Insulation Coordination

The transformer is properly coordinated if the protective ratio achieved is at least the required value:

$$PR = \frac{\text{BIL of transformer}}{\text{arrester residual (discharge) voltage}} \ge 1.2$$

Arrester characteristic $V_d = kI_d^{\beta}$

From the two table points:

$$\begin{aligned} \frac{360}{300} &= \left(\frac{20}{10}\right)^{\beta} \;\Rightarrow\; \beta = \frac{\ln 1.2}{\ln 2} = 0.2630 \\ k &= \frac{300}{10^{0.2630}} = 163.71 \end{aligned}$$

So $V_d = 163.71\,I_d^{0.263}$ (kV, kA).

Residual voltage at 40 kA

$$\begin{aligned} V_d &= 163.71 \times 40^{0.263} \\ &= 300 \times \left(\frac{40}{10}\right)^{0.263} = 300 \times 1.44 = 432.0\ \text{kV} \end{aligned}$$

(Doubling the current each time multiplies the voltage by 1.2: 300 → 360 → 432 kV.)

Protective ratio achieved

$$PR = \frac{500}{432.0} = 1.157$$

Check

• Required $PR = 1.2$, so the maximum allowed residual voltage is $500/1.2 = 416.7$ kV.
• Actual residual voltage is 432.0 kV > 416.7 kV.
• Achieved $PR = 1.157 < 1.2$.

The transformer is NOT properly coordinated for a 40 kA discharge current.

Remedies

• Use an arrester with a lower protective level (better characteristic, for example a metal-oxide arrester).
• Raise the transformer BIL to at least $1.2 \times 432 = 518.4$ kV (the next standard level, for example 550 kV).
• Reduce the expected discharge current with better line shielding and lower tower footing resistance.

The arrester is at the end of the line, next to the transformer. When it conducts, the line's Thevenin equivalent is $2V_w$ behind $Z$, so

$$2V_w = V_d + I_dZ, \qquad V_d = kI_d^{\beta}$$

The surge impedance of the line is not given. A typical 132 kV overhead line value, $Z = 400\ \Omega$, is assumed.

Arrester characteristic $V_d = kI_d^{\beta}$

From the two table points:

$$\begin{aligned} \frac{360}{300} &= \left(\frac{20}{10}\right)^{\beta} \;\Rightarrow\; \beta = \frac{\ln 1.2}{\ln 2} = 0.2630 \\ k &= \frac{300}{10^{0.2630}} = 163.71 \end{aligned}$$

So $V_d = 163.71\,I_d^{0.263}$ (kV, kA).

Discharge current and residual voltage

$$2 \times 8000 = 163.71\,I_d^{0.263} + 400\,I_d$$

Solving by trial (or iteration):

So $I_d = 38.93$ kA and $V_d = 428.9$ kV.

Protective ratio

$$PR = \frac{\text{BIL}}{V_d} = \frac{500}{428.9} = 1.166$$

Required $PR = 1.2$, which means $V_d \le 500/1.2 = 416.7$ kV. Since $428.9 > 416.7$ kV, i.e. $1.166 < 1.2$:

The transformer is NOT properly coordinated with this arrester for an 8000 kV incident surge.

Notes

• For $V_d = 416.7$ kV the arrester current must be at most $I_d = (416.7/163.71)^{1/0.263} = 34.87$ kA. That needs $Z \ge (16000 - 416.7)/34.87 = 447\ \Omega$. So with any usual line surge impedance (300–450 Ω) the result is the same: not coordinated. With $Z = 300\ \Omega$, $I_d = 51.8$ kA and $PR = 1.08$.
• Remedy: choose a transformer BIL of at least $1.2 \times 428.9 = 515$ kV (standard 550 kV), or use an arrester with a lower residual voltage.

A surge reaching the arrester rises at $S = 500$ kV/μs until the arrester sparks over at $V_a = 420$ kV. It reaches the transformer, which is $d$ metres beyond, after $T = d/v$. The transformer reflects it (open end) and the reflection returns to the arrester after $2T$. Until then, the transformer voltage continues to rise. The standard result is

$$V_t = V_a + 2S\frac{d}{v}, \quad v = 300\ \text{m}/\mu\text{s}$$

The arrester sparks over at $t = 420/500 = 0.84\ \mu$s.

(i) Coordination check (arrester right at the transformer)

$$PR = \frac{\text{BIL}}{V_a} = \frac{500}{420} = 1.19$$

The required PR is 1.25, which means the protective level must not exceed $500/1.25 = 400$ kV. Since $420 > 400$ kV, i.e. $1.19 < 1.25$:

Not properly coordinated, even with zero separation.

(ii) Transformer voltage with the arrester 40 m away

$$\begin{aligned} T &= \frac{40}{300} = 0.1333\ \mu\text{s} \\ V_t &= 420 + 2 \times 500 \times 0.1333 \\ &= 420 + 133.3 = 553.3\ \text{kV} \end{aligned}$$

This exceeds the BIL of 500 kV, so the transformer insulation would fail. The achieved PR is $500/553.3 = 0.90$.

(iii) Maximum allowed distance

With PR = 1.25: the allowed transformer voltage is 400 kV, which is already below the arrester level of 420 kV. So no distance works ($d_{max}$ is negative). The arrester rating must be lowered or the BIL raised.

