1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

FInding distance from top of diving bell to lake surface

Tags:
  1. May 27, 2015 #1
    1. The problem statement, all variables and given/known data
    A cylindrical diving bell with open bottom and closed top 12.0m high is lowered into a lake until water within the bell rises 8.0m from the bottom end. Determine the distance from the top of the bell to the surface of the lake.

    2. Relevant equations
    I actually solved this problem through trial and error. the fundamental issue I have is defining the ratio volumes and why does this work despite it seeming to make sense or does it?
    4Va/12 =8Vw/12; (Va=Volume of air, Vw=Volume of Water.)<<<Why does this work? Because it makes more sense to have 2Vw=Va


    3. The attempt at a solution
    PaVa=PwVw
    Plugging in "equivalency" in part 2;
    Pa2Vw=PwVw
    2Pa=Pw
    2(100kpa)=Pw
    Pw=200kpa

    Pw=pGh
    h=Pw/pG
    h=200kpa/(1000kg/m3)(9.8ms-2)
    h=20.6
    Height from bell surface to top of lake
    20.6 - 4-=16.6m

    Schaums College physics problem 16.37
     
    Last edited by a moderator: May 28, 2015
  2. jcsd
  3. May 28, 2015 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    A bit hard to follow without definitions of the variables (Pw? Whereabouts?).
    Clearly the air is at 3 atmospheres of pressure.
    Wouldn't it be more surprising if it worked but didn't make sense?
     
  4. May 28, 2015 #3
    Pw=pressure of water
    Pa=pressure of air
    p=density of water
    Va=volume air
    Vw=Volume water

    How is it at 3 atm?
     
  5. May 28, 2015 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, but pressure of water where, exactly?
    How has the air volume changed since leaving the surface?
     
  6. May 28, 2015 #5
    yes it makes NO sense is what i meant ton extent.
     
  7. May 28, 2015 #6
    In the tube.

    The air volume has decreased
     
  8. May 28, 2015 #7
    The tube is submeged open end down and acquires an 8m influx of water
     
  9. May 28, 2015 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The water pressure will not be the same everywhere in the bell.
    In what ratio has the air volume decreased?
     
  10. May 28, 2015 #9
    A third?

    The pressure changes with depth
     
  11. May 28, 2015 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Right. So what will that do to the air pressure? And how would you now define your Pw?
     
  12. May 28, 2015 #11
    Vw=2/3 Pressure of water decreases by 1/3
     
  13. May 28, 2015 #12

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Eh? The pressure of what water where decreases by a third? And why should it decrease at all? Water is virtually incompressible, and if you could reduce its volume by a third by compression the pressure would be something to be found inside a star.
    How do you normally calculate the pressure at a given depth in a liquid? Why should it be any different here?
     
  14. May 28, 2015 #13
  15. May 28, 2015 #14
    pressure of liquid is pgh
     
  16. May 28, 2015 #15
    So volume of air decreases by a third. Pw=pgh

    PaVa/3=pgh(Vw)
     
  17. May 28, 2015 #16

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Not by a third.
    Not sure what your equation is saying, but it looks wrong to me. Why should it be true? What principle are you applying?
     
  18. May 28, 2015 #17
    The one I applied was this 4Va/12 =8Vw/12; I don't know why it is right it may not be though.

    The tube is 12m long the water goes up 8m leaving 4m of empty air at the top. I figured 4/12 is air and 8/12 is water?

    LOL See I told u I was lost.
     
  19. May 28, 2015 #18
    Pressure of Water (Pw)= pgh
    So using Boyles Law

    PwVw=PaVa

    pgh(Vw)=PaVa?
     
  20. May 28, 2015 #19
    Hmm we can't apply boyles law really b/c like you said we cant compress water like a gas?

    I know that;

    Ptotal=Patm+ pgh
     
  21. May 28, 2015 #20

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That's better. But what would be a useful value for h in the present context, i.e. whereabouts would you like to know what the pressure in the water is?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: FInding distance from top of diving bell to lake surface
  1. Diving bell (Replies: 5)

  2. Diving Bell Problem? (Replies: 1)

Loading...