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Thermodynamics - Saturated Mixture to Superheated Vapor

  1. Sep 14, 2012 #1
    1. The problem statement, all variables and given/known data
    A rigid tank contains 5 lb of a two-phase, liquid-vapor mixture of H2O, initially at T = 260 F with a quality of 0.6. Heat transfer to the contents of the tank occurs until the temperature is T = 320 F. Show the process on a p-v diagram. Determine the mass of vapor, in lb, initially in the tank in the final pressure, in lbf/in^2.


    2. Relevant equations

    V = Vf + X ( VG - VF

    Various thermodynamic charts found in any thermodynamics book.

    3. The attempt at a solution

    I can get the first part of the question fine. I took a look at a solution for this problem, and they randomly make the statement that while in phase 1 the mixture is a saturated vapor, in phase 2 the substance is a superheated vapor. This is what I need help with. I am not exactly sure why the substance is a superheated vapor in phase 2.

    I understand that since the the system is within a "rigid tank" that the specific volume must remain constant between the two phases. I've calculated the specific volume of phase one (using the quality, temperature, and the specific volume of both fluid and gas) and I've come up with V = 7.0688 ft3/lb. I know that the specific volume must be the same in phase two, but I do not have a Quality in which to calculate it.

    I used the same quality from phase 1 (although I know it would not be the same) and I've also used an assumptive quality of 1 (MAX). In both cases the specific volume of phase two is less than that of phase 1. Is this why I can justify in calling the substance a "superheated vapor in phase 2?

    This is my first time posting here, I hope I am not breaking any of your rules. I'm not looking for the answer to the question, I just need help understanding how I can determine why the substance became a "superheated vapor."

    Thank you.
     
    Last edited: Sep 14, 2012
  2. jcsd
  3. Sep 16, 2012 #2

    Q_Goest

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    Hi, and welcome to the board.
    Your relevant equations are fine and I agree you will need your steam table. I also agree with your calculation for specific volume = 7.0688 (though my database provides a very slightly different number, 7.0695) so all that is fine.

    It might help to consider density instead of specific volume. Either is fine of course, one is just the inverse of the other. But for density, you can think of there being so much mass in a given volume. So instead of a specific volume of 7.0688, consider that a density of 0.14147 lbm/ft3.

    For a temperature of 320 F, saturated gas has a density of aproximately 0.203 lbm/ft3. So if it were saturated at that temperature (pressure is roughly 89.6 psia) then you would have more water in the container than you originally had. Since there's less mass in the vessel at 320 F than there is for saturated vapor at 320 F, do you see any way for the water to be saturated? Note that if the quality of saturated vapor is 1.0 and a density of 0.203, then decreasing the quality (ie: decreasing from 1.0 to something between 0 and 1) will only increase density.

    If the density is less than the saturated vapor line, is that superheated or does something else happen?

    If it is superheated, then how can you use the steam tables to find the temperature?

    Hopefully this helps answer your question even if it doesn't answer the question that the problem poses directly. The final product is superheated of course, and you need to consider how to show that on a p-v diagram. Part of that will be to find the pressure and temperature where the water has a quality of 1.0. Have you determined that state yet?
     
  4. Sep 16, 2012 #3

    rude man

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    Look at the p-v diagram for water. Go to the isotherm for T = 260F and follow it until it flattens out in the mixed-phase region. Stop when the quality is 60%. Now, how do you move on the chart to correspond with heating the mixture until T= 320F?
     
  5. Sep 16, 2012 #4
    I just want to thank you both tremendously for your help with this. I spent a lot of time mulling over what you've said and it is pretty clear to me. It isn't very clear when looking at it at first, but it needs a little intuition and logic. Simply enough, the volume needs to remain constant because of the rigid tank, but V1 =/= V2 if V2 remains a saturated vapor.

    The T - V diagram is a clearer way of visualizing it. If I draw a vertical line (since V remains constant) the position 2 will fall out of the bounds of the saturated vapor and into those of superheated vapor.

    I also found an easier way of evaluating these things. According to a diagram I found online, If V1 < V2f, it will be supercooled liquid. For V2f < V1 < V2g it will be saturated mixture. For V1 > V2g it will be superheated vapor. I haven't tried this method out on any other examples, but if it works it will definitely help simplify things for me. Thanks again guys, I really appreciate it.
     
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