Without margin, $V_t \le$ BIL = 500 kV:

$$\begin{aligned} 420 + 2 \times 500 \times \frac{d}{300} &\le 500 \\ d &\le \frac{(500 - 420) \times 300}{2 \times 500} = 24\ \text{m} \end{aligned}$$

  500 kV/us                  d
 ---------->  Arrester  ----------- Transformer
             (420 kV)               (BIL 500 kV)

Answer: (i) not coordinated (PR = 1.19); (ii) 553.3 kV; (iii) the arrester can be at most 24 m away to keep the voltage within the BIL. With the 1.25 margin, no distance satisfies the requirement.

Significance of the volt–time characteristic in insulation coordination

The volt–time (V–t) curve of an insulation or a protective device plots the peak voltage at breakdown (flashover) against the time to breakdown, for impulses of a standard shape but increasing amplitude. With a steeper or higher impulse, breakdown happens sooner and at a higher voltage, so the curve rises at short times.

 V |\
   | \  equipment insulation (upper)
   |  \___________________
   |\
   | \__ protective device (lower)
   |    \_________________
   +-------------------------> t (us)

Significance:

1. Basis of coordination. The V–t curve of the protective device (arrester, rod gap) must lie below that of the protected insulation at all times, with a margin of about 15–25%. Then the device always operates first.
2. Short-time behaviour. For very steep fronts, a device with a rising V–t curve (such as a rod gap) may operate too late. The curves show whether it gives protection for steep waves.
3. Selection of BIL and arrester rating. Comparing the curves fixes the transformer BIL, the arrester protective level and the protective ratio.
4. Coordination of different insulations in a substation (bushings, line insulators, breakers) so that flashover, if any, happens at a harmless place.

General working principle of a surge arrester

A surge arrester (lightning arrester) is a non-linear device connected between line and earth near the equipment.

• Normal voltage: the arrester has a very high resistance (series gap open, or a metal-oxide block at low current). Only a tiny leakage current flows.
• Surge arrives: when the voltage exceeds the spark-over (or reference) level, the gap sparks over or the ZnO blocks become highly conductive. The surge current is diverted to earth.
• Clamping: because of the non-linear characteristic $V = kI^{\beta}$ (with $\beta \approx 0.02$–$0.05$ for ZnO), the voltage across the arrester (the residual voltage) stays almost constant even for large currents. The equipment sees only this clamped voltage.
• Resealing: after the surge, the voltage returns to normal. The resistance becomes high again and the power-follow current is interrupted (by the gap in gapped types), so the arrester is ready for the next surge.

 line ----+---- to equipment
          |
        [ZnO]  non-linear
          |
        earth

The transformer is properly coordinated if the arrester residual voltage at the given discharge current is low enough to give the required protective ratio:

$$PR = \frac{\text{BIL}}{V_d} \ge 1.2$$

Residual voltage at 24 kA

$$\begin{aligned} V_d &= 200\,I_d^{0.25} = 200 \times 24^{0.25} \\ 24^{0.25} &= \sqrt{\sqrt{24}} = \sqrt{4.899} = 2.2134 \\ V_d &= 200 \times 2.2134 = 442.7\ \text{kV} \end{aligned}$$

Protective ratio

$$PR = \frac{500}{442.7} = 1.129$$

Check

• Maximum allowed residual voltage $= 500/1.2 = 416.7$ kV.
• Actual $V_d = 442.7$ kV > 416.7 kV.
• Achieved $PR = 1.13 < 1.2$.

The transformer is NOT properly coordinated.

The largest discharge current for which coordination holds is

$$I_d = \left(\frac{416.7}{200}\right)^4 = 18.8\ \text{kA}$$

Remedies

• Raise the BIL to at least $1.2 \times 442.7 = 531$ kV (standard 550 kV).
• Use an arrester with a lower residual voltage.
• Reduce the stroke current reaching the arrester with better shielding and footing resistance.

Answer: $V_d$ = 442.7 kV and PR = 1.13 < 1.2, so the transformer is not properly coordinated.

Note on the data: a surge impedance of 3000 Ω is not realistic for an overhead line (typical values are 300–500 Ω). The same problem in other papers uses $Z = 300\ \Omega$, so it is solved here with $Z = 300\ \Omega$. Results for 3000 Ω are given at the end. The reflection and refraction coefficients are the same in both cases.

Data: $V_w = 3000$ kV, $V_p = 1700$ kV (arrester protective level), $Z = 300\ \Omega$.

When the wave reaches the arrester (line end), the Thevenin equivalent is $2V_w$ behind $Z$, and the arrester holds the voltage at $V_p$.

(i) Current in the line before reaching the arrester

$$I_w = \frac{V_w}{Z} = \frac{3000}{300} = 10\ \text{kA}$$

(ii) Current through the arrester

$$\begin{aligned} I_a &= \frac{2V_w - V_p}{Z} = \frac{2 \times 3000 - 1700}{300} \\ &= \frac{4300}{300} = 14.33\ \text{kA} \end{aligned}$$

(iii) Arrester resistance

$$R = \frac{V_p}{I_a} = \frac{1700}{14.33} = 118.6\ \Omega$$

(iv) Reflection and refraction coefficients

Treat the arrester as a termination $R = 118.6\ \Omega$:

$$\begin{aligned} \text{Voltage reflection: } \Gamma_v &= \frac{R - Z}{R + Z} = \frac{118.6 - 300}{418.6} = -0.4333 \\ \text{Voltage refraction: } T_v &= \frac{2R}{R + Z} = \frac{V_p}{V_w} = 0.5667 \\ \text{Current reflection: } \Gamma_i &= -\Gamma_v = +0.4333 \\ \text{Current refraction: } T_i &= \frac{2Z}{R + Z} = 1.4333 \end{aligned}$$

Check: reflected voltage $= -0.4333 \times 3000 = -1300$ kV, and $3000 - 1300 = 1700$ kV ✓. Reflected current $= 0.4333 \times 10 = 4.33$ kA, and $10 + 4.33 = 14.33$ kA ✓.

Discussion

• The arrester clamps the voltage at 1700 kV instead of letting it double to 6000 kV at the open end. A negative voltage reflection of −1300 kV travels back along the line.
• The arrester current (14.3 kA) is larger than the line current (10 kA), because current reflection at a low-resistance termination is positive.
• The arrester must be rated to carry this current and to absorb the energy.

The transformer is properly coordinated only if the arrester residual voltage at the given discharge current is low enough that $\dfrac{\text{BIL}}{V_{res}} \ge$ the required protective ratio (1.25).

Data

• BIL of transformer = 500 kV
• Required protective ratio, $PR = 1.25$
• Arrester discharge current, $I_d = 24$ kA
• Residual voltage characteristic: $V_{res} = 200\, I_d^{0.25}$ kV ($I_d$ in kA)

Step 1: Residual voltage of the arrester

$$\begin{aligned} V_{res} &= 200 \times (24)^{0.25} \\ &= 200 \times 2.2134 \\ &= 442.67\ \text{kV} \end{aligned}$$

Step 2: Protective ratio actually obtained

$$PR_{actual} = \frac{\text{BIL}}{V_{res}} = \frac{500}{442.67} = 1.13$$

Step 3: Compare with the requirement

The maximum residual voltage allowed for $PR = 1.25$ is

$$V_{res,max} = \frac{\text{BIL}}{PR} = \frac{500}{1.25} = 400\ \text{kV}$$

Since $442.67\ \text{kV} > 400\ \text{kV}$ (or $1.13 < 1.25$), the protective margin is not enough.

Step 4: Largest discharge current the arrester can handle with PR = 1.25

$$\begin{aligned} 200\, I_d^{0.25} &= 400 \\ I_d &= \left(\frac{400}{200}\right)^4 = 16\ \text{kA} \end{aligned}$$

So the arrester keeps the required margin only for discharge currents up to 16 kA; a 24 kA discharge exceeds this.

Remedies

• Use an arrester with a lower protective level (lower rating or better non-linear blocks, e.g. ZnO).
• Choose a transformer with higher BIL (at least $1.25 \times 442.67 = 553.3$ kV, i.e. the next standard level, 550 kV or above).
• Reduce the stroke current reaching the arrester by better shielding of the line near the substation.

Answer: $V_{res} = 442.67$ kV, giving $PR = 1.13 < 1.25$, so the transformer is NOT properly coordinated (the required margin holds only up to $I_d = 16$ kA).

When the surge reaches the arrester, the arrester sparks over and holds the line voltage at its protective level $V_p$. It behaves as a resistance $R_a = V_p / I_a$ shunting the line, and the arrester current follows from the travelling-wave relations.

Data

• Incident surge voltage, $V_w = 3000$ kV
• Protective level of arrester, $V_p = 1700$ kV
• Line surge impedance, $Z = 300\ \Omega$
• The arrester is treated as the end (termination) of the line.

Incident current

$$I_w = \frac{V_w}{Z} = \frac{3000}{300} = 10\ \text{kA}$$

(i) Current through the arrester

The voltage at the arrester is $V_p = V_w + V_r$, so the reflected voltage is

$$V_r = V_p - V_w = 1700 - 3000 = -1300\ \text{kV}$$

The reflected current is $I_r = -V_r/Z = 1300/300 = 4.333$ kA. The arrester current is

$$\begin{aligned} I_a &= I_w + I_r = \frac{2V_w - V_p}{Z} \\ &= \frac{2 \times 3000 - 1700}{300} \\ &= 14.33\ \text{kA} \end{aligned}$$

(ii) Arrester resistance

$$R_a = \frac{V_p}{I_a} = \frac{1700}{14.333} = 118.6\ \Omega$$

(iii) Reflection and refraction coefficients

With termination $R_a = 118.6\ \Omega$ on a line of $Z = 300\ \Omega$:

$$\begin{aligned} \beta_v &= \frac{R_a - Z}{R_a + Z} = \frac{118.6 - 300}{418.6} = -0.433 \\ \alpha_v &= \frac{2R_a}{R_a + Z} = \frac{2 \times 118.6}{418.6} = 0.567 \\ \beta_i &= \frac{Z - R_a}{R_a + Z} = +0.433 \\ \alpha_i &= \frac{2Z}{R_a + Z} = \frac{600}{418.6} = 1.433 \end{aligned}$$

Check: $\alpha_v V_w = 0.567 \times 3000 = 1700$ kV $= V_p$, and $\alpha_i I_w = 1.433 \times 10 = 14.33$ kA $= I_a$.

(iv) Optimal current rating of the arrester

The arrester must discharge at least $I_a = 14.33$ kA while keeping its residual voltage at 1700 kV. The optimal (minimum) discharge current rating is therefore about 14.3 kA; in practice the next standard nominal discharge current class, 20 kA (IEC 60099-4 classes: 5, 10, 20 kA), is selected.

