Insulation Coordination

Probabilistic Line Design and the Switching Surge Flashover Rate (SSFOR)

How to calculate the probability that a transmission line flashes over from phase-to-ground switching overvoltages — the stress-versus-strength concept, the histogram and continuous (integral) methods, the many-tower and voltage-profile effects, and the design quantity SSFOR (flashovers per 100 breaker closings).

Reading time ≈ 35 min

Section 1

What SSFOR Means and Why Probabilistic Design

This note covers how to calculate the probability that a transmission line will flash over due to phase-to-ground switching overvoltages. The key quantity is SSFOR — the Switching Surge Flashover Rate — which tells us how often a line is expected to flash over due to switching surges, usually expressed as flashovers per 100 switching operations (per 100 breaker closings).

The previous note designed line insulation with a deterministic method, equating the maximum overvoltage to the statistical withstand voltage \((E_m = V_3)\) — simply comparing the maximum stress with the minimum acceptable strength. Modern EHV design uses a better approach — probabilistic design — because both the switching overvoltage and the insulation strength are random.

Section 2

The Stress-Strength Concept

The chapter rests on the classical idea: failure occurs when stress exceeds strength. For switching-surge insulation, both are random.

Stress — the switching overvoltage

The stress is the switching overvoltage (SOV) applied to the line. It is random because each breaker closing can produce a different overvoltage, depending on the pole-closing instant, trapped charge, voltage angle, line length, source impedance, shunt reactors, compensation, and any pre-insertion/closing resistors or controlled switching. So the SOV is not a single value — it is a probability distribution.

Strength — the tower insulation

The strength is the tower insulation strength, which is also random: at the same voltage a tower may flash over in one test and withstand in another. It is represented by a cumulative Gaussian distribution described by CFO and \(\sigma\), with \(\sigma/\text{CFO} \approx 5\%\) for tower switching-impulse insulation.

Section 3

Why Deterministic Design Was Replaced

The deterministic method was used for the early 500 kV and 765 kV lines because SOV distributions were not yet well known, the probabilistic theory had not been developed for this application, and engineers did not initially trust the new method.

Later, especially for line uprating, probabilistic design became necessary. To answer a question like “Can an existing 138 kV line be operated at 230 or 345 kV?” it is not enough to check only the maximum possible surge — the engineer must know how often high SOVs occur, how likely flashover is at each level, how many towers are exposed, and how the voltage varies along the line. That is exactly what the probabilistic method provides.

Section 4

The Switching-Overvoltage Distribution

When a breaker closes, the overvoltage depends on the point on the voltage wave and on pole-scatter, so it varies. Engineers therefore simulate many switching operations — historically on a Transient Network Analyzer (TNA), today with EMTP®, PSCAD, ATP or EMTP®. Each random case produces one maximum SOV; after many runs, a distribution emerges. For example:

Table 1 — Example switching-overvoltage distribution (the stress).

SOV Level	Probability of Occurrence
1.9 pu	1%
1.8 pu	5%
1.7 pu	10%
1.6 pu	21%

Section 5

The Histogram Method

The simplest approach uses the histogram. For each voltage level, multiply the probability that the voltage occurs by the probability of flashover at that voltage, then sum. Take \(1\ \text{pu} = 450\) kV (crest line-to-neutral for a 550 kV system), with tower insulation \(\text{CFO} = 900\) kV and \(\sigma = 45\) kV \((\sigma/\text{CFO} = 5\%)\).

Consider the highest level, \(1.9\ \text{pu} = 855\) kV, occurring with probability 0.01. The flashover probability uses the reduced normal variable:

\[ Z = \frac{V - \text{CFO}}{\sigma} = \frac{855 - 900}{45} = -1.0 \qquad F(-1.0) = 0.1587 \]

\(Z\): reduced (standard) normal variable
\(F(-1.0)\): probability of flashover at 855 kV (~15.9%)

The contribution of this level is the product of “voltage occurs” and “flashover given that voltage”:

\[ P(V)\times P(\text{FO}\mid V) = 0.01 \times 0.1587 = 0.001587 \]

Repeating for the other significant levels and adding gives, for one tower:

\[ \text{SSFOR} = 0.00288 = \frac{0.288}{100} \quad\Rightarrow\quad 0.288\ \text{flashovers per 100 closings} \]

That is roughly one flashover per \(100/0.288 \approx 347\) switching operations for a single tower in this simplified example.

