Cable Properties: Computation of Cable Parameters and ATP
simulations.
By Aditya Upadhye and Mohamed El-Sharkawi
A) Calculating the cable parameters using the theory of flux linkages:
The ALCATEL cable, which is a candidate cable that may be used in the NEPTUNE network, has
a number of peculiar features. Firstly, the core is hollow. Secondly, the sheath and the core are not
insulated from each other. Also, the NEPTUNE network will be operated under single-phase DC
conditions. Thus when we model the cable in ATP (Alternative Transient Program) we have to
consider the special features of this cable.
We need to get the following information about the cable:
1) The cable data provided by the cable manufacturer ALCATEL.
2) The cable parameters calculated using the fundamentals of transmission lines, electromagnetics
and electrostatics.
3) The ATP-computed cable parameters.
To be sure we are using the correct cable parameters, and understand the basics; the ATP-computed
data is taken and matched with the previous two. This way we can be sure of having the exact cable
model.
In this report, firstly the basic derivation for the flux linkages in tubular conductors will be
explained. Then this basic derivation will be applied to the ALCATEL cable in order to find the
cable inductance theoretically. This will be followed by an explanation of how the cable is modeled
in ATP.
Calculation of cable inductance:
1. Basic derivation for flux linkages in tubular conductors:
The NEPTUNE cable has 2 conductors: the steel core (the 2 layers of bundled wires can be
assumed to be a single conductor) and the copper sheath. Both these conductors along with the
insulation are tubular in nature. Hence finding a generic formula for flux linkages associated with a
tubular conductor can be used in finding the flux linkages for both core and sheath.
b
a
r
Fig 1: Cross-section of tubular conductor.
The tubular conductor in the Fig 1 has an inner radius of a, an outer radius of b. r is the radius of
any annular ring within the conductor and may take values from a to b.
Any current carrying conductor has a magnetic field associated with it, which is customarily
pictured in terms of lines of magnetic flux encircling the current or what we call flux linkages.
For an annular conductor,
Total flux linkage associated with conductor
= Flux linkage internal to conductor + Flux linkage external to conductor
Thus,
T  i  e.............................1
To find i:
For DC conditions, there is uniform current density in the conductor. If ‘ir’ is the current enclosed
by the annular element of radius ‘r’ and ‘i’ is the total current enclosed by the conductor, then
ir
i
 2
2
r  a
b  a 2
2
ir 
r 2  a2
i
b2  a2
Now,
‘The line integral of the magnetic field intensity over a closed loop is equal to the current enclosed
by that loop’.
  H  ds  i
For an element of radius r such that
( a  r  b)
H  ds 
r
r 2  a2
i
b2  a2
r 2  a2
i
H 2

2
b  a 2r
B
r 2  a2
i

2
2
b  a 2r
Where  is the conductor permeability and B is the magnetic flux density within the conductor.
If ‘r’ is the radius of the flux line and for uniform current distribution, the fraction of the total
current enclosed within the flux line is given by:
r2  a2
b2  a2
This fraction is the value of ‘N’ for flux line of radius ‘r’ within the conductor.
b
   B  da
a
r 2  a2
r 2  a2
i




dr
2
2
2
2
2

r
b

a
b

a
a
b
i  
Solving we get, the flux linkage internal to the conductor as follows:
i 
i
2
 
 b 2  3a 2

a4

* ln b ...................2
 2
2
2
a
 4b  a  b 2  a 2 

To find e:
Select any annular element with radius ‘r’ such that
(b  r )
It encloses the full conductor current ‘i’.
Thus
ir = i
H  ds  i
r
H
i
2r
B  out 
i
2r
where,
out is the permeability of the material external to the conductor.
Let us assume that ‘c’ is the external radius of the material surrounding the conductor.
c
e   out 
b
 e 
i
2r
dr
 
