Post on 18-Nov-2014
BEE3133 Electrical Power Systems
Chapter 3Transmission Line Parameters
Rahmatul Hidayah Salimin
INTRODUCTION
• All transmission lines in a power system exhibit the electrical properties of resistance, inductance, capacitance and conductance.
• Inductance and capacitance are due to the effects of magnetic and electric fields around the conductor.
• These parameters are essential for the development of the transmission line models used in power system analysis.
• The shunt conductance accounts for leakage currents flowing across insulators and ionized pathways in the air.
• The leakage currents are negligible compared to the current flowing in the transmission lines and may be neglected.
RESISTANCE
• Important in transmission efficiency evaluation and economic studies.
• Significant effect– Generation of I2R loss in
transmission line.– Produces IR-type voltage drop
which affect voltage regulation.
RESISTANCE
• The dc resistance of a solid round conductor at a specified temperature is
Where :ρ = conductor resistivity (Ω-m),
l = conductor length (m) ; and
A = conductor cross-sectional area (m2)
dc
lR
A
RESISTANCE
• Conductor resistance is affected by three factors:-
• Frequency (‘skin effect’)
• Spiraling
• Temperature
RESISTANCE
Frequency – Skin Effect• When ac flows in a conductor, the
current distribution is not uniform over the conductor cross-sectional area and the current density is greatest at the surface of the conductor.
• This causes the ac resistance to be somewhat higher than the dc resistance. The behavior is known as skin effect.
RESISTANCE
• The skin effect is where alternating current tends to avoid travel through the center of a solid conductor, limiting itself to conduction near the surface.
• This effectively limits the cross-sectional conductor area available to carry alternating electron flow, increasing the resistance of that conductor above what it would normally be for direct current
RESISTANCE
RESISTANCE
• Skin effect correction factor are defined as
Where R = AC resistance ; and
Ro = DC resistance.
O
R
R
RESISTANCE
Spiraling• For stranded conductors, alternate
layers of strands are spiraled in opposite directions to hold the strands together.
• Spiraling makes the strands 1 – 2% longer than the actual conductor length.
• DC resistance of a stranded conductor is 1 – 2% larger than the calculated value.
RESISTANCETemperature• The conductor resistance increases as
temperature increases. This change can be considered linear over the range of temperature normally encountered and may be calculated from :
Where: R1 = conductor resistances at t1 in °C R2 = conductor resistances at t2 in °C T = temperature constant (depends on the conductor material)
22 1
1
T tR R
T t
RESISTANCE
• The conductor resistance is best determined from manufacturer’s data.
• Some conversion used in calculating line resistance:-
1 cmil = 5.067x10-4 mm2 = 5.067x10-
6 cm2
= 5.067x10-10 m2
Resistivity & Temparature Constant of Conductor Metals
Material
ρ20ºC T
Resistivity at 20ºC Temperature Constant
Ωm×10-8 Ωcmil/ft ºC
Copper
Annealed 1.72 10.37 234.5
Hard-drawn 1.77 10.66 241.5
Aluminum
Hard-drawn 2.83 17.00 228
Brass 6.4 – 8.4 38 – 51 480
Iron 10 60 180
Silver 1.59 9.6 243
Sodium 4.3 26 207
Steel 12 – 88 72 – 530 180 – 980
RESISTANCE
• Example:-A solid cylindrical aluminum conductor 25km long has an area of 336,400 circular mils. Obtain the conductor resistance at (a) 20°C and
(b) 50°C.
The resistivity of aluminum at 20°C is
ρ = 2.8x10-8Ω-m.
RESISTANCE
• Answer (a)
25
8 3
4
6
2.8 10 25 10
336,400 5.076 10
4.0994 10
l km
lR
A
RESISTANCE
• Answer (b)
5050 20
20
6
6
228 504.0994 10
228 20
4.5953 10
CC C
C
T tR R
T t
RESISTANCE
• Exercise 1A transmission-line cable consists of 12 identical strands of aluminum, each 3mm in diameter. The resistivity of aluminum strand at 20°C is 2.8x10-8Ω-m. Find the 50°C ac resistance per km of the cable. Assume a skin-effect correction factor of 1.02 at 50Hz.
RESISTANCE
• Exercise 2:-A solid cylindrical aluminum conductor 115km long has an area of 336,400 circular mils. Obtain the conductor resistance at:(a) 20°C
(b) 40°C
(c) 70°C
The resistivity of aluminum at 20°C is
ρ = 2.8x10-8Ω-m.
