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Superheated steam and water together

  1. Aug 2, 2016 #1
    Lets start with an imaginary experiment. There is an enclosed container having a wide (i.e. having sufficient surface area) bowl filled with water at the bottom. The container is in vacuum and filled only with saturated steam at that specific temperature. Both water and steam are at the same temperature and pressure.
    Now, some amount of superheated steam has been injected into that container. What will happen then?
    As per my guess, a part of the hot superheated steam will enter the water and will raise its temperature and soon the pressure and temperature inside will rise and the superheated stem will start evaporating water from the bowl until the condition inside will come to an equilibrium and all the steam will become saturated.
    By this process, superheated steam will loose some heat that will be used to evaporate some water from the bowl to make all the steam insdie saturated.
    Am I right?
  2. jcsd
  3. Aug 2, 2016 #2


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    Yes. Within certain limits. In equilibrium the vapour is saturated.
    With a simple enthalpy balance you can calculate the final p and T.
  4. Aug 2, 2016 #3
    Thanks! But, I am curious to know how long it will take and what kind of factors can affect this time span.
  5. Aug 2, 2016 #4


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  6. Aug 2, 2016 #5
    OK. Lets discuss it and try to find out the time necessary.
    Actually, I want to say that both steam will condense and water will evaporate in the process. That may look contradictory but it will be fact in such a case. High energy steam molecules from the superheated steam will enter the water and will raise its temperature. At the same time, comparative low energy steam molecules will rise and will mix up with supherheated steam to drag it down to saturated state. Am I right?
    Last edited: Aug 2, 2016
  7. Aug 2, 2016 #6
    Why speculate? Why don't you just do the calculation (and thereby remove all the uncertainty)?
  8. Aug 2, 2016 #7
    It's not speculation, but rather trying to envisioning the whole process first and then go for the next step. And, by the way, can you suggest how to calculate the time necessary for evaporation?
  9. Aug 2, 2016 #8
    Before calculating the time required for the equilibration, I strongly recommend that you first determine (quantitatively) the final equilibrium state. Why? If you are unable to model that simpler situation, you will never be able to analyze the much more complicated problem of the time-dependent behavior (in my judgment). Plus, the solution for the final equilibrium state might suggest some ideas on how the system will behave quantitatively in the transient situation. Are you prepared to work with me on solving for the final equilibrium state?
  10. Aug 2, 2016 #9
    No problem!
  11. Aug 2, 2016 #10
    OK. To get us started, I need you to specify a focus problem for us to analyze:

    Initial Conditions in Tank
    1. Volume of tank
    2. Temperature in tank
    3. Mass fraction of liquid water in tank

    Conditions of Superheated Injectate
    1. Temperature
    2. Pressure
    3. Mass of superheated steam injected into tank
  12. Aug 3, 2016 #11
    this whole process that you have said what is the surety level that this will happen exactly as you said ? it depends on the amount of water in the container the mass of superheated steam the degree of super heat, and mass of liquid etc, you can't say this before even solving for the final conditions
  13. Aug 3, 2016 #12


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    Are you planning to model this step as a constant-pressure process? Adding mass will have to increase pressure as well for a fixed-volume container.

