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Finding vapour pressure using compressibility chart

  1. Mar 10, 2015 #1
    1. The problem statement, all variables and given/known data
    I was asked to find the vapour pressure and saturated liquid molar density of propane at 263.15K using a generalised compressibility chart.
    (not allowed to use NIST or steam tables either, the chart i was given does not have reduced volume lines)

    2. Relevant equations
    Tr=T/Tc Pr=P/Pc z=Pv/nRT

    3. The attempt at a solution
    The reduced temperature is 0.712. I calculated the critical point compressibility to be 0.276 but that probably has nothing to do with what i was asked to find. Compressibility is a function of reduced temperature and pressure....... Tried using the chart with reduced temperature line at 0.7, followed the line to where it meets the line ends, got a reduced pressure of about 0.1, compressibility of about 0.88. But using the z=Pv/nRT equation above i got a density of 198 mol m^3 which is the saturated vapour density, not the saturated liquid density i was asked to find.
    I was also told that "you look at the figure for the compressibility factor Z=Pv/(RT) as a function of P/Pc you will find that it is a unique universal curve where Z is not need to given in reduced units" which i do not understand sadly.
    Last edited: Mar 10, 2015
  2. jcsd
  3. Mar 10, 2015 #2
    Have you been learning about the acentric factor?

  4. Mar 10, 2015 #3
    we were told to assume that the assentric factor w for propane is low so Z=Z0
  5. Mar 10, 2015 #4
    According to Smith and Van Ness, Introduction to chemical engineering thermodynamics, the acentric factor for propane is 0.152. The reduced saturation vapor pressure of a gas at a reduced temperature of 0.7 is related to the acentric factor by:
    You also know that, at the critical point, the reduced saturation vapor pressure = 1 and the reduced temperature is equal to 1.

    The slope of a graph of log of the reduced saturation vapor pressure vs 1/Tr is (a) approximately constant or (b) not approximately constant?

  6. Mar 10, 2015 #5
    the slope is approximately constant?
    but what does that have to do with my problem?
  7. Mar 10, 2015 #6
    If the slope is constant, and you know the equilibrium vapor pressure at two temperatures, then you know it at all temperatures. In your problem, you know it at the critical point and at a reduced temperature of 0.7. So either interpolate or fit an equation to log Psat vs 1/T.

  8. Mar 10, 2015 #7
    what does that have to do with the compressibility chart im supposed to use though?
  9. Mar 10, 2015 #8
    Let me guess. The compressibility factor chart you have goes from Pr = 0 to Pr = 1, and it shows reduced temperatures running from 0.7 to 4.0, correct?

  10. Mar 10, 2015 #9
    yes thats it
  11. Mar 10, 2015 #10
    On your figure, there is a lower curve that all the reduced temperature lines intersect (for reduced temperatures less than 1.0). This is the equilibrium vapor pressure vs temperature line.

  12. Mar 10, 2015 #11
    and how do i use this?
  13. Mar 10, 2015 #12
    I also found the Psat value for Tr = 0.712 to be 3.50 bar
  14. Mar 10, 2015 #13
    It looks like you used the graph correctly. I really don't know how to get the liquid molar density from this chart. Of course, the gas molar density is no problem.

  15. Mar 10, 2015 #14
    i asked the lecturer and he said i can "calculate the liquid density using the compressibility" and something to do with the zo value and deviation from an ideal gas.......
    but thank you very much for helping
  16. Mar 10, 2015 #15
    I found a graph online with a saturated liquid line also shown, but the compressibility factor along this line was very low (<0.1), and the value was very inaccurate to read from the graph at Pr = 0.1.

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