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Bernoulli's equation and a fountain

  1. Nov 28, 2009 #1
    A fountain shoots a vertical jet of water at a rate of 0.33 L/s to a height of 1.6m.
    What is the initial speed of the jet and what is the radius of the hole out of which the jet passes?
    What pressure must the pump of the fountain supply? (Assuming that it sits just below the emerging jet).
    At height 1.2 m., what is the speed of the jet, and what is the radius of the column of water? Ignore effects of turbolence and assume that the jet does not break up.

    -My main difficulty with this problem comes when I apply Bernoulli equation to answer the second question and I get the clearly impossible result that the pump pressure is 0.

    1) Should I consider atmosferic pressure?
    2) Is the final pressure 0?
    3) Can I calculate the initial speed by simply equating initial kinetic energy and final potential energy?

    I am missing something and I don't know what it is. Can someone help me?
    Thanks for paying attention to this thread.
  2. jcsd
  3. Nov 28, 2009 #2


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    Hi eversor,

    Can you show how you got this result?

    Remember that the pump takes water that is not moving and "gives" that water the initial speed of the jet.

  4. Nov 29, 2009 #3
    Hi Alphysicist,

    Thanks a lot for your precious suggestion.
    If I interpret it well,

    Pressure pump = patm + (density*v^2)/2 = patm + density*g*h

    where for both patm is approximately equal to 1 atm. What if h were very large? would
    this equation still hold? I know that the atmosferic pressure decreases with height but I find that (density*v^2)/2 is always equal to density*g*h where v is the initial speed. What do I do wrong? This thing is driving me crazy :)) Thanks for your help
  5. Nov 29, 2009 #4
    Reviewing what I wrote, I am thinking that I should include patm in the first of the three members. That would explain everything because all the patm cancel out. Another question.. Pressure is a scalar but force is a vector.. Why? Also, Why is pressure always represented as a vector on physics books.. and the last thing.. does pressure have orientation? For example, in the problem above it would be natural to think that patm has an opposite orientation to pressure pump. However, this would imply that the first term of the equation above looks like Pressure pump - patm , which is probably wrong because patm does not cancel out.
  6. Nov 30, 2009 #5


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    Your equation looks right to me, if I'm imagining the setup correctly. What do you get for the pump pressure?

    I think your equation is correct as it is written, and that you do not want the patm to cancel.

    I'm not sure what you mean by pressure always represented as a vector in the books, but if you are talking about diagrams for Bernoulli equations, then what we are interested in is the effect of pressure on the streamline at two points (the two points that we use in Bernoulli's equation). Since the streamline has one velocity direction at that point, to see the effect of the pressure you would think about its effect along that direction.
  7. Nov 30, 2009 #6
    If my equation is correct as it is written, how do you explain that patm decreases with height? I mean if (density*v^2)/2 = density*g*h because v^2= 2*g*h the equation above for large values of h is clearly wrong!!

    By pressure represented as a vector in books I mean that it is shown as an arrow with direction and orientation in explanatory drawings.
  8. Nov 30, 2009 #7


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    We are assuming that the air pressure is not changing over the height that is given in the problem.

    If you could not approximate the pressures as equal, then yes, you would not set up Bernoulli's equation like that. Most equations are only accurate within a certain range.

    For example, notice that you are doing the same thing by using the term (density*g*h); by using that term you are assuming that the gravitational force is constant over the height the jet travels. We know that the gravity force decreases as height increases, but over the distance we care about that effect is negligible. If it was not negligible, then the equation would have to be changed.

    The arrows show the effect of the pressure at one particular point (that is, it shows the direction of the force that the pressure creates at that point). But at other points, the pressure might push in a totally different direction. So pressure does not have one specific direction, but it does create a specific force at one point.
  9. Dec 1, 2009 #8
    Does Bernoulli equation work only in proximity of Earth's surface?
  10. Dec 1, 2009 #9
    What is an acceptable range for which Bernoulli's equation is valid?
  11. Dec 2, 2009 #10


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    The form of Bernoulli's equation that you are using:

    [tex]P+\frac{1}{2}\rho v^2+\rho g h=\mbox{constant}[/tex]

    assumes that the gravitational force is mg, so that form is only valid near the earth's surface (where that assumption is true).

    You have to follow the derivation of Bernoulli's equation to see which assumptions are being made. (There are quite a few equations that are different forms of Bernoulli's equation or are analogous to Bernoulli's.) Off the top of my head, your book probably says to assume incompressible fluids, no viscosity, steady flow, near the earth's surface, etc.

    Those would all be approximations made for the derived equation. There are also approximations you make for the problem itself: that the atomspheric pressure and gravitational force is constant over the dimensions of the problem, that the difference in heights of the pump and nozzle is negligible, etc.
  12. Dec 3, 2009 #11
    Thanks a lot for your help!!
  13. Dec 4, 2009 #12


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    You're welcome, glad to help!
  14. Dec 13, 2009 #13
    I did an experiement to measure flow rate through a pipe, varying radius and pressure difference. I used Bernoulli's equation (v=√2gh) to compute the velocity of the fluid flow through the pipe, and multiplied the cross sectional area of the pipe (πr^2) by this flow velocity to compute volume flow, Q. I did this at each of 10 different pressures, for 4 different pipe sizes.

    I thought I could also use the Pouseuille equation (∆P*πr^4/8μL), to double-check the results, thinking that both results should be at least in the same ballpark.

    It didn't work, and I'm not sure why it didn't. My initial thought was an error in the viscosity constant. We figured that 1 Pa= .01207 cm H2O at 20 degrees celsius. This correction makes the units work, however, the values for Q are still off.

    Thanks for any thoughts you may have. Am I correct that the two equations should produce similar numbers for flow (if we do the math correctly)? What am I missing?
  15. Dec 13, 2009 #14


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    Was this an experiment where you have a container, fill it up with water, and punch a small hole somewhere near the bottom? If so, Poiseuille's equation doesn't apply. Take a look at its derivation: http://en.wikipedia.org/wiki/Poiseuille_equation#Derivation

    The derivation assumes a horizontal pipe of uniform cross-section and laminar flow. In your experiment, the "pipe" is vertical, the water isn't flowing through a pipe with a uniform cross-section (it's flowing through the hole, which is much smaller than the pipe), and the water flowing through the hole is almost certainly turbulent.
  16. Dec 13, 2009 #15
    That's exactly what we did... used a gallon milk jug with a hose attached, connected to hypodermic needles of various gauges. So, by using the Bernoulli equation, I guess we relied on the average flow velocity, rather than separately considering the more complex equations needed to deal with turbulent flow in the smaller needles. I guess my actual measurements can tell me how much this simplification causes my results to differ from the results predicted strictly by the Bernoulli equation?
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