This is the best video I've ever seen on the subject. Finally, something I can give to trainees that they stand a hope of understanding. Thanks hugely!
I didn't realise that if i run the same size cpc (16mm) alongside a 2c x 16mm SWA it need not be calculated. Thats almost 30 mins of my life I won't get back. Good video though. Never had luxury information like this in the late 70's, struggled like hell at college in the days when the lecturer guy would clip you round the head if you didn't get it right first time.
I'm trying to brush up on my cable calculations and this video is concise and clear. It answers my question re. plucking a t value out of the table on the time current characteristics for the adiabatic calculation. Not easy information to come by it seems. Shame it has so few viewings/ thumbs up. It makes me think electricians are familiar with the theory already or they ignore it... Mmm, I wonder which?
Many thanks, the values are all rounded up/down. But thankyou all the same - hopefully you find it useful, please share with like minded people or maybe even like and subscribe.
What if your designing the installation before it is done and don’t have the Zs value? Can you not work out the fault current by checking which breaker you use for instance c50 would require 10 times the rated current. Or b50 would require 5 times the rated current
Hi All new buildings are designed (electrically) before the buildings are built and therefore the electrical supply has been installed, assumptions based on experience have to be made or you take the maximum values of Ze as supplied from the DNO (you would be required to tell them what size supply you would like for the new build). Hope that helps it is a big subject, Please share with like minded people or maybe even like and subscribe.
No you use Uo as it is Zs=(Uo X 0.95{Cmin})/Ia - this is used to find the Max Zs of protective devices. Hope that helps. Please share with like minded or maybe even like and subscribe.
Hi following on from Zs assistance ( Thank you) If Zs is 0.396 ohms Then If is 230/0.3960 = 580.81A If we apply the equation as above With a 0.1 disconnection time we end up with 183.67/115 = 1.597 mm This exceeds the 1.5mm CPC selected (2.5/1.5 was selected ) Would cable choice be 2.5mm/2.5mm
Check what your protective device log says about the fault current & how quick it operates. When you divide your viltage by Zs, check your answer against the current requirement for disconnection to find your t value in the equation.
This is a great video, I have one query - what happens to the I^2 value if the circuit breaker or fuse is current limiting. Does the device operating in 0.1 second not limit the fault current in the circuit to the value shown in the table? Apologies if this has been asked already but I’ve read 2 books to revise for a 2396 exam a they don’t seem to answer the question.
Thankyou, please share with like minded people. Protective devices do not limit current - on the contrary they will disconnect quicker (for a thermal device that is) the more current that passes through (let through energy = I^2xt). It will be the ohmic value that will limit current. Hope you liked and subscribed. Again many thanks.
The limiting of the current is being done for the conductors by the disconnection of supply before the fault current can damage the cables. If your exam is on the 13th, I’ve been told to brush up on your earth & bonding knowledge. & there might be a LV transformer diagram to draw. Don’t know which one tho. All this is rumours of course. Good luck.
Hi Because the fault current that flows will disconnect the protective device in 0.1Second (see graph App 3) Hope that helps, Please Like, Share & Subscribe.
It's a circuit breaker type B & C will disconnect in 0.1 sec if they meet the design fault current threshold see this video to help your understanding: th-cam.com/video/caWAevix9a0/w-d-xo.html
i am very baffled about his formula...so...in order to ascertain fault current we would need to have an existing earth conductor...so why are we calculating the size when we already install the circuit...in reality we would have to ascertain the fault current...and thats only possible for the distribution circuit...could u correct me if im wrong
You would be required to design the circuit before installation, this will ascertain if the designed cable is adequate before going to the expense of buying and then installing the wrong cable. Its all in the design. hope that helps.
Please Like Share & Subscribe. If you use 5 seconds this will show the conductors heating up in the longest disconnection time (presuming you circuit does disconnect at 5 seconds). Presuming your fault current is accurate and it does disconnect in the time indicated and the size determined is larger than the design size, then all will be good. Or you could determine the actual fault current based on your design and calculate the fault current to determine the size of conductor. Choice is yours - all academic if the conductors are oversized.
Could you tell me if it is acceptable to use this equation to ascertain wether or not an existing 6mm csa bonding conductor on an TNS supply system is adequate?
This is the best video I've ever seen on the subject. Finally, something I can give to trainees that they stand a hope of understanding. Thanks hugely!
Many thanks, I have produced it for the guys I teach, I just need them to look at it... Please share like and subscribe.
hello sir, im curious, why you choose .1 for disconnection time? not 5 seconds? can you help to explain sir or anybody know
@@thesilentgamer4210He uses the figure in the chart and lines it up in the graph against seconds.
Without a doubt the best detailed explanation of this I have seen. Thanks!
You're very welcome! Please Like, Share and Subscribe.
@@sparkyhelp3997 Done 👍👍
I didn't realise that if i run the same size cpc (16mm) alongside a 2c x 16mm SWA it need not be calculated. Thats almost 30 mins of my life I won't get back.
Good video though. Never had luxury information like this in the late 70's, struggled like hell at college in the days when the lecturer guy would clip you round the head if you didn't get it right first time.
Glad you liked it, happy days at college...
