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Determine force from flow on diffuser

  1. Sep 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Water exits the 3-in.-diameter pipe at a velocity of 12 ft/s and is split by the wedge diffuser. Assume water is ideal fluid, that is, incompressible and frictionless. Determine the force the flow exerts on the diffuser. Take θ = 30 ∘

    2. Relevant equations
    F = ρAV

    3. The attempt at a solution
    So what I've seen others do, and what tools I've been given so far in the course, here's the numbers I plugged in, and the equation I used:

    ρ = 62.4
    A = πr2
    R = (3/24) <- 3 inch diameter, and this just eliminates one step in the calculator
    V = 12
    θ = 15 (half of 30)

    F = ρAV(Vcosθ). But I'm getting 426, which isn't correct . . .
     

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  3. Sep 29, 2015 #2
    The force, which affected the water flow can be calculated as the difference of the momentum flows before and after the diffusor. It seems that you've mixed up the velocities of these two flows (you combined the velocities in horizontal direction of both states in one term).
     
  4. Sep 29, 2015 #3
    What do you mean? ρAV = (ρAV)1 - (ρAV)2?
     
  5. Sep 29, 2015 #4
    What direction is the fluid velocity when it is flowing over the diffuser? (a) parallel to the axis (b) parallel to the diffuser surface

    Is the mass rate of flow constant in going over the diffuser?

    Is the velocity of the fluid on the diffuser constant or changing with axial position along the diffuser?

    Is the thickness of the fluid layer on the diffuser constant or changing with axial position along the diffuser?

    Chet
     
  6. Sep 29, 2015 #5
    mass flow rate = ρVA = 62.4 * -12 * π(3/24)2
    That's density times velocity dotted with the normal (that's why it's negative) times the area of the circle the water is traveling through.

    Would it be constant over the diffuser, but different than the above calculation? Including the velocity? And nothing is changing I don't think with respect to position. The force is different after the wedge, but still constant.
     
  7. Sep 29, 2015 #6
    What does Bernoulli's equation tell you about the velocity of the fluid as it is passing over the diffuser (i.e., the magnitude of the velocity vector) compared to its value before it hits the diffuser?
     
  8. Sep 29, 2015 #7
    No idea. We haven't gotten that far . . .
     
  9. Sep 30, 2015 #8
    EDIT: I misread the statement and therefore cancelled this post, sorry.
     
    Last edited: Sep 30, 2015
  10. Sep 30, 2015 #9
    You are giving very good advice here, but you seem to be assuming that the geometry is flat. According to the problem statement, the actual geometry is cylindrical. This is why I was trying to get the OP to use Bernoulli to determine that the magnitude of the fluid velocity along the diffuser is the same as the magnitude of the velocity out of the pipe. As the flow passes over the diffuser, it is the thickness of the water layer that changes (decreases) to conserve mass flow rate. I'm hoping you will continue with your help and advice, modifying it as necessary to address the cylindrical geometry.

    Chet
     
  11. Sep 30, 2015 #10
    Dear Chet,

    as the statement said "[]...is split by the wedge diffuser." I thought it is a two dimensional problem. Additionally in the drawing the pipe is cut (at the upper side), but it seemed to me that it ends at the lower side without changing its diameter (but where is the diffusor then?). Of course, if it would be 3-dimensional without the diffusor (only the cone/wedge), the thicknesses of the split water flows must decrease along the cone. However there are some additional hints that it is 3-dimensional, as the flows keep in touch with the wedge/cone even after it ends. I probably misunderstood the statement/drawing, I'm sorry for that.
     
  12. Sep 30, 2015 #11
    No problem. Your analysis can still be continued, taking into account, from Bernoulli, that the fluid velocity tangent to the cone is still the same as in the pipe. Please continue.

    Chet
     
  13. Oct 1, 2015 #12
    Wait . . . so are you saying to use ρAV2 to find the flow at the inlet and each outlet (but for the outlets, use the cos of half the angle)? And just take the difference of inlet/outlet flows as the force?
     
  14. Oct 1, 2015 #13
    Not exactly. That's called the rate of momentum, not the flow. But otherwise, OK.

    Chet
     
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