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About TNPE

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    Project Engineer

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  1. Means nothing. Don't get your hopes up! I'd wager they haven't even began the "grading" process.
  2. Good luck! Just use what works for you, and as I stated previously, I prefer the pu method cause errors will jump off the page at you. If you have fractional impedances but end up with an odd ball 10 pu, you know you've made a mistake somewhere. With that said, it wouldn't be unheard of to have a fault magnitude less than the full load current of the XFMR, depending on impedances, fault location and fault type. However, you most likely won't see it, but it is possible and fair game. The biggest hurdle with this exam is the concepts, ideally with regards to motor applications. If you have a good motor book, be sure to take it with you! I would recommend Wilde or Chapman (if you have each, take both of them).
  3. 820 is the closest and it wouldn't be unheard of to see something similar on the test. Remember, this exam will test your gut (and comprehension) as much or more than your subject knowledge and engineering competency. That said, I didn't have/use the Camara text when I took it, and it seems the rest of you should be cautious about some topics it presents. Ideally, this very problem, for one, and I've seen other erroneous representations on this very forum (power factor correction comes to mind... how the hell do you mess that up? - these are the softballs)
  4. The Vnew/old would come into affect if you had a generator producing at, say, 13.2 kV, but the XFMR was rated at 12.47 kV on that side (i.e. the primary side connected to the generator). In this case, we aren't given any info about the voltage of the generator, therefore, it's safe to assume it is producing voltage at the rated input of the XFMR.
  5. Yes, the generator counts. You have to sum all impedances up to the point of the fault. However, as previously mentioned, the generator and XFMR are on different power bases. This has to be resolved, whereby the Znew for G = 0.05, not 0.1. This comes from the change of base formula, Znew=Zold(Vold/Vnew)^2(Snew/Sold)=0.1(11/11)^2(25/50). Get everything in terms of the XFMR base cause it has a "let through" of 25 MVA, and the line is connected to the XFMR of the same base (i.e. Zbase=(33kV)^2/(25 MVA)). Therefore, you have: 1pu/(0.05+0.06+0.4364) = 1.83 pu Ifault=(1.83)(437.3866A)=800.4879A Per the question and what's asked for, this is how you solve it (but can be solved multiple ways, I just prefer the pu method due to it being more uniform and easier to spot errors). If I'm wrong, someone correct me; but I am confident in what's presented above. I hammered hard on fault current problems before I took the exam and feel it is one of my strongest areas.
  6. All banks are generally connected in parallel to the grid (usually Y-GND). I may be misunderstanding you, but in AC circuits, all values of impedance (reactance or vector sum of reactance and resistance) are combined exactly like resistors in DC circuits. Recall that impedance and capacitance/inductance are not the same (i.e. XL=jwL and XC=1/(jwC)). As for analyzing PFC problems, implementing the impedance triangle is the safest and most fail-proof approach, at least IMHO. Not to sound discouraging, but these types of problems are the proverbial "softballs" and should be gimme points. Don't overthink it. The bank is connected in a 3-phase configuration, but the bank itself is in parallel with the grid. Hope this helps!
  7. I got 800.5 A using the pu method. Keep in mind, the generator and XFMR are not on the same power base. This has to be corrected, in turn, adjusting the impedance of the generator. It seems to be somewhat of a consensus here that the answer in the book is incorrect. I'd agree. If someone has anything to interject to validate the veracity of the answer, please do.
  8. No, what @TNSparky said is not exactly true, but I know what he means. Without a source, the fault current would be goose eggs! You have to get the generator and transformer on the same power base, then all other pu values will adjust accordingly. After obtaining all pu values, use the appropriate formula and multiply this value by the FLA of the transformer.
  9. Customarily, an electromechanical 51 element (reasonable assumption) is a time over current device, hence B is the appropriate answer for the given question/scenario.
  10. I believe you are right on all except #3. A 51 element is a time over current device (does not look in both directions), therefore, I feel confident in B, not C, being the correct answer. However, the 67 element is indicated to be a non-time delay (i.e. instantaneous) and is a device used for looking in both directions (therefore, you can eliminate a few of the given answers).
  11. This problem can be solved using PF, but there is certainly no need to complicate it. However, I will illustrate this below so that you can use this approach if you get hung up: From our power equations, we know P=VIcos(theta) Conversely, Q=VIsin(theta) and S=VI or VI* We are interested in the magnitude here, therefore, PF is not necessary. However, if PF were used, recall that P=Spf, thereby, P=(500KVA)(0.85). Once you have determined P, then you can use the above solution with PF, namely: 425kW/(13.2kVx0.85)=37.9A (or I = P/(VLLxPF) Easy enough, but entirely unnecessary to solve this problem. As others have mentioned, know what you're solving for and analyze what data is given (write everything down and put a question mark beside the unknown...sounds remedial, but hey, if it helps to pass, so what). For instance, to carry this another step further, rarely ever will you have to use HP to calculate electrical power, amps, etc. (generally speaking, the HP that is usually given is in terms of mechanical/shaft HP, not electrical HP). It can be done with generalized formulas and some elbow grease, but this test is about how well do you understand concepts, not how well can you regurgitate numbers and manipulate a problem algebraically to arrive at a solution.
  12. Spent dang near $100 to frame mine. Still awaiting the certificate. But hey, I'll take this wait over the one endured from about 4:00PM Oct. 28, 2016 till 7:57AM Dec. 9, 2016.
  13. Any word on when certificates will be delivered? Glad you finally received your license.
  14. Sparky- I moved to the panhandle of FL in early Jan. and I received my license/pocket card almost two weeks ago. Even stranger, I received the packet, mentioned above by others, about a week or so after I received the license. I would think you should've gotten yours by now...? Has Ms. Wanda given you any indication as to what the delay is? The license is what matters, but I'm looking forward to the certificate; however, I've heard it takes 2-3 months after you become active to receive it if you do not attend the NSPE/TSPE banquet. Good luck and hopefully it comes soon! -TNPE
  15. You may be right. I'm just so used to processes like this taking much longer than they should, at least superficially. I will say the TN State Board has been pleasant to work with and they do expedite better in relation to how other Boards appear to operate.