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Lowering a tube into seawater

  1. Jan 19, 2014 #1
    1. The problem statement, all variables and given/known data
    The assignment is not in english but I'll try to translate it. I hope I've posted it in the right thread.
    You have a tube which is 48 meters long and one end is closed while the other end is open. You take the tube with the open end first into seawater (1025 kg/m3) and drop it down to 48 meter depth. The atmospheric pressure is 1020 hPa. Do not take into account temperature differences.
    How high will the water rise inside the tube?


    2. Relevant equations
    Fluid pressure = Density * 9,81 * height
    Total pressure = Air pressure + Fluid Pressure
    Pressure * Volume = Pressure * Volume (Constant)


    3. The attempt at a solution
    To find a solution I think I need to find out when the fluid pressure is equal to the pressure inside the tube. And to be honest I'm not sure how to figure this out. I've tried with the different equations but the answer seems way out of line. Hope someone can help me and at the same time try and tell me why it is the way it is since I really want to learn. It's probably an easy question but to me it's difficult.
     
  2. jcsd
  3. Jan 19, 2014 #2

    phinds

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    Your "attempt at a solution" is nothing but a statement that you don't know how. That's not acceptable on this forum. Show one of your attempts and explain why you think it should work but doesn't and we can go from there.
     
  4. Jan 19, 2014 #3
    Okey, I'll try but it probably won't make any sence.

    Fluid pressure: 483 kPa
    Air pressure: 102 kPa
    Inside the tube there is a pressure of 102 kPa, outside there is a pressure of 483 kPa. The water needs to compress the pressure inside the tube by: 483 - 102 = 380 kPa.
    This equals a height off:
    h = Pressure / (9,81 * density)
    h = 37,8 meters
     
  5. Jan 19, 2014 #4

    gneill

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    Pressure changes with depth in the fluid. The air inside the tube will compress until the its pressure matches the water pressure at the depth of the air/fluid boundary inside the tube. Here's a diagram to help you think about it:

    attachment.php?attachmentid=65816&stc=1&d=1390148716.jpg
     

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  6. Jan 19, 2014 #5

    haruspex

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    The tube is 48m long, not 40m, so the closed end is at the surface.
    Other than that, nice diagram :approve:
     
  7. Jan 19, 2014 #6

    gneill

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    Huh. Could've sworn I saw 40m for the length :redface: Ah well, the diagram should still serve to identify the relevant physical concepts. Thanks haruspex.
     
  8. Jan 19, 2014 #7
    Start thinking about using the ideal gas law for the air trapped in the tube. If the new length of the trapped air is L, and it started off with a length of 48 m, what is the new pressure? What is the water pressure at the bottom of the tube? What is the water pressure at the interface with the air if the height of the column of water in the tube is 48 - L?

    Chet
     
  9. Jan 20, 2014 #8
    Let me see if I get this right.
    p * V = p * V (Constant)
    But since the tube doesn't expand in width I only have to consider the height.
    So equation will be:
    p * h = p * h
    The pressure inside the tube before it gets filled with water is 102 kPa at 48 meters.
    The water pressure at the bottom of the tube is: 48 * 1.025 * 9.81 = 482.652 kPa
    The water will rise: 102 * 48 / 482.652 = 10,144 meters
    The water pressure now is: (48 - 10.144) * 1.025 * 9.81 = 380.652 kPa
    Am I on the right track here? Sorry about the bad english...
     
    Last edited: Jan 20, 2014
  10. Jan 20, 2014 #9

    SteamKing

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    Be careful and check your pressure units. A pressure of 380,000 kPa would damage, if not destroy, the tube.
     
  11. Jan 20, 2014 #10

    gneill

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    In many countries, a comma is the decimal point.
     
  12. Jan 20, 2014 #11
    Sorry about that! Where I'm from it means 380 (decimal) 000 kPa.
    The right way is 380.000 or?
     
  13. Jan 20, 2014 #12

    haruspex

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    Not really, though I suppose you could iterate that way.
    The trouble is that, as gneill showed in the diagram, the final pressure in the trapped air is equal to the water pressure at its new base. You computed the pressure at depth 48m, which will be too much.
    Let the final length of the trapped air column be h. (This will work out a little more simply then using x = 48-h.) What is the water pressure at at depth? What will the air pressure be having been compressed from length 48m to length h?
     
  14. Jan 20, 2014 #13
    Good start. But, the water won't rise 10.14 meters because the air in the tube is compressed, and it is pressing down with an absolute pressure of more than 102 kPa. If L is the length of the air column in the tube, the absolute pressure of the air is 102*48/L. The absolute pressure at the bottom of the tube is 102 + 483. The column of water inside the tube above the base is 48-L, so the pressure at the air interface in the column is 102 + 483-(48-L)*1.025*9.81 = 102 +L*1.025*9.81. This must match the absolute pressure of the compressed air at the interface.
     
  15. Jan 22, 2014 #14
    I'm completely toasted and the assignment is due tomorrow but I gave it anothere shot.
    I tried and I tried and I tried what you guys said but since I'm not English it's hard for me to understand everything.

    I came up with approx. 8 meters at my last try. It's probably wrong though.. And I can't say I've learned how this works when I can't tell if the answer is right or wrong..

    Thank you all for the help. I really appreciate it! Sorry if I've ''wasted'' your time.. That was certanly not my intention. I went over and over again through the help you guys gave you and when I finally thought I'd figured it out, something popped up saying ''This can't be right''..

    My teacher will probably tell me how to do it since I'm the only one who really gave this assignment a shot. And if he does, I will post it here :)

    Thanks all, again!
     
  16. Jan 22, 2014 #15

    haruspex

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    That is not close.
    Sorry if you could not understand what I posted. Let me try again:
    Let the length of the trapped air column be h. That is, if the answer to the question is x then h = 48-x.
    What is the pressure at a depth of h in the seawater?
    The trapped air was originally length 48m and at atmospheric pressure. When it has been compressed to length h, what will the pressure be?
    Please post your working, not just a numerical answer.
     
  17. Jan 23, 2014 #16
    This is what my teachers says:

    p*L=(Pt+Pa)*L

    102*48=583*L

    L = 102*48 / 583

    L = 8,4 meters

    48 - 8,4 = Approx. 40 meters is the length the water will rise

    Because: The air pressure at 8,4 meters = 102*48 / 8.4 = 583 (Which is the same pressure as the water pressure by the opening in the tube)

    What do you guys think?
     
  18. Jan 23, 2014 #17
    With all due respect to your teacher, the air pressure in the bubble is not equal to the pressure of the water at the opening in the tube. It is equal to the pressure of the water at the top of the water column in the tube (obviously, since the pressure is continuous at the interface). I get L = 17.74 M, so the length of the water column is about 30 M. I get a pressure in the air bubble of about 276 kPa. I obtained these results following the approach recommended by haruspex and myself.

    Chet
     
  19. Jan 23, 2014 #18

    haruspex

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    I agree. Olara, you can check this by thinking about the pressure at the base of the tube. This will be the sum of the pressure of the air in the tube plus the pressure due to the weight of water in the tube. This must equal the pressure either side, i.e. at the depth of the base of the tube but outside the tube.
     
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