Results 1 to 6 of 6

Thread: An exam question from a mike holt book.

Hybrid View

  1. #1
    Join Date
    May 2012
    Location
    kamiah, idaho
    Posts
    2

    An exam question from a mike holt book.

    Just prepping for a masters exam and want to get this right. Appreciate some clarification if you can figure out what I have typed in Thanks!

    Master exam #4 from the 2008 calcs prep book, #4:

    20-800 sq.ft. apartments
    20-12 kw ranges
    40- small appliance circuits
    20 - laundry ckts.

    Size the Neutral conductor to ______ amperes. 230/115v single phase.

    Answer according to book: 20x 800 sq. ft. x 3 va = 48000va
    small app. & laundry ckts. 60 x 1500va = 90000va
    138000va
    Table 220.42 Demand :
    First 3000va @ 100% = 3000va
    Next 117000va @ 35% = 40950va
    Final 18000va @ 25% = 4500va
    48450va


    Table 220.55 Column C : 35kw x .70% = 24500va
    Up to this point everything seems good, but I have a problem with adding it to the 48450va and then taking an additional demand deduction as shown below:

    ...its added onto the the 48,450 which equals 72,950va, and then he further says, 72,950va/230v = 317amps
    220.61: First 200a @ 100% and the next 117A @ 70% = 82A for a total of 282 A on the neutral"

    As I read 220.61(b) it says " A service or feeder supplying the following loads shall be permitted to have an additional d.f. of 70% applied to the amount in 220.61b1 OR portion of the amoutn int 220.61b2 determined by the basic calculation:

    1) a feeder or service supplying household electric ranges, w.m. ovens, c.m. cooking units, & elect. dryers, where the max. unbal. ld. has been determined in accordance with T220.55 and T220.54

    2) That portion of the unbal. ld. in excess of 200a where the feeder or servaice is supplied from .... etc...

    I would figure that the 48,540va \ 230 would be 211a - first 200 at 100% and the 11A at 70% making it 208A and the Ranges at the 35,000KW x's the 70% and divided by 230v would be 107A and the two added together would be 315A. Is Tom right (hes a pretty smart guy, but sometimes I run into clerical errors in his books, Mikes too--- no one is perfect! Or would I use the second method when testing?

  2. #2
    Join Date
    May 2012
    Location
    kamiah, idaho
    Posts
    2

    Clarification--- Its a Tom Henry book, not a Mike Holt book.

    Calculations for the Electricians Exam 2008 NEC

  3. #3
    Join Date
    Aug 2011
    Location
    south texas
    Posts
    284
    i can't believe no on e has chimed in on this one yet. i ran the numbers and i think ( i may be wrong) that it's 315 amps
    Last edited by PEDRO ESCOVILLA; 07-11-12 at 05:27 PM. Reason: typo

  4. #4
    Join Date
    Jun 2012
    Location
    Dumas Tx. 79029
    Posts
    42
    I came up with a neutral load of 304a is there a reason why you are dividing by a nominal voltage of 230 and not 240

  5. #5
    Join Date
    Jun 2012
    Location
    Dumas Tx. 79029
    Posts
    42
    Sorry, I see in the question now my bad

  6. #6
    Join Date
    Jun 2012
    Location
    Dumas Tx. 79029
    Posts
    42
    I believe the answer is correct. There is a good example of this in annex d

Tags for this Thread

Posting Permissions

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