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engineer123

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About engineer123

  • Rank
    Project Engineer

Previous Fields

  • License
    Working on it!
  • Discipline
    Enviro

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  1. @GirlsCanDesignAWESOME!! Congrats!!! 🎉🎈✨
  2. I'd like to know this as well! lol I just tried searching on amazon & ebay for books but didn't find one. I know Chegg has textbook solutions so that might be an option.
  3. @MaryamWhere did you find them for that amount? I think it might be the older books, but trying check the ISBN number.
  4. Thank you both! @txjennah PE@GirlsCanDesign
  5. @waternerdThese are the two CBT Schneiter books on PPI that @GirlsCanDesignhad recommended to me (thanks again!) https://ppi2pass.com/pe-environmental-practice-exams-peenpx.html https://ppi2pass.com/pe-environmental-practice-pack.html I'm definitely going to get the practice book with 500 problems. I'm still debating on the exam booklet with two exams (80 questions) though... I just hope it doesn't have the same questions as the practice book since it's the same author 😅
  6. Hey all I'm getting really hung up on what length (distance) to use in the hydraulic gradient equation. In the solution below, they use the "diagonal" distance between the wells. I would have just used 175 ft. The example from this link also used the "horizontal" distance https://www.quora.com/What-is-hydraulic-gradient Which one is the correct value to use?!
  7. @GirlsCanDesignI downloaded the handbook probably back in January so I didn't think they'd update it so fast haha I started studying last month and nowhere near ready to take the exam due to my work schedule. I definitely need many more months of studying, so I won't know when to take the exam until I feel ready lol Best of luck to you though! I hope this exam will be "fair" enough with just using the NCEES handbook Perhaps it might be better than the paper version!
  8. @GirlsCanDesignyou're right. looks like they just updated it last month lol thanks!
  9. I'm having trouble understanding the storm return period equation on Page 32 of the NCEES reference handbook. It says: The probability that a T-year return period event will occur at least once in n years is P = 1 - (1/T)^n However, in my Intro to Environmental Eng textbook, the risk that the event will be equaled or exceeded at least once in n years is: R = 1 - ( 1 - 1/T)^n You get different answers obviously if you plug in the variables, but essentially it seems both are asking for the probability that the event will occur at least once in n years. What is the difference between these two equations? Is the one in the NCEES handbook incorrect? o.O
  10. @waternerdI still havn't signed up for a course yet. I think I'd rather wait to see how the cbt goes for everyone. Is there anything else you're doing to prepare for the exam? I'm so nervous as to how different this exam will be with it being closed book lol
  11. Got another eng econ question if anyone can assist me A new computer system will cost $25,000. Using the MACRS, the computer system has a useful life of 5 years. What is the estimated salvage value at the end of five years? Heres my approach: I looked at the MACRS table to find the recovery rate percent which is 11.52%, and calculated the depreciated value in year 5, D = (percent) x Cost = 0.1152 x 25,000 = 2880. I then tried using the straight line equation and plugged in all my values to solve for the salvage value: D = (Cost - Salvage value) / useful life ….but after doing some research, apparently there is no salvage value for the MACRS method because the asset is always depreciated down to 0. Zero was not given as a choice and the answer chosen was $1,450. So I'm wondering if this is a mistake?
  12. @GeoDudeThanks I think I get it now! If I recall, when comparing different options - the life value n should be the same for both so its an equal comparison (in this case n = 10). Would I be able to solve this using Annual Worth analysis? I wasn't sure why they chose Present Worth. Thanks again.
  13. Hi All, I wasn't sure which board to post this question but I'm assuming all disciplines will have engineering economics on their exam? Anyways I'm studying this topic right now to refresh my memory and I was stuck on the problem/solution below: Two coatings are under consideration for a liquid chemical storage tank. A tank coating identified as Material A will have a first cost of $50,000 and a 10 year life, if repaired at the end of year five. Material B will cost $20,000 initially and $5,000 per year through its five year life. At an interest rate of 10% per year, the amount that could be spent for repairing Material A that would make the two breakeven is closest to: Solution: Equate present worth relations for 10 year service life. Let x = repair cost in year 5 for breakeven. 50,000 + x(P/F, 10%, 5) = 20,000 + 20,000(P/F, 10%, 5) + 5,000(P/A, 10%,10) 50,000 + x(0.6209) = 20,000 + 20,000(0.6209) + 5,000(6.1446) 0.6209x = 13,141 x = $21,164 On the right side of the equation, I don't know where 20,000(P/F, 10%, 5) came from? And for 5,000(P/A, 10%,10) , isn't n supposed to be 5? Also, instead of doing a present worth analysis, could this be solved with an annual worth equation instead? Also generally speaking, how difficult can engineering economics be on the exam? I hope they are straight forward. Often times, it's the wording of the question that messes up your equation. The math itself is easy though lol.
  14. @Slay the P.E. I don't have that reference but do you mind posting the equation 17.83? thanks for your help!
  15. Hi All, I'm trying to solve this problem about a tank draining: A 3 inch diameter orifice on the side of a 5 ft diameter tank draws the surface down from 7 ft to 5 ft above the orifice in 53 seconds. Calculate the discharge coefficient. I started using the equation Q = C A sqrt 2gh, but the solution manual shows a different equation using time. I don't see that equation anywhere in the PE formula handbook and don't know how I could even derive that! so I'm wondering is there another way I can solve this? If anyone can please explain, please let me know! Thanks (p.s. the answer is supposed to be C = 0.77)
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