Ok, that makes sense they all come to around 13, so would this be more accurate than just using 12.9 for K copper? Could I plug in cmil X OhmKft / 1000 = K? For a more accurate method?
If you plug it in the cmil's cancel with one being a numerator, the other being a denominator.
Vd = 2KIL/cmil = 2RIL/1000
...where R is the resistance (or impedance) per 1000ft from the tables.
I use "D" for distance instead of "L" for length and call it the "2 RID" formula as a mnemonic.
I will have achieved my life's goal if I die with a smile on my face.
Accuracy is a fleeting concept in voltage drop calculations. You get closer by using the actual current versus the calculated current. You get closer by using the IEEE exact formula. And even then, you have to know the exact temperature of the conductor. Just a change of a few degrees in temperature will change the impedance. The K value of 12.9 is probably closer to 60°C. The table values are at 75°C. If the conductor is only conducting 60% of the calculated current on average, the conductor may never reach those temperatures, and thus the impedance would be lower... on average.
I will have achieved my life's goal if I die with a smile on my face.
imho using a basic calc like (assume 3 ph)
1.732 x L x i (rated ampacity not load) x Z (nec tables) / rated v x 100% will keep it simple and safe
also considering
load i is usually < ampacity
the sf of 1.25 for cont loads/motors
the utility v is usually a few % > rate nominal
many loads will operate at less than nominal, eg, a 480 motor is rated at 460
no right, wrong or absolute
multiple ways to skin this cat lol
The units of K are actually cmil-Ohms/ft, which are dimensionally the same as Ohm-meters, the unit for resistivity.
K is simply another version of rho for resistivity, that you would look up in a Physics handbook. The difference is that it is expressed with cmil instead of m^2 as the unit for area, and ft instead of meters as the unit for length.
Conductivity is the reciprocal of resistivity, just like conductance is the reciprocal of resistance. The word ending in "-ivity" is a material property, independent of the geometry of what you make out of the material. The word ending in "-ance" is a specific component's property, which does depend on geometry. Conductance has units of Siemens (originally mhos). Conductivity has units of Siemens per meter.
Usually the electrical world thinks in terms of resistivity, while the thermal world thinks in terms of conductivity. Thermal conductivity has nothing to do with Ohms, Volts, or Amps, but instead uses power and temperature units.
You always use I-rms for the purpose of thermal rating but what I am wondering how you should calculate for capacitored rectifier loads that draw near the peak. The resulting DC link voltage is dependent on the peak and these devices are most concerned with the Vpk and not so much about RMS voltage. There is no phase shift, but the PF in the range of 0.5-0.6.
A 208v input power supply that consumes 1kW with 0.6PF draws 8A RMS. These things often have a crest factor of around 3 meaning that it draws a peak of 3x the RMS value or 24A which happens right at the voltage peak.
When you use Vpk 295v and Ipk 24A, the percent volt drop at the peak is a bit more than double what you'd get calculating using RMS values.
Using 200ft of #12
208v RMS
8A RMS
202.9v RMS
2.4%
295v Pk
24A pk
280v pk
5.2%
Last edited by Electric-Light; 09-09-17 at 07:59 AM.
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