8

Overvoltages Caused by Indirect Lightning Strokes

Pritindra Chowdhuri

Tennessee Technological University

8.1    Inducing Voltage

8.2    Induced Voltage

8.3    Green’s Function

8.4    Induced Voltage of a Doubly Infinite Single-Conductor Line

Evaluation of Green’s Function • Induced Voltage Caused by Return-Stroke Current of Arbitrary Waveshape

8.5    Induced Voltages on Multiconductor Lines

8.6    Effects of Shield Wires on Induced Voltages

8.7    Stochastic Characteristics of Lightning Strokes

8.8    Estimation of Outage Rates Caused by Nearby Lightning Strokes

8.9    Estimation of Total Outage Rates

8.A    Appendix A: Voltage Induced by Linearly Rising and Falling Return-Stroke Current

References

A direct stroke is defined as a lightning stroke when it hits a shield wire, a tower, or a phase conductor. An insulator string is stressed by very high voltages caused by a direct stroke. An insulator string can also be stressed by high transient voltages when a lightning stroke hits the nearby ground. An indirect stroke is illustrated in Figure 8.1.

During the early years of power transmission, protection against a direct lightning stroke was thought to be impossible. Therefore, power lines were designed with overhead ground wires to protect against transient voltages induced from nearby lightning strokes to ground. As the transmission voltages became higher and higher with the consequent increase in the insulation levels of overhead lines, it was realized that the transient voltages induced from nearby lightning strokes were not the chief source of disturbance, and that protection against direct lightning strokes is indeed technically feasible and economically justifiable (Fortescue, 1930). Now emphasis is put on protection against both direct and indirect lightning strokes. As the reliability of distribution power lines has become increasingly important, many researchers have been studying the effects of lightning-induced voltages on power distribution lines. These various researches have been reviewed by Chowdhuri et al. (2001).

The voltage induced on a line by an indirect lightning stroke has four components:

1.  The charged cloud above the line induces bound charges on the line while the line itself is held electrostatically at ground potential by the neutrals of connected transformers and by leakage over the insulators. When the cloud is partially or fully discharged, these bound charges are released and travel in both directions on the line giving rise to the traveling voltage and current waves.

FIGURE 8.1  Illustration of direct and indirect lightning strokes.

2.  The charges lowered by the stepped leader further induce charges on the line. When the stepped leader is neutralized by the return stroke, the bound charges on the line are released and thus produce traveling waves similar to that caused by the cloud discharge.

3.  The residual charges on the upper part of the return stroke induce an electrostatic field in the vicinity of the line and hence an induced voltage on it.

4.  The rate of change of current in the return stroke produces a magnetically induced voltage on the line.

If the lightning has subsequent strokes, then the subsequent components of the induced voltage will be similar to one or the other of the four components discussed earlier.

The magnitudes of the voltages induced by the release of the charges bound either by the cloud or by the stepped leader are small compared with the voltages induced by the return stroke. Therefore, only the electrostatic and the magnetic components induced by the return stroke are considered in the following analysis. The initial computations are performed with the assumption that the charge distribution along the leader stroke is uniform, and that the return-stroke current is rectangular. However, the result with the rectangular current wave can be transformed to that with currents of any other waveshape by the convolution integral (Duhamel’s theorem). It was also assumed that the stroke is vertical and that the overhead line is loss free and the earth is perfectly conducting. The vertical channel of the return stroke is shown in Figure 8.2, where the upper part consists of a column of residual charge that is neutralized by the rapid upward movement of the return-stroke current in the lower part of the channel.

Figure 8.3 shows a rectangular system of coordinates where the origin of the system is the point where lightning strikes the surface of the earth. The line conductor is located at a distance y_o meters from the origin, having a mean height of h_p meters above ground and running along the x-direction. The origin of time (t = 0) is assumed to be the instant when the return stroke starts at the earth level.

FIGURE 8.2  Return stroke with residual charge column.

FIGURE 8.3  Coordinate system of line conductor and lightning stroke.

8.1    Inducing Voltage

The total electric field created by the charge and the current in the lightning stroke at any point in space is

$E_{\text{i}}=E_{\text{ei}}+E_{\text{mi}}=-\nabla \phi -\frac{\partial A}{\partial t}$

(8.1)

where φ is the inducing scalar potential created by the residual charge at the upper part of the return stroke and A is the inducing vector potential created by the upward-moving return-stroke current (Figure 8.2). φ and A are called the retarded potentials, because these potentials at a given point in space and time are determined by the charge and current at the source (i.e., the lightning channel) at an earlier time; the difference in time (i.e., the retardation) is the time required to travel the distance between the source and the field point in space with a finite velocity, which in air is c = 3 × 10⁸ m/s. These electromagnetic potentials can be deduced from the distribution of the charge and the current in the return-stroke channel. The next step is to find the inducing electric field (Equation 8.1). The inducing voltage, V_i, is the line integral of E_i:

$V_{\text{i}}={\displaystyle \underset{0}{\overset{h_{\text{p}}}{\int }}{E}_{\text{i}}\cdot \text{d}z}=-{\displaystyle \underset{0}{\overset{h_{\text{p}}}{\int }}{E}_{\text{ei}}}\cdot \text{d}z-{\displaystyle \underset{0}{\overset{h_{\text{p}}}{\int }}{E}_{\text{mi}}}\cdot \text{d}z=V_{\text{ei}}+V_{\text{mi}}$

(8.2)

As the height, h_p, of the line conductor is small compared with the length of the lightning channel, the inducing electric field below the line conductor can be assumed to be constant and equal to that on the ground surface:

$V_{\text{i}}=\left(\nabla \phi +\frac{\partial A}{\partial t}\right)\cdot h_{\text{p}}$

(8.3)

The inducing voltage will act on each point along the length of the overhead line. However, because of the retardation effect, the earliest time, to, the disturbance from the lightning channel will reach a point on the line conductor would be

$t_{\text{o}}=\frac{\sqrt{x^2+y_{\text{o}}^2}}{c}$

(8.4)

Therefore, the inducing voltage at a point on the line remains zero until t = t_o. Hence,