Answer: $I_a = 14.33$ kA, $R_a = 118.6\ \Omega$, $\beta_v = -0.433$, $\alpha_v = 0.567$, $\beta_i = 0.433$, $\alpha_i = 1.433$, rating ≈ 14.3 kA (choose the 20 kA class).

An arrester protects a transformer fully only if it sits at the transformer terminals. When it is a distance $L$ away, the voltage at the open-circuited transformer rises above the arrester protective level because of travelling-wave reflections. The relation between $L$, the surge steepness and the transformer BIL is the coordination expression.

Assumptions

• Surge with a linearly rising front reaches the arrester at $t = 0$.
• The transformer behaves as an open circuit to the surge (its surge impedance is much larger than the line's), so the reflection coefficient there is $+1$.
• The arrester holds a constant voltage $V_p$ once it sparks over.
• Velocity of the wave is $v$ (300 m/μs for an overhead line); travel time from arrester to transformer $T = L/v$.

 surge -->   A (arrester)        Tr (transformer)
 ------------+-------------------+
             |<------- L ------->|  open end
            [LA]                 = 
             |                  [Tr]
            ===                  |
                                ===

Steepness in terms of current

For a lightning current wave on a line of surge impedance $Z_c$, voltage and current are related by $e = Z_c\, i$. If the current rises at $di/dt$ (kA/μs), the voltage steepness is

$$S = \frac{de}{dt} = Z_c \frac{di}{dt}$$

Derivation

1. The incident ramp $e = St$ passes the arrester and reaches the transformer at $t = T$.
2. At the open end it doubles, so the transformer voltage is $e_t = 2S(t - T)$, and a reflected ramp $S(t - 2T)$ travels back.
3. The reflected wave reaches the arrester at $t = 2T$. After that the arrester voltage is

$$e_a = St + S(t - 2T) = 2St - 2ST$$

4. The arrester sparks over when $e_a = V_p$:

$$t_s = \frac{V_p}{2S} + T$$

5. The clamping effect of the arrester needs a further time $T$ to reach the transformer. Until then the transformer voltage keeps rising, so its peak is

$$\begin{aligned} V_t &= 2S\big[(t_s + T) - T\big] = 2S\, t_s \\ &= V_p + 2ST = V_p + \frac{2SL}{v} \end{aligned}$$

6. Putting $S = Z_c\, di/dt$:

$$\boxed{V_t = V_p + \frac{2 Z_c L}{v}\,\frac{di}{dt}}$$

Coordination condition

The transformer is protected if $V_t$ does not exceed its BIL (with protective ratio $k$, $V_t \le \text{BIL}/k$):

$$V_p + \frac{2 Z_c L}{v}\frac{di}{dt} \le \frac{\text{BIL}}{k}$$

The maximum allowed separation is

$$L_{max} = \frac{v\left(\dfrac{\text{BIL}}{k} - V_p\right)}{2 Z_c \,(di/dt)}$$

With $v = 300$ m/μs, $Z_c$ in Ω, $di/dt$ in kA/μs and voltages in kV, $L$ comes out in metres.

Remarks

• The expression is valid when sparkover occurs after the reflection returns ($V_p > 2ST$). If the arrester sparks earlier, $V_t$ is limited to about $2V_p$.
• Steeper surges, higher $Z_c$ or longer leads reduce $L_{max}$, so arresters are placed as close as possible to the transformer.

Result: $V_t = V_p + 2Z_c L\,(di/dt)/v$, and coordination requires $L \le v(\text{BIL}/k - V_p)/[2Z_c\,(di/dt)]$.

When the travelling wave reaches the arrester at the line end, the arrester conducts and holds the voltage at its rated (protective) level of 1000 kV. The arrester acts as a resistance $R_a$, so part of the wave is reflected.

Data

• Incident voltage, $V_i = 4200$ kV
• Arrester voltage (BIL rating / protective level), $V_a = 1000$ kV
• Surge impedance, $Z = 300\ \Omega$

Incident current

$$I_i = \frac{V_i}{Z} = \frac{4200}{300} = 14\ \text{kA}$$

(i) Reflected voltage and current

Voltage at the arrester = incident + reflected:

$$\begin{aligned} V_r &= V_a - V_i = 1000 - 4200 = -3200\ \text{kV} \\ I_r &= -\frac{V_r}{Z} = \frac{3200}{300} = 10.67\ \text{kA} \end{aligned}$$

The reflected voltage is negative (it cancels most of the incident wave), and the reflected current is positive (it adds to the incident current).

(ii) Arrester discharge current and resistance

$$\begin{aligned} I_a &= I_i + I_r = \frac{2V_i - V_a}{Z} \\ &= \frac{2 \times 4200 - 1000}{300} = 24.67\ \text{kA} \end{aligned}$$ $$R_a = \frac{V_a}{I_a} = \frac{1000}{24.667} = 40.54\ \Omega$$

Check with reflection coefficient

$$\beta = \frac{R_a - Z}{R_a + Z} = \frac{40.54 - 300}{340.54} = -0.762, \quad \beta V_i = -0.762 \times 4200 = -3200\ \text{kV}$$

Answer: $V_r = -3200$ kV, $I_r = 10.67$ kA, $I_a = 24.67$ kA, $R_a = 40.54\ \Omega$.

The arrester limits the voltage only at its own terminals. Because the transformer is 100 m away and acts as an open end, the surge doubles there, so the transformer voltage can be much higher than the arrester sparkover voltage.

