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Faucet connected to water tower

  1. Nov 26, 2014 #1
    1. The problem statement, all variables and given/known data
    A 1 cm radius pipe connects a 3 m tall water tower to a faucet.
    (a) What is the gauge pressure at the faucet when the faucet is closed
    (b)12 What is the gauge pressure at the faucet if the faucet is opene
    d so that there is a flow of 0.2 Liter/s? Assume that the water flow is laminar.

    2. Relevant equations
    P_1 + .5d(v_1)^2 + d(y_1) = P_2 + .5d(v_2)^2 + d(y_2)
    d = density , P = pressure ,y = height, v = velocity
    (A_1)(v_1) = (A_2)(v_2)
    A = area v = velocity

    3. The attempt at a solution
    for part a)
    if the faucet is closed, gauge pressure at the faucet would simply be
    P = dgy = (1000 kg/m^2)(9.8 m/s^2)(3 m)

    for part b)

    bernoullis-png.75869.png
    i must use bernoulli's equation. looking at the picture i drew I can say P_2 is 0, v_2 is 0 (because it is negligible), y_2 = 3 m , and y_1 = 0

    to find v_1 :
    use Av where v = l/t
    Av = A(l/t) = Vol/t (which they give us as .2 liter/s = 2E-4 m^3/s)
    A = pi r^2 = pi (.01 m) ^2 = 3.14 E-4
    Av = Vol/t
    (3.14E-4)v = 2E-4
    v = 2/3.14 = .637 = v_1

    using bernoullis:
    P_1 + .5d(v_1)^2 + d(y_1) = P_2 + .5d(v_2)^2 + d(y_2)
    P_1 + .5d(.637)^2 + d(0) = 0 + .5d(0)^2 + d(3)
    P_1 + .5(1000)(.637)^2 = 3(1000)
    P_1 = 3000 - .5(1000)(.637)^2

    im not as worried about the actual numbers but is my method correct?
     

    Attached Files:

  2. jcsd
  3. Nov 26, 2014 #2
    You left out a factor of g in Bernoulli's Equation, but besides that your method looks good
     
  4. Nov 26, 2014 #3

    haruspex

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    ##(1000 kg/m^3)##
    What dimension does ##d(y_1)## have? Something missing?
     
  5. Nov 26, 2014 #4
    oops! it should be d(y_1)g

    and good catch i meant 1000 kg/m^3
    sorry for the sloppiness =[
     
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