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 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 is operated at single-phase DC conditions. Thus when we model the cable in
ATP we have to consider the special features of this cable.
We need to get the following data 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.
The ATP-computed data is taken and matched with the previous two. This way we can be sure of
having the exact cable model.
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.
r
a
b
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.
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
H
r 2  a2
i

2
2
b  a 2r
B
Where  is the conductor permeability.
r 2  a2
i
b2  a2
r 2  a2
i

2
2
b  a 2r
If ‘r’ is the radius of the flux line and for uniform current distribution, the total current enclosed
within the flux line is
r2  a2
b2  a2
This is the value of ‘N’ for flux line of radius ‘r’ within the conductor.
r 2  a2
r 2  a2
i




dr
2
2
2
2
2

r
b

a
b

a
a
b
i  
Solving we get,
i
i 
2
 
 b 2  3a 2

a4
b ...................2

*
ln
 2
2
a
2
2 2

 4b  a  b  a 
To find e:
For any annular element with radius ‘r’ such that
(b  r )
It encloses 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
2. Derivation for inductance of NEPTUNE cable:
Insulating sheath Ø 17 mm
Optical fibers
Thixotropic Jelly
Steel wires strand
Steel tube
Ø: 2.3 mm
Composite conductor
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.
1. Core:
Let steel be the sum of the flux linkages due the ‘core current’ in all 3 regions of
a) Core.
b) Sheath
c) Insulator.
Using the results in equations (1),(2) and (3),
 
 
 

st  ist  b 2  3a 2
a4
b   cu  ist ln c .  ins  ist ln d ............(4)
 steel 

*
ln
 2
a
b
c
2  4b  a 2  b 2  a 2 2
2
2

Where,
ist = current in core.
st = permeability of steel.
cu = permeability of copper.
ins = permeability of insulator.
2) 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),

2
 
 

 3b 2
b4
c   ins  icu ln d ............(5)

*
ln
2
2
2
b
c
2
 4c  b  c 2  b 2 

 cu  cu  icu  c
2
Here, icu = current in sheath.
Also note that in (5), (N)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 is always zero.
Now, the total flux linkages associated with the cable are given by:
total = 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,
total = 1.8236*10-6 ist + 8.8337*10-9 icu………………….(6)
We know the fact that
itotal = 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 = (N) = 
Thus,
Rst * ist  jst  Rcu * icu  jcu
Where,
st
Rst 
 b  a
Rcu 
 c  b
2
2

 5.607 * 10 3 

 1.263 *10 3 
cu
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 for equation (7) we get,
ist = 0.1838* itotal
(8)
icu = 0.8162* itotal
Substituting (8) in (6),
total = (3.3521*10-7) itotal + (7.21*10-9) itotal ………………..
Thus the composite inductance of the cable is given as,
L
total
itotal
L = 0.3424 H/m
(9)
2) Computation of cable R,L and C by ATP:
The 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.
That is why, it was allowing the entire cable current as 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

comp
 c 2  a 2 


cu
 b  a   c  b 2 
st
cu

2
2
2
 b  a   c  b 2 
2
2
2
Solving,
comp = 5.1753*10-8 m.
Calculation of comp :
The flux linkages associated with the cable are given by:
total = 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  itotal  c 2  3a 2
a4
c   ins  itotal 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 * itotal …………………………(10)
Equating (9) and (10),
comp = 7.869
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
0.3424
0.179
ATP values
1.03
0.3422
0.179
ALCATEL values
1
0.128
0.2
3) Calculating the ground impedance using the Bessel function approach:
To calculate the earth-return impedance 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
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.
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=R.
Simplifying,
Zearth 
m K 0mr 

2R K 1mR
R: 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
R = 1
The above method and the Bessel function approach is 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.
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