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Zach Stone, P.E.

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Zach Stone, P.E. last won the day on October 12 2019

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About Zach Stone, P.E.

  • Rank
    Lead Instructor for Electrical PE Review (.com)

Previous Fields

  • Engineering Field
    Electrical Power, Industrial Controls, Field Troubleshooting, Construction Management, Turbo Generators and Power Generation, 24hr Production Facilities
  • License
    PE
  • Calculator
    TI
  • Discipline
    Electrical

Contact Methods

  • Website URL
    http://www.electricalpereview.com

Profile Information

  • Gender
    Male
  • Location
    USA
  • Interests
    Teaching Electrical Engineering concepts

Recent Profile Visitors

4,215 profile views
  1. @speakeelsy Great job! You'll find yourself scoring higher and higher on practice exams the closer you get to the actual PE exam date. It's good to get used to the "shock factor" of attempting to solve questions that you are unfamiliar with now, and learning to use your references as tools.
  2. Here is the full length solution to the problem for anyone else that wanted to work through it. The solution explains the why D is the correct answer, and why answers A, B, and C are incorrect: (you can click the image of the solution to make it bigger)
  3. To be fair it's also a really tough problem. We put a lot of effort into making the qualitative questions in the practice exam challenging and thought provoking since qualitative questions are the hardest to prepare for.
  4. Bingo! It's a bit counter intuitive at first thought. It's also why transformers are commonly referred to as current limiting devices under fault conditions. Another way to think of it is that inductors by nature slow, or "limit", the rate of change of current and are commonly used in filters and power conditioners for this very reason. Transformers are really just inductors with the primary and secondary coils manufactured inside of each other to allow for electromagnetic induction.
  5. Hi @Swift Fox, do you not have the solution? This question is from the Electrical PE Review Practice Exam.
  6. Here's the trick: Most books do not explain this well and gloss over it entirely. If you use a KCL equation to solve for the neutral current (In) from the three line currents (Ia, Ib, Ic) of a wye wye system, it looks like this: In = Ia + Ib + Ic However, the direction of the neutral current is typically shown from the source to the load, just like the line currents. This is where the negative sign comes from. How do we change the direction of the neutral current to express it mathematically as flowing from the source to the load? We multiply it by negative one, or lead/lag the phase angle by 180 degrees (both have the same affect). The neutral current formula then becomes: In = -(Ia + Ib + Ic) or In = -Ia - Ib - Ic or -In = Ia + Ib + Ic In this particular problem since there is no C line current, the following happens once we substitute Ic = 0 into the previous formula: In = -(Ia + Ib + 0) In = -(Ia + Ib) However, since the problem is only asking for the magnitude of the neutral current, and not the complex neutral current with magntidue and angle, the negative sign does not matter. Multiplying a complex number by negative one is really just leading/lagging the phase angle by 180 degrees, it has no affect on the magnitude. So even if you solved it without the negative sign for the neutral current, you'll still end up with the same magnitude of 29.3A that the author did. Try it:
  7. Hi @BebeshKing, the problem is essentially asking for the amount of fault power that originates from the generator that makes it through the transformer all the way to the fault bus. Transformers are known as "current limiting devices" because during fault conditions there is only so much current (or fault power) that they will "let through" to the other side, even if the transformer is being fed from an infinite bus. Because the transformer is in series with the generator, you'll have to take both into consideration when solving this question.
  8. Because the impedance of the line is also being taken into consideration here. The approach you are trying: "Why can't I just do the normal S(xfmr)/Z%, then use Sf to find Isc by doing Sf/sqrt3*VL?" would be perfectly fine if the line impedance was not being taken into consideration. You can still use your approach (MVA Method) by finding the fault duty of the transmission line in MVA, and taking the reciprocal sums of the transformer and T line: Sf = S_XFMR//S_Line Below is a video example of a Fault Current Analysis practice problem solved using the MVA method with a line impedance in series with a transformer. is a little more in-depth than the NCEES practice problem, but it should give you a good idea of how to calculate the fault duty of the transmission line in the NCEES practice problem and then how to solve it using the MVA method (the video should start at 7:46 when the fault duty of the line impedance is being calculated): The only difference is that the NCEES problem gives you the impedance of the line in ohms/1,000 ft. You will need to multiply the Ω/1,000 ft by the total length of the T-line in order to convert to the total amount of impedance in the line before calculating the fault duty of the line.
  9. Thanks for the mention @Saul Good. I'm happy to hear that our online study program for the Electrical PE Exam helped you pass.
  10. Happy to hear that our online program for the Electrical PE Exam helped you pass on your first try!
  11. Happy to hear that our free videos on our YouTube channel helped you pass on your first try!
  12. Hi @pigking8190 thanks for showing an interest in our program. For those interested, I always recommend by starting with the free trial of our online program. No credit card is required and you can study from the material contained in the Free Trial for as long as you like: Electrical PE Review - FREE Trial
  13. Thanks for the mention @MadamPirate and @MEtoEE. Our material is located at www.electricalpereview.com if anyone is looking for more help with the electrical power PE exam.
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