RESISTANCE
• Exercise 3A transmission-line cable consists of 15 identical strands of aluminum, each 2.5mm in diameter. The resistivity of aluminum strand at 20°C is 2.8x10-8Ω-m. Find the 50°C ac resistance per km of the cable. Assume a skin-effect correction factor of 1.015 at 50Hz.
INDUCTANCE :A SINGLE CONDUCTOR
• A current-carrying conductor produces a magnetic field around the conductor.
• The magnetic flux can be determined by using the right hand rule.
• For nonmagnetic material, the inductance L is the ratio of its total magnetic flux linkage to the current I, given by
where λ=flux linkages, in Weber turns.L
I
INDUCTANCE : A SINGLE CONDUCTOR
• For illustrative example, consider a long round conductor with radius r, carrying a current I as shown.
• The magnetic field intensity Hx, around a circle of radius x, is constant and tangent to the circle.
2x
x
IH
x
INDUCTANCE :A SINGLE CONDUCTOR
• The inductance of the conductor can be defined as the sum of contributions from flux linkages internal and external to the conductor.
Flux Linkage
INDUCTANCE :A SINGLE CONDUCTOR
INDUCTANCE :A SINGLE PHASE LINES
INDUCTANCE :3-PHASE TRANSMISSION LINES
INDUCTANCE :3-PHASE TRANSMISSION LINES
What and How to Calculate:-• Lint , Lext @ L?
• L1 , L2 @ L?
• L11 , L12 @ L22?
• GMR?• GMD?
INDUCTANCE :A SINGLE CONDUCTOR
• INTERNAL INDUCTANCE– Internal inductance can be express as
follows:-
– Where
µo = permeability of air (4π x 10-7 H/m)
– The internal inductance is independent of the conductor radius r
70int
110 /
8 2L H m
INDUCTANCE :A SINGLE CONDUCTOR
• INDUCTANCE DUE TO EXTERNAL FLUX LINKAGE– External
inductance between to point D2 and D1 can be express as follows:
7 2
1
2 10 ln /ext
DL H m
D
INDUCTANCE :A SINGLE PHASE LINES
• A single phase lines consist of a single current carrying line with a return line which is in opposite direction. This can be illustrated as:
INDUCTANCE :A SINGLE PHASE LINES
• Inductance of a single-phase lines can be expressed as below with an assumption that the radius of r1=r2=r.
7 7 2int
1
7 7 7
17 74
1
4
70.25
110 2 10 ln /
2
1 110 2 10 ln / 2 10 ln /
2 4
12 10 ln ln / 2 10 ln ln /
2 10 ln /
ext
DL L L H m
D
D DH m H m
r r
D De H m H m
r re
DH m
re
SELF AND MUTUAL INDUCTANCES • The series inductance per phase can be
express in terms of self-inductance of each conductor and their mutual inductance.
• Consider the one meter length single-phase circuit in figure below:-
– Where L11 and L22 are self-inductance and the mutual inductance L12
SELF AND MUTUAL INDUCTANCES
Dx
DxL
DxL
erxL
ILLID
xer
xIL
ILL
ILL
mHD
xer
xL
mHD
xer
xL
1ln102
1ln102
1ln102
1ln102
1ln102
1ln102
/1
ln1021
ln102
/1
ln1021
ln102
7712
712
25.01
711
1121117
25.01
7111
222212
112111
725.0
2
72
725.0
1
71
SELF AND MUTUAL INDUCTANCES
• L11, L22 and L12 can be expressed as below:-
711 0.25
1
722 0.25
2
712 21
12 10 ln
12 10 ln
12 10 ln
Lre
Lr e
L LD
SELF AND MUTUAL INDUCTANCES
• Flux linkage of conductor i
ijD
Ier
Ixn
j ijj
iii
1
ln1
ln1021
25.07
INDUCTANCE :3-PHASE TRANSMISSION LINES
• Symmetrical Spacing– Consider 1 meter length of a three-phase
line with three conductors, each radius r, symmetrically spaced in a triangular configuration.
