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Slay the P.E.

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Slay the P.E. last won the day on October 7

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About Slay the P.E.

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  1. Slay the P.E.


  2. Slay the P.E.


  3. Slay the P.E.

    Best of luck tomorrow!

    A shout-out to the men and women that have visited these boards and either just read along or actively seeked advice and/or help. I hope you found value in the group effort we all put around here in helping out. Be sure to come back later to share stories from the trenches!
  4. Slay the P.E.

    NCEES TFS practice exam problem 533

    Correct. If T5 had not been given, then you would need to calculate it.
  5. Slay the P.E.

    NCEES TFS practice exam problem 533

    The table of temperature values are all actual values. When they say "Station 5" they mean that is what a probe is registering there. The provided turbine efficiency is unnecessary for this problem and is a "distractor".
  6. Slay the P.E.

    TFS Lindeburg #33

    I think their 17 feet and your 23 feet are within the uncertainty of these problems in the real world, given the approximate nature of these values for loss coefficients (especially for valves). The problem in the PPI book is a throwback from the olden days when a human being actually laid eyes on your hand-written essay solution. Those days are obviously long-gone since the scantron and 100% multiple choice format took over the exam. Back in the day, you would have received full credit for your solution (I think) when graded by a human. In modern times, we cannot have this much wiggle room and NCEES has to craft these problems so that you land squarely on the one right answer. How do they do that? Well, they make it quite obvious what values of K (or L_eq) you have to use. For examples, look at these in the NCEES official practice exam: Problem 516. They are explicitly telling you the equivalent lengths of the elbows and the globe valve. You then use those values. DO NOT go all nuts trying to look up values of K for these fittings. Use what they're giving you in the problem statement. Problem 519. They are explicitly telling you the values of K to use for the elbow and the entrance. Use that. Don't go trying to get equivalent lengths in some table. Note that the problem you are posting about doesn't provide Ks or L_eqs in the problem statement; hence the ambiguity. You will not (or shouldn't) have this in the exam.
  7. Slay the P.E.

    TFS Lindeburg #33

    Ok. 4.6ft for the elbows makes more sense, but K is still not 0.6
  8. Slay the P.E.

    TFS Lindeburg #33

    Well, for starters 40ft + 5(6.4ft) + 2.5ft is actually 74.5 ft, not 65.5 -- but that's not all. There also seems to be something going on with the "conversion" from K to L_eq. For example, according to your data, the elbows have an equivalent length L_eq = 6.4 feet, and a loss coefficient K = 0.6, correct? Well, the relationship is L_eq = K*(D/f). If we use K=0.6, f=0.019, and D=4.026 in, we get L_eq = 0.6*[(4.026/12 ft)/0.019] = 10.6 ft (i.e. not 6.4 feet). You're only going to get the same answer if the Ks and the L_eqs are related by L_eq = K*(D/f). Also, what book is this?
  9. Slay the P.E.

    NCEES TFS 508 Problem

    If the given conditions (T,p) are such that the thermodynamic state is “compressed liquid” then you could use the compressed liquid table. That table, however, typically only lists moderately high pressures. 1 MPa (10 bar) is too low and is not listed (at least in the one in MERM Appendix 23.Q in MERM13 where the lowest pressure is 25 bar) Therefore, for compressed liquids we use the approximations; h(T,p) ~ h_f(T) v(T,p) ~ v_f(T)
  10. Slay the P.E.

    Help with cooling problem

    Interesting observation. The dry bulb T and the relative humidity at the inlet are also given, so giving h_1 makes the problem over-specified. Luckily 33.6 Btu/lbm is the right enthalpy for the given values of T and r.h. I think it is more likely that the h_1 in the handwritten notes was looked up in a psych chart and was not part of the given data.
  11. Slay the P.E.

    TFS practice problem of the week...

    We also advocate for this approach. The very first post in this thread is a good example showing how much time can be saved. Of course, Mollier diagrams are useful for superheated steam and high quality water-steam mixtures, that is, the typical steam turbine problem.
  12. Slay the P.E.

    Help with cooling problem

    Close. m*Cp*DELTA-T is accurate for sensible cooling/heating. This would be somewhat accurate even for moist air as long as there is no condensation or humidification (i.e., the process is a horizontal line on a psych chart). If there is condensation (like in this problem) or humidification, then you need to account for latent heat effects associated with phase change. Thus, it is best to use the psych chart to get DELTA-H, as m*Cp*DELTAT fails to capture this effect.
  13. Slay the P.E.

    TFS practice problem of the week...

    No. the nitrogen is flowing through an external jacket around the chamber. Look at the figure. One surface of the jacket is at -320F the other is adjacent to insulation. the 80F is the surface temperature of a heat source (because it has a heating element inside) inside the vacuum chamber. There is no nitrogen in the heater. There is no media between the heater and the jacket. heat is transferred by radiation between the heater surface and the nitrogen-cooled jacket. This heat must equal the latent heat of vaporization for nitrogen.
  14. Slay the P.E.

    TFS practice problem of the week...

    The nitrogen is changing phase because it is being heated by convection het transfer occurring from the chamber walls to the nitrogen. It cannot be neglected. It is just not necessary to calculate it, because you can do an energy balance on the nitrogen. For example; suppose I ask to calculate the heat transfer rate required for m_dot amount of R-134a to enter an evaporator at x=0.3 and T=-20F and be discharged as saturated vapor at -20F. You could do two things: a) perform an energy balance and figure out that the heat transfer rate is simply the change in enthalpy for the given end states, or b) try to do a heat transfer analysis inside the evaporator pipe: find the appropriate Nusselt number correlation, get the convection coefficient inside the pipe and from there get the heat transfer rate. I think what you're trying to do in our problem is something like approach (b) above.
  15. Slay the P.E.

    TFS practice problem of the week...

    If in that example they had specified the inlet and outlet conditions of the air in the duct, then you wouldn't need to calculate the heat loss with a heat transfer analysis. A simple energy balance for the air in the duct should give you the heat loss. In our problem, the inlet and outlet conditions of the nitrogen are given, therefore the heat transfer rate into the nitrogen can be calculated with an energy balance on the nitrogen alone.