$V_{\text{i}}=\psi \left(x,t\right)u\left(t-t_{\text{o}}\right)$

(8.5)

where u(t – t_o) is the shifted unit step function. The continuous function, ψ(x,t), is the same as Equation 8.3, and is given, for a negative stroke with uniform charge density along its length, by (Rusck, 1958)

$\psi (x,t)=-\frac{60I_o{h}_p}{\beta }\left[\frac{1-{\beta }^2}{\sqrt{{\beta }^2{c}^2{(t-t_o)}^2+(1-{\beta }^2)}{r}^2}-\frac{1}{\sqrt{h_c^2+r^2}}\right]$

(8.6)

where

I_o is the step-function return-stroke current (A)

h_p is the height of line above ground (m)

β is the v/c, v is the velocity of return stroke (m/s)

r is the distance of point x on line from point of strike (m)

h_c is the height of cloud charge center above ground (m)

The inducing voltage is the voltage at a field point in space with the same coordinates as a corresponding point on the line conductor, but without the presence of the line conductor. The inducing voltage at different points along the length of the line conductor will be different. In the overhead line, these differences will tend to be equalized by the flow of current, as it is a good conductor of electricity. Therefore, the actual voltage between a point on the line and the ground below it will be different from the inducing voltage at that point. This voltage, which can actually be measured on the line conductor, is defined as the induced voltage. The calculation of the induced voltage is the primary objective.

8.2    Induced Voltage

Neglecting losses, an overhead line may be represented as consisting of distributed series inductance L (H/m) and distributed shunt capacitance C (F/m). The effect of the inducing voltage will then be equivalent to connecting a voltage source along each point of the line (Figure 8.4). The partial differential equation for such a configuration will be

$-\frac{\partial V}{\partial x}\Delta x=L\Delta x\frac{\partial I}{\partial t}$

(8.7)

FIGURE 8.4  Equivalent circuit of transmission line with inducing voltage.

and

$-\frac{\partial I}{\partial x}\Delta x=C\Delta x\frac{\partial }{\partial t}\left(V-V_{\text{i}}\right)$

(8.8)

Differentiating Equation 8.7 with respect to x, and eliminating I, the equation for the induced voltage can be written as

$\frac{{\partial }^{2}V}{\partial x^2}-\frac{1}{c^2}\frac{{\partial }^{2}V}{\partial t^2}=-\frac{1}{c^2}\frac{{\partial }^2{V}_{\text{i}}}{\partial t^2}=F\left(x,t\right)$

(8.9)

where

$c=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{{\text{μ}}_{\text{o}}{\epsilon }_{\text{o}}}}=3\times {10}^8\text{m/s}$

(8.10)

In Laplace transform,

$\frac{{\partial }^{2}V\left(x,s\right)}{\partial x^2}-\frac{s^2}{c^2}V\left(x,s\right)=-\frac{s^2}{c^2}{V}_{\text{i}}\left(x,s\right)=F\left(x,s\right)$

(8.11)

Equation 8.9 is an inhomogeneous wave equation for the induced voltage along the overhead line. It is valid for any charge distribution along the leader channel and any waveshape of the return-stroke current. Its solution can be obtained by assuming F(x,t) to be the superposition of impulses that involves the definition of Green’s function (Morse and Feshbach, 1950).

8.3    Green’s Function

To obtain the voltage caused by a distributed source, F(x), the effects of each elementary portion of the source are calculated and then integrated for the whole source. If G(x; x′) is the voltage at a point x along the line caused by a unit impulse source at a source point x′, the voltage at x caused by a source distribution F(x′) is the integral of G(x; x′)F(x′) over the whole domain (a,b) of x′ occupied by the source, provided that F(x′) is a piecewise continuous function in the domain a ≤ x′ ≤ b:

$V\left(x\right)={\displaystyle \underset{a}{\overset{b}{\int }}G\left(x;x^{\prime }\right)F\left(x^{\prime }\right)d{x}^{\prime }}$

(8.12)

The function G(x; x′), called Green’s function, is, therefore, a solution for a case that is homogeneous everywhere except at one point. Green’s function, G(x; x′), has the following properties:

1.  $G\left(x;x^{\prime }+0\right)-G\left(x;x^{\prime }-0\right)=0$

(8.13)

2.  ${\left(\frac{\text{d}G}{\text{d}x}\right)}_{x^{\prime }+0}-{\left(\frac{\text{d}G}{\text{d}x}\right)}_{x^{\prime }-0}=1$

(8.14)

3.  G(x;x′) satisfies the homogeneous equation everywhere in the domain, except at the point x = x′

4.  G(x;x′) satisfies the prescribed homogeneous boundary conditions.

Green’s function can be found by converting Equation 8.11 to a homogeneous equation and replacing V(x, s) by G(x; x′,s):

$\frac{{\partial }^{2}G\left(x;x^{\prime },s\right)}{\partial x^2}-\frac{s^2}{c^2}G\left(x;x^{\prime }s\right)=0$

(8.15)

The general solution of Equation 8.15 is given by

$G\left(x;x^{\prime },s\right)=A{e}^{sx/c}+B{e}^{-sx/c}$

(8.16)

The constants A and B are found from the boundary conditions and from the properties of Green’s function.

8.4    Induced Voltage of a Doubly Infinite Single-Conductor Line

The induced voltage at any point, x, on the line can be determined by invoking Equation 8.12, where G(x; x′) F(x′) is the integrand. F(x′) is a function of the amplitude and waveshape of the inducing voltage, V_i (Equation 8.5), whereas Green’s function, G(x; x′), is dependent on the boundary conditions of the line and the properties of Green’s function. In other words, it is a function of the line configuration and is independent of the lightning characteristics. Therefore, it is appropriate to determine Green’s function first.

