UNDERGROUND CABLES

1. UNDERGROUND CABLES
UNDERGROUND CABLES

2. ELECTRICAL CHARACTERISTICS OF
CABLES
• Electric Stress in Single-Core Cables
• Capacitance of Single Core Cables
• Charging Current
• Insulation Resistance of Single- Core Cables
• Dielectric Power Factor & Dielectric Losses
• Heating of Cables: Core loss ; Dielectric loss and
intersheath loss

3. ELECTRICAL
CHARACTERISTICS OF
CABLES

4. Electric Stress in Single-Core Cables
D= q/(2πx)
E = D/ε = q/(2πεx)
q: Charge on conductor surface (C/m)
D: Electric flux density at a radius x (C/m2)
E: Electric field (potential gradient), or electric stress, or dielectric
stress.
ε: Permittivity (ε= ε0. εr)
εr: relative permittivity or dielectric constant.

5. r
R
x
V
x
q
E
r
R
q
dx
E
V
R
r
ln
.
.
2
ln
2
.



 



6. r: conductor radius.
R: Outside radius of insulation or inside radius of sheath.
V: potential difference between conductor and sheath (Operating
voltage of cable).
Dielectric Strength: Maximum voltage that dielectric can
withstand before it breakdown.
Average Stress: Is the amount of voltage across the insulation
material divided by the thickness of the insulator.

7. Emax = E at x = r
= V/(r.lnR/r)
Emin = E at x = R
= V/(R.lnR/r)
For a given V and R, there is a conductor radius that gives the
minimum stress at the conductor surface. In order to get the
smallest value of Emax:
dEmax/dr =0.0
ln(R/r)=1 R/r=e=2.718

8. Insulation thickness is:
R-r = 1.718 r
Emax = V/r (as: ln(R/r)=1)
Where r is the optimum conductor radius that satisfies
(R/r=2.718)

9. EXAMPLE
A single- core conductor cable of 5 km long has a conductor
diameter of 2cm and an inside diameter of sheath 5 cm. The
cable is used at 24.9 kV and 50 Hz. Calculate the following:
a- Maximum and minimum values of electric stress.
b- Optimum value of conductor radius that results in smallest
value of maximum stress.

10. a- Emax = V/(r.lnR/r) = 27.17 kV/cm
Emin = V/(R.lnR/r) = 10.87 kV/cm
b- Optimum conductor radius r is:
R/r = 2.718
r= R/2.718= 0.92 cm
The minimum value of Emax:
= V/r = 24.9/0.92=27.07 kV/cm

11. GRADING OF CABLES
Grading of cables means the distribution of dielectric stress
such that the difference between the maximum and minimum
electric stress is reduced. Therefore, the cable of the same size
could be operated at higher voltages or for the same operating
voltage,
a cable of relatively small size could be used.

12. 1. CAPACITANCE GRADING
This method involves the use of two or more layers of
dielectrics having different permittivities, those with higher
permittivity being near the conductor.
Ex =q/(2 πεo.εr .x)
The permittivity can be varied with radius x such that (ideal
case):
εr = k/x
Then Ex =q/(2 πεo. k)
Ex is constant throughout the thickness of insulation.

13. ε1
ε3
ε2
r1
r
R
r2
r < r1 < r2
ε1 > ε2 > ε3

14. In the figure shown
At x=r Emax1 =q/(2 πεo. ε1r)
At x=r1 Emax2 =q/(2 πεo. ε2r1)
At x=r2 Emax3 =q/(2 πεo. ε3r2)
If all the three dielectrics are operated at the same maximum
electric stress (Emax1=Emax2=Emax3=Emax) , then:
(1/ ε1r) = (1/ ε2r1) = (1/ ε3r2)
ε1r = ε2r1 = ε3r2, get r1 , r2

16. The operating voltage V is:














   
2
2
1
2
1
1
max
2
3
1
2
2
1
1
ln
ln
ln
ln
2
ln
2
ln
2
.
.
.
1 2
1 2
r
R
r
r
r
r
r
r
r
E
V
r
R
q
r
r
q
r
r
q
dx
E
dx
E
dx
E
V
o
o
o
r
r
r
r
R
r
x
x
x







17. CABLE CAPACITANCE
2
3
1
2
2
1
1
ln
1
ln
1
ln
1
2
C
r
R
r
r
r
r
V
q
o









18. 2. INTERSHEATH GRADING
V1
V
V2
R
r1
r2
r ε ε
ε
V=0

19. Intersheath Grading is a method of creating uniform voltage gradient across the
insulation by means of separating the insulation into two or more layers by thin
conductive strips. These strips are kept at different voltage levels through the
secondary of a transformer.

