r1 + r2 for 3 core 10mm XLPE

I,m trying to design a  submain for an outbuilding.The circuit comprises 50m of 3 core 10mm xlpe

on a bs88 40A fuse.Have worked out R1+R2 for the cores as 50x3.66x1.2/1000 =0.0216 ohms

Can the armour be included in the calcs to lower ZS and how would it be done?The supply is TNS 0.8 ohms (understand we have to work on that value)

                                                                                     Thanks for any help,

                                                                                                   Regards,Hz

                                                                         

Parents
  • Indeed, an all copper round loop 0f  100m of 10mm2 running hot will be about 200 milliohms.  One might hope the earth core is colder, but that does not make much difference.

    The inductance per unit length of the internal CPC is largely cancelled  compared to a wire in free space we can estimate this from the dielectric constant of the insulation and the capacitance per unit length of the cores, which is something the better multi-meters can measure directly.

    If we know the insulation is polyethylene  say, (and XPLE is just polyethylene abused a bit  by radiation or nasty chemicals to toughen it up ) the dielectric constant  is about 2.3 - so the capacitance between the wires is about 2.3 times what it would be if the wires were in the same layout in free space. This means the wave velocity is c*= c /√2.3 where c is the speed of EM waves (light, radio waves etc ) in vaccuum. and c* is the velocity in the dielectric. So c~ 300m/microsecond and  c* is ~ 200m/microsecond..

    If you know the wave velocity and the capacitance, the inductance follows as c*= 1/ √(L.C)  where L and C are the inductance and capacitance  per unit length respectively.

    So if you measure 1000pF core to core for a 10m length, that is 100pF per metre, and we can go to L= 1 / (c*)2 *C
    1/ ((200,000 000) *100 pico) = 0.00000025 H or in units we can chew more usefully 250nH per metre..

    A CPC on the outside does a lot worse than this, and will look like more like 1000 nH per metre.

    However, even at that level, at 50Hz, a nano Henry is ~ 314 nano ohms reactive at 50 Hz, so we re looking at 1/3 of a milliohms  per metre of inductive reactance for the loop of live and the external CPC - the simple resistance still dominates with an external CPC but the inductance is not totally negligible - with the internal core, it pretty much is.

    Mike.

Reply
  • Indeed, an all copper round loop 0f  100m of 10mm2 running hot will be about 200 milliohms.  One might hope the earth core is colder, but that does not make much difference.

    The inductance per unit length of the internal CPC is largely cancelled  compared to a wire in free space we can estimate this from the dielectric constant of the insulation and the capacitance per unit length of the cores, which is something the better multi-meters can measure directly.

    If we know the insulation is polyethylene  say, (and XPLE is just polyethylene abused a bit  by radiation or nasty chemicals to toughen it up ) the dielectric constant  is about 2.3 - so the capacitance between the wires is about 2.3 times what it would be if the wires were in the same layout in free space. This means the wave velocity is c*= c /√2.3 where c is the speed of EM waves (light, radio waves etc ) in vaccuum. and c* is the velocity in the dielectric. So c~ 300m/microsecond and  c* is ~ 200m/microsecond..

    If you know the wave velocity and the capacitance, the inductance follows as c*= 1/ √(L.C)  where L and C are the inductance and capacitance  per unit length respectively.

    So if you measure 1000pF core to core for a 10m length, that is 100pF per metre, and we can go to L= 1 / (c*)2 *C
    1/ ((200,000 000) *100 pico) = 0.00000025 H or in units we can chew more usefully 250nH per metre..

    A CPC on the outside does a lot worse than this, and will look like more like 1000 nH per metre.

    However, even at that level, at 50Hz, a nano Henry is ~ 314 nano ohms reactive at 50 Hz, so we re looking at 1/3 of a milliohms  per metre of inductive reactance for the loop of live and the external CPC - the simple resistance still dominates with an external CPC but the inductance is not totally negligible - with the internal core, it pretty much is.

    Mike.

Children
  • Thanks for all the help.,thought I could use a 40A bs88 fuse because of the higher zs it allows on a submain and might be a cheaper option than a type S rcd.

                                                        Regards,

                                                            Hz

  • Maybe you can, Is the Zs really as high as 0.8 ohms at the origin ? - if you can get a PSCC of a couple of hundred A when cable is hot and supply is low you can make the 5 seconds needed for sub-main protection. Achieving the half a heartbeat of  0,4 seconds is not so likely, needing more like 300-400 A.

