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B Expansion of a gas

  1. Feb 28, 2016 #1
    Hi .I'm a beginner at thermodynamics,very limited in my understanding of math.I have a handbook on CO2 properties.According to it the properties for CO2 at 47.3 bar are T=12C,H= -3792,S= -42.6,and for 34.85 bar,
    T= 0 C,H= -3339.8,S=-39.33.If the gas expands from 47.3 to 34.85 bar,the gas will liquify because the enthalpy at the pressure of 47.3 is less than that for 34.85 bar.Can you please answer the following question
    Will the temperature be 0 C,or will the liquification cause the temperature to fall below 0 C?If so,to determine the temperature would you simply divide the difference between the enthalpies by the heat capacity and subtract the quotient from 0 C (-3339.8 - -3792 =452.2 / 90 = 5.2 C. 0 - 5.2 = -5.2 C.According to the handbook the entropy for CO2 at this temperature (-5.2 C)and pressure (34.85 bar)is -74.64,considerably lower than at the vapor temperature.
  2. jcsd
  3. Feb 28, 2016 #2
    How are you dropping the pressure: (a) at constant enthalpy (say through a throttling valve) or (b) at constant entropy?
  4. Feb 29, 2016 #3
    It would be at constant enthalpy.
  5. Feb 29, 2016 #4
    OK. The final temperature is going to be 0 C, and the system is going to be comprised partly of liquid and partly of vapor. It will probably be about 90% vapor, and about 10% liquid. You need to take into account the enthalpy of the liquid and enthalpy of the vapor, and set the total final enthalpy equal to the initial enthalpy. This will enable you to determine the precise proportions of liquid and vapor.
  6. Mar 1, 2016 #5
    Thank you for the help.What I am trying to do is how to use this handbook on CO2 properties in order to determine the entropy of CO2 when it expands from 47.3 bar
    and 12 C to 34.85 bar and 0 C.I think I have it figured out.The entropy values for 47.3 bar are -71.82 (liquid).-42.60 (vapor) and 29.22 (vaporisation),and for 34.85 bar
    they are;-76.53.-39.33 and 37.20.The final entropy should be -76.53 + 29.22 = -47.31 j / mol.K.Is this correct?If so it would mean that the entropy has gone from -42.6
    -47.31,a reduction of 4.71.I thought that in all natural processes entropy must increase.This is a closed system.If I am wrong can you explain how I should be using the
    handbook to determine the correct entropy.
  7. Mar 2, 2016 #6
    The entropy change should definitely be positive. I checked on a P-H diagram for CO2 and at all conditions, the entropy increases when the pressure decreases at constant enthalpy. So there must be something wrong with your calculation.

    Before we start getting into that, please tell me the units of enthalpy (and entropy) in your posts. You seemed to indicate that the units of s are J/molK, but please check on that. Your enthalpies don't seem to be consistent with your entropies.

    I can help you, but only if the data are all consistent in terms of units.
  8. Mar 2, 2016 #7
    OK. I checked the units on your enthalpies and entropies, and they seem to be J/mol and J/mol-K, respectively. So, at 0C, the molar enthalpy of the saturated liquid should be -3340-(37.2)(273.2)=-13503 J/mol. If x is the mole fraction liquid and (1-x) is the mole fraction of vapor at 0 C, then, in order for the enthalpy of the mixture to be the same as the saturated vapor at 12 C, we must have $$-13503x-3340(1-x)=-3792$$

    The solution to this equation is x = 0.0445. So the fraction saturated liquid at 0 C is 4.45%, and the fraction saturated vapor is 95.55%.

