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Bernoulli's Principle Question

  1. Mar 2, 2006 #1

    mrjeffy321

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    This isn’t really homework, but it might as well be with all the calculations involved.
    I am trying to figure out the difference in pressures inside a pipe carrying a fluid when the pipe goes from a ¼” diameter to 1/8” diameter when the constant flow rate is 9 Liter per minute. I got an answer, but it seems way to large to be reasonable.

    I am using these two equations,
    A_1 * V_2 = A_2 * V_2
    P_1 + ½ (V_1)^2 = P_2 + ½ (V_2)^2

    If I have some fluid flowing through a system of pipes at a constant rate of 9 Liters per minute, this would be equal to 1.5 E-4 m^3 / s.
    The starting diameter of the pipe is ¼ inches (radius of 3.2 E-4 m^2) and the smallest diameter is 1/8 inches (radius of 1.6 E-4 m^3).

    I know that A_1 * V_1 = 1.5 E-4 m^2/s and that A_1 = 3.17 E-7 m^2,
    So that means that V_1 must equal 473.2 m/s. This seems very fast, but OK, I will accept it. When the pipe diameter is halved, the velocity of the fluid is quadrupled. So V_2 must equal 1894 m/s. This seems extremely high (over 5 times the speed of sound), but I keep going.

    Now that I know the two velocities of the fluid in both parts of the pipe, I want to find the pressure difference.

    I know,
    P_1 + ½ (V_1)^2 = P_2 + ½ (V_2)^2, plugging in gives me,
    P_1 + ½ (473.2)^2 + P_2 + ½ (1894)^2, solving for the difference in pressure,
    P_1 – P_2 = ½ (473.2)^2 - ½ (1894)^2 = 1.69 E6 N/m^2

    A pressure difference of 1.69 E6 N/m^2 = 1690 kPa = 16.7 atmospheres, this has got to be wrong, right? Where did I screw up?
     
  2. jcsd
  3. Mar 2, 2006 #2

    mrjeffy321

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    Oh, wait, I figured out the problem, I missplaced the decimal on the conversions between inches and meters on the pipe diameter, causing them to be a factor of 10 too small.

    When I rework everything, I get a pressure difference of 163.8 Pa which seems much more reasonable.
     
  4. Mar 2, 2006 #3
    Your work looks correct (I admit I didn't check the numbers.) Consider that your original flow rate of 9 L/min is pretty sizable given the diameter of your pipe, so I would imagine the numbers would be rather large. And consider, if you are pumping liquid at 9 L/min though a pipe 1/8 of an inch diameter at Mach 5, you're not exactly going to be merely blowing through a straw to do it! :rofl:

    -Dan
     
  5. Mar 2, 2006 #4

    mrjeffy321

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    Assuming the 163.8 Pa pressure difference is correct, the I can take the calculations a bit further.

    If the small pipe opens up into the atmosphere, then we know the pressure of the fluid in the small pipe (~1 atm). Since we know the pressure difference, then we also know the pressure in the large pipe (1 atm + 163.8 Pa). if the large pipe is being fed directly by some other very large pipe like in the figure below,

    _________
    ..............|
    ..............|
    ..............|
    ..............|
    ..............|
    ..............---------
    ...........................|__________
    ...........................|-----------
    ..............----------
    ..............|
    ..............|
    ..............|
    ..............|
    ..............|
    -----------
    Then I could figure out the pressure inside the original, very large pipe.

    But what if this very large pipe is actually a PVC pipe with a small hole cut in the side and the big pipe (from the orignal question) is fitted into the side, what would be the very large pipe's diameter? Would it be the height of the pipe or would it be the actual pipe diameter?

    ____
    |.....|
    |.....|----
    |............=======
    |......____
    |.....|
    |.....|
    |.....|
    |.....|
    |.....|
    |.....|
    |.....|
    |.....|
    -----
     
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