Evaluation of Surge Voltages on the Overhead Lines due to Direct and Indirect Lightning Impulse

In order to identify the Induced effect of lightning on the transmission line system we have to identify the effect of electromagnetic fields in the overhead cables CC lightning often serves as an indirect lightning stroke in certain high-altitude regions, influencing system parameters.

2.1 Calculation of EMF across the line

In order to calculate the electromagnetic fields (EMF) the current channel is considered as a function of altitude and time i(z',t). The Induced lightning current is a function of current obtained on the basis of current channel i(o,t) As it is the only form of current which can be measured directly [11].

$i(0, t)=I_{0}\left(e^{(-\alpha t)}-e^{(-\beta t)}\right)$              (1)

I₀ - magnitude of surge currents produced due to Induced effect of lightning.

α, β are constants,

$i\left(z^{\prime}, t\right)=i\left(0, t-z^{\prime} / v\right) e^{\left(-z^{\prime} / \lambda\right)}$              (2)

The model that describes the relationship between indirect lightning and phase line fields is used to calculate the emf equation of the lightning return stroke. We must consider the Maxwell theory in both scalar and vector potentials in order to obtain the horizontal and vertical components of the emf equation.

$\varphi=\frac{1}{4 \pi \varepsilon_{0}} \int_{(V)} \frac{t-R / C}{R} d V$             (3)

$\bar{A}=\frac{\mu}{4 \pi} \int_{(V)} \frac{\bar{J}(t-R / C)}{R} d V$       (4)

when there is a known potential in the system, the electric field is calculated by using the following equation:

$\bar{E}=-\operatorname{grad} \varphi-\frac{\partial \bar{A}}{\partial t}$           (5)

1.png

Figure 1. Field component coordinate system

X axis - transmission line.

Y axis - to determine distance between point of strike to cable.

Z axis - to analyse vertical dipole.

By considering the ground as a reference conductor, we derive the equations of electromagnetic fields of horizontal and vertical dipoles with help of coordinate system as shown in Figure 1. The Figure 1 describes the information about field component coordinate ordination system in which we consider a 3D axis in order to identify the point of strike which lies at a particular distance from transmission line. The overall fields consist of three different components of fields they are:

Electrostatics: It consists of the field effects of static charged particles and it is directly proportional to integral of current.

Electrostatics $\propto \int$ current.

Induction: This effect changes with the change in currents of the system.

Induction $\propto$  current.

Radiation: This effect changes with the rate of change of current with respect to time.

Radiation $\propto \frac{\partial(\text { current })}{\partial t}$.

The overall vertical dipole EMF (E_Z) is written as:

E_z = E_z1 + E_z2+E_z3           (6)

where,

$E_{z 1}(r, z, t)=\frac{1}{4 \pi \varepsilon_{0}} \int_{-z^{\prime}}^{z^{\prime}}\left(\frac{2\left(z-z^{\prime}\right)^{2}-r^{2}}{R^{5}} e^{\left(-z^{\prime} / \lambda\right)} \cdot \int_{0}^{t} i\left(0, \tau-z^{\prime} / V-R / c\right) d \tau\right) d z^{\prime}$        (7)

$E_{z 2}(r, z, t)=\frac{1}{4 \pi \varepsilon_{0}} \int_{-z^{\prime}}^{z^{\prime}}\left(\frac{2\left(z-z^{\prime}\right)^{2}-r^{2}}{c R^{4}} e^{\left(-z^{\prime} / \lambda\right)} \cdot i\left(0, t-z^{\prime} / V-R / c\right)\right) d z^{\prime}$      (8)

$E_{z 3}(r, z, t)=\frac{1}{4 \pi \varepsilon_{0}} \int_{-z^{\prime}}^{z^{\prime}}\left(\frac{r^{2}}{c^{2} R^{3}} e^{\left(-z^{\prime} / \lambda\right)}\right.$.