Data

• Arrester sparkover (and residual) voltage, $V_p = 320$ kV
• Surge steepness, $S = 500$ kV/μs
• Separation, $L = 100$ m; velocity $v = 300$ m/μs
• BIL = 500 kV, required protective ratio = 1.3
• Transformer treated as an open circuit (reflection coefficient +1)

Time quantities

$$\begin{aligned} T &= \frac{L}{v} = \frac{100}{300} = 0.333\ \mu\text{s} \\ 2T &= 0.667\ \mu\text{s} \\ t_s &= \frac{V_p}{S} = \frac{320}{500} = 0.64\ \mu\text{s} \end{aligned}$$

 t=0   ramp 500 kV/us reaches arrester A
 t=T   (0.333 us) ramp reaches Tr, doubles
 t=ts  (0.64 us) A sparks over at 320 kV
 t=2T  (0.667 us) reflection from Tr would
       have reached A -- too late

(i) Voltage at the arrester location

The arrester voltage reaches 320 kV at $t_s = 0.64$ μs, which is before the reflected wave from the transformer returns ($2T = 0.667$ μs). So the arrester sparks over on the incident wave alone and holds

$$V_A = 320\ \text{kV}$$

(ii) Voltage at the transformer terminal

The wave travelling towards the transformer is cut off (clamped) at 320 kV. At the open end it doubles:

$$V_T = 2 \times 320 = 640\ \text{kV}$$

(The simple textbook formula $V_T = V_p + 2SL/v = 320 + 2 \times 500 \times 0.333 = 653.3$ kV gives nearly the same, slightly conservative, value. It strictly applies when sparkover occurs after $2T$.)

Coordination check

Maximum voltage allowed at the transformer:

$$V_{T,allowed} = \frac{\text{BIL}}{PR} = \frac{500}{1.3} = 384.6\ \text{kV}$$

Since $640\ \text{kV} > 384.6\ \text{kV}$ (it even exceeds the BIL of 500 kV), the transformer is NOT properly coordinated.

Condition for proper coordination

(a) Reduce the separation. For a short lead, sparkover occurs after the reflection returns and $V_T = V_p + 2SL/v$:

$$\begin{aligned} 320 + \frac{2 \times 500 \times L}{300} &\le 384.6 \\ L &\le \frac{(384.6 - 320) \times 300}{1000} = 19.4\ \text{m} \end{aligned}$$

Check: $T = 19.4/300 = 0.065$ μs, and $V_p = 320 > 2ST = 64.6$ kV, so the formula is valid.

(b) Or, keeping 100 m, use an arrester with lower sparkover voltage so that $2V_p \le 384.6$ kV, i.e. $V_p \le 192.3$ kV.

Answer: $V_A = 320$ kV, $V_T = 640$ kV (≈653 kV by the simple formula) > 384.6 kV, so not coordinated. It becomes coordinated if the arrester is within about 19.4 m of the transformer (or its sparkover is reduced to ≤ 192 kV).

When the travelling wave reaches the arrester at the end of the line, the arrester conducts and holds the voltage at its rated (protective) level of 1000 kV. It acts as a resistance $R_a$ terminating the line, so a large negative voltage wave is reflected.

Data

• Incident voltage, $V_i = 4200$ kV
• Arrester voltage (BIL rating / protective level), $V_a = 1000$ kV
• Surge impedance, $Z = 350\ \Omega$

Incident current

$$I_i = \frac{V_i}{Z} = \frac{4200}{350} = 12\ \text{kA}$$

(i) Reflected voltage and current

$$\begin{aligned} V_r &= V_a - V_i = 1000 - 4200 = -3200\ \text{kV} \\ I_r &= -\frac{V_r}{Z} = \frac{3200}{350} = 9.14\ \text{kA} \end{aligned}$$

(ii) Arrester discharge current and resistance

$$\begin{aligned} I_a &= I_i + I_r = \frac{2V_i - V_a}{Z} \\ &= \frac{8400 - 1000}{350} = 21.14\ \text{kA} \end{aligned}$$ $$R_a = \frac{V_a}{I_a} = \frac{1000}{21.143} = 47.30\ \Omega$$

Check

$$\beta = \frac{R_a - Z}{R_a + Z} = \frac{47.30 - 350}{397.30} = -0.762, \quad \beta V_i = -3200\ \text{kV}$$

Compared with a 300 Ω line, the higher surge impedance gives a smaller arrester current and a larger equivalent resistance.

Answer: $V_r = -3200$ kV, $I_r = 9.14$ kA, $I_a = 21.14$ kA, $R_a = 47.30\ \Omega$.

Insulation failure is the loss of the insulating property of a material so that it can no longer withstand the applied voltage. It usually results from several stresses acting together over time.

Factors responsible for insulation failure

1. Electrical stress: lightning and switching overvoltages, temporary overvoltages, and long-term partial discharges in voids that slowly erode the insulation (treeing).
2. Thermal stress: overloading, poor cooling and hot spots raise temperature; insulation ages faster (roughly half life for every 8–10 °C rise) and thermal runaway may occur.
3. Mechanical stress: vibration, short-circuit forces, thermal expansion and bending cause cracks and voids.
4. Environmental factors: moisture, dust, salt and industrial pollution, UV radiation, which reduce surface resistance and cause tracking and flashover.
5. Chemical action: oxidation of oil, acids and sludge formation, attack by ozone and nitric acid produced by corona.
6. Manufacturing defects and ageing: voids, impurities, poor impregnation, and the natural deterioration of material with time.

