Regarding Marks question (1:11:25) on cable rating and why not use 90°C cable instead of 60°C cables. One thing that was not mentioned was that 90°C cables are only suitable for wiring to devices that are rated at 90°C. See Reg 512.1.5 (page 128 BGB): "Switchgear, protective devices, accessories and other types of equipment shall not be connected to conductors intended to operate at a temperature exceeding 70 °C at the equipment in normal service unless the equipment manufacturer has confirmed that the equipment is suitable for such conditions, or the conductor size shall be chosen based on the current ratings for 70 °C cables of a similar construction."
Cheers guys. Awesome video. I qualified as an electrician in the Army 9 years ago, but am a little out of practice. This was great as a refresher to get my head back into cable calcs.
Great videos as always, something I have noticed is you said that if overload is not a factor eg, not using Cf, then we use the equation without it, later on in the video when you use examples of an immersion element where oveload is not required, you still show the equation with Cf but just use Ib instead of In, could you clarify this please. Appreciate the effort you all put in to the videos given the current situation, am sure it is a strain on everyone in one way or another. Thanks
Hi Sparky, I'm currently in 4th year of university and I have a large project on to design the electrical installation for a 4 story health centre with office space on the top floor. I love doing some cable calcs but my main issue is in regards to a design project where you are unsure of exactly what equipment will be used, how do you suggest to carry out load estimations in order to figure out your Design current. There's a little guidance of average loads of certain devices in Brian Scaddan's Electrical Installation for designers and in the past in college I would have just for example chose a particular computer/cooker etc... but I just wondered if there was any good rules of thumb or a better system for the load/current estimations.
Great vid,im trying to find if you've done any vids on Calc's or Scenarios which involve on how to work out/calc from C.U to a Sub main for an out building,and, how to calc from different incomming supplies to see if a TT would be needed for the supply of Sub-main to an outbuilding. I cant seem to find anything on the web other than vids on software programs that seem to do all the work, it would be useful to know as to how it works. Much appreciated if you have time to do this.Thanks
@@SparkyNinja Thanks for replying so soon,im trying to look at how you would do a cable calculation from the main C.U. from the house to install a Sub-main to a Summerhouse for about a 4Kw load (for sockets and lighting). I dont know whether you would calc each individual load i.e working out the lighting circuit (applying the Method and correction factors) and then the same for the ring or radial for the sockect's. I have only managed to calc sofar based on the 4Kw load in total from the house to where the sub main would go would be a distance of 14m buried direct for 10m (2m internal from house to outside, then re-entering from outside to the summer house 2m). The overall distance and radius of the circuit would be 14m - from house to sub main 10m - for sockets (from sub main) 6m - lighting (from sub main) Im calc Ze =0.8 Ca - 1.04 Cd - 1.02 Cc - 0.9 20/1.04x1.02x0.9= 20.94 Bs7671 Installation method 70 ref method D I think im partially on the right track but i think im overthinking it. Im not sure if i need to break down the 4Kw into smaller calcs rather than looking at the overall. Again any help would be grate. Thanks again.
Hi guy's, great video again! Please could you clarify one piece of information - just add some clarity to the wording 'a circuit carrying less than 30% of it's grouped rating may be ignored' what does that actually mean?
thanks. I would like to ask a question. The calculations apply for a single dwelling. We use single core wire in conduits in wall and plaster. The supply given by DEWA is a 3-phase and then the load is balanced for each phase, it is really complicated here not to mention the HVAC and ventilation system. Here, in Dubai, we have high rises which have a sub main MDBs and DBs, power factor correction unit, switchgear. Basically, a substation on the ground floor. My question is, does the BS 7671 includes regulations for substation design? I actually bought the BS 7671 : 2018, the onsite-guide and guidance note 1: selection and erection to help me with the calculations of large projects
Hi Sorry to sound dumb here but if I have 10 single phase circuits each = 7.2kw connected into a 3 phase board how would you work out the Ib for the main cable feeding the board. if it was a single phase supply then the lb would be 313A but as the board is 3 phase this lb will be a lot less
Hello great video, could you clarify a query? At 42.24 in the video you select reference method E for the 1.5mm? Is this a mistake? Should it have been reference method C as discussed earlier (at 38.45) in this cable calculation?? I’m Confused 🤷♂️
I thought the same while working through the question which brought me to the comments....I can only assume it should have been 2.5mm @ 31A under ref method C...
Great video! but why is the correction factor calculation not... In/Ca*Cg=Ita rather than the one your teaching? That is what i'm currently been taught in college.
