Thanks for watching, please share with like minded people. Voltage drop is set in the uk as 3 & 5% depending on the circuit. Or do you mean mV drop for cable? In which case use resistances of the cable (from manufacturer or standard table) to work out volt drop, but all manufactures of cables will publish technical literature for their cables. Hope that helps Please like share and subscribe.
Hello I am a follower of your lessons and a big fan of them. I hope that you will show videos explaining the method of calculating the power factor, as well as the correct way to calculate the main breaker, and whether it should be larger than the service provider’s breaker, equal to it, or less in size than it. sincerely Khalid
You explained cable grouping excellently. In effect this is worse than thermal insulation put over cables as each cable can be at 70 degrees so when grouped they are heating each other potentially if they are at full load.
from Jordan , i am truly thankful for you to sharing this valuable guides with us and looking forward to seeing more about circuit breakers and fuses selection too according to bs 7671
This is a very well put video ( and great channel that I have just come across ). I am now designing electrical circuits as part of my job and something I never thought about, until now is - when we design, as you in this video with the step by step guide, how can we be sure that our ZS when verifying after installation will be acceptable? Thanks - great wok. 😊
You need to elaborate on your use of Ib & In in your voltage drop & It calculations. Both can be used depending on the nature of your circuit. If the circuit can be overloaded, you would use the In value. If for a fixed load, you would use Ib.
This is True, but in this video it has not been covered. One for another video maybe. If I keep adding new ideas the videos are too long, so I have to stop somewhere. But thank you.
Great explanation but a slight correction on the way you represent numbers in standard form in your calculations e.g. 2.8mV should be represented as (2.8 * 10^-3)V instead of (2.8^-3)V which is actually 4.6mV
Yes I know but a pain to go back and add an addendum now. Well done for spotting it though. My copy has already been corrected. Please Like, Share & Subscribe.
That is good question, the same principles apply, you would just need to find the actual current carry capacity of the type of cables used from the manufacturer. Hope that helps.
Thanks for sharing. In real life, I wonder if electricians do actually do all these calculations, especially for Ca, Cg and voltage drop. Normally, rules of thumb are applied, so for example 20A mcb equals 2.5mm2 cable, regardless of Ca, Cg and voltage drop. Just like to hear what real life electricians do or do not do in practice.
A good designer / electrician will take the method of installation and type of cable into consideration. Obviously we can all choose not to take these into consideration, but we are hoping things do not go wrong (and they may not, most of the time) - remember BS7671 can be used in a court of law to provide evidence. The OSG has tables to take some of the design out of installations - you just need to stay in those parameters.
You explain much better then my college teacher that he is supose to be an Engineer, thanks a lot for making this explanation!! If i could I would marry you but unfortunately I'm straight 😄
Thanks for th video, can you confirm that in cases where there is no possibility of overload in respect to usage you would use "Ib" (design current) rather than "In" (.size of protection) when calculating for "It". (tabulated current) e.g Ib/Cg x Ca x Ci etc
Thank you very much, although some tables are not complete and also in 12:01 the text ended with " size is " and the rest of words are not there. would you please help us with your reference to let us get the complete knowlege and also help those who are making a research on that topic to get soft copy of the document.
Hi the reference material used is the 'IET Wiring Regulation 2018', but a new one is out now dated 2022 Amendment 2. Hope it helps please Like, Share & Subscribe.
Hi the Cable is copper, calculations can be carried out for Aluminium, but the current carrying capacity of aluminium is lower than copper. Please Like, Share & Subscribe.
How about multicore cables like SOCAPEX? 6 circuits on 1 multicore cable. Does that mean that a 2.5mm SOCA cable is downrated from 16A to 9A per circuit just by being bunched up??
Hi good question, that would be a little more detailed to answer as you would need to read the regulations further. you would also need to know the % load of each circuit as it may be neglected in in certain circumstances. Have fun determining the rating - happy designing. Please Like, Share & Subscribe.
