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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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  • 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.

     

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  • 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.

     

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