Line parameters
EEE 2413: POWER SYSTEM I

Transmission Line Parameters - RL

By
K Kaberere




An electric transmission line has four parameters;
resistance (R), inductance (L), capacitance (C), and
shunt conductance (G).
The parameters are uniformly distributed along
the line and they are given per unit length of the
line.
The parameters are functions of line geometry,
construction material and operational frequency.
The line resistance and inductance form series
impedance whereas the capacitance and
conductance form the shunt admittance.
R determines the losses and current rating.
L determines fault level, power transfer capability.
2
Types of overhead conductors
▪ Aluminum and its alloys offer the following
benefits:
1.It is cheaper than copper
2. It offers larger diameter for the same amount of
current which reduces corona.
▪ In early days, copper was used for overhead
conductors due to its excellent conductivity.
▪ It was later replaced with much lighter, but less
conductive aluminum.
▪ Over time, various aluminum alloys were
introduced that offered improved strength (with
lower conductivity), and steel core strands were
added to increase the conductors’ overall strength
to accommodate greater spans with reduced sag
An ideal conductor has following features:
1. It has maximum conductivity
2. It has high tensile strength
3. It has least specific gravity i.e. weight / unit volume
4. It has least cost without sacrificing other factors.
3
Categories of overhead conductors
1.



Homogeneous Conductors
Copper
All Aluminum Conductor (AAC)
All Aluminum Alloy Conductor (AAAC)
Non Homogeneous Conductors
Aluminum Conductor Steel Reinforced
(ACSR)
 Aluminum Conductor Alloy Reinforced
(ACAR)
 Aluminum Alloy Conductor Steel Reinforced
(AACSR)
2.

4
Choice of conductors
Power delivery requirements
 Current carrying capacity (ampacity)
 Electrical losses
Line design requirements
 Distances to be spanned
 Sag and clearance requirements
Environmental considerations
 Ice and wind loading
 Ambient temperatures
1
RESISTANCE
This is the major cause of power loss in a
transmission line.
 The dc resistance at temperature T is given by
T L

Temperature
R2 =
Rdc ,T =

A
Factors affecting conductor resistance
1. The resistivity and hence the resistance of a
conductor/material varies with temperature as
shown in Fig.1. However, the relationship
between temperature and resistivity is linear
and a correction may thus be made for varying
temperature using the following expression
(T + t2 ) R
T + t1
1
where T is a constant e.g.
T= 234.5 for annealed copper
= 241 for hard drawn copper
= 228 for aluminium
Resistance
Fig. 1: Variation of resistance
with Temperature
The ac or effective
resistance of a
conductor is given by
Rac =
Ploss
I
2

7
Stranded conductors are normally spiralled thus
making them longer than the solid conductor.
Thus, the DC resistance of a stranded
conductor of a specified length is 1-2% more
than the value calculated from the equation
given earlier.
Skin effect
2.
3.
8
The alternating flux induces higher voltages at
the centre and by Lenz’s law this opposes the
change of current producing it → higher
current density on the surface→ therefore,
higher effective resistance.
▪ Conductor power loss increases with
increased frequency and hence resistance also
increases.
▪ Skin effect is a significant factor in the large
conductors.
▪ The AC resistance is always higher than the
DC value.
▪
⚫ This is the phenomenon of non-uniform distribution
of AC current within a conductor with higher
current densities being experienced on the
conductor surface.
⚫ This is caused by flux inside the conductor which is
stronger at the centre than on the periphery.
⚫ The effect is felt more as the frequency increases.
9
10
INDUCTANCE OF A CONDUCTOR
The flux linking the conductor consists of
internal and external fluxes.
 Internal flux due to current in the conductor;
external flux due to conductor current and
current of other conductors in the vicinity.
 To obtain the inductance of a transmission
line, both the internal and external fluxes must
be taken into consideration.
 Total inductance per metre length of the
conductor is given by

 Consider a solid conductor length 1 m radius r
and carrying current I.
Ignore skin effect and assume the conductor is
non-magnetic i.e. µ = μ0= 4π x 10-7 H/m.
 The internal inductance per unit length
Lint = 1 x10-7 H / m
2
Lt = Lint + Lex H / m
11
12
2
External inductance
1. Inductance of a single phase two-wire line
The points P1 and P2 lie r and R metres
respectively, away from the centre of a
conductor carrying current I.
 The inductance due to the flux included
between P1 and P2


