Page 13 of 13 FirstFirst ... 3111213
Results 121 to 129 of 129

Thread: Can someone please explain this to me?

  1. #121
    Join Date
    Oct 2009
    Location
    Austin, TX, USA
    Posts
    8,483
    Quote Originally Posted by Smart $ View Post
    Actually it is not a constant though many call it such.

    It's a value based on the conductivity of a material. Units are Siemens...
    Sometimes Mhos, right?

  2. #122
    Join Date
    Dec 2012
    Location
    Placerville, CA, USA
    Posts
    18,377
    Quote Originally Posted by ggunn View Post
    Sometimes Mhos, right?
    Used to be Mhos until the ceremonial namers decided it was not right to waste a naming opportunity on a pun.

  3. #123
    Join Date
    Apr 2017
    Location
    New England
    Posts
    67
    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?

  4. #124
    Join Date
    Jun 2006
    Location
    Ohio
    Posts
    21,385
    Quote Originally Posted by crispysonofa View Post
    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.

  5. #125
    Join Date
    Jun 2006
    Location
    Ohio
    Posts
    21,385
    Quote Originally Posted by crispysonofa View Post
    ... For a more accurate method?
    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.

  6. #126
    Join Date
    Oct 2009
    Location
    Austin, TX, USA
    Posts
    8,483
    Quote Originally Posted by Smart $ View Post
    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.
    And that's not to mention that the loads are usually only an estimate, anyway.

  7. #127
    Join Date
    Jan 2016
    Location
    PA
    Posts
    4,046
    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



  8. #128
    Join Date
    May 2013
    Location
    Massachusetts
    Posts
    2,269
    Quote Originally Posted by Smart $ View Post
    Actually it is not a constant though many call it such.

    It's a value based on the conductivity of a material. Units are Siemens, and the reciprocal of resistivity. A conductor of uniform cross section has a resistance proportional to its resistivity and length and inversely proportional to its cross-sectional area.

    Go to Chapter 9 Table 8. Pick any conductor and multiply its cmil area by its resistance per 1,000 ft and divide by 1,000.
    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.

  9. #129
    Join Date
    Jun 2010
    Posts
    1,923
    Quote Originally Posted by steve66 View Post
    Actually, I think the Southwire calculator assumes 3 phase. That makes a bigger difference than the power factor. I don't seen anywhere on the Southwire calculator where I can choose 1 phase.

    But back to the power factor - it makes some difference in the calculated impedance of the wire. If you know R and X for a wire, then Z depends on R and X and the power factor.

    Z = X*PF+R*Sin(Acos(PF))

    (I think that's right - I'm trying to read it from an excel formula.)

    Just as an example, my Square D calculator lists L-L voltage drop per amp per 100' in magnetic copper wire: (trying to make a table, which isn't working)

    ....................... ..... #8 Wire #10 wire
    1 phase: 90% PF. . 0 .15 ....... . 0 .23
    1 phase: 80% PF ...0.13 ........ 0 .20
    3 phase: 90% PF .. 0 .13 ........ 0 .20
    3 phase: 80% PF... 0 .11 ........ 0 .17
    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 08:59 AM.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •