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142 A.J. ONAH
Figure 1: 33-kV line conductors and their mirror images.
from a space-charge might be responsible for electro-static
hazards such as the explosion of a super tank,gasoline tank and so
on [4].
It is required to determine the electric field strengthat the
surface of the gas pipeline and then decidewhether it is harmful or
not. If the value obtainedexceeds the acceptable value as per
InternationalElectro-technical Commission (IEC) standard,
elec-trostatic screening would have to be provided for thepipeline.
The effect of earth on the capacitance ofthe transmission line may
be compensated for by themethod of images. The presence of ground
below a
charged conductor may be replaced by a fictitious con-ductor
having equal and opposite charge and locatedas far below the
surface of ground as the overheadconductor above, as shown in
Figure 1. The fictitiousconductor is known as the mirror image of
the over-head conductor [5].
2. Determination of Voltage Gradient
Basically, according to Gausss law, the total elec-tric charge
on a conductor equals the total electric fluxemerging from the
conductor. In other words, the to-tal charge within the closed
surface equals the integralover the surface of the normal component
of the elec-tric flux density. To start with, let us consider a
longstraight cylindrical conductor in air - Figure 2. Theradius of
the conductor is r [m]. The flux density perunit length at a
distancex meters from the axis of theconductor is given as:
Dx= q
2x [C/m2] (1)
q= the charge on the conductor in C/m. The poten-tial gradient
is,
Ex= q
2xo[V/m] (2)
Figure 2: Electric field x meters around a conductor.
o = permittivity of free space.
The conductor is an equipotential surface with uni-formly
distributed charge on the wire, equivalent tocharge concentrated at
the center for calculating fluxexternal to the wire Positive charge
on the wire ex-erts repelling force on a positive charge placed in
thefield. Thus energy is expended in moving a chargefrom the
conductor to a point P which is x[m] fromthe conductor. This energy
is the amount of workdone per coulomb of charge moved and it is
numeri-cally equal to the potential difference between the
twopoints [6]. Therefore the potential difference betweenthe
conductor and P is given by:
Vrx = q2o
lnxr
[V (3)
In the case of the three conductors in Figure 1, con-ductors a,
b and c are carrying charges qa, qb and qc
respectively. The mirror images of the conductors arecarrying
charges qa
, qb
and qc
By the principle ofsuperposition, the total potential on
conductor a dueto the sinusoidal charges,qa,qb andqc, as well as
theirmirror images qa
, qb
and qc
is given as:
Va =
1
2o [qaln ra+qbln d+qcln 2d+qaln 2h+qbln y+qcln z]
(4)
Nigerian Journal of Technology Vol. 32, No. 1, March 2013.
