Fluid mechanics - is there enough information for this problem?

  • Thread starter pyroknife
  • Start date
  • #1
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3
Attached the problem.

The TA for my class makes up all the damn homework problems, but she has a habit of leaving out important information on every single homework assignment. Sorry for the rant, just had to vent a bit.

Unless they want all this in variables, I don't see a way to get a # for the velocity. We're not given any densities...am i supposed to assume that the density for both inlets are the same?
Cross section for the exit is not given...


my approach was to use the conservation of linear momentum
as stated in the problem, the horizontal and vertical forces are 0.

m'=mass flow (kg/s)
m'1=mass flow at inlet 1
m'2=mass flow at inlet 2
m'3=mass flow at outlet 3
V1=velocity1



Sum of forces in the x gives: m'1*V1=(m'1+m'2)*V3*cos(theta)
sum of forces in the y gives: m'2*V2=(m'1+m'2)*V3*sin(theta) << solving this equation for V3 and substituting back into other equation gives
m'1*V1=m'2*V2*cos(theta)/sin(theta)

That's 3 unknowns

I guess I can do CONSERVATION of mass flow as well

m'1+m'2=m'3
density1*Area1*velocity1+density2*area2*velocity2=density3*area*3*velocity*3
This doesn't really help me at all.
 

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Answers and Replies

  • #2
1,198
5
Assuming the densities are the same, separately equate the forces in the x and y directions. For instance:

Q1 * V1 = Q3 * V3 * cos(theta)

where Q = V * A

so

A1 * V1^2 = A3 * V3^2 * cos(theta)

etc...
 

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