    What you're describing sounds a little like a flash evaporation process in a vapor-liquid separator.
  14. Aug 3, 2016 #13
    As per my guess, more steam will be added to the water than the amount of water evaporated. Because, only in that case the temperature of water can rise. You are right that it will certainly be dependent on the degree of superheating and also the amount of superheated steam being injected. If the temperature is too high (above triple point) and the amount too is high and more than the amount of water inside, then the condition inside will be above triple point after the steam being injected. Lets consider that the temperature of superheated steam is above 30C of its saturation level and the amount is far less than the amount of water inside.
    No. The pressure inside can't be the same after the superheated steam being injected. Both temperature and pressure inside will rise until an equilibrium point will come when all the steam inside will become saturated.
    Again No. This is just opposite to flash evaporation. In flash evaporation process, pressure above water/liquid is suddenly reduced and a part of the liquid/water will become steam. In this process, the pressure will rise and part of steam will be added to the water inside so that its temperature can rise.
  15. Aug 3, 2016 #14
    I guess when you said "No problem," what you really meant was "Yes problem." All I asked from you in post #10 was 6 numbers. After all the thinking you have done on this problem, it is hard to imagine that you would have trouble coming up with these 6 numbers.
  16. Aug 3, 2016 #15
    1. 1 cubic meter
    2. 20C
    3. 1 kg
    1. 60C
    2. 7.37 kPa (saturation pressure of water at 40C)
    3. 10 gm
  17. Aug 3, 2016 #16
    Good. Now, from the steam tables, the specific volume of saturated liquid water at 20 C is 0.001002 m^3/kg and the specific volume of saturated water vapor at 20 C is 57.79 m^3/kg. So, can you please calculate what part of the 1 cubic meter tank volume is occupied by the liquid water (in cubic meters)? What part of the 1 cubic meter tank volume is occupied by water vapor (in cubic meters)? What is the mass of the water vapor in the tank under these initial conditions? What is the total mass of water (both liquid water and water vapor) initially in the tank?
  18. Aug 4, 2016 #17
    Well, if the amount is 1 kg, then certainly the volume occupied by water is 0.001002 cubic meter and rest 0.998998 cubic meter has been occupied by saturated steam at 20C (Assuming the pot is very thin and its volume is negligible). As per steam table, the specific density of steam at 20C is 0.01728 kg/m^3 and therefore the mass of steam occupying 0.998998 cubic meter is 0.01726268544 kg. Therefore, the total mass of water and steam inside the tank initially was 1.01726268544 kg.
  19. Aug 4, 2016 #18
    Excellent. So the mass of saturated liquid water in the tank is 1.000 kg and the mass of saturated water vapor in the tank is 0.0173 kg. Here is an excerpt from a more complete steam table than the one is Wiki for saturated steam: Steam.PNG
    It shows internal energy, enthalpy, and entropy of saturated water (for both liquid and water vapor) relative to liquid water at 0 C and 1 atm pressure. From the table you can see that the internal energy of saturated liquid water at 20 C is ##u_L=##83.94 kJ/kg, and the internal energy of saturated water vapor at 20 C is ##u_V=##2403 kJ/kg. Now, given the amounts of liquid water and water vapor in our tank initially, I would like you to calculate the total internal energy of the water in the tank initially.

    Also, for later reference, these same more complete steam tables give a value of ##h_v=2610## kJ/kg for the enthalpy of suprerheated water vapor at 60 C and 7.37 kPa.

    Now for the most important part. Are you familiar with the "open system" control volume version of the first law of thermodynamics (given in every serious thermodynamics text)? This version of the first law of thermodynamics applies to systems like ours in which mass is entering the system between the initial and final states of the system. If you are familiar with the open system version of the 1st law, please write the equation down. Thanks.

  20. Aug 4, 2016 #19
    The version just says that the mass at the beginning and at the end and also the gross amount of energy at the beginning and at the end will be same. I can't write the formula here as I don't know how to insert symbols in the thread. Whatsoever, you can continue.
  21. Aug 4, 2016 #20
    I guess you didn't want to calculate the total internal energy of the water and steam in the tank prior to the superheated steam injection (even though I requested that you do this). If you are uncomfortable with doing this calculation, I can do it for you. We will need the initial internal energy later in our calculations. Do you prefer that I evaluate the initial internal energy myself?
    This is not precise enough for what we are doing. We need to identify the precise quantities in the various terms of the equation.

    Here is the open system control volume version of the First Law of Thermodynamics that is presented in Fundamentals of Engineering Thermodynamics by Moran, shapiro, Boettner, and Bailey (note that this was derived directly from the usual closed system version of the First Law):

    $$\frac{dE_{cv}}{dt}=\dot{Q}_{cv}-\dot{W}_{cv}+\sum_i{\dot{m}_i(h_i+\frac{V_i^2}{2}+gz_i})-\sum_e{\dot{m}_e(h_e+\frac{V_e^2}{2}+gz_e})$$where the total energy in the control volume ##E_{cv}## is given by:$$E_{cv}=\int_V{\rho(u+\frac{V^2}{2}+gz})dV$$
    In these equations, the subscript i refers to inlet streams to the control volume and e refers to the exit streams from the control volume. In addition h represents enthalpy per unit mass, u represents internal energy per unit mass, V represents velocity, g is the acceleration of gravity, z represents elevation above a specified datum, ##\dot{m}## represents the mass flow rate of an inlet stream or exit stream from the control volume, ##Q_{cv}## represents rate of heat addition to the control volume, and ##\dot{W_{cv}}## represents the rate of doing work on the boundary of the control volume, excluding the work to push material into and out of the control volume (the latter is included in the summation terms for the inlet and exit streams).

    I think I'll stop here and give you a chance to digest these (standard) relationships before proceeding to show how they are to be applied to our specific problem.
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