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Bernoullis equation

  1. Nov 30, 2016 #1
    1. The problem statement, all variables and given/known data
    The presure inside the pipe is 6.0 atm and the speed of the water flowing past the inlet valve is 3.00 m/. On a floor 5.0m above the inlet pipe, the single pipe has only a 2.5 cm diameter. Assume that water is an ideal fluid.

    C) If the smaller pipe ruptures, propelling water vertically upward as in the figure to what further height h will the water rise? neglect air resistance.

    Figure: http://imgur.com/42vlUCu
    2. Relevant equations
    Bernoullis equation
    velocity at top of figure=12m/s
    pressure at top=491400pascals

    3. The attempt at a solution
    .5(1000kg/m^3)(12m/s^2)=1000kg/m^3(9.81m/s^2)H+.5(1000kg/m^3)(12m/s^2)
    I get a weird H of 706.32
     
  2. jcsd
  3. Nov 30, 2016 #2
    pls send help :,(
     
  4. Nov 30, 2016 #3

    Delta²

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    Can you tell us how exactly (more analytically) you apply Bernoulli's equation for the water ejection, because what you wrote as attempt for solution doesn't make much sense, it simplifies to ##0=\rho gH##. What pressure and velocity do you take at the very top of water?
     
  5. Nov 30, 2016 #4
    At the very top the pressure should be 491400 pascals and the velocity should be 12 m/s. This was calculated during part a) and b) of the problem. A)What is the speed of the water in the smaller pipe? B)What is the pressure in the smaller pipe at the 5.0m height.
     
  6. Nov 30, 2016 #5
    At the 5.0m height the speed should still be 12 m/s since energy is conserved right?
     
  7. Nov 30, 2016 #6

    Delta²

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    I don't mean at the top pipe, I mean at the top of the ejected water (at height h).
     
  8. Nov 30, 2016 #7

    Delta²

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    This follows from continuity equation and that the fluid is non compressible, not from energy conservation.
     
  9. Nov 30, 2016 #8
    Oh sorry, I misread that. So i need to use bernoullis equation to find the pressure and speed at the top of the ejected water. My notes give me this equation when we talked about geysers so the professor derived this form bernoullis original equation. 23d.GIF . Im assuming to find the height could I just plug the velocity at the top of the pipe in here to find the height?
     
  10. Nov 30, 2016 #9
    Because the pressure should be the same at both places, the start of the ejection and the end of the ejection
     
  11. Nov 30, 2016 #10

    Delta²

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    Ok well use that equation you say your professor gave you , at post #8. I don't think you get H=706.32 from that equation.
     
  12. Nov 30, 2016 #11
    Think about the values of the pressure and velocity at the highest point the water rises? Would the pressure be the same as it was in the pipe?

    What would the velocity at that highest point be? Think about at ball, when you throw it up what is its velocity at the higest point it reaches?
     
  13. Nov 30, 2016 #12
    Yea I got a much reasonable answer from that. The answer i got for the height is approx 7.34 m. way better that in the 700m zone. Can a water ejection from a pipe really go 7.34m though? I feel like it doesnt pass a common sense check.
     
  14. Nov 30, 2016 #13
    That is correct
     
  15. Nov 30, 2016 #14

    Delta²

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    yes maybe it doesn't pass a common sense check but neglecting air resistance and the fact that water speed is not so small (calculate in what height a point mass will reach if it is ejected vertically with a speed of 12m/s) plays a role here.
     
  16. Nov 30, 2016 #15
    thank you for your guidance!
     
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