$\left.\frac{\partial i\left(0, t-z^{\prime} / V-R / c\right)}{\partial t}\right) d z^{\prime}$                                  (9)

For E_r, the EMF of horizontal dipole is written as:

$E_{r}=E_{r 1}+E_{r 2}+E_{r 3}$          (10)

where,

$E_{r 1}(r, z, t)=\frac{1}{4 \pi \varepsilon_{0}} \int_{-z^{\prime}}^{z^{\prime}}\left(\frac{3 r\left(z-z^{\prime}\right)}{R^{5}} e^{\left(-z^{\prime} / \lambda\right)} \cdot \int_{0}^{i} i\left(0, \tau-z^{\prime} / V-R / c\right) d \tau\right) d z^{\prime}$                (11)

$E_{r 2}(r, z, t)=\frac{1}{4 \pi \varepsilon_{0}} \int_{-z^{\prime}}^{z^{\prime}}\left(\frac{3 r\left(z-z^{\prime}\right)}{c R^{4}} e^{\left(-z^{\prime} / \lambda\right)} \cdot i\left(0, t-z^{\prime} / V-R / c\right) d z^{\prime}\right.$                  (12)

$E_{r 3}(r, z, t)=\frac{1}{4 \pi \varepsilon_{0}} \int_{-z^{\prime}}^{z^{\prime}}\left(\frac{r\left(z-z^{\prime}\right)}{c^{2} R^{3}} e^{\left(-z^{\prime} / \lambda\right)}\right.$.

$\frac{\partial i\left(0, t-z^{\prime} / V-R / c\right)}{\partial t} d z^{\prime}$          (13)    

where, i(0,t) is the current at channel base.

C = $3 \times 10^{8}$ m/s (speed of light).

$R=\sqrt{\left(r^{2}+\left(z-z^{\prime}\right)^{2}\right)}$              (14)

The value of z is altitude (taking ground as reference) where the field is calculated (z=h).

In order to obtain the value of time t we integration limit is set w.r.to z.

$z^{\prime} / v+\sqrt{\left(r^{2}+\left(z-z^{\prime}\right)^{2}\right) / c}=t$            (15)

2.2 Identification of overvoltage magnitude using EMF

In order to design a coupling system which calculates the induced over voltages due to indirect lightning we have to project the components of Er (11) to (13) on the x-Er axis. The purpose of getting x-Er axis is to shift the point of strike to the cable point as we are supposed to calculate emf at cable not on the point of strike. This is done by the following equation which provides relation between Ex and Er [12].

$E_{x}(r, z, t)=E_{r}(r, z, t)(L / 2-x) / r$                 (16)

2.png

Figure 2. Geometry of surge calculations

The field parameters are calculated for ground with high conductive property with the steps.

Step 1: Vertical parameters of the field.

Step 2: Frequency domain vertical components using DFT.

Step 3: Frequency domain horizontal components using transfer function.

$W(j \omega)=\frac{E_{r}(j \omega)}{E_{z}(j \omega)}=\left(\varepsilon_{r}+\frac{\sigma_{g r}}{j \omega \varepsilon_{0}}\right)^{-\frac{1}{2}}$              (17)

The above equation includes permittivity and conductivity of the ground which is displayed in Figure 2.

Step 4: Time domain horizontal components using IDFT.

The induced voltage due to indirect lightning stroke is calculated with the help of modifications in the normal form of travelling wave equation, this considers only a certain portion of the transmission line, this makes a modification in the system basic equations.

$\frac{\partial u^{s}(x, t)}{\partial x}+L^{\prime} \frac{\partial i(x, t)}{\partial t}=E_{x}^{i}(x, h, t)$             (18)

$\frac{\partial i(x, t)}{\partial x}+C^{\prime} \frac{\partial u^{s}(x, t)}{\partial t}=0$              (19)   

The equation in time domain are as follows,

$u^{s}(t, x)=v\left(t-\tau_{x}\right)+w\left(t-\tau_{l-x}\right)$               (20)

$Z_{c} i(t, x)=v\left(t-\tau_{x}\right)-w\left(t-\tau_{l-x}\right)$

$+Z_{c} \frac{1}{L} \int_{0}^{t} E_{x}^{i}(t) d t$              (21)

The surge calculation is done in MATLAB which intakes the following parameters as an input to the code.

1. Io- magnitude of current channel, alpha and beta, damping constant;
2. H- height of surge channel, Yo, L, h, r;

$r=\sqrt{Y_{0}^{2}+\left(\frac{L}{2}-(p-1) \cdot \frac{L}{N}\right)^{2}}$           (22)

3. N- number of sections of partition line, description line;
4. T- time required to calculate the parameters;
5. Calculation of magnitude of overvoltage.