20. In this method only one dielectric is used but the dielectric is
separated into two or more layers by thin metallic intersheaths.
Emax1 = (V-V1)/(r. ln(r1/r))
Emax2 = (V1 –V2)/(r1. ln(r2/r1))
Emax3 = V2/(r2.ln(R/r2))
For the same maximum electric strength:
(r1/r) =(r2/r1) =(R/r2) = α
R/r = α3
Then: (V-V1)/(r.ln α) =(V1-V2)/(r1.ln α)=(V2/r2.ln α)
(V-V1)/r =(V1-V2)/r1= V2/r2

21. If the cable does not have any intersheath, the maximum stress
is:
Emax = V/(r.ln(R/r)) = V/(3r.ln α)
The intersheath radius can be found from
R/r = α 3
(r1/r) =(r2/r1) =(R/r2) = α
The voltages V1, V2 can be found from:
(V-V1)/r =(V1-V2)/r1= V2/r2
Emax /Emax without intersheath =3/(1+ α + α 2)
where === α > 1

22. SHEATH INTERNAL RADIUS
(R)
R =1.881V / Emax
Will be proved in sections.

23. Difficulties of Grading
a-Capacitance grading :
1- non-availability of materials with widely varying
permittivities.
2- The permittivities of materials will be change with time, so
the electric field distribution may change and lead to insulation
breakdown.

24. b- Intersheath Grading
1- Damage of intersheaths during laying
operation.
2- The charging current that flows through
the intersheath for long cables result in
overheating.
3- The setting of proper voltages of
intersheaths.

25. EXAMPLE
A single core cable for 53.8 kV has a conductor of 2cm
diameter and sheath of inside diameter 5.3 cm. It is
required to have two intersheaths so that stress varies
between the same maximum and minimum values in
three layers of dielectric. Find the positions of
intersheaths, maximum and minimum stress and
voltages on the intersheaths. Also, find the maximum
and minimum stress if the intersheaths are not used.

26. R/r = a3
a= 1.384
(r1/r) =(r2/r1) =(R/r2) = a
r1= 1.384 cm, r2= 1.951 cm
(V-V1)/(r.lna) =(V1-V2)/(r1.lna)=(V2/r2.lna)
(V-V1)/(1.lna) =(V1-V2)/(1.384.lna)
=(V2/1.915.lna)

27. V= 53.8 kV
V1= 41.3 kV, V2=23.94 kV
Emax = (V-V1)/(r. lna)=38.46 kV/cm
Emin = (V-V1)/(r1. lna)= 27.79 kV/cm
If Intersheaths are not used:
Emax = V/(r.ln(R/r)) = 55.2 kV/cm
Emin = V/(R.ln(R/r)) = 20.83 kV/cm

28. EXAMPLE
Find the maximum working voltage of a single core
cable having two insulating materials A and B and the
following data. conductor radius 0.5 cm, inside sheath
radius 2.5cm. The maximum working stress of A 60
kV/cm, maximum working stress of B 50 kV/cm, relative
permittivities of A and B, 4 and 2.5 respectively.

29. 60=(q/2πεo. εAr)
q/(2πεo)=120
50=(q/2πεo. εBr1) = 120/(2.5 r1)
r1= 0.96 cm
V=q.ln(r1/r)/(2πεo. εA) + q.ln(R/r1)/(2πεo. εB)
=(120/4). ln(0.96/0.5) +(120/2.5). ln(2.5/0.96)
= 65.51 kV

30. ELECTRICAL CHARACTERISTICS OF
CABLES
• Electric Stress in Single-Core Cables
• Capacitance of Single Core Cables
• Charging Current
• Insulation Resistance of Single- Core Cables
• Dielectric Power Factor & Dielectric Losses
• Heating of Cables: Core loss ; Dielectric loss and
intersheath loss

31. CAPACITANCE OF SINGLE CORE
CABLES
Assume that the potential difference between conductor an
sheath is V, then
a charge of conductor and sheath will be +q and –q (C/m)
C= q/V
C= 2 πε/ln(R/r) F/m

