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18th Edition Cable Calculations

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.
Parents
  • 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
Reply
  • 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
Children
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