It takes the numerical value of sections present in the transmission line system.

3.png

Figure 3. The induced voltages at x = 0m

44.png

Figure 4. The induced voltage at x = 250m

2.3 Direct formula for magnitude of induced voltage

Whenever an indirect lightning strikes nearby the transmission line system, the surge U_i will be injected into the system due to the coupling of magnetic field H. The induced voltage waveform for different values of x is shown in Figure 3 and Figure 4. Here, x is the distance between point of strike to the line, at x = 0 meters it is almost a direct lightning strike so the distribution Ex is positive, and at x = 250 meters, it is an indirect lightning strike, so as time increases it has to compensate. There is a direct formula which provides us with the approximate value of the induced voltage surge, it was given by Rusck [13].

$U_{\max , i}=Z_{0} i_{b} \frac{h}{x}$      (23)

where,

Z₀ - mutual impedance (between line and strike point);

h - height of line with respect to the ground;

x - displacement between line and strike point;

i_b- surge current induced in the transmission line.

Modified version of this formula is given as,

$U_{\max }=\frac{z_{0} I_{0} h}{d}\left(1+\frac{1}{\sqrt{2}} \frac{v}{c} \frac{1}{\sqrt{1-\frac{1}{2}\left(\frac{v}{c}\right)^{2}}}\right)$             (24)

where,

d - displacement between line and strike point;

v - return stroke velocity;

c = $3 \times 10^{8}$ m/s (speed of light).

2.4 RBF-FDTD method

This approach is used to solve the numerical equations in the coupling of the field produced by the lightning to the transmission line model. This method's output is the precise value of the lightning-induced overvoltage. The Radial Basis Function (RBF) and the Finite Difference Time Domain (FDTD) method are integrated in this method [2]. It provides comprehensive data on electromagnetic coupling in different phases of the power system (distribution and transmission phase). Advantages of FDTD is you can use to analyse electromagnetic fields for a time varying field, RBF is used to make nonlinear data point into linear.

2.4.1 Introduction to FDTD assumptions

The Maxwell’s equations are used in the finite difference time domain system to formulate the upcoming electric and magnetic field values. Based on the initial parameters, these formulae are used to predict the upcoming electric and magnetic field values [14]. The initial x-t plane must be used to construct a finite difference grid, which divides the overall plane into equipotential grids of steps, respectively $\Delta x$ and $\Delta t$. By considering The first and second order spatial derivatives of the function f(x,t) at a grid point, as well as the second order estimates of centred difference, are provided below.

$\frac{\partial}{\partial x} f(x, t)=\frac{f_{k+1}^{n}-f_{k-1}^{n}}{2 \Delta x}$             (25)

The second derivative of the above equation is given as:

$\frac{\partial^{2}}{\partial x^{2}} f(x, t)=\frac{f_{k+1}^{n}-2 f_{k}^{n}+f_{k-1}^{n}}{\Delta x^{2}}$              (26)

The first order time related derivative is given as:

$\frac{\partial}{\partial t} f(x, t)=\frac{f_{k}^{n+1}-f_{k}^{n-1}}{2 \Delta t}$             (27) 

2.4.2 RBF FDTD Formulation

The m^th order derivative of a function with respect to x, at a point x_i, is approximated by the summation given below.

$\left.\frac{d^{m} f}{d x^{m}}\right|_{x_{i}}=\sum_{j=1}^{n} w_{i, j}^{m} f\left(x_{j}\right)$           (28)

The radial basis function is given as:

$f(x)=\sum_{j=1}^{n} \lambda_{j} \phi\left(\left\|\underline{x}-\underline{x}_{j}\right\|\right)+\beta$           (29)

The above equation includes RBF, $\|.\|$  is considered to be the norm.

We use multiquadric RBF to get the derived approximations in a one-dimensional (1D) domain as shown in Figure 5.

