What is the definition of rated voltage of a lightning arrester?
As per IS 3070 the definition of the “rated Voltage of a Lightning Arrester” as given below.
The maximum permissible r.m.s. value of power-frequency voltage between line and earth terminal under temporary over-voltage conditions for a duration of 10 seconds as established in the operating duty tests.
In simple terms, the rated voltage of a lightning arrester is the maximum normal voltage that the arrester can handle without activation or allowing current to pass through. This voltage can be applied to the arrester for up to 10 seconds during temporary overvoltage conditions. The arrester is designed to remain inactive at this voltage and only work when there is a higher surge, like a lightning strike or switching surge.
For example, suppose an arrester has a rated voltage of 120 kV. It means the arrester remains inactive at voltages below this level and it will be inactive atleast for 10 seconds for this voltage. It activates only when the system voltage exceeds safe limits to protect equipment.
How to calculate the rated voltage of a lightning arrester?
This is calculated with the formula for effectively earthed system \[ V_r = \frac{\sqrt{2} \times V_{max}}{\sqrt{3}} \] For non – effectively earthed system \[ V_r = \frac{2 \times V_{max}}{\sqrt{3}} \] Where,
\( V_r \) = Rated Voltage of the arrester (kV)
\( V_{max} \) = Highest System Voltage (kV)
\( \sqrt{2} \) = Factor accounting for transient over voltages
\( \sqrt{3} \) = Conversion factor for three-phase systems.
Examples of Rated Voltage of Lightning Arrester
For effectively earthed 220KV system \[ V_r = \frac{\sqrt{2} \times 245}{\sqrt{3}} \approx 198KV \]
For effectively earthed 132KV system \[ V_r = \frac{\sqrt{2} \times 145}{\sqrt{3}} \approx 120KV \]
For non – effectively earthed 33KV system \[ V_r = \frac{2\times 36}{\sqrt{3}} \approx 42KV \]
What is Maximum Continuous Operating Voltage (MCOV) of a lightning arrester?
As per IS 3070 the definition of MCOV of a Lightning Arrester” as given below.
The highest power-frequency voltage that may be continuously applied between line and earth terminals of an arrester without causing it to conduct.
MCOV is the voltage level that a lightning arrester can withstand continuously without being activated for conduction. The Rated Voltage of an arrester can be withstood for only 10 seconds. In contrast, the Maximum Continuous Operating Voltage (MCOV) can be applied continuously without the arrester being activated.
How to calculate the Maximum Continuous Operating Voltage (MCOV) of a lightning arrester?
The Maximum Continuous Operating Voltage (MCOV) is typically 80-85% of the Rated Voltage of the arrester.
\[ MCOV = k \times V_r \]
Where, \( MCOV \) = Maximum Continuous Operating Voltage (kV)
\( V_r \) = Rated Voltage of the arrester (kV)
\( k \) = MCOV factor, typically 0.80 to 0.85 (as per IEC 60099-4 and IS 3070)
For examples, if the rated voltage of a lightning arrester is 42 kV: \[ MCOV = 0.85 \times 42 = 35.7 \text{ kV} \approx 36 \text{ kV} \]
If the rated voltage of a lightning arrester is 120 kV: \[ MCOV = 0.85 \times 120 = 102 \text{ kV} \]
If the rated voltage of a lightning arrester is 198 kV: \[ MCOV = 0.85 \times 198 = 168.3 \text{ kV} \approx 168 \text{ kV} \]
What is the Nominal Discharge Current of a lightning arrestor?
Nominal Discharge Current is the peak current that a lightning arrester can safely discharge without a permanent damage. It is rated with a standard 8/20 µs impulse wave. It is measured in kiloampere (peak) (kAp). The common range of this current is from 5 kA to 20 kA.
8/20 µs waveform means 8 µs is the time taken for the impulse current to reach its peak value. Whereas, 20 µs is the time taken for the current to decay to 50% of its peak value. This waveform simulates a typical lightning strike, as defined by IEC 60099-4 and IS 3070.
For example, suppose 10 kA nominal discharge current rating is given for an arrester. That means, the arrester can discharge 10 kA of surge current for an 8/20 µs impulse without damage.
What is the Line Discharge Class of a lightning arrestor as per IS 3070?
The Line Discharge Class of a lightning arrester, as per IS 3070, categorizes arresters based on their energy handling capability and discharge current levels during high-energy surges. It represents the ability of a lightning arrester to dissipate energy during lightning. The classification is based on the amount of energy and current the arrester can handle without degradation and damage. Line Discharge Classes as per IS 3070
| Line Discharge Class | Nominal Discharge Current (kA) | Application Voltage Range (kV) |
|---|---|---|
| Class 1 | 2.5 kA | Up to 66 kV |
| Class 2 | 5 kA | 66 kV – 132 kV |
| Class 3 | 10 kA | 132 kV – 220 kV |
| Class 4 | 20 kA | 220 kV – 400 kV |
| Class 5 | 20 kA and above | Above 400 kV |
What is Minimum Energy Discharge capability (KJ/KV) at rated voltage of a lightning arrester?
The Minimum Energy Discharge Capability of a lightning arrester refers to the amount of energy the arrester can safely absorb and dissipate per unit of rated voltage without failure. This value quantifies the durability of LA when exposed to repeated high-energy surges. A higher energy discharge capability means the arrester can withstand stronger and more repeated overvoltage events. For example, say an arrester has 7.5 KJ/kV, and its rated voltage is 120 kV. Then, the total energy discharge capacity is\[ 7.5 \times 120 = 900\; kJ \] This means the arrester can absorb 900 kilojoules of energy before damage.
| Line Discharge Class | Nominal Discharge Current (kA) | Minimum Energy Discharge Capability (kJ/kV) |
|---|---|---|
| Class 1 | 2.5 kA | 2.5 kJ/kV |
| Class 2 | 5 kA | 5 kJ/kV |
| Class 3 | 10 kA | 7.5 kJ/kV |
| Class 4 | 20 kA | 10 kJ/kV |
| Class 5 | 20 kA and above | 10 kJ/kV and above |
Suppose, a 120KV lightning arrester is of class 3. Let us calculate, the energy the arrester shall dissipate during a surge. \[ E = V \times I \times t \]
Where,
\( E \) = Energy discharge capability (J or kJ)
\( V \) = Rated voltage of the arrester (kV) = 120KV
\( I \) = Discharge current (kA) =10kA
\( t \) = Time duration of the surge (typically 8/20 µs, in seconds) = \( 20\times 10^{-6} seconds\)
\[ Therefore,\; E = 120KV \times 10kA \times 20\times 10^{-6} = 24 KJ \]
So, the LA can handle 900/24 or 37 repeated surge events of 10kA or 24KJ.
What is the High Current Impulse withstand for 4/10 micro second wave?
The High Current Impulse Withstand rating of a lightning arrester is the maximum peak current (kA) it can withstand when subjected to a 4/10 microsecond impulse wave without failing. 4 µs is the time taken for the current to reach its peak value.10 µs is time taken for the current to decay to half of its peak value. This waveform represents a more intense impulse wave than the 8/20 µs standard wave used for nominal discharge testing.