I have 2 questions. Firstly if you have a cable ran in a studwall between plasterboard and thermal insulation can we just used method 102 without applying a derating factor for thermal insulation?
Hi. A question about carrying out the adiabatic equation and PSCC, selecting t from either the graphs in BS7671 or manufacturers data I have a scenario where an MCCB passes on max Zs confirming the protective device will trip in the required time but the fault of negligible impedance is 900A which when on the x axis comes out as 8 seconds on the y axis. Do i use this number in the equation? i have been thrown off track as in my head i have the maximum permitted disconnection time of 5 seconds but this is saying it could take 8 seconds. TYIA
If I have a DB feeding multiple EV chargers all with a rating of 22kW would I assume simultaneous overload is liable (likely) to occur? My initial thoughts where I wouldn't as it is a fixed load, it shouldn't exceed the 22kw under normaly operating conditions
If you mean they all add up to 22kW then that would be fine. Some charging points will even allow you to limit the current demand depending upon the rest of the system.
1. The webinar is very good, however, I thought you would have included along with the adiabatic equation to confirm the minimum CPC size an example of phase to neutral fault showing the max time the fault could remain on the conductors; not all a max fault currents are 0.1 sec. Very good webinar all the same thank you.
The adiabatic is used for the purpose of protective conductor sizing and the regulations only goes is quick as 0.1 seconds. We do go in more detail to mention manufacturers more rapid times in another video.
Hi I’m currently doing a level CNG 2365 and I’m really struggling with continuity I understand it’s so we know there is a correct path for the circuit no interconnection + break in the cable. However il really struggling to get my head around the formula R1 + R2 thing and installation resistance any help on either would be brilliant thankyou ( iv bought insepection and testing book )
Great webinar, however based on the first question , Phil chose 2.5 mm square... And As you mentioned Iz must be larger than In so in that case Iz= 18*0.94*0.70 = 11.8 A which is smaller than In as In was 16 A so isnt it something to consider during cable calculation?
@@peterwalker1962 sorry I am in a learning stage aswell but just wondering 19.76 is tabulated current carrying capacity It Iz is current carrying capacity of the selected cable which is Iz=It X Ca X Cg which is 18X 0.94X 0.70 in this case hence 11.8A ?
krazylad84 where are you getting 18 from mate ? I,t is the value AFTER correction factors have been applied, Iz is before. 2.5mm cable Iz is 23 after applying Ca and Cg the It is 19.76. 👍🏻
@@peterwalker1962 Thanks Peter for the reply however my understanding is that "It" is tabulated value of the selected conductor which Phil chose from table 4D2A and for 2.5 mm it was 23A and I accidently wrote 18 sorry ... Iz is the current carrying capacity of the cable in the given condition which includes those factors such as groupings and Ambient temperature... Now to check whether the chosen cable which is 2.5mm with 23 A current carrying capacity is suitable once installed in the conditions such as groupings and Ambient temperature ... Thats when I used Iz = It * Ca* Cg Iz = 23 * 0.94 * 0.70 = 15.1 A So my argument was once the cable 2.5mm will be introduced in these factors conditions , It wont be carrying 23A but 15.A so it wont be suitable? Sorry correct me if I am wrong as I am only applying what my college text book says. I wish I could print screen here.
krazylad84 yeah It and Iz are quite confusing, I think where you are going wrong is you need to take Ib which is 13 and divide that by Ca and Cg not It*Ca*Cg
Excellent and valuable video , many thanks to take the time guys.
Regarding Marks question (1:11:25) on cable rating and why not use 90°C cable instead of 60°C cables. One thing that was not mentioned was that 90°C cables are only suitable for wiring to devices that are rated at 90°C. See Reg 512.1.5 (page 128 BGB): "Switchgear, protective devices, accessories and other types of equipment shall not be connected to conductors intended to operate at a temperature exceeding 70 °C at the equipment in normal service unless the equipment manufacturer has confirmed that the equipment is suitable for such conditions, or the conductor size shall be chosen based on the current ratings for 70 °C cables of a similar construction."
Good point and a mistake that could be easily made while thinking it would give increased safety factor
Superb, chaps - thank you 🙏
As far as soil thermal resistivity goes, I find plucking a figure out of thin air usually works.
Many Thanks. This video was very useful. Still revising for my 18th Edition. Exam centre will be open when things get better. That means more time..
This was very helpful especially the worked examples. Thank you
Nice one chaps, great video, it’s good to keep the grey matter ticking over during this time 🙏
Cheers guys. Awesome video. I qualified as an electrician in the Army 9 years ago, but am a little out of practice. This was great as a refresher to get my head back into cable calcs.
