akyip - Engineer Boards Jump to content
Engineer Boards

akyip

Member
  • Content Count

    102
  • Joined

  • Last visited

Community Reputation

35 Excellent

1 Follower

About akyip

  • Rank
    Project Manager

Recent Profile Visitors

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

  1. Hey guys, I have a question about transformer short circuit tests specifically for 3-phase transformers. This comes from Cram for Exam Vol. 2 question 58. In this problem, there is a 3-phase transformer given rated as 30 MVA, 130 KV/22 KV, wye-delta. The given short circuit test results are: Short Circuit Test: 13 KV, 130 A, 110 KW The 13 KV given in the short-circuit test is not specified as L-L or L-N, but I assume this to be L-L. The question asks what is the total leakage impedance. In the given solution: V SC = 13000 V / √3 = 7514 V Leakage impedance Z =
  2. I just did the PPI practice exams... And sadly even with much practice from other practice sets (e.g. Cram for Exam, Electrical PE Review, Eng Pro Guides, Complex Imaginary, Spin-Up), there were still some questions & topics in the PPI exam that I just was not familiar or rusty with. The earliest I can take the CBT exam is January 2021...
  3. I just practiced the PPI exams. Even with all the practice from other sets (e.g. Cram for Exam, Electrical PE Review, Eng Pro Guides, Complex Imaginary, Spin-up), there were still some questions that I just did not know in the PPI exams. A small list: -DC motors (didn't see much of these in the other practice exams... Some were in Cram for Exam though...) -Op Amp circuits (I was surprised this was in PPI exam, and I am rusty with op amps since I haven't dealt much with DC circuit problems in other practice sets...) -Hysteresis curve -Ferromagnetism, Diamagnetism, and Parama
  4. Hey guys, I want to confirm my understanding on the differential protection relay scheme, from a Cram for Exam Vol. 2 problem. In problem 4 of the Cram for Exam Vol. 2 book, the solution basically states that a differential protection relay scheme will not trip for a turn-to-turn fault in the machine winding, an open-circuit in the machine winding, and high currents caused by an external short circuit. It doesn't elaborate fully on why, but I think I understand it and I just want to confirm my understanding. In a turn-to-turn fault, the current flowing on both sides of the windi
  5. So I just ordered the PPI 2 Pass practice exams (3rd edition... that was the latest edition I could find). With all the practicing I've done on other practice exams... there are still some questions in the PPI exams that I have no idea how to answer (at least from a quick glance). Man, I feel like the wide range of topics in the PE Power exam will probably get to me the most. And I feel that the provided reference handbook is definitely not sufficient. I haven't actually done the PPI 2 Pass practice exams yet... currently working on other practice exams (just finished Complex Im
  6. That sheds some light. Thank you for your responses in this and my other topic questions!
  7. Hey guys, On Shorebrook PE Power Exam question 77, is this circuit a typical thumper (the device used to locate faults)? I haven't seen this circuit before in other PE exam prep materials, and the only other reference to a thumper is in one of the Cram for PE Power exam questions. From what I understand, the left side of this circuit is basically a Wheatstone Bridge (balanced, since the measured current in the middle (A) is zero). On the right side of this circuit, it seems that: Distance to fault = Total cable length / One-way cable resistance = Total cable length / (Round trip
  8. Hey guys, In this Shorebrook Exam question 74 about two-wattmeter power factor, my initial attempt to solve for the power factor was using: tan(theta) = sqrt(3) * (Phigh - Plow) / (Phigh + Plow) = 1.732 * (12 KW - 8.7 KW) / (12 KW + 8.7 KW) = 0.276 power factor angle: theta = tan^-1 (0.276) = 15.4295 degrees power factor: p.f. = cos(15.4295 degrees) = 0.964 However, the given solution calculates pf = P/S, with: P = P WA + P WB = 8.7 KW + 12 KW = 20.7 KW S = sqrt(3) * V LL * I L = 1.732 * 300 V * 50 A = 25.98 KVA P/S = 20.7 KW / 25.98 KVA = 0.8
  9. Hi guys, On Shorebrook PE Power Exam question 71, it briefly mentions that on a resonant-grounded system, the resonant point is located at the middle of the transmission line, regardless of where the ground fault is located at. What is the explanation behind this? Is there more literature regarding resonant-grounded system and the resonant point? This is not something that I have encountered in other sequence components or grounding materials or practice PE exams other than Shorebrook. So I'm just trying to understand this better. Thanks for any input on this!
  10. Hi guys, I have a few autotransformer questions from Shorebrook PE Exam question 65. 1. What is the difference between actual load and equivalent load? From what I see, it seems like actual load is the load that is actually connected to the transformer's secondary... but then what is the equivalent load? Does the equivalent load have to do with the autotransformer ratings? Or is the equivalent load the load of a 1-phase standard transformer that is somehow equivalent to the autotransformer? 2. Where does this equation (Actual Load VA) / (Equivalent Load VA) = (V high - V lo
  11. Hi guys, On Shorebrook PE Power Exam problem 66 solution, where does the formula (Ca + a^2 Cb + a Cc)/(Ca + Cb + Cc) come from when calculating the zero-sequence voltage En in the grounded Y system? Is there an explanation behind this formula? Can this same formula also be applied to the ungrounded delta zero-sequence voltage En in Shorebrook Problem 62? Thanks for any input! I'm just trying to understand this, since I haven't seen this in other practice PE exams or materials...
  12. Hey guys, I just want to confirm this regarding a PLC/relay interlocking scheme question from Shorebrook PE Power Exam, question 54. Shouldn't the solution to this problem actually be D? At least, that's how I learned it from Zach Stone's Electrical PE Review class. If you use the given solution C : When you press the PB-A push button, this de-energizes Relay Coil RA and light LX, since it opens the normally closed contact RA leading to these two. When you press the PB-B push button, this de-energizes Relay Coil RB and light LY, since it opens the normally closed con
  13. Hey guys, Typically when solving for voltage drop percentage, it is as a percentage of sending or source voltage, correct? Is the voltage drop solution for this problem incorrect? The solution uses receiving-end voltage Vr in the denominator... I always thought it was sending-end voltage in the denominator for voltage-drop percentage. Thanks for any help or input on this!
  14. Hi guys, Regarding preventing overcurrent damage to a diode, I understand the reasoning behind using a 2nd diode in parallel (to divide the current). But is there a reason why adding resistors to the diode branch is not an appropriate solution to reducing current? Is because of resistor I^2 * R losses? The solution in Shorebrook PE Power Exam Question 46 does not elaborate on this. Thanks!
  15. Hey guys, Typically when you are asked to calculate for the max power that can be sent on a transmission line given Vs, Vr, and theta, this should really be the 3-phase power, right? The reason I ask is that in Shorebrook PE Power Exam question 30, the solution only uses the 1-phase power. Generally speaking, in a transmission line, max power should really be the 3-phase power, correct? So: Pmax, line 3-ph = 3 * (Vs * Vr / X) * sin(theta) Whenever I am asked to calculate the max power possible sent in synchronous generators, it usually refers to the 3-phase max power. So I
×
×
  • Create New...