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18th Edition Cable Calculations

Steve Brown
Hi guys I need some help calculating cable sizes for a 24v DC supply. I'm a student doing a job report and I want to explain cable sizing basically, and whilst on site my mentor, with his vast experience, knew what cable sizes would work so I just listened to him. Now I want to calculate it properly.


So it's the nitrate monitor requires a 230v supply but 4 cables, 2 pairs each for 2 different alarms, I want to install are on the 24v DC side. (They're just going to a junction box that already has cables feeding back to the PLC).

The cables will be installed in conduit on the wall less than 0.5m away. (distance between monitor and junction box).

The monitor they're being connected to is only 20W.


So the farthest I've got, (unless I'm wrong), using the 18th is this:

Ib, (Design Current), = 20W/24v = 0.83A.


I read in the 18th edition, (correctly I hope), that I will have a rating factor of 0.65 because there will be 4 cables enclosed. (Am I right?) But I've no idea what to do with that figure.


I've really no idea where to go next or even if I'm on the right lines thus far. Can someone explain it to me? I don't necessarily want the answer, but a formula would be brilliant.


Thankyou.
  • 0
    The rating factor is used for de-rating the cable from what it would carry in ideal conditions to what it will carry under the bunching/enclosed conditions in a real installation. Say the cable has a normal rating of 10A, then with a rating factor of 0.65 this would come down to 6.5A

    It can also be used in reverse - if the cable needs to carry a certain current then dividing that current by the rating factor will give the normal cable capacity that is needed for this installation. Therefore with a load of 1A and a rating factor of 0.65 you would need a cable normally rated at just over 1.5A

    I am sure you can work out the ratings for the situation you have, but remember that you may want larger cables for different reasons  - voltage drop being an obvious candidate.

    Alasdair
  • 0
    Thanks for your answer Alasdair but I'm still lost with this calculation. The rating factor now makes sense so I multiplied my Ib by rated factor 0.65 and now I'm at 0.56. I'm lost in the 18th trying to find the right table to carry out this cable size calculation.
  • 0
    Stephen,  You are doing it the wrong way round.

    If you have a cable which can carry 1A then with the rating factor it can only carry 1 * 0.65 = 0.65A

    If you have an Ib of 0.83A then you will need a cable which can normally carry 0.83 / 0.65 = 1.23A

    Alasdair
  • 0
    You say the monitor is 20 Watt rated. So what current will the probe/sensor cables carry, not 20 W presumably. If they are connected to some type of sensor they may only carry milli-Amps. The monitor may take 20 W from the 230 Volt mains supply but the extra low Voltage sensor cables may carry much less than that. You need to check. The length of 0.5 m is very short and probably can be ignored. The sensor cables will not get hot in such a short length of conduit and be damaged. If carrying heavy mains' loads they may get a bit warm though.


    If I have got the wrong end of the stick then just use a readily available size of electrical cable above the current drawn by tthe load, say a 1.0mm2 cable for the short run and ignore any factors as that cable will be able to carry much more than you need. It is compact, inexpensive and readily available and nominally rated rated at 13.5 Amps, see B.S. 7671 Table 4D1A. You can ignore Volt drop as well here. Conduit factors etc. apply more to mains cables carrying heavy loads. See Appendix 4 of B.S. 7671. Table 4C1 and notes apply, mainly to mains cables carrying heavy loads.
  • 0
    P.S. Table 4C1 supplies rating factors for ONE circuit, OR for ONE multicore cable, OR  for a GROUP of circuits, OR a GROUP of multicore cables, used in conjunction with Tables 4D1A to 4J4A.


    The factors are applicable to uniform groups of cables equally loaded.


    If you have four cables, (two CIRCUITs) then your rating factor will be 0.8 if they run in conduit. See item 1 column. 


    See also notes 2 to 9 in this confusing section.


    As I said before small currents in detector circuits from probes/sensors over 0.5 m can probably be ignored. There will be virtually no heating effect at all in the cables.


    If by "alarms" you mean sounders rather than probes/sensors, then the same principle applies. Just calculate the load current of the devices and chose a cable that will carry the load current, again 1.0mm2 will be more than equal to the task. 


    Whereas several cables all carrying say 10 Amps in conduit will be of concern because the heating effect may be significant and insulation damage may occur with constant loads.


    C.




  • 0
    It is interesting to note that in the IET publication "Electrical Installation Design Guide" 4th Edition on page 17 this useful formula is shown:


    Iz= current carrying capacity of a cable for continuous service in its particular installed conditions.



    Iz = It Cg Ca Cs Cd Ci Cc.


    Just remove any correction factors not needed.


    C.
  • Former Community Member
    0 Former Community Member

    Stephen Brown:

    Hi guys I need some help calculating cable sizes for a 24v DC supply. I'm a student doing a job report and I want to explain cable sizing basically, and whilst on site my mentor, with his vast experience, knew what cable sizes would work so I just listened to him. Now I want to calculate it properly.


    So it's the nitrate monitor requires a 230v supply but 4 cables, 2 pairs each for 2 different alarms, I want to install are on the 24v DC side. (They're just going to a junction box that already has cables feeding back to the PLC).

    The cables will be installed in conduit on the wall less than 0.5m away. (distance between monitor and junction box).

    The monitor they're being connected to is only 20W.


    So the farthest I've got, (unless I'm wrong), using the 18th is this:

    Ib, (Design Current), = 20W/24v = 0.83A.


    I read in the 18th edition, (correctly I hope), that I will have a rating factor of 0.65 because there will be 4 cables enclosed. (Am I right?) But I've no idea what to do with that figure.


    I've really no idea where to go next or even if I'm on the right lines thus far. Can someone explain it to me? I don't necessarily want the answer, but a formula would be brilliant.


    Thankyou.






    This may or may not come under BS 7671, but the science is the same, just the method and terms may need tweaking slightly if you were to illustrate it under the alternative standards.

    The monitor is 20W of load, but is this supplied by 230V or by 24V, if so, then it is doubtful that the monitor will be powered by the signal cables, it will likely have a power supply, and separate signal wires.

    As far as the number of circuits goes, if you they are signal wires then you will have two circuits, and I very much doubt that the signal load is 20W, as the input impedance of things like monitors & PLC's is normally massive by design to carry the minimum of current so that it has the minimum affect on the devices being monitored.

    If they are two power circuits then you still only have two, 2 wire circuits.

     

  • 0
    The OP was concerned with the method of cable sizing according to the 18th edition of BS 7671 as he stated. He was interested to know how to apply correction factors when cables are bunched. This primarily applies to mains' cables  (LV) carrying largish currents as I have already said in my previous posts. There are insufficient details in his post to accurately answer his question regarding his monitor, but I believe that he has been put on the right track to undertake the application of correction factors in the future using BS 7671. I tried to provide simple straightforward answers to help him, and not get too bogged down with too much technical detail as he is a student. I have directed him to the relevant sections of BS 7671 for further information.


    C.
  • 0
    Questioning whether BS 7671:2018 should be used is actually quite valid.


    If the cabling is part of Machinery, as defined in the Supply of Machinery (Safety) Regulations, then it's worth also checking whether the particular correction factors line up with BS EN 60204-1, because:
    • Machinery (as defined in the Regulations) is outside the scope of BS 7671, and BS 7671 accordingly points to BS EN 60204-1; and

    • Whilst the approach in BS EN 60204-1 is based on the approach used in BS 7671 (i.e. based on the IEC 60364-series ), there are some minor differences when it comes to correction factors and ambient temperature effects etc.


    In other words, sometimes, you come up with the same answers using BS EN 60204-1, sometimes there are minor differences.


  • 0
    Hi everyone sorry it's been a while just been busy!

    Thanks for all of your input. Appreciate it.

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