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Late reply, the answer is yes, you will find major part of the course that includes the short circuit protective devices, the conductors, feeders, and disconnect switches design using NEC 2017 and NEC 2023.
this is EXACTLY what I needed. Basically a full course of the concept. Its so hard to find any proper video on this. Thank you my good man. I owe you a beer.
I tried to find the solution which ends here. Thank you do much fir such nice explanation . Also suggest if possible go get the PPT which you shown here for future
Mr For earth termination system The value of their resistance Is it 10 ohm maximum or minimum or exactly 10 ?? Same for the distance between to earthing conductors
A Slam dear sir Hope your fine shine.best teacher ever in my life i learn Practical Engineering From you sir.Allah bless u .and once again Very informative video for students thank u dear sir
NICE EXPLANATION BUT I HAVE A QUESTION WE CALCULATE CROSS SECTION AREA OF DOWN CONDUCTOR FROME EQUATION a(mm2) = 9*SQRT(T)*Isc ?? AND WHAT IS THE VALUE OF SHORT CIRCUIT CURRENT ( I SEE ITS 3 KA MIN AND 200 KA MAX FOR TYPE I) , THANKS ALOT :D
First of all very good video. But How many lighting protection earthing pit required in a building? Its not confirmed at your video? Can you please explain it.
Hi Ahmedmahdy, informative video. refer to time 23:00, distance between 2 rods =2 x sqrt( 2RH-H2) basically this distance gives angle of protection, ie, 2 times radius of protected area. so from the graph, if we select height and angle, with equation H*Tan(alpha) we can find protected area radius. Then what is significance of radius of rolling sphere? Suppose we take same height H from graph, and take radius of rolling sphere (R) from protection class table (given at left of graph), and apply in formula, 2 x sqrt( 2RH-H2), we should get same value which is protected area radius. But when i tried both ways the result is2 different value. Can you have a look and clarify it?
I have one question I m from India, Good information👌, Brother if the building height is morethan 60mtr height means how we are going to protect building for side flash ? Top 20% of the building height for side flash protection is it sufficient ?
Hello, I can tell you about it. The design on Burj Khalifa, 832 mtrs- it is above lower cumulus cloud cover where thunderstorm occurs. We had to provide steel 50X6mm strip as equipotential bonding all across , going round n round , connected to steel core, that is earthed. For more details www.ampereprotection com
Dear sir.. Have a nice day .. please and kindly , can you send me the PDF lecture ? I would like to make a video in Arabic Lang , please your permission for this.. Many thanks' for you in advance.
🌟 Access 100+ courses, 1,600+ videos, and 300+ hours of professional lessons!
Master electrical engineering, renewable energy, and much more with Khadija Academy Student Membership. Learn real-world skills through expertly designed courses on topics like:
✅ Solar Energy, Electrical Design, MATLAB Simulink, Power Electronics, Wind Energy, Wave Energy, and more!
🔗 Explore it all here: www.khadijaacademy.com/p/khadija-academy-student-membership
💡 Your future starts here-click to join now!
I am interested in taking the course. My intension is to design according to NEC. Let me know if your course follows the NFPA70 & NFPA780.
Late reply, the answer is yes, you will find major part of the course that includes the short circuit protective devices, the conductors, feeders, and disconnect switches design using NEC 2017 and NEC 2023.
this is EXACTLY what I needed. Basically a full course of the concept. Its so hard to find any proper video on this. Thank you my good man. I owe you a beer.
You're welcome 😂💪💪
Excellent, Really appreciated the way you are presenting 🎁, I ❤️ to in touch with you
This is great.
Need your contact number?
This is so far the BEST video on LPS. Thanks for putting up this video.
You're so welcome!
Thank you for sharing this essential courses
I tried to find the solution which ends here. Thank you do much fir such nice explanation . Also suggest if possible go get the PPT which you shown here for future
Very nicely explained... If u include iec references for all tables / formulas then it will be 100%perfect...
Did you get them you help me out
Superb explanation for risk assessment.
I love it from the very beginning to the end. Thank you very much.
You are welcome 🙏 happy to help
Thanks for the clear explanation.
Mr
For earth termination system
The value of their resistance
Is it 10 ohm maximum or minimum or exactly 10 ??
Same for the distance between to earthing conductors
A Slam dear sir Hope your fine shine.best teacher ever in my life i learn Practical Engineering From you sir.Allah bless u .and once again Very informative video for students thank u dear sir
Thank you Engr Luqman for your kind words :)
Wonderful explanation. Thank you for sharing this valuable information with so much of ease. May you be blessed.
With affectionate regards
You are welcome 😍❤❤
Thank You Sir ! Love from Nepal.
Most welcome!
