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Water Pressure in One Pipe Flowing to Another

  1. Dec 17, 2011 #1

    TMO

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    1. The problem statement, all variables and given/known data

    A water pipe tapers down from an initial radius of R1 = 0.22 m1 to a final radius of R2 = 0.08 m1. The water flows at a velocity v1 = 0.85 m1s-1 in the larger section of pipe. The water pressure in the center of the larger section of the pipe is P1 = 283640 Pa. Assume the density of water is 1000 kg1m-3.

    2. Relevant equations

    (1) A ∝ r2
    (2) p + ½·ρ·v2 + p·g·h = C
    (3) A1·v1 = A2·v2

    3. The attempt at a solution

    I assumed that since the water remains at roughly the same height, and that the only thing changing is the velocity, and that pressure in Bernoulli's equation is conserved, that one should try to compute the velocity from this.

    ½·ρ·v2 = P1
    ρ·v2 = 2·P1
    v2 = 2·P1
    v = √(2·P1/ρ)

    Using Equation 3 in conjunction with the fact that A ∝ r2, I concluded that the final velocity must be (R1)2·√(2·P1/ρ)/(R2)2, however this isn't giving the correct numerical answer. Any thoughts?
     
  2. jcsd
  3. Dec 18, 2011 #2
    You are using Bernoulli's equation incorrectly. Write the equation as

    P1 + .5*rho*(V1)^2 = P2 + .5*rho*(V2)^2

    and evaluate for P2.
     
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