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BigWheel

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BigWheel last won the day on June 9

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About BigWheel

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    Leeloo Dallas; Multi-Pass

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  • Gender
    Male
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    Alabama
  • Interests
    Guitar, Music Recording, FPGA development, Microcontrollers,

Previous Fields

  • Engineering Field
    Architectural Engineering
  • License
    PE
  • Calculator
    TI
  • Discipline
    Electrical

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  1. Engineering Economy

    I'm not trying to give anything away - I'm just trying to say that Engineering Economy questions on the PE are kind of "softball" questions compared to the FE exam where they were wanting to make sure you understood the underlying principles of "Engineering Economics." I don't know how to speak to your specific experience, as you're an examinee who is trying to figure out where to spend your valuable study time, but I would say if you think you could pass an FE-type style of Engineering Economics question, then I think you could pass a PE-style type of Engineering Economic question. The FE is looking to make sure you understand the underlying principles while the PE is looking to ensure that you understand how the underlying principles can be used to make something beneficial to society... Does this make sense?
  2. Engineering Economy

    HA! I thought the FE Engineering Economics was more comprehensive, while the the PE was much simpler. If you did well on the FE Engineering Economics problems, you'll have no trouble with the PE.
  3. Help explain Lagging/Leading PF

    It might help to think of "leading" vs. "lagging" in terms of current and voltage instead of positive and negative. Power factor is generally used in calculations as an absolute value, so the sign of the PF is irrelevant. A leading power factor means the current is "leading" the voltage (i.e., current crosses the zero-axis first) while a lagging power factor means that the current is "lagging" behind the voltage (i.e., current crosses the zero axis second).
  4. Graffeo - Single Phase Problem

    Yep, that's true. The three phases serve as return paths for each other, but I think dwelling on this too much will just create more confusion than necessary for people trying to prepare for the exam when all they want to know is why a single phase VD calculation counts the conductor length twice while the 3-phase VD calculation only counts it once. There are two types of 3-phase configurations: delta and wye. In a delta configuration, there is no neutral, so that was easy - just one conductor length is considered. In a wye configuration, there is a neutral. The return current on the neutral in a grounded-wye three-phase system is a vector sum of the three (3) phase currents. In a balanced, grounded-wye configuration, the vector sum of the currents will result in 0 amps in the neutral, so no voltage drop in the neutral; therefore, only one conductor length is considered. In an unbalanced, grounded-wye configuration, the current in the neutral depends on what the unbalanced conditions are. It can be very close to zero, meaning we're almost (but not quite) balanced (typical), or it can be as high as one of the phase currents alone (atypical, and is just one of the reasons why the NEC requires that the neutral conductor in a grounded-wye configuration is the same gauge/size as the phase conductors). But the total three-phase voltage drop is a vector sum of the voltage drop across each phase and the neutral. The sqrt3 simply falls out of the 3-phase formulae derivations. When performing 3-phase voltage drop calculations in practice, a balanced three-phase load is usually assumed because you're working with something like a motor, pump, transformer, VFD (intrinsically balanced), or you're hanging a new three-phase panelboard, and if you're doing that (hanging a new panelboard), a good engineer recognizes that it's good practice to arrange the circuits such that the anticipated loads are balanced across all three phases of the new panelboard.
  5. Graffeo - Single Phase Problem