8.4.1    Evaluation of Green’s Function

As Green’s function is finite for x → –∞ and x → + ∞,

$G_1=A{e}^{sx/c}\text{for}x<x^{\prime };G_2=B{e}^{-sx/c}\text{for}x>x^{\prime }$

From Equation 8.13,

$A{e}^{s{x}^{\prime }/c}=B{e}^{-\left(s{x}^{\prime }/c\right)},\text{i}\text{.e}.,B=A{e}^{2s{x}^{\prime }/c}$

From Equation 8.14,

$A=\frac{c}{2s}{e}^{-sx/c}$

Hence,

$\begin{array}{c}B=-\frac{c}{2s}{e}^{sx/c}\\ G_1\left(x;x^{\prime },s\right)=-\frac{c}{2s}\exp \left(-\frac{s\left(x^{\prime }-x\right)}{c}\right)\text{for}x<x^{\prime }\end{array}$

(8.17)

and

$G_2\left(x;x^{\prime },s\right)=-\frac{c}{2s}\exp \left(\frac{s\left(x^{\prime }-x\right)}{c}\right)\text{for}x>x^{\prime }$

(8.18)

By applying Equation 8.12,

$V\left(x,s\right)=-\frac{c}{2s}{\displaystyle \underset{-\infty }{\overset{x}{\int }}{e}^{s\left(x^{\prime }-x\right)/c}F\left(x^{\prime },s\right)\text{d}{x}^{\prime }-\frac{c}{2s}{\displaystyle \underset{x}{\overset{\infty }{\int }}{e}^{-s\left(x^{\prime }-x\right)/c}F\left(x^{\prime },s\right)\text{d}{x}^{\prime }=V_1\left(x,s\right)+V_2\left(x,s\right)}}$

(8.19)

8.4.2    Induced Voltage Caused by Return-Stroke Current of Arbitrary Waveshape

The induced voltage caused by return-stroke current, I(t), of arbitrary waveshape can be computed from Equation 8.11 by several methods. In method I, the inducing voltage, V_i, due to I(t) is found by applying Duhamel’s integral (Haldar and Liew, 1988):

$V_{\text{i}}=\frac{\text{d}}{\text{d}t}{\displaystyle \underset{0}{\overset{t}{\int }}I\left(t-\tau \right)V_{\text{istep}}\left(x^{\prime },\tau \right)\text{d}\tau }$

(8.20)

where V_istep is the inducing voltage caused by a unit step-function current. In other words,

$V_{\text{istep}}\left(x^{\prime },\tau \right)={\psi }_{\text{o}}\left(x^{\prime },\tau \right)u\left(\tau ,t_{\text{o}}\right)$

(8.21)

where

${\psi }_{\text{o}}\left(x^{\prime },\tau \right)=\frac{\psi \left(x^{\prime },\tau \right)}{I_{\text{o}}}$

and ψ(x′, τ) is given in Equation 8.6. Inserting Equation 8.21 in Equation 8.20 and taking Laplace transform of V_i in Equation 8.20,

$V_{\text{i}}\left(x^{\prime },s\right)=sI\left(s\right){\psi }_{\text{o}}\left(x^{\prime },s\right)e^{-s{t}_{\text{o}}}$

(8.22)

and

$F\left(x^{\prime },s\right)=-\frac{s^2}{c^2}{V}_{\text{i}}\left(x^{\prime },s\right)=-\frac{s^2}{c^2}I\left(s\right){\psi }_{\text{o}}\left(x^{\prime },s\right)e^{-s{t}_{\text{o}}}$

(8.23)

Replacing F(x′,s) in Equation 8.19 by Equation 8.23, the induced voltage, V(x,s), is

$V\left(x,s\right)=\frac{1}{2c}\left[sI\left(s\right)\left\{s{\displaystyle \underset{-\infty }{\overset{x}{\int }}{\psi }_{\text{o}}\left(x^{\prime },s\right)e^{-s\left(t_{\text{o}}-\frac{x^{\prime }-x}{c}\right)}}\text{d}{x}^{\prime }+s{\displaystyle \underset{x}{\overset{\infty }{\int }}{\psi }_{\text{o}}\left(x^{\prime },s\right)e^{-s\left(t_{\text{o}}+\frac{x^{\prime }-x}{c}\right)}\text{d}{x}^{\prime }}\right\}\right]$

(8.24)

Inverting to time domain by convolution integral,

$\begin{array}{ll}V\left(x,t\right)\hfill & =\frac{1}{2c}{\displaystyle \underset{0}{\overset{t}{\int }}\frac{\text{d}}{\text{d}t}I\left(t-\tau \right)}\left[\frac{\text{d}}{\text{d}\tau }{\displaystyle \underset{-\infty }{\overset{x}{\int }}{\psi }_{\text{o}}\left(x^{\prime },\tau +\frac{x^{\prime }-x}{c}\right)u\left(\tau -t_{\text{o}}+\frac{x^{\prime }-x}{c}\right)\text{d}{x}^{\prime }}\right]\text{d}\tau \hfill \\ \hfill & +\frac{1}{2c}{\displaystyle \underset{0}{\overset{t}{\int }}\frac{\text{d}}{\text{d}t}I\left(t-\tau \right)\left[\frac{\text{d}}{\text{d}\tau }{\displaystyle \underset{x}{\overset{\infty }{\int }}{\psi }_{\text{o}}}\left(x^{\prime },\tau -\frac{x^{\prime }-x}{c}\right)u\left(\tau -t_{\text{o}}-\frac{x^{\prime }-x}{c}\right)\text{d}{x}^{\prime }\right]\text{d}\tau =V_1\left(x,t\right)+V_2\left(x,t\right)}\hfill \end{array}$

(8.25)

Because of the shifted unit step function in V₁(x,t),

$\tau \ge t_{\text{o}}-\frac{x^{\prime }-x}{c}$

In the limit,

$\tau =t_{\text{o}}-\frac{x_{\text{o}1}-x}{c}=\frac{\sqrt{x_{\text{o}1}^2+y_{\text{o}}^2}}{c}-\frac{x_{\text{o}1}-x}{c}$

or

$x_{\text{o}1}=\frac{y_{\text{o}}^2-{\left(c\tau -x\right)}^2}{2\left(c\tau -x\right)}$

(8.26)

Similarly,

$\text{for}{V}_2\left(x,t\right):x_{\text{o}2}=\frac{{\left(c\tau +x\right)}^2-y_{\text{o}}^2}{2\left(c\tau +x\right)}$

(8.27)

