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(Thermodynamics) A tank with water & air heated

  1. Dec 11, 2015 #1
    1. The problem statement, all variables and given/known data
    For a rigid, impervious, closed tank - The volume of tank is filled with 70% water and the rest is with air- The initial temperature of the tank is 20 °C, and the absolute pressure is 1 bar. - What would be the final pressure if we raise the temperature to 200 °C?

    2. Relevant equations

    3. The attempt at a solution
    I tried finding some values at the water-vapour table but I just dont know what to do about it because I cant be sure if the water will become vapour or will stay in its liquid form. ANY help is appreciated
  2. jcsd
  3. Dec 11, 2015 #2
    Do not be afraid!
    It is not dangerous to start calculating with an assumption that turns out to be wrong.
    (The vessel will not explode, you won't get burns, the only risk is that you would have to try something else.)
  4. Dec 11, 2015 #3
    The problem is that I cant even be sure of my assumption. How am i supposed to know if the water will stay liquid or will turn into gas?
  5. Dec 11, 2015 #4


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    Water can't stay liquid at 200° C unless the internal pressure in the tank is greater than the vapor pressure of water at that temp.
  6. Dec 11, 2015 #5
    Are you currently learning how to use the Steam Tables in your course, or are you learning about the ideal gas law and vapor pressure?

  7. Dec 11, 2015 #6
    Yes and any way to find out if it will or not?
  8. Dec 11, 2015 #7
    Its a off course question, asked by mentor at a company. He said it has the same principle for calculations for extruders which contain screw and barrel. He has given me time for me to solve it until the end of 1st semester and said I could ask anybody if I cant make it myself. I asked it to my teacher but she didnt help at all not sure if she wants me to think, study and learn by myself or she also doesnt know it either. So its probably higher than thermodynamics 1 course
  9. Dec 11, 2015 #8
    Take as a basis a 1 cubic meter tank (the actual volume you take doesn't matter). First focus on the initial conditions. What is the specific volume of liquid water at 20 C and 1 atm? What is the equilibrium vapor pressure of water at 20 C? What is the specific volume of saturated water vapor at 20 C? Based on these answers, what is the initial mass of liquid water in the tank? What is the initial mass of water vapor in the tank? What is the initial total mass of water in the tank. What is the partial pressure of air initially? How many moles of air are in the tank initially?

    When you have the answers to these questions, let's see what you get.

  10. Dec 11, 2015 #9
    You know that by the outcome of the calculation. When you numbers are consistent, that is the answer.

    The water does not "know" either.
  11. Dec 11, 2015 #10
    If you know the total mass M of water in the tank, you let the final amount of water vapor be x and the final amount of liquid water be M-x. You then look up the specific volume of saturated liquid water at 200 C and the specific volume of saturated water vapor at 200 C. Then you solve for x under the constraint that the total volume of liquid water and water vapor must be equal to the volume of the tank.

  12. Dec 11, 2015 #11
    There is no need to solve for x. The only question is what the pressure will be.

    If there is any water in the liquid phase, you just look at the vapor pressure curve.
  13. Dec 11, 2015 #12
    Well, if there is a change in the volume of liquid, the volume of the air also changes, and this can affect the partial pressure of the air and total pressure.
  14. Jan 11, 2016 #13
  15. Jan 11, 2016 #14
    Still looking for answers. Need help with the calculation part mostly
  16. Jan 11, 2016 #15
    You still haven't answered my questions in posts #5 and #8. How are we supposed to help you if you don't answer our leading questions? I know you can answer the question in post #5, because it doesn't require any analysis.
  17. Jan 11, 2016 #16
    for your fifth post my reply was

    and for your 8th post you already pointed out the fact that makes it very complicated and that doesnt give me a clue what are the constants in here. Everything in this system are effected by the parameters
  18. Jan 11, 2016 #17
    OK. Let's do it without using the steam tables. Let's take as the basis of the calculations a tank with a volume of 1 cubic meter. That means that there is 0.7 cubic meters of liquid water plus 0.3 cubic meters of a gaseous mixture of air an water vapor at 20 C and 1 bar. The first step is to figure out the mass of water and the mass of air in the tank to start with (this won't change when the contents is heated)? Do you know how to figure this part out?

  19. Jan 11, 2016 #18
    I expect you to look up the data asked for in post #8 on your own. This is what is required to start solving this problem. In real life, people don't spoon feed you data like they do in school. Welcome to the real world.

    If you follow the steps I am leading you through, starting with the answers to the questions in post #8, I can guarantee you will get your answer. Otherwise, good luck.

  20. Jan 12, 2016 #19
    Ok so the volume density of saturated water at 20 C is 0.001002 m^3/kg ( neglecting the small difference between compressed water at 20 C and 1 bar) and the volume density of air at 20 C and 1 bar is around 0,833 m^3 / kg

    Assuming we have a 1 m^3 tank we have

    0,7 / 0,001002 = 698,6 kg of water and
    0,3 / 0,8333 = 0,36 kg of air in the tank
  21. Jan 12, 2016 #20
    Chestermiller already told you that the 0.3 m^3 also contains water. Yet you refuse to calculate how much.
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