Voltage Drop Calculation

[andytang]

Per NEC table 310.15B2a, you need to use a derating factor of 80% for 4-6 current carrying conductors in the same conduit. In that case, how do you deal with the voltage drop? In other words, is there any differnce in voltage drop when you use one big conductor or two smaller equivalent conductors running in a conduit? Please advice.

 

[jbartos]

Suggestion: Normally, if one conductor is replaced by two subconductors forming the same cross-sectional area equal to the big conductor, the voltage drop is approximately the same. Two subconductors will have a marginally smaller impedance than one big conductor; therefore, there will be the marginally smaller voltage drop. See for example the "skin effect" factor.

 

[Marke]

Hello andytang

The rating of cables is a function of the temperature rise when rated current is passed through them. Although many assume that the important parameter is the cross sectional area, this is not strictly correct. The temperature rise in a cable or busbar is a function of the power dissipated (I2R) and the ability to dissipate that heat. The power dissipated is dependant on the current and the electrical resistance where the resistance is inversly proportional to the cross sectional area, but the ability to dissipate that heat is dependant on the thermal resistance of the conductor and that is inversely proportional to the surface area of the cable.
The electrical resistance is also influenced by skin effect where the current tends to flow on the outer region of the cable.
If you take two cables in parallel as opposed to one single conductor, the surface area is effectively higher on the two cables and so they can have a lower cross sectional area increasing the resistance and therfore increasing the voltage dop. Derating the cables to 80% will reduce the resistance a little and there fore reduce the voltage drop, but I would expect to see a higher voltage drop with parallel cables operated at their parralel rating, than with a single cable operated at its rating.

Best regards, Mark Empson

http://www.lmphotonics.com

 

[jbartos]

Suggestion to the previous posting marked ///\\\:

If you take two cables in parallel as opposed to one single conductor, the surface area is effectively higher on the two cables
///Agree\\ and so they can have a lower cross sectional area increasing the resistance
///Agree\\ and therfore increasing the voltage dop.
///This remains to be seen, since the heat dissipation Hd is dependent on conductor resistance, surface and current flowing through the conductor. Therefore, approximately:
Hd1 ~ R x I**2
now considering two subconductors and the current divided evenly to 1/2
Hd2 ~ 2 x R x (I/2)**2 = (1/2) x R x I**2
Therefore:
Hd2 ~ (1/2)Hd1
There will be smaller heat dissipation and smaller voltage drop.\\\