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Homework Help: Bernoulli's stuff

  1. Feb 8, 2005 #1


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    Ok this stuff is confusing. Can someone give me a dumbified overview of this? I don't get when/how to use the equation. There are so many different forms and variations of pressure, bernoulli's equations it gets really confusing. I don't know what to equate what to what... ugh i don't know how to explain. I looked it up in wolfram and wikipedia and I read through the section numerous times and i still do not get it.... :mad:

    http://k.ebzlo.com/prob.htm [Broken] is an example of some problem i don't know how to solve. It shows I got #1 right but i don't know how to get it, i just copied it from the textbook.

    help is greatly appreciated.
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Feb 8, 2005 #2
    If you consider a current flow starting from the liquid surface through the straw, you must have that the sum of the static, dynamic and positional pressures must be the same along the flow. At the liquid surface, in the glass, you have
    [tex]p_0+\rho g h_0[/tex]
    considering the lower end of the straw as zero gravitational energy level. The dynamic pressure can be neglected here because the initial velocity [tex]v_{surface}=0[/tex].

    At the lower straw end you have the total pressure
    [tex]p_0+\frac{1}{2}\rho v^2[/tex].

    These two terms must be equal (Bernoulli's law) and v=..........
    Last edited: Feb 8, 2005
  4. Feb 8, 2005 #3


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    Bernoulli says that there are 3 terms that can add up to give you the total pressure (energy) of the fluid at a given point. That sum should be the same no matter where in your system you are looking. Those 3 items are:

    1) [tex]P_{o}[/tex] The fluid's static pressure (pressure created by molecules just bumping into each other)

    2) [tex]\frac{1}{2}\rho v^2[/tex] The fluid's dynamic pressure (pressure created by the fluid moving)

    3) [tex]\gamma z[/tex]The potential or positional pressure (similar to potential energy, it's pressure due to height above a selected level)

    You can use this to calculate pressures, velocities, etc... at other points in a a system if you know that data in another part.

    Does that help?
  5. Feb 8, 2005 #4


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    how do you know where to set height=0?
    why is the dynamic pressure in the cup 0? there's fluid going into the straw isn't there?
    why doesn't the lower end of the straw have any positional pressure in it? isn't it falling down so there should be force pulling the fluid down the straw? even both sides?

    thanks for the explaination thogh.
  6. Feb 8, 2005 #5
    1. you can set the zero gravitational energy level where you want (because in uniform gravitational field you deal with [tex]\Delta E_p[/tex] and not with the absolute value of [tex]E_p[/tex])
    2. Because of [tex]S_{glass}\cdot v_1=S_{straw}\cdot v_2[/tex] when [tex]S_{glass}\gg S_{straw}[/tex] you have [tex]v_1\ll v_2[/tex].
  7. Feb 8, 2005 #6


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    ohhh i see.. thanks. that made sense.

    can you help me with this one? I got the answer to this question already but I don't really get it.
    http://k.ebzlo.com/prob2.htm [Broken]

    Can I get an explaination of the summary below part A's answer? the part about how fluid in part A is not pushed up by kenetic energy.. etc. Why doesn't the pressure involve the dynamic pressure of the flow beneath both tubes?
    Last edited by a moderator: May 1, 2017
  8. Feb 8, 2005 #7
    Now you must be able to solve this problem yourself.....the sum of the three pressures must be the same in all the points of each streamline of the flow. If you chose the horizontal central streamline, you have the same positional pressure but different dynamic ([tex]\frac{1}{2}\rho v_1^2[/tex] and [tex]\frac{1}{2}\rho v_2^2[/tex]) and static pressures ([tex]p_0+\rho g h_1[/tex] and [tex]p_0+\rho g h_2[/tex]).

    Last edited: Feb 8, 2005
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