@@sparkyhelp3997 ah ok. So it depends on the load for each circuit? I'll do some more research. Love the channel it has helped me a lot since I discovered it! Cheers
Thanks for your sharing . Could you help me to advice that when we calculated the Design Current Ib , the power factor is required to include or can omit as this power factor make the result of the Design Current into 1.25 Time when the power factor is included in Design Current Calculation Step .
If there is a power factor it must be included in the calculation as the design current will be higher, and therefore the cable will have to carry the extra current. Hope this helps.
Hi, 60.79A is the It rating for the cable not the circuit breaker. The Load has remained unchanged, therefore the Circuit Breaker is still acceptable. Please Like, Share & Subscribe.
Hi Sparky, doing an electrical installation design for a health centre at the moment and I'm curious in regards to design current. What would I use for the design current for sockets where I'm not really expecting much permanently connected equipment? such as a changing room/circulation areas. The protective device would still have to be 32A or 20A for radials to comply with appendix H surely? even though the load will never reach anywhere near this? The client(lecturer) hasn't really provided any guidance on what electrical devices are intended for use. I mean I can spec for a typical industrial hoover being plugged in & a few mobile chargers but current is something negligible like 7A for example but Appendix H states it would have to be on a 20 or 32A breaker for radials? Maybe I'm missing something or maybe 20A or 32A radials with negligible load is fine? Hope to hear back from you soon, submission is in 2 weeks lol.
Hi Part H of OSG explains that diversity has already been taken into account from historical usage. keep the floor area as per 100m^2 for 32A (A1) RFC providing you do not know the load and the load is not reasonably expected to rise above 32A at any one time. So what I would do is measure the floor area of the design and provide enough A1, A2 or A3 circuits as per the floor area. Areas to give special consideration would be rooms such as kitchens, utility rooms or any with room with a lot of appliances which could be on at the same time. Remember diversity is not an exact science (just read the first couple of paragraphs from OSG maximum demand and diversity). Good luck - Please like, share and subscribe. Many thanks.
@@sparkyhelp3997 Thanks alot mate, after a bit more digging around I seen I should be Using my cpd rating as design current Ib=In I'm only using radials, no rings. I ended up with something like 31 seperate socket circuits (4 Storey building with 8 on bottom 3 floors and 7 on the 4th) with almost all except for 2 or 3 being A2s 75m squared radials on 32A cpds and the other 2 or 3 Being A3s 50m squared on 20A cpds, haven't fully specced into the cabling yet as it must've took me about 3/4 hours of looking back and forth between the model and drawing my circuits on CAD. I also have to wait for my team members to feed back to me info regarding the lighting and LV stuff such as CCTV and security systems and fire alarms. That way I can get my groupings. I only noticed today when reading appendix A that A1 is for your initial circuits and A2 is diversity to be applied after all circuits have been spec'd to get your main incoming fuse size. I swear I done my college HND coursework from 3 years ago wrong because I'm sure I was applying A2 diversity to my lighting d'oh. I've recently gotten a job as an electrical engineering consultant so it's key I keep refreshing my knowledge and keeping up to date. Thanks again for your help mate!
Can you explain why you didn't consider power factor 17:55? For single phase it is V*I*COSPHI rigth ?.....how much load can single phase carry? For beyond 8kw it should be 3phase right ? @sparkyhelp
Power Factor only applies if the load is inductive or Capacitive. If power factor is not given (maybe for example an electric heater) it is more than likely a resistive load and therefore power factor is 1.
