DilutedAr18 - Engineer Boards Jump to content
Engineer Boards

DilutedAr18

Member
  • Content Count

    102
  • Joined

  • Last visited

Community Reputation

57 Excellent

About DilutedAr18

  • Rank
    Engineer Intern

Previous Fields

  • Discipline
    Electrical

Recent Profile Visitors

The recent visitors block is disabled and is not being shown to other users.

  1. Correct. The other answers are somewhat correct, but D is most correct. That is what I elaborated on in the first paragraph.
  2. Not sure where an actual answer is in a book, but logically it is correct. When you size the overcurrent protection, you have to size the conductors to meet your load. The insulation has a temperature rating that determines how many amps the conductor is rated for. The overcurrent protection has to be sized so that if the ampacity is greater than the load it is sized to handle, the circuit breaker will trip and protect the insulation of the conductor. For thermal/magnetic circuit breakers, the magnetic portion protects against faults and the thermal portion protects against overloads; both to
  3. Correct. Although, that would have actually been relevant to the question, since they listed that there were two motors. I would say that if that answer were also included, I would select that one as it would be most correct. That’s why I say it is a bad question. The answer isn’t really relevant when it only lists motors on the circuit.
  4. Those are the rules listed in 430.24. Notice the title “Several Motors or a Motor(s) and Other Load(s). I only have my 2014 version here at home, but those rules haven’t changed in the newer versions.
  5. The question is a bad question. The answer is not how you would actually size this load for a two motor circuit, as it does not indicate that there are any other loads on the circuit beyond those motors. Answer B is correct b/c it is the only one that would comply with the rules for sizing a multimotor circuit regardless of what loads were on the circuit. Answer A is not correct b/c it is 125% of the continuous non-motor load, not 100%. Answer C is not correct b/c it is 125% of the FLA of the highest rated motor, not 100%. Answer D is not correct b/c it is 100% of the FLA of other motors
  6. Did the specify a maximum or minimum? If not, how would you know what value to use?
  7. That table is only valuable for getting a possible range of KVA based on the design “letter” (in this case C), not Locked Rotor Current. Read the table column labels closely. In practical use, the LRC is used for sizing disconnects, so Table 430.251(A) or (B) would be the useful table. If you look back at question 2-34, that is the type of a question you could expect to use Table 430.7(B). They would need to ask for a maximum or minimum KVA or could spin it by giving you a number and have you find the letter to fit the “KVA/HP with Locked Rotor”.
  8. Not that I’m aware of. I’ll use Bluebeam at work tomorrow to check for where there are changes. It can compare documents for differences. In glancing at it briefly, I didn’t notice what changes were made.
  9. The question is vague enough that I would argue that you could reasonably assume that it is a single load especially when the question is asking for the maximum overcurrent protective device. I use the NEC everyday and I find that many of the people who write practice tests with NEC questions don’t know the intricacies of the code. There are plenty of instances where things are misapplied. Where a table will indicate specific requirements to use it and the question will not meet these requirements, but the solution will use it.
  10. It is because the 300% for fuses or the 250% is the maximum overcurrent protective device. You can size the fuse or circuit breaker as small as you want, but it might trip depending upon your loads and inrush currents.
  11. I’m not using any other literature at this point. Any time I run into a practice problem that I know I have the answer in one of my resources, I do my best to recall it. Then go back to that section and read it over a few times to try to memorize it. It is so stupid to have to resort to memorizing stuff, but that is where we are now with this ridiculous handbook.
  12. You are correct if you are looking at the older version of the handbook . Instead of , it should be . And Instead of , it should be . They have corrected it in version 1.1.
  13. I think you understand how to use the formula. They have attempted to simply the formula from a form with no angle and a formula with an angle. The formula without the cos(Θ) is the inverse square law. With the cos(Θ) is the inverse square cosine law. When Θ is 0 degrees, the cos(Θ) is 1 and the distance would be D. If Θ doesn’t equal 0, then you would use D’ as your distance. I am going to send in a comment to NCEES to have them remove the “x” from the formula, because that is really only meaning multiply, but “x” is used as a variable in many other instances. I have attac
  14. Just found that. Why would you define Vline as Vmax? There is no reason to redefine variables where it makes your formula shorter. Use the typical formulas found in textbooks.
  15. *Edit: I see this was already discussed. No idea why they would define Vline as Vmax. I don’t know of any other cases where you would use Vline as the maximum value. It just makes no sense...* I discovered an error in one of the formulas this morning, so I thought I would start a thread for discussion of any errors that anyone else comes upon. There is an error in the power PE reference handbook. The formula for capacitance required to change the reactive power on page 65 is missing a “2” in the denominator as the whole formula’s denominator should be “ω*(Vline)^2”, where ω=2*π*f as
×
×
  • Create New...