5.png

Figure 5. Support points for i in 1D domain

The first and second order spatial derivatives of finite difference and the first order derivative of temporal finite difference are given as:

$\frac{\partial}{\partial x} f(x, t)=k_{x}^{1}\left(f_{k+1}^{n}-f_{k-1}^{n}\right)$        (30)

$\frac{\partial^{2}}{\partial x^{2}} f(x, t)=k_{x}^{2}\left(f_{k+1}^{n}+f_{k-1}^{n}-2 f_{k}^{n}\right)$         (31)

$\frac{\partial}{\partial t} f(x, t)=k_{t}^{1}\left(f_{k}^{n+1}-f_{k}^{n-1}\right)$         (32)

The approximations of the radial basis function corresponding to the geometrical parameters of c are:

• Spatial Domain

$k_{x}^{1}=\frac{\Delta x}{\left(\sqrt{4 \Delta x^{2}+c^{2}}-c\right) \sqrt{\Delta x^{2}+c^{2}}}$         (33)

$k_{x}^{2}=\frac{\frac{c^{2}}{\left(\sqrt{\Delta x^{2}+c^{2}}\right)^{3}}-\frac{1}{c}}{3 c-\sqrt{\Delta x^{2}+c^{2}}+\sqrt{4 \Delta x^{2}+c^{2}}}$           (34)

• Temporal Domain

$k_{t}^{1}=\frac{\Delta t}{\left(\sqrt{4 \Delta t^{2}+c^{2}}-c\right) \sqrt{\Delta t^{2}+c^{2}}}$               (35)

when c tends to ∞ the approximations generated are same in both the methods [15]. In order to select a parameter for the spatial domain, we consider different values which provide a better result than the FDTD method [16]. The induced voltage calculation geometry is depicted in Figure 6.

We consider the coupling equation FTL (field to transmission lines (lossless)).

$\frac{\partial V^{s}(x, t)}{\partial x}+L \frac{\partial}{\partial t} I(x, t)=E_{x}^{i}(x, t)$               (36)

$\frac{\partial I(x, t)}{\partial x}+C \frac{\partial}{\partial t} V^{s}(x, t)=0$               (37)

where, $V^{s}(x, t)$ and $I(x, t)$ are the surges induced into the line due to FTL coupling.

6.png

Figure 6. Induced voltage calculation geometry

L and C are the inductance and capacitance (per unit length).

By differentiating the Eqns. (36) and (37) with respect to x, we get the second order equations as shown below:

$\frac{\partial^{2} V^{s}(x, t)}{\partial x^{2}}-L C \frac{\partial^{2} V^{s}(x, t)}{\partial t^{2}}=\frac{\partial E_{x}^{i}(x, t)}{\partial x}$            (38)

$\frac{\partial^{2} I(x, t)}{\partial x^{2}}-L C \frac{\partial^{2} I(x, t)}{\partial t^{2}}=-C \frac{\partial E_{x}^{i}(x, t)}{\partial t}$               (39)

By considering the Taylor series for the surges generated in the system up to second order is given as:

$\begin{aligned} V^{s}(x, t)=V^{s}(x,&\left.t_{0}\right)+\Delta t \frac{\partial V^{s}(x, t)}{\partial x} \\+& \frac{\Delta t^{2}}{2 !} \frac{\partial^{2} V^{s}(x, t)}{\partial t^{2}}+O\left(\Delta t^{3}\right) \end{aligned}$               (40)

$\begin{aligned} I(x, t)=I\left(x, t_{0}\right) &+\Delta t \frac{\partial I(x, t)}{\partial x}+\frac{\Delta t^{2}}{2 !} \frac{\partial^{2} I(x, t)}{\partial t^{2}} \\ &+O\left(\Delta t^{3}\right) \end{aligned}$         (41)

By substituting the values of first and second order derivatives of surges (time-based domain).

$V^{s}(x, t)=V^{s}\left(x, t_{0}\right)-\Delta t C^{-1} \frac{\partial I(x, t)}{\partial x}+\frac{\Delta t^{2}(L C)^{-1}}{2}\left(\frac{\partial^{2} V^{s}(x, t)}{\partial x^{2}}-\frac{\partial E_{x}^{i}(x, t)}{\partial x}\right)+O\left(\Delta t^{3}\right)$           (42)