INDUCTANCE :3-PHASE TRANSMISSION LINES
• Assume balance 3-phase current
Ia+ Ib+ Ic = 0
• The total flux linkage of phase a conductor
• Substitute for Ib + Ic=-Ia
DI
DI
erIx cb
aaa
1ln
1ln
1ln102
25.07
25.07
25.07 ln102
1ln
1ln102
er
DIx
DI
erIx
aaa
aaa
INDUCTANCE : 3-PHASE TRANSMISSION LINES
• Because of symmetry, λa=λb=λc
• The inductance per phase per kilometer length
kmmHre
Dx
IL /ln102
25.07
INDUCTANCE :3-PHASE TRANSMISSION LINES
• Asymmetrical Spacing– Practical transmission lines cannot maintain
symmetrical spacing of conductors because of construction considerations.
– Consider one meter length of three-phase line with three conductors, each with radius r. The conductor are asymmetrically spaced with distances as shown.
INDUCTANCE :3-PHASE TRANSMISSION LINES
– The flux linkages are:-
231325.0
7
231225.0
7
131225.0
7
1ln
1ln
1ln102
1ln
1ln
1ln102
1ln
1ln
1ln102
DI
DI
reI
DI
DI
reI
DI
DI
reI
bacc
cabb
cbaa
INDUCTANCE : 3-PHASE TRANSMISSION LINES
– For balanced three-phase current with Ia as reference, we have:-
ao
ac
ao
ab
aIII
IaII
120
240 2
INDUCTANCE :3-PHASE TRANSMISSION LINES
• Thus La, Lb and Lc can be found using the following equation:-
23
225.0
12
7 1ln
1ln
1ln102
Da
reDa
IL
b
bb
1312
225.0
7 1ln
1ln
1ln102
Da
Da
reIL
a
aa
25.0
2313
27 1ln
1ln
1ln102
reDa
Da
IL
c
cc
INDUCTANCE : 3-PHASE TRANSMISSION LINES• Transpose Line
– Transposition is used to regain symmetry in good measures and obtain a per-phase analysis.
INDUCTANCE :3-PHASE TRANSMISSION LINES• This consists of interchanging the phase
configuration every one-third the length so that each conductor is moved to occupy the next physical position in a regular sequence.
• Transposition arrangement are shown in the figure
INDUCTANCE :3-PHASE TRANSMISSION LINES
• Since in a transposed line each phase takes all three positions, the inductance per phase can be obtained by finding the average value.
0.2512 13
7
0.2523 12
0.2513 23
7
0.2512
3
1 1 1ln 1 240 ln 1 120 ln
2 10 1 1 1ln 1 240 ln 1 120 ln
3
1 1 1ln 1 240 ln 1 120 ln
2 10 1 1 13ln ln ln
3
a b cL L LL
re D D
re D D
re D D
re D D
23 13
312 23 137
0.25
1ln
2 10 ln
D
D D D
re
• Since in a transposed line each phase takes all three positions, the inductance per phase can be obtained by finding the average value.
3cba
a
LLLL
• Noting a + a2 = -1
• Inductance per phase per kilometer length
25.0
3
1
1323127
3
1
132312
25.07
13231225.0
7
ln102
1ln
1ln102
1ln
1ln
1ln
1ln3
3
102
re
DDD
DDDre
DDDreL
kmmH
re
DDDL /ln2.0
25.0
3
1
132312
What and How to Calculate:-• Lint , Lext @ L?
• L1 , L2 @ L?
• L11 , L12 @ L22?
• GMR?• GMD?
Inductance of Composite Conductors
In evaluation of inductance, solid round conductors were considered. However, in practical transmission lines, stranded conductors are used.
Consider a single-phase line consisting of two composite conductors x and y as shown in Figure 1. The current in x is I referenced into the page, and the return in y is –I.
Inductance of Composite Conductors
Conductor x consist of n identical strands or subconductors, each with radius rx. Conductor y consist of m identical strands or subconductors, each with radius ry.
The current is assumed to be equally divided amon the subconductors. The current per strands is I/n in x and I/m in y.
Inductance of Composite Conductors
b d'
a
c
d
n
b'
a'
c'
m'
x y
nncnbnax
mnmncnbnan
n
nanacabx
mamacabaaa
a
nanacabx
mamacabaa
a
amacabaa
anacabxa
DDDr
DDDDn
nIL
DDDr
DDDDn
nIL
DDDr
DDDDI
or
DDDDm
I
DDDrn
I
...'
...ln102
/
...'
...ln102
/
...'
...ln102
1ln...
1ln
1ln
1ln102
1ln...
1ln
1ln
'
1ln102
'''7
'''7
'''7
'''
7
7
'...
