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Saturated Liquid-Gas constant pressure expansion/contraction

  1. Sep 19, 2008 #1
    A substance is contained within a simple piston-cylinder arrangement, perfectly insulated from the external environment, at a pressure,temperature,volume combination where it exists as a saturated liquid-vapour.

    If I now expand the volume using the piston, more of the liquid phase should boil and change to vapour in order to occupy the expanding volume.
    Is this a constant pressure, consant temperature process? , how is it effected by the rate of change of volume?

    Similarly, if I now recompress the volume, will the gaseous phase re-condense? , or will the gaseous phase be inclined to compress adiabatically and become a locally superheated vapour until it has time to give up heat to the liquid?

    I can handle basic thermodynamics but I am confused on this one as I think there are internal heat transfers required between the phases and the rate of expansion/compression must have implications on this being a constant temperature, constant pressure process.

    Can anyone shed light on this, or have experimental experience on what might happen?
     
  2. jcsd
  3. Sep 22, 2008 #2

    Q_Goest

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    Hi fitz,
    Welcome to the board.
    You might think this is true for all cases, but it’s not. Whenever you see “adiabatic” and “work being done” on a fluid, you need to think, “isentropic process”. Imagine a TS diagram and the process being a vertical line from one pressure to another under this dome. Now imagine you’re on the far left side of the dome where reducing pressure (and temperature) results in the quality (vapor fraction) of the 2 phase mixture increasing. In this area, you’ll increase volume as pressure drops and liquid will boil.

    Now imagine you’re over on the far right side of the dome where reducing pressure and temperature results in CONDENSING of some of the gas in the 2 phase mixture. That’s actually what will happen if we make the assumption that the entire fluid contents is in thermal equilibrium (ie: constant temperature throughout).

    As stated before, this is an isentropic process. Temperature will drop as work is done by the fluid on the piston. Note that this assumes there can be sufficient time for heat transfer to maintain a constant temperature throughout. However, this isn’t realistic as you have implied. The rate of change will affect how fast heat can flow from gas to vapor or visa versa. There’s no good method I know of to model this. Geometries and volumes inside the piston/cylinder arrangement will greatly affect the rate of heat transfer between the gas and liquid. To do this realistically, I’d probably keep them separate and calculate the states individually.



    Again, assuming heat flow has time to maintain a constant temperature throughout, the product will either recondense or boil again, depending on where ‘under the dome’ you are.

    This is actually a not so unusual case. In cryogenic tanks for example, these kinds of processes are common, and it is much more ‘normal’ (for cryogenic tanks) for the gas to be superheated above the liquid. Unless the compression is extremely/agonizingly slow, you should expect to find the gas becomes superheated and the liquid to simply become subcooled. The best solution is to calculate the state of the gas and liquid individually for almost any 'normal' rate of expansion or compression and then consider how the heat transfer between the two would affect each separate phase.
     
  4. Sep 22, 2008 #3
    Realistically, this is neither a constant pressure nor constant temperature process assuming its an adiabatic process. By increasing the specific volume, the pressure will decrease along with temperature. How these affects are related depends on the type of model you want to use. At low temperatures the ideal gas law works well in conjunction with a polytropic process.

    pv^n = constant
    Pv = NRT

    Also, there is no such thing as "saturated liquid-vapor", that doesn't make any sense. A saturated liquid is all liquid on the edge of the left side of the vapor dome (quality = 0) while a saturated vapor is all vapor (quality=1). That is why the term "saturated" is used.

    If you were to start pushing the piston in, or decreasing the specific volume, yes the vapor would start to condense assuming you are still on the right edge or in the P-v vapor dome.

    Assuming this is a quasi-static process there will be absolutely no heat transfer among anything.

    These are very important concepts that you will have to be second nature to you even if you want to understand the very basics of thermodynamics. As for experimental data, have a look at the property tables in the back of any thermo text book there is tons of experimental results there.
     
  5. Sep 23, 2008 #4

    stewartcs

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    Sure there is. Another commonly used name is saturated mixture. It's the region between the two you described (i.e. where the quality is between 0 and 1). It's simply the state where a saturated liquid and a saturated vapor exist in equilibrium.

    CS
     
  6. Sep 23, 2008 #5

    LURCH

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    I've noticed that Q and Topher both refer to this as "assuming" that it is an "adiabatic process," but is this posible given the conditions in the OP, in which the system is said to be, "...perfectly insulated from the external environment..."?
     
  7. Sep 23, 2008 #6

    stewartcs

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    That's what adiabatic means, no transfer of heat to or from the system to the surroundings. If it is perfectly insulated, one may assume that it is an adiabatic process.

    CS
     
  8. Sep 23, 2008 #7

    Q_Goest

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    Thanks for the backup stewartcs. :smile:

    Hi Topher,
    Note that a polytropic process is only an isentropic one if the exponent (n) is equal to the ratio of specific heats, and then only for processes where the fluid is superheated gas. The polytropic process and the ideal gas law can't be used for 2 phase fluids. One could however, use these equations to calculate the changes of state in the gas phase if there is no change in phase of either the liquid or gas.

    As pointed out also by stewartcs, one can talk about a saturated mixture of liquid and vapor. Nothing wrong there.

    On the right side, under the dome (on a TS diagram) compression of the mixture (assuming isothermal conditions throughout) is represented by a verticle line moving upwards, so rather than the vapor condensing, the saturated liquid is actually boiling and the quality is increasing. It only condenses during compression if the compression process is under the left hand half of the dome. As I'd mentioned earlier however, this assumes isothermal conditions which often times is not realistic. Instead, the gas will typically heat up as it's being compressed and the liquid will then become subcooled, and it won't be untill there is sufficient heat transfer between the warmer gas and cooler liquid that we'll actually see this process return to the 'verticle line' under the dome of the TS diagram.
     
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