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Ideal Gas Entropy Equation Conceptual Question

  1. Dec 7, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm having a little trouble knowing when to use the ideal gas equations for entropy vs just the ones like this: (T2/T1)=(p2/p1)^((k-1)/k).
    I've noticed a pattern in the solutions for my homework( where you're finding isentropic efficiency of turbines and compressors) they tend to use the ideal gas equations for entropy to find the real values for the properties and then they use the ones like this:(T2/T1)=(p2/p1)^((k-1)/k) to find the 2s values.

    Is my assumption correct? Sorry if its poorly worded, I'm still trying to figure this stuff out :)
    If it is correct, can you please explain why? Thanks!

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Dec 7, 2015 #2
    Ideal gas laws are only good for ideal gases, i.e. gases that don't interact with anything, including itself (like noble gases). Otherwise it's only an approximation. Your best bet is to find the appropriate equations, or derive them yourself using statistical mechanics.
     
  4. Dec 7, 2015 #3
    Sorry, I should have included that the problems I'm working on right now are assuming that the gas is ideal. So what exactly is the difference between those equations? What are the conditions for using them?
     
  5. Dec 7, 2015 #4
    I'm not so sure, but there is something non-ideal occurring if T2/T1 ≠ p2/p1.
     
  6. Dec 7, 2015 #5
    In my book next to the set of equations that have k(constant specific heat ratio) in them is says entropy must be constant to use those equations?
     
  7. Dec 7, 2015 #6
    Okay, I think I figured out a better phrasing for my question. Which of those ideal gas formulas provide you with the ideal property values (i.e. s2s, h2s, ect.) and which give you the actual properties(i.e. s, h, ect.)?
     
  8. Dec 7, 2015 #7
    The equation (T2/T1)=(p2/p1)^((k-1)/k) applies to an adiabatic reversible expansion or compression of an ideal gas. For such a path, the temperature, pressure, and volume all vary along the path. The ideal gas law is satisfied at all states along the path. The entropy change for an adiabatic reversible path is zero. Such a path is called isentropic. If you take the general equation for the entropy change of an ideal gas and set the entropy change to zero, the above equation follows directly. Is this close to what you were asking?

    Chet
     
  9. Dec 7, 2015 #8
    That clarifies things a bit! I'm still a little confused on how you know if you're getting an ideal property or the real value for the property. There have been some problems that I work where they give me T1, P1, P2, and k, then they plug those values into the formula. Sometimes they will denote it as T2 and sometimes as T2s. I guess my question is how do you know if its going to turn out to be a T2 value or an ideal T2s value?
     
  10. Dec 7, 2015 #9
    I don't know what an "ideal property" means, and I don't know what the notation 2s means (unless it implies a constant entropy path). If you are talking about ideal gases, then the results you obtain for all the thermodynamic parameters apply to changes in which deviations from ideal gas behavior are negligible. So the results accurately describe the actual gas behavior. Ideal gas behavior is typically observed in the limit of low pressures (typically less than 10 atm).

    Maybe it would help if we focused on a specific problem (or problems).

    Chet
     
  11. Dec 7, 2015 #10
    Oops sorry, my book uses T2s, h2s, ect. to refer to isentropic expansion and compression through the turbine or compressor. They use 2s for the ideal case where the process is isentropic. I guess calling it ideal is bad wording when you're also talking about ideal gases.
     
  12. Dec 7, 2015 #11
    I guess a better terminology would be adiabatic reversible expansion and compression.

    Chet
     
  13. Dec 7, 2015 #12
    I found a problem that shows what I'm confused about (#3). I hope its ok that I just posted the file, I thought it would be easier to read from there than me typing it out.

    I don't understand why using (T2/T1)=(p2/p1)^((k-1)/k) gives you T2s instead of T2? And then I don't understand how he found T2 for part B.
     

    Attached Files:

  14. Dec 7, 2015 #13
    The words "well-insulated compressor" and "minimum theoretical work needed to compress the gas" implies that you should be solving part (a) for the case in which each parcel of gas passing through the compressor experiences an adiabatic reversible compression. An adiabatic reversible compression results in the minimum amount of work required to compress the gas to a specified pressure. An adiabatic reversible compression is the same thing as an isentropic compression. That's the reason for the s on the 2s.

    Regarding the T2 for part B, the question is saying that the experimentally measured outlet temperature of the compressor is 397 C. So, the way he found T2 is that he just specified it as part of the problem statement, assuming that it was a measured value on the actual process.

    Chet
     
  15. Dec 7, 2015 #14
    Haha I can't believe I missed that on part B, I wrote it down and everything. I think I finally get it, thanks so much for your help!
     
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