Replacing –∞ by x_o1 in V₁(x,t) and +∞ by x_o2 in V₂(x,t) in Equation 8.25,

$V_1\left(x,t\right)=\frac{1}{2c}{\displaystyle \underset{0}{\overset{t}{\int }}\frac{\text{d}}{\text{d}t}I\left(t-\tau \right)\frac{\text{d}}{\text{d}\tau }\left\{{\displaystyle \underset{x_{\text{o}1}}{\overset{x}{\int }}{\psi }_{\text{o}}\left(x,\tau +\frac{x^{\prime }-x}{c}\right)\text{d}{x}^{\prime }}\right\}u\left(\tau -t_{\text{o}}\right)\text{d}\tau }$

(8.28)

and

$V_2\left(x,t\right)=\frac{1}{2c}{\displaystyle \underset{0}{\overset{t}{\int }}\frac{\text{d}}{\text{d}t}I\left(t-\tau \right)}\frac{\text{d}}{\text{d}\tau }\left\{{\displaystyle \underset{x}{\overset{x_{\text{o}2}}{\int }}{\psi }_{\text{o}}\left(x,\tau -\frac{x^{\prime }-x}{c}\right)\text{d}{x}^{\prime }}\right\}u\left(\tau -t_{\text{o}}\right)\text{d}\tau$

(8.29)

A lightning return-stroke current can be represented by a linearly rising and linearly falling wave with sufficient accuracy (Figure 8.5) (Chowdhuri, 2004):

$I\left(t\right)={\text{α}}_{1}tu\left(t\right)-{\text{α}}_2\left(t-t_{\text{f}}\right)u\left(t-t_{\text{f}}\right)$

(8.30)

where

${\text{α}}_1=\frac{I_{\text{p}}}{t_{\text{f}}}\text{and}{\text{α}}_2=\frac{2t_{\text{H}}-t_{\text{f}}}{2t_{\text{f}}\left(t_{\text{H}}-t_{\text{f}}\right)}{I}_{\text{p}}$

(8.31)

FIGURE 8.5  A linearly rising and falling lightning return-stroke current.

It will be evident from Equation 8.30 that V₁(x,t) in Equation 8.28 will have two components: one component, V₁₁(x,t), will be a function of I₁(t), and the other component, V₂₁(x,t), will be a function of I₂(t), that is, V₁(x,t) = V₁₁(x,t) + V₂₁(x,t). Similarly, V₂(x,t) = V₁₂(x,t) + V₂₂(x,t). After integration and simplifying Equation 8.28, V₁₁(x,t) can be written as

$V_{11}\left(x,t\right)=-\frac{{\text{α}}_1{h}_{\text{p}}}{\text{β}}\times {10}^{-7}u\left(t-t_{\text{o}}\right)\left[\left(1-{\text{β}}^{\text{2}}\right)\ell n\frac{f_{11}\left(\tau =t\right)f_{21}\left(\tau =t_{\text{o}}\right)}{f_{11}\left(\tau =t_{\text{o}}\right)f_{21}\left(\tau =t\right)}+\ell n\frac{f_{31}\left(\tau =t\right)}{f_{31}\left(\tau =t_{\text{o}}\right)}\right]$

(8.32)

where

$\begin{array}{c}{f}_{11}\left(\tau \right)=m_{11}+\sqrt{m_{11}^2+a_{11}^2;}{f}_{21}\left(\tau \right)=m_{21}+\sqrt{m_{21}^2+a_{11}^2;}{f}_{31}=x_{\text{o}1}+\sqrt{x_{\text{o}1}^2+y_{\text{o}}^2+h_{\text{c}}^2};\\ m_{11}=x+{\beta }^2\left(c\tau -x\right);m_{21}=x_{\text{o}1}+{\beta }^2\left(c\tau -x\right);a_{11}^2=\left(1-{\beta }^2\right)\left[y_{\text{o}}^2+{\beta }^2{\left(c\tau -x\right)}^2\right]\end{array}$

The expression for V₂₁(x,t) is similar to Equation 8.32, except that is replaced by (–α₂), and t is replaced by (t – t_f). The computation of V₂(x,t) is similar; namely,

$V_{12}\left(x,t\right)=-\frac{{\text{α}}_1{h}_{\text{p}}}{\beta }\times {10}^{-7}u\left(t-t_{\text{o}}\right)\left[\left(1-{\beta }^2\right)\ell n\frac{f_{12}\left(\tau =t\right)f_{22}\left(\tau =t_{\text{o}}\right)}{f_{12}\left(\tau =t_{\text{o}}\right)f_{22}\left(\tau =t\right)}-\ell n\frac{f_{32}\left(\tau =t\right)}{f_{32}\left(\tau =t_{\text{o}}\right)}\right]$

(8.33)

where

$\begin{array}{c}{f}_{12}\left(\tau \right)=m_{12}+\sqrt{m_{12}^2+a_{12}^2;}{f}_{22}=m_{22}+\sqrt{m_{22}^2+a_{12}^2;}{f}_{32}=x_{\text{o2}}+\sqrt{x_{\text{o2}}^2+y_{\text{o}}^2+h_{\text{c}}^2};\\ m_{12}=x_{\text{o}2}-{\beta }^2\left(c\tau +x\right);m_{22}=x-{\beta }^2\left(c\tau +x\right);a_{12}^2=\left(1-{\beta }^2\right)\left[y_{\text{o}}^2+{\beta }^2{\left(c\tau -x\right)}^2\right]\end{array}$

V₂₂(x,t) can be similarly determined by replacing α₁ in Equation 8.33 by (–α₂) and replacing t by (t – t_f). The second method of determining the induced voltage, V(x,t), is to solve Equation 8.19, for a unit step-function return-stroke current, and then find the induced voltage for the given return-stroke current waveshape by applying Duhamel’s integral (Chowdhuri and Gross, 1967; Chowdhuri, 1989a). The solution of Equation 8.19 for a unit step-function return-stroke current is given by (Chowdhuri, 1989a):

$V_{\text{step}}\left(x,t\right)=\left(V_{11}+V_{12}+V_{21}+V_{22}\right)u\left(t-t_{\text{o}}\right)$

