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  1. #1
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    R134a Compression



    Hi there!

    I'm new to this forum..found it by coincidence and have to say that its really well done.

    Well I have a problem that i've been struggling with for a few days and hope that you guys might help find a solution.

    I want to calculate the compression of R134a.

    The formula known for isentropic power is:
    W_isentr=m_dot_refr*[p1/(rho1*(k-1)/k)]*{[(p2/p1)^((k-1)/k)]-1}
    where, rho is the density of refrigerant, m_dot the massflow and p the pressure and k the isentropic exponent.

    For calculating the real power we can utilize following equation
    W_real = m_dot_refr*cp_refr*(T2-T1)

    Combining previous equations we get the isentropic efficiency
    eta_isentr=W_isentr/W_real

    For determining the discharge tempereature of the refrigerant T2 we can utilize mentioned equatios and get

    T2/T1=1+[p1/(T1*eta_isentr*cp_refr*rho1*(k-1)/k)]*{[(p2/p1)^((k-1)/k)]-1}


    Assuming ideal gas behaviour past equation can further be simplified to:

    T2/T1=1+(1/eta_isentr)*{[(p2/p1)^((k-1)/k)]-1}


    I hope the equations are clear. You might need to write it down a paper to recognize them.

    Anyhow, I did my calculations using the simplified equation and I got far too high temperatures for an isentropic compression.
    To clarify this I did the calculation for a compression from a saturation pressure of 5°C (p1=3,5bar) to that of 70°C (p2=21,2bar). The discharge Temperature I got was T2=202°C which doesn't at all seem to be realistic considering an eta_isentropic of 100%.

    Therefore I did some research and I got to know that we can't consider R134a an ideal gas at these conditions.

    Then I made my calculations using the equation without the simplification and the result was (as far as I can judge) too low. The discharge temperature was only 43,5°C which is remarkably lower that saturation Temperature of 70°C. This would mean that the compression was "better" than an isentropic compression.

    I can't imagine that this is correct, but I can't find the mistake.

    Maybe it's the isentropic exponent k. I considered it to be k=cp/cv. But maybe thats only for ideal gases and for real gases you need to use another formula.

    I really don't know right now.

    Maybe some of you guys had similar problems to face.

    Anyways, thank you in advance and best regards!

    Baker



  2. #2
    Eng_Baker's Avatar
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    Re: R134a Compression

    Hi again,

    now that I didn't get a reply so far, I guess the way I presented my question wasn't clear.

    Let me try to ease:
    How would you calculate the (isentropic) compressor discharge temperature of e.g. R134a without referring to an enthalpy data base?

    Thanks,
    Baker

  3. #3
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    Re: R134a Compression

    Hi there,

    I have not checked your equations but as you need the discharge temperature then it is nearly 76°C at the conditions you mentioned.
    I don't know where you got it wrong but try it on P-h diagram and it is fairly straight forward. Also you can check it on T-s diagram.
    For realistic figure which you have to know the isentropic efficiency of the compressor then you must select the equipment and then do the balancing procedures to find the actual discharge temperature.

    Cheers
    Even Einstein Asked Questions

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    Re: R134a Compression

    You might also use an average of the specific heat ratios between the actual suction and discharge pressures.

    Since the normally listed specific heat ratio is for a constant pressure it tends to give mixed results when the conditions are different from the basis of the ratio.

    At best, all I think you will get is an approximate value. Trying to obtain a precise value is more difficult, which calls for test data on the compressor.
    If all else fails, ask for help.


  5. #5
    Eng_Baker's Avatar
    Eng_Baker Guest

    Re: R134a Compression

    Thank you both!

    Iceman, in order to get precise (and real) values for the discharge temperature we need to have compressor data, as you mentioned. But if I only want to get the isentropic (ideal) discharge temperature, I think there should be an equation which meets this demand.
    I mean there has to be an equation that presents the p,h or T,s diagram of a fluid.

    I guess the formula is the one I have mentioned in my first post.

    Using this formula I tried many different variations of the specific heat ratio, but I didn't get the correct isentropic work (resp. isentropic power) that would result by applying the p,h diagram, as lana mentioned in her reply.

    Therefore, I guess that there might be no correct (and at the same time easy) equation to calculate the correct isentropic work, so that we might always need to refer to p,h diagrams or equivalent Look-up tables!?

    What do you say?

  6. #6
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    Re: R134a Compression

    I think what you are finding is similar to the same problem I ran into years ago. This is just a difficult task to accomplish.

    The isentropic efficiency is fairly easy if you calculate the ideal power input and compare that to the listed/tested power input.

    Unfortunately, you can calculate an ideal discharge temperature but it has very little factual substance in the real world.

    You always find the actual discharge temperature shifts to the right of the constant entropy lines from the operating suction pressure. As you mentioned in one of your posts, the pressure ratio version should get you about as close as I know how to.

    With this I always use the specific heat ratios based on the actual suction and discharge pressures though as a simple average value.

    Using the specific heat ratio commonly found in refrigerant data is usually only listed for 1 atm. at a moderate ambient temperature. This will definitely cause the calculated discharge temperature to be lower than what you might expect.
    If all else fails, ask for help.


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