Hi the conductor temperature is 70 degrees, the ambient temperature is the surrounding air temperature. Put the 0.94x0.7 in brackets 40/(0.94x0.7)= Hope this helps. Please like Share & Subscribe - Good luck
@@sparkyhelp3997 sorry I ment is ambient temperature 30 degrees if wrapped in thermal insulation or would it be greater Na still not getting the calcs but il try work it out lol
After calculating the cable size and when attempting to calculate the short circuit current am I correct in saying you need to calculate the resistance and reactance of the cable which is separate from the values in the voltage drop calculations? Do you know where to find the tables for resistance and reactance values? The do not appear to be in BS7671
For the vast majority of circuits reactance of the cable will be ignored as it is such a small value it would be insignificant. If you are talking long distribution or transmission cables then the information from BS7671 would not cover this. BS7671 covers low voltage installations. I would suggest seeking information from the cable manufacturer (this is where I go to find the resistance of cables or armouring of cables - BS7671 does not provide resistance of conductors, the On-Site-Guide does but only up to 50mm^2) . I hope this goes some way to answering your question, but the answer is not that straight forward as we could write an entire paper on this subject.
Hello and good day to you! I have been looking on all the internet, TH-cam, and I just happen to come a cross your video, my question (if you can able to help me in this little matter!!) I just bought a Fish Finder, and I just want to splice the cable to make it longer, the cable's of the unit they are very thin, could i able to spliced it with a bigger gauge,??? so I can able to reach the 12v Battery. thank you for your expertise, and for your little help on this!! Sincerely the Dummy.
In order to do this you will have to know the current passing through the cable (it will be small). You could just put a larger cable on (it should not cause a problem - just may look unsightly) and try it. A bit of trial and error, just means your warranty will be invalid. Good luck, hope this has helped in some way.
It is from the wiring regulations IET 18th Edition (used in the uk), 3% or 5% of the nominal voltage for particular circuits. Hope that answers your question.
sorry to sound dumb here but if I have 10 single phase circuits each 7.2kw and I wanted to find the design current for a sub main to feed a board and the supply is 3 phase how do I work that out as obviously I don't just add all 10 circuits up (72kw) as you would if the supply was single phase.
Good Question. Assuming it will be split over the three phases, you could divide by three, (this is crude - approx. 100A per phase), or design which phase you want them to be on to be more accurate. If you want to design the sub-main, you may with your more in-depth knowledge what these circuits are, you may want to apply diversity. Really hope that helps.
@@sparkyhelp3997 they are for EV single phase 7.2kw car chargers so can't use diversity. I looked at it as I have 10 circuits each 7.2kw on a 3 phase DB so that's 4 + 3 + 3 so did 4 x 7.2kw = 28.8kw power factor of 0.95 which gave me a design current of 132A.
In the end I took the total power of 9 of the single phase circuits as these will be balanced over 3 phases, 3 - brown, 3 - black, 3 grey then used a pF correction of 0.95. So 400v x 0.95 = 380v. 7200 x 9 = 64800w / 380v / Sq route 3 = 98.5A. Then the final circuit 230v x 0.95=218.5v. 7200/218.5v = 32.9A + 98.5A So lb = 131.4A
Hi you need to know the resistance of the cable conductor over a given length and apply ohm's law. Hope that helps, please like share and subscribe to enable me to continue doing this. Many thanks.
I get this lesson last week, and I didn't understand anything until I saw this wonderful video. Everything became clear and easy. Many thanks.
Wonderful, Glad it was useful. Please Share with your colleagues.
Please Like, Share & Subscribe.
Many thanks.
This video is the best. It explained it so well & a lot better than my tutors at college.
Thank you very much, I'm glad you found it useful - I only hope my students look at it and find it helpful!
Please, Like, Share & Subscribe
Heyyy you explained it very well. Can you please explain the calculations when the voltage drop is not given??
Thanks for watching, please share with like minded people. Voltage drop is set in the uk as 3 & 5% depending on the circuit. Or do you mean mV drop for cable? In which case use resistances of the cable (from manufacturer or standard table) to work out volt drop, but all manufactures of cables will publish technical literature for their cables. Hope that helps Please like share and subscribe.