Section 6

Only the Tails Matter

The heart of probabilistic design

Only the upper tail of the SOV distribution and the lower tail of the insulation-strength distribution matter — because flashover happens only when stress is high and strength is low. It is the overlap of the two distributions that controls the risk.

So the average SOV and the CFO alone are not the whole story. Suppose the average overvoltage is only 1.5 pu — seemingly safe — but a small probability of 1.9 or 2.0 pu can dominate the flashover risk. Likewise a high CFO still has a lower tail where some towers flash over at lower voltage. In short:

Risk ≠ (average stress vs average strength) — risk depends on the overlap of the distribution tails.

Section 7

Many Towers in Parallel

A line has many towers, and a switching surge exposes them all. If one tower has flashover probability \(p\), its withstand probability is \(q = 1-p\); for \(n\) towers the probability that none flash over is \(q^n\), so the probability of at least one flashover is:

\[ P(\text{FO}) = 1 - q^{\,n} = 1 - (1-p)^{\,n} \]

For two towers this is \(1 - q^2\), correctly capturing all cases (tower 1 only, tower 2 only, or both). Applying the histogram method to a line of \(n = 100\) towers (all seeing the same SOV) raises the rate dramatically:

\[ \text{SSFOR}: \quad \frac{0.288}{100} \;\;\longrightarrow\;\; \frac{7.09}{100} \quad (1\text{ tower} \rightarrow 100\text{ towers}) \]

From a risk view, a line is many insulation points in parallel: if any one tower flashes over, the line has failed for that operation. So flashover probability grows with the number of towers, line length, and how uniformly the stress is shared — which is why long EHV lines need probabilistic treatment.

Section 8

The Voltage Profile Along the Line

The 100-tower calculation assumed every tower saw the same SOV, but the voltage is not uniform — it is lower near the switched breaker and higher near the open/receiving end. A simplified profile uses \(V_S/V_R = 0.9\) and \(V_{mid} = 0.95\,V_R\), and groups the line into three zones of 33 towers each:

Table 2 — Simplified three-zone voltage profile (for \(V_R = 1.9\) pu).

Zone	Voltage	Example Value
Receiving / open end (33 towers)	\(V_R\)	1.90 pu
Midpoint (33 towers)	\(V_{mid}=0.95\,V_R\)	1.805 pu
Switched end (33 towers)	\(V_S=0.90\,V_R\)	1.71 pu

For each zone compute the no-flashover probability \((q_R,\ q_{mid},\ q_S)\); the whole-line no-flashover probability is the product, so:

\[ P(\text{FO}) = 1 - q_R^{\,33}\,q_{mid}^{\,33}\,q_S^{\,33} \]

Including the profile makes the result more realistic — only some towers see the maximum — so the rate falls:

\[ \text{SSFOR}: \quad \frac{7.09}{100} \;\;\longrightarrow\;\; \frac{4.28}{100} \quad (\text{uniform} \rightarrow \text{with profile}) \]

Section 9

From Histogram to Continuous Integral

In a continuous distribution the probability of exactly one voltage is zero; we use areas under the SOV probability density \(f_s(V)\), with total area 1. The strength is the cumulative \(F_S(V)\) = probability that the strength is below the applied voltage = \(P(\text{FO}\mid V)\). The incremental flashover probability is \(dP = f_s(V)\,F_S(V)\,dV\), and integrating gives the continuous form of the histogram sum:

\[ \text{SSFOR} = \frac{1}{2}\int_{E_1}^{E_m} f_s(V)\,F_S(V)\,dV \]

\(f_s(V)\): probability density of the switching overvoltage (stress)
\(F_S(V)\): cumulative strength distribution \(= P(\text{FO}\mid V)\)
\(E_1,\ E_m\): minimum (~1.0 pu) and maximum SOV limits

Why the factor of ½?

The SOV distribution contains both positive and negative polarity surges. Because the negative-polarity switching strength is significantly higher than the positive, negative-polarity surges are not critical and are neglected. With roughly half the surges positive, only that half is counted — hence the \(\tfrac{1}{2}\).