(8.34)

where

$V_{11}=\frac{30h_{\text{p}}\left(1-{\beta }^2\right)}{{\beta }^2{\left(ct-x\right)}^2+y_{\text{o}}^2}\left[\beta \left(ct-x\right)+\frac{\left(ct-x\right)x-y_{\text{o}}^2}{\sqrt{c^2{t}^2+\left(\left(1-{\beta }^2\right)/{\beta }^2\right)\left(x^2+y_{\text{o}}^2\right)}}\right]$

(8.35)

$V_{12}=-\frac{30h_{\text{p}}}{\beta }\left[1-\frac{1}{\sqrt{k_1^2+1}}-{\beta }^2\right]\frac{1}{ct-x}$

(8.36)

$V_{21}=\frac{3o{h}_{\text{p}}\left(1-{\beta }^2\right)}{{\beta }^2{\left(ct+x\right)}^2+y_{\text{o}}^2}\left[\beta \left(ct+x\right)-\frac{\left(ct+x\right)x+y_{\text{o}}^2}{\sqrt{c^2{t}^2+\left(\left(1-{\beta }^2\right)/{\beta }^2\right)\left(x^2+y_{\text{o}}^2\right)}}\right]$

(8.37)

$V_{22}=-\frac{30h_{\text{p}}}{\beta }\left[1-\frac{1}{\sqrt{k_2^2+1}}-{\beta }^2\right]\frac{1}{ct+x}$

(8.38)

$k_1=\frac{2h_{\text{c}}\left(ct-x\right)}{y_{\text{o}}^2+{\left(ct-x\right)}^2}$

(8.39)

and

$k_2=\frac{2h_{\text{c}}\left(ct+x\right)}{y_{\text{o}}^2+{\left(ct+x\right)}^2}$

(8.40)

The expressions for the induced voltage, caused by a linearly rising and falling return-stroke current, are given in Appendix 8.A.

The advantage of method II is that once the induced voltage caused by a step-function return-stroke current is computed, it can be used as a reference in computing the induced voltage caused by currents of any given waveshape by applying Duhamel’s integral, thus avoiding the mathematical manipulations for every given waveshape. However, the mathematical procedures are simpler for method I than that for method II.

A third method to solve Equation 8.9 is to apply numerical method, which bypasses all mathematical complexities (Agrawal et al., 1980). However, the accuracy of the numerical method strongly depends upon the step size of computation. Therefore, the computation of the induced voltage of long lines, greater than 1 km, becomes impractical.

8.5    Induced Voltages on Multiconductor Lines

Overhead power lines are usually three-phase lines. Sometimes several three-phase circuits are strung from the same tower. Shield wires and neutral conductors are part of the multiconductor system. The various conductors in a multiconductor system interact with each other in the induction process for lightning strokes to nearby ground. The equivalent circuit of a two-conductor system is shown in Figure 8.6. Extending to an n-conductor system, the partial differential equation for the induced voltage, in matrix form, is (Chowdhuri and Gross, 1969; Cinieri and Fumi, 1979; Chowdhuri, 1990, 2004):

$\frac{{\partial }^2\left[V\right]}{\partial x^2}-\frac{1}{c^2}\frac{{\partial }^2\left[V\right]}{\partial t^2}=-\left[L\right]\left[C_{\text{g}}\right]\frac{{\partial }^2\left[V_{\text{i}}\right]}{\partial t^2}=-\left[M\right]\frac{{\partial }^2\left[V_{\text{i}}\right]}{\partial t^2}$

(8.41)

FIGURE 8.6  Equivalent circuit of a two-conductor system.

where [L] is an n × n matrix whose elements are

$L_{rr}=2\times {10}^{-7}\ell n\frac{2h_r}{r_r};L_{rs}=2\times {10}^{-7}\ell n\frac{d_{r^{\prime }s}}{d_{rs}}$

[C_g] is an n × n diagonal matrix whose elements are C_jg = C_j1 + C_j2 + + C_jn, where C_jr is an element of an n × n matrix, [C] = [p]⁻¹ and

$p_{rr}=18\times {10}^9\ell n\frac{2h_r}{r_r};p_{rs}=18\times {10}^9\ell n\frac{d_{r^{\prime }s}}{d_{rs}}$

where

h_r and r_r are the height above ground and the radius of the rth conductor

d_r′s is the distance between the image of the rth conductor below earth and the sth conductor

d_rs is the distance between the rth and sth conductors

From Equation 8.41, for the jth conductor,

$\frac{{\partial }^2{V}_j}{\partial x^2}-\frac{1}{c^2}\frac{{\partial }^2{V}_j}{\partial t^2}=-\left(M_{j1}\frac{{\partial }^2{V}_{j1}}{\partial t^2}+\cdots +M_{jj}\frac{{\partial }^2{V}_{jj}}{\partial t^2}+\cdots +M_{jn}\frac{{\partial }^2{V}_{jn}}{\partial t^2}\right)$

(8.42)

If the ratio of the inducing voltage of the mth conductor to that of the jth conductor is k_mj (m = 1, 2,…, n), then

$\frac{{\partial }^2{V}_j}{\partial x^2}-\frac{1}{c^2}\frac{{\partial }^2{V}_j}{\partial t^2}-c^2\left(M_{j1}{k}_{1j}+\cdots +M_{jj}+\cdots +M_{jn}{k}_{nj}\right)\frac{1}{c^2}\frac{{\partial }^2{V}_{ij}}{\partial t^2}=\left(c^2{M}_j\right)F_j\left(x,t\right)$

(8.43)

where

$M_j=M_{j1}{k}_{1j}+\cdots +M_{jj}+\cdots +M_{jn}{k}_{nj}\text{and}{F}_j\left(x,t\right)=\frac{1}{-c^2}\frac{{\partial }^2{V}_{ij}}{\partial t^2}$

(8.44)

If the jth conductor in its present position existed alone, the partial differential equation of its induced voltage, V_js, would be the same as Equation 8.9, that is,

$\frac{{\partial }^2{V}_{js}}{\partial x^2}-\frac{1}{c^2}\frac{{\partial }^2{V}_{js}}{\partial t^2}=F_j\left(x,t\right)$

(8.45)

Therefore, the ratio of the induced voltage of the jth conductor in an n-conductor system to that of a single conductor at the same position would be

$\frac{V_j}{V_{js}}=M_j{c}^2$

(8.46)

The inducing voltage being nearly proportional to the conductor height and the lateral distance of the stroke point being significantly larger than the separation distance between phase conductors, the presence of other conductors in a horizontally configured line will be minimal. On the other hand, for a vertically configured line, the induced voltage of the highest conductor will be lower than that for the same conductor without any neighboring conductors. Similarly, the lowest conductor voltage will be pulled up by the presence of the neighboring conductors of higher elevation, and the middle conductor will be the least affected by the presence of the other conductors (Chowdhuri, 2004).