I'm trying to brush up on my cable calculations and this video is concise and clear. It answers my question re. plucking a t value out of the table on the time current characteristics for the adiabatic calculation. Not easy information to come by it seems. Shame it has so few viewings/ thumbs up. It makes me think electricians are familiar with the theory already or they ignore it... Mmm, I wonder which?
Many thanks, glad it helped.
Please Like, Share & Subscribe - maybe the viewing figure will go up in time!
Great video, very clear and precise only thing I spotted and can be quite confusing 9:27 there’s a typo in the calculation
Many thanks, the values are all rounded up/down. But thankyou all the same - hopefully you find it useful, please share with like minded people or maybe even like and subscribe.
very simple to follow, thank you
You are welcome! Please Like Share & Subscribe.
Great videos thanks alot
Glad you like them! Please share with like minded or maybe even like and subscribe.
What if your designing the installation before it is done and don’t have the Zs value? Can you not work out the fault current by checking which breaker you use for instance c50 would require 10 times the rated current. Or b50 would require 5 times the rated current
Hi All new buildings are designed (electrically) before the buildings are built and therefore the electrical supply has been installed, assumptions based on experience have to be made or you take the maximum values of Ze as supplied from the DNO (you would be required to tell them what size supply you would like for the new build). Hope that helps it is a big subject, Please share with like minded people or maybe even like and subscribe.
Awesome
Many Thanks, Glad you liked it, please share with like minded people or like and subscribe.
Great video. Thanks
Glad you liked it please like, share and subscribe. Thankyou.
At 4:40 when calculating the fault current, wouldn't you multiply Uo by Cmax?
No you use Uo as it is Zs=(Uo X 0.95{Cmin})/Ia - this is used to find the Max Zs of protective devices. Hope that helps. Please share with like minded or maybe even like and subscribe.
Hi following on from Zs assistance ( Thank you)
If Zs is 0.396 ohms
Then If is 230/0.3960 = 580.81A
If we apply the equation as above
With a 0.1 disconnection time we end up with 183.67/115 = 1.597 mm
This exceeds the 1.5mm CPC selected
(2.5/1.5 was selected )
Would cable choice be 2.5mm/2.5mm
Yes it is indeed, but you would need to verify it as the fault current would go up wit ha 2.5/2.5
Check what your protective device log says about the fault current & how quick it operates.
When you divide your viltage by Zs, check your answer against the current requirement for disconnection to find your t value in the equation.
This is a great video, I have one query - what happens to the I^2 value if the circuit breaker or fuse is current limiting. Does the device operating in 0.1 second not limit the fault current in the circuit to the value shown in the table?
Apologies if this has been asked already but I’ve read 2 books to revise for a 2396 exam a they don’t seem to answer the question.
Thankyou, please share with like minded people. Protective devices do not limit current - on the contrary they will disconnect quicker (for a thermal device that is) the more current that passes through (let through energy = I^2xt). It will be the ohmic value that will limit current. Hope you liked and subscribed. Again many thanks.
The limiting of the current is being done for the conductors by the disconnection of supply before the fault current can damage the cables.
If your exam is on the 13th, I’ve been told to brush up on your earth & bonding knowledge.
& there might be a LV transformer diagram to draw. Don’t know which one tho.
All this is rumours of course.
Good luck.
@@NaciMahmut-rf5xf Good Luck
At 6:55 in table it says for 0.1 to 5 secs. Why do we pick 0.1 secs ?
Hi Because the fault current that flows will disconnect the protective device in 0.1Second (see graph App 3)
Hope that helps, Please Like, Share & Subscribe.
And what fault current gives 5 seconds . Thats what confuses me. Table says 0.1 to 5
It's a circuit breaker type B & C will disconnect in 0.1 sec if they meet the design fault current threshold see this video to help your understanding:
th-cam.com/video/caWAevix9a0/w-d-xo.html
i am very baffled about his formula...so...in order to ascertain fault current we would need to have an existing earth conductor...so why are we calculating the size when we already install the circuit...in reality we would have to ascertain the fault current...and thats only possible for the distribution circuit...could u correct me if im wrong
You would be required to design the circuit before installation, this will ascertain if the designed cable is adequate before going to the expense of buying and then installing the wrong cable. Its all in the design. hope that helps.
I'm sizing for a distribution circuit , do I not use standard 5 secs or do what you have done with app 3 ?
Please Like Share & Subscribe.
If you use 5 seconds this will show the conductors heating up in the longest disconnection time (presuming you circuit does disconnect at 5 seconds). Presuming your fault current is accurate and it does disconnect in the time indicated and the size determined is larger than the design size, then all will be good. Or you could determine the actual fault current based on your design and calculate the fault current to determine the size of conductor.
Choice is yours - all academic if the conductors are oversized.
Why are you going across at 6:34?
Nvm i know now
Hi, Glad it has helped, and you figured it out.
Could you tell me if it is acceptable to use this equation to ascertain wether or not an existing 6mm csa bonding conductor on an TNS supply system is adequate?
HI see this video - as bonding is partly sized in accordance with the incoming supply size - th-cam.com/video/TkL8vBAeC_c/w-d-xo.html