Fluid Dynamics final one

  1. Water flows steadily from an open tank as in the figure below. The elevation of point 1 is 10.0m, and the elevation of points 2 and 3 is 2.00m. The cross-sectional area at point 2 is 0.0480m^2; at point 3 it is 0.0160m^2. The area of the tank is very large compared with the cross-sectional area of the pipe.


    Part A is "Assuming that Bernoulli's equation applies, compute the discharge rate in cubic meters per second."
    Which I've solved using Bernoulli's principle

    so Q_3 = 0.200m^3/s

    Part B is "What is the gauge pressure at point 2?"
    Now this one I'm stuck with, but here's what I've done:

    A_2*v_2 = A_3*v_3
    (0.048)v_2 = (0.2)(0.016)

    v_2 = 0.0666667m/s

    so using Bernoulli's principle..(and cancelling some stuff)

    P_2 + 0.5*rho*v_2^2 = P_3 + 0.5*rho*v_3^2
    which becomes
    P_2 + (0.5)(1000)(0.0666667)^2 = 101300 + (0.5)(1000)(0.2)^2
    thus P_2 = 101317.7778

    And I figured since gauge pressure meant P_G = P - atm
    I subtracted 101317.7778 by 101300
    which got me 17.7778 as my gauge pressure...
    So what am I doing wrong? I don't seem to be getting the right answer here...
  2. jcsd
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