)...)...(...(
)...)...(...(
/ln102
2
''''
7
xnnbbaa
nnnnbnaanabaax
mnnmnbnaamabaa
xx
rDDD
where
DDDDDDGMR
DDDDDDGMD
where
mHGMR
GMDL
GMR of Bundled Conductors
d
d d
d
d
d
d
d
Extra high voltage transmission lines are usually constructed with bundled conductors. Bundling reduces the line reactance, which improves the line performance and increases the power capability of the line.
GMR of Bundled Conductors
4 316 42/1
3 29 3
09.1)2(
)(
dDdddDD
bundleorsubconductfourthefor
dDddDD
bundleorsubconductthreethefor
ssbs
ssbs
dDdDD
bundleorsubconducttwothefor
DDDDDDGMR
ssbs
nnnnbnaanabaax
4 2)(
)...)...(...(2
Inductance of Three-phase Double Circuit Lines
A three-phase double-circuit transmission line consists of two identical three-phase circuits. To achieve balance, each phase conductor must be transposed within it group and with respect to the parallel three-phase line.Consider a three-phase double-circuit line with relative phase positions a1b1c1-c2b2a2.
Inductance of Three-phase Double Circuit Lines
c1
a1
a2
b1 b2
c2S11
S22
S33
GMD between each phase group
422122111
422122111
422122111
cacacacaAC
cbcbcbcbBC
babababaAB
DDDDD
DDDDD
DDDDD
Inductance of Three-phase Double Circuit Lines
The equivalent GMD per phase is then
3ACBCAB DDDGMD
Similarly, GMR of each phase group is
214 2
21
214 2
21
214 2
21
)(
)(
)(
ccb
ccb
SC
bbb
bbb
SB
aab
aab
SA
DDDDD
DDDDD
DDDDD
ss
ss
ss
where is the geometric mean radius of bundled conductors.
bsD
Inductance of Three-phase Double Circuit Lines
The equivalent GMR per phase is then
3SCSBSAL DDDGMR
The inductance per-phase is
mHGMR
GMDL
Lx /ln102 7
INDUCTANCE :3-PHASE TRANSMISSION LINES
• Question 4
A three-phase, 50 Hz transmission line has a reactance 0.5 Ω per kilometer. The conductor geometric mean radius is 2 cm. Determine the phase spacing D in meter.
INDUCTANCE :3-PHASE TRANSMISSION LINES
• Question 4
A three-phase, 60 Hz transmission line has a reactance 0.25Ω per kilometer. The conductor geometric mean radius is 5 cm. Determine the phase spacing D in meter.
INDUCTANCE :3-PHASE TRANSMISSION LINES
• Question 4
A three-phase, 50 Hz transmission line has
Xc = 0.5 Ω per kilometer. The conductor geometric mean radius is 2 cm. Determine the phase spacing D in meter.
CAPACITANCE
• Transmission line conductors exhibit capacitance with respect to each other due to the potential difference between them.
• The amount of capacitance between conductors is a function of conductor size, spacing, and height above ground.
• Capacitance C is:-
qC
V
LINE CAPACITANCE
• Consider a long round conductor with radius r, carrying a charge of q coulombs per meter length as shown.
• The electrical flux density at a cylinder of radius x is given by:
2
q qD
A x
LINE CAPACITANCE• The electric field intensity E is:-
Where permittivity of free space, ε0 = 8.85x10-12 F/m.
• The potential difference between cylinders from position D1 to D2 is defined as:-
The notation V12 implies the voltage drop from 1 relative to 2.
0 02
D qE
x
212
0 1
ln2
q DV
D
CAPACITANCE OF SINGLE-PHASE LINES • Consider one meter length of a single-
phase line consisting of two long solid round conductors each having a radius r as shown.
• For a single phase, voltage between conductor 1 and 2 is:-
120
ln /q D
V F mr
CAPACITANCE OF SINGLE-PHASE LINES
• The capacitance between the conductors:-
012 /
lnC F m
Dr
CAPACITANCE OF SINGLE-PHASE LINES
• The equation gives the line-to-line capacitance between the conductors
• For the purpose of transmission line modeling, we find it convenient to define a capacitance C between each conductor and a neutral line as illustrated.
