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Slay the P.E.

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Slay the P.E. last won the day on October 7 2018

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About Slay the P.E.

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    Principal in Charge

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  • Engineering Field
    CONSULTING
  • License
    PE
  • Discipline
    Mechanical

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    www.slaythepe.com

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    Prospect Heights, IL
  • Interests
    Mechanical Engineering Exam Prep

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  1. No. Just never use ideal gas assumptions for steam. It’s safer that way. Use tables or a Mollier diagram. For this particular problem: If you start at 3 psia and compress, there is some liquid present at the “inlet” therefore that liquid-vapor mixture is far from being an ideal gas. So that’s a huge no-no. Even if you start with saturated or slightly superheated vapor at 3 psia (a condition in which the steam arguably behaves like an ideal gas) you are taking it to a condition of pressure high enough so the deviation from ideal gas behavior is significant, so whatever it is you’re calculating is likely to be wrong. Just please drop this idea of wanting to use Pv=RT for steam in power problems. Regarding psychrometrics: in typical HVAC applications, the partial pressure of water vapor is very small so moist air is modeled as a mixture of two ideal gases: dry air and water vapor. If you read in a thermo book the derivation of the equations used to plot the psychrometric chart you will see they model the water vapor as an ideal gas. But, you don’t have to worry about this, just use your psychrometric chart. So let me repeat: become familiar with the Mollier diagram and the steam tables. Use those for steam power problems.
  2. 1. With the inlet conditions, find the enthalpy at the inlet, h1, and entropy at the inlet, s1 2. The entropy at the discharge for the isentropic process (s2s) is the same as s1, so for the isentropic process you have s2s and pressure. 3. Is s2s < sg at 3 psia? If yes, you have a saturated mixture at the end of the isentropic process. Calculate quality at the end of the isentropic process, x2s 4. With x2s find the enthalpy at the exit of the isentropic process h2s 5. With h2s and the definition of isentropic efficiency find the enthalpy at the discharge of the actual process.
  3. Dense gases such as steam in power plants and refrigerants in refrigeration cycles in general cannot be modeled as ideal gases. Only when the pressure is VERY low can steam be treated as an ideal gas (this is done in psychrometrics, where the partial pressure of water vapor is in the order of less than 2 psia or so). You have comprehensive steam tables available. Use them and avoid catastrophic mistakes like the one you described.
  4. Really hard to say. Maybe the end of this fall.
  5. Thanks! Yes, it was left out for the sake of brevity. I think we will have a short tutorial just on linear interpolation. Oh, absolutely yes. We’ve been advocating for Mollier diagrams here for the longest time. That video (just on how to use h-s and P-h diagrams) will come after entropy and isentropic efficiency are introduced.
  6. We have been developing video tutorials for the TFS exam. When we finish all of them, we will put them behind a paywall and offer a course based on these videos. For now, as we develop them they will be free. There will be roughly about 50 videos to cover all of TFS. As of the writing of this post, the first 7 are ready and in our website. No video will ever be longer than 10 minutes, to keep your attention focused. The dynamic will be to watch a short video and immediately solve problems associated with the video before moving on to the next one. We're looking for honest constructive feedback, especially from our friends @Audi driver, P.E. and @MikeGlass1969 Go here for the videos: https://www.slaythepe.com/thermo-superheated-vapors-video.html
  7. Yeah, we're working on it. Probably ready by late 2020. Hopefully by then you will not need it
  8. This is correct. Hence our book would cover all you need for the supporting theoretical principles questions, but you will still need another resource to prepare for the applications part. If you do get our book, I think you would be OK if you skipped the following: In Part I: THERMODYNAMICS, skip the following sections: 09. The Rankine Cycle 10. The Brayton Cycle 12. Otto and Diesel Cycles 13. Gas Mixtures 15. Combined Power Cycles Also, in this part our coverage of ideal gases might be overkill (HVAC&R peeps may not need to navigate the variable Cp air tables) In Part II: FLUID MECHANICS, skip the following sections: 04: Conservation of Momentum 05: Flows over Immersed Bodies 09: Compressible Flow: Isentropic Flow & Nozzles 10: Compressible Flow: Normal Shockwaves 11: Compressible, Adiabatic Flow with Friction (Fanno Flow) In Part IV: HEAT TRANSFER & HEAT EXCHANGER ANALYSIS 04: Transient Conduction -- The "Late" Regime 05: Transient Conduction -- The ";Early" Regime 06: External Forced Convection 08: Natural Convection 11: Radiation Heat Transfer In Part V: GENERAL ENGINEERING AND SUPPORTIVE KNOWLEDGE, maybe skip the whole thing except for Section 01: Economic Analysis Most definitely go through all of Part III: PSYCHROMETRICS & HVAC Get a free preview of the book here: https://www.slaythepe.com/tfs-cbt-preview.html
  9. They each have their pros and cons. In the real world, most steam power plants have a combination of both. Here's something I found in one of my thermo textbooks, which explains the pros and cons better than I ever could: Open feedwater heaters are simple and inexpensive and have good heat transfer characteristics. They also bring the feedwater to the saturation state. For each heater, however, a pump is required to handle the feedwater. The closed feedwater heaters are more complex because of the internal tubing network, and thus they are more expensive. Heat transfer in closed feedwater heaters is also less effective since the two streams are not allowed to be in direct contact. However, closed feedwater heaters do not require a separate pump for each heater since the extracted steam and the feedwater can be at different pressures.
  10. Hello. FYI, we don't have (yet) products targeted to the HVAC&R exam; only TFS. There is some overlap, but we also cover a LOT of stuff you don't need.
  11. That would be a poorly crafted problem. Can you share the problem statement (don't reveal the source)?
  12. See this post http://engineerboards.com/topic/34191-pe-thermal-and-fluids-system-study-pal-october-2019/?do=findComment&comment=7569055
  13. We just received the 3rd successful entry, so the contest is now closed. Thanks!
  14. Great question. For ideal gases it can be shown that both internal energy, u, and enthalpy, h depend on temperature only (you can find this proof in any Thermo textbook). Since u and h depend only on temperature for an ideal gas, the specific heats Cv and Cp also depend on temperature only. Thus, for ideal gases, the partial derivatives in the definitions provided in your post can be replaced by ordinary derivatives: In other words: For ideal gases Cv = du/dT and Cp = dh/dT. Furthermore, if you assume Cv and Cp are constant then these relationships can be easily integrated to yield: Δu = (Cv)ΔT and Δh = (Cp)ΔT So, where does this leave us? Well... if you need to calculate a change in internal energy, then you do it by using (Cv)ΔT -- no matter what the process is. If you need Δu for a constant pressure process then Δu is still (Cv)ΔT. If you need to calculate a change in enthalpy, then you do it by using (Cp)ΔT -- no matter what the process is. If you need Δh for a constant volume process then Δh is still (Cp)ΔT. If you are given an air flow heater, then you are dealing with an OPEN system. That is a system which has mass flow through it (unlike, say a tank with closed inlet and outlet valves which would be a CLOSED system). Now, take a look at the energy balance for a generic open system: You will note that what comes into play for such systems is the enthalpy at the inlet, hi and the enthalpy at the exit, he. So, in OPEN systems (such as your steady-flow heater), you are usually interested in how much the enthalpy changes as the gas flows from the inlet to the exit (he - hi), regardless of the nature of the process. Since you are working with an ideal gas then (he - hi) = (Cp)(Te - Ti) regardless of how the heater is accomplishing its task. Let me know if this isn't clear. I can come up with some easy examples to walk through if needed.
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