$I(x, t)=I\left(x, t_{0}\right)-\Delta t L^{-1}\left(\frac{\partial V^{s}(x, t)}{\partial x}-E_{x}^{i}(x, t)\right)+\frac{\Delta t^{2}(L C)^{-1}}{2}\left(\frac{\partial^{2} I(x, t)}{\partial x^{2}}+C \frac{\partial E_{x}^{i}(x, t)}{\partial t}\right)+O\left(\Delta t^{3}\right)$              (43)   

By representing the above equations in the spatial and time related derivatives using RBF-FDTD approximation are given as:

$V_{k}^{n+1}=V_{k}^{n}-\Delta t C^{-1} k_{x}^{1}\left(I_{k+1}^{n}-I_{k-1}^{n}\right)+\frac{\Delta t^{2}(L C)^{-1}}{2}\left[k_{x}^{2}\left(V_{k+1}^{n}+V_{k-1}^{n}-2 V_{k}^{n}\right)-k_{x}^{1}\left(E_{k+1}^{n}-E_{k-1}^{n}\right)\right]$             (44)

$I_{k}^{n+1}=I_{k}^{n}+\Delta t L^{-1}\left(E_{k}^{n}-k_{x}^{1}\left(V_{k+1}^{n}-V_{k-1}^{n}\right)\right)+\frac{\Delta t^{2}(C L)^{-1}}{2}\left[k_{x}^{2}\left(I_{k+1}^{n}+I_{k-1}^{n}-2 I_{k}^{n}\right)+C k_{t}^{1}\left(E_{k}^{n+1}-E_{k}^{n-1}\right)\right]$         (45)

The approximations are given as:

$V_{k}^{n} \equiv \mathrm{V}^{s}((k-1) \Delta x, n \Delta t)$        (46)

$I_{k}^{n} \equiv \mathrm{I}((\mathrm{k}-1) \Delta x, n \Delta t)$           (47)

where, $\Delta x, \Delta t$  are spatial and time steps (equal grid steps).

k, n are k^th and n^th integral steps (k is spatial and n is temporal).

The conditions to the boundaries of the surge voltage with a load at two-line closing is given as follows:

$V_{1}^{n}=\int_{0}^{h} E_{z}^{i}(0,0, t) d z-Z_{0} I_{1}^{n}$             (48)

$V_{N x+1}^{n}=\int_{0}^{h} E_{z}^{i}(L, 0, t) d z-Z_{L} I_{N x+1}^{n}$               (49)

where, $E_{z}^{i}$  refers to the incident electrical field (with respect to vertical field lines).

Z refers to the impedance of the termination of boundary lines.

The terminating conditions are designed in order to avoid the reflected waves on the boundaries [12].

The overall surge voltage induced into the transmission line system due to the indirect lightning stroke is given as a sum of the voltage scattered and the incident vertical electric field with a finite integral ranging from 0 to h.

$V^{T}(x, t)=V^{s}(x, t)-\int_{0}^{h} E_{z}^{i}(x, h, t) d z$             (50)

The above equation can be expressed as the sum of scattered and incident voltage.

$V^{T}(x, t)=V^{s}(x, t)+V^{i}(x, t)$          (51)

7.png

Figure 7. Distance between line and stroke

A perpendicular is drawn from the point of incidence of lightning to the transmission line considering the observation point at 500m, the 1D varies in the range [-500, 500] as shown in Figure 7. The results of magnitude of surge voltage due to lightning is shown in the Figure 8. 3D surface plot (voltage vs distance vs time).

8.png

Figure 8. Induced voltage plot

Outage rate gives the relation between the lightning shielding success and its failure. It mostly describes about the shielding success in the transmission line system. If the lightning shielding works efficiently, outages can be eradicated from the system. The lightning outage in a transmission line system causes back flashover voltages. The lightning outage at different conductors is shown in Figure 11.

11.png

Figure 11. Examples of lightning outage at different conductors

4.1 Outage rate calculation

The IEEE standards provide us with the formulae applied to find the lightning density on the ground. Generally, ground flash density is calculated based on LLC (lightning location system) and LFC (lightning flash counters) but these are inaccurate, So instead we calculated N_g as below (Only one location per flash us considered).

$N_{g}=0.04 T_{d}^{1.25}$       (54)

$N_{g}=0.054 T_{h}^{1.1}$             (55)  

where, $N_{g}$  is the Lightning density on the ground (GLD).