−1
4
-I
+I
R
Lex = 2  10-7 ln H / m
r
R
1
Lt =   10-7 + 2  10-7 ln  H / m
2
r

R
-7
Show Lt = 2  10 ln H / m
r
where r  = re
Consider a single phase line consisting of two
conductors a and b each of radius r, that are D
metres apart as shown in Fig. 2 (r << D)
La ,int = 1 x10-7 H / m
2
r
a
13
where r  = re
−1
4
Fig. 2: Two-wire line
= 0.7788r
14
2. Inductance of composite-conductor lines
❑ r´ is the geometric mean radius (GMR).
❑ Gives equivalent radius of a thin-walled
conductor with no internal flux linkage but
with the same inductance as a conductor of
radius r.
❑ Similarly for conductor b
D
Lb = 2  10-7 ln H / m
r
D
L = La + Lb = 4  10-7 ln H / m
r



Composite conductors have two or more
wires/elements/strands which are electrically in
parallel.
Assume the strands are identical and that they
share the current equally.
Consider a single phase circuit as shown in Fig. 3.
N strands
M strands
k
2
Dk1
1
k´
2´
Dk1´
X
L is the inductance of the two-wire line
considering currents in both conductors.
1´
Y
Fig. 3: single phase circuit with composite conductors
15


b
D
D
La ,ex = 2 x10 ln H / m
r
D
-7
La = 2  10 ln H / m
r
-7
= 0.7788r
r
Each strand in X carries current I/N and that in Y
carries –I/M.
The inductance of strand k, Lk in conductor X is
given by
16
Lav =
L1 + L2 +
N
+ LN
Since the strands are in parallel, the resultant
inductance of conductor X, LX is given by
1
  M
 M
   Dkm  


 H /m
Lk = 2  10-7. N  ln  m =1
1 
N
 N
 
D

  m =1 km  
Lav L1 + L2 + + LN
=
N
N2
1


  M
 M 



D

km 

-7
N
 2  10   m =1
 
= 
.  ln
1 
N
N
k =1 

 N 

Dkm 


  
m =1
  

LX =
The average inductance of the strands of
conductor X Lav is given by
17
18
3
N
= 2  10-7 ln 
k =1
LX = 2  10-7 ln
 M

  Dkm 
 m =1

1


  Dkm 
 m =1

1
N
Dxy
Dxx
N
NM
NM
N
N2
N
 D
km
k =1 m =1
= 2  10-7 ln
km
k =1 m =1
= 2  10-7 ln
N2
M
 D
Dm
H / m per conductor
Ds
❑ Similarly for conductor Y
D
LY = 2  10-7 ln xy H / m per conductor
D yy
The total inductance L of the single phase
circuit is given by
L = LX + LY H / m
Dxy – Geometric Mean Distance (GMD) between
conductors X and Y (also called mutual GMD)
Dxx – Geometric Mean Radius (GMR) of conductor X
(also called self GMD)
19
20
Example 1
Solution
N = 3; M= 2
A circuit of a single phase transmission line is
composed of three solid 0.25 cm radius wires.The
return circuit has two 0.5 cm radius wires.The
arrangement of the conductors is as shown in Fig. 4.
Find the circuit inductance per metre.
9m
6m
DmX = DXY =
3
D
DXY
= 2  10-7 ln mX =6.21  10-7 H / m
DXX
DsX
= DXY
LX = 2  10-7 ln
2’
DmX = DmY
3
DsY = DYY = 4 D12 D21 ( rY ) = 0.153
2
2
1’
LY = 2  10-7 ln
1
Fig. 4: Single phase transmission line
21

where r  = re
−1
4
= 0.7788r
b


r
a
D
H /m
r

Unsymmetrical lines commonly used in practice;
cheap and convenient in design and construction.
The phases have different inductances; different
voltage drops even under balanced current
conditions.
Receiving end voltages are unbalanced.
a
D
D
D12
r
22
4. Inductance of unsymmetrical 3-phase line
Symmetrical or equilateral spacing; solid conductors are at
the vertices of an equilateral triangle as in Fig. 5
Assuming the system is balanced i.e. Ia + Ib + Ic = 0, the
inductance of each conductor is given by
L = 2  10-7 ln
D
DXY
= 2  10-7 ln mY =8.5  10-7 H / m
DYY
DsY
L = LX + LY = 1.47  10−6 H / m
3. Inductance of symmetrical 3-phase line

