ECE 421/599
Electric Energy Systems
4 – Transmission Line Parameters
Instructor:
Kai Sun
Fall 2014
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 or corona discharge
caused by ionization of air)
•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
(Source: wikipedia.org and EPRI dynamic tutorial)
3
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, 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
t1 and t2 are in
T≈228 for aluminum
oC
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 = Lint + Lext= λint / I + λext / I
• Ampere’s law
Hx
x
µ0 I x
=
Bx µ=
H
0
x
2π x
Hx(Ix=I)
Ix: current enclosed at radius x
Hx: magnetic field intensity
Bx: magnetic flux density
µ0=4π×10-7H/m: permeability of free space
6
Inductance due to the internal flux linkage
• Assume uniform current density
r
Ix
I
=
2
π x π r2
⇒ =
Hx
dφx
Ix
I
x
=
2π x 2π r 2
=
Bx µ=
0Hx
µ0 I
x
2
2π r
1
Ix
• Differential flux and flux linkage:
The portion of I is linked to dφx
• Total internal flux linkage and the inductance:
7
Inductance due to the external flux linkage
• Ix=I for x>r
µ0 I x µ0 I
=
2π x 2π x
=
Bx µ=
0Hx
dφx
D1
µ0 I
r2
d λ=
dφ=
Bx dx=
dx
⋅1
x
x
2
r
2π x
• External flux linkage and inductance
between two points:
µI
λext= 0
2π
∫
D2
D1
D2
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
Phase current
Return current
I1= -I2 D1=r1 D2=D (why?)
L1 = L1(int) + L1( ext ) =
1
D
×10−7 + 2 ×10−7 ln H/m
2
r1
def
r1′ = r1e −1/ 4
D
H/m
r1′
D
L2 = 2 ×10−7 ln
H/m
r2′
L1= 2 ×10−7 ln
• If two conductors are identical: L= 2 × 10−7 ln D H/m
r ′ = r e −1/4
GMR (Geometric mean radius):
DS=r’ for a single conductor
r′
r1=r2=r
−7
Compared to Lext = 2 ×10 ln
D
H/m
r
L = 0.2 ln
D
mH/km
Ds
GMR is the radius of a fictitious conductor without
internal flux but the same inductance as the actual
conductor
9
Self- and Mutual Inductances
• Consider the flux linkages for 1-meter length of the
single-phase circuit
I1+I2=0
λ1
L=
( L11 − L12 ) I1 =
L1 I1
11 I1 + L12 I 2
L21 I1 + L22 I 2 =
(− L21 + L22 ) I 2 =
L2 I 2
λ2 =
Compare to
L1 =
2 × 10−7 ln
D12
1
1
=
2 × 10−7 ln − 2 × 10−7 ln
H/m
r1′
r1′
D12
L2 =
2 × 10−7 ln
D12
1
1
H/m
=
−2 × 10−7 ln
+ 2 × 10−7 ln
r2′
D12
r2′
 λ1   L11
λ =  L
 2   21
λ = LI
 1
 ln r ′
L12   I1 
1
= 2 × 10−7 



L22   I 2 
 1
ln D
12

L11= 2 ×10−7 ln
1
r1′
L22 = 2 ×10−7 ln
1
r2′
L12 L21 = 2 ×10−7 ln
=
1
D12
1 
D12   I1 
 
1  I
ln   2 
r2′ 
ln
10
•Consider n conductors
I1 + I 2 + ⋅⋅⋅ + I i + ⋅⋅⋅ + I n = 0
I1+I2=0
L11= 2 ×10−7 ln
1
r1′
L22 = 2 ×10−7 ln
1
r2′
L12 = L21= 2 ×10−7 ln
n
Lii = 2 × 10−7 ln
1
ri′
Lij = L ji = 2 × 10−7 ln
1
D12
1
Dij
1 n
1
λi =
Lii I i + ∑ Lij I j =
2 × 10 ( I i ln + ∑ I j ln
)
′
ri j 1
Dij
j 1=
−7
j ≠i
λ=
1
1 
 1
ln
ln

ln
 r′
LI
D12
D1n 
1


1
1 
 1
ln
ln
ln
r2′
D2 n 
L= 2 ×10−7  D12

 

 


