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Homework Help: Bernoulli's Principle homework

  1. Sep 8, 2005 #1
    This isn't really a homework problem, but my physics teacher was going over a test review and he made up some numbers on the spot for a problem involving Bernoulli's equation and a building. The specifics were,

    A water pipeline with the water under a pressure of 5 atm is being pumped into a building through a pipe of radius .2m with a velocity of 5 m/s. The pipe then proceeds to climb the 20m tall building and narrow to a radius of .02m. What is the pressure and velocity of the water at the top of the building?

    Now it makes sense to me to take the ratio of the areas, which ends up being 100:1 and applying it to the equation of continuity, giving me a velocity of 500 m/s at the top. When I plus this into Bernoulli's equation,

    506500 + .5(1000)(5)^2 + 0(9.8)(1000) =
    P2 + (.5)(1000)(500)^2 + (20)(9.8)(1000)

    And I end up with the wonderful pressure of around -1.2 x 10^8 Pascals, which obviously can't be the real pressure for the water. I've been racking my brain trying to figure out why exactly this situation is impossible, assuming that water is an ideal nonviscous incompressible fluid.

    So I can't figure out why in the real world, the pressure should turn out to be some negative number, and it seems like this would happen for any large differences in velocity in Bernoulli's equation. I've been looking around for an explanation of this, as our AP Physics class spent half an hour trying to reason out logically why this would happen, and our Physics teacher couldn't explain it either.

    Any help would be greatly appreciated.

    -J Stoncius
  2. jcsd
  3. Sep 9, 2005 #2


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    The first thing that came to mind when you found a negative pressure was that the water couldnt get that high under those conditions, and after a few calculations, I might have been onto something.

    Assuming you filled the entire 20 meter pipe with a radius of .2 meters with water, that comes out to be 2.52E6 mL of water. That water is heavy, it has a mass of 2520 kg! So it weighs 24621.2 Newtons.
    This weight is distributed on the new water trying to come up the pipe, the water on top is exerting a pressure downward on the water on bottom.
    To find the downward pressure, P = F/ A,
    24721.2 Newtons / .126 meters squared = 196.2 kPa = 19.4 atmospheres of pressure downward. Remember, the water at the bottom is only being pushed up at 5 atmosphers.
    So using my logic, the water is too heavy to reach all the way up the pipe to squrit out the top if it is only being pumped up at 5 atm.
  4. Sep 9, 2005 #3


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    That is probably correct after a quick run through of your equation. Like Mrjeffy stated, that negative pressure is needed to obtain that flowrate under those conditions. Since the pressures are stated in gauge pressures, the negative sign indicates that you would need a slight vacuum source to help pull the water to those conditions at the other end, i.e. atmospheric pressure minus your answer.
  5. Sep 9, 2005 #4
    101.3kPa = 1 atm...not 10.13. The downward pressure of the water is only about 2 atm. (Water increases 1 atm in pressure about every 10.34m in depth). So it's not that there isn't enough pressure, the pressure actually runs around 3atm gauge...

    So I guess we're back to the drawing board...
    -J Stoncius
  6. Sep 9, 2005 #5


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    Blast, I misplaced the decimal place and just kept going without catching it.
  7. Sep 9, 2005 #6
    I guess the thing that bothers me is that there seems to be some sort of inherent problem in solving Bernoulli's equation for pressure using velocity. Even given that the fluids are at the same height, if there is some sort of large velocity increase, it seems that the pressure could end up being negative.

    Solving for P2 in general with no height change seems to give

    P2 = P1 + .5pv1^2 - .5 pv2^2
    Or factoring,
    P2 = P1 + .5p(v1^2 - v2^2)

    Which says to me that if v2 is somewhat larger than v1, we have a problem.

    Also, with regards to negative pressure, a negative pressure has to be totally bogus. A vacuum has a pressure of 0Pa. I can see how a negative pressure could possibly be a negative force applied to an area, but it doesn't make much sense to me when applied, since I can't think of one instance in reality where one would use this negative pressure. Most pressure-related forces are caused by a difference in pressure (usually air and a vacuum), so I can't think of anything that creates a negative force...

    I'm trying to picture this problem in my mind with the pipe going straight up and having an equal radius all throughout. If I cut off the top of the pipe, water spews a nice 31.02m into the air (assuming the 5 atm on the pipe is absolute pressure). Measuring the gauge pressure at the point at which the water comes out, I get 3 atms or there about. Now, if I replace that section of pipe that I cut with one that narrows down to a smaller radius, one that is a tenth of the larger one, I measure gauge pressure at that point and I end up with -1230 atms. That's what seems to be happening, and from a logical viewpoint, that shouldn't happen...

    I'm still really confused..
    -J Stoncius
  8. Sep 10, 2005 #7


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    A good problem, but I have a few confusions.

    1. Is it required to find the pressure at the other opening of the tube or it is just atmospheric pressure?

    2. Whether the velocity of pumping water is independent of the pressure difference, rather I should say that is it possible to push water at our desired velocity?
  9. Sep 10, 2005 #8
    The question only asked for the atmospheric pressure at the top of the building, nothing was said about the other end (I assume it's hooked up to a water tower or something).

    Now, as far as your second question goes, I don't really know. If the fluid is ideal, the velocity of the water should be independent of the pressure since for the water to move at constant velocity, it shouldn't take any force through Newton's first law...

    So for number two, I'm going to assume yes, that it is possible to push water with our desired velocity of 5 m/s in the low pipe...and by conservation of mass, it has to be 500m/s in the second pipe..

    I guess the problem could be that I'm idealizing the equation too much so that it starts to not really work in the real world, but in theory, the problem should work out to something reasonable. I don't really need an answer to this problem, I actually just need more or less an explanation of why Bernoulli's equation seems to fail in my case...

    -J Stoncius
  10. Sep 10, 2005 #9


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    I think taht the velocity and the rate of flow of the water through any tube depends on the pressure difference. In any case with a given pressure difference across a given tube we can't change the rate of flow. Say, with fully open tape if you want more rate of flow of water, how will you get it?
  11. Sep 12, 2005 #10
    The Bernoulli's principle does work in all most all real world cases but the numericals given in non engineering text books are ridiculous, sometimes. Only correction to your above method is that you should use absolute static pressure head. 506500 on LHS is actually 607968 Pa.

    In many cases, the final gauge pressure goes below 1 bar if the design is improper. Th This actually happens when the velocity increases at the expense of static pressure head or elevational head. That is why it is a good practice to draw hydraulic gradient line for a fluid system to avoid ingression of air into the system.

    In your case, there are many solutions. Actually, such a system is never possible (as the final pressure is below 1 bar gauge).

    There are few things which may be beyond the scope of your cirriculum, like frictional losses and pump performance etc. Any flowing fluid first tries to overcome the resistance in the piping system. To match this, the flow automatically gets reduced. So, the velocities will be much lower than what you anticipate.

    You can never work, in real life, with any initial pressure (like the arbitrary 5atm). The inlet pressure should be sufficient enough to create the water flow upto the top most point.
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