out  i c
ln
...............................3
b
2
This is the flux linkage external to the tubular conductor.
2. Deriving the inductance of the ALCATEL cable:
Insulating sheath Ø 17 mm
Optical fibers
Thixotropic Jelly
Steel wires strand
Composite conductor
Steel tube
Ø: 2.3 mm
Figure 2: ALCATEL cable
Let, a = inner radius of steel core
b = outer radius of steel core and inner radius of copper sheath.
c = outer radius of copper sheath and inner radius of insulator.
d = outer radius of insulator.
The flux linkages associated with the cable are the sum of the flux linkages due to the core current
and the flux linkages due to the sheath current.
a) Core:
Let steel be the sum of the flux linkages due the ‘core current’ in all 3 regions of
a) Core (material: steel).
b) Sheath (material: copper).
c) Insulator (material: polyethylene).
Using the results in equations (1), (2) and (3),
steel 
 
 
 
 cu  ist
st  ist  b 2  3a 2
a4
ins  ist d

* ln b  
ln c . 
ln
............(4)
 2
2
2
a
b
c
2  4b  a  b 2  a 2 
2
2

Where,
ist = current in core.
st = permeability of steel.
cu = permeability of copper.
ins = permeability of insulator.
b) Sheath:
Let cu be the sum of the flux linkages due the ‘sheath current’ in 2 regions of
a) Sheath
b) Insulator.
Using the results in equations (1), (2) and (3),
cu 
 
 

cu  icu  c 2  3b 2
b4
c   ins  icu ln d ............(5)

*
ln
 2
b
c
2  4c  b 2  c 2  b 2 2
2

Here, icu = current in sheath.
Also note that in (5), cu does not have a term relating to the flux linkages within the core. This is
because, the sheath current enclosed by any annular element within the core will always be zero.
Now, the total flux linkages associated with the cable are given by:
cable = st + cu
Using the physical dimensions of the cable provided by ALCATEL,
a = 0.00115 m
b = 0.0036 m
c = 0.00416 m
d = 0.0085 m.
Now solving,
cable = 2.1081*10-6 ist + 8.9849*10-9 icu………………….(6)
We know the fact that
=
icable
ist + icu
So the question now is to find the current distribution within the core and sheath. This is done as
follows,
Voltage drop per meter of steel core = Voltage drop per meter of copper sheath.
Rst * ist  jLst * ist  Rcu * icu  jLcu * icu.........................(7)
Now,
L*i = 
Thus,
Rst * ist  jst  Rcu * icu  jcu
Where,
st
Rst 
 b  a 2 
 5.607 * 10 3 
Rcu 
cu
 c  b 2 
 1.263 *10 3 
2
2
m
m
st and cu are the resistivities of the steel core and the copper sheath materials respectively.
For DC conditions,  = 0.
Solving equation (7) we get,
ist = 0.1838* icable
(8)
icu = 0.8162* icable
Substituting (8) in (6),
cable = (3.874*10-7) icable + (7.333*10-9) icable ………………..
cable = (3.947*10-7) icable
Thus the composite inductance of the cable is given as,
L
cable
icable
L = 0.3947 H/m
(9)
3) Calculation of capacitance:
The cable capacitance per unit length can be calculated by the formula:
C
2
F /m
ln d
c
 
Where,
 is the permittivity of the insulator.
d is the outer radius of insulator
c is the inner radius of insulator.
The values are as follows:
 = 0R = (2.3)*(8.854*10-12)
d = 0.0085 m
c = 0.00416 m
Solving we get,
C = 0.179 nF/m
4) Computation of cable R,L and C by ATP:
Initial results presented for the ATP calculations of R and L of the cable gave very large values.
There was discrepancy between these ATP-generated values and the values provided by
ALCATEL. Even using accurate physical dimensions of the cable could not solve the problem.
It turned out that ATP was not treating the core and sheath of the cable as a composite conductor,
the entire cable current was passing through the core. The high resistivity and high permeability of
the cable core combined with the problem that the cable core was carrying the entire cable current,
resulted in the large values of both cable inductance and cable resistance.
To resolve this issue, we need to consider the cable core and cable sheath as a composite conductor
in the ATP model.
The parameters that really matter when we try to model the composite conductor are the composite
resistivity (comp) and the composite permeability (comp).
Calculation of comp :
The steel core and the copper sheath are conductors in parallel and the current will be distributed
depending on their resistances.
Rcomp  Rst || Rcu
st

