underground cables part 2. electrical characteristics of cables electric stress in single-core...

UNDERGROUND CABLESpart 2

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

Electrical Characteristics of Cables

Electric Stress in Single-Core Cables p. 408D= 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.

rR

x

V

x

qE

r

RqdxEV

R

r

ln..2

ln2

.

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.

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

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)

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.

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 cmThe minimum value of Emax:

= V/r = 24.9/0.92=27.07 kV/cm

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.

1. Capacitance Grading p.414 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.

ε1 ε3ε2

r1

rR

r2

r < r1 < r2

ε1 > ε2 > ε3

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

The operating voltage V is:

22

1

21

1max

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

2. Intersheath Grading p.419

V1V

V2R

r1r2

r ε ε ε

V=0

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.

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

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 fromR/r = α 3

(r1/r) =(r2/r1) =(R/r2) = αThe 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

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.

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.

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.

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)

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

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.

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

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

Capacitance of Single Core Cables p.433

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

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

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

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

Insulation Resistance of Single- Core Cables p.431

R =ρ l /A

r

R

lR

x

dx

lR

dxlx

R

i

R

r

i

i

ln.2

.2

.2

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.

Dielectric Power Factor and Dielectric Losses p.442

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.

Id

Id

Power factor of dielectric :

= Cos фd = Cos (90-δ) = Sin δ

This provides a useful measure of the quality

of the cable dielectric.

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

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.

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.

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.

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 MΩe- Pd = Ich.V.Cos фd =622 W

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

Capacitance of a 3-core Cable p.434

=

Cy = Cs + 2 Cc

Measurement of Capacitance of 3-core Cables p.436

Cx = 3 Cs

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)

Measurement of Capacitance of 3-core Cables

Examples p.438

Heating of Cables p.441 - p.447

• Core loss • Dielectric loss• Intersheath loss

See You on

Third Year