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UNDERGROUND CABLES

part 2

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Electrical Characteristics of Cables

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Electric Stress in Single-Core Cables p. 408

D= q/(2x)E = D/= q/(2x)

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.

: Permittivity (= 0. r)

r: relative permittivity or dielectric constant.

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rRx

V

x

qE

r

RqdxEV

R

r

ln..2

ln2

.

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r: conductor radius.

R: Outside radius of insulation or inside radiusof 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 acrossthe insulation material divided by the

thickness of the insulator.

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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 conductorradius 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

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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)

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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

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Grading of Cables p.414

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.

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1. Capacitance Grading p.414

This method involves the use of two or more

layers of dielectrics having differentpermittivities, those with higher permittivitybeing near the conductor.

Ex=q/(2 o.r .x)

The permittivity can be varied with radius x suchthat (ideal case):

r= k/x

Then Ex=q/(2 o. k)Exis constant throughout the thickness ofinsulation.

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In the figure shown

At x=r Emax1=q/(2 o. 1r)At x=r1 Emax2=q/(2 o. 2r1)

At x=r2 Emax3=q/(2 o. 3r2)

If all the three dielectrics are operated at thesame maximum electric stress

(Emax1=Emax2=Emax3=Emax) , then:

(1/ 1r) = (1/ 2r1) = (1/ 3r2)1r = 2r1= 3r2, get r1 , r2

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The operating voltage V is:

2

2

1

2

1

1

m ax

231

2

2

1

1

lnlnln

ln2

ln2

ln2

...

1 2

1 2

r

Rr

r

rr

r

rrEV

r

Rq

r

rq

r

rq

dxEdxEdxEV

ooo

r

r

r

r

R

r

xxx

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2. Intersheath Grading p.419

V1V

V2R

r1r2r

V=0

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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= (V1V2)/(r1. ln(r2/r1))

Emax3= V2/(r2.ln(R/r2))

For the same maximum electric strength:

(r1/r) =(r2/r1) =(R/r2) =

R/r = 3

Then: (V-V1)/(r.ln ) =(V1-V2)/(r1.ln )=(V2/r2.ln )

(V-V1)/r =(V1-V2)/r1= V2/r2

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If the cable does not have any intersheath,

the maximum stress is:Emax= V/(r.ln(R/r)) = V/(3r.ln )

The intersheath radius can be found from

R/r = 3(r1/r) =(r2/r1) =(R/r2) =

The voltages V1, V2can be found from:

(V-V1)/r =(V1-V2)/r1= V2/r2Emax/Emaxwithout intersheath=3/(1+ +

2)

where === > 1

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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.

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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 maximumand 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.

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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)

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V= 53.8 kVV1= 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

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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 2.5respectively.

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60=(q/2o. Ar)

q/(2o)=120

50=(q/2o. Br1) = 120/(2.5 r1)r1= 0.96 cm

V=q.ln(r1/r)/(2o. A) + q.ln(R/r1)/(2o. B)

=(120/4). ln(0.96/0.5) +(120/2.5). ln(2.5/0.96)= 65.51 kV

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Electrical Characteristics of Cables

Electric Stress in Single-Core Cables p. 408

Capacitance of Single Core Cables p.433

Charging Current Insulation Resistance of Single- Core Cables p.431

Dielectric Power Factor & Dielectric Losses p.442

Heating of Cables: Core loss ; Dielectric loss and

intersheath loss p.441

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Capacitance of Single Core Cables p.433

Assume that the potential differencebetween conductor an sheath is V, then

a charge of conductor and sheath will be +q

andq (C/m)

C= q/V

C= 2 /ln(R/r) F/m

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Since = 0.

r

C = 20. r /ln(R/r) F/m

Where: 0= 8.854x10-12

r dielectric constant of insulation.

C= 10-9r/(18.ln(R/r)) F/m

C= r/(18.ln(R/r)) F/km

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Ich= V/Xc= .C.V = 2f.C.V

It is observed that as cable length andoperating voltage increase, Capacitance (c)

and the charging current will be increase.

So, it is not recommended to transmit powerfor a long distance using underground cables

(Overvoltage problems)

Charging Current

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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 /ln(R/r) and Ich= .C.V

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Insulation Resistance

of Single- Core Cables p.431

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r

R

lR

x

dx

lR

dxlx

R

i

R

ri

i

ln.2

.2

.

2

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Where:

Ri: insulation resistance in ohms.

: insulation (dielectric) resistivity in .m

: Cable length (m).

It is observed that the insulation resistance is

inversely proportional to the cable length.

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Id

Id

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Power factor of dielectric :

= Cos d= Cos (90-) = Sin

This provides a useful measure of the qualityof the cable dielectric.

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For a good dielectric insulation,dis close to 90o.

Pd=I. V. Cosd

Cos d= Sin= tan = (rad)

is called dielectric loss angle.

The dielectric Losses: Pd

Pd= Id.V = Ic.tan.V = Ic.V. == Ic= CV

Pd = CV2 is in radians

C: Cable capacitance.

V: operating voltage

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Since = 90- dand < 0.5ofor most cables.

Here Cos d should be very small under alloperating conditions.

If it is large, the power loss is large and the

insulation temperature rises. The rise intemperature causes a rise in power lossin the dielectric which again results in additionaltemperature rise. If the temperature continues

to increase, the cable insulation will bedamaged.

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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 (=Cosd) and

dielectric resistivity of the insulation is 1.3x107

M.cm.

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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 endcurrent and power factor.

S l i

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Solution

a- Emax = V/(r.ln(R/r))

= 6.55 kV/cm

b- C= k/(18.ln(R/r)) F/km

= 0.176x6 = 1.0619 F

c- Ich= V/Xc = .C.V = 2.88 A

d- Ri=.ln(R/r)/(2l)= 3.79 M

e- Pd= Ich.V.Cos d=622 W

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f- load current:

I= 20 ( Cosj sin) =12 - j16

Ich= j2.88

Is= I + Ich=12- j13.12 = 17.78 A

s = 47.55o

Cos s= 0.67 lag

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Capacitance of a 3-core Cable p.434

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=

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Cy = Cs + 2 Cc

Measurement of Capacitance of 3-core Cables p.436

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Cx = 3 Cs

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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)

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Measurement of Capacitance of 3-core Cables

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Heating of Cables p.441- p.447

Core loss

Dielectric loss

Intersheath loss

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