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Earthing size.

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  • 24-05-2018 11:08pm
    #1
    Registered Users Posts: 1,051 ✭✭✭


    With the change on the T&E manufacture. CPC to match the line conductor so on...

    If you run single pvc/pvc 16mm^2 I take it the Earth must be 16mm^2 also.

    ET101. Pg120 table 54A is this table redundant now so?
    Must the cpc match the line conductor in every case now.


Comments

  • Registered Users Posts: 2,567 ✭✭✭Risteard81


    Tuco88 wrote: »
    With the change on the T&E manufacture. CPC to match the line conductor so on...

    If you run single pvc/pvc 16mm^2 I take it the Earth must be 16mm^2 also.

    ET101. Pg120 table 54A is this table redundant now so?
    Must the cpc match the line conductor in every case now.
    No. There is nothing to preclude the use of the adiabatic equation to size the cpc instead of doing it by selection. However the old type T&E no longer complies with I.S. 201 and therefore is prohibited. The cable standard no longer recognises this cable, and ET101 makes reference to I.S. 201 cables. If you are doing a proper installation using singles in containment then the world's your oyster.


  • Registered Users Posts: 1,051 ✭✭✭Tuco88


    Thanks for clearing that up Risteard.

    So there is no issue with the use if 10mm^2 Earth used with 16mm^2 single pvc/pvc meter tails.


  • Closed Accounts Posts: 56 ✭✭thecribber


    Incorrect information being given here

    The wiring rules require that meter tails are a min of 16sq. Also the Main Protective Conductor (neutralising link) must have the same CSA as the live and neutral conductor. Therefore the Main Protective conductor must be a min of 16sq also. There is a dispensation for the old type T&E cable written into the rules but the 2016 number 2 amendment renderes this dispensation redundant.
    A 10sq Earth used as the main protective conductor would not comply with the current rules.
    Bonding conductors on the other hand are required to be 50% the size of the mains live conductor but a min of 10sq.


  • Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,595 Mod ✭✭✭✭2011


    thecribber wrote: »
    The wiring rules require that meter tails are a min of 16sq.

    Agreed.
    Also the Main Protective Conductor (neutralising link) must have the same CSA as the live and neutral conductor.

    Have you a link that shows this? Or can you point to where you got this information from?

    All I can find is this and this.

    Nobody is disputing that twin and earth with a reduce bare conductor is no longer permitted. I am not saying you are wrong, but what you are saying does not align with anything I can find. I only work in the industrial sector and have not worked on my tools in a long time, so this is not something that I normally deal with.

    I would have thought that Risteard81's position is correct i.e. if tails were wired in singles (16 mm sq. phase and neutral with a 10 mm sq. earth) that this would be acceptable once the adiabatic equation is satisfied and when testing the earth fault loop impedance is recorded as being sufficiently low. Let's not forget the aim is to ensure that the minimum disconnection time is met.


  • Registered Users Posts: 1,051 ✭✭✭Tuco88


    I cant find any amendment to table 54A so it must still be vaild. I would think its confusion with the change to T&E cable.


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  • Closed Accounts Posts: 56 ✭✭thecribber


    Rule 543.4 requires that the Main Protective Conductor complies with rule 543.1. If you look at Table 54C in the current rules (Book with blue stripe) you will see that if the supply cable is less than 35sq the protective conductor must be the same size as the phase conductor. Look at first line if S is 16 sq or less the protective conductor must be S (the same) There are two notes under this table.
    If you look at the same table in the older rule books (yellow stripe) there are three notes under table 54C. Note 2 which has been removed in the 2016 No. 2 Amendment states.
    “Table 54C does not apply to PVC/PVC twin core cable with bare protective conductor type N05VVH4”
    Removing this note has the effect of requiring that the protective conductors in all installations where the phase conductor is 16sq or less must be the same size. Therefore using a 10sq earth with two single 16sq PVCs does not comply with Table 54C and is a breach of the current rules.


  • Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,595 Mod ✭✭✭✭2011


    thecribber wrote: »
    Rule 543.4 requires that the Main Protective Conductor complies with rule 543.1


    Agreed.