cu
 b  a   c  b 2 


st
cu
 c 2  a 2 

2
2
2
 b  a   c  b 2 
comp
2
2
2
Solving,
comp = 5.1753*10-8 m.
Calculation of comp :
The flux linkages associated with the cable are given by:
cable = st + cu = comp
The flux linkages associated with the composite conductor current will be in the 2 regions of:
1) Internal to composite conductor.
2) Insulator.
Using equations (2) and (3), we get
 
 

comp  icable  c 2  3a 2
a4
c   ins  icable ln d
comp 

*
ln
 2
2
a
c
2
2 2
2
2
 4c  a  c  a 

Solving,
comp = 4.348*10-8 * comp * icable …………………………(10)
Equating (9) and (10),
comp = 9.0788
This is the relative permeability of the composite conductor.
Using these values of resistivity and permeability in the ATP model for the composite conductor
we get the following results for the cable parameters:
R (/km)
L (mH/km)
C (F/km)
Theoretical values
1.03
0.3947
0.179
ATP values
1.03
0.3948
0.179
ALCATEL values
1
0.128
0.2
5) Calculating the ground impedance using the Bessel function approach:
To calculate the internal impedance of any tubular conductor ATP uses the generalized formula for
tubular conductors, which is as follows:
Ztube  in 
m
I 0 mq  K 1 mr   K 0 mq  I 1 mr  m ........................(11)
2qD    
In the above equation,
Ztube-in : Internal impedance of tubular conductor.
 : Resistivity of earth.
q : inner radius of tubular conductor.
r : outer radius of tubular conductor.
I and K are the ‘Modified Bessel functions’ of the first and second kind [1].
m
j

Where ‘m’ is the reciprocal of the ‘complex depth of penetration’.
 : Permeability of earth and
D  I 1 mr  K 1 mq  K 1 mr  I 1 mq
Applying this generalized formula for the earth-return impedance, we have to let outer radius ‘r’
tend to infinity and q=d.
Simplifying,
Zearth 
m K 0mr 

.......................................(12)
2R K 1md 
d: outside radius of tubular conductor and insulation.
For the case of NEPTUNE cable and for any type of submarine cable, the seawater has a resistivity
of 0.2m.
 = 0.2m.
Also, the relative permeability of seawater is given by
sea = 1
The above method and the Bessel function approach are discussed in [1] and [2].
ATP calculates the ground impedance using the above formula and comes up with the following
values:
R = 1.30698*10-7 /m. = 0.130698 m/km.
X = 1.44254*10-6 /m @ freq.=0.1hz.
So, L = 2.2958 mH/km.
We need to know whether these ATP-computed values match with the actual values.
[3] provides a simplified version of equation (12) under some assumptions. The paper suggests that
provided,
m Di
2
where in our case,
Di: Outer diameter of the cable.
 1
Then in (12) K0(x) and K1(x) can be replaced by their asymptotic expansions for x->0 and that
gives
Zsea 

8
 j


 2 Re
ln
......................................(13)
Di
2
Substituting values,
m  1.9869 *10 3
And
Di
 d  0.0085m
2
Thus we get,
m Di
2
 1
Now solving (13) we get,
R = 0.098 m/km.
L = 2.221 mH/km.
Note that the calculation in (13) has been done at a frequency of 0.1hz.
R (m/km)
L (mH/km)
Theoretical values
0.098
2.221
ATP values
0.131
2.295
References:
[1] Handbook of mathematical functions
edited by M.Abramowitz and I.A.Stegun, publ. By US Dept. of Commerce, 1964.
[2] EMTP Theory Book
[3] Bianchi, G. and Luoni, G.
Induced Currents and Losses in Single-core Submarine Cables.
IEEE Transactions on PAS, Vol.- PAS-95, no.1, January/February 1976