    In a single-phase system, current travels to the load on the line conductor and returns on the neutral conductor, so this "series" circuit voltage is dropped across both the line and neutral conductors. In a three-phase system, you have two conditions: balanced and unbalanced. In a balanced three-phase system, there is no current in the neutral conductor (each phase's current cancels the other two phase's currents). Some examples of balanced 3-phase loads include 3-phase motors, pumps, and VFDs. Conversely, with an unbalanced 3-phase system, current is produced in the neutral; however an unbalanced 3-phase system is rarely (if ever) calculated considering the line and neutral impedances because the effort to perform the calculations isn't worth the results. I'm happy to talk to this and prove what I just claimed, but I'll spare the community until pressed to do so, but I will conclude with this: a three-phase system is not a series system; it's a parallel system, so the math is considerably more complicated than a single phase analysis, and the kicker is that you can do the math and calculate the true 3-phase voltage drop of an unbalanced 3-phase system, but the results of these calculations will leave you wondering why you even bothered. Finally, it helps to understand why "Voltage Drop" is such a hot topic. Is it because it's a code requirement? It's not a code requirement, but rather a recommendation. It is absolutely true that too much voltage drop will cause the load to behave erratically, and can even be a very dangerous (and even deadly) situation for craftspersons working at the load; nevertheless, it's not a "CODE REQUIREMENT." The NEC recommends no more than a 3% voltage drop on the branch circuit and no more than 2% on the feeder. Voltage drop calculations are essential to the difference between a true engineered design and a classroom assignment.
  6. I can tell English is a second language for you, but the mathematics language is universal. You post good work on here, @rg1...I have no doubts about your passing. @cos90 is also mathematically strong. Both of you guys will do well. Just don't make the mistake of overthinking things. The actual PE exam questions have been vetted through-and-through before they appear in the actual exam, so if you miss the question, well, you missed it fair-and-square. In other words, you should not have to "figure out what they mean;" it will be obvious to the casual observer what they're asking for. You will see questions that are obviously "test questions;" these are questions that they are "testing" for viability, but won't be actually included in the final scoring. I recognized one right away because the grammar, spelling, and setup was so utterly confusing that I couldn't figure out what it was EXACTLY they were looking for. Honestly, the questions are worded so carefully that there will be no room for interpretation. You either know the answer or you don't. This is why I know you both will pass. Both of you find mistakes in your reference material and are able to deduce why it's wrong, what should have been asked based on what the "right" answer is, and can rationalize both why the question and answer conflict. I would wish you both good luck, but neither of you need it. Instead, I'll wish you both a foregone conclusion of "CONGRATULATIONS!"
  7. Agreed. I predict both of you will pass your PE Exams, and will have plenty of time to play with this later.
  8. Complex Imaginary Test 1 Problem 1

    You're right. I missed the connection. I interpreted the question as what would be the peak voltage of the given phase voltage at Vab (perhaps that was the intended question), but the question simply asks what what is the peak voltage at Vab, so your answer is right. I think you meant 400 * sqrt2 = 565V, though. Thanks.
  9. Complex Imaginary Test 1 Problem 1

    Here is a brief explanation: https://electronics.stackexchange.com/questions/12453/if-a-standard-three-phase-400v-ac-connection-is-rectified-what-dc-voltage-comes
  10. Complex Imaginary Test 1 Problem 1

    Where you have your voltmeter in your sketch will read sqrt3 * 125V, or 216.5V...that is not where Vab is measured from. Vab is measured from the outputs of the rectifier, as shown in my attachment.
  11. Graffeo - Single Phase Problem

    The "1 is 2 and 3 is 1" monicker means "single phase counts 2 times the conductor length and 3 phase counts 1 times the conductor length." The OP was confused about why the conductor length was doubled.
  12. Complex Imaginary Test 1 Problem 1

    I guess what I'm trying to say is that the load isn't seeing the line-to-line rectified voltage...it's seeing the sum of three single-phase voltages that have been fully rectified, so each rectified phase voltage is simply sqrt 2*Vrms. There could be 6 or 1,000 phases that were fully rectified and it wouldn't make a difference to what the load sees.
  13. Autotransformer Problem

    Probably a better way to say this is that a standard transformer couples voltage from the primary side to the secondary side by electromagnetic induction. This electromagnetically coupled voltage induces a current on the secondary. Since power in must equal power out, the amount of current induced is inversely proportional to the voltage on the secondary (i.e., big voltage/small current on primary = small voltage/big current on secondary). All of this happens based on transformer impedances. An autotransformer doesn't rely on emag coupling; the primary and secondary windings are physically wired together, so the current produced in your secondary follows KCL while the voltage follows KVL, so the power "adds up."
  14. Autotransformer Problem

    No, not overloaded...the single phase transformer used as an Autotransformer allows the total capacity to be fully realized. You started out with a single-phase transformer rated for 15kVA, but through a clever use of the law conservation of energy, you unlocked the true and total capacity of a 75kVA power source. A single-phase XFMR has a primary voltage and secondary voltage winding ratio and results in apparent power equality that defines the primary and secondary currents (i.e., 15 kVA in = 15 kVA out). On the other hand, an Autotransformer kind of works in reverse (i.e., primary maximum apparent power in = secondary maximum apparent power out) primary current available is related to the primary voltage, while secondary current available is related to the secondary voltage. Once the voltages and currents are known, one can calculate the "true" apparent power on the primary and secondary sides of the Autotransformer, which will be equal (per the conservation of energy law).
  15. Graffeo - Single Phase Problem

    All of these explanations are great and completely valid. To make this easy to remember for instant recall during the exam, use this: "Single-Phase is Two Times, and Three-Pase is One Times." ...or, "1 is 2 and 3 is 1." Kinda clumsy, but it got me through Circuits II. ?
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