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Homework Help: Force on a chamber

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    A chamber has water entering horizontally with a velcoity of 80m/s with an inlet of an area of [itex] 0.1m^[/itex] and leaving through an opening of area [itex]0.15m^2[/itex]. The exiting flow makes an angle of 30 degrees with respect to the entering flow. What force is needed to hold the chamber in place

    2. The attempt at a solution
    Well the total force on the chamber is
    [tex] \vec{F}=\dot{m_{2}}\vec{v_{2}}-\dot{m_{1}}\vec{v_{1}} [/tex]
    and [tex] \dot{m_{2}}=\rho A_{2} [/tex]
    and [tex] \dot{m_{1}}=\rho A_{1} [/tex]

    the component of the force is

    [tex]F_{x}=\rho A_{2} v_{2} \cos\theta - \rho A_{1} v_{1}[/tex]
    [tex]F_{y}=\rho A_{2} v_{2}\sin\theta[/tex]

    Now heres the thing, we dont know v2...
    But since the mass is conserved the volume flow rate in is equal to the volume flow rate out
    [tex]\int_{S_{inlet}} \vec{v_{1}}\cdot\hat{n}dA=\int_{S_{outlet}} \vec{v_{2}}\cdot\hat{n}dA[/tex]
    [tex]v_{1}A_{1}=v_{2}A_{2}[/tex]

    From this we can find the X and Y components and hence the force and the direction. Is this all correct?

    Thank you for your input and suggestions!
     
  2. jcsd
  3. Nov 11, 2008 #2

    Q_Goest

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    Hi stunner,
    Works for me. To be nit picky, the mass flow in = mass flow out (not volume flow) unless there's mass being stored inside the control volume, so we can equate VA(in) = VA(out) only if density(in) = density(out). Note that if this were a gas for example, and density changed, we'd have to determine velocity from the change in density and mass flow. But yea, you got it right.
     
  4. Nov 12, 2008 #3
    thanks a lot

    you can mark is solved
     
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