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F=MA 2012 Exam #19

  1. Jan 26, 2013 #1
    1. The problem statement, all variables and given/known data
    19. A 1,500 Watt motor is used to pump water a vertical height of 2.0 meters out of a flooded basement through a
    cylindrical pipe. The water is ejected though the end of the pipe at a speed of 2.5 m/s. Ignoring friction and
    assuming that all of the energy of the motor goes to the water, which of the following is the closest to the radius
    of the pipe? The density of water is ρ = 1000 kg/m3
    .
    (A) 1/3 cm
    (B) 1 cm
    (C) 3 cm
    (D) 10 cm ← CORRECT
    (E) 30 cm


    2. Relevant equations
    Pressure = ρgh
    Bernoulli's Eq: p + ρgh + 1/2ρv^2 = Constant
    Power = F dot v, Power = Work / t = dW/dT
    Flow Rate Continuity:
    A_0v_0 = Av

    3. The attempt at a solution
    First, I said that:
    A_0v_0 = pir^2*2.5
    Thus, v_0 = pir^2*2.5 / A_0
    Then, I used Bernoulli's Eq, but I'm confused as to how I make use of the power? I need some guidance on how to use the power in the problem or how to turn one of these equations into
    one I can handle.
     
  2. jcsd
  3. Jan 26, 2013 #2

    Simon Bridge

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    How much energy does it take to lift a mass through a given height?
    What is the expression that tells you the mass of water that is ejected through the pipe each second?
     
  4. Jan 26, 2013 #3
    It takes mgh to lift an object height h. Are you talking about the mass flow rate equation:
    That's m/t which is equal to (rho)Av.
    Here's an approach:
    W = PE + KE
    W = mgh + 1/2mv^2
    W/t = Power =m/t(gh + 1/2v^2)
    1500 = pAv(gh+ 1/2v^2)
    A = pi*r^2
    1500 = r^2(pi)(rho)v(gh), since the velocity of the basement is negligible ( I would think )
    r = sqrt (1500 /(pi*rho*v*gh)
    so r = .099 m = 9.9 cm so
    10cm -- This is the correct answer.
     
    Last edited: Jan 26, 2013
  5. Jan 26, 2013 #4

    Simon Bridge

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    How good is that assumption?
    What happened to the KE term?
     
  6. Jan 26, 2013 #5
    Yep, I saw that error in the radius and fixed it. How can we do the problem without assuming v is negligible?
     
  7. Jan 26, 2013 #6

    Simon Bridge

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    what happens to the step you made the assumption if you don't?
    i.e. what problem did the assumption solve exactly?

    Try writing down the rate mass flows through the pipe as ##\frac{dm}{dt}## and then ##P=\frac{dE}{dt}## ... remembering that we are told that only mass changes with time.
     
  8. Jan 26, 2013 #7
    so:
    dE/dt = dm/dt(gh + 1/2v^2)
    if power is constant, does that suggest that
    gh + 1/2v^2 = 0,
    or that v= root(2gh)?
    No I don't think I see how to use that.
     
  9. Jan 27, 2013 #8
    Here's a thought that disregards the velocity:
    Power = Fv
    F in this case is the gravitational force,mg because that is what must be overcome to pump the water at a height of 2m
    v is given as 2.5 m/s - that is the speed of pumping.
    Thus:
    P = mgv
    m = ρV
    P = ρVgv
    V = h*A, where A is cross-sectional area
    P = ρhAgv
    A =∏r^2, sorry I can't find the little pi
    P = r^2ρ∏hgv
    r = √(P / ρ∏hgv)
    All these are given
    r ≈ 10 cm
     
    Last edited: Jan 27, 2013
  10. Jan 29, 2013 #9

    Simon Bridge

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    Can you assume the speed of the water is constant?
     
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