HIGH-VOLTAGE TRANSMISSION LINES
Losses on high-voltage transmission lines are higher than normal duriny foggy
or rainy weather. One reason for this is that water collects on the wires and
forms droplets. Since a droplet carries a charge of the same sign as that of
the wire, electrostatic repulsion forces the droplet to elongate and form a
sharp point where the electric field intensity can be high enough to ionize
the air. The result is a corona discharge in the surrounding air, in which the
ions are accelerated by the electric field and collide with neutral molecules,
forming more ions and heating the air.
Corona discharges are objectionable because they dissipate energy, mostly by
heating the air. Over long distances, they can cause heavy losses. They also
cause, radio and television interference.
Droplets collect on the wire in drifting by. They are also attracted by the
nonuniform electric field.
You are asked to evaluate the importance of the non-uniform electric field. To
do this, assume that the air is stagnant and calculate within what distance of
a wire the electrostatic force will be larger than the gravitational force.
Clearly, if that distance is much larger than the conductor radius, then
electrostatic attraction is important.
The transmission line under consideration consists of a pair of conductors
\(11.7\) millimeters in diameter, separated by a distance of 2 meters and
operating at 100 kilovolts ( \(\pm 50\) kilovolts).
You can find an approximate value for the field at a distance \(r\) near one
wire as follows. If \(\lambda\) is the charge per meter,
$$
E \approx \frac{\hbar}{2 \pi \epsilon_{0} r}
$$
The expression on the right is the field near a single isolated wire, which is
a good approximation as long as \(r\) is much less than the distance to the
other wire. Now the capacitance per meter for this line is approximately \(\pi
\epsilon_{0} / 5^{\prime}\) and
$$
i=\left(\pi \epsilon_{0}, 5\right) 10^{5} \text { coulombs/meter. }
$$
The expression for the force per unit volume given in Sec. \(7.4\) involves
\(\epsilon_{r}\) and the value of \(E\) inside a water droplet. For water, at low
frequencies, \(\epsilon_{r}=81\) (Table 6-1). To estimate the \(E\) inside a
droplet, one may use the formula for a dielectric sphere in a unilorm field.
\(+\) This is not a bad approximation, because even a raindrop is small,
compared to the conductor radius. so that \(E\) does not change much over one
drop diameter. Then
$$
E_{\text {u?lcr }} \approx\left(\frac{3}{\epsilon_{r}+2}\right)
\frac{\lambda}{2 \pi \epsilon_{0} r}
$$
Solution: Let us first calculate \(E\) inside a water droplet situated at a
distance \(r\) from the center of one of the wires:
$$
E_{\text {uatic }}=\left(\frac{3}{81+2}\right) \frac{\left(\pi \epsilon_{0} /
5\right) 10^{5}}{2 \pi \epsilon_{0} r}=361.4 / r \text { volts/meter }
$$
The electrostatic force per cubic meter on a droplet is
$$
\begin{aligned}
F^{\prime} &=\left(1-\frac{1}{81}\right) \frac{d}{d r}\left[\frac{1}{2} \times
81 \times 8.85 \times 10^{-12} \times\left(\frac{361.4}{r}\right)^{2}\right]
\\\
&=4.625 \times 10^{-5} \frac{d}{d r}\left(\frac{1}{r^{2}}\right)=-9.250 \times
10^{-5} \frac{1}{r^{3}} \text { ncwtons/meter }^{3}
\end{aligned}
$$
The negative sign indicates that the force is attractive.
Now the gravitational force per cubic meter is \(1000 \mathrm{~g}\) newtons, so
$$
\frac{F^{-}}{1000 g}=\frac{9.250 \times 10^{-5}}{9800 r^{3}}
$$
This ratio is equal to unity for \(r=2.113\) millimeters. which is smaller than
the radius of the wire.
The electrostatic foroe of attraction on a droplet is therefore negligible.
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