    I have a copy of ET101 in front of me, here is a word for word extract of rule 543.1 from my copy:
    The cross-sectional area of a protective conductor shall be either:

    - selected in accordance with 543.1.1, or
    - calculated in accordance with 543.1.2.


    543.1.2 (calculation of cross-sectional area of a protective conductor) provides the adiabatic equation. So in my opinion we have a choice, comply with 543.1.1 (table 54C) or comply with 543.1.2 (adiabatic calculation)
    Am I missing something?


  • Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,595 Mod ✭✭✭✭2011


    Tuco88 wrote: »
    I cant find any amendment to table 54A so it must still be vaild. I would think its confusion with the change to T&E cable.



    Wrong table!
    Table 54A page 120 is part of section 542, therefore it applies to "earthing arrangements".
    You need to be looking in section 543 if you want to size the protective conductor for your tails.


  • Registered Users Posts: 1,051 ✭✭✭Tuco88


    Cheers 2011.


  • Closed Accounts Posts: 56 ✭✭thecribber


    2011 wrote: »
    Agreed.


    I have a copy of ET101 in front of me, here is a word for word extract of rule 543.1 from my copy:



    543.1.2 (calculation of cross-sectional area of a protective conductor) provides the adiabatic equation. So in my opinion we have a choice, comply with 543.1.1 (table 54C) or comply with 543.1.2 (adiabatic calculation)
    Am I missing something?

    Can’t argue with any of the above, why do they make it all so complicated.
    Can someone who is good at maths work out the protective conductor CSA size for a 16sq phase conductor using

    (A). IEC 60949
    (B). The Formula for disconnection times not exceeding 5s. S=etc etc


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  • Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,595 Mod ✭✭✭✭2011


    thecribber wrote: »
    Can’t argue with any of the above, why do they make it all so complicated.

    No idea.
    Can someone who is good at maths work out the protective conductor CSA size for a 16sq phase conductor using

    (A). IEC 60949
    (B). The Formula for disconnection times not exceeding 5s. S=etc etc

    A or B? No thanks.
    I always go with option C.


  • Registered Users Posts: 2,567 ✭✭✭Risteard81


    2011 wrote: »
    thecribber wrote: »
    Rule 543.4 requires that the Main Protective Conductor complies with rule 543.1


    Agreed.


    I have a copy of ET101 in front of me, here is a word for word extract of rule 543.1 from my copy:
    The cross-sectional area of a protective conductor shall be either:

    - selected in accordance with 543.1.1, or
    - calculated in accordance with 543.1.2.


    543.1.2 (calculation of cross-sectional area of a protective conductor) provides the adiabatic equation. So in my opinion we have a choice, comply with 543.1.1 (table 54C) or comply with 543.1.2 (adiabatic calculation)
    Am I missing something?
    That is also how I see it. Selection can be used if you do not wish to calculate with the adiabatic equation, however its use is not precluded should it allow more cost effective installation etc. Selection is simply the easy way of compliance - not the only way.


  • Registered Users Posts: 2,567 ✭✭✭Risteard81


    thecribber wrote: »
    2011 wrote: »
    Agreed.


    I have a copy of ET101 in front of me, here is a word for word extract of rule 543.1 from my copy:



    543.1.2 (calculation of cross-sectional area of a protective conductor) provides the adiabatic equation. So in my opinion we have a choice, comply with 543.1.1 (table 54C) or comply with 543.1.2 (adiabatic calculation)
    Am I missing something?

    Can’t argue with any of the above, why do they make it all so complicated.
    Can someone who is good at maths work out the protective conductor CSA size for a 16sq phase conductor using

    (A). IEC 60949
    (B). The Formula for disconnection times not exceeding 5s. S=etc etc
    You can't calculate it without knowing the prospective fault current etc. as these will influence disconnection times etc. So you would need the actual installation to calculate this or information from the ESB as to what their design assumptions are for the supply.


  • Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,595 Mod ✭✭✭✭2011


    Risteard81 wrote: »
    You can't calculate it without knowing the prospective fault current etc.

    Agreed.
    So you would need the actual installation to calculate this or information from the ESB as to what their design assumptions are for the supply.