8.6    Effects of Shield Wires on Induced Voltages

If there are (n + r) conductors, of which r conductors are grounded (r shield wires), then the partial differential equation for the induced voltages of the n number of phase conductors is given by Chowdhuri and Gross (1969), Cinieri and Fumi (1979), and Chowdhuri (1990, 2004):

$\frac{{\partial }^2\left[V_n\right]}{\partial x^2}-\frac{1}{c^2}\frac{{\partial }^2\left[V_n\right]}{\partial t^2}=-\left[L^{\prime }\right]\left[C_{gn}\right]\frac{{\partial }^2\left[V_{in}\right]}{\partial t^2}=-\left[M_g\right]\frac{{\partial }^2\left[V_{in}\right]}{\partial t^2}$

(8.47)

The matrix [L′] is obtained by partitioning the (n + r) × (n + r) inductance matrix of the (n + r) conductors, and putting [L′] = [L_nn] – [L_nr][L_rr]⁻¹[L_rn], where

${\left[L\right]}_{\left(n+r\right),\left(n+r\right)}=\left[\begin{array}{cc}{L}_{nn}& L_{nr}\\ L_{rn}& L_{rr}\end{array}\right]$

(8.48)

[C_gn] is an n × n diagonal matrix, each element of which is the sum of the elements of the corresponding row, up to the nth row, of the original (n + r) × (n + r) capacitance matrix of the (n + r) conductors, [C] = [p]⁻¹, where [p] is the matrix of the potential coefficients of the (n + r) conductors. The jth element of [C_gn] is given by

$C_{jgn}\left(j\le n\right)={\displaystyle \sum _{k=1}^{n+r}{C}_{jk}}$

(8.49)

From Equation 8.47, the induced voltage of the j-th conductor is

$\frac{{\partial }^2{V}_j}{\partial x^2}-\frac{1}{c^2}\frac{{\partial }^2{V}_j}{\partial t^2}=-c^2\left(M_{gj1}{k}_{1j}+\cdots +M_{gjj}+\cdots M_{gjn}{k}_{nj}\right)\frac{1}{c^2}\frac{{\partial }^2{V}_{ij}}{\partial t^2}-\left(c^2{M}_{gi}\right)F_j\left(x,t\right)$

(8.50)

Defining the protective ratio as the ratio of the induced voltages on the jth conductor with and without the shield wires in place,

$\text{Protective}\text{ratio=}\frac{M_{gi}}{M_j}$

(8.51)

where M_j is given by Equation 8.44.

8.7    Stochastic Characteristics of Lightning Strokes

The voltage induced on an overhead line is caused by the interaction between various lightning return-stroke parameters and the parameters of the line. The most important return-stroke parameters are (a) peak current, I_p, (b) current front time, t_f, (c) return-stroke velocity, v, and (d) ground flash density, n_g. These parameters are stochastic in nature.

Analysis of field data shows that the statistical variation of the peak, I_p, and the time to crest, t_f, of the return-stroke current fit lognormal distribution (Anderson and Eriksson, 1980). The probability density function, p(I_p), of I_p then can be expressed as

$p\left(I_{\text{p}}\right)=\frac{{\text{e}}^{-0.5f_1}}{I_{\text{p}}\cdot {\text{σ}}_{\ell n{I}_{\text{p}}}\cdot \sqrt{2\pi }}$

(8.52)

where

$f_1={\left(\frac{\ell n{I}_{\text{p}}-\ell n{I}_{\text{pm}}}{{\text{σ}}_{\ell n{I}_{\text{p}}}}\right)}^2$

(8.53)

and σ_ℓnIp is the standard deviation of ℓnI_p, and I_pm is the median value of I_p. Similarly, the probability density function of t_f can be expressed as

$p\left(t_{\text{f}}\right)=\frac{e^{-0.5f_2}}{t_{\text{f}}\cdot {\text{σ}}_{\ell n{t}_{\text{f}}}\cdot \sqrt{2\pi }}$

(8.54)

where

$f_2={\left(\frac{\ell n{t}_{\text{f}}-\ell n{t}_{\text{fm}}}{{\partial }_{\ell {\text{nt}}_{\text{f}}}}\right)}^2$

(8.55)

The joint probability density function, p(I_p, t_f), is given by

$p\left(I_{\text{p}},t_{\text{f}}\right)=\frac{\text{e}\left(0.5/\left(1-{\text{ρ}}^2\right)\right)\left(f_1-2\text{ρ}\sqrt{f_1\cdot f_2}+f_2\right)}{2\pi \left(I_{\text{p}}\cdot t_{\text{f}}\right)\left({\text{σ}}_{\ell n{I}_{\text{p}}}\cdot {\text{σ}}_{\ell n{t}_{\text{f}}}\right)\sqrt{1-{\rho }^2}}$

(8.56)

where ρ is the coefficient of correlation. The statistical parameters of return-stroke current are as follows (Anderson and Eriksson, 1980; Eriksson, 1986; IEEE = PES Task Force 15.09, 2005):

Median time to crest, t_fm = 3.83 μs; Log (to base e) of standard deviation, σ(ℓnt_f) = 0.553.
For I_p < 20 kA: Median peak current, I_pm1 = 61.1 kA; Log (to base e) of standard deviation, σ(ℓnI_p1) = 1.33.
For I_p > 20 kA: Median peak current, I_pm2 = 33.3 kA; Log (to base e) of standard deviation, σ(ℓnI_p2) = 0.605.