Hello
I am a follower of your lessons and a big fan of them. I hope that you will show videos explaining the method of calculating the power factor, as well as the correct way to calculate the main breaker, and whether it should be larger than the service provider’s breaker, equal to it, or less in size than it.
sincerely
Khalid
HI Thank you, Power Factor is on my to do list - keep watching. Many thanks. Please Like, Share & Subscribe.
You explained cable grouping excellently. In effect this is worse than thermal insulation put over cables as each cable can be at 70 degrees so when grouped they are heating each other potentially if they are at full load.
Thank you - please read the notes below the table for grouping in BS7671 for other variations on its application.
How then are multiple (more than 16) circuits sized to fit in a single trunking?
Earned my subscription, very clear example, I will defiantly pass this on to the apprentices as i think this would help them a lot.
Many thanks, others in the pipe line - they are made for the apprentices I teach, if only they would look at them... Enjoy.
from Jordan , i am truly thankful for you to sharing this valuable guides with us and looking forward to seeing more about circuit breakers and fuses selection too according to bs 7671
My pleasure!
how this smart man couldn't have more subscribers. sadly i hope to have more subscribe
thank you for sharing an amazing video
I appreciate that!
Excellent explanation! Really helped me in my Masters electrical design module
Excellent - well done
This is a very well put video ( and great channel that I have just come across ). I am now designing electrical circuits as part of my job and something I never thought about, until now is - when we design, as you in this video with the step by step guide, how can we be sure that our ZS when verifying after installation will be acceptable? Thanks - great wok. 😊
Best of luck! Please Like Share & Subscribe.
I'm doing c&g 2396 in may, so any videos regarding this would be super cool. Great video 👍
Good luck!
You need to elaborate on your use of Ib & In in your voltage drop & It calculations. Both can be used depending on the nature of your circuit. If the circuit can be overloaded, you would use the In value. If for a fixed load, you would use Ib.
This is True, but in this video it has not been covered. One for another video maybe. If I keep adding new ideas the videos are too long, so I have to stop somewhere. But thank you.
Great explanation but a slight correction on the way you represent numbers in standard form in your calculations e.g. 2.8mV should be represented as (2.8 * 10^-3)V instead of (2.8^-3)V which is actually 4.6mV
Yes I know but a pain to go back and add an addendum now. Well done for spotting it though. My copy has already been corrected. Please Like, Share & Subscribe.
I can use these techniques to calculate the cable of an automotive wiring harness.
That is good question, the same principles apply, you would just need to find the actual current carry capacity of the type of cables used from the manufacturer. Hope that helps.
Thanks for sharing. In real life, I wonder if electricians do actually do all these calculations, especially for Ca, Cg and voltage drop. Normally, rules of thumb are applied, so for example 20A mcb equals 2.5mm2 cable, regardless of Ca, Cg and voltage drop. Just like to hear what real life electricians do or do not do in practice.
A good designer / electrician will take the method of installation and type of cable into consideration. Obviously we can all choose not to take these into consideration, but we are hoping things do not go wrong (and they may not, most of the time) - remember BS7671 can be used in a court of law to provide evidence. The OSG has tables to take some of the design out of installations - you just need to stay in those parameters.
great vid! example at the end was a huge help.
Glad it helped! Please like and share...
You explain much better then my college teacher that he is supose to be an Engineer, thanks a lot for making this explanation!! If i could I would marry you but unfortunately I'm straight 😄
I'm glad it helped. Good luck!
great video tutorial, thank you for sharing your knowledge💌
You are so welcome, Please Like, Share & Subscribe.
Very helpful.Many thanks.
Glad it was helpful! Please share with like minded people or maybe even like and subscribe.
good explanation sir
thank you, please like, share and subscribe.
Just perfect.. Very informative ❤
Many Thanks, Glad you found it useful.
Please Like, Share & Subscribe.