    Yes, either you measure it or you work off a worst case calculation. The ESB must work within certain parameters, if their earth fault loop impedance is beyond a value (I don't know what this is off hand) they must resolve.


  • Registered Users Posts: 2,567 ✭✭✭Risteard81


    2011 wrote: »
    Yes, either you measure it or you work off a worst case calculation. The ESB must work within certain parameters, if their earth fault loop impedance is beyond a value (I don't know what this is off hand) they must resolve.

    Agreed. At the design stage realistically you would be doing this off declared values from the ESB. For periodic inspection and testing (or initial verification to verify that the design assumptions are correct) you can actually measure what is there.


  • Registered Users Posts: 1,051 ✭✭✭Tuco88


    Thanks for the feeback gents.


  • Closed Accounts Posts: 56 ✭✭thecribber


    ESB Networks check that the loop impedance value at their cutout is less than 0.5 of an ohm. If it exceeds this they will increase the cable size to bring it down. The normal values achieved are much less than 0.5 ohm. 0.1 to 0.3 ohm wouldn't be unusual. When you take into account that the maximum loop allowed for a 63amp MCB (B) or a 63amp neozed fuse is a little over 0.7 of an ohm the 0.5 ohm target is a realistic one.

    To get back to the Origional question, if the incoming loop was a worst case 0.5 ohm what size cable would the calculation produce. I am not up to speed with the formula and can't do it myself


  • Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,595 Mod ✭✭✭✭2011


    thecribber wrote: »
    To get back to the Origional question, if the incoming loop was a worst case 0.5 ohm what size cable would the calculation produce. I am not up to speed with the formula and can't do it myself

    There is no single answer. It depends on the length of the cable run. A longer cable run will require a cable with a larger cross sectional area.


  • Closed Accounts Posts: 56 ✭✭thecribber


    2011 wrote: »
    There is no single answer. It depends on the length of the cable run. A longer cable run will require a cable with a larger cross sectional area.

    Assuming a cable run of 10 meters and a loop of .03 ohms what cable size would be required. ???


  • Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,595 Mod ✭✭✭✭2011


    thecribber wrote: »
    Assuming a cable run of 10 meters and a loop of .03 ohms what cable size would be required. ???

    Let's start with the equation:
    Zs = Ze + (R1 + R2 )

    So I take it that the total earth fault loop impedance (Zs) must not exceed 0.7Ω for a 63amp MCB with a B type characteristic or a 63amp neozed fuse (I got this value from your post #18 above).

    From your last post the external earth fault loop impedance (Ze) is 0.3Ω (I assume that 0.03Ω is a typo). This means that:

    R1 + R2 < or = 0.4Ω

    The phase and neutral conductors must be at least 16 mm sq. to comply with ET101:2008

    Based on that let's say that we are using this 16mm sq. tri-rated panel flex made by Anixter for the phase and neutral conductors. From the linked data sheet we know that it has a resistance of 0.0121Ω per meter. This means that a 10m run will have a resistance of 0.121Ω

    Therefore R1 = 0.121Ω

    ...and let's suppose that we select this 10mm sq. tri-rated panel flex as the CPC made by Anixter. From the same linked data sheet we know that it has a resistance of 0.0191Ω per meter. This means that a 10m run will have a resistance of 0.191Ω

    Therefore R2 = 0.191Ω

    So back to our equation:

    Zs = Ze + (R1 + R2 )
    Zs = 0.3 + (0.121 + 0.191) = 0.612Ω which is less than the value required (of 0.7Ω)

    So in this example the earth fault loop impedance is sufficiently low when using a 10 mm sq. CPC with 16mm sq. phase and neutral conductors.

    That is the theoretical design part done, the next bit is to prove it by installing it and measuring the actual earth fault loop impedance and documenting the test.

    This is back to first principles, not the way it is generally done but the theory is sound. In work I use a trusty graduate engineer that has been trained to use a cable sizing software package to size all cables :D:):D


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  • Closed Accounts Posts: 56 ✭✭thecribber


    Well done for working it out, so would it be fair too say that for any mains run in excess of approx 10 to 12 meters you would need to increase the size of the Main Protective Conductor to 16sq .


  • Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,595 Mod ✭✭✭✭2011


    That depends, mainly on the Ze value.


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