Correlation coefficient, ρ = 0.47.

Field data on the return-stroke velocity are limited. Lundholm (1957) and Rusck (1958) proposed the following empirical relationship between the return-stroke peak current and its velocity from the available field data:

$v=\frac{c}{\sqrt{1+\left(500/I_{\text{p}}\right)}}\left(\text{m}/\text{s}\right)$

(8.57)

where

c is the velocity of electromagnetic fields in free space = 3 × 10⁸ m/s

I_p is the return-stroke peak current, kA

The ground flash density, n_g (number of flashes to ground per km² per year), varies regionally and seasonally. However, the average ground flash density maps around the globe are available to estimate lightning-caused outages.

8.8    Estimation of Outage Rates Caused by Nearby Lightning Strokes

The knowledge of the basic impulse insulation level (BIL) is essential for estimating the outage rate of an overhead power line. With this knowledge, the electrogeometric model is constructed to estimate the attractive area (Figure 8.7). According to the electrogeometric model, the striking distance of a lightning stroke is proportional to the return-stroke current. The following relation is used to estimate this striking distance, r_s:

$r_{\text{s}}=8I_{\text{p}}^{0.65}\left(\text{m}\right)$

(8.58)

where I_p in kA is the peak of the return-stroke current. In the cross-sectional view of Figure 8.7, a horizontal line (representing a plane) is drawn at a distance of r_s meters from the ground plane corresponding to the return-stroke current, I_p. A circular arc is drawn with its center on the conductor, P, and r_s as radius. This represents a cylinder of attraction above the line conductor. The circular arc and the horizontal line intersect at points A and B. The strokes with I = I_p falling between A and B will strike the conductor, resulting in direct strokes; those falling outside AB will hit the ground, inducing voltages on the line. The horizontal projection of A or B is y_o1, which is given by

$y_{\text{o}1}=r_{\text{s}}\text{for}{r}_{\text{s}}\le h_{\text{p}}$

(8.59a)

$y_{\text{o}1}=\sqrt{r_{\text{s}}^2-{\left(r_{\text{s}}-h_{\text{p}}\right)}^2}\text{for}{r}_{\text{s}}>h_{\text{p}}$

(8.59b)

FIGURE 8.7  Electrogeometric model for estimating the least distance of ground strike.

and y_o1 is the shortest distance of a lightning stroke of the given return-stroke current from the overhead line, which will result in a flash to ground.

To compute the outage rate, the return-stroke current, I_p, is varied from 1 to 200 kA in steps of 0.5 kA (Chowdhuri, 1989c, 2004). The current front time, tf, is varied from 0.5 to 10.5 μs in steps of 0.5 μs. At each current level, the shortest possible distance of the stroke, y_o1, is computed from Equation 8.59. Starting at t_f = 0.5 μs, the induced voltage is calculated as a function of time and compared with the given BIL of the line. If the BIL is not exceeded, then the next higher level of current is chosen. If the BIL is exceeded, then the lateral distance of the stroke from the line, y, is increased by Δy (e.g., 1 m), and the induced voltage is recalculated and compared with the BIL of the line. The lateral distance, y, is progressively increased until the induced voltage does not exceed BIL. This distance is called y_o2. For the selected I_p and t_f, the induced voltage will then exceed the BIL of the line and cause line flashover, if the lightning stroke hits the ground between y_o1 and y_o2 along the length of the line. For a 100 km sector of the line, the attractive area, A, will be (Figure 8.8)

$A=0.2\left(y_{\text{o}2}-y_{\text{o}1}\right){\text{km}}^2$

(8.60)

The joint probability density function, p(I_p, t_f), is then computed from Equation 8.56 for the selected I_p – t_f combination. If n_g is the ground flash density of the region, the expected number of flashovers per 100 km per year for that particular I_p – t_f combination will be

$\text{nfo}=\text{p}\left(I_{\text{p}},t_{\text{f}}\right)\cdot \Delta I_{\text{p}}\cdot \Delta t_{\text{f}}\cdot n_{\text{g}}\cdot A$

(8.61)

where

ΔI_p is the current step

Δt_f is the front-time step

The front time, t_f, is then increased by Δt_f = 0.5 μs to the next step, and nfo for the same current but with the new t_f is computed and added to the previous nfo. Once t_f = 10.5 μs is reached, the return-stroke current is increased by ΔI_p = 0.5 kA, and the whole procedure is repeated until the limits I_p = 200 kA and t_f = 10.5 μs are reached. The cumulative nfo will then give the total number of expected line flashovers per 100 km per year for the selected BIL.

The lightning-induced outage rates of the horizontally configured three-phase overhead line of Chapter 7 are plotted in Figure 8.9. The effectiveness of the shield wire, as shown in the figure, is optimistic, bearing in mind that the shield wire was assumed to be held at ground potential. The shield wire will not be held at ground potential under transient conditions. Therefore, the effectiveness of the shield wire will be less than the idealized case shown in Figure 8.9.

FIGURE 8.8  Attractive area of lightning ground flash to cause line flashover. A = 2A₁ = 0.2(y_o2 − y_o1) km².

FIGURE 8.9  Outage rates of overhead line by indirect strokes vs. BIL.

FIGURE 8.10  Total outage rates of overhead line vs. BIL. (a) Unshielded line, and (b) shielded line.

8.9    Estimation of Total Outage Rates

Outages of an overhead power line may be caused by both direct and indirect lightning strokes, especially if the voltage withstand level of the line is low, for example, power distribution lines. Outages caused by direct strokes are caused by either backflash (i.e., lightning hitting the tower or the shield wire) or by shielding failure if lightning hits one of the phase conductors. Outages caused by direct strokes were discussed in Chapter 7. Outages caused by indirect strokes are caused by lightning strokes hitting the nearby ground, which is discussed in this chapter. Therefore, information from both these chapters will be required to estimate the total outage rates of overhead lines caused by lightning. The outage rates of the three-phase horizontally configured overhead line of Chapter 7 are shown in Figure 8.10a for the unshielded line and in Figure 8.10b for the shielded line.

