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

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    Fluids Momentum
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Homework Help Overview

The discussion revolves around incorporating the angle theta into the momentum equation within the context of fluid dynamics. The original poster is exploring how to visualize and apply the conservation of momentum principle to a scenario involving fluid flow around a wedge.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster considers dividing the wedge into upper and lower right triangles to resolve the momentum equation into Cartesian coordinates. They express uncertainty about how to incorporate theta and question the implications of given units of force per unit length.

Discussion Status

Participants are engaging with the original poster's ideas, providing affirmations and suggesting considerations for mass flow rate. There is an ongoing exploration of how to interpret the problem setup, particularly regarding the dimensions and characteristics of the fluid flow.

Contextual Notes

Participants are discussing the implications of treating the inlet as a "sheet" rather than a pipe, which affects the calculation of mass flow rate. There is also mention of specific angles and dimensions that are relevant to the problem.

Saladsamurai
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Homework Statement



Picture1-35.png


Homework Equations


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


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 [itex]\frac{\theta}{2}[/itex].

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:

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

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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