Possible remedies

Regular condition monitoring (insulation resistance, tan δ, DGA of oil, PD measurement) helps detect weakening insulation before a breakdown happens.

The transformer BIL is fixed from the arrester protective level and the transformer protective ratio. The maximum separation is then the distance at which the extra voltage built up between arrester and transformer just uses up the margin between BIL and the arrester protective level.

Assumptions

• Highest system voltage of a 400 kV system = $1.05 \times 400 = 420$ kV (line-to-line, rms).
• Effectively grounded system: coefficient of earthing 0.8, so arrester rating = 80% of highest system voltage.
• "PR = 3" of the arrester is the ratio of its protective level (peak) to its rated voltage (rms).
• The figure gives no surge steepness, so a lightning surge current of 10 kA with 1 μs front is assumed (as in the similar 2070 question). On the 300 Ω line its steepness is $S = 10 \times 300 / 1 = 3000$ kV/μs.
• Velocity $v = 300$ m/μs. The transformer is treated as an open circuit (worst case).

Step 1: Arrester rating and protective level

$$\begin{aligned} V_{rating} &= 0.8 \times 420 = 336\ \text{kV (rms)} \\ V_p &= PR \times V_{rating} = 3 \times 336 = 1008\ \text{kV} \end{aligned}$$

Step 2: BIL of the transformer

$$\text{BIL} = TPR \times V_p = 1.25 \times 1008 = 1260\ \text{kV}$$

The next standard level, 1300 kV, would be chosen in practice.

Step 3: Maximum distance

Voltage at an open-circuited transformer a distance $L$ from the arrester:

$$V_t = V_p + \frac{2SL}{v}$$

Coordination requires $V_t \le \text{BIL}$:

$$\begin{aligned} 1008 + \frac{2 \times 3000 \times L}{300} &\le 1260 \\ 20L &\le 252 \\ L_{max} &= 12.6\ \text{m} \end{aligned}$$

(With the standard BIL of 1300 kV: $L_{max} = 292/20 = 14.6$ m.)

At the 30 m shown in the figure: $V_t = 1008 + 20 \times 30 = 1608$ kV $> 1260$ kV, so the arrester is too far away.

Note on the 350 Ω transformer impedance

If the 350 Ω shown for the transformer is used literally, the coefficients at the transformer are

$$\alpha = \frac{2 \times 350}{300 + 350} = 1.077, \quad \beta = \frac{350 - 300}{650} = 0.077$$

and the transformer voltage cannot exceed about $\alpha V_p = 1085.5$ kV, which is below the BIL at any distance. Real transformer windings have surge impedances of several thousand ohms, so the open-circuit (β ≈ 1) case above is the one used for design.

Answer: BIL = 1260 kV (choose 1300 kV standard); maximum arrester–transformer distance ≈ 12.6 m for the assumed 3000 kV/μs surge (30 m is not acceptable).

Volt-time curve

A volt-time (V–t) curve of an insulation or a gap is the graph of the breakdown (flashover) voltage against the time to breakdown for impulse voltages of a given wave shape (normally 1.2/50 μs). Insulation does not break down instantly; for very fast impulses it withstands a higher voltage, so the curve rises steeply at short times and flattens at longer times.

In insulation coordination the V–t curve of the protective device (arrester or rod gap) must lie below the V–t curve of the protected equipment over the whole time range, with a safety margin of about 20–25%. Then the protective device always flashes over first.

 V |
   | \  equipment (transformer)
   |  \___
   |      \______________
   |  \                   margin
   |   \__  arrester
   |      \______________
   +-------------------------- t
     2 us              20 us

Step-wise plotting of the V–t curve

1. Set up the test: Connect the test object (gap, insulator or equipment) to an impulse generator. Measure voltage with a CRO and a voltage divider.
2. Fix the wave shape: Use a standard impulse, e.g. 1.2/50 μs, and keep the same shape for all shots; only the peak value is changed.
3. Start low: Apply an impulse below the expected flashover level. No breakdown occurs.
4. Increase the peak: Raise the peak in steps until flashover occurs on the tail of the wave (just above the critical flashover voltage). Record the time from the start of the wave to flashover.
5. Plot the point correctly:
   • If flashover occurs on the tail, plot the peak voltage of the applied wave against the time of flashover.
   • If flashover occurs on the front, plot the actual voltage at the instant of flashover against that time.
6. Repeat at higher voltages: Increase the applied peak further. Flashover occurs earlier, moving onto the front of the wave. Record voltage and time for each shot.
7. Repeat shots: Because breakdown is statistical, apply several impulses at each level and use mean values.
8. Join the points: Draw a smooth curve through the plotted points. This is the V–t characteristic.

 V |      x  flashover on front
   |     /  .   (plot V at instant)
   |    /     .
   |   /   o    .   o = peak, flashover
   |  /         ..   on tail (plot peak
   | /    applied     at time of FO)
   |/     waves
   +------------------------- t

Use in insulation coordination

• It shows whether the arrester or gap protects the equipment for both steep (fast) and slow surges.
• It is used to select the arrester and the BIL of equipment with proper margins.
• Two curves that cross mean protection fails over part of the time range, so devices are chosen whose curves do not cross.

When the surge reaches the arrester, the arrester conducts and holds the voltage at its protective level $V_p$. It acts as a shunt resistance $R_a$ at the end of the line, and the travelling-wave relations give the currents.