The effects of the variation of different parameters of the power line and also of the lightning stroke should also be considered in estimating the total outages of the line (Chowdhuri, 1989b). The power lines will be of finite length, terminated by transformers, underground cables, and lightning arresters. These terminations will generate multiple reflections. Therefore, power lines of finite lengths should also be considered for specific applications (Chowdhuri, 1991).

8.A   Appendix A Voltage Induced by Linearly rising and Falling return-Stroke Current

$V\left(x,t\right)=V_1\left(x,t\right)u\left(t-t_{\text{o}}\right)+V_2\left(x,t\right)u\left(t-t_{\text{of}}\right)$

where

$\begin{array}{c}{V}_1\left(x,t\right)=\frac{30{\alpha }_1{h}_{\text{p}}}{\beta c}\left[b_{\text{o}}\cdot \ell n\frac{f_{12}}{f_{11}}+0.5\ell n\left(f_{13}\right)\right];V_2\left(x,t\right)=-\frac{30{\alpha }_2{h}_{\text{p}}}{\beta c}\left[b_{\text{o}}\cdot \ell n\frac{f_{12\text{a}}}{f_{11a}}+0.5\ell n\left(f13\text{a}\right)\right]\\ b_{\text{o}}=1-{\text{β}}^2;t_{\text{of}}=t_{\text{o}}+t_{\text{f}};t_{\text{tf}}=t-t_{\text{f}}\\ f_1=m_1+{\left(ct-x\right)}^2-y_{\text{o}}^2;f_2=m_1-{\left(ct-x\right)}^2+y_{\text{o}}^2\\ f_3=m_0+{\left(ct-x\right)}^2+y_{\text{o}}^2;f_4=m_0+{\left(c{t}_{\text{o}}-x\right)}^2-y_{\text{o}}^2\\ f_5=n_1+{\left(ct+x\right)}^2-y_{\text{o}}^2;f_6=n_1-{\left(ct+x\right)}^2+y_{\text{o}}^2\\ f_7=n_{\text{o}}-{\left(c{t}_{\text{o}}+x\right)}^2+y_{\text{o}}^2;f_8=n_0+{\left(c{t}_{\text{o}}+x\right)}^2-y_{\text{o}}^2\\ f_9=b_{\text{o}}\left({\beta }^2{x}^2+y_{\text{o}}^2\right)+{\beta }^2{c}^2{t}^2\left(t+{\beta }^2\right);f_{10}=2{\beta }^{2}ct\sqrt{{\beta }^2{c}^2{t}^2+b_{\text{o}}\left(x^2+y_{\text{o}}^2\right)}\\ f_{11}=\frac{c^2{t}^2-x^2}{y_{\text{o}}^2};f_{12}=\frac{f_9-f_{10}}{b_{\text{o}}^2{y}_{\text{o}}^2};f_{13}=\frac{f_1\cdot f_3\cdot f_5\cdot f_7}{f_2\cdot f_4\cdot f_6\cdot f_8}\\ f_{1\text{a}}=m_{1\text{a}}+{\left(c{t}_{\text{tf}}-x\right)}^2-y_{\text{o}}^2;f_{2\text{a}}=m_{1\text{a}}-{\left(c{t}_{\text{tf}}-x\right)}^2+y_{\text{o}}^2\\ f_{3\text{a}}=f_3;f_{4\text{a}}=f_4;f_{7\text{a}}=f_7;f_{8\text{a}}=f_8\\ f_{5\text{a}}=n_{1\text{a}}+{\left(c{t}_{\text{tf}}+x\right)}^2-y_{\text{o}}^2;f_{6\text{a}}=n_{1\text{a}}-{\left(c{t}_{\text{tf}}+x\right)}^2+y_{\text{o}}^2\\ f_{9\text{a}}=b_{\text{o}}\left({\text{β}}^2{x}^2+y_{\text{o}}^2\right)+{\beta }^2{c}^2{t}_{\text{tf}}^2\left(1+{\beta }^2\right);f_{10\text{a}}=2{\beta }^{2}c{t}_{\text{tf}}\sqrt{{\beta }^2{c}^2{t}_{\text{tf}}^2+b_{\text{o}}\left(x^2+y_{\text{o}}^2\right)}\\ f_{11\text{a}}=\frac{c^2{t}_{\text{tf}}^2-x^2}{y_{\text{o}}^2};f_{12\text{a}}=\frac{f_{9\text{a}}-f_{10\text{a}}}{b_{\text{o}}^2{y}_{\text{o}}^2};f_{13\text{a}}=\frac{f_{1\text{a}}\cdot f_{3\text{a}}\cdot f_{5\text{a}}\cdot f_{7\text{a}}}{f_{2\text{a}}\cdot f_{4\text{a}}\cdot f_{6\text{a}}\cdot f_{8\text{a}}}\\ m_0=\sqrt{{\left[{\left(c{t}_{\text{o}}-x\right)}^2+y_{\text{o}}^2\right]}^2+4h_{\text{c}}^2{\left(c{t}_{\text{o}}-x\right)}^2;}{m}_1=\sqrt{{\left[{\left(ct-x\right)}^2+y_{\text{o}}^2\right]}^2+4h_{\text{c}}^2{\left(ct-x\right)}^2}\\ n_0=\sqrt{{\left[{\left(c{t}_{\text{o}}+x\right)}^2+y_{\text{o}}^2\right]}^2+4h_{\text{c}}^2{\left(c{t}_{\text{o}}+x\right)}^2;}{n}_1=\sqrt{{\left[{\left(ct+x\right)}^2+y_{\text{o}}^2\right]}^2+4h_{\text{c}}^2{\left(ct+x\right)}^2}\\ m_{1\text{a}}=\sqrt{{\left[{\left(c{t}_{\text{tf}}-x\right)}^2+y_{\text{o}}^2\right]}^2+4h_{\text{c}}^2{\left(c{t}_{\text{tf}}-x\right)}^2;}{n}_{1\text{a}}=\sqrt{{\left[{\left(c{t}_{\text{tf}}+x\right)}^2+y_{\text{o}}^2\right]}^2+4h_{\text{c}}^2{\left(c{t}_{\text{tf}}+x\right)}^2}\end{array}$

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