    The entropy in the final state is S = (0.0445)(-76.53)+(0.9555)(-39.33)=-40.98. So the entropy has gone from -42.6 to -41.0, for a change of +1.6 J/mol-K.
  9. Mar 2, 2016 #8
    Thank you for your help.Does this method work for compressing a gas say from 47.3 bar at 12 C to 56.6 bar at 19.5 C?The values I have for liquid enthalpy at

    56.6 bar is -11116.5 and for vapor it is -4308.65.So,-11116.5x-4308.65(1-x) = 3792.This gives .076 liq and .924 vap.The entropy values for 56.5 bar are -68.51(liq)

    and -45.24 (vap),so .924 x -45.24 = -41.8 +- 5.2 (.076 x -68.51 = -5.2) = -47.Once again the entropy is lower.I must have missused the formula.Can you tell me where I went wrong?
  10. Mar 2, 2016 #9
    I don't think that this works for compressing. You can't flow from a lower pressure on one side of a valve or porous plug to a higher pressure at the other side.
  11. Mar 3, 2016 #10
    An undergraduate gave me a similar formula for isentropic compression (S1=S2).S (vapor) for 47.3 bar is -42.6,the S for 56.6 bar is- 45.2 (vapor) and- 68.51 (liquid).
    So, -42.6 = (1-q) -68.51+q-45.2,q = 1.111 superheated vapor.Now,is it permissible to do what you did with adiabatic expansion?,namely use the liquid/vapor fraction to
    calculate the real entropy.If so then 1.111 x -45.2 = -50.22.This appears illogical,but the S1 = S2 formula is just a method to compute the liquid / vapor fraction,the two pressures could not really have the same entropies.I have read that increasing pressure decreases entropy.Applying the ideal gas equation PV = nRT,should not the superheat increase the pressure (perhaps just for an instant),so for that instant,the real pressure and hence the real entropy would be higher than the values from the
    manual (which are at vapor pressure and temperature)?
  12. Mar 3, 2016 #11
    Sorry,in the last sentence I meant to say ,applying the ideal gas equation,the pressure should be higher,and hence the entropy should be lower,than the values at the vapor pressure and temperature (meaning neither subcooled nor superheated).
  13. Mar 3, 2016 #12
    You need to look at a pressure-enthalpy diagram for CO2 to understand better. This is a very valuable type of diagram. There is a p-H diagram for CO2 in Fundamentals of Chemical Engineering Thermodynamics by Moran et al. You can probably also find the diagram online. The diagram even shows constant entropy lines, for your convenience.

    If you start out 47.3 bar on the saturation curve, and move along the constant entropy line to higher pressure, the diagram shows that you move into the superheated vapor region. So you only have one phase. For a single phase gas, when you compress it isentropically, both the temperature and the pressure increase. The increase in entropy from the increased temperature exactly cancels the decrease in entropy from the increased pressure, so the entropy stays the same. The ideal gas law applied to an isentropic compression predicts this too, although, in this case, the pressures are a little too high for the ideal gas law to accurately apply.
  14. Mar 3, 2016 #13
    Thanks for your help
  15. Mar 4, 2016 #14
    May I ask one final question?What if CO2 gas at pressures of 47.3 bar, S=-42.6 and 56.6 bar,S= -45.2 were mixed?,as follows.There are two cylinders of the same volume,one open,full of CO2 at 47.3 bar (cylinder A),the other closed (cylinderB).A pipe connects the two cylinders.As a piston pushes the gas in cylinder A through
    the pipe,cylinder B opens up so there is no compression.On the way,the gas in the pipe is joined by gas coming from a reservoir,much larger than the cylinders.The pressure of the gas in the reservoir is 56.6 bar.This gas pressurizes the gas in the pipe going from cylinder A to B so that the pressure in cylinder B is 56.5 bar.The question is; what would be the entropy of the gas in cylinder B? Would it be higher than -42.6 or an average of the two?
    ( - 42.6 + -45.2 / 2 = -43.9 ).My guess is -43.9.My reasoning is,assuming isentropic compression and adiabatic expansion of the two gasses.The lower pressure gas
    experiences no expansion and its compression is isentropic,while the higher pressure gas experiences no expansion,so also no change in entropy.So the final entropy would be an average of the two.
  16. Mar 4, 2016 #15
    Sorry, I'm unable to picture what you are describing. Maybe you can provide a diagram(s).

  17. Mar 4, 2016 #16
    I don't have a picture available at this time.Assuming this scenario is as I have described,do you whether whether the final entropy would be greater than the initial
    value of -42.6,or somewhere in between -42.6 and 45.2?
  18. Mar 4, 2016 #17
    As I said, I am unable to figure out the geometry and the scenario. You can draw a diagram in Powerpoint, capture it with the Windows Snipping Tool, save it to your desktop, and upload it rather easily to Physics Forums by hitting the UPLOAD button.