Data

• Incident surge, $V_w = 3000$ kV (crest)
• Protective level, $V_p = 1700$ kV
• Line surge impedance, $Z = 300\ \Omega$

(a) Current in the line before reaching the arrester

$$I_w = \frac{V_w}{Z} = \frac{3000}{300} = 10\ \text{kA}$$

(b) Current through the arrester

Reflected voltage: $V_r = V_p - V_w = 1700 - 3000 = -1300$ kV. Reflected current: $I_r = -V_r/Z = 4.33$ kA.

$$\begin{aligned} I_a &= I_w + I_r = \frac{2V_w - V_p}{Z} \\ &= \frac{6000 - 1700}{300} = 14.33\ \text{kA} \end{aligned}$$

(c) Arrester resistance

$$R_a = \frac{V_p}{I_a} = \frac{1700}{14.333} = 118.6\ \Omega$$

Reflection and refraction coefficients of voltage

$$\begin{aligned} \beta &= \frac{R_a - Z}{R_a + Z} = \frac{118.6 - 300}{418.6} = -0.433 \\ \alpha &= \frac{2R_a}{R_a + Z} = \frac{237.2}{418.6} = 0.567 \end{aligned}$$

Check: $\alpha = 1 + \beta = 0.567$, and $\alpha V_w = 0.567 \times 3000 = 1700$ kV $= V_p$.

The arrester current (14.33 kA) is larger than the line current (10 kA) because the negative reflected voltage wave carries a positive current. The arrester must be rated for this current.

Answer: (a) 10 kA, (b) 14.33 kA, (c) $R_a = 118.6\ \Omega$; $\beta = -0.433$, $\alpha = 0.567$.

The transformer BIL is fixed from the arrester protective level and the transformer protective ratio. Then, with the arrester 30 m away, the travelling-wave rise between arrester and transformer is checked against this BIL.

Assumptions

• Highest system voltage = $1.05 \times 400 = 420$ kV (rms, line-to-line).
• Effectively grounded system: arrester rating = 80% of highest system voltage.
• Arrester PR = (protective level, peak) / (rated voltage, rms).
• Lightning current wave on the line: $e = Z_c\, i$; velocity $v = 300$ m/μs; transformer acts as an open circuit.

Step 1: Arrester protective level

$$\begin{aligned} V_{rating} &= 0.8 \times 420 = 336\ \text{kV (rms)} \\ V_p &= 3 \times 336 = 1008\ \text{kV} \end{aligned}$$

Step 2: BIL of transformer

$$\text{BIL} = TPR \times V_p = 1.25 \times 1008 = 1260\ \text{kV}$$

(Next standard value: 1300 kV.)

Step 3: Surge steepness

$$\begin{aligned} V_{surge} &= I Z_c = 10 \times 350 = 3500\ \text{kV} \\ S &= \frac{3500}{1\ \mu s} = 3500\ \text{kV}/\mu\text{s} \end{aligned}$$

Step 4: Voltage at the transformer (L = 30 m)

$$\begin{aligned} T &= \frac{L}{v} = \frac{30}{300} = 0.1\ \mu\text{s} \\ V_t &= V_p + 2ST = 1008 + 2 \times 3500 \times 0.1 \\ &= 1008 + 700 = 1708\ \text{kV} \end{aligned}$$

Validity check: $V_p = 1008 > 2ST = 700$ kV, so the arrester sparks over after the reflection returns and the formula holds.

Since $1708\ \text{kV} > 1260\ \text{kV}$, the insulation is NOT properly coordinated with the LA at 30 m.

Step 5: Maximum separation

$$\begin{aligned} 1008 + \frac{2 \times 3500 \times L}{300} &\le 1260 \\ 23.33\, L &\le 252 \\ L_{max} &= 10.8\ \text{m} \end{aligned}$$

(If the standard BIL of 1300 kV is adopted, $L_{max} = 292/23.33 = 12.5$ m.)

Answer: BIL = 1260 kV; at 30 m the transformer sees 1708 kV > BIL, so not coordinated; the LA must be within about 10.8 m of the transformer.

TRUE.

Justification

The protective ratio (PR) is the ratio of the equipment insulation level (BIL) to the protective level of the arrester:

$$PR = \frac{\text{BIL of equipment}}{\text{Protective level of arrester}}$$

The arrester protective level depends on the arrester rating, and the rating depends on how the system neutral is grounded:

• During a line-to-ground fault, the voltage of the healthy phases rises. The arrester must not conduct (and must reseal) at this power-frequency voltage, so its rating is chosen as

$$V_{rating} = C_e \times V_{max(L-L)}$$

where $C_e$ is the coefficient of earthing.

• An effectively grounded system allows a lower-rated arrester with a lower protective level. For the same BIL this gives a higher protective ratio (more margin), or for the same margin it allows a reduced BIL.
• An ungrounded system needs a higher-rated arrester, so the protective level is higher and the protective ratio for the same equipment is smaller.

So the protective ratio obtainable depends on the type of grounding employed. This is why EHV systems are effectively grounded: it permits reduced insulation levels.

The transformer BIL is obtained from the arrester protective level and the required margin. The arrester resistance is then found from the travelling-wave current through the arrester during the stroke.

Assumptions

• Highest system voltage = $1.05 \times 500 = 525$ kV (rms, line-to-line).
• Effectively grounded: arrester rating = 80% of highest system voltage.
• Arrester protective ratio = protective level (peak) / rated voltage (rms).
• The 5 kA stroke current travels on the line, producing a voltage wave $V_w = I Z$.