32. Since ε = ε0 . εr
C = 2πε0. εr /ln(R/r) F/m
Where: ε0= 8.854x10-12
εr dielectric constant of insulation.
C= 10-9 εr /(18.ln(R/r)) F/m
C= εr /(18.ln(R/r)) μF/km

33. Ich = V/Xc = ω.C.V =
2πf.C.V
It is observed that as cable length and operating voltage
increase, Capacitance (c) and the charging current will be
increase.
So, it is not recommended to transmit power for a long
distance using underground cables (Overvoltage problems)
Charging Current

34. The charging current and the capacitance are relatively greater
for insulated cables than in O.H.T Lines because of closer
spacing and the higher dielectric constant of the insulation of
the cables. The charging current is negligible for O.H circuits at
distribution voltage (Short Lines).
Since C= 2 πε/ln(R/r) and Ich =
ω.C.V

35. INSULATION RESISTANCE
OF SINGLE- CORE CABLES

36. R =ρ l /A

37. r
R
l
R
x
dx
l
R
dx
l
x
R
i
R
r
i
i
ln
.
2
.
2
.
2












38. Where:
Ri : insulation resistance in ohms.
ρ: insulation (dielectric) resistivity in Ω.m
l: Cable length (m).
It is observed that the insulation resistance is inversely
proportional to the cable length.

39. DIELECTRIC POWER FACTOR
AND DIELECTRIC LOSSES
When a voltage is applied across a perfect dielectric, there is no
dielectric loss because the capacitor current Ic is at 90o ahead
of the voltage V.
In practice, there is a small current component Id (leakage
current) that in phase with voltage V, so, the total current I
leads the voltage V by an angle less than 90 as shown in figure.

40. Id
Id

41. Power factor of dielectric :
= Cos фd = Cos (90-δ) = Sin δ
This provides a useful measure of the quality of the cable
dielectric.

42. For a good dielectric insulation, фd is close to 90o.
Pd =I. V. Cosфd
Cos фd = Sinδ = tan δ = δ (rad)
δ is called dielectric loss angle.
The dielectric Losses: Pd
Pd = Id.V = Ic.tanδ.V = Ic.V.δ == Ic = ωCV
Pd = ωCV2δ δ is in radians
C: Cable capacitance.
V: operating voltage

43. Since δ = 90- фd and δ < 0.5o for most cables.
Here Cos фd should be very small under all operating
conditions.
If it is large, the power loss is large and the insulation
temperature rises. The rise in temperature causes a rise in
power loss in the dielectric which again results in
additional temperature rise. If the temperature continues to
increase, the cable insulation will be damaged.

44. EXAMPLE
A single-core cable has a conductor diameter of
2 cm, inside diameter of sheath is 6 cm and
a length of 6 km. The cable is operated at 60 Hz and 7.2 kV. The
dielectric constant is 3.5, the dielectric power factor is 0.03
(δ=Cosфd) and dielectric resistivity of the insulation is 1.3x107
MΩ.cm.

45. Calculate the following:
a- Maximum electric stress.
b- Capacitance of the cable.
c- Charging current.
d- Insulation resistance.
e- Total dielectric losses.
f- If the cable feeds a load at receiving end of 20A at 0.6 power
factor lag, find sending end current and power factor.

46. Solution
a- Emax = V/(r.ln(R/r))
= 6.55 kV/cm
b- C= k/(18.ln(R/r)) μF/km
= 0.176x6 = 1.0619 μF
c- Ich = V/Xc = ω.C.V = 2.88 A
d- Ri =ρ.ln(R/r)/(2πl)
= 3.79 MΩ
e- Pd = Ich.V.Cos фd =622 W

47. f- load current:
I= 20 ( Cosф – j sinф) =12 - j16
Ich= j2.88
Is= I + Ich =12- j13.12 = 17.78 A
фs = 47.55o
Cos фs = 0.67 lag

48. CAPACITANCE OF A 3-CORE
CABLE

50. =

51. Cy = Cs + 2 Cc
Measurement of Capacitance of 3-core Cables

52. Cx = 3 Cs

53. C0 = Cs + 3Cc =(Cx/3) + 3((Cy /2) - (Cx /2))
C0 = 3 (Cy /2) - (Cx /6)
The capacitance per phase is given
by:
In case the test are not available the
following empirical formulas can be
used (p. 347)

54. MEASUREMENT OF CAPACITANCE OF 3-CORE
CABLES

55. HEATING OF CABLES
Core loss
Dielectric loss
Intersheath loss