Ask Me Help Desk

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)
-   Electrical & Lighting (https://www.askmehelpdesk.com/forumdisplay.php?f=105)
-   -   Largest Motor calculation (https://www.askmehelpdesk.com/showthread.php?t=488970)

  • Jul 16, 2010, 04:17 PM
    donf
    Largest Motor calculation
    Greetings,

    I'm taking on a class for the NEC 2008. Right now we are doing Article 220.

    Most of this is straight forward but one trivial nit is just eluding me.

    I understand how to get to the largest motor. What is eluding me is finding in code where we are allowed to take a 25% demand factor against the "Largest Motor" so that you can then use the calculation for the calculated load.

    Also, where is it supported in code that an additional 25% can be taken off this calc. when you are reducing the neutral load.
  • Jul 17, 2010, 03:08 AM
    tkrussell

    Not sure what your referring to.

    Exactly what section are you working on?

    Motor loads are to be calculated at 125%, not 25% of the load.
  • Jul 17, 2010, 07:41 AM
    donf

    T.K,

    I'm working on Article 220. Part III (220.40). Article 220.50 Motors.

    The Gen. Use, Heat, Range, Dryer and appliance VAs have been calculated.

    Next the instructor told us to take the largest motor, (for our example it was the dishwasher at 1kVA. The multiplier is .25) Than the resultant 250 VA are added to the other VAs.

    My problem is that I cannot find anywhere in the article 220 that allows such a deduction or even puts that into the calculated load.

    I've read through Articles 430.24,25,26 because they are referenced by 230.50.

    I understand how to calculate the Full Load Current from T430.248. The ampacity calculation derived from T430.248 is a not factor (at least as far as I know) in determining the Feeder/Service Load.

    But I have no idea where the Lg. Motor calc. is coming from. Or more important for me as to why it is needed since I've already applied the 1kVA for the D/W in with the appliance load.
  • Jul 17, 2010, 08:53 AM
    tkrussell
    OK a little confusing, but I think I got what your problem is.

    The instructor.

    Did I say that out loud?

    A DW is an appliance load, not a motor load. I think this is where the issue is.

    A DW has both motor and heat load. The motor is not 1 KVa. Most of the load is heat.

    In any case, a DW is not considered a motor load, it has a combination of loads and is considered an appliance.

    If the instructor is looking for "the largest motor" to include in the service calcs, I would think a refrigerator motor is larger, but it does not apply either.

    Why is he worried about a large motor on an simple residential service rating calc?

    Read Sec 220.82, then follow the example in Annex D # D2a.

    If your DW is 1Kva, it is added to the general load calculation, same as a range, dryer, water heater.

    Next, the 25% issue. What are you trying to size? The 125% is used to size a branch circuit. Nothing is increased by 25% when calculating service load.
  • Jul 17, 2010, 01:47 PM
    donf

    No, what we were told is that we are applying a demand factor of 25%. So that means the 1000kW is multiplied by 25% (or 0.25) to reduce the load to 250 VA.

    The load is not increased.
  • Jul 18, 2010, 06:02 AM
    tkrussell
    Please tell me exactly where in the Code this is stated to reduce a load by 25%.

    Also, show me how you reduce a 1000 watt load by 25% and get it down TO 250 watts.

    I still not sure what your trying to calculate or size, a branch circuit or a service.
  • Jul 18, 2010, 06:27 AM
    donf

    Thank you, you are now where I am.

    I cannot find this deduction anywhere in code. That's the reason I posed this question.

    As to the math, 1000 X.25 = 250.
  • Jul 18, 2010, 06:37 AM
    tkrussell
    You cannot find it because it does not exist.

    Think about it, we need to increase load of branch circuit, motors , etc, by 25%, to size branch circuits, etc.

    To size services, we add up the loads, and apply either 35% or 40 % derating factors, which means reduce the load, but this is only to specific loads.

    Only ranges and dryers can be reduced if there are more than one unit.

    So the instructor is telling you to reduce a load by 25%? He needs to tell you exactly where he is getting this from.

    I suggest you review the examples at the back of the book.
  • Jul 18, 2010, 07:41 AM
    donf

    T.K.

    I know you are a stickler for accuracy, that's why this is driving me balmy.

    It started a simple homework asignment that I decided to add code supported references to. Just for my reference.

    I hit this calc. and then hit a wall!

    Thanks for the help.
  • Aug 5, 2010, 02:47 PM
    donf

    TK,

    Well Part I of the NEC class is over and I was just kicking around this gnat of a problem yet again, when all of a sudden the light bulb went on. As you promised me, if I did the due diligence I would finally understand my own question. Please let me know if I am finally correct in my thinking. I was looking at this from a pure mathematical description, not code calculation.

    To me, when you multiply a number by 25%, you are reducing that number to 1/4th. Of its original value. However, because the code requires that you size the ampacity of a motor to 125% of the FLC that you then need to calculate what the 25% is and then add it back into the calculation. What you are actually doing is completing the calculation 125% of the FLC.

    For example, if the largest motor is an 8 A, 240V motor, the base VA would be 1920 VA. This figure is then included in with the general appliance load calc. After all of the fixed appliances totaling is completed and then you can reduce the total VA by 75%, if allowed. Now you add the 25% (480 VA) back to the total appliance load.

    Did I finally get this correct or am I back in Brooklyn, NYC trying to find a way back to the Bronx, by way of New Jersey.
  • Aug 6, 2010, 07:06 AM
    tkrussell
    Motor circuits, etc are size by amperage, not VA.

    To size a branch circuit, for example Section 430.22 states , paraphrased, "Conductors shall be shall have an ampacity not less than 125% of the motor's full load current"

    If, as in your example, a motor has a RLA of 8 amps, 8 x 1.25 = 10 works for me.

    Elsewhere the Code may state" increase", again 1.25 works.

    The Code is very specific, each word is carefully placed. Each statement must be read in it's entirely to undertand what the statement is trying to say.


    However, I need you to show me where the 10 amps needs to be calculated in a general appliance load calc.
  • Aug 6, 2010, 09:48 AM
    donf

    TK,

    At the time we were doing Chapter 2 - 220.50.

    Regardless, the initial question is best illustrated by Annex D, example D1(B).

    You have completed all of the general calculations as well as Dryer, Range, Heat or Air (Largest of the two).

    Next, the Appliance load is calculated. If you have four or more fixed in place appliances, you total up the VA for the appliances and then take 70%.

    Correct so far?

    Now after all of those calculations are completed, you take 25% of the largest Motor's VA and add that to the sub-total of all the other calculations.

    This becomes the final VA calculation that you divide by 240 VAC to determing the required amperage for the ungrounded conductors. You then size the conductors according to 310.15 (B)(6).

    At that time, you can also size you Neutral based on 220.61 (not for discussion here.

    That's my synopsis, did I follow the logical process (Annex D - Example D1(B)) correctley?

  • All times are GMT -7. The time now is 01:06 AM.