Step 1: Arrester rating and protective level

$$\begin{aligned} V_{rating} &= 0.8 \times 525 = 420\ \text{kV (rms)} \\ V_p &= 2.83 \times 420 = 1188.6\ \text{kV} \end{aligned}$$

Step 2: BIL of the transformer

$$\text{BIL} = 1.4 \times 1188.6 = 1664.0\ \text{kV}$$

The next standard BIL, 1675 kV, would be chosen.

Step 3: Arrester resistance for a 5 kA stroke

Voltage wave on the line:

$$V_w = I Z = 5 \times 360 = 1800\ \text{kV}$$

Arrester current (line terminated by the arrester at voltage $V_p$):

$$\begin{aligned} I_a &= \frac{2V_w - V_p}{Z} \\ &= \frac{2 \times 1800 - 1188.6}{360} \\ &= 6.70\ \text{kA} \end{aligned}$$

Arrester resistance:

$$R_a = \frac{V_p}{I_a} = \frac{1188.6}{6.698} = 177.4\ \Omega$$

(If 5 kA is taken directly as the arrester discharge current, $R_a = 1188.6/5 = 237.7\ \Omega$.)

Answer: BIL ≈ 1664 kV (use 1675 kV); arrester resistance ≈ 177.4 Ω (237.7 Ω if 5 kA is the arrester current itself).

When the travelling wave reaches the arrester at the line end, the arrester conducts and holds the voltage at its rated level of 1600 kV. It behaves as a resistance $R_a$ terminating the line.

Data

• Incident voltage, $V_i = 4200$ kV
• Arrester voltage (BIL rating / protective level), $V_a = 1600$ kV
• Surge impedance, $Z = 300\ \Omega$

Incident current

$$I_i = \frac{V_i}{Z} = \frac{4200}{300} = 14\ \text{kA}$$

i) Reflected voltage and current

$$\begin{aligned} V_r &= V_a - V_i = 1600 - 4200 = -2600\ \text{kV} \\ I_r &= -\frac{V_r}{Z} = \frac{2600}{300} = 8.67\ \text{kA} \end{aligned}$$

ii) Arrester discharge current and resistance

$$\begin{aligned} I_a &= I_i + I_r = \frac{2V_i - V_a}{Z} \\ &= \frac{8400 - 1600}{300} = 22.67\ \text{kA} \end{aligned}$$ $$R_a = \frac{V_a}{I_a} = \frac{1600}{22.667} = 70.59\ \Omega$$

Check

$$\beta = \frac{R_a - Z}{R_a + Z} = \frac{70.59 - 300}{370.59} = -0.619, \quad \beta V_i = -0.619 \times 4200 = -2600\ \text{kV}$$

Answer: $V_r = -2600$ kV, $I_r = 8.67$ kA, $I_a = 22.67$ kA, $R_a = 70.59\ \Omega$.

An arrester placed some distance from the transformer cannot limit the transformer voltage to its own protective level. The surge doubles at the transformer, and the voltage there exceeds the arrester level by $2SL/v$.

Data

• BIL of transformer = 500 kV
• Separation, $L = 30$ m; velocity $v = 300$ m/μs
• Surge = 3500 kV peak, reached in 1 μs
• Line surge impedance $Z = 350\ \Omega$ (so the surge current is $3500/350 = 10$ kA)
• Transformer treated as an open circuit

Step 1: Steepness and travel time

$$\begin{aligned} S &= \frac{3500\ \text{kV}}{1\ \mu s} = 3500\ \text{kV}/\mu\text{s} \\ T &= \frac{L}{v} = \frac{30}{300} = 0.1\ \mu\text{s} \end{aligned}$$

Step 2: Voltage at the transformer

$$\begin{aligned} V_t &= V_p + 2ST \\ &= V_p + 2 \times 3500 \times 0.1 \\ &= V_p + 700\ \text{kV} \end{aligned}$$

The separation term alone (700 kV) exceeds the BIL of 500 kV. So whatever the arrester protective level $V_p$, $V_t > 500$ kV.

Check with a typical arrester

The arrester level is not given. Assume a typical 120 kV (132 kV system) arrester with residual voltage $V_p \approx 300$ kV at 10 kA.

• Sparkover time $t_s = 300/3500 = 0.086$ μs, which is before the reflection returns ($2T = 0.2$ μs).
• So the clamped wave doubles at the transformer: $V_t = 2V_p = 600$ kV.
• The simple formula gives $300 + 700 = 1000$ kV.

Both values are above 500 kV.

Conclusion: the transformer is NOT properly coordinated.

Distance needed (with $V_p = 300$ kV)

$$\begin{aligned} 300 + \frac{2 \times 3500 \times L}{300} &\le 500 \\ L &\le \frac{200 \times 300}{7000} = 8.57\ \text{m} \end{aligned}$$

(For a protective ratio of 1.2, the limit drops further: $(416.7 - 300) \times 300/7000 = 5.0$ m.)

Remedies

• Mount the arrester at (or within a few metres of) the transformer terminals.
• Reduce surge steepness with shield wires and a lower footing resistance near the substation.

Answer: $V_t = V_p + 700$ kV > 500 kV BIL, so not coordinated. With a typical 300 kV arrester, the LA must be within about 8.6 m of the transformer.