Transcript UNDERGROUND CABLES - موقع الدكتور مجدى

UNDERGROUND CABLES
part 2
Electrical Characteristics of Cables
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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. 408
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.
R
R
V   E.dx 
ln
2

r
r
E
q
q
2 .x

V
R
x. ln
r
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 cm
The 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.
r < r 1 < r2
ε1 > ε2 > ε3
r2
R
r1
r
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:
r1
r2
R
r
r1
r2
V   Ex .dx   Ex .dx   E x .dx
q
r1
q
r2
q
R

ln 
ln 
ln
2 o1 r 2 o 2 r1 2 o 3 r2
 r1
r2
R
V  Emax r ln  r1 ln  r2 ln 
r
r1
r2 

2. Intersheath Grading p.419
r2
r1
r
ε ε
V
R
V1
ε
V2
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 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
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
Charging Current
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)
Since
C= 2 πε/ln(R/r)
and
Ich = ω.C.V
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).
Insulation Resistance
of Single- Core Cables p.431
R =ρ l /A

Ri 
.dx
2 x l
 dx
Ri  
.
2

l
x
r

R
Ri 
. ln
2 l
r
R
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
C: Cable capacitance.
V: operating voltage
δ is in radians
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 MΩ
e- 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
=
Measurement of Capacitance of 3-core Cables p.436
Cy = C s + 2 C c
Cx = 3 C s
The capacitance per phase is given by:
C0 = Cs + 3Cc =(Cx/3) + 3((Cy /2) - (Cx /2))
C0 = 3 (Cy /2) - (Cx /6)
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
• Core loss
• Dielectric loss
• Intersheath loss
p.441 - p.447
See You
on
Third Year