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At what height will the flow stop?

  1. Nov 28, 2014 #1
    1. The problem statement, all variables and given/known data
    A large tank of water has a 'z-shaped' hose connected to it. The tank is sealed at the top and has compressed air between the water surface and the top. When the water height, h, has the value 3.50m, the absolute pressure p of the compressed air is 4.20x10^5 Pa. Assume that the air above the water expands at a constant temperature, and take the atmospheric pressure to be 1.00x10^5 Pa. a)What is the speed with which the water flows out of the hose when h=3.50m? b)As water flows out of the tank, h decreases. Calculate the speed of flow for h=3.00m and h=2.00m c)At what value of h does the flow stop?

    2. Relevant equations
    Bernoulli's eq.
    P1 + (1/2)ρ(v1)^2 + ρgh1 = P2 + (1/2)ρ(v2)^2 + ρg(h2)
    P1 + ρgh1 = P2 + (1/2)ρ(v2)^2 + ρg(h2)

    assuming viscosity is zero
    water is incompressible


    3. The attempt at a solution
    p=4.2*105 Pa
    p2=1*105 Pa
    and h2=1m
    h=3.5m
    a.) v=√(2((p-1.0*105)/ρ)+g(h-h2))=.....≈26.2m/s

    b.) p=(0.5p1)/(4m-h) (pV=constant as the temperature is constant) and now h=3m
    v=√(2((p-1.0*105)/ρ)+g(h-h2))=√(2((p=(0.5p1)/(4m-h)-1.0*105)/ρ)+g(h-h2))

    By substituting in new values for h I got 16.m/s and 5.44m/s both of which are right according to my book.
    However, the part c is confusing me a little. I have tried setting v=0 and solving for h but I didn't get the right answer which is h=1.78meters according to my book. I know somebody will ask me to show the calculations so I ended up having equation of second degree like gρh2-5gρ+(4gρ-0.5p1+p2)=0 I got h>4m or h<0 which make no sense at all. Any piece of advice will be appreciated.
     
    Last edited: Nov 28, 2014
  2. jcsd
  3. Nov 28, 2014 #2
    What is exact solution to these issues? I mean numbers. I have some discrepancy about yours solutions. I probably dont understand because I mean the speed of out flow depend on diameter of hose no? If you use h2 in Bernulli eq. it means speed in high h2, no in hose.
    Thank you.
     
    Last edited: Nov 28, 2014
  4. Nov 28, 2014 #3
  5. Nov 28, 2014 #4
    Here are my calculations for h when v2=0 I forgot to write the sub index 2. Am I missing something?
    2ppbtb5.jpg
     
    Last edited: Nov 28, 2014
  6. Nov 28, 2014 #5
    Ahaaaa :) Now it is bit clearer :) And the Z hose is 1m high? I mean that it is important fact, it arises new discrepancies, because if you have no pressure case (P1=0) than the h should stop at same high as high of Z hose. So if there is pressure is for me strange that result should be 1.78m (high than 1m). And what is exact solution for A) and B). I have to recalculate it and if I will have it same I will believe that I have good all.
     
    Last edited: Nov 28, 2014
  7. Nov 29, 2014 #6
    bump
     
  8. Nov 30, 2014 #7
    Anyone?
     
  9. Nov 30, 2014 #8
    First assume that the 1.78 is right and calculate the velocity the same way you did it in the previous cases. See if that give you a velocity of zero.

    Chet
     
  10. Nov 30, 2014 #9
    There is a typo in my first post, the right answer according to my book is h=1.74m It gives me ≈0.6m/s≠0 So is my formula for v wrong in part c?
     
    Last edited: Nov 30, 2014
  11. Nov 30, 2014 #10

    haruspex

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    Since we're not told the length of the pipe, I assume we don't need to take into account the impetus from water that's already moving.
    You don't really need to consider an equation for the velocity for this part. What will the pressure be at height 1m inside the tank when the flow stops?
     
  12. Nov 30, 2014 #11
    1atm
     
  13. Nov 30, 2014 #12
    According to my calculations, 1.74 m should satisfy your starting equation. So you must have an algebra or arithmetic error. (I told you to plug the answer into your starting equation and see if it satisfies it, but apparently, you had no value for doing that).

    Chet
     
  14. Nov 30, 2014 #13
    I did plug the answer(=1.74) into my equation of velocity and it didn't satisfy it. I will check my calculations.
     
  15. Nov 30, 2014 #14
    I got:
    [tex]\frac{(0.5)(4.2)}{4-h}+0.1(h-1)=1[/tex]
    This is satisfied by h = 1.74
     
  16. Nov 30, 2014 #15
    Thank you Chestermiller! I got h1=1.7363m
     
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