$T_{d}$  is the count of thunderstorm days.

$T_{h}$  is the count of lightning hours.

For every 100 kilometres line being hit by lightning surges each year is described as:

$N=N_{g}\left(\frac{28 H_{T}^{0.6}+b}{10}\right)$          (56)

$N=N_{g}\left(\frac{4 h_{p}^{1.09}+b}{10}\right)$           (57)

$N=N_{g}\left(\frac{38 h_{p}^{0.45}+b}{10}\right)$            (58)

where, N is the count of lightning strikes.

$h_{p}$ - height of line with respect to ground.

$H_{T}$ - tower height with respect to ground.

b - ground wire spacing.

The function of return stroke current is given as:

$f_{i}(I)=\left(\frac{1}{\sqrt{2 \pi} \sigma_{l n} I}\right) e^{\frac{-(\ln I / \bar{I})^{2}}{2 \sigma_{l n}^{2}}}$              (59)

where, $\sigma_{l n}$  for current less than 20kA.

$\bar{I}$  is the reference current.

If the amplitude of the return stroke > the collective, requires $f_{i}(I)$.

The probability can also be given as:

$P\left(I_{f}>I\right)=\frac{1}{1+\left(\frac{I}{\bar{I}_{\text {first }}}\right)^{2.6}}$         (60)

4.2 Geometrical analysis

In order to understand the concept of lightning strike in the transmission line environment, we consider the geometrical analysis which provides the protection angle value in order to increase the shielding efficiency. This analysis is known as Electrical shielding geometry. We use this model in order to analyse the failure in shielding which consists of ground wire on the top of the tower and reference ground component. Let us consider, α as a protection angle, the striking distance is given by S and its factor is given as β. The model of the considered case is shown in Figure 12.

12.png

Figure 12. Geometrical model (Xs ≠ 0)

where, DPER- path of lightning pilot.

a- line of lightning.

b - point of lightning strike.

c - describes the phase line.

Due to increase in the magnitude of surge current (due to lightning) the distance of striking will also increase. Due to this, the distance PE will gradually decrease and becomes 0 eventually.

13.png

Figure 13. Geometrical model (Xs = 0)

where, point P is known as critical point.

The Figure 13 describes the model where X_sis absent, at this point, the conductor is completely shielded due to maximum surge current shielding and maximum strike distance.

We do observe sag in the conductor which results in the new height of the phase line with respect to the ground. We need to identify the average height of the phase line which is provided by the given formula.

$H_{p}=h_{p}-\frac{2}{3} S_{a p}$             (61)

where, $S_{a p}$  denotes sag observed in the phase line.

If the strike distance is identified and the product of the strike distance factor and the strike distance is less than the height of phase line (considering sag).

$X_{s}=S\left[\cos \theta+\sin \left(\alpha_{s}-\omega\right)\right]$           (62)

$\theta=\sin ^{-1}\left(\frac{\beta S-H_{p}}{S}\right)$            (63)

$\omega=\cos ^{-1}\left(\frac{\sqrt{\left(x_{p}-x_{g}\right)^{2}+\left(H_{p}-H_{g}\right)^{2}}}{2 S}\right)$             (64)

$\alpha_{s}=\tan ^{-1}\left(\frac{x_{p}-x_{g}}{H_{g}-H_{p}}\right)$            (65)   

These are obtained with the help of horizontal coordinates of phase and ground wires. Under this condition.

$X_{s}=S\left[1+\sin \left(\alpha_{s}-\omega\right)\right]$            (66)

whenever there is a minimum amount of surge current the exposure distance is maximum and the exposure distance is 0 corresponding to maximum surge current [18].

$N_{S F}=0.1 N_{g} \frac{X_{s}}{2}\left(P_{\min }-P_{\max }\right)$            (67)

where, P_min is the probability of the surge current amplitude being above I_min.

P_max is the probability of the surge current amplitude being above I_max.

In HV transmission lines, whenever a direct lightning strikes on the top of the pole it produces a back flash if the insulation failure occurs.

Let us consider a voltage calculation model with respect to resistance and inductance.

14.png

Figure 14. Electrical circuitry of tower

In Figure 14, we have considered tower or the pole as a transmission line.

where,

$Z_{S}^{\prime}=\frac{2 Z_{S} Z_{T}}{Z_{S}+2 Z_{T}}$            (68)

$R^{\prime}=\frac{R Z_{T}}{Z_{T}-R}$              (69)

By considering the time of propagation and the attenuation coefficient we get the values.

$Z_{W}=\frac{2 Z_{T} Z_{S}^{2}}{\left(2 Z_{T}+Z_{S}\right)^{2}} \frac{Z_{T}-R}{Z_{T}+R}$           (70)

$\psi=\frac{2 Z_{T}-Z_{S}}{2 Z_{T}+Z_{S}} \frac{Z_{T}-R}{Z_{T}+R}$             (71)

where, $\psi$  is the coefficient of attenuation.

With these parameters, we can find out the total inductance of the tower.

$L=\left(\frac{Z_{S}^{\prime}+2 R^{\prime}}{Z_{S}^{\prime}}\right) \frac{2 Z_{w} \tau_{T}}{(1-\psi)^{2}}$           (72)

We use volt time characteristics of lightning current to calculate the value of voltage at which the critical breakdown takes place.

Whenever the lightning strike takes place, we have to calculate the voltage on the insulation at only two points. Let us consider the time divisions as 2-6 micro seconds. Let’s consider two cases in calculating the value of voltage at insulator.

Case 1 - Non reflective case

$V_{T 2}=\left[Z_{1}-\frac{\mathrm{Z}_{w}}{1-\psi}\left(1-\frac{\tau_{T}}{1-\psi}\right)\right] I$           (73)

The impedance is given as:

$Z_{i}=\frac{Z_{s} Z_{T}}{Z_{s}+2 Z_{T}}$             (74)

Case 2 - Reflective case

$V_{T 2}^{\prime}=\left[Z_{1}-\frac{-4 \mathrm{~K} V_{T 2}}{Z_{s}}\left(1-\frac{\tau_{T}}{Z_{s}}\right)\right]\left(1-\tau_{s}\right)$               (75)

This includes the coefficient of attenuation (considering K_s = 0.85 in this case).

If the current propagation time is greater than one microsecond then it come under a non-reflective case (at time = 2 micro seconds). When the propagation time is less than one microsecond.

$V_{T 2}=V_{T 2}+V_{T 2}{ }^{\prime}$          (76)

The grounding voltage at time is equals 2+τ_T is given as:

$V_{R 2}=\left[\frac{\alpha_{1} Z_{1}}{1-\psi}\left(1-\frac{\psi \tau_{T}}{1-\psi}\right)\right] I$       (77)

$\alpha_{1}=\frac{2 R}{z_{T^{+R}}}($ Grounding current related $)$                     (78)

$V_{p n 2}=V_{R 2}+\frac{\tau_{T}-\tau_{p n}}{\tau_{T}}\left(V_{T 2}-V_{R 2}\right)$                 (79)

$\tau_{p n}$ is the time of transmission from pole top to the n bars of magnitude of surge current.

The voltage of the n^th number phase line is given as the difference between the voltage of insulator strings to the phase voltage.

$V_{s n 2}=V_{p n 2}-K_{n} V_{T 2}$        (80) 

At a time, t = 6 micro seconds, the pole has lost its impact resistance as the peak of surge current has passed. So, the voltage is given as:

$V_{T 6}=\left(\frac{Z_{s} R}{Z_{s}+2 R}\right) I$            (81)

For a reflecting case, the voltage expression is given as:

$V^{\prime}{ }_{T 6}=-4 K_{s} Z_{s}\left(\frac{R}{Z_{s}+2 R}\right)^{2}\left(1-\frac{2 R}{Z_{s}+2 R}\right) I$          (82)   

The insulator voltage is given as:

$V_{s n 6}=\left[V_{T 6}+V_{T 6}^{\prime}\right]\left(1-K_{n}\right)$            (83)

The voltage of insulators at two microseconds and 6 microseconds is given as:

$V_{I 2}=820 l$           (84)

$V_{16}=585 l$             (85)  

where, l is the length of insulation.

The flash over currents is given as:

$I_{c n 2}=\frac{V_{I 2}}{V_{s n 2}}$             (86)

$I_{c n 6}=\frac{V_{I 6}}{V_{s n 6}}$               (87)