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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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  • 0

    gkenyon:




    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.


    (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.


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


     




    Oh dear. I have the 17th edn, Amd 3 (2015), version of the EIDG. Has the first paragraph of section 8.4.2: Circuit-breakers, been removed in the 18th edition version (which states, for those who don't have the book, 'If it is assumed that the circuit will be designed so that there is instantaneous operation in no less than 0.1s, that is a low fault levels, simple calculations of minimum protective conductor sizes can be carried out, see table 8.2 for a type D circuit-breaker...* However, at high fault levels, say 3kA and above, Regulation 434.5.2 requires a check that the energy let-through of the device does not exceed the k2S2 of the cable...')?


    * the adiabatic equation used in Table 8.2 uses values of fault current derived from:If=CminUo/Zs  



    F
Reply
  • 0

    gkenyon:




    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.


    (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.


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


     




    Oh dear. I have the 17th edn, Amd 3 (2015), version of the EIDG. Has the first paragraph of section 8.4.2: Circuit-breakers, been removed in the 18th edition version (which states, for those who don't have the book, 'If it is assumed that the circuit will be designed so that there is instantaneous operation in no less than 0.1s, that is a low fault levels, simple calculations of minimum protective conductor sizes can be carried out, see table 8.2 for a type D circuit-breaker...* However, at high fault levels, say 3kA and above, Regulation 434.5.2 requires a check that the energy let-through of the device does not exceed the k2S2 of the cable...')?


    * the adiabatic equation used in Table 8.2 uses values of fault current derived from:If=CminUo/Zs  



    F
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