1
1
1
ln
ln
 ln 
 D1n
D2 n
rn′ 
j ≠i
 1
 ln D
∆
11

Dii = ri '
 1
ln
−7 
= 2 ×10  D12
 

ln 1
 D1n
ln
1
D12
ln
1
D22
ln
1
D2 n
1 
D1n 

1 
ln
D2 n 


 

1 
 ln
Dnn 
 ln
11
Inductance of 3-Phase Transmission Lines (Asymmetric Spacing)
• Consider a three-phase line with 3 identical conductors
 1
 ln r ′

 1
L= 2 × 10−7 ln
 D12
 1
 ln
 D13
λ = LI
ln
1
D12
ln
ln
1
r′
1
D23
1 
D13 
1 
ln
D23 
1 
ln 
r′ 
ln
1
1
1
+ I c ln
2 × 10 ( I a ln + I b ln
)
λ a=
r′
D12
D13
−7
1
1
1
2 × 10 ( I a ln
)
λ b=
+ I b ln + I c ln
D12
r′
D23
−7
λ c=
2 × 10−7 ( I a ln
1
1
1
+ I b ln
+ I c ln )
D13
D23
r′
For balanced three-phase current:
(D11=D22=D33=r’)
λa
1
1
1
=
+ a ln
L a=
2 × 10−7 (ln + a 2 ln
)
Ia
r′
D12
D13
λ
1
1
1
L b =b =
2 × 10−7 ( a ln
)
+ ln + a 2 ln
′
Ib
D12
r
D23
λc
1
1
1
L c=
=
2 × 10−7 ( a 2 ln
+ a ln
+ ln )
Ic
D13
D23
r′
Note: La, Lb and Lc may have imaginary terms
I b = I a ∠ − 120°= I a ∠240°= a 2 I a
a=
1∠120°
I c = I a ∠ − 240°= I a ∠120°= aI a
a + a 2 = 1∠120° + 1∠240° = −1
12
Transpose Line: Mitigation of Asymmetry
GMD
GMD
GMD
L=
GMD: geometric mean distance
La + Lb + Lc
3
2 × 10−7
1
1
1
1
=
(3ln − ln
− ln
− ln
)
3
r′
D12
D23
D13
−7
= 2 ×10 ln
3
D12 D23 D13 =GMD
r′
For symmetric spacing:
GMD
mH/km
L = 0.2 ln
Ds
D12=D23=D13=D
D
L = 0.2 ln
mH/km
Ds
13
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
D12
r′
D23
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
14
Inductance of Bundled Conductors
•Single-phase with two bundled conductors
n conductors
m conductors
GMD
Lx = 2 × 10 ln
H/m
GMRx
−7
GMRx
GMD =
mn
∏
i∈x , j∈ y
Dij
GMD
…
GMRx = n2 ∏∏ Dij
i∈x
j∈x
= n2 ∏ Dii
i∈x
λa = 2 × 10−7
∏
i , j∈x ,i ≠ j
Dij2
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
15
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 =
2n ×10−7 ln
L=
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
GMD
H/m
GMRx
GMR
GMD
for m=1, n=1:
GMR=r’, GMD=D
( Daa ' Dab ' Dac ' ⋅ ⋅ ⋅ Dam ) ⋅ ⋅ ⋅ ( Dna ' Dnb ' ⋅ ⋅ ⋅ Dnm )
( Daa Dab ⋅ ⋅ ⋅ Dan ) ⋅ ⋅ ⋅ ( Dna Dnb ⋅ ⋅ ⋅ Dnn )
∆
Daa= Dbb= Dnn
= rx=' Ds , x
16
GMR of Typical Bundled Conductors
GMRx = n2 ∏∏
Dij
=
i∈x
j∈x
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
17
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
4
( 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
18
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 (Voltage drop from D1 to D2)
2πε 0 x
2πε 0 D1
19
Capacitance of Single Phase Lines
•Consider 1-meter length of a single-phase line
V12 =
q
2πε 0
ln
D2
D1
(q2 =
− q1 =
−q)
– Conductor 1 carries a charge of q1(c/m)
V12( q1 ) =
q1
2πε 0
ln
D
