applepieordie - Engineer Boards
Jump to content
Engineer Boards


  • Content Count

  • Joined

  • Last visited

Community Reputation

0 Neutral

About applepieordie

  • Rank

Previous Fields

  • Engineering Field
    Power Engineering
  • License
  • Calculator
  • Discipline

Recent Profile Visitors

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

  1. @Stephen2awesome @rg1 i thought so. Thank you for your replies.
  2. I've seen 2 different equations from 2 different references for finding the GMD for 4 conductors. -one used (D12 x D23 x D13 x D14 x D24 x D34)^(1/4) -the other used (D12 x D23 x D13 x D14 x D24 x D34)^(1/6) i assume one of them is wrong? which one is correct? im leaning towards the 6 root one to be correct, can someone please confirm? what is the general eqn for n number of conductors? i assume nth root of the product of n possible distances. Thanks in advance for help.
  3. @rg1 OH I see now. this question really had me confused but I understand why i had so much trouble solving it now. Thank you for your help, really appreciate it!
  4. @Stephen2awesome @rg1 thanks for your explanations. I think i understand now. I initially tried to solve for Ipri in the Sauto equation under fig. 17 then I tried to solve for Icomb by plugging in Ipri and Isec into the equation above it - it is from a reference i purchased from a website. can you tell me is the equation wrong? here is a pic of the the reference:
  5. I was able to solve the problem by solving for low side current, I2, first. But I was unable to solve the problem when I try to solve the problem by finding the high side current first. I am doing: I1 = (500MVA)/(200/sqrt(3) KV) = 3765A. What current did I just solve for? when I solve for this current I assume it is the current from the high side into the series winding. @rg1 im sorry i will have to ask you to further explain the 4 KVAs you mentioned. I agree with input KVA = output KVA. i havent fully understood why the series and common wdg KVA is equal.
  6. Can someone please explain the solution to me? Question: A wye-wye, 230 kV, 3-ph transmission line is being stepped down to 200 kV using three autotransformers connected between the neutral and the line. For each phase, if the autotransformer is rated for 500 MVA, and the turns ratio is 5:2 (common:series) what is the apparent power in the common coil? Their Solution: (Nc/Ns) = (5/2), S = 500 MVA (S/Sc) = (Ns+Nc)/Ns = 1+(Nc/Ns) Sc = S/ (1+(Nc/Ns)) = 500MVA/ (1+(5/2)) = 143 MVA To me it seems like they solved for the apparent power in the series coil. Or am I thinking about this wrong? please help.
  7. Please explain to me why sin(delta2) = (P1/P2)*sin(delta1) I don't understand how it got to that step. can someone please show the steps leading up to how sin(delta2) is determined? Thanks!
  8. I didn't purchase if from Brightwood. I got this exam (in pdf format) from a colleague and was told it is the Kaplan exam. My guess is it is probably an older sample exam as it has many errors. I found this thread dated 2010 in the forum where a poster called cableguy made an errata list and it matches the errors I see in the pdf I have:
  9. Hello, What equation is being used in the solution? (question from Kaplan sample exam) Is max torque synonymous to max power? In Graffeo's book I see the formula: Slip at which Pmax occurs at = Sp = Rrotor/(Rrotor + Zeq), where Zeq = Zmotor + Zstator. But this equation seems slightly different than the one used int he solution. Also, if you have good machine references (such as a formula sheet) to share it would be greatly appreciated. Thanks!
  10. I often find myself making the mistake of using Vpp rather than Vpn and vice versa. After I see the solution and realized the mistake I made, it seems so obvious from the start that I should have used Vpn. Question 1: are there any techniques or rules to follow to be sure what voltage potential to use and minimize mistakes? or is it just trial and error learning process? I dont want to make such a simple mistake on exam day. Question 2: When converting from Vpp to Vpn I usually follow the formula Vpn = Vpp/(sqrt(3)∠30°). In the example below, the ∠30° is left out of the denominator. Can someone explain to me why? When to include ∠30° and when to exclude? Thanks!
  11. Thanks for the clarification. I didn't take any power courses in college and have been learning power through books and online material so I really appreciate the effort other forum members put in to answering my questions and explaining the fundamentals.
  12. Yes, this makes sense...though I'm more used treating positive P or Q as generating/outputting power rather than as loads because I work for a utility company. But the concept is the same but reversed. My confusion was addressed when I took a look at the cheat sheet on your website, particularly rule #2 “The phase current angle and apparent power angle will always be opposite in polarity (when one is negative, the other is positive and vice versa)” I was confused why the theta in Graffeo's solution was negative while I thought theta should be positive as I mentioned in the rule of thumb I follow. It is because the theta in the solution is the phase current angle while I was thinking of the apparent power angle. S = 3(Vp)(Ip*) ... I finally know why the complex conjugate is there lol. Thank you for the great resource. And thank you to everyone for the replies. I now have a more profound understanding of leading and lagging.
  13. I see. I misinterpreted the CT ratio as the transformer turn ratio but the CT ratio is the ratio (Ipri/Isec) which makes a lot more sense. Thanks for the explanations.
  14. turn ratio = (Npri/Nsec) = (Vpri/Vsec) = (Isec/Ipri) shouldn't Ipri in the example be Ipri = (Irelay)/(CT ratio)??
  15. Hi, I've always used this rule of thumb: lagging generator (over-excited): exporting Watt and exporting VAR leading generator (under-excited): exporting Watt and importing VAR lagging motor (under-excited): importing Watt and importing VAR leading motor (over-excited): importing Watt and exporting VAR so along this line of thinking, lagging generator θ is positive, leading generator θ is negative, lagging motor θ is negative, leading motor θ is positive. where θ is the impedance phase angle in the "power triangle." Is this correct? I'm getting confused with the phase difference between voltage and current (see attached photos from graffeo book). in his lagging generator explanation, the θ is negative. is the θ between voltage and current different than the θ in the power triangle? I thought these were the same. Or is the rule of thumb i've been using wrong from the start? can someone please shed some light here for me. thank you.
  • Create New...