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Conservation of Momentum Question

  1. Sep 2, 2016 #1

    joshmccraney

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    1. The problem statement, all variables and given/known data
    An incompressible fluid of density ##\rho## flows steadily through a 2D infinite row of fixed shapes. The vertical distance between shapes is ##a##. Define station 1 as the space where velocity enters and station 2 where it exits. Also, velocity and pressure are constant along stations 1 and 2, and are given by ##v_1##, ##v_2##, ##p_1##, ##p_2##. The angle the flow makes with the horizontal direction at station 1 is ##\beta_1##, and it is ##\beta_2## at station 2. Compute the reactions ##R_x## and ##R_y## necessary to keep one vane in place.


    2. Relevant equations
    Conservation of momentum

    3. The attempt at a solution
    Horizontal forces at station 1 are ## F_{x1} = ap_1+\rho a^2v_1^2 \cos^2 \beta_1## (since 2D I assume pressure is force per unit line and density is mass/square unit). I only wrote ##a^2## to make sure the dimensions agree. I have no intuition as to why it should be there though, although I do understand it for the pressure. Horizontal forces at station 2 are ## F_{x2} = -ap_2+\rho a^2v_2^2 \cos^2 \beta_2##. Then the reaction would be ##R_x = F_{x1} - F_{x2}##. How does this look so far?
     
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  3. Sep 3, 2016 #2

    haruspex

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    I find the description much too vague. Is there a diagram with this? The horizontal is mentioned, so is one of the two dimensions vertical or are they both horizontal?
     
  4. Sep 3, 2016 #3

    joshmccraney

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    Here's the picture. Please let me know if I can be clearer.
     

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  5. Sep 3, 2016 #4

    haruspex

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    Yes, that certainly clarifies it.
    Your work so far looks right to me, except that I see no reason for the extra factor of a. The dimensions were fine without it, on the basis that, as you say, force is per unit depth. I think your mistake is in also saying density is per unit area. It might help to introduce some arbitrary length in the direction perpendicular to the page.
    I believe some simplification can be done. Since the flow is incompressible, there is a relationship between v1, v2, β1 and β2.
     
  6. Sep 4, 2016 #5

    joshmccraney

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    Okay, so let's let ##W## be the width coming out of the page. Then a force balance implies ##R_x+p_1aW + \rho V_1^2 a W \cos \beta_1 = p_2aW + \rho V_2^2 a W \cos \beta_2##. Does this look correct? Here ##R_x## is the force exerted on the shapes.

    Is the relationship you're referring to Bernoulli's?
     
  7. Sep 4, 2016 #6

    haruspex

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    The way you have written the equation, it would be the force the shapes exert on the flow. Other than that, it looks right.
    No, it is the equation which expresses non-compressibility, i.e. volumetric flow out = volumetric flow in.
     
  8. Sep 4, 2016 #7

    joshmccraney

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    Hmmm can you explain how you knew this? I'm wanting to know so I can do this on my own next time. Also, looking at the equation I wrote, shouldn't the pressure force at station two be negative when placed on the right side of the equation, since it is acting to the left, in the negative-x direction?

    Do we need this since density, angles, and both velocities are given? Isn't conservation of volume ##\rho a W v_1 \cos \beta_1=\rho a W v_2 \cos \beta_2##?
     
  9. Sep 5, 2016 #8

    haruspex

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    Just look at the two pressure terms. p1 applies a force left to right, p2 applies a force right to left. The net force to the right is therefore the p1 term minus the p2 term.
    Yes, that's the equation. As I wrote, it provides a simplification to the net horizontal force.
     
  10. Sep 5, 2016 #9

    joshmccraney

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    Shoot, so I guess what I should have written is the total force exerted on the shapes equals the momentum force and the pressure force. Then we would have ##R_x = p_1 a W - p_2 a W + a W v_1^2 \cos \beta_1-a W v_2^2 \cos \beta_2##? This seems like I've written something wrong though since the momentum force at station 2 is going in the opposite direction as the pressure force at station 2. What are your thoughts?

    Gotcha!
     
  11. Sep 5, 2016 #10

    haruspex

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    It is? The station 1 terms there are both positive, and the station 2 terms both negative. It all looks consistent to me.
     
  12. Sep 5, 2016 #11

    joshmccraney

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    Hmmmm, at station 2 I thought momentum was leaving to the right yet pressure was acting to the left. Is this true? If so, wouldn't they have opposite signs since they are acting in opposite directions?

    Thanks for your help by the way!
     
  13. Sep 6, 2016 #12

    haruspex

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    First, I only just noticed that in going from post #1 to post #5 you dropped the powers of 2 on the cos terms. Is that a typo? I believe they should be squared. (If not squared, the simplification I mentioned is not that dramatic.)
    With the above corrected, there is no change to horizontal momentum, so it becomes academic - the terms cancel.
    Overlooking that, the drop in horizontal momentum, will be momentum1-momentum2. That would be a result of a leftward force from the vanes. The corresponding rightward force the flow exerts on the vanes would therefore equal momentum1-momentum2.
     
  14. Sep 6, 2016 #13

    joshmccraney

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    Yes, that is a typo. Thanks for pointing this out.

    Okay, so that I understand you we have ##\Sigma {F} = m {a}## (since mass is not time dependent here) which implies the x-direction balance becomes ##-R_x + \rho a W v_1^2 \cos^2\beta_1-\rho a W v_2^2 \cos^2\beta_2 + p_1aW-p_2aW=0##. Conservation of mass implies flow in = flow out, which implies ##\rho a W v_1^2 \cos^2\beta_1=\rho a W v_2^2 \cos^2\beta_2##. This simplifies the equation to ##-R_x + p_1aW-p_2aW=0 \implies R_x=p_1aW-p_2aW##. This is what you were saying, right?
     
  15. Sep 6, 2016 #14

    haruspex

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    Yes.
     
  16. Sep 6, 2016 #15

    joshmccraney

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    Cool! So how would we deal with the vertical force? I don't think pressure is considered, and conservation of mass implies vertical volumetric flow is zero too, right?
     
  17. Sep 6, 2016 #16

    haruspex

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    There is a change to the vertical momentum of the fluid.
     
  18. Sep 6, 2016 #17

    joshmccraney

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    So I am now wondering if the correct conservation of volume constraint should instead be written as ##v_1 \cos \beta_1 + v_1 \sin \beta_1=v_2 \cos \beta_2 + v_2 \sin \beta_2##? If so then ##F_x## is not as simple, but then we have ##-F_y +a W \rho v_1^2 \sin^2 \beta_1-a W \rho v_2^2 \sin^2 \beta_2=0##, where I did not include pressure since I think it acts only horizontally. What do you think?
     
  19. Sep 6, 2016 #18

    haruspex

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    By what reasoning?
     
  20. Sep 6, 2016 #19

    joshmccraney

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    Shoot, now I'm confusing myself. Perhaps you could explain to me why this is not correct? Sorry for dragging this out, I just really want to understand this.
     
  21. Sep 7, 2016 #20

    haruspex

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    Consider the flow in. If the speed is v1, how would you calculate the volumetric flow? What do you need to multiply by?
     
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