Thanks for th video, can you confirm that in cases where there is no possibility of overload in respect to usage you would use "Ib" (design current) rather than "In" (.size of protection) when calculating for "It". (tabulated current) e.g Ib/Cg x Ca x Ci etc
This is true BS7671 Appendix 4 equation 3,4 & 5 could be used as you have put above. Hope that helps.
Hi, thanks very much for your video. My question is, would the calculated Ib be divided by 3 if the circuit were to be a 3 phase supply?
No use the three phase formula at the start to calculate. Please Like Share & Subscribe.
I’m struggling so so much with this!! Dammmit
Thanks for watching, keep going it will get easier, share with like minded people or maybe even like and subscribe.
Thank you very much, although some tables are not complete and also in 12:01 the text ended with " size is " and the rest of words are not there. would you please help us with your reference to let us get the complete knowlege and also help those who are making a research on that topic to get soft copy of the document.
Hi the reference material used is the 'IET Wiring Regulation 2018', but a new one is out now dated 2022 Amendment 2.
Hope it helps please Like, Share & Subscribe.
@@sparkyhelp3997 Thank you very much
Thanks for your great help.. much appreciated.
Glad to help
Hello are we talking about a a copper or aliminum cable here? Could you tell how does cable size change according to type
Hi the Cable is copper, calculations can be carried out for Aluminium, but the current carrying capacity of aluminium is lower than copper.
Please Like, Share & Subscribe.
Very well explained!
Glad it was helpful!
How about multicore cables like SOCAPEX? 6 circuits on 1 multicore cable. Does that mean that a 2.5mm SOCA cable is downrated from 16A to 9A per circuit just by being bunched up??
Hi good question, that would be a little more detailed to answer as you would need to read the regulations further. you would also need to know the % load of each circuit as it may be neglected in in certain circumstances.
Have fun determining the rating - happy designing.
Please Like, Share & Subscribe.
@@sparkyhelp3997 ah ok. So it depends on the load for each circuit?
I'll do some more research.
Love the channel it has helped me a lot since I discovered it!
Cheers
Thanks for your sharing . Could you help me to advice that when we calculated the Design Current Ib , the power factor is required to include or can omit as this power factor make the result of the Design Current into 1.25 Time when the power factor is included in Design Current Calculation Step .
If there is a power factor it must be included in the calculation as the design current will be higher, and therefore the cable will have to carry the extra current.
Hope this helps.
Thankyou very much sir!
I really appreciate.
Most welcome! Please like and share. Many thanks.
😮What about the 40A circuit breaker after having 60.79A? Do we still keep the In=40A?
Hi, 60.79A is the It rating for the cable not the circuit breaker. The Load has remained unchanged, therefore the Circuit Breaker is still acceptable. Please Like, Share & Subscribe.
Many thanks for your explanation!
Hi Sparky, doing an electrical installation design for a health centre at the moment and I'm curious in regards to design current. What would I use for the design current for sockets where I'm not really expecting much permanently connected equipment? such as a changing room/circulation areas. The protective device would still have to be 32A or 20A for radials to comply with appendix H surely? even though the load will never reach anywhere near this? The client(lecturer) hasn't really provided any guidance on what electrical devices are intended for use. I mean I can spec for a typical industrial hoover being plugged in & a few mobile chargers but current is something negligible like 7A for example but Appendix H states it would have to be on a 20 or 32A breaker for radials? Maybe I'm missing something or maybe 20A or 32A radials with negligible load is fine? Hope to hear back from you soon, submission is in 2 weeks lol.
Hi Part H of OSG explains that diversity has already been taken into account from historical usage. keep the floor area as per 100m^2 for 32A (A1) RFC providing you do not know the load and the load is not reasonably expected to rise above 32A at any one time. So what I would do is measure the floor area of the design and provide enough A1, A2 or A3 circuits as per the floor area. Areas to give special consideration would be rooms such as kitchens, utility rooms or any with room with a lot of appliances which could be on at the same time. Remember diversity is not an exact science (just read the first couple of paragraphs from OSG maximum demand and diversity). Good luck - Please like, share and subscribe. Many thanks.