D11 D12 D21 D22 D31 D32 = 10.7
DsX = DXX = 9 D12 D13 D21 D23 D31 D32 ( rX ) = 0.48
9m
6m
6
r
c
D
b
Fig. 5: Symmetrical 3-phase line
23
r
D31
r
r
D23
c
Fig. 6: Unsymmetrical 3-phase line
24
4
To balance the inductances, the line is
transposed; the position of the conductor is
changed so that each phase occupies each
conductor position for the same length as in
Fig. 7.
Transposition also helps reduce the inductive
interference with nearby communication lines.
c
a
b
a
b
c
L = 2  10-7 ln
3
D
D12 D23 D31
= 2  10-7 ln eq H / m
r
r
Deq is the geometric mean distance between the
phases.
L is the positive (or negative sequence) inductance
of a fully transposed three phase line.
b
c
a
l/3
l/3
l/3
Assuming the system currents are balanced i.e. Ia +
Ib + Ic = 0, the average inductance per phase of a
transposed line is given by

Fig. 7: Transposition cycle of unequally spaced line conductors
25
26
5. Inductance of double circuit single phase line
N = M= 2
Consider the single phase lines shown in Fig. 8. The
conductors are identical with radius r.
d2
1
DmX = DXY = 4 D11 D12 D21 D22
DsX = DXX =
1’
4
LX = 2  10-7 ln
d1
Y
X
2
D
DXY
= 2  10-7 ln mX
DXX
DsX
DmX = DmY = DXY
2’
2
D12 D21 ( r  )
Fig. 8: Double circuit single phase line
27
Since the conductors are identical and the
arrangement similar for X and Y
DsY = DsX
D
L = LX + LY = 4  10-7 ln m H / m
Ds
28
6. Inductance of double circuit three phase line
 Two three-phase circuits, that are connected electrically
in parallel are carried on the same tower.
 Inductance is the same for identical circuits –
construction and operation.
Dc’a
a
c’
Db’c’
Dab
b
Dca
Dc’a’
Dbc
c
As in Fig. 9, one circuit consists of conductors a, b,
and c and the other a’, b’, and c’.
 Phase A; a connected in parallel with a’
 Phase B; b connected in parallel with b’
 Phase C; c connected in parallel with c’
 Transposition cycles are as shown in Fig. 10

b’
Da’b’
Dca’
a’
Fig. 9: Double circuit three phase line
29
30
5
a
b
a’
b’

A
b
c
b’
c’
L = 2  10-7 ln
B
c
a
c’
a’
The average inductance per phase of a
transposed line is given by
Dm
H / m where Dm =
DsL
3
DAB DBC DCA
• DsL is the self GMD of a conductor over a
complete transposition cycle.
• Use same formula for three phase double circuit
C
DsL = 3 DsLA DsLB DsLC
Fig. 10: Transposition cycle of a double circuit three phase line
31
DsLA = GMR of phase A
DAB = mutual GMD between phases A and B
= 4 Daa Daa Daa Daa = r Daa
= 4 Dab Dab Dab Dab
Similarly, DsLB and DsLC GMRs of phases B and C,
respectively are given by
DBC = mutual GMD between phases B and C
= Dbc Dbc Dbc Dbc
4
DCA = mutual GMD between phases C and A
=
4
32
DsLB =
4
Dbb Dbb Dbb Dbb = r Dbb
DsLC =
4
Dcc Dcc Dcc Dcc = r Dcc
Dca Dca Dca Dca
Equivalent self GMD DsL
DsL =
3
DsLA DsLB DsLC =
6
Daa Dbb Dcc ( r  )
3
33
b) Flat vertical spacing
Consider conductor arrangement shown in Fig. 11
 The inductance of the double-circuit line per
phase is given by
L = 2  10-7 ln
34
x
a
Dm
H /m
DsL
v
 Since each phase has two conductors connected
in parallel, the inductance of each conductor is
double the phase inductance.
 For low inductance value, Dm should be low and
DsL high.
 Thus, phases should be spaced as close as
permissible and conductor in same phase as far
as possible.
z
b
c’
y
b’
v
c
a’
Fig. 11: Double circuit three phase line
with flat vertical spacing
35
 1  v  12  y  13 
L = 2  10-7 ln  2 6      H / m
 r    z  

