Transmission Lines and Transmission Line Parameters Robert R. Krchnavek Rowan University Glassboro, New Jersey
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Transmission Linesand
Transmission Line Parameters
Robert R. KrchnavekRowan 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 ModelRecall Engineering Electromagnetics
R�z L�z
C�zG�z
�z
+
�
V (z) V (z +�z)
I(z +�z)I(z) +
�Note: R, G, L and C are per meter values.
V (z)� I(z) [R�z + |!L�z]� V (z +�z) = 0
dV
dz= � [R+ |!L] I(z)
Transmission Line ModelRecall Engineering Electromagnetics
R�z L�z
C�zG�z
�z
+
�
V (z) V (z +�z)
I(z +�z)I(z) +
�Note: R, G, L and C are per meter values.
�I(z) + I(z +�z) + V (z +�z) [G�z + |!C�z] = 0
dI
dz= � [G+ |!C]V
Transmission Line ModelRecall Engineering Electromagnetics
dI
dz= � [G+ |!C]V
dV
dz= � [R+ |!L] I(z)
These two equations are the coupled, time-harmonic, transmission-line equations.
d2V
dz2= (R+ |!L) (G+ |!C)V = �2V
V (z) = A1e�z +A2e
��z
Design Considerations• Conductors
• Insulators
• Support Structures
• Shield Wires
• Electrical Factors
• Mechanical Factors
• Environmental Factors
• Economic Factors
Conductors
http://en.wikipedia.org/wiki/Overhead_power_line
• 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.
Insulators
http://en.wikipedia.org/wiki/Insulator_(electricity)
275 kV
cap and pin insulator
http://en.wikipedia.org/wiki/Corona_ring
380 kV
400 kV
Support Structures
Dead-End Tower Max. distance between dead-
end 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.
• 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.
• Series impedance and shunt admittance requirements.
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.
Resistance• Temperature
• Frequency
• Spiraling
• Current magnitude for magnetic conductors
Rdc, T =⇢T l
A⌦
lA
⇢T2 = ⇢T1
✓T2 + T
T1 + T
◆
J. Duncan Glover, M. S. Sarma, T. J. Overbye, Power System - Analysis and Design,
Cengage Learning
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.)
• Usually small compared to losses in the phase conductors.
• Often ignored for high-tension wires.
I2R
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.
• 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:
• In a wire, .
• 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.
L =�
I
� = N�
N = 1
Inductance• The self-inductance is defined as the magnetic
flux linkage per unit current in a current loop. For a wire, .
• 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.
• Find B, magnetic flux density ( ).
• Find flux linkages ( ).
• Inductance is given by: L =�
I
N = 1
~B = µ ~H
� = N�
Inductance Internal Self-Inductance
cross-section of the conductorcarrying a total current of I.
• Assume the current is uniformly distributed across the cross-section of the conductor. (No skin effect)
• Calculate H.
• Calculate B.
• Calculate total internal flux.
r
x
J =I
⇡r2I~H · d~l = I
enclosed
H(x) = I⇡x2
⇡r21
2⇡x=
Ix
2⇡r2
B(x) =µIx
2⇡r2
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 ) is given by:d m
The differential flux linkage is the differential flux that is linked by the fraction of the current it encloses
d� = d mx
2
r
2=
x
2
r
2
µIx
2⇡r2dx
The total flux linkages inside the conductor is given by
�int =
Z r
0d� =
Z r
0
x
2
r
2
µIx
2⇡r2dx =
µI
8⇡
d� = ~
B(x) · d~s
d� =
Z 1
z=0
µIx
2⇡r2a� · a�dzdx =
µIx
2⇡r2dx
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.
�
ext
=
Z 1
rd� =
Z 1
r
µI
2⇡xdx =
µI
2⇡
Z 1
r
dx
x
�ext
=µI
2⇡ln
1r
Problem: Flux linkages go to infinity!
d� = ~
B(x) · d~s
d� =
Z 1
z=0
µI
2⇡xa� · a�dzdx =
µI
2⇡xdx
d� = d� =µI
2⇡xdx
Inductance Self-Inductance
After we resolve the problem of infinite flux linkages, the self-inductance is given by:
�total
= �in
+ �ext
How do we reconcile the infinite external flux linkages?