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
D
ln
r
F/m
D
Compare to the
L= 2 ×10−7 ln
H/m
inductance per conductor:
r′
20
• 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)
ρAl(20oC)=2.82×10−8 Ω⋅m
C=0.0171 µF/km → 1/(ωC)= 0.155MΩ/km
L=0.7018 mH/km → ωL = 0.2646Ω/km ~ 34.4×R
R=ρAl(20oC)×1000/(πr2) =0.00769Ω/km
21
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
Dkj
q1 + q2 + ⋅ ⋅ ⋅ + qn = 0
Dki
22
Capacitance of Three-Phase Lines
Vij =
V=
ab ( I )
1
2πε 0
n
1
2πε 0
( qa ln
∑q
k =1
k
ln
Dkj
Dki
D
D12
r
+ qb ln
+ qc ln 23 )
r
D12
D13
qa + qb + qc =
0
D23
D13
r
V=
(
q
ln
+
q
ln
+
q
ln
)
ab ( II )
a
b
c
2πε 0
r
D23
D12
1
Vab
=
( III )
1
2πε 0
( qa ln
D13
D
r
+ qb ln
+ qc ln 12 )
r
D13
D23
D12 D23 D13
D12 D23 D13
r3
1
Vab
q
q
( qa ln
ln
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 )
1
GMD
r
)
(
ln
ln
)
q
q
=
+
a
b
1
2
r
GMD
πε
0
3
23
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
1
GMD
r
+ ( qb + qc ) ln
]
2πε
r
GMD
1
GMD
r
(2qa ln
) qb + qc =
− qa ln
− qa
2πε
r
GMD
3qa
GMD
Compared to
ln
=
2πε 0
r
0.0556
C=
µ F/km
D
ln
2πε 0
qa
0.0556
r
=
C =
F/m =
µ F/km
=
Van
ln
[2qa ln
GMD
r
ln
GMD
r
L = 0.2 ln
(for singlephase line)
GMD
mH/km (for 3-phase
line)
r′
24
Effect of Bundling
GMD
L = 0.2 ln
mH/km
b
Ds
=
Dsb
=
Dsb
Ds × d
3
Ds × d 2
=
Dsb 1.09 4 Ds × d 3
GMRL
C=
0.0556
µ F/km
GMD
ln b
r
b
r=
b
r=
r×d
3
r×d2
=
r b 1.09 4 r × d 3
GMRC
25
A Summary: GMD, GMRL and GMRC → L and C
A
a1
Db
b1
Da1,b1
s
B
rb
Da1,bm
…
…
Da1,an
Db1,bm
Dan,b1
bm
an
Dan,bm
•
rb=r and Dbs=Ds for single conductors
GMD = nm Da1b1 Da1b2 ...Danbm −1 Danbm
GMRL, A = n2 ( Dsb Da1a2 ...Da1an )...( Dan a1 ...Dan an−1 Dsb )
GMRC , A = n2 ( r b Da1a2 ...Da1an )...( Dana1 ...Danan −1 r b )
LA = 0.2 ln
CA =
GMD
mH/km
GMRL, A
0.0556
µ F/km
GMD
ln
GMRC , A
L↓ if GMRL↑.
C↑ if GMRC↑ 26
Three-Phase Double-Circuit Lines
A
• GMD between 3 phase groups
DAB = 4 Da1b1 Da1b2 Da2b1 Da2b2
DBC = 4 Db1c1 Db1c2 Db2c1 Db2c2
B
DAC = 4 Da1c1 Da1c2 Da2c1 Da2c2
• GMD per phase (consider transposition)
C
GMD = 3 DAB DBC DAC
• GMRL of each phase group
( Dsb Da1a2 ) 2
=
Dsb Da1a2
( Dsb Db1b2 ) 2
=
Dsb Db1b2
DSA
4
DSB
4
=
DSC
=
( D Dc1c2 )
4
b
s
2
3
rA = r b Da1a2
D Dc1c2
• Equivalent GMRC
GMRC =
DSA DSB DSC
• Inductance and
Capacitance:
rC = r b Dc1c2
rB = r b Db1b2
b
s
• Equivalent GMRL
GMRL =
• GMRC of each phase group
C=
0.0556
µ F/km
GMD
ln
GMRC
L = 0.2 ln
3
rArB rC
GMD
mH/km
GMRL
27
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
Ground
Problem 4.15
28
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
29
Example 4.3
ACSR (7 steel and 24 aluminum strands)
30