@@sparkyhelp3997 Thanks alot mate, after a bit more digging around I seen I should be Using my cpd rating as design current Ib=In I'm only using radials, no rings. I ended up with something like 31 seperate socket circuits (4 Storey building with 8 on bottom 3 floors and 7 on the 4th) with almost all except for 2 or 3 being A2s 75m squared radials on 32A cpds and the other 2 or 3 Being A3s 50m squared on 20A cpds, haven't fully specced into the cabling yet as it must've took me about 3/4 hours of looking back and forth between the model and drawing my circuits on CAD. I also have to wait for my team members to feed back to me info regarding the lighting and LV stuff such as CCTV and security systems and fire alarms. That way I can get my groupings. I only noticed today when reading appendix A that A1 is for your initial circuits and A2 is diversity to be applied after all circuits have been spec'd to get your main incoming fuse size. I swear I done my college HND coursework from 3 years ago wrong because I'm sure I was applying A2 diversity to my lighting d'oh. I've recently gotten a job as an electrical engineering consultant so it's key I keep refreshing my knowledge and keeping up to date. Thanks again for your help mate!
@@sparkyhelp3997 How then are multiple (more than 16) circuits sized to fit in a single trunking?
How can I get that book 📖 excellent work there
Many thanks, the book is by the IET, it is the 18th edition Electrical Wiring Regulations (New one out soon) available from Amazon etc
thank you for your video it it very good but calculation voltage drop formula minus -3 not gave correct answer please clearly gave formula thank you
Hi, my apologies it is raised to the power of x10^-3 or (milli). hope that helps
@@sparkyhelp3997 yes thank you very much now it work many many thank do more Video for calculation please thank you
When calculating Ib do I include diversity? Or the value before I do diversity calculation
I would tend to apply diversity after the design of the circuit.
Hi great videos
Can I just confirm, when calculating vd , the original Ib of 36.52v is used and not the new It of 60.79v .
Is this correct 👍
Yes that is correct, it is Ib that is actually flowing through the cable to cause the volt drop.
Thank you for your explanation, could you please tell me how I can download appendix tables of BS 7671
I'm afraid I don't think you can legally, they are from BS7671. Please Like Share & Subscribe.
@@sparkyhelp3997 Thank you for your response, so how I can get it legally please ?
@@adamAziz683 IET web site (Wiring Regulations BS7671 - 2022 {Brown Cover})
Can you explain why you didn't consider power factor 17:55? For single phase it is V*I*COSPHI rigth ?.....how much load can single phase carry? For beyond 8kw it should be 3phase right ? @sparkyhelp
Power Factor only applies if the load is inductive or Capacitive. If power factor is not given (maybe for example an electric heater) it is more than likely a resistive load and therefore power factor is 1.
Is ambient temperature still at 70 degrees in thermal insulation ?
Also 40 divided by 0.94 x 0.7 = 29.78 or am I wrong ? 🤔
Hi the conductor temperature is 70 degrees, the ambient temperature is the surrounding air temperature. Put the 0.94x0.7 in brackets 40/(0.94x0.7)= Hope this helps.
Please like Share & Subscribe - Good luck
@@sparkyhelp3997 sorry I ment is ambient temperature 30 degrees if wrapped in thermal insulation or would it be greater
Na still not getting the calcs but il try work it out lol
@@user-yh7bk7dx7c Ambient Temperature is what ever the designer deems it to be (or the question setter!)
Keep going...
