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Water pressure question

  1. Apr 9, 2006 #1
    Could anyone offer me some guidance on parts b and c of the following question?

    The pressure in a uniform pipe of flowing water is enough to hold up 4.00 m of water in an open, vertical tube as shown in Figure P.60. Assume that water is an ideal fluid. The pipe leads 3.00 m upstairs and empties into the open air through an opening with an area one-fourth that of the uniform pipe.

    (a) At the exit opening of the pipe, what is the pressure?
    (b) What is the pressure at the point where the vertical tube is connected to the pipe?
    (c) What is the exit speed of the water from the pipe?


    this is the picture of the problem:
    http://img346.imageshack.us/img346/5326/p11607ht.gif

    for part (a), i figured b/c the water is open to the air, its pressure is 1 atm (approx. 1.01E5 Pa).

    for parts (b) and (c), i think i need to somehow use the equation:
    P1 + (1/2)*density*v1^2 + density*g*y1 = P2 + (1/2)*density*v2^2 + density*g*y2

    but i'm not sure how to get the velocity or pressure at the point under the 4m opening.

    any ideas/hints for me?

    thanks!!
     
  2. jcsd
  3. Apr 9, 2006 #2
    i figured this one out, looks like i was on the right track afterall!
     
  4. Apr 2, 2009 #3
    I am doing this exact same problem and am also unsure of how to use Bernoulli's principle for b) and c).
    Help please.
     
  5. Apr 2, 2009 #4

    rl.bhat

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    Homework Helper

    but i'm not sure how to get the velocity or pressure at the point under the 4m opening.
    Pressure P1 = Pa + rho*g*h where Pa atmospheric pressure, rho is the density of water and h = 4m. P2 = Pa
     
  6. Apr 2, 2009 #5
    I've gotten the pressure at the opening= P(atmosphere)
    I've gotten the pressure under the 4 m opening= 140565=P(atmosphere)+1000(9.81)(4 m)

    What is the exit speed of the water from the pipe?? In my case,
    The pipe leads 3.00 m upstairs and empties into the open air through an opening with an area one-third that of the uniform pipe."

    so we know that A1v1=A2v2 so A1v1=1/3A1v2 and thus 3v1=v2 where v2 is the exit speed and v1=speed in the pipe

    how do I set up

    Bernoulli's principle to get v1 and thus v2?

    P1 + (1/2)*density*v1^2 + density*g*y1 = P2 + (1/2)*density*v2^2 + density*g*y2
     
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