physics

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The intake in the figure has cross-sectional area of 0.75 m2 and water flow at 0.41 m/s. At the outlet, distance D = 180 m below the intake, the cross-sectional area is smaller than at the intake and the water flows out at 9.6 m/s. What is the pressure difference between inlet and outlet?

  • physics -

    p + rho g h + (1/2)rho v^2 = constant
    Vin = .41 m/s
    Vout = 9.6 m/s
    hinlet = 180
    houtlet = 0

    rho = 1000 kg/m^2
    g = 9.8 m/s^2

    so
    Pinlet + 1000(9.8)(180)+(1/2)1000(.41)^2
    = Poutlet +1000(9.8)(0)+(1/2)1000(9.8)^2
    solve for Poutlet - Pinlet

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