Lets say your doing a 3-phase delta system with no line-to -neutral loads but your pulling a grounded/neutral conductor for fault conditions when sizing this grounded conductor, and all the conductors are in the same conduit you would add the circular mils of the conductors of any 1 phase and then divide by 12.5 % for your grounded conductor. So my question is how can you derate the ungrounded conductors when your calculating them as 1 larger conductor for the neutral?
derating of parallel runs?
- Thread starter nunu161
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Those are two totally different issues. We are derating because of the I squared R losses that produce heat. Each of the ungrounded conductors produce heat and so you must count each of them for derating purposes.
Also there is no neutral conductor in a delta system. The EGC is not sized based on the circular mil area of the ungrounded conductors, it is sized based on the size of the OCPD that protects the conductors. The method you show would be used to find the minimum size of an unused grounded conductor for a service. It is sized this way to provide an adequate fault clearing path.
Also there is no neutral conductor in a delta system. The EGC is not sized based on the circular mil area of the ungrounded conductors, it is sized based on the size of the OCPD that protects the conductors. The method you show would be used to find the minimum size of an unused grounded conductor for a service. It is sized this way to provide an adequate fault clearing path.
i did state that the neutral was only being pulled for ground fault conditions but if you are combining the phase conductors to calculate the neutral size your considering them to act as 1 large conductor i don't see how u can add them and then have to subtract them for derating purposes
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- Illinois
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- retired electrician
Like I said, they all produce heat and heat is the reason we derate. The two issues are in no way related to each other.
- Location
- Illinois
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- retired electrician
How is it a double standard? One issue is how much current could be supplied into a fault and having a fault return conductor large enough to clear that fault. The other issue is how much heat is produced by the load current and keeping that heat low enough to prevent damage to the conductor insulation.
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