  19. Mar 4, 2016 #18
    As I am at the public library that might take some time.I was just reading a statement by a member that when gases mix the entropy always increase.My example
    is contrived.Lets say the gas does no work on the piston which is activated by an outside force (electrical).I just want to know whether when the lower pressure gas is joined by the higher pressure gas
    would the increase in entropy be more than both the higher and lower pressure gas or just the higher pressure?
  20. Mar 4, 2016 #19
    I don't follow your question, but maybe this will help. Suppose you have an adiabatic chamber with a partition that divides the chamber in half, and you have gas at a high pressure on one side of the partition and the same gas with a lower pressure on the other side of the partition. You poke a hole in the partition and you allow the chamber to re-equilibrate. The final entropy of the system will be higher than original entropy of the system. Does that help?
  21. Mar 5, 2016 #20
    When you say higher than the original entropy of the system do you mean higher than both the higher or lower pressure gas,or just higher than the higher pressure gas.(I am assuming that the higher pressure gas has a lower entropy than the lower pressure gas.Using my example the 47.3 bar gas is -42.6 and the 56.6 bar gas is -45.2).In my example the 47.3 bar gas coming from cylinder A through the pipe is experiencing no expansion and is compressed up to 56.6 by the gas coming from
    the high pressure reservoir.It too is experiencing no expansion.The gas flowing through the pipes is travelling at a high velocity so the pressure is much less than 47.3 bar,so there is more space between the molecules.This means the increase in pressure (up to 56.6 bar when the gas enters cylinder B ) is due to the increase in the density of the gas.Would this not mean that this would be an isentropic compression of the 47.3 bar gas,and since there is no expansion of either gas would not this
    be isentropic for both gases?If so,would it also imply that the final entropy of the gas in cylinder B would be somewhere between -42.6 and -45.2?
  22. Mar 5, 2016 #21
    Entropy is an extensive property. What I'm saying is that the final entropy of the combined gases (at the final equilibrium pressure) will be higher than the sum of the original entropies of the higher pressure and lower pressure gases.
  23. Mar 6, 2016 #22
    Unfortunately, you still haven't provided a diagram, so I am at a loss for understanding what this system consists of, and what is happening physically.

  24. Mar 6, 2016 #23
    Do you mean higher than the average of the sum of the entropies (in my case it would be higher than -42.6+-45.2 =- 87.8 / 2 = -43.9),and is it proportional,so
    that if the proportions were 1/4 gas -42.6 and 3/4 at -45.2 the final entropy would be more than -44.55.If so could not the final entropy be less than -42.6?
  25. Mar 6, 2016 #24
    Yes, if you are talking about the entropy per unit mass, then the -44.55 would be correct. But, it is better to think of it as Sfinal>-42.6+3(-45.2) Regarding the final entropy per unit mass being less than -42.6, how could the weighted average of two numbers be less than or equal to either one of them?
  26. Mar 7, 2016 #25
    I was thinking per unit mass.No more questions but a clarification regarding isentropic compression.Using the compression of 47.3 bar isentropically and adiabatically up to 65bar as
    an example.Using my manual of CO2 properties the temperature,entropy and enthalpy at 47.3 bar (vapor pressure) is 12C,-42.6, and -3792.At 65 bar (vapor pressure) the temperature is 25.5 C,entropy -48.25 and enthalpy is -5026.4 and density 249.3.At 65 bar and entropy -42.65,the temperature is 35 C.and the enthalpy is -3334.Now,since this is a
    closed system would not the increase in temperature (above the 25.5 C at the vapor pressure) to 35 C also increase the pressure (if only for an instant).Would not applying the ideal gas equation (PV = nRT) give an approximate figure for the real pressure?Also I used the figure for the density at the vapor pressure (249.3) and
    the higher temperature (35 C) to find the pressure to be about 73 bar.If this is correct,are you saying that this increase in pressure,which would decrease entropy would
    be balanced by the increase in entropy resulting from the 9.5 C increase in temperature caused by the work of compression, so the final entropy figure would still be -42.6?
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