Section 10

Normal Stress and Normal Strength

A useful theoretical case: stress and strength both normal, for one tower. With strength \(S\) and stress \(s\), failure occurs when \(S < s\), i.e. \(Z = S - s < 0\). If both are normal, \(Z\) is normal with:

\[ \mu_Z = \text{CFO} - \mu_0 \qquad \sigma_Z = \sqrt{\sigma_f^{\,2} + \sigma_0^{\,2}} \qquad P(\text{FO}) = P(Z < 0) \]

\(\text{CFO},\ \sigma_f\): mean and standard deviation of strength
\(\mu_0,\ \sigma_0\): mean and standard deviation of stress (SOV)

With \(\text{CFO}=900\), \(\sigma_f=45\), \(\mu_0=675\), \(\sigma_0=90\) kV:

\[ \mu_Z = 225\ \text{kV} \qquad \sigma_Z = \sqrt{45^2 + 90^2} = \sqrt{10125} \approx 100.6\ \text{kV} \qquad \frac{\mu_Z}{\sigma_Z} = 2.236 \]

\[ \text{SSFOR} = \tfrac{1}{2}\bigl[\,1 - F(2.236)\,\bigr] = 0.0064 = \frac{0.64}{100} \]

That is 0.64 flashovers per 100 breaker closings for this single-tower normal-normal case.

Section 11

The Complete SSFOR Equation

Building up from one tower to a full line with a voltage profile gives the three forms below. For \(n\) towers all at the same voltage, \(P(\text{FO}\mid V) = 1 - q^n\) with \(q = 1 - F_S(V)\):

\[ \text{SSFOR} = \tfrac{1}{2}\int_{E_1}^{E_m} f_s(V)\,\bigl[\,1 - (1 - F_S(V))^{\,n}\,\bigr]\,dV \]

For a real line, each tower sees a different (but correlated) voltage \(V_j\) from the profile, so the whole-line no-flashover probability is the product \(\prod_j q_j(V)\). The most complete form is:

\[ \text{SSFOR} = \tfrac{1}{2}\int_{E_1}^{E_m} f_s(V)\,\Bigl[\,1 - \prod_{j=1}^{n} q_j(V)\,\Bigr]\,dV \qquad q_j(V) = 1 - F_{S,j}(V_j) \]

What the equation says, in words

For every possible maximum SOV \(V\): find how likely that voltage is; use the profile to get the voltage at every tower; compute each tower’s flashover probability; multiply the no-flashover probabilities to get “no flashover anywhere”; subtract from 1 for “at least one flashover”; weight by the SOV density; integrate over all SOVs; and halve it because only positive polarity is critical.

Section 12

Practical EMTP® Workflow

In modern practice the procedure is:

SSFOR calculation workflow

Run many EMTP® switching simulations with random breaker-closing times.
Record the maximum phase-to-ground SOV along the line in each case.
Build the SOV probability distribution \(f_s(V)\).
Define the insulation-strength distribution (CFO, \(\sigma\)) — typically \(\sigma/\text{CFO}=5\%\) for tower switching impulse.
For each SOV, compute the tower flashover probability.
Account for the number of towers and the voltage profile along the line.
Compute the SSFOR (per 100 operations).
Compare the SSFOR against the acceptable design criterion.

Section 13

Why It Beats Deterministic Design — Summary

The deterministic method only asks “is \(E_m \le V_3\)?” The probabilistic method asks “how often will flashover occur?” This matters because two lines can share the same maximum surge yet have very different probabilities of high overvoltage:

Line A: max SOV = 2.0 pu, but 2.0 pu occurs extremely rarely.
Line B: max SOV = 2.0 pu, and values near 2.0 pu occur often.

The deterministic method treats them identically; the probabilistic method does not.

Key messages

Switching overvoltage is random — represent it as a distribution, not one value.
Tower strength is also random — a cumulative Gaussian (CFO, \(\sigma\)).
Risk depends on the overlap of stress and strength — only the upper SOV tail and lower strength tail matter.
More towers / longer line → higher flashover probability; the voltage profile reduces it versus assuming all towers see the maximum.
The design quantity is SSFOR — expected switching-surge flashovers per operation (commonly per 100), found by integrating stress against strength over all towers and halving for positive polarity.

Part One Reading now

Probabilistic Line Design and SSFOR

The stress-strength concept, the histogram and continuous (integral) methods, many-tower and voltage-profile effects, and the design quantity SSFOR — flashovers per 100 breaker closings.

Series progress 1 of 3

Part Two

Switching-Overvoltage Distributions

How the switching-overvoltage stress is collected and represented — the brute-force, case-peak and phase-peak methods, and fitting a Gaussian or extreme-value distribution via the 2% statistical overvoltage E₂.