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!

Thermodynamics - Regenerator Efficiency

  1. Nov 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Hey I just have a question about regenerator efficiency. I have a problem where the efficiency of the regenerator in a Regenerative Brayton cycle is 0.75, the regenerator takes air in from a compressor which I called h2=636.109 kJ/kg, air then leaves the regenerator and is now at h3=590.1625 kJ/kg. The air then travels through a combustion chamber and then to a turbine. After this the air heads back to the regenerator at an enthalpy of h5=574.847 kJ/kg. Once the air passes through the regenerator is is expelled to the ambient at some T6 and h6. I am thinking if I can find the h6 I can simply interpolate to find T6.
    If some of my numbers do not make sense here is the question,


    2. Relevant equations
    e = regenerator efficiency = 0.75

    e = (h6 - h5) / (h6s - h5)

    From this website, http://www.mhtl.uwaterloo.ca/courses/me354/lectures/pdffiles/c8.pdf, on page 8 I have used this equation, ( The numbers used in the following equation correspond to the picture here )


    e = (h3 - h2) / (h5 - h2)

    3. The attempt at a solution

    I have already solved the question up to this point but don't know what to do for this part. From the equation above h6s is assuming the system is isentropic and there is no change in enthalpy between the exit and entrance to the regenerator, could I simply use

    h2 = h6s

    and since I know h2 I can solve for h6 ?
    Last edited: Nov 26, 2014
  2. jcsd
  3. Dec 2, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. Dec 3, 2014 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    This is the part I don't understand. According to your calculations, the regenerator is acting like a refrigerator, since the air exiting the compressor has a higher enthalpy than when it exits the regenerator and enters the combustor. It seems to me that in order to increase cycle efficiency, the air leaving the regenerator should have a higher enthalpy, which energy it has recovered from the exhaust gases, and which energy would normally be lost when the exhaust leaves the engine.

    In other words, why are you using compressed intake air to heat the exhaust?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted