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Bernoulli's Equation & Fluid Dynamics LAB

  1. Dec 6, 2009 #1
    Before I describe the problem, I would like to apologize for my lack of comprehension
    of this subject. It is the first semester of Physics, but the end, and has become quite
    complicated.

    1. The problem statement, all variables and given/known data

    There is a 2.0 L soda bottle used, with a hole a certain height (hhole) from
    the bottom (and a small mini-hose to assist in the motion of the exiting of the fluid from the
    bottle). The bottle is aligned with the opening of a sink, and a hose is attached to the top
    so as to fill the bottle to a certain height and remain at that height constantly (h1).
    Then the distance the water travels out the hole is measured and recorded as a value
    of x1.
    Subsequently, this is done one more time with the water level at a new higher height (h2) and a new value of x2.

    Okay, hopefully that makes sense.

    Next, we are given two equations that are used to derive some other new equations
    meant to discover the velocity (I'm uncertain if I should include the messy method of
    proving these equations weren't pulled out of my ear, or just cut to the chase).

    THEN... we are to solve for velocity.
    Then we are asked to comment on our results, giving probable reasons for difference
    in values obtained from equations 1 & 2.

    2. Relevant equations

    Now for the equations.

    Equation 1:

    Point 1_____Point 2_____Point 3
    ___P0___= P0 + [tex]\rho[/tex]gh1 = P0+([tex]\rho[/tex]vH2O2)/2

    OR

    [tex]\rho[/tex]vH2O2 = [tex]\rho[/tex]gh1 ; vH2O = [tex]\sqrt{(2gh[sub1])}[/tex]
    ___2

    Equation 1: vH2O = [tex]\sqrt{(2gh[sub1])}[/tex]

    Equation 2:

    h2 = ½ gt2 OR t = [tex]\sqrt{[2h(sub2)]/g}[/tex] ; X = vH2Ot ; OR t = x/vH2O = [tex]\sqrt{(2gh[sub1])}[/tex]

    THEN

    vH2O = X*[tex]\sqrt{g/(2h(sub2))}[/tex] ; vH2O = 2.21 * X/[tex]\sqrt{h(sub2)}[/tex]

    Equation 2: vH2O = 2.21 * X/[tex]\sqrt{h(sub2)}[/tex]


    3. The attempt at a solution

    hhole = h2 = 6.9 cm
    h1 = 8.4 cm
    x1 = 8.6 cm
    v1H2O=________?

    hhole = h2 = 6.9 cm
    h1 = 11.1 cm
    x2 = 9.8 cm
    v2H2O = _______?

    Using Equation 1:

    v1H2O = [tex]\sqrt{2gh(sub1)}[/tex] = [tex]\sqrt{(2)(9.8)(0.084m)]}[/tex] = 1.6464 m/s

    v2H2O = [tex]\sqrt{2gh(sub2)}[/tex] = [tex]\sqrt{(2)(9.8)(0.11m)]}[/tex] = 2.156 m/s


    Using Equation 2:

    v1H2O = (2.21) (X/[tex]\sqrt{h(sub2)}[/tex] = [(2.21)(0.086m)]/[tex]\sqrt{0.084m}[/tex] = 0.655769301 m/s

    v2H2O = (2.21) (X/[tex]\sqrt{h(sub2)}[/tex] = [(2.21)(0.098m)/[tex]\sqrt{0.11m}[/tex] = 0.65302 m/s

    So to discuss these differences and probably reasons, I came to the conclusion
    that the 1st equation is lacking the value of X. I purport that because we normally
    consider volume (not cross-sectional area) when we are dealing with pressure,
    and also because the X value is not taken into consideration within the first equation,
    that perhaps it is wrong for those reasons.

    However, the truth is, I don't even know what I am talking about, and am simply
    left in a complete state of perplexity. Any and all help is appreciated.

    P.S.: I have even asked my class mate and my father for help, both of whom
    are equally confused.
    P.P.S.: My roommate mentioned something along the lines of the potential for
    the two Bernoulli equations in question to be from two different relatives from
    the Bernoulli family (the roommate has read many a Mathematician's biography,
    which is how they know about the relations)... I believe this to be nonsense. ;)
     
  2. jcsd
  3. Dec 7, 2009 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    What does Bernouilli's equation predict about the value of X as a function of H? That is the question you have to answer.

    Since the hole is at the same height above the bottom of the bottle and bottom of the sink, the distance of fall of the exiting water is constant for all H (height of the water in the bottle above the hole). This means that the time of fall is the same. So X is proportional to the horizontal speed of the exiting water: [tex]X \propto v_{H_2O}[/tex]

    What does Bernouilli say the relationship is between H and v? So what is the relationship that you should find between H and X? Does your data support that relationship?

    AM
     
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