36
6
7. Bundled conductors
Special cases of double circuit lines
At EHV, transmission lines usually have more
than one conductor per phase.
 These are called bundles which may be 2, 3, or 4
arranged as shown in Fig. 13.

a) Conductors at the vertices of a regular hexagon
Consider conductor arrangement shown in Fig. 12
D
a
c’
D
D
d
d
d
d
d
d
b
3D
b’
D
c
D
D
L = 10-7 ln
a’
d
d
Fig. 13: Arrangement of bundled conductors
3D
H /m
2r 
 The 3-conductor bundle has its sub-conductors at the
vertices of an equilateral triangle whereas those of a 4conductor bundle are at the vertices of a square.
Fig. 12: Double circuit three phase line with hexagonal spacing
37
38
Bundled conductors
4-bundle conductor
8-bundle conductor
39
40

Bundling is normally done to reduce the
electric field strength on a conductor
surface, which in turn reduces/eliminates
corona and its resulting effects i.e. power
loss, communication interference and audible
noise.
 Bundling also results in reduced line
reactance due to increase in GMR of the
bundle.
 The current carrying capacity of the lines
also increases since the load is shared by
multiple (2, 3, 4) sub-conductors per phase.

41
The line inductance is given by
L = 2  10-7 ln
Deq
Ds
H /m
 Ds is the GMR of the bundle and the equation of a
stranded conductor is applied.
 If the conductors are stranded and bundle spacing
d is much greater than the conductor radius, each
stranded conductor is replaced by a solid
conductor with GMR = Ds.
 The bundle is replaced by one equivalent
conductor whose GMR = DSL
where DSL =
N
N2
N
 D
k =1 m =1
km
(N is number of bundled
sub-conductors)
42
7

Example 2
Thus, for a 2 sub-conductor bundle
A three phase 50 Hz transmission line has twin
conductors as shown in Fig. 14 with horizontal spacing
of 6 m. Each sub-conductor of the bundle has a
diameter of 25 mm and spacing between the subconductors is 0.3m. Each phase group shares the total
load current and charge equally. The line is completely
transposed.
a) Determine the line inductance per kilometer and the
line inductive reactance per phase per kilometer.
b) If the sub-conductors of each phase are replaced by
single conductors, calculate the percentage decrease in
inductive reactance due to bundling.Assume the crosssectional area of each single conductor is equal to the
total area of the two sub-conductors of a phase.
DSL = Ds .d
For a 3 sub-conductor bundle
DSL = 3 Ds .d 2
For a 4 sub-conductor bundle
DSL = 1.09  4 Ds .d 3
d
d
d
d
In general, for an N sub-conductor bundle
DSL = 1.09  N N . Ds . A N −1
A=
d
Radius of a circle through the

2  sin   centres of the bundle sub N  conductors
43
d
d
D
44
GMD between sub-conductors of one bundle and
those of another
d
D
Dmb = 3 DAB DBC DCA = 7.56 m
Fig. 14: Twin conductor three phase line
Inductance of the bundled-conductor line
SOLUTION
Lb = 2  10−7 ln
D = 6m; d = 0.3 m; r = 12.5 mm
Dmb
= 9.882  10−7 H/m
b
DsL
The inductive reactance of the bundledconductor line per phase
XLb = 2fLb = 0.3105 /km
a) GMR of each bundle
D SLb = r d = 5.404  10−2 m
45
Use of tables to determine stranded conductor GMR
b) Equivalent line with single conductor per phase
For equal cross-sectional areas
2r2 = r12  r1 = 2 x 12.5 x 10-3 m
The GMR of stranded conductors is
usually provided by manufacturers.
 If the distance between conductors are
large compared to the distance between
sub-conductors of each conductor, then
GMD is approximately equal to the
distance between the conductor centres.
READ ON USE OF MANUFACTURERS
TABLES

GMR of each conductor
DsL = r1 = 13.767  10−3 m
GMD, DmL = 3 DAB DBC DCA = 7.56 m
L = 2  10−7 ln
46
DmL
= 12.616  10−7 H/m
DsL
The inductive reactance of the line per phase
XL = 2fLb = 0.3963 /km  21.7% reduction in
reactance
47
48
8