Calculate Lightning Protection for Building / Structure
March 12, 2016 26 Comments
Example: Calculate Whether Lightning Protection is required or not for following Building. Calculate No of Down Conductor for Lightning Protection
Area of Building / Structure:
- Length of Building (L) = 60 Meter.
- Width of Building ( W ) = 28 Meter.
- Height of Building (H) = 23 Meter.
Lightning Stock Flushing Density
- Number of Thunderstorm (N)= 80.00 Days/Year
- Lightning Flash Density (Ng)=69 km2/Year
- Application of Structure (A)= Houses & Buildings
- Type of Constructions (B)= Steel framed encased without Metal Roof
- Contests or Consequential Effects (C)= Domestic / Office Buildings
- Degree of Isolation (D)= Structure in a large area having greater height
- Type of Country (E)= Flat country at any level
- Maximum Acceptable Overall Risk Factor =0.00000001
Reference Table As per IS:2309 | |
Thunder Storm Days / Year | Lightning Flash Density (Flashes to Ground /km^{2}/year) |
5 | 0.2 |
10 | 0.5 |
20 | 1.1 |
30 | 1.9 |
40 | 2.8 |
50 | 3.7 |
60 | 4.7 |
80 | 6.9 |
100 | 9.2 |
Application of Structure | Factor |
Houses & Buildings | 0.3 |
Houses & Buildings with outside aerial | 0.7 |
Factories / workshop/ Laboratories | 1 |
Office blocks / Hotel | 1.2 |
Block of Flats / Residences Building | 1.2 |
Churches/ Hall / Theaters / Museums, Exhibitions | 1.3 |
Departmental stores / Post Offices | 1.3 |
Stations / Airports / Stadium | 1.3 |
Schools / Hospitals / Children’s Home | 1.7 |
Others | 1.2 |
Type of Constructions | Factor |
Steel framed encased without Metal Roof | 0.2 |
Reinforced concrete without Metal Roof | 0.4 |
Steel framed encased with Metal Roof | 0.8 |
Reinforced concrete with Metal Roof | 1 |
Brick / Plain concrete or masonry without Metal Roof | 1.4 |
Timber framed or clad without Metal Roof | 1.7 |
Brick / Plain concrete or masonry with Metal Roof | 2 |
Timber framed or clad with Metal Roof |
Contests or Consequential Effects | Factor |
Domestic / Office Buildings | 0.3 |
Factories / Workshop | 0.3 |
Industrial & Agricultural Buildings | 0.8 |
Power stations / Gas works | 1 |
Telephone exchange / Radio Station | 1 |
Industrial key plants, Ancient monuments | 1.3 |
Historic Buildings / Museums / Art Galleries | 1.3 |
Schools / hospitals / Children Homes | 1.7 |
Degree of Isolation |
Factor |
Structure in a large area having greater height | 0.4 |
Structure located in a area of the same height | 1 |
Structure completely Isolated | 2 |
Calculation:
Collection Area (Ac)=(L x W) + 2 (L x H) + 2(W x H) +(3.14 x H2)
- Collection Area (Ac) = (60×28)+2x(60×23)+2x(28×23)+(3.14x23x23)
- Collection Area (Ac) =7389 Meter2
Probable No of Strikes to Building / Structure (P)= Ac x Ng x 10^{-6} No’s / Year
- Probable No of Strikes to Building / Structure (P)= 7389x69x10^{–}^{6} No’s / Year
- Probable No of Strikes to Building / Structure (P)= 05098 No’s / Year
Overall Multiplying Factor (M) =A x B x C x D x E
- Application of Structure (A)= Houses & Buildings as per Table Multiplying Factor = 0.3
- Type of Constructions (B)= Steel framed encased without Metal Roof as per Table Multiplying Factor =0.2
- Contests or Consequential Effects (C)= Domestic / Office Buildings as per Table Multiplying Factor =0.3
- Degree of Isolation (D)= Structure in a large area having greater height as per Table Multiplying Factor =0.4
- Type of Country (E)= Flat country at any level so as per Table Multiplying Factor =0.3
- Overall Multiplying Factor (M) =0.3×0.2×0.3×0.4×0.3
- Overall Multiplying Factor (M) =0.00216
Overall Risk Factor Calculated (xc)= M x P
- Overall Risk Factor Calculated (xc)= 0.00216 x0.05098
- Overall Risk Factor Calculated (xc)= 000110127
Base Area of Structure (Ab) = (LxW)
- Base Area of Structure (Ab)=60×28
- Base Area of Structure (Ab)=1680 Meter2
Perimeter of Structure (P) =2x (L+W)
- Perimeter of Structure (P)=2x(60+28)
- Perimeter of Structure (P)=176 Meter
Lightning Protection Required or Not
- If Calculated Overall Risk Factor Calculated > Maximum Acceptable Overall Risk Factor than only Lighting Protection Required
- Here Calculated Overall Risk Factor is 0.000110127 > Max Acceptable Overall Risk Factor is 00000001
- Lightning Protection is Required
No of Down Conductor
- Down Conductors As per Base Area of Structure (s) =1+(Ab-100)/300
- Down Conductors As per Base Area of Structure (s) =1+(1680-100)/300
- Down Conductors As per Base Area of Structure (s) =6 No’s
- Down Conductors As per Perimeter of Structure (t)= P/30
- Down Conductors As per Perimeter of Structure (t)= 176/30
- Down Conductors As per Perimeter of Structure (t)= 6 No’s
- Minimum No of Down Conductor is 6 No’s
Results:
- Lightning Protection is Required
- Down Conductors As per Base Area of Structure (s) =6 No’s
- Down Conductors As per Perimeter of Structure (t)= 6 No’s
- Minimum No of Down Conductor is 6 No’s
Very useful datas
Thanks a lot. Could you explian about selection of air terminal , surge arrester, please?
