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adiabatic equation - 2.5 2c SWA

Stewart Mason
I don't get involved with adiabatic equations all to often, hence I'm a bit rusty! I'm trying to calculate the R1+R2 for a 2 core 2.5mm SWA. Obviously the R1 is the easy bit, I'm struggling with the SWA as the R2.  I saw a link to a table on the old forum but I can't find it now. Any ideas?


Thanks Stewart
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    Guidance Note 1 Tables D6 to D8 tell you min. sizes for adiabatic.You don't actually need to know R2 to work it out for any sizes not listed (see Chapter 8 of the IET's Electrical Installation Design Guide).
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    11bff9ed5083c238e20bdbbd2e775838-huge-20190425_205001.jpg

    Here's a R1+R2 table used to calculate the earth loop impedance using SWA armour for R2, (Zs), You can then calculate the prospective short circuit current and hence the adiabatic calculation.


    Legh
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    Thank you both for your input.
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    Legh Richardson:


    Here's a R1+R2 table used to calculate the earth loop impedance using SWA armour for R2, (Zs), You can then calculate the prospective short circuit current and hence the adiabatic calculation.


    Legh




    Hang on - this isn't the correct method.


    (a) There's no need to calculate Zs for adiabatic for R2 even for fuses. You can simply plot k2S2 on the time-current rating curves. And this is important - see below.


    (b) If you're using a circuit breaker (mcb, RCBO, mccb) you cannot use this method, you must use the let-through energy from either the product standard, or let-through energy provided by the manufacturer, and simply compare against k2S2 - again, no need to calculate the fault current.


    The absolute simplest way for most final circuits with SWA, is to look it up in GN1.


    So, why did I say it's not the correct method?


    Well, the worst-case condition for electric shock is indeed at the remote end of the cable - the furthest point in the circuit.


    This is NOT the case for protection against overcurrent - the adiabatic criterion. The worst-case is often further up the cable, at a point where, if you choose too small a CSA, the line corresponding to k2S2 meets the overcurrent protective device curve, i.e. at higher fault currents / shorter circuit length.


    See Chapter 8 of Electrical Installation Design Guide for a complete explanation.

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    With what Graham has said I'm slightly confused. 


    So with a Ze of 0.35, a 16A 60898 type B, and 30 metres of 2c 2.5mm using the SWA as the earth would this comply? 


    Thank you.
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    Without doing any calculations and just using previous "experience knowledge" I venture to say that the proposed installation as described will comply.


    But I may be concerned about Volt drop. What is the design load and installation method?


    Z.
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    Does this help?

    https://professional-electrician.com/technical/armoured-cable-supplies-outbuildings/


    Also, what about installing a 100mA S type R.C.D. at the origin?


    Also Table 1 and its preceding notes are very useful here:

    https://gadsolutions.biz/regulations/swa-as-the-cpc


    Z.
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    Zoomup:

    Without doing any calculations and just using previous "experience knowledge" I venture to say that the proposed installation as described will comply.


    But I may be concerned about Volt drop. What is the design load and installation method?


    Z.




    CORRECTION. Now, after doing some quick sums I now believe that the disconnection time will be nearer to 10 seconds with an earth fault of negligible impedance at the far end of the S.W.A. That just shows us what guessing can achieve.......bad results. The C.P.C. needs to be bigger, supplemented with an additional copper C.P.C. or an R.C.D. installed at the origin of the circuit.


    Z.

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    Stewart Mason:

    With what Graham has said I'm slightly confused. 


    So with a Ze of 0.35, a 16A 60898 type B, and 30 metres of 2c 2.5mm using the SWA as the earth would this comply? 


    Thank you. 




    If you mean with the adiabatic equation for protection against fault current, then yes... quite simply, there's a note to the bottom of table D7, and para 1 of D3,  in GN1 that says so.


    No calculation required. unless it's an unusual installation method, or a different than usual ambient temperature.


    Will it provide fault protection as well as overload protection? That would require a knowledge of R1+R2, but with a B16. Again, assuming no correction factors apply, you're well in at 30 m, you could probably go over 3 times as far and still achieve 0.4 s disconnection times - but I think volt-drop will limit you to about 45 m with 2.5 mm2 conductors in any case.

  • 0

    °Zoomup:




    Zoomup:

    Without doing any calculations and just using previous "experience knowledge" I venture to say that the proposed installation as described will comply.


    But I may be concerned about Volt drop. What is the design load and installation method?


    Z.




    CORRECTION. Now, after doing some quick sums I now believe that the disconnection time will be nearer to 10 seconds with an earth fault of negligible impedance at the far end of the S.W.A. That just shows us what guessing can achieve.......bad results. The C.P.C. needs to be bigger, supplemented with an additional copper C.P.C. or an R.C.D. installed at the origin of the circuit.


    Z.


     




    How so Z?


    A B16 as Max Zs of 2.73 Ω for 0.4 s disconnection

    Less Ze of 0.35 Ω leaves 2.38 Ω

    So, using even the over-excessive final temperature resistances leaves over 90 m to achieve disconnection time of 0.4 s (although for Type B mcb this is the same as 5 s) ???

    The limiting factor as you initially said would be volt-drop at 45 m (if it's a final circuit conductor, less if you want to drop short circuits off the other end).

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