CAPACITANCE OF SINGLE-PHASE LINES
• Voltage to neutral is half of V12 and the capacitance to neutral is C=2C12 or:-
02/
lnC F m
Dr
Potential Difference in a Multiconductor configuration
• Consider n parallel long conductors with charges q1, q2,…,qn coulombs/meter as shown below.
ki
kjn
kkij D
DqV ln
2
1
10
• Potential difference between conductor i and j due to the presence of all charges is
qjqi
q1
q3q2
qn
CAPACITANCE OF THREE-PHASE LINES
• Consider one meter length of 3-phase line with three long conductors, each with radius r, with conductor spacing as shown below:
qc
qb
qa
D13
D23
D12
CAPACITANCE OF THREE-PHASE LINES
For balanced 3-phase system, the capacitance per phase to neutral is:
1/3
12 23 13
2F/m
ln
a o
an
qC
V D D D
r
CAPACITANCE OF THREE-PHASE LINES
1/3
12 23 13
0.0556F/km
ln
CD D D
r
The capacitance to neutral in µF per kilometer is:
Effect of bundling
mF
rGMD
C
b
/ln
2 0
• The effect of bundling is introduce an equivalent radius rb. The radius rb is similar to GMR calculate earlier for the inductance with the exception that radius r of each subconductor is used instead of Ds.
Effect of bundling
• If d is the bundle spacing, we obtain for the two-subconductor bundle
drrb
• For the three-subconductor bundle
3 2drrb
• For the four-subconductor bundle
4 309.1 drrb
Capacitance of Three-phase Double Circuit Lines
mF
GMRGMD
C
c
/ln
2 0
• The per-phase equivalent capacitance to neutral is obtained to
• GMD is the same as was found for inductance calculation
422122111
422122111
422122111
cacacacaAC
cbcbcbcbBC
babababaAB
DDDDD
DDDDD
DDDDD
Capacitance of Three-phase Double Circuit Lines
• The equivalent GMD per phase is then
3ACBCAB DDDGMD
• The GMRC of each phase is similar to the GMRL, with the exception that rb is used instead of
bsD
• This will results in the following equ…
21
21
21
ccb
C
bbb
B
aab
A
Drr
Drr
Drr
3CBAC rrrGMR
EFFECT OF EARTH ON THE CAPACITANCE • For isolated charged conductor the
electric flux lines are radial and orthogonal to cylindrical equipotential surfaces, which will change the effective capacitance of the line.
• The earth level is an equipotential surface. Therefore flux lines are forced to cut the surface of the earth orthogonally.
• The effect of the earth is to increase the capacitance.
EFFECT OF EARTH ON THE CAPACITANCE
• But, normally, the height of the conductor is large compared to the distance between the conductors, and the earth effect is negligible.
• Therefore, for all line models used for balanced steady-state analysis, the effect of earth on the capacitance can be negligible.
• However, for unbalance analysis such as unbalance faults, the earth’s effect and shield wires should be considered.
MAGNETIC FIELD INDUCTION
• Transmission line magnetic fields affect objects in the proximity of the line.
• Produced by the currents in the line.
• It induces voltage in objects that have a considerable length parallel to the line (Ex: telephone wires, pipelines etc.).
MAGNETIC FIELD INDUCTION
• The magnetic field is effected by the presence of earth return currents.
• There are general concerns regarding the biological effects of electromagnetic and electrostatic fields on people.
ELECTROSTATIC INDUCTION
• Transmission line electric fields affect objects in the proximity of the line.
• It produced by high voltage in the lines.
• Electric field induces current in objects which are in the area of the electric fields.
• The effect of electric fields becomes more concern at higher voltages.
ELECTROSTATIC INDUCTION
• Primary cause of induction to vehicles, buildings, and object of comparable size.
• Human body is effected to electric discharges from charged objects in the field of the line.
• The current densities in human cause by electric fields of transmission lines are much higher than those induced by magnetic fields!
CORONA
• When surface potential gradient exceeds the dielectric strength of surrounding air, ionization occurs in the area close to conductor surface.
• This partial ionization is known as corona.
• Corona generate by atmospheric conditions (i.e. air density, humidity, wind)
CORONA
• Corona produces power loss and audible noise (Ex: radio interference).
• Corona can be reduced by:– Increase the conductor size.– Use of conductor bundling.
Review
• Transmission Line Parameters:– Resistance
• Skin effect
– Inductance• Single phase line• 3 phase line equal & unequal spacing
– Capacitance• Single phase line• 3 phase line equal & unequal spacing
– Conductance• Neglected• Corona
Review
• Effect of Earth on the Capacitance
• Magnetic Field Induction
• Electrostatic Induction
• Corona