Dear Sir,
Nice Information
Thanks Tushar Mahajan.
Hello, do you have any software to simulate the artistic lighting of bridges? – Thank you!
Any one interested to undertake insulation coordination study of HV andEHV Transmission Lines? If so please do contact. Though we performed in the past and performing at present, we wish to oursource this study, if find viable.
Dear Sir,
Thanks for the information. I humbly request you to provide me with a similar working on substations 220 KV and above DSLP calculation. urgently required for an interview. Please help
Maximum Acceptable Overall Risk Factor =0.00000001 value is fixed or may change as par building type?
Hi Sir, Maximum Acceptable Overall Risk Factor =0.00000001 value is fixed or may changed according to building type?
there is a missing table
you referred to 5 factors we already have 4 above
where is the table of E factor (type of country ) your choice was ( flat country at any level = 0.3 ) its not stated above where is the fifth table
before i also put a statement about the equation of voltage drop in wires and asked why we use (r cos phi + x sin phi ) instead of sqrt (( rcos phi) 2 + ( x sin phi ) 2 ) and still no answer
anyways thanks alot for your efforts
there is something make me confuse about the data input
in the start of the data input you put
Maximum Acceptable Overall Risk Factor =0.00000001 and this is 1 to the power -8
and after the example near the end you put
Here Calculated Overall Risk Factor is 0.000110127 > Max Acceptable Overall Risk Factor is 00000001
in the excel sheet we have from your site as you are the author of it you put the overall risk factor is .000001 that means only 1 to the power of -6
so please advice is it 0.00000001 or 0.000001
thanks for the efforts in this blog
Dear sir how to calculate the cable size of lightning arrestor ?
Sir, Thanks for your guidelines. One thing i need to confirm this is for rectangular, square types of building. How to calculate No of Down Conductor for Lightning Protection for circular building?
Is earthing of pure metallic structure required?
Dear sir,
1. Can we connect static earthing system and power [electrical] earthing system?
2. Which of the earthing pits [Power, instrument, lightning, static, telecommunication] we can interconnect without hazard?
Thank for your information.
very informative..
Could you explain about the rolling sphere method
Thank you for your information.
how to calculate the lightning protection calculation for tall metal structures? ex: Telecom towers
please give some suggestions.
what is the size of bare copper wire for 800 square meter area of a building and 10 meters in height?
Dear Sir ,
I would like to say thank you so much
your website very useful
God bless you
how much the height of building required lightning protection
Dear Sir
Thanking for your information kindly provide the information about number of lighting arrester requirement
sir this is sagar, I want to how many lights using in one bed room, L-50 MTR B-30 MTR H-40 MTR,, PLZ CAN U GIVE THE EXPLAIN THIS PROBLEM
Jinesh
Your notes compilation is excellent and I have downloaded that.
However I want to highlight that this calculation as per is 2309 is no longer applicable now since this IS has become obsolete now. And is replaced by ICE 62305.
Veena Gore
Dear Sir,
Thanks for these nice information. Sir, what size of cable must be used if my Down conductor increases and whether earth pits are required for each DC. And the distance between these earth pits if applicable.
Kindly do the needful.
Sunil.
how many lightening arrestor required if the area of the building is 500 m2
what will happen if earth pit of electronic equipment and telecommunication system is not seperated.