ECE 421/521
Electric Energy Systems
Power Systems Analysis I
4 – Transmission Line Parameters
Instructor:
Kai Sun
Fall 2013
1
Introduction
• Transmission lines
– Overhead lines
– Underground Cables (less than 1%)
• Properties
Corona discharge on insulator string of a 500 kV line
(source: wikipedia.org)
– Series Resistance (stranding and skin effect)
– Series Inductance (magnetic & electric fields; flux linkages within the
conductor cross section and external flux linkages)
– Shunt Capacitance (magnetic & electric fields; charge and discharge due
to potential difference between conductors)
– Shunt Conductance (due to leakage currents along insulators and corona
loss)
• Line-to-line voltage levels
– 69kV, 115kV, 138kV and 161kV (sub-transmission)
– 230kV, 345kV, and 500kV (EHV)
– 765kV (UHV)
2
Overhead Transmission Lines
Shield wires (ground wires) are ground conductors used to
protect the transmission lines from lightning strikes
3
(Source: wikipedia.org and EPRI dynamic tutorial)
Overhead Transmission Lines
• Materials
–
–
–
–
–
AAC (All Aluminum Conductor),
AAAC (All Aluminum Alloy Conductor)
ACSR (Aluminum Conductor Steel Reinforced)
ACAR (Aluminum Conductor Alloy Reinforced)
ACCC (Aluminum Conductor Composite Core)
ACSR (7 steel and 24 aluminum strands)
• Why not copper?
– Relative lower costs and higher strength-toweight ratios than copper
• Bundle conductors
24/7 ACSR and modern ACCC conductors
– Preferred for high voltages to reduce power
loss and line reactance
– e.g. 2 conductor bundles for 230kV, 3-4 for
345-500kV, and 6 for 765kV
4
A bundle of 4 conductor
Line Resistance
Consider a solid round conductor at a specific temperature:
• DC resistance
Rdc =
ρl
A
ρ = conductor resistivity
l = conductor length
A = conductor cross-sectional area
• AC resistance
– Current is not uniformly distributed over the cross-sectional
area; current density is greatest at the surface (skin effect)
– 2% higher than DC resistance at 60Hz
• Temperature impact
oC
t1 and t2 are in
T≈228 for aluminum
Skin effect: circulating eddy
current IW cancelling the
current flow in the center of
the conductor.
(source: wikipedia.org)
5
Inductance of a Single Conductor
• Inductance for nonmagnetic material:
L=λ/I
Hx
• Ampere’s law
Hx: magnetic field intensity
Ix: current enclosed at radius x
• Assume uniform current density throughout the conductor
Ix
I
=
2
πr
π x2
⇒ Hx =
I
2π r
2
x
• Magnetic flux density:
=
Bx µ=
0Hx
µ0 I x
µ0 I
=
x
2π x 2π r 2
µ0=4π×10-7H/m: permeability of free space
6
Inductance of a Single Conductor (cont’d)
• Differential flux:
r
dφx
1
• Differential flux linkage:
The portion of Ix is linked to dφx
Ix
• Total flux linkage:
• Inductance due to the internal flux linkage:
7
Inductance of a Single Conductor (cont’d)
• Due to external flux linkage
Ix=I for x>r
D1
µ0 I x µ0 I
=
2π x 2π x
=
Bx µ=
0Hx
µ0 I
r2
d λ=
d
B
dx
dx
=
=
⋅
1
φ
x
x
x
2
r
2π x
D2
External flux linkage and inductance between two points:
µI
λext= 0
2π
∫
D2
D1
D
1
dx= 2 × 10−7 I ln 2 Wb/m
x
D1
Lext= 2 × 10−7 ln
D2
H/m
D1
8
Inductance of Single-Phase Lines
• 1-meter length of a single-phase line with two solid round conductors
I1= -I2 D1=r1 D2=D
Phase current
Return current
The image part with relationship ID rId6 was not found in the file.
L1 = L1(int) + L1( ext ) =
1
D
×10−7 + 2 ×10−7 ln H/m
2
r1
1
D
L1 =
2 ×10−7 ( + ln )
r1
4
−7
1
4
2 ×10 (ln e + ln
=
1
D
+ ln )
r1
1
1
D
=
2 ×10−7 (ln −1/4 + ln )
r1e
1
If two conductors are identical: r1=r2=r
def
r1′ = r1e −1/ 4
1
D
L1 = 2 ×10 ln + 2 ×10−7 ln H/m
r1′
1
1
D
L2 = 2 ×10−7 ln + 2 ×10−7 ln
H/m
r2′
1
−7
L= 2 ×10−7 ln
=
r ′ r=
e −1/4 DS
Lext= 2 × 10−7 ln
9
D2
H/m
D1
D
H/m
r′
L = 0.2 ln
D
mH/km
Ds
GMR (Geometric mean radius)
GMR is the radius of a fictitious conductor with no
internal flux but the same inductance as the actual
conductor
Self- and Mutual Inductances
• Consider the flux linkages for 1-meter length of
the single-phase circuit
I1+I2=0
λ1 =L11 I1 + L12 I 2 =( L11 − L12 ) I1 =L1 I1
λ2 =
L21 I1 + L22 I 2 =
(− L21 + L22 ) I 2 =
L2 I 2
Compare to
L1 = 2 × 10−7 ln
1
1
H/m
− 2 × 10−7 ln
r1′
D12
L2 =−2 × 10−7 ln
 λ1   L11
λ =  L
 2   21
1
1
+ 2 × 10−7 ln
H/m
D12
r2′
L11= 2 × 10−7 ln
1
r1′
L22= 2 × 10−7 ln
1
r2′
L12= L21= 2 × 10−7 ln
 1
 ln r ′
L12   I1 
1
−7

