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Fluids - Squeezing flow - How to find force, work, and flux

  1. Apr 26, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-4-26_12-13-11.png
    The initial separation of the pistons is 2cm. They are submerged in water at room temperature.

    I need to calculate the work required to have two pistons touch, find the volumetric flux of the water as a function of time, and the force at a given time
    2. Relevant equations
    upload_2015-4-26_12-12-30.png
    upload_2015-4-26_12-12-36.png
    upload_2015-4-26_12-12-41.png
    3. The attempt at a solution

    [tex]F= \int p(r,z) 2 \pi r dr [/tex] (from 0 to R)

    F=(3/2) (pi u *V)/(H^3) * [z^2*R^2+R^4/2-R^4/4+Po*R^2/2]


    This is as far as I got with it.

    Is my solution for force correct? I'm not sure of the integration I need to do for work and flux from here. How do I find the force as a function of time?

    I tried googling for lectures on axisymmetric squeezing flow, but I only found research papers on the topic that did not help me in solving this problem. What other keywords should I be searching for?
     
  2. jcsd
  3. Apr 26, 2015 #2
    No. The p0 should not be inside the parenthesis. In fact, it should not be in there at all; you forgot to include the force of the water pushing down on the piston from the top. Also, you should have used the pressure at z = H to get the force on the piston, and not at arbitrary z. I didn't check the rest of your integration. Also, your equations are hard to read because you don't use Latex.

    What is H as a function of time?

    Chet
     
    Last edited: Apr 26, 2015
  4. Apr 26, 2015 #3
    Here's what i meant to enter earlier:
    [itex]F=\frac{3\pi \mu V}{2H^{3}}*[z^2R^2+\frac{R^4}{2}-\frac{R^4}{4}]+\pi P_{o}*R^2[/itex]

    Besides replacing "z" with H, Is Po the only correction I need to make? Just dropping it? In thinking about the problem, I had a feeling it would cancel out.

    For work, would I just integrate the force over the distance "z" the pistons move?

    The initial gap between the two pistons is 2cm and the velocity of each piston is 0.1cm/s, so I'm taking H(t) to be
    [itex]H(t)=2cm-2*0.1cm/s*t[/itex]

    Thank you!
     
  5. Apr 26, 2015 #4
    No. I think the R4 term should have an 8 in the denominator, not a 4.
    As I said, you neglected to include the downward force from the water on the top of the piston, which exactly cancels out the pressure term that you got in the integration.
    Your intuition was correct.

    Sure, but I don't know whether they want you to include the work on both the pistons or on just one of them.
    The diagram looks incorrect. From your equations for the velocities, it is obvious that H is half the distance between the pistons (and z is measured upward from the center-plane). So, H = 1 - (0.1) t

    Chet
     
    Last edited: Apr 26, 2015
  6. Apr 26, 2015 #5
    Thanks - I wasn't too sure about H(t).

    As for where you said the 4 should be an 8.

    Here's how I got it:

    [itex]F=\int_{0}^{R}[\frac{3 \mu V}{4H^{3}}*[2z^2+R^2-r^2]]*2\pi r dr[/itex]

    becomes:

    [itex]F=\frac{3\pi \mu V}{2H^{3}}*[z^2 r^2+\frac{R^2*r^2}{2}-\frac{r^4}{4}][/itex]

    Then substituting in R:
    [itex]F=\frac{3\pi \mu V}{2H^{3}}*[z^2 R^2+\frac{R^4}{2}-\frac{R^4}{4}][/itex]

    I don't think I'm making any mistakes -- please let me know if there is something you see that I'm overlooking!

    Thanks again!
     
  7. Apr 26, 2015 #6
    In addtion - for work, would taking the work done by 1 piston and multiplying it by 2 provide a valid answer? or even just integrating from -1cm to +1cm would provide the same result as integrate from 0 to +1 or -1 and multipling by two, I think. I always doubt myself on these situations and this is where I make my mistakes.
     
  8. Apr 26, 2015 #7
    Should I be replacing "H" with either the function "H(t)" or z? At first I thought it should be the size of the initial gap or distance from z=0 (in this case, 1cm), but I'm starting to question if that assumption was correct.

    For the volumetric flux Q(t), would it be the circumference of the pistons (2*pi*r) * ur*H(t)?

    Thanks!
     
  9. Apr 26, 2015 #8
    Oooops. My mistake. Sorry. That's what I get for trying to do it in my head.

    You should mention to your professors about that error in the figure regarding H. Another issue is that the platens are immersed in water. So, unless the platens are neutrally buoyant, they are going to contribute to the force, and the bottom force is going to differ from the upper force. So, I guess you need to assume that they are neutrally buoyant.

    Chet
     
  10. Apr 26, 2015 #9
    You want to do the upper piston and integrate from 0 to H(0), then multiply by 2 to get the work by both the pistons. Or, do the lower piston and integrate from 0 to -H(0), then multiply by 2 to get the work by both pistons. But neither piston moves over the full gap, so, even though the result might give the right answer mathematically (I haven't tried it to make sure), it doesn't make sense physically.

    Chet
     
  11. Apr 26, 2015 #10
    It doesn't matter what you use. Either way, you are going to be integrating from 0 to H(0).
    No. You have to integrate over z, from z = -H(t) to z = +H(t). Here's another (simpler) way to do it:

    What is the volume of fluid between the plattens at time t when their separation is 2H(t)? What is the rate of change of that volume with t? That must be equal to minus the volumetric flow rate out. (It has nowhere else to go).

    Chet
     
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