Page 11 of 11 FirstFirst ... 91011
Results 101 to 107 of 107

Thread: Can someone please explain this to me?

  1. #101
    Join Date
    Feb 2003
    Location
    Post Falls, ID
    Posts
    74
    Quote Originally Posted by jrohe View Post
    I suspect they are expecting you to use the method that uses DC-constant resistance for determining voltage drop to calculate the minimum wire area, in cmils, required.

    For single phase circuits (assumed because that information is missing in the test in the question), the equation is:

    VD = 2 x K x I x D / CM

    Rearranging the equation to solve for CM results in the equation:

    CM = 2 x K x I x D / VD

    Making assumptions for copper conductors and VD% cannot exceed 3 percent because the information is missing in the test question:

    K = 12.9 ohms
    I = 30 amps
    D = 100 feet
    VD = 230 volts x 0.3 = 6.9 volts

    Therefore:

    CM = 2 x 12.9 x 30 x 100 / 6.9 = 11217.4 cmils.

    Looking at Chapter 9, Table 8, the smallest conductor with this area is #8 AWG conductor at 16510 cmils. A #10 conductor has an area of 10380 cmils, which is too small, at least according to this calculation method.

    I agree completely with this reply by Jason; he is spot-on correct.

  2. #102
    Join Date
    Jun 2003
    Posts
    6,601
    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
    Quote Originally Posted by oldsparky52 View Post
    So, ... as PF goes farther away from 1, the resistance in the circuit becomes less? So the worse the PF the less resistance?
    Not quite. First of all, we are talking about the wire impedance, not just resistance.

    And a low PF makes the voltage drop a little less for small wire where the wires resistance is fairly large compared to the wires reactance.

    But for larger wires, where reactance is larger, the voltage drop is less for higher power factors, with 2/0 wire actually having about the same drop for 95% PF and 80% PF.

  3. #103
    Join Date
    Jan 2009
    Location
    Maryland
    Posts
    11,421
    Quote Originally Posted by Ingenieur View Post
    perhaps
    but he is correct
    Agree with both.
    "Electricity is really just organized lightning." George Carlin


    Derek

  4. #104
    Join Date
    Feb 2009
    Location
    Durango, CO, 10 h 20 min without traffic from winged horses.
    Posts
    8,319
    Quote Originally Posted by jumper View Post
    Agree with both.
    Don't push me.
    Once in a while you get shown the light
    In the strangest of places if you look at it right. Robert Hunter

  5. #105
    Join Date
    Jan 2016
    Location
    PA
    Posts
    3,453
    Quote Originally Posted by jumper View Post
    Agree with both.


    Without being too snarky not sure how he could disagree lol



  6. #106
    Join Date
    Aug 2017
    Location
    Orange tx
    Posts
    5

    8 awg is correct!

    Mike uses the formula 2xKxIxD/Vd to find the circular mils of the wire.
    nec recommends no more the 3% so..
    vd=6.9 voltage drop (3%of 230)
    D=100' distance
    I=30a current
    K=12.9 for copper wire.

    cmils= 2x12.9x30x100/6.9 = 11,217cmils

    Chapter 9 table 8 back of book
    10awg = 10380 =too small to limit vd to 3%
    8awg = 16510= correct

  7. #107
    Join Date
    Dec 2007
    Location
    NE Nebraska
    Posts
    32,508
    Quote Originally Posted by Daniel malack View Post
    I just looked on several breakers GE 40°, Sq D home line 40°, Sq D QO 40°, Siemens 40°, Cutler Hammer Ch 60-75° Cutler Hammer bolt in 60-75° most are 40°. I got no reason to lie to ya.

    Sent from my SM-T350 using Tapatalk
    40 deg rating is not terminal temp rating it is the ambient temp that the trip rating was calibrated for.

Posting Permissions

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