=
×
2
10
L22   I 2 
 1
ln D
12

1 
D12   I1 
 
1  I
ln   2 
r2′ 
ln
10
λ = LI
1
D12
• Consider n conductors
Lii = 2 × 10−7 ln
1
ri′
Lij = L ji = 2 × 10−7 ln
1
Dij
I1 + I 2 + ⋅⋅⋅ + I i + ⋅⋅⋅ + I n = 0
n
1 n
1
λi =
Lii I i + ∑ Lij I j =
2 × 10 ( I i ln + ∑ I j ln
)
ri′ j 1
Dij
j 1=
j ≠i
−7
λ = LI
j ≠i
1
1 
1
1 
 1
 1


ln
ln
ln
ln
ln
ln
 r′
 D
D12
D1n 
D12
D1n 
1
11




1
1 
1
1 
 1
 1
ln
ln
ln
ln
ln
ln
−7 
r2′
D2 n  =
D12
D22
D2 n 
L=
×
2 ×10−7  D12
2
10



 
 

 

 




1
1
1
1
1
1
ln
ln

 ln 
 ln
ln
ln
 D1n
 D1n
D2 n
rn′ 
D2 n
Dnn 
11
∆
Dii = ri '
Inductance of Three-Phase Transmission Lines:
Symmetric Spacing
• Consider 1-meter length of a three-phase line with 3 identical conductors
spaced symmetrically:
D12=D23=D13=D
r1=r2=r3=r
Ia + Ib + Ic =
0
λa =
2 × 10−7 ( I a ln
1
1
1
+ I b ln + I c ln )
r′
D
D
Ib + Ic =
−Ia
2 × 10−7 ( I a ln
λa =
=
2 ×10−7 ln
1
1
− I a ln )
r′
D
D
× Ia
′
r
λ=
λ=
λc
a
b
L=
2 × 10−7 ln
D
D
=
0.2 ln
mH/km
r′
Ds
12
Inductance of Three-Phase Transmission
Lines: Asymmetric Spacing
λ = LI
 1
 ln r ′

