1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Entropy Generation

  1. Jan 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Air at 100 kPa, 300 K is to be delivered to a pipeline at 500 kPa, 500 K. The scheme involves reversible adiabatic compression of the air and then internally reversible isobaric heating. Assume that, heat is exchanged with a reservoir at 600 K.

    Determine the work and entropy generation.

    2. Relevant equations

    v = RT/P

    3. The attempt at a solution

    • For the reversible adiabatic compression part. (Between 1-2)
    I have found both the work and the entropy generation.
    • For the internally reversible isobaric heating (Between 2-3)
    I have found boundary work from PdV but i'm not clear on how to find the entropy generation for this heating process.
     
  2. jcsd
  3. Jan 9, 2016 #2
    I'm interested in what you got for the entropy change in the adiabatic reversible compression. Do you know the general equation for the entropy change of an ideal gas between two thermodynamic equilibrium states? Please show your work for step 2 also.

    Chet
     
  4. Jan 9, 2016 #3

    • For the reversible adiabatic compression part. (Between 1-2)
    v1 = 0.861 m3/kg


    From T2/T1 = (P2/P1)(k-1)/k

    T2 = 475.15 K


    - W = U = cv (T2-T1) W = - 126 kJ


    Reversible adiabatic means isentropic. Sgen = 0

    • For the internally reversible isobaric heating (Between 2-3)
    W = PdV = 500 kPa . (v3-v2)


    From v=RT/P

    v2 = 0.273 m3/kg

    v3 = 0.287 m3/kg


    W = 7 kJ

    q - w = ΔH = cp (T3-T2)

    q = 32 kJ/kg

    Sgen = cpln(T3/T2) - Q/T

    This leads me to negative entropy generation. So, I must be making some wrong turns on the way.
     
  5. Jan 9, 2016 #4
    The final temperature at the end of step 1 (according to your calculations) is 475 K. The final temperature at the end of step 2 is 500K. So the entropy change of the system is Cp ln(500/475). What is the amount of heat removed from the reservoir is step 2? I get Cp (500-475).
     
  6. Jan 9, 2016 #5
    From q - w = ΔH = cp (T3-T2)
    I get q = 32 kJ/kg

    sgen = Δs - q/T = 0,052 - 32 kJ / 600 K = 0.052 - 0.053 = - 0.01

    Edit: Why didn't you involve work in your first law equation? Isn't there some boundary work PdV also. That is why I get q = w + cp(500-475) = 7 + 25 = 32 kJ/kg
     
    Last edited: Jan 9, 2016
  7. Jan 9, 2016 #6
    It's q-w = ΔU, not ΔH. The constant pressure work w is included in ΔH=q=ΔU+w.

    Try again. You almost have it. With this change, you will get your desired increase in entropy.
     
  8. Jan 9, 2016 #7
    Oh, I see. That helped a lot. Thank you so much.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted