Calculating Force on Chamber in Water Flow

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Homework Statement


A chamber has water entering horizontally with a velcoity of 80m/s with an inlet of an area of 0.1m^ and leaving through an opening of area 0.15m^2. 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
\vec{F}=\dot{m_{2}}\vec{v_{2}}-\dot{m_{1}}\vec{v_{1}}
and \dot{m_{2}}=\rho A_{2}
and \dot{m_{1}}=\rho A_{1}

the component of the force is

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

Now here's the thing, we don't know v2...
But since the mass is conserved the volume flow rate in is equal to the volume flow rate out
\int_{S_{inlet}} \vec{v_{1}}\cdot\hat{n}dA=\int_{S_{outlet}} \vec{v_{2}}\cdot\hat{n}dA
v_{1}A_{1}=v_{2}A_{2}

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!
 
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Hi stunner,
stunner5000pt said:
Now here's the thing, we don't know v2...
But since the mass is conserved the volume flow rate in is equal to the volume flow rate out
\int_{S_{inlet}} \vec{v_{1}}\cdot\hat{n}dA=\int_{S_{outlet}} \vec{v_{2}}\cdot\hat{n}dA
v_{1}A_{1}=v_{2}A_{2}

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

you can mark is solved
 
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