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Water Pressure

  1. Apr 30, 2005 #1
    I have no idea how to solve this problem I know bernoulli's equaiton and the equation of continuity, but I don't know how to apply them to this problem...If someone could get me started I would be very appreciative. My final exam is monday and I fear this problem is going to be on the exam... thanks for the help...

    Through a pipe of diameter 15.0 cm, water is pumped from the Colorado River up to Grand Conyon Village, on the rim of the canyon. The river is at 564m elevation and the village is at 2096 m... (a)at what minimum pressure must the water be pumped to arrive at the village? (b)If 4500 cubic meters are pumped per day, what is the speed of the water in the pipe? (c) What additional pressure is necessary to deliver this flow?
     
  2. jcsd
  3. May 1, 2005 #2

    xanthym

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    SOLUTION HINTS:
    {River Elevation} = HA = (564 m)
    {Village Elevation} = HB = (2096 m)
    {Area of Pipe} = A =(π/4){Diameter}2 = (π/4){0.15 m}2 = (0.017671 m^2)
    {Flow Velocity} = v
    {Volume Flow Rate} = V = A*v = (4500 m^3/day) = (0.05208 m^3/sec)
    {Water Density} = ρ = (1 gram/cm^3) = (1000 kg/m^3)

    The basic equation relating conditions at Point A (River) and Point B (Village) is:
    PA + (1/2)*ρ*(vA)2 + ρ*g*HA = PB + (1/2)*ρ*(vB)2 + ρ*g*HB

    For your information, the continuity equation is usually required when the pipe crossection Area changes. Since the pipe Area doesn't change in this problem, the continuity equation is not used. If the pipe Area changed from point "1" to point "2", this equation would state:
    ρ*A1*v1 = ρ*A2*v2 ::: <---- not needed for this prob since A never changes

    a) Considering only change in elevation, the minimum gauge pressure required to deliver water from "A" to "B" is:
    {Minimum Gauge Pressure} = PA - PB =
    = ρ*g*HB - ρ*g*HA =
    = (ρ*g)*(HB - HA) =
    ={(1000 kg/m^3)*(9.81 m/sec^2)}{(2096 m) - (564 m)} =
    = (1.50289e(+7) N/m^2)

    b) From the definition and given value of Volume Flow Rate:
    {Volume Flow Rate} = V = A*v
    ::: ⇒ v = V/A
    = (0.05208 m^3/sec)/(0.017671 m^2)
    = (2.947 m/sec)

    c) The ADDITIONAL gauge pressure needed to attain flow velocity "v" determined in "b)" above is given by:
    {Additional Gauge Pressure} = PA - PB =
    = (1/2)*ρ*(vB)2 - (1/2)*ρ*(vA)2 =
    = (1/2)*ρ*{(vB)2 - (vA)2} =
    = (1/2)*(1000 kg/m^3)*{(2.947 m/sec)2 - 0}
    = (4.3424e(+3) N/m^2)


    ~~
     
    Last edited: May 1, 2005
  4. May 1, 2005 #3
    Thank You very much for showing me how to work that problem...maybe you would be willing to take the final for me :smile:
     
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