Great videos as always, something I have noticed is you said that if overload is not a factor eg, not using Cf, then we use the equation without it, later on in the video when you use examples of an immersion element where oveload is not required, you still show the equation with Cf but just use Ib instead of In, could you clarify this please. Appreciate the effort you all put in to the videos given the current situation, am sure it is a strain on everyone in one way or another. Thanks
Hi Sparky, I'm currently in 4th year of university and I have a large project on to design the electrical installation for a 4 story health centre with office space on the top floor. I love doing some cable calcs but my main issue is in regards to a design project where you are unsure of exactly what equipment will be used, how do you suggest to carry out load estimations in order to figure out your Design current. There's a little guidance of average loads of certain devices in Brian Scaddan's Electrical Installation for designers and in the past in college I would have just for example chose a particular computer/cooker etc... but I just wondered if there was any good rules of thumb or a better system for the load/current estimations.
How would you work out the Ca is you had an Ambient Temperature of 20 degrees?
Where do we see that cables in a trunking give a lower cable rating than if clipped I got lost there.i can understand why but where doni see this?
Thanks guys. This really helps after listening a few times.
Great vid,im trying to find if you've done any vids on Calc's or Scenarios which involve on how to work out/calc from C.U to a Sub main for an out building,and, how to calc from different incomming supplies to see if a TT would be needed for the supply of Sub-main to an outbuilding. I cant seem to find anything on the web other than vids on software programs that seem to do all the work, it would be useful to know as to how it works. Much appreciated if you have time to do this.Thanks
Hi Marc.
If you can explain for me exactly what it is you are looking to achieve in it, then I will see what I can do.
@@SparkyNinja Thanks for replying so soon,im trying to look at how you would do a cable calculation from the main C.U. from the house to install a Sub-main to a Summerhouse for about a 4Kw load (for sockets and lighting). I dont know whether you would calc each individual load i.e working out the lighting circuit (applying the Method and correction factors) and then the same for the ring or radial for the sockect's. I have only managed to calc sofar based on the 4Kw load in total from the house to where the sub main would go would be a distance of 14m buried direct for 10m (2m internal from house to outside, then re-entering from outside to the summer house 2m). The overall distance and radius of the circuit would be
14m - from house to sub main
10m - for sockets (from sub main)
6m - lighting (from sub main)
Im calc Ze =0.8
Ca - 1.04
Cd - 1.02
Cc - 0.9
20/1.04x1.02x0.9= 20.94
Bs7671 Installation method 70 ref method D
I think im partially on the right track but i think im overthinking it. Im not sure if i need to break down the 4Kw into smaller calcs rather than looking at the overall. Again any help would be grate. Thanks again.
Hi guy's, great video again! Please could you clarify one piece of information - just add some clarity to the wording 'a circuit carrying less than 30% of it's grouped rating may be ignored' what does that actually mean?
Great video! Thank you for your time! One question (so far) 18:00 . Table 4A2 flat cables t&e, shouldn’t this also include flat 3c&e?
thanks. I would like to ask a question. The calculations apply for a single dwelling. We use single core wire in conduits in wall and plaster. The supply given by DEWA is a 3-phase and then the load is balanced for each phase, it is really complicated here not to mention the HVAC and ventilation system. Here, in Dubai, we have high rises which have a sub main MDBs and DBs, power factor correction unit, switchgear. Basically, a substation on the ground floor. My question is, does the BS 7671 includes regulations for substation design? I actually bought the BS 7671 : 2018, the onsite-guide and guidance note 1: selection and erection to help me with the calculations of large projects
Hi Sorry to sound dumb here but if I have 10 single phase circuits each = 7.2kw connected into a 3 phase board how would you work out the Ib for the main cable feeding the board.
if it was a single phase supply then the lb would be 313A but as the board is 3 phase this lb will be a lot less
Hello great video, could you clarify a query? At 42.24 in the video you select reference method E for the 1.5mm? Is this a mistake? Should it have been reference method C as discussed earlier (at 38.45) in this cable calculation?? I’m Confused 🤷♂️
I thought the same while working through the question which brought me to the comments....I can only assume it should have been 2.5mm @ 31A under ref method C...
same
Great video! but why is the correction factor calculation not... In/Ca*Cg=Ita rather than the one your teaching? That is what i'm currently been taught in college.
If you're referring to the use of Ib, then it is with regards to the whether there is a need for overload protection.
Thanks for the video lads very appreciated
I have 2 questions. Firstly if you have a cable ran in a studwall between plasterboard and thermal insulation can we just used method 102 without applying a derating factor for thermal insulation?
