How do I incorporate theta into the momentum equation for Fluids homework?

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Incorporating theta into the momentum equation involves visualizing the flow around a wedge and dividing it into upper and lower triangles, each with an angle of theta/2. The momentum equation can be resolved into Cartesian coordinates, considering the exit velocities are equal and the inlet flow splits symmetrically. The discussion clarifies that the problem can be simplified by treating the flow as a wide sheet of water, allowing for the use of momentum per unit length. The mass flow rate should be calculated based on the sheet's dimensions, leading to the formula \(\dot{m} = \rho V (h \cdot W)\). Overall, the key takeaway is to ensure clarity on the dimensions and flow characteristics to apply the momentum conservation principle correctly.
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Homework Statement



Picture1-35.png


Homework Equations


\sum(\dot{m}\vec{v})_{exit}-\sum(\dot{m}\vec{v})_{inlet}=\sum F_{ext}


I only know that this is cons of momentum because my prof told us. I am having a hard time visualizing how to incorporate THETA into the above equation since it is not a nice right triangle.

I am thinking that since the exit velocities are equal, than the inlet flow must be being split about the wedges horizontal axis of symmetry.

Thus, I think I can divide the wedge into an "upper" and "lower" right triangle whose angle wrt to the horizontal is \frac{\theta}{2}.

Then I can resolve the momentum eq into Cartesian coordinates.

Sound good?
 
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Sounds lovely!
 
Nice! I guess that what is confusing me now, is why they gave me
force/unit length into page

But I am just going to start plugging in and see what happens.
 
Saladsamurai said:
I guess that what is confusing me now, is why they gave me
force/unit length into page
Consider it to be an arbitrarily wide sheet of water. Use momentum per unit length as well.
 
Doc Al said:
Consider it to be an arbitrarily wide sheet of water. Use momentum per unit length as well.

Okay, so the m-1 just drops out anyway, and this problem reduces to regular cons of mom
 
Yep.
 
Thanks! 83.4 degrees sounds reasonable I think :smile:
 
Saladsamurai said:
83.4 degrees sounds reasonable I think
Show how you got that answer.
 
Picture2-21.png
 
  • #10
Rethink your result for m, the mass flow rate. You want it to be mass flow per unit width (or depth).
 
  • #11
I am not sure that I follow. Errr... Okay. So that inlet with the 4 cm dimension is NOT a PIPE...right?

It is a "sheet" with area 4cm*WIDTH. Thus my, mass flow rate should be, with h=4cm:

\dot{m}=\rho V (h*W)\Rightarrow \frac{\dot{m}}{W}=\rho Vh

Am I with you now?
 
  • #12
Now you're cooking.
 
  • #13
Typical 'not-paying-attention' mistake. Perhaps I should turn off Band of Brothers while I do my studies? :smile:

Thanks Doc!
 
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