Page 1 of 2 12 LastLast
Results 1 to 10 of 13

Thread: Load Calculation: 480V vs 460V

  1. #1
    Join Date
    Feb 2008
    Posts
    3

    Question Load Calculation: 480V vs 460V

    When calculating three phase motor loads I've always used 480V times 1.732 to get VA. I've recently seen this done with 460V by an extremely respected engineer. Can anyone provide clarification on this.

    Thanks

  2. #2
    Join Date
    Feb 2003
    Location
    Seattle, WA
    Posts
    17,199
    This is a debatable issue. Part of the problem comes from the fact that we are supposed to take the "running current" values from table 430.250 (instead of from the nameplate), and the fact that that table shows the voltage as 460. OK, we all know that the power system is rated for 480, and that (for voltage drop considerations, perhaps) the motors are rated at 460. My method is to take the tabulated current times 480 (then times 1.732), for two reasons. First, that is the voltage at which the system will operate. Secondly, it gives me a slightly higher (i.e., more conservative) value for the motor load in KVA. I figure that it is better to provide a little more capacity than to provide too little capacity.

    That said, do I think that the other engineer is doing the math incorrectly? I do not. It is a valid method as well.
    Charles E. Beck, P.E., Seattle
    Comments based on 2014 NEC unless otherwise noted.

  3. #3
    Join Date
    Sep 2009
    Location
    Los Angeles, Ca
    Posts
    646
    Many motor's are "rated" at 460 but of course allow of 10% voltage tolerance. I would say using 460V is not a bad idea since it takes into account possible voltage drop (depending on how far the motor loads are located) and also would result in a larger current (more conservative). Having said that, doing the calculation using 480V is not really incorrect.

    Edit: Haha, Charlie beat me to the punch.

  4. #4
    Join Date
    May 2005
    Location
    Richmond, Virginia
    Posts
    21,838
    Quote Originally Posted by charlie b View Post
    My method is to take the tabulated current times 480 (then times 1.732), for two reasons. First, that is the voltage at which the system will operate. Secondly, it gives me a slightly higher (i.e., more conservative) value for the motor load in KVA.
    Quote Originally Posted by skeshesh View Post
    I would say using 460V is not a bad idea since it takes into account possible voltage drop (depending on how far the motor loads are located) and also would result in a larger current (more conservative).
    Well, both of you can't be correct about arriving at a more conservative (i.e., "better to big than too small") value for current by using 480v vs. 460v, unless you're each doing the math differently.

    It seems that Charlie is using V and I to derive KVA, while Skeshesh is using V and KVA to derive A. I certainly agree that using the larger valu ofr ampacity is important when its close.
    Last edited by LarryFine; 08-27-10 at 03:19 PM.
    Code references based on 2005 NEC
    Larry B. Fine
    Master Electrician
    Electrical Contractor
    Richmond, VA

  5. #5
    Join Date
    Feb 2008
    Posts
    3

    Question

    Appreciate the responses. Do any of you have any thoughts on standardizing a voltage so there isn't confusion.

    Also, wouldn't an engineer utilize the higher of the two voltages that would increase service size (thus creating a system that was inherently safer) due to the their responsibility to "...hold paramount the safety, health, and welfare of the public."

    Thanks again.

  6. #6
    220.5 Calculations.
    (A) Voltages. Unless other voltages are specified, for purposes of calculating branch-circuit and feeder loads, nominal system voltages of 120, 120/240, 208Y/120, 240, 347, 480Y/277, 480, 600Y/347, and 600 volts shall be used.

  7. #7
    ANSI C84.1-2006, American National Standard for Electric Power Systems and Equipment—Voltage Ratings (60 Hertz)

  8. #8
    Join Date
    Sep 2009
    Location
    Los Angeles, Ca
    Posts
    646
    You know Larry, now that I think about it maybe Charlie is right afterall. Most motors do have nameplate info (or the info can be requested from mfg) that shows the current information (FLA, MCA, etc.) so having an accurate account of the current or being given the size of the conductor by the mfg. negates my argument of using 460V to arrive at higher amps. Since the current is given using 480V will yeild higher KVA and thus be more conservatives when considering total load for the system.

  9. #9
    Join Date
    Feb 2003
    Location
    Seattle, WA
    Posts
    17,199
    Quote Originally Posted by LarryFine View Post
    It seems that Charlie is using V and I to derive KVA, while Skeshesh is using V and KVA to derive A.
    From the wording of the question, I think that the mission is to calculate load for the purpose of sizing branch circuits, feeders, and services. If that is the case, then "amps" is an input value, obtained from a table, rather than a result.

    Charles E. Beck, P.E., Seattle
    Comments based on 2014 NEC unless otherwise noted.

  10. #10
    Join Date
    Jul 2004
    Location
    Texas by the Gulf, geat crab, great shrimp
    Posts
    5,407
    Quote Originally Posted by charlie b View Post
    This is a debatable issue. Part of the problem comes from the fact that we are supposed to take the "running current" values from table 430.250 (instead of from the nameplate), and the fact that that table shows the voltage as 460. OK, we all know that the power system is rated for 480, and that (for voltage drop considerations, perhaps) the motors are rated at 460. My method is to take the tabulated current times 480 (then times 1.732), for two reasons. First, that is the voltage at which the system will operate. Secondly, it gives me a slightly higher (i.e., more conservative) value for the motor load in KVA. I figure that it is better to provide a little more capacity than to provide too little capacity.

    That said, do I think that the other engineer is doing the math incorrectly? I do not. It is a valid method as well.
    The voltage drop itself is a load (I^2*R), so indeed, it is appropriate to use the supply voltage, not the utilization voltage.

    Example: a 12AWG single phase circuit will produce ~200VA @ 16A, 250'

Posting Permissions

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