.
• Bundling.
• Transposition
Inductance
• Inductance is an important quantity in high-power
transmission lines.
• We will begin by calculating the self-inductance of
a length of wire due to internal and external
inductance.
• We will then calculate the inductance due to
coupling.
• We will then generalize to the transmission line
case.
Inductance
• Self-inductance is defined as the ratio of the total flux
linkages (λ) to the current which they link.
L=
I
• A flux linkage is the number of times the flux (Φ) links
the wire containing the current.
• In a coil, it is usually given by: = N
• In a wire, N = 1 .
• Self-inductance consists of internal inductance and
external inductance.
• In addition to self-inductance, there is mutual
inductance which includes flux linkages from the
current of a nearby circuit.
Inductance
•
The self-inductance is defined as the magnetic
flux linkage per unit current in a current loop.
For a wire, N = 1 .
•
Since we only have a wire and not a loop, we can
only calculate the inductance per unit length.
•
Begin by assuming a current in the wire.
•
•
•
•
Find H using Ampere’s Law or Biot-Savart.
~ = µH
~ ).
Find B, magnetic flux density ( B
Find flux linkages (
).
=N
Inductance is given by: L =
I
Inductance
Internal Self-Inductance
r
x
•
cross-section of the conductor
carrying a total current of I.
I
~ · d~l = Ienclosed
H
⇡x2 1
Ix
H(x) = I 2
=
⇡r 2⇡x
2⇡r2
µIx
B(x) =
2⇡r2
•
•
•
Assume the current is
uniformly distributed
across the cross-section
of the conductor. (No
I
skin effect)
J=
⇡r2
Calculate H.
Calculate B.
Calculate total internal
flux.
Inductance
Internal Self-Inductance
r
x
First, calculate the inductance inside the wire.
Consider a thin shell between x and x+dx.
The flux (really d m) is given by:
~
= B(x)
· d~s
Z 1
µIx
µIx
d =
â · â dzdx =
dx
2
2
2⇡r
z=0 2⇡r
d
The differential flux linkage is the differential flux that is linked by the fraction of the current
it encloses
2
2
x
x µIx
d =d m 2 = 2
dx
r
r 2⇡r2
The total flux linkages inside the conductor is given by
Z r
Z r 2
x µIx
µI
d =
dx =
int =
2
2
8⇡
0
0 r 2⇡r
Inductance
External Self-Inductance
For the external self-inductance, we are concerned with the flux linkages that are
outside the conductor. In this case, the current enclosed is constant.
d
d
~
= B(x)
· d~s
Z 1
µI
µI
=
â · â dzdx =
dx
2⇡x
z=0 2⇡x
µI
d =d =
dx
2⇡x
Z 1
Z
d =
ext =
r
ext
µI 1
=
ln
2⇡
r
1
r
µI
µI
dx =
2⇡x
2⇡
Z
1
r
dx
x
Problem: Flux linkages go to infinity!
Inductance
Self-Inductance
After we resolve the problem of infinite flux linkages, the
self-inductance is given by:
total
Lself =
total
I
=
in
+
I
ext
=
in
+
ext
µ
µ
1
=
+
ln
(H/m)
8⇡ 2⇡
r
How do we reconcile the infinite external flux linkages?
Inductance
Self-Inductance
Some texts use the following concept. The total self
inductance, due to internal and external flux linkages,
out to a distance, D, where we assume a return path
occurs such that no flux linkages occur, is given by:
Lself
µ
µ
D
=
+
ln
(H/m)
8⇡ 2⇡
r
Note: This problem on infinite inductance only
arrives because we are neglecting the return path.
We can work around this.
Inductance
Mutual-Inductance
In addition to selfinductance, we have the
mutual inductance to
consider. Mutual
inductance is due to the
flux produced by Ib (and Ic)
which links the filament
that carries Ia and vice
versa.
a, Ib , mutual
µ0 Ib 1
=
ln
2⇡
D
Inductance
Mutual-Inductance
For 3 equal-spaced conductors, with a spacing of D,
we have a flux linking phase a due to Ib and Ic.
Inductance
Per-Phase Inductance
The per-phase inductance is the total flux linking a
particular phase (e.g., phase a) divided by the current
in phase a (Ia).
La =
a, total
La =
a, total
La =
Ia
a, total
Ia
Ia
1
=
(
Ia
a, Ia
1
=
(
Ia
a, Ia , self
1
=
(
Ia
a, Ia , int
+
+
a, Ib
+
a, Ia , ext
+
a, Ic )
a, Ib , mutual
+
+
a, Ib , mutual
a, Ic , mutual )
+
a, Ic , mutual )
Inductance
Per-Phase Inductance
La =
a, total
Ia
1
La =
Ia
1
=
(
Ia
✓
a, Ia , int
+
a, Ia , ext
+
a, Ib , mutual
+
a, Ic , mutual )
µ0 Ia
µ0 Ia 1 µ0 Ib 1 µ0 Ic 1
+
ln
+
ln
+
ln
8⇡
2⇡
r
2⇡
D
2⇡
D
Ia + Ib + Ic = 0
Ib + Ic = Ia
✓
◆
1 µ0 Ia
µ0 Ia D
La =
+
ln
Ia
8⇡
2⇡
r
µ0
µ0 D
La =
+
ln
(H/m)
8⇡ 2⇡
r
◆
Capacitance
•
Two conductors, separated with a
dielectric, and with a voltage difference
between them, will yield a capacitance.
• In transmission lines, there will be line-toline capacitance as well as line-to-neutral
capacitance.
Capacitance
•
•
•
Using Gauss’s law,
determine the electric
field, E.
From E, calculate the
voltage difference
between the two
conductors.
Calculate the
capacitance.
Q
C=
V
~
E
Single, charged, cylindrical conductor.
~ , will have
The electric flux density, D
cylindrical symmetry.
Capacitance
Q
C=
V
~
E
I
Single, charged, cylindrical conductor.
~ , will have
The electric flux density, D
cylindrical symmetry.
ZZZ
~ · d~s = Qenclosed =
D
⇢v dv
S
vol
Z 2⇡ Z 1
D
⇢d dz = Qenclosed
0
0
D2⇡⇢ = Qenclosed
Q
enclosed
~
E=
â⇢ (V/m)
✏2⇡⇢
~ = ✏E
~
D
Note: Assumes no
longitudinal E-field, i.e., no
voltage drop along the
length of the conductor.
Capacitance
Q
C=
V
⇢init
⇢final
Q
enclosed
~
E=
â⇢ (V/m)
✏2⇡⇢
Z final
~ · d~l
V =
E
init
V =
Z
final
init
Qenclosed
â⇢ · â⇢ d⇢
✏2⇡⇢
Qenclosed
⇢init
V =
ln
✏2⇡
⇢final
~
E
Single, charged, cylindrical conductor.
~ , will have
The electric flux density, D
cylindrical symmetry.
Note: Equipotential surfaces
are coaxial cylinders. The
line integral then simply
because a change in radial
direction.
If the two conductors were at ⇢init
and ⇢final , then C is given by:
Qenclosed
⇢final
C=
= 2⇡✏ ln
V
⇢init
Capacitance
Three-phase, three-wire, with equal phase spacing
Q
C=
V
c
⇢init
⇢final
Qenclosed
⇢final
C=
= 2⇡✏ ln
V
⇢init
~
E
a
b
Neglecting C to ground, calculate C between the conductors.
Qa, Qb, and Qc are charges on the 3 phase lines.
These charges are related to the voltage through the capacitance.
✓
◆
Qa
D
Vab,Qa =
ln
where r is the radius of the conductor.
2⇡✏
r
Capacitance
Three-phase, three-wire, with equal phase spacing
Q
C=
V
c
⇢init
⇢final
Qenclosed
⇢final
C=
= 2⇡✏ ln
V
⇢init
~
E
a
Similarly,
b
✓
◆
Qa
D
Vab,Qa =
ln
2⇡✏
r
✓
◆
Qb
D
Vba,Qb =
ln
2⇡✏
r
and Qc does not produce a voltage drop between a and b.
Capacitance
Three-phase, three-wire, with equal phase spacing
Q
C=
V
c
Vab = Vab,Qa + Vab,Qb + Vab,Qc
a
b
Vab =
✓
Qa
2⇡✏
Vab = (Qa
◆
D
ln
r
✓
Qb
2⇡✏
◆
D
ln + 0
r
1
D
Qb )
ln
2⇡✏
r
Vab is the total potential drop from a to b due to charges on a, b, and c.
Capacitance
Three-phase, three-wire, with equal phase spacing
c
We also know:
Van + Vab + Vbn
a
b
c
a
Ecn
Ean
Eca
n
Ebn
b
Vab = Van
Vbn
Qa
=
Ca
Qb
Cb
Qa
Qb
Vab
Eab
Ebc
Q
C=
V
=0
Vab =
C
Capacitance
Three-phase, three-wire, with equal phase spacing
c
Putting it together:
Vab = (Qa
a
1
D
Qb )
ln
2⇡✏
r
and Vab =
Qa
Qb
C
b
r
C = 2⇡✏ ln
D
(F/m)
Note: This is the simplest case for 3-phase. The textbook briefly mentions
unequal phase spacing and bundled conductors and gives a couple
of references.
Lumped Representations
of Transmission Lines
Surge Impedance and
Surge Impedance Loading