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Isentropic efficiency of a pump

  1. Feb 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the isentropic efficiency of the pump: effpump=(h4s-h3)/(h4-h3)

    state 3: p3=1.5 bar, h3=467.11, x3=0, v3=1.0528/1000 m3/kg ,s3=1.4336 kJ/kg*K, T3=111.4 deg C

    state 4: p4=60 bar, h4=474.14, compressed liquid
    state 4s: p4=60 bar, s3=s4s

    The values in red are the ones I wasn't given, I looked them up in the tables.
    How do I find the h4s value?? I tried going to the compressed liquid tables and looking up the enthalpy at 60 bar, but my tables only have 50 bar and 75 bar. So am I supposed to interpolate between those two tables?

    Do I need to find the temperatures at states 3 and 4 before I start trying to interpolate? I'm pretty lost on this problem.


    2. Relevant equations


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Feb 12, 2016 #2

    SteamKing

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    25 bar sounds like a big step between tabular values, even for compressed liquid. I would recommend you get a better table to use for the values of compressed liquid at 60 bar.

    The NIST publishes some good tables on-line for this very purpose:

    http://www.nist.gov/srd/upload/NISTIR5078-Tab3.pdf

    Try this table and see if it helps. If you have any further questions, please post them.
     
  4. Feb 12, 2016 #3
    That table is a lot better! So s3=s4s=1.4336 kJ/kg*K. So I went and I looked at the 6MPa table and the lowest entropy value is 6.0703. I thought I could just find the entropy value of 1.4336 and find the corresponding enthalpy in that pressure table. Did I maybe look up the entropy at state 3 incorrectly?
     
  5. Feb 12, 2016 #4

    SteamKing

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    You're looking at the wrong page. Go back to the previous page, and you'll find lower entropy values for compressed liquid at P = 6.0 MPa.
     
  6. Feb 12, 2016 #5
    Oh ok! So I tried to interpolate between s1=1.4139 h1=465.68 and s2=1.4686 and h2=486.77
    I found h4s=458.1

    I plugged all my values into the isentropic efficiency formula and wound up with a negative percentage... Did I maybe interpolate incorrectly?
     
  7. Feb 12, 2016 #6

    SteamKing

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    Yes, you messed up. h = 458.1 is less than h = 465.68. The correct value for h is going to be between 465.68 and 486.77.
     
  8. Feb 12, 2016 #7
    Hmm... What you said makes total sense. I keep getting 458.1 for my h4s value though.

    y=y1+(x1-x)[(y2-y1)/(x2-x1)]
    letting enthalpy=y and entropy=x
    h4s=465.68+(1.4139-1.4336)[(486.77-465.68)/(1.4686-1.4139)]

    I'm almost positive I'm using all the correct values...
     
  9. Feb 12, 2016 #8
    Haha oops... I think it was my formula, let me see if that fixes it.
     
  10. Feb 12, 2016 #9
    That was it!! Thanks so much for your help!
     
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