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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.
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  • 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
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  • 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
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