 1
L= 2 × 10−7 ln
 D12
 1
 ln
 D13
ln
1
D12
ln
ln
1
r′
1
D23
1 
D13 
1 
ln
D23 
1 
ln 
r′ 
ln
λ a=
2 × 10−7 ( I a ln
1
1
1
+ I b ln
+ I c ln
)
r′
D12
D13
λ b=
2 × 10−7 ( I a ln
1
1
1
)
+ I b ln + I c ln
D12
r′
D23
λ c=
2 × 10−7 ( I a ln
1
1
1
+ I b ln
+ I c ln )
D13
D23
r′
λa
1
1
1
=
+ a ln
L a=
2 × 10−7 (ln + a 2 ln
)
Ia
r′
D12
D13
λ
1
1
1
+ ln + a 2 ln
L b =b =
2 × 10−7 ( a ln
)
Ib
D12
r′
D23
λc
1
1
1
2 × 10−7 ( a 2 ln
L c=
=
+ a ln
+ ln )
Ic
D13
D23
r′
For balanced three-phase current:
I b = I a ∠ − 120°= I a ∠240°= a 2 I a
I c = I a ∠ − 240°= I a ∠120°= aI a
a=
1∠120°
13
La, Lb and Lc may have imaginary terms
Transpose Line: Mitigation of Asymmetry
L=
La + Lb + Lc
3
2 × 10−7
1
1
1
1
=
(3ln − ln
− ln
− ln
)
3
r′
D12
D23
D13
= 2 × 10−7 ln
L = 0.2 ln
( D12 D23 D13 )
r′
a + a 2 = 1∠120° + 1∠240° = −1
1
3
GMD
mH/km
Ds
Geometric mean distance:
14
GMD = 3 D12 D23 D13
A three-phase line has three conductors with r=1.345 in. Determine the
inductance per phase. What if transposition is adopted?
D12=D23=0.889m D13=1.778m
Ds=r’=1.345e-1/4=1.0475in=0.0266m
L a =0.2 × (ln
1
1
1
+ a 2 ln
+ a ln
) =0.7711 − j 0.1201 =0.7804∠-8.85° mH/km
r′
D12
D13
L b =0.2 × (a ln
1
1
1
+ ln + a 2 ln
) =0.7018 mH/km (i.e. inductance/phase
D12
r′
D23
for symmetric spacing)
L c = 0.2 × (a 2 ln
1
1
1
+ a ln
+ ln ) = 0.7711 + j 0.1201 =0.7804∠8.85° mH/km
D13
D23
r′
With transposition:
1
3
L
La + Lb + Lc
( D12 D23 D13 )
= 0.2 ln =
0.7480 mH/km
3
Ds
15
Inductance of Bundled Conductors
• Single-phase with two bundled conductors
n conductors
m conductors
…
λa = 2 × 10−7
1
1
1
1
I
1
1
1
1
−7 − I
(ln
)
+ ln
+ ln
+ ⋅⋅⋅ + ln
+ ln
+ ⋅ ⋅ ⋅ + ln
(ln + ln
) +2 × 10
n
rx′
Dab
Dac
Dan
m
Daa '
Dab '
Dac '
Dam
16
λa= 2 × 10−7 I ln
m
Daa ' Dab ' Dac ' ⋅ ⋅ ⋅ Dam
n
rx′Dab Dac ⋅ ⋅ ⋅ Dan
m D D D ⋅⋅⋅ D
λa λa
aa ' ab ' ac '
am
−7
L=
=
= 2n ×10 ln
a
n r ′D D ⋅⋅⋅ D
Ia I / n
x ab ac
an
…
Average inductance of each conductor:
Lav =
L=
2n ×10−7 ln
n
m
Dna ' Dnb ' Dnc ' ⋅⋅⋅ Dnm
n
r'x Dnb Dnc ⋅⋅⋅ Dnc
La + L b + Lc + ⋅⋅⋅ + Ln
n
• Inductance of bundle x:
Since all conductors are connected in parallel,
Lav La + L b + Lc + ⋅ ⋅ ⋅ + Ln
=
Lx =
n
n2
Lx = 2 ×10−7 ln
=
GMD
=
GMRx
mn
n2
GMR
GMD
H/m
GMRx
GMD
for m=1, n=1:
GMR=r’, GMD=D
( Daa ' Dab ' Dac ' ⋅ ⋅ ⋅ Dam ) ⋅ ⋅ ⋅ ( Dna ' Dnb ' ⋅ ⋅ ⋅ Dnm )
( Daa Dab ⋅ ⋅ ⋅ Dan ) ⋅ ⋅ ⋅ ( Dna Dnb ⋅ ⋅ ⋅ Dnn )
17
∆
Daa= Dbb= Dnn
= rx=' Ds , x
GMR of Typical Bundled Conductors
=
GMRx
n2
( Daa Dab ⋅ ⋅ ⋅ Dan ) ⋅ ⋅ ⋅ ( Dna Dnb ⋅ ⋅ ⋅ Dnn )
GMR =
GMR=
GMR
=
4
9
16
( Ds × d ) 2 =
( Ds × d × d )3=
Ds × d
3
Ds × d 2
1
2 4
( Ds × d × d × d × 2 =
)
1.09 4 Ds × d 3
18
Inductance of Stranded Conductors
• Example 4.1: Determine GMR
A special case of the bundled conductors:
1 r
7
6
D=
D=
D=
2r
12
16
17
2
3
D14 = 4r
D13 = D15 = D − D = 2 3r
2
14
GMR=
49
5
2
45
( r ′ ⋅ 2r ⋅ 2 3r ⋅ 4r ⋅ 2 3r ⋅ 2r ⋅ 2r )6
−
1
4
6
7
= r ( e) (2)6 (3)6 (3) (2)
