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Fluid Mechanics- inviscid fluid pressure in a pipe

  1. Dec 4, 2015 #1
    1. The problem statement, all variables and given/known data

    Water flows through a circular pipe with a 180° horizontal elbow and exits to the atmosphere through a nozzle as shown in Fig. Q3. The diameter of the pipe is 300 mm and the diameter of the nozzle exit is 160 mm. The density of water is 999 kg/m3 . The mass flow rate of water is 140 kg/s. Incompressible, inviscid flow may be assumed. Determine:
    (i) the velocity of the water at sections 1 and 2 respectively,
    (ii) the gauge pressure at section 1, and
    (iii) the magnitude and direction of the force Rx exerted by the water on the elbow



    2. Relevant equations
    Bernoulli's Equation?

    3. The attempt at a solution
    I have determined that Vin=1.98m/s and Vout=6.97m/s and i know how to determine Rx, but i have no idea how to get the gauge pressure

    Do you use bernoulli's equation and let P2 equal atmospheric pressure? But does that mean you let the exit velocity equal zero?
     

    Attached Files:

  2. jcsd
  3. Dec 4, 2015 #2

    SteamKing

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    I don't see any other way to analyze this pipe than to use the Bernoulli equation. I would not assume that if the gauge pressure is zero, the exit velocity is zero, which you have found is clearly not the case. If the velocity were zero, where would all the water go which enters the pipe at location 1?

    Remember, gauge pressure = absolute pressure - atmospheric pressure, if that helps.
     
  4. Dec 4, 2015 #3
    ok, so using bernoulli's equation:

    pin = ρ((Vout2 - Vin2)/2) ?

    And that's the gauge pressure?
     
  5. Dec 4, 2015 #4
    Also for the next part, do you continue to assume that pout=0 ?
     
  6. Dec 4, 2015 #5

    SteamKing

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    It would be better if you showed your complete calculations.
     
  7. Dec 4, 2015 #6
    converting mass flow rate to volume flow rate and using Q=VA:
    Vin=1.98m/s ; Vout=6.97m/s

    pin=999((6.972 - 1.982)/2)=22308 Pa
     
  8. Dec 4, 2015 #7

    SteamKing

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    Yeah, but where's your Bernoulli equation for this problem?
     
  9. Dec 4, 2015 #8
    sorry, being lazy

    pin/ρ + Vin2/2 + gz = pout/ρ + Vout2/2 + gz

    pout=0
    pin/ρ + Vin2/2 = + Vout2/2

    isolating pin
    pin = ρ((Vout2 - Vin2)/2)
     
  10. Dec 4, 2015 #9

    SteamKing

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    Since you are not exhausting to a vacuum, Pout is clearly a gauge pressure.

    In the Bernoulli equation, you should use absolute pressures in your calculations. Any results can be converted to gauge readings afterwards.
     
  11. Dec 4, 2015 #10
    I'm not sure what you mean, so your saying let Pout equal to Patm as there is no gauge pressure on exit?

    My understanding is that if there is a gauge pressure on exit you would need to have been told that in order to calculate Pin
     
  12. Dec 4, 2015 #11

    SteamKing

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    In the absolute pressure scale, a perfect vacuum has P = 0 psia and atmospheric pressure is P = 14.7 psia.

    P = 0 psig (psi gauge) is equivalent to P = 14.7 psia (psi absolute)

    If you are dealing with pressure differentials, it makes no difference if both pressures are absolute or gauge, but Bernoulli's equation is written assuming the pressures are measured on the absolute scale.
     
  13. Dec 4, 2015 #12
    Your answers for the gauge pressures in and out of the section of pipe (22308 and 0) are correct. Now, what are your ideas on how to do part iv?

    Chet
     
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