Lself
=�total
I=
�in
+ �ext
I=
µ
8⇡+
µ
2⇡ln
1r
(H/m)
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
8⇡+
µ
2⇡ln
D
r(H/m)
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 self-inductance, 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 =µ0Ib2⇡
ln1D
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
Ia
=1
Ia
(�a, Ia + �
a, Ib + �a, Ic)
La
=�a, total
Ia
=1
Ia
(�a, Ia, self + �
a, Ib,mutual
+ �a, Ic,mutual
)
La
=�a, total
Ia
=1
Ia
(�a, Ia, int + �
a, Ia, ext + �a, Ib,mutual
+ �a, Ic,mutual
)
Inductance Per-Phase Inductance
La
=�a, total
Ia
=1
Ia
(�a, Ia, int + �
a, Ia, ext + �a, Ib,mutual
+ �a, Ic,mutual
)
La =1
Ia
✓µ0Ia8⇡
+µ0Ia2⇡
ln1r
+µ0Ib2⇡
ln1D
+µ0Ic2⇡
ln1D
◆
Ia + Ib + Ic = 0
Ib + Ic = �Ia
La =1
Ia
✓µ0Ia8⇡
+µ0Ia2⇡
lnD
r
◆
La =µ0
8⇡+
µ0
2⇡ln
D
r(H/m)
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-to-line 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.
C =Q
V
~E
Single, charged, cylindrical conductor.The electric flux density, , will have cylindrical symmetry.
~D
CapacitanceC =
Q
V
~ESingle, charged, cylindrical conductor.The electric flux density, , will have cylindrical symmetry.
~D
I
S
~D · d~s = Qenclosed
=
ZZZ
vol
⇢vdv ~D = ✏ ~E
D
Z2⇡
0
Z1
0
⇢d�dz = Qenclosed
D2⇡⇢ = Qenclosed
~E =Q
enclosed
✏2⇡⇢a⇢ (V/m)
Note: Assumes no longitudinal E-field, i.e., no voltage drop along the length of the conductor.
CapacitanceC =
Q
V
Single, charged, cylindrical conductor.The electric flux density, , will have cylindrical symmetry.
~D~E =
Qenclosed
✏2⇡⇢a⇢ (V/m)
V = �Z final
init
~E · d~lNote: Equipotential surfaces are coaxial cylinders. The line integral then simply because a change in radial direction.
V = �Z
final
init
Qenclosed
✏2⇡⇢a⇢ · a⇢d⇢
~E
⇢init
⇢final
V =Q
enclosed
✏2⇡ln
⇢init
⇢final
C =Q
enclosed
V= 2⇡✏ ln
⇢final
⇢init
If the two conductors were at and , then C is given by:
⇢init⇢final
Capacitance Three-phase, three-wire, with equal phase spacing
~E
⇢init
⇢final
C =Q
V
C =Q
enclosed
V= 2⇡✏ ln
⇢final
⇢init
a b
c
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.
Vab,Qa =
✓Qa
2⇡✏
◆ln
D
rwhere r is the radius of the conductor.
Capacitance Three-phase, three-wire, with equal phase spacing
~E
⇢init
⇢final
C =Q
V
C =Q
enclosed
V= 2⇡✏ ln
⇢final
⇢init
a b
c
Similarly, Vab,Qa =
✓Qa
2⇡✏
◆ln
D
r
Vba,Qb =
✓Qb
2⇡✏
◆ln
D
r
and Qc does not produce a voltage drop between a and b.
Capacitance Three-phase, three-wire, with equal phase spacing
C =Q
V
a b
c
Vab = Vab,Qa + Vab,Qb + Vab,Qc
Vab =
✓Qa
2⇡✏
◆ln
D
r�✓
Qb
2⇡✏
◆ln
D
r+ 0
Vab = (Qa �Qb)1
2⇡✏ln
D
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 =Q
V
a b
c
We also know:
� �
�
Ean EabEbc
Eca
Ebn
Ecnn
a
b
c
�Van + Vab + Vbn = 0
Vab = Van � Vbn
Vab =Qa
Ca� Qb
Cb
Vab =Qa �Qb
C
Capacitance Three-phase, three-wire, with equal phase spacing
a b
c
Putting it together:
Vab =Qa �Qb
CVab = (Qa �Qb)
1
2⇡✏ln
D
r and
C = 2⇡✏ lnr
D (F/m)
Note: This is the simplest case for 3-phase. The textbook briefly mentionsunequal phase spacing and bundled conductors and gives a coupleof references.
Lumped Representations of Transmission Lines
Surge Impedance and Surge Impedance Loading
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