7
6
7
= 2.1767r
19
4
Line Capacitance
• Consider a long round conductor carrying a charge of q (c/m)
– Capacitance:
q
C=
V
– Electric flux density at a cylinder of radius x:
D=
q
q
=
A 2π x ×1
q
x
– Electric field intensity
E=
D
ε0
＝
q
2πε 0 x
– Electric potential difference between cylinders at D1 and D2
Defined as the work done in moving a unit charge from D1 to D2
=
V12
D2
Edx
∫=
D1
∫
D2
D1
D
q
q
=
dx
ln 2
2πε 0 x
2πε 0 D1
20
(Voltage drop from D1 to D2)
Capacitance of Single Phase Lines
• Consider 1-meter length of a single-phase line
– Conductor 1 carries a charge of q1(c/m)
V12( q1 )
(q2 =
−q1 =
−q)
q1
D
=
ln
2πε 0 r
– Conductor 2 carries a charge of q2(c/m)
V21( q2 ) =
q2
2πε 0
ln
D
r
V12 = V12( q1 ) + V12( q2 ) =
V12( q2 ) = −
q1
2πε 0
ln
q2
2πε 0
ln
D
r
D −q2
D
q
D
+
ln =
ln
r 2πε 0 r πε 0 r
– Line-to-line capacitance between the conductors
C12 =
πε 0
ln
D
r
L= 2 ×10−7 ln
F/m
D
H/m
r′
Compare to the inductance per conductor
21
• Define C, the capacitance per conductor, as the capacitance between
each conductor and a neutral
=
C
2πε 0
q
= 2=
C12
F/m
D
V12 / 2
ln
r
C=
0.0556
µ F/km
D
ln
r
D
D
L 0.2ln= 0.05 + 0.2ln mH/km
Compared to =
r′
r
• For an all aluminum conductor r=1.345 in (0.0342m) and D=35 in (0.889m)
C=0.0171 µF/km
L=0.7018 mH/km ~ 0.2646Ω/km
R=ρAl(20oC)×1000/(πr2) =0.00769Ω/km
22
ρAl(20oC)=2.82×10−8 Ω⋅m
Multi-Conductors
• Consider n parallel long conductors with charges of qk c/m
Since for one single conductor with qk:
Vij =
qk
2πε 0
ln
Dkj
Dki
Dii=Djj=r
for n conductors:
Vij =
1
2πε 0
n
∑q
k =1
k
ln
q1 + q2 + ⋅ ⋅ ⋅ + qn = 0
Dkj
Dki
23
Capacitance of Three-Phase Lines
Vij =
V=
ab ( I )
V=
ab ( II )
Vab
=
( III )
1
2πε 0
1
2πε 0
qk
2πε 0
ln
Dkj
Dki
( qa ln
D
D12
r
+ qb ln
+ qc ln 23 )
r
D12
D13
( qa ln
D23
D
r
+ qb ln
+ qc ln 13 )
r
D23
D12
1
2πε 0
( qa ln
qa + qb + qc =
0
D13
D
r
+ qb ln
+ qc ln 12 )
r
D13
D23
D12 D23 D13
D12 D23 D13
r3
1
=
Vab
( qa ln
+
q
ln
+
q
ln
)
b
c
3
r
D12 D23 D13
D12 D23 D13
3 × 2πε 0
1
3
(D D D )
1
( qa ln 12 23 13 + qb ln
Vab =
2πε 0
r
r
( D12 D23 D13 )
24
1
GMD
r
)
(
ln
ln
)
q
q
=
+
a
b
1
2
r
GMD
πε
0
3
Vab
1
2πε 0
( qa ln
GMD
r
+ qb ln
)
r
GMD
Vca
How to calculate the capacitance per phase?
Vac
1
2πε 0
( qa ln
GMD
r
)
+ qc ln
r
GMD
Vab + Vac =
3Van
GMD
r
+ ( qb + qc ) ln
]
r
GMD
2πε
GMD
r
1
− qa ln
(2qa ln
) qb + qc =
− qa
r
GMD
2πε
3qa
GMD
Compared to
=
ln
r
2πε 0
0.0556
C=
µ F/km
D
ln
qa
2πε 0
0.0556
r
C =
F/m =
µ F/km
=
=
1
Van
ln
[2qa ln
GMD
r
ln
GMD
r
L = 0.2 ln
25
GMD
mH/km
′
r
Effect of Bundling
GMD
mH/km
L = 0.2 ln
b
Ds
∆
b
GMR
= D=
s
Ds × d
∆
b
GMR
= D=
s
3
Ds × d 2
∆
C=
0.0556
µ F/km
GMD
ln b
r
b
r=
b
r=
r×d
3
r×d2
b
r 1.09 r × d
GMR
= D=
1.09 4 Ds × d 3 =
s
b
GMRL
26
4
GMRC
3
A Summary: GMD, GMRL and GMRC → L and C
A
a1
Db
b1
Da1b1
s
B
rb
Da1bm
…
…
Da1an
Db1bn
Danb1
bm
an
Danbm
•
rb=r and Dbs=Ds for single conductors