After calculating the cable size and when attempting to calculate the short circuit current am I correct in saying you need to calculate the resistance and reactance of the cable which is separate from the values in the voltage drop calculations? Do you know where to find the tables for resistance and reactance values? The do not appear to be in BS7671
For the vast majority of circuits reactance of the cable will be ignored as it is such a small value it would be insignificant. If you are talking long distribution or transmission cables then the information from BS7671 would not cover this. BS7671 covers low voltage installations. I would suggest seeking information from the cable manufacturer (this is where I go to find the resistance of cables or armouring of cables - BS7671 does not provide resistance of conductors, the On-Site-Guide does but only up to 50mm^2) . I hope this goes some way to answering your question, but the answer is not that straight forward as we could write an entire paper on this subject.
Hello and good day to you! I have been looking on all the internet, TH-cam, and I just happen to come a cross your video, my question (if you can able to help me in this little matter!!) I just bought a Fish Finder, and I just want to splice the cable to make it longer, the cable's of the unit they are very thin, could i able to spliced it with a bigger gauge,??? so I can able to reach the 12v Battery. thank you for your expertise, and for your little help on this!! Sincerely the Dummy.
In order to do this you will have to know the current passing through the cable (it will be small). You could just put a larger cable on (it should not cause a problem - just may look unsightly) and try it. A bit of trial and error, just means your warranty will be invalid. Good luck, hope this has helped in some way.
How do we get the 11.5 Volt drop?
It is from the wiring regulations IET 18th Edition (used in the uk), 3% or 5% of the nominal voltage for particular circuits. Hope that answers your question.
@@sparkyhelp3997 yes, it does completely answer my question.
Thank you!
Thanku very much
Most welcome, Please Like Share & Subscribe
I'm going to take a wild guess here, is this from the NEC or the CEC 2018 BOOK?
It comes from the 'IET wiring regulation 2018', but a new one is out now dated 2022 Amendment 2.
Hope it helps please Like, Share & Subscribe.
Any reason why you didn't include thermal constraints?
Topic to come soon - watch this space
@@sparkyhelp3997 That wasn't a dig by the way, I was merely curious.
sorry to sound dumb here but if I have 10 single phase circuits each 7.2kw and I wanted to find the design current for a sub main to feed a board and the supply is 3 phase how do I work that out as obviously I don't just add all 10 circuits up (72kw) as you would if the supply was single phase.
Good Question.
Assuming it will be split over the three phases, you could divide by three, (this is crude - approx. 100A per phase), or design which phase you want them to be on to be more accurate. If you want to design the sub-main, you may with your more in-depth knowledge what these circuits are, you may want to apply diversity.
Really hope that helps.
@@sparkyhelp3997 they are for EV single phase 7.2kw car chargers so can't use diversity.
I looked at it as I have 10 circuits each 7.2kw on a 3 phase DB so that's 4 + 3 + 3 so did 4 x 7.2kw = 28.8kw power factor of 0.95 which gave me a design current of 132A.
The dilemma of the modern age (EV chargers), it will just need a large supply and a dis-board rated to cope.
In the end I took the total power of 9 of the single phase circuits as these will be balanced over 3 phases, 3 - brown, 3 - black, 3 grey then used a pF correction of 0.95. So 400v x 0.95 = 380v.
7200 x 9 = 64800w / 380v / Sq route 3 = 98.5A. Then the final circuit 230v x 0.95=218.5v. 7200/218.5v = 32.9A + 98.5A So lb = 131.4A
I agree with the answer (easier to write down than type in - dont you think) - Good luck
Thanks mate :)
Happy to help, please like, subscribe a share.
without using tabular form, how to calculate mV ?
Hi you need to know the resistance of the cable conductor over a given length and apply ohm's law.
Hope that helps, please like share and subscribe to enable me to continue doing this. Many thanks.
like and share
Thankyou
Thanks ❤️👍
No problem 👍 Please Like Share & Subscribe.
🎉
Thank you.
👍👍👍
Any thing in particular you think would be helpful?
@@sparkyhelp3997 that's the problem, reading examiners past reports earthing, bonding, explaining diversity all seem a problem. 👍👍👍