Homework Help: Fluids Cons of Momentum

1. May 23, 2009

1. The problem statement, all variables and given/known data

2. Relevant 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?

2. May 23, 2009

Staff: Mentor

Sounds lovely!

3. May 23, 2009

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.

4. May 23, 2009

Staff: Mentor

Consider it to be an arbitrarily wide sheet of water. Use momentum per unit length as well.

5. May 23, 2009

Okay, so the m-1 just drops out anyway, and this problem reduces to regular cons of mom

6. May 23, 2009

Staff: Mentor

Yep.

7. May 23, 2009

Thanks! 83.4 degrees sounds reasonable I think

8. May 23, 2009

Staff: Mentor

Show how you got that answer.

9. May 23, 2009

10. May 23, 2009

Staff: Mentor

Rethink your result for m, the mass flow rate. You want it to be mass flow per unit width (or depth).

11. May 23, 2009

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. May 23, 2009

Staff: Mentor

Now you're cooking.

13. May 23, 2009