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Bernoulli, Gauge pressure in hose?

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data

    A village maintains a large tank with an open top, containing water for emergencies. The water can drain from the tank through a hose of diameter 6.60cm. the hose ends with a nozzle of diameter 2.20 cm. A rubber stopper is inserted into the nozzle. the water level in the tank is kept 7.50 m above the nozzle. a) Calculate the friction force exerted on the stopper by the nozzle. b) the stopper is removed. What mass of water flows from the nozzle in 2.00hours? c)Calculate the Gauge pressure of the flowing water in the hose just behind the nozzle.

    2. Relevant equations
    Bernoulli's equation, Torricelli's equation, Pressure= Force/Area, Area = (∏/4)(Diameter)2, P-P0=ρgh, Continuity of flow


    3. The attempt at a solution

    A) The pressure pushing the plug out is atmospheric pressure and the pressure of the water in the tank, PTank=P0+ρgh
    the pressure keeping the plug in is the atmospheric pressure and the force of friction acting over the area of the nozzle, P0+F/A
    They must be equal, otherwise the plug would pop out, so
    [STRIKE]P0[/STRIKE] + ρgh =[STRIKE] P0[/STRIKE] + F/A
    Aρgh = F = 27.94

    B) Torricelli's equation, Vhose=sqrt(2gh)
    Use it to find the flow rate, VhoseAhose= .004607 m3/s
    Multiply the flow rate by 2*60*60 seconds to get 33.17 m3
    Multiply by the density of water to get the total mass, m= 33170 kg

    C) This is where I'm having trouble.
    Vhose = 12.12 m/s, I can take that from the last part.
    I can find Vnozzle = 1.3467 m/s using the continuity of flow A1V1=A2V2

    I should plug the velocities, density, and pressures into the Bernoulli equation, P1 + (.5)ρV12 + ρgh1 = P2 + (.5)ρV22 + ρgh2, but I'm not sure where to put which pressures and if I should only be considering the hose or if I should include the tank.

    I found some other explanations and solutions but none of them have helped me very much, so if someone could set up the equation for me and explain how you did, that would be amazing!

    Thanks,
    Ian
     
    Last edited: Sep 22, 2013
  2. jcsd
  3. Sep 22, 2013 #2

    SteamKing

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    Are you sure about this calculation? The water flows at 12.12 m/s in the hose and then slows down to 1.35 m/s going out of the nozzle? It seems that it would be silly to put a nozzle on a hose when the nozzle would slow the water stream to a trickle. Slowing the water flow down for emergencies seems a bit counter intuitive.

    For your Bernoulli equation, you can calculate the entrance conditions where the hose exits the tank. You know the head and you have calculated the velocity in the hose.

    BTW, your continuity equation is incorrect. It's A1 * Velocity1 = A2 * Velocity2
     
  4. Sep 22, 2013 #3
    I'll triple check in the morning; it does seem really weird that Vnozzle would be smaller, but I found a solution key that said that was correct.

    Thank you for pointing out the continuity equation. I only wrote it down incorrectly, I used it properly in the work.

    For the Bernoulli equation, I'm not sure if I understand correctly.
    On the left side of the equation I'd have the Pressure of the water in the tank, the atmospheric pressure, the water moving at near 0 speed, and ρgh?
    Then for the right, the atmospheric pressure, ρVnozzle2 and pressure head of 0?
    Ptank + P0 + 0 + ρgh = P0 + Vnozzle2 + 0?
     
  5. Sep 22, 2013 #4

    SteamKing

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    For your Bernoulli equation, I think you want to take point 1 at the entrance to the hose. V1 = 12.12 m/s there and the total pressure would be Po + ρgh. The wording of the question is sort of vague, but I take 'just behind the nozzle' to mean 'just before the exit from the nozzle'. A picture would have clarified this problem greatly.
     
  6. Sep 22, 2013 #5
    So the left side is P0 + ρgh + .5ρV12? Or should it be 2ρgh?

    How would I set the right side to be able to find the gauge pressure? The only pressure on the outside of the hose is P0, so wouldn't they just cancel?

    Or would the left side be P + .5ρV12 since P = P0 + ρgh

    then P + .5ρV12 = P0 + .5V22 and solve for P - P0?

    By the way, you're absolutely correct about Vnozzle being low, both of the solutions I looked at confused the areas and flipped them the wrong way, the actual speed is 109 m/s, if I didn't mess it up again. I'm skeptical of the solution now. Would you mind taking a look at it and seeing if it's set up correctly or if I should ignore it completely?
     
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