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Bernouli's equation Problem

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data

    On a bet, you try to remove water from a glass by blowing across the top of a vertical straw immersed in the water

    What is the minimum speed you must give the air at the top of the straw to draw water upward through a height of 2.0cm?

    2. Relevant equations

    Bernoulli's eq. p1+1/2pv1^2+ pgh1 = p2+1/2pv22 + pgh2

    3. The attempt at a solution

    Tried to rearrange for velocity, I guess I did it wrong.

    v= √2gh*rho2/rho1

    ( The square root should be a whole root, of like all the terms, not sure how do fix that part)

    Did i rerarrange it wrong?

    Umm . I got 17.5, and 49 (somehow) , maybe im pllugging things in right, but im not sure what the right answer is , 17.5 isnt for sure.
  2. jcsd
  3. Sep 29, 2012 #2
    You need to consider two equations. One for water, another for air in the straw. The water equation has no speed, just the height difference. The air equation has no speed in the straw, and there is no height difference.
  4. Sep 29, 2012 #3
    So I have to have two seperate bernoulli equations? I still don't get it.
  5. Sep 29, 2012 #4
    You have two fluids. So you have two equations.
  6. Sep 29, 2012 #5
    Solving for the velocity of both then subtracting..?
  7. Sep 29, 2012 #6
    p1+1/2pv1^2+ pgh1 = p2+1/2pv22 + pgh2

    I dont get what im using both fluids for.. I dont have velocities so I dont use them right?

    I dont get how to encorporte athis into the eq? so this means i have two velocities but i am not given any variable for either so how i can solve for two variables..

    p1+1/2pv1^2+ pgh1 = p2+1/2pv22 + pgh2

    I dont see any relation except if i get rid of velocities and just do 1/2p + pgh1 = 1/2p + ph2 . But i have all of those values.. Unless i solve for pressure by doing P= p(atmosphereic) + pgh, then i can get the pressure of the water in the straw
  8. Sep 29, 2012 #7
    Start with the equation for the air in the straw. What is the velocity just below the edge?
  9. Sep 29, 2012 #8
    It is 17.5 i think. I applied this formula v= √2gh*rho2/rho1
  10. Sep 29, 2012 #9
    How can the velocity in the straw be non-zero?
  11. Sep 29, 2012 #10
    air moves thru a straw wouldnt that make it non zero
  12. Sep 29, 2012 #11
    If air moves through the straw then the straw can't lift any water - it will have bubbles of air come through the water instead. You are supposed to move the air parallel to the edge.
  13. Sep 29, 2012 #12
    ok so V in the straw is 0..now what? :/ I really dont get it and this question is taking like hours..
  14. Sep 29, 2012 #13
    Now there is one equation for air. On one side, air is just above the edge. On the other it is just below. What is the difference in pressure?
  15. Sep 29, 2012 #14
    So I woud use the same formula for the bottom, and calculate the top.. and divide the bottom by the top? Or subtract the two values?
  16. Sep 29, 2012 #15
    I do not understand what you are saying.
  17. Sep 29, 2012 #16
    So I have one equation for air in the straw, Now i want another equation for water. Do I divide the eq for water by te one of straw? or plug in the values on the LS for air and the RS for water and solve for v1?
  18. Sep 29, 2012 #17
    You are going too fast. Answer my question in #13 first.
  19. Sep 29, 2012 #18


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    Homework Helper

    Blowing the air implies creation of a flow by increasing the static pressure of the blown air, which ideally remains higher than ambient until it's velocity goes to zero. The problem would be better stated that a vacuum is sucking air across the top of a straw, reducing the air's pressure and increasing it's speed.

    Another detail being ignored here is that a vortice is created by air flowing over the exposed end of a straw, resulting in a futher reduction in pressure at the end of the straw. This could be elminated by having the end of the straw flush mounted to a flat plate perpendicular to the straw, so that the plate and straw act as a static port, where the pressure at the opening is the same as the static pressure of the air flowing across the port. I'm pointing this out in case someone decides to actually do an experiment with air blowing over the open end of a straw.
    Last edited: Sep 30, 2012
  20. Sep 29, 2012 #19
    Calculate pressure needed to lift water by 2 cm. Air pressure must drop by that value due to its velocity.
  21. Sep 29, 2012 #20
    really? Can someone help me solve this. its been like 7 hours.
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