    Fuse curves for BS88 part C

    Mike

  • Thanks for all the help.,thought I could use a 40A bs88 fuse because of the higher zs it allows on a submain and might be a cheaper option than a type S rcd.

    You may well be able to "get away" with a BS 88-3 fuse at 45 A ... without using the armour

    (but if you use that you're well in).

    Disconnection time of sub-main is 5 s on TN-S.

    Table in Fig 3A1 says we need 220 A for a 45 A cartridge fuse.

    Ze of 0.8 ohms + (R1+R2) of 0.2196 = 1.0196 ohms.

    230 V / 1.0196 ohms = 225 A.

    (so, if you parallel the armour, you're well above that, but not at the 380 A required for 0.4 s required for a final circuit).

    Check adiabatic for the copper means S > 4.4 sq mm

    So, again, if you parallel the armour with the copper cpc (bond at both ends), you'll be well in for ADS and adiabatic for a sub-main. If necessary, you could use RCBOs for final circuits if ADS were considered an issue (or you didn't want to do more calculations).

    The down-side is that you will have a calculated 8.8 V volt-drop at 40 A down the sub-main ... which means lighting circuits in the outbuilding can't comply with (i) of Table 4Ab. Not knowing your design currents and arrangement of final circuits, I don't know if that's a problem for you.

  • Hi Mike,

    The measured ZE was 0.15 ohms,but I thought we had to work on the published value

    of 0.8 in case there was a change in the network.

                                                                      Regards,

                                                                           Hz

  • Batt Cables publishes a useful table on SWA armour resistance - No need to buy the BS book.

    www.batt.co.uk/.../Armour-Resistance

  • No need to buy the BS book

    You still need the reactance formulas from PD CLC/TR 50480, as even with small csa SWA, it makes quite some difference over using resistance alone (between 5 and 10 % for SWA used as armour alone, but a greater difference when considering a parallel cpc).

    But still, no need for the standard and the cable construction BS ... the formulas from PD CLC/TR 50480 are available in Guidance Note 6 and the Electrical Installation Design Guide, and tables of armour resistance available in a few places as you say.

  • Or as above when it gets more complicated even if you cannot measure the inductance, if you have a meter that does capacitance you can deduce the surge impedance and therefore the inductance of a line pair from first principles for your own specific layout, rather than rely on a tabulated value that relates to an idealized hairpin inductor geometry where core to core spacing is twice the cable diameter, as that may well not be quite what you have. A measurement done well beats paying for an estimate based on the work of a distant committee.

    Like the OSG, it is useful as a first cut cookbook, but not in situations where things are very near the limit, or indeed with non 50Hz or non- sinusoidal waveforms.

    All TR 50480 does is scale for groups of conductors from these figures..


    For that twin wire layout, as a more general form than the 2:1. for wire diameters 'd' spaced centre to centre by any distance  'D'

    the L and C per metre of a line pair are given as  (ref)

    Because the geometry factor is the same you can deduce the short circuit inductance from a measurement of open circuit capacitance (or vice versa) and knowledge of the material properties to get the e and u factors.

    Mike.

  • GN 6 is not as helpful as initially anticipated -

    See comments from 43:46 onwards for a prime example of it's use 'out in the field'

    www.youtube.com/watch

  • hah no he does not like it or mince his words does he. Mind you, I'm not afraid of the maths, and I'm inclined to agree with the sentiment.

    The problem is that unlike pure copper at ~ 16 milliohms per meter length per mm square, the word  'steel' covers a multitude of different  alloys and it's resistance is not predictable without knowing the exact composition. It is typically about ten times more resistive than the same wire diameter in copper but it may be more or less, and is far from exact. 
    The cable makers are the only folk who actually know for sure what they have used, so  if they do not publish it, then a measurement beats 4 aces.
    Mike

  • I do think there is something in the response he quoted from the NICEC though. Sometimes you can overthink it and make it needlessly more complex than is actually required. As a scholarly exercise there is nothing wrong with chasing it down to the last decimal point, but with the practical limitations ever present on the job, it's a different ballgame out in the field.

    For example, I recently did a job whereby I ran 19 metres of 16mm 3 core SWA for a submain, protected by a type s 100A DP RCD in the tails, then a switch fuse with a 63A BS88 fuse. The system is TT with a electrode resistance of 14 ohms, so I just paralleled the armour with the third core and exported the TT earth. I'm getting 14.8 ohms at the destination consumer unit end.

    I initially thought about doing a 'island' but there wasn't anywhere pratical to sink another rod so I just exported the existing instead.

    I was going for a 40A or 45A fuse but I could not get one lower than 63A with the correct form factor for the switchfuse.