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Using Bernoulli's equation

  1. Sep 30, 2012 #1
    Part a is pretty easy

    m'1+m'2=m'3
    m'=mass flow rates
    density1*velocity1*Area1+(analogous terms for inlet 2)=(analogous terms for inlet 3)
    The 3 densities are unknown
    but can be found from equation pressure/(RT)
    R=286.9 J/(kgk)

    This gives:
    density 1=1.66 kg/m^3
    density 2=1.432kg/m^3
    density 3=2.35 kg/m^3

    solving for mass flow rate gives 7.128 kg/s





    not for part B our professor told us to use the equation
    Q_dot_in - W_dot_out = m_dot_3 (cp T3 + V3^2/2 + gx3) - (analogous terms for 2) - (analogous terms for 1)

    cp=specific heat =1004 J/kgK
    h=cpT so that's where those terms came from

    V1=Q1/A1
    V2=Q2/A2
    V3=Q3/A3

    Where Q=volume flow rate given in the problem
    this gives:
    v1=86.61 m/s
    v2=66.1m/s
    v3=33.41 m/s

    The equation also requires mass flow rates
    mass flow rate = density*Q or density*velocity*area
    m'1=4.98 kg/s
    m'2=2.15 kg/s
    m'3=7.128 kg/s

    Now we have all the unknowns and the energy equation becomes
    67-W_dot_out = 7.128 (1004*311+33.41^2 / 2 + 9.81*.5) - 2.15(1004*365 + .5*66.1^2+9.81*0.5) - 4.98(1004*294+.5*86.615^2+9.81*0)
    W_dot_out = -118 kJ/s



    W_dot_in = 118 kW
    However the answer is supposed to be 223 kW or something like that.
    I checked our my answer a few times and haven't found any mistakes. Can you guys check?
     

    Attached Files:

  2. jcsd
  3. Oct 1, 2012 #2
    Hmmmm odd. Is the "correct" answer given in the back of the book or given by your professor? I too came up with 118kW.
     
  4. Oct 1, 2012 #3
    Sorry, never mind, he just told us it was the wrong answer. UGHhhh
     
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