Which textbook you use for this formulae and presentation?
what is the difference b/n radius protection and radius b/n rods
i have issues understanding the amount of lighting rods required after i got the distances between two rods ,
Congratulations. Excellent presentation!
Hello any video for earthing system ???? To bring values below 1 ohm
Excellent explanation thanks
How we get the map for finding the value of Ng ? Please share the Webpage
Whats standard did you use? I think the risk assessement by IEC 62305 is not so simple.
NICE EXPLANATION BUT I HAVE A QUESTION WE CALCULATE CROSS SECTION AREA OF DOWN CONDUCTOR FROME EQUATION
a(mm2) = 9*SQRT(T)*Isc ?? AND WHAT IS THE VALUE OF SHORT CIRCUIT CURRENT ( I SEE ITS 3 KA MIN AND 200 KA MAX FOR TYPE I) , THANKS ALOT :D
Thank you very much.
You are welcome anytime ☺🙏
For finding C1, what does it mean “within a distance of 3H”? Is it the distance between the highest building and the building we are designing?
Just what I needed. Thank you Sir.
Great! Specialized!
Thank you ☺🙏
High quality video!
Thank you 🤗
could you inform me what standards you used for this subject?
Thank you so much for the valuable information. Excellent 💯
Appreciated your efforts. Cheers.
Recommended to engineers.
Thank you for your really nice comment, really appreciate it :)
Is this as per IEC 62305 ?
in Nd formulae why we are multiplying 10^-6
Hi Friend , does the (H) here means the length of the rod or it means the length of the rod plus the building height
Brother, you are making really good videos.
That is so nice from you :) I try to do my best
Thank you for your sharing... May God bless you ❤🌼
You are welcome :)
Very good video, thank you!
You are welcome :)
First of all very good video.
But How many lighting protection earthing pit required in a building? Its not confirmed at your video? Can you please explain it.
Which standard did you refer?
Great job bro 👏👏👏👏👏👌👌👌👌
You are welcome brother
Nice explanation, can you do presentation for ELV system
What is corresponding standard you referring IEC or NFPA ?
Based on NFPA 780
Excellent information. Thank you.
You are welcome :)
Sir can you upload lightning calculations example video?
Sir how to choose class??
Very nice video
Nice explanation. Is this based from the IEC 62305? Thanks.
This is not based on IEC 62305 which is very complicated
Excellent job Mr. Can you send the this Presentation . It is better.
Where do I get the map of the average flash density in the region per year ?
Hey Tako, you can find it here: www.weather.gov/pub/lightningFlashDensityMaps
How can you derived the Collective area?
LPS effciency is 1 - Nc/Nd??? or 1-Nd/Nc?
Good class sir
Thank you
Hi Ahmedmahdy, informative video.
refer to time 23:00, distance between 2 rods =2 x sqrt( 2RH-H2)
basically this distance gives angle of protection, ie, 2 times radius of protected area.
so from the graph, if we select height and angle, with equation H*Tan(alpha) we can find protected area radius.
Then what is significance of radius of rolling sphere?
Suppose we take same height H from graph, and take radius of rolling sphere (R) from protection class table (given at left of graph), and apply in formula, 2 x sqrt( 2RH-H2), we should get same value which is protected area radius.
But when i tried both ways the result is2 different value. Can you have a look and clarify it?
Very informative video. Thank you
Can I get your ppt you explained from it ?
WHAT IS YOUR SOURCE TO SUPPORT THIS PLEASE SPECIFY, Just for clarifications and standards ;-)
thanks
❤🙏
Great explanation! Highly appreciating your effort, mate.
Btw, is this video included in your above course? Thanks
Hi Mobashsher, I didn't add this to the course, I love sharing knowledge through my channel and my courses.
Thank you. GOD bless you.
Excellent
Thank you :)
can I have this PPT?
I have one question I m from India, Good information👌, Brother if the building height is morethan 60mtr height means how we are going to protect building for side flash ?
Top 20% of the building height for side flash protection is it sufficient ?
Hello, I can tell you about it. The design on Burj Khalifa, 832 mtrs- it is above lower cumulus cloud cover where thunderstorm occurs. We had to provide steel 50X6mm strip as equipotential bonding all across , going round n round , connected to steel core, that is earthed. For more details www.ampereprotection
com
Dear sir.. Have a nice day .. please and kindly , can you send me the PDF lecture ? I would like to make a video in Arabic Lang , please your permission for this.. Many thanks' for you in advance.
Thanks
You are welcome anytime :)
Hello brother , Can you give me your email i have a very important question regarding simulink , i hope this message reaches you ...
You can contact me at ahmedmahdy@khadijaacademy.com
nice content
want to be friend?