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Tap water flow and water diameter at end of it (Bernoulli eqzn )

  1. Feb 4, 2008 #1


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    1. The problem statement, all variables and given/known data
    At a faucet, the diameter of the stream is 0.960cm. The stream fills a [tex]125cm^3[/tex] container in 16.3s.
    Find the diameter of the stream 13.0cm below the opening of the faucet.

    2. Relevant equations
    Bernoulli's eqzn: [tex]P+ 1/2\rho v_1^2 +\rho gy = constant [/tex]
    or.....[tex]P_1-P_2 = \rhog(y_2-y_1)= \rho gh [/tex]

    continuity eqzn: [tex]Av_1= Av_2 [/tex]

    3. The attempt at a solution

    um....I'm not sure how to go about doing this Q...

    I did do:

    Flow rate= [tex]Av_1= 125cm^3 / 16.3s = 7.67cm^3/s [/tex]

    after that I'm not sure about how to find the diameter.
    I think I need the Area since Av= 7.67cm^3/s that I found and I guess I would go and find v for the first v1 at least.

    I am confused as to:

    1. is P1 the same as P2 (bottom of stream) ? If it is then would it cancel out?

    2. would the v be the same? I think not...but if it isn't then how would I find v1 for when the water comes out of the faucet?

    If someone could help me out I'd appreciate it alot.

    Thanks :smile:
  2. jcsd
  3. Feb 5, 2008 #2


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    well I worked on it for awhile...since yesterday and now I think.

    Flow rate= Av1= 125cm^3/ 16.3s = 7.67cm^3/s

    [tex]A_1=\pi r^2 =\pi d^2/4= \pie (0.960cm^2)/4= 0.007238cm^2 [/tex]

    since A1V1= 0.007238cm^2
    then V1= (7.67cm^3/s)/ (0.007238cm^2) = 0.10597m/s

    I'm not sure how to find V2 though. (the v of the water at the bottom of the stream of water) It's open to the air so I would think gravity but I'm not sure..

    Info I know:

    v1= 0.10597m/s (from A1V1 and area)
    A1= 7.67cm^3/s
    y2-y1= 13cm => 0.13m

    [tex]P_1 + 1/2 \rho v_1^2 + \rho gy = P_2 + 1/2 \rho v_2^2 + \rho gy [/tex]
    P1 and P2 cancel out I think so then I have:

    [tex] 1/2 \rho v_1^2 + \rho gy = 1/2 \rho v_2^2 + \rho gy [/tex]

    don't know:
    v2= ?
    A2=? => question asks for this

    I wanted to get v2 through the A1V1= A2V2
    but if I don't have V2 how can I find A2? UNLESS I have to find it using the Bernoulli's equation.....I'm confused here

    Can anyone help me out Please??
  4. Feb 5, 2008 #3

    Doc Al

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    Staff: Mentor

    You've got it. First find V2 using Bernoulli (which just tells you the increase in speed due to gravity), then you can use the continuity equation to find A2.
  5. Feb 5, 2008 #4


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    Hm...I went and used this equation ([tex]V= Av\Delta t [/tex]) (the thing is that I'm not sure it's the continuity equation.....it includes time and the Volume.

    If I used the continuity eqzn (A1v1= A2v2)
    with what I found from the Bernoulli's eqzn which is:

    [tex] v_2= \sqrt{2g(y2-y1)} [/tex]
    [tex] v_2= \sqrt{2(-9.8)(0.13m)} [/tex]
    [tex] v_2= 1.596m/s [/tex]

    now...I'm debating wheter to use [tex]V= Av_2 \Delta t[/tex] or [tex]A_1v_1= A_2v_2 [/tex]

    I used the other equation when working on the question and got:

    [tex]V=Av_2 \Delta t [/tex]
    t= 16.3s
    V= 125 cm^3 => 1.25m^3
    v2= 1.596m/s

    [tex]1.25m^3= \pie r^2(1.596m/s)(16.3s) [/tex]
    r= 0.12367

    d= 0.2473m

    I'm not sure how I'd do the question with A1v1=A2v2
    since I do have
    but do I know v1?

    Thanks alot :smile:
  6. Feb 6, 2008 #5

    Doc Al

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    Staff: Mentor

    In the continuity equation, what does A1v1 represent? (Hint: They give you the flow rate for a reason!)

    Looks like you assumed v1 = 0 when using Bernoulli. That can't be right. You need to figure out the initial speed from the given data.

    The analysis you did here is exactly what you have to do to find the initial speed. You are given the fow rate (Volume and time). Since you know the diameter (thus the area) at the top, you can use it to find the speed at the top (v1).

    You're almost there. :wink:
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