Yes as the deration is included in the reference method. Assuming its installed to 102 and using T&E.
Great video guys, thanks a lot!
if 3 core cable the grouping factor 3 right?
Hi. A question about carrying out the adiabatic equation and PSCC, selecting t from either the graphs in BS7671 or manufacturers data I have a scenario where an MCCB passes on max Zs confirming the protective device will trip in the required time but the fault of negligible impedance is 900A which when on the x axis comes out as 8 seconds on the y axis. Do i use this number in the equation? i have been thrown off track as in my head i have the maximum permitted disconnection time of 5 seconds but this is saying it could take 8 seconds. TYIA
Many thanks guys - really useful - very clear explanation
how did you calculate the earth fault current is 245A? brushed over quite a key bit of the calculation there.
If I have a DB feeding multiple EV chargers all with a rating of 22kW would I assume simultaneous overload is liable (likely) to occur? My initial thoughts where I wouldn't as it is a fixed load, it shouldn't exceed the 22kw under normaly operating conditions
If you mean they all add up to 22kW then that would be fine. Some charging points will even allow you to limit the current demand depending upon the rest of the system.
Thanks for the video, Can anyone recommend a good work book with questions and and worked answers in electrical design? Thanks
This one is pretty good: www.amazon.co.uk/Electrical-Installation-Designs-Bill-Atkinson/dp/1119992842
1. The webinar is very good, however, I thought you would have included along with the adiabatic equation to confirm the minimum CPC size an example of phase to neutral fault showing the max time the fault could remain on the conductors; not all a max fault currents are 0.1 sec. Very good webinar all the same thank you.
The adiabatic is used for the purpose of protective conductor sizing and the regulations only goes is quick as 0.1 seconds.
We do go in more detail to mention manufacturers more rapid times in another video.
Hi I’m currently doing a level CNG 2365 and I’m really struggling with continuity I understand it’s so we know there is a correct path for the circuit no interconnection + break in the cable. However il really struggling to get my head around the formula R1 + R2 thing and installation resistance any help on either would be brilliant thankyou ( iv bought insepection and testing book )
Did you work it out in the end?lol
Brilliant thanks,
Where did you get 11.5 voltage for voltage drop on that initial equation?
That the max volt drop allowed, which is 5% of 230
@@sachb3374 3% for a lighting circuit.
@@chrissmith7259 6.9V
Great video thank you.
how can I calculate volt drop on a ring circuit...
Iet design guide gives an equation lvd= 4xvdx1000/mvam x ib x ct
Great webinar, however based on the first question , Phil chose 2.5 mm square... And As you mentioned Iz must be larger than In so in that case Iz= 18*0.94*0.70 = 11.8 A
which is smaller than In as In was 16 A so isnt it something to consider during cable calculation?
krazylad84 i thought Iz was 19.76 amps On question 1 ?
@@peterwalker1962 sorry I am in a learning stage aswell but just wondering 19.76 is tabulated current carrying capacity It
Iz is current carrying capacity of the selected cable which is Iz=It X Ca X Cg which is 18X 0.94X 0.70 in this case hence 11.8A ?
krazylad84 where are you getting 18 from mate ? I,t is the value AFTER correction factors have been applied, Iz is before.
2.5mm cable Iz is 23 after applying Ca and Cg the It is 19.76. 👍🏻
@@peterwalker1962 Thanks Peter for the reply however my understanding is that "It" is tabulated value of the selected conductor which Phil chose from table 4D2A and for 2.5 mm it was 23A and I accidently wrote 18 sorry ...
Iz is the current carrying capacity of the cable in the given condition which includes those factors such as groupings and Ambient temperature...
Now to check whether the chosen cable which is 2.5mm with 23 A current carrying capacity is suitable once installed in the conditions such as groupings and Ambient temperature ...
Thats when I used Iz = It * Ca* Cg
Iz = 23 * 0.94 * 0.70 = 15.1 A
So my argument was once the cable 2.5mm will be introduced in these factors conditions , It wont be carrying 23A but 15.A so it wont be suitable?
Sorry correct me if I am wrong as I am only applying what my college text book says. I wish I could print screen here.
krazylad84 yeah It and Iz are quite confusing, I think where you are going wrong is you need to take Ib which is 13 and divide that by Ca and Cg not It*Ca*Cg
Folks, any circuit can potentially get overloaded if some cowboy adds a socket or something to the existing circuit in the future
green to green, brown to brown and blue to bits...
Just re watched this sorry im a nightmare lol
C max that's new!
Can’t believe you made an error……….. Penguins in the Artic? 🤔😉
😯