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Bernoulli's equation

  1. Jan 11, 2012 #1

    Femme_physics

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    Again-- I'm not entirely sure if this goes to physics or engineering, but the word thing that comes in mind when I hear Bernoulli equation is PHYSICS, so I decided to post it here...

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

    So I have this hydraulic system

    http://img689.imageshack.us/img689/3596/clipboard01rvt.jpg [Broken]


    With the figures

    H1 = Height of suctioning = 3m
    H2 = Height of delivery = 25m
    D1 = diameter of the suctioning pipe = 150mm
    D2 = diameter of the delivery pipe = 100mm
    Q = 0.02 [m^3/s]
    y1 = pressure lost at the suctioning line
    y2 = pressure lost at the delivery line
    Energy conversation efficency of the pump = 75%

    Find the pressure at the pump's entrance (p1).

    2. Relevant equations
    In the solution


    3. The attempt at a solution

    So I build the equation:

    http://img196.imageshack.us/img196/2145/ber1j.jpg [Broken]

    http://img21.imageshack.us/img21/9740/ber2h.jpg [Broken]

    HOWEVER, the result in my manual is different. It's p1 = -0.05064x10^5 [Pa]

    I don't get how could we have gotten such hugely different results!
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 11, 2012 #2
    What is the density of the fluid that is pumped? For water I get a pressure reduction of about 30,000 Pa from potential energy term and a pressure reduction of about 500 Pa from the velocity term.
     
  4. Jan 12, 2012 #3

    Femme_physics

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    The fluid is water, the density of water is 1 g/cu.cm. We work in Newton/Meter so it's 10 N/m.

    How does it change Bernoulli's equation?
     
  5. Jan 12, 2012 #4
    From

    http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html

    Bernoulli's equation has the density in it and I don't think you mentioned water?

    Good luck!
     
  6. Jan 13, 2012 #5

    I like Serena

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    Hi Fp! :smile:

    I can't match your relevant equation with Bernoulli's equation.
    Perhaps you can explain what the symbols in your equation stand for?
     
  7. Jan 13, 2012 #6

    Femme_physics

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    Hsa = static pressure at a
    Hva = kinematic pressure at a
    Za = height difference from zero point

    Hsb = static pressure at b
    Hvb = kinematic pressure at b
    Zb = height different from zero point
    Sigma Ya-b = pressure losses throughout point a to b
     
  8. Jan 13, 2012 #7

    I like Serena

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    Do you have formulas for static pressure Hs and also for kinematic pressure Hv then?

    And while we're at it, also for Z and Sigma Y?

    I ask, because they are not simply pressure, speed, and height.
     
  9. Jan 13, 2012 #8

    Femme_physics

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    Oh, they are.

    Static pressure is just static pressure.

    Kinematic pressure is V^2/2g

    The height is just height, in our case a difference of 3.
     
  10. Jan 13, 2012 #9

    I like Serena

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    Okay...

    But that kinematic pressure is not what you calculated in your scan...


    To match with your kinematic pressure and your height, static pressure has to be:
    $$H_s = {p \over \rho g}$$
    where ##p## is just pressure, ##\rho## is the density of water (1000 kg/m3), and ##g## is the regular acceleration of gravity.




    Where does the pressure loss come in?
    In your scan you mention that it is 2, but where does that come from?
     
    Last edited: Jan 13, 2012
  11. Jan 21, 2012 #10

    Femme_physics

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    ILS -- I had consulted classmates and solve it and I will post the solution later.

    Our static pressure is just a value, we don't put it in a fraction. Pressure lost comes in due to the friction in pipe, length of pipe, diameter of pipe, and non-uniformity in the pipe (90 degrees change, valves in between... etc)....
     
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