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
You need to look up the manufacturer's specification, which will give you the CSA of the armour. Then you adjust for the resistivity of steel compared with copper.
That calculation seems very low. I have obtained 0.07 Ω (by calculation and measurement) for a 50 m drum of 25 mm² cable. Have you slipped a decimal place?
well the 10mm copper cores will give you ~ 1.6 milliohms per meter to begin with, so 80 milliohms in the live core. (cold copper - if you are sweating it the 1,6 becomes more like 1,9)
Now are these 3 cores 3phases, or single phase with one core as earth ?
If the latter the armour is in shunt with the copper core, so 80 milliohms parallel the steel wire If 3 phases, the steel wire is on its own
the CSA of the steel on 10mm 3 core is about 40-45mm2 but of course steel is ~ 10 times more resistive than copper, so it is more like 4- 4.5mm2 copper equivalent so if you have a copper core CPC the copper takes almost 2/3 of the current..
The usual solution to a supply with a high earth loop feeding armoured cables is a 300mA RCD, (or maybe a 100mA type ) either instant or delay type to trap the earth faults, and leave the fuse to do the L-N shorts.
Mike
.Have worked out R1+R2 for the cores as 50x3.66x1.2/1000 =0.0216 ohms
I think I get that at 0.2196 ohms?
The figures on the left hand side look correct for copper DC resistance 1.83 mΩ/m per core at 20 deg C, and 1.2 factor to correct to 70 deg C.
Can the armour be included in the calcs to lower ZS and how would it be done?
There isn't a published formula, at least that I'm aware of, to use SWA in parallel with an internal copper conductor (there is with an external conductor). However, the following formula for an external cpc, from NA.4.5 of PD IEC/TR 50480, is likely to give you an over-estimate of the impedance for the internal copper cpc in parallel with the armour (because the loop reactance for the portion of current returning down the other copper internal conductor ought to be zero, but the formula below accounts for external conductor separated from internal copper conductor by armour):
Resistive component of (Z_{1}+Z_{2}), r = R_{1} + R_{2} //1.1R_{2a}
where
R_{1} and R_{2} are the DC copper conductor resistances impedance we have used already ({1.83 × 1.2} mΩ/m)
R_{2a} is the DC resistance of the armour (I think {4 × 1.225} mΩ/m for 3 c 10 mm^{2} SWA corrected to 70 deg C)
Reactive component of (Z_{1}+Z_{2}), x = 0.4 (you'd have to take reactance into account even for small csa armour, as for copper csa 2.5 sq mm and above, armour csa > 16 sq mm - in the case of 10 sq mm 3 c SWA, the armour csa is 39 sq mm).
and overall impedance is then (Z_{1}+Z_{2}) = √(r^{2}+x^{2})
If you work this through, for 3 c 10 sq mm you might get a loop impedance per metre around 85 % of that assuming copper conductors only.
This broadly supports what Mike says about the fact you won't get huge reductions taking the armour into account.
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) ^{2 }*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.
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.
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.
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.
I don't think BS 7671 says we have to - it's certainly a good idea in the general case (although I think it unlikely that modern changes to the network would likely raise Ze much above 0.35 Ohms if it's below that already). In some (rare) situation that actual Ze can exceed the published figures - so you can't completely ignore the actual Ze. In others (e.g. sites with own transformers) the chances of significant increases in Ze are extremely unlikely.
- Andy,
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