GMD = nm Da1b1 Da1b2 ...Danbm −1 Danbm
LA = 0.2 ln
GMRL, A = n2 ( Dsb Da1a2 ...Da1an )...( Dana1 ...Danan −1 Dsb )
GMRC , A = n2 ( r b Da1a2 ...Da1an )...( Dana1 ...Danan −1 r b )
27
CA =
GMD
mH/km
GMRL, A
0.0556
µ F/km
GMD
ln
GMRC , A
L↓ if GMRL↑.
C↑ if GMRC↑
Three-Phase Double-Circuit Lines
• GMD between 3 phase groups
A
DAB = 4 Da1b1 Da1b2 Da2b1 Da2b2
B
DBC = 4 Db1c1 Db1c2 Db2c1 Db2c2
DAC = 4 Da1c1 Da1c2 Da2c1 Da2c2
C
• GMD per phase (consider transposition)
GMD = 3 DAB DBC DAC
28
Three-Phase Double-Circuit Lines
• GMRL of each phase group
DSA
( Dsb Da1a2 ) 2
=
Dsb Da1a2
4
( D Db1b2 )
=
D Db1b2
( Dsb Dc1c2 ) 2
=
Dsb Dc1c2
DSB
4
DSC
4
2
b
s
A
b
s
B
• Equivalent GMRL
C
GMRL = 3 DSA DSB DSC
• GMRC of each phase group
rA = r Da1a2
rB = r Db1b2
b
b
• Inductance and Capacitance
L = 0.2 ln
rC = r b Dc1c2
• Equivalent GMRC
GMRC =
3
C=
rArB rC
29
GMD
mH/km
GMRL
0.0556
µ F/km
GMD
ln
GMRC
Effect of Earth on the Capacitance
• The presence of earth alters the distribution of electric flux lines and
equipotential surfaces
– The earth level is an equipotential surface → Image Charges Method
– The effect of the earth is to increase the capacitance
• Negligible for balanced steady-state analysis if conductors are high
Problem 4.15
30
Example 4.2
1 mil = 0.001 inch = 0.0254 mm
1 cmil (circular mil, i.e. the area of a circle with a
diameter of 1 mil)
= π/4 x mil2 =5.067 x 10-10 m2 = 5.067 x 10-4 mm2
31
Example 4.3
ACSR (7 steel and 24 aluminum strands)
32
Induction and Corona Issues
• Magnetic field induction
– Transmission line magnetic fields affect objects in the proximity of the line
– Related to the currents in the line
– Voltage is induced in objects that have a considerable length parallel to the line, like
fences, pipelines, phone lines
– Reported to affect human health (not proven, not fully agreed in research community)
• Electrostatic induction
–
–
–
–
Transmission line electric fields affect objects in the proximity of the line
Becomes increasing at higher voltage (stronger electric field)
Induction to vehicles, buildings, etc.
Human body may be affected by the electric discharge
• Corona
– Partial ionization when surface potential gradient exceeds the dielectric strength of the
surrounding air (30kV/cm at 25oC and standard atmospheric pressure =76cm of Hg)
– Corona produces power losses, audible hissing sound, O3, and radio and television
interference
– Shunt conductance to model corona losses (normally negligible)
– Factors: conductor diameter, line configuration, type of conductor, and surface
condition
33
Magnetic Field Induction – Example 4.6
34
Homework #5
• Read through Saadat’s Chapter 4
• ECE521: 4.2, 4.5, 4.6, 4.8, 4.11, 4.14, 4.15
• ECE421: 4.2, 4.5, 4.6, 4.8, 4.11
• Due date: 10/14 (Monday) submitted in class or by email
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