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I Radial flow between two circular discs (fluid mechanics)

  1. Oct 24, 2016 #1

    So I was reading wikipedia the other day, as I do from time to time. I came across a rather interesting sample problem posed in the article, but seeing as Wikipedia is horrible in some of their physics articles on explaining whats the hell they're doing, I became lost. Here is the problem, I'll be using cylindrical coordinates for simplicity here (r,z,Theta).

    Suppose we have two discs of equal radius. Disc one is placed at z = +h, and disc two is placed at z = -h. Let's assume that the flow is going to be purely radial, so the only non zero velocity component is the radial velocity component. The fluid that will be flowing is in-compressible, viscous, and newtonian. Following the no-slip condition, we demand that the radial velocity component go to zero when Z = +h and when Z = -h. The mass continuity equation simplifies to d/dr(r*V) = 0, where r is the radial coordinate and V is the radial velocity component. To satisfy this I assume the form V(r,z) = F(z)/r. Plugging this guess into the continuity equation we find that mass continuity is satisfied given V(r,z) = F(z)/r. Using this ansantz we can simplify the momentum equations into the following form. dP/dr = p*F(z)^2/r^3 + u/r*d^2(F(z))/dz^2, or r*dP/dr = p*(F/R)^2 + u*d/dz(dF/dz).

    According to wikipedia, we should somehow be able to get this into the form F''(z) + R*F^2 = -1, F(-1) = 0, F(1) = 0, where R is the reynolds number in the flow. I'm struggling to see how they manage to deal with the pressure term in the momentum equations that I derived.

    Here is the link to the actual wikipedia article : https://en.wikipedia.org/wiki/Navier–Stokes_equations#Application_to_specific_problems

    The problem I'm talking about is problem b.) in the "applications to specific problems" sections.

    Thanks for your help

    PS: sorry about not having the equations in latex. I've not learned how to use latex yet. Thanks for getting through them and trying to decipher them.
  2. jcsd
  3. Oct 24, 2016 #2
    I got the same result that you did, but, if the approach is supposed to be exact, there shouldn't be any r's in the equation. So they must be obtaining an approximate solution. I'm guessing that they integrate the radial variation out of the equation to obtain an equation that applies on average over the range of r's.
  4. Oct 25, 2016 #3
    Thanks for confirming that I'm not insane or something. I initially saw this problem and became interested in it. I quickly realized that an analytical solution to the ODE (F" + R*F^2 = -1, F(-1) = 0, F(1) = 0) was not possible or outside the range of my abilities, so I decided to try to find an approximate solution whose error is O(R) as R approaches 0, and I did infact find such a solution. I then realized that computing anything based on these values was gonna be difficult as the ODE was scaled and it was somehow put into dimensionless form. So I tried to actually derive the ODE they gave in dimensional form and was unable to do it.

    If you have any time could you explain how exactly they would've gone about integrating the radial variation out of the momentum equations?
  5. Oct 25, 2016 #4
    If you solved the problem without the inertial term present, this should suggest what to do. I think I can help you through this. What did you get for the solution without the inertial term? (I've already done this, but I would like to see the form of the solution that you obtained)
  6. Oct 26, 2016 #5
    Well, given that the inertial term goes to zero we arrive at the following : r*dP/dr = u*d^2F(z)/dz^2. In order for this to hold both of these must be equal to some constant, let's call it C. So we get C = r*P'(r), and we get that C = u*F"(z). These are easy ODE's to solve. We find that F(z) = C/2 * z^2 + A*z + B, where A and B are determined by the boundary conditions. Applying the noslip condition that F(-H) = 0 and F(H) = 0, we find that F(z) = C/2 * (z^2 - h^2). The ODE for pressure is solvable as well, but since they're now uncoupled I don't need to do that to solve for F(z).

    The issue that I see is that I cannot use a similar procedure to solve for the pressure when the inertial term remains. I'll think on it and see if I've thought of anything in the morning. I'll post an update tomorrow (at the moment it's about 10:35 pm where I live). Thanks so much for your continued help :).
  7. Oct 26, 2016 #6
    For the case of negligible inertia, if Q is the volumetric throughput rate, then $$v=\frac{3Q}{8\pi rh}\left[1-\left(\frac{z}{h}\right)^2\right]$$ and $$F=\frac{3Q}{8\pi h}\left[1-\left(\frac{z}{h}\right)^2\right]$$So, $$\frac{dp}{dr}=-\frac{3Q\mu}{4\pi r h^3}$$

    How does this work for you so far?
  8. Oct 26, 2016 #7
    That certainly makes sense for the low Reynolds limit. I'm also interested though in deriving the differential equation they derived when inertial contributions are not neglected. I just drove to school and I have a few hours to kill, I'm gonna see if I can work on it and get anywhere. I'll post an update soon.

    EDIT : Oh, oh, oh. Now I'm excited. I found this lovely little article here to get my brain going on some ideas of what to do. http://iopscience.iop.org/article/10.1088/1757-899X/78/1/012021/pdf. I'll keep you posted.
    Last edited: Oct 26, 2016
  9. Oct 26, 2016 #8
    I completed the analysis that I had in mind. So here goes:
    The fluid velocity is represented by $$v=\frac{Q}{4\pi rh}f(\eta)\tag{1}$$where ##\eta=z/h##
    Substitution into the differential equation yields:$$-\rho\left[\frac{Q}{4\pi rh}\right]\left[\frac{Q}{4\pi r^2h}\right]f^2=-\frac{dp}{dr}+\mu\left[\frac{Q}{4\pi rh^3}\right]f''$$where f'' represents the 2nd derivative of f with respect to ##\eta##. Combining terms yields:
    $$f''+\frac{\rho}{\mu}\frac{Qh}{4\pi r^2}f^2=\frac{4\pi h^3}{Q\mu}r\frac{dp}{dr}\tag{2}$$Here again we have r in the inertial term. So we are going to need to take an integrated average of the equation to get rid of the r variation. This, of course, is necessary if we are going to approximate the velocity by an equation of the form given by Eqn. 1. Multiplying Eqn. 2 by r and integrating between ##r_i## and ##r_o## yields:
    $$f''+Rf^2=-C\tag{3}$$where the Reynolds number R is given by:
    $$R=\frac{\rho Qh}{4\pi \mu}\frac{\frac{1}{r_i^2}-\frac{1}{r_o^2}}{\ln{\left(\frac{1}{r_i^2}\right)}-\ln{\left(\frac{1}{r_o^2}\right)}}\tag{4}$$and where $$C=\frac{4\pi h^3}{Q\mu \ln{(r_o/r_i)}}(p(r_i)-p(r_o))\tag{5}$$
    Note that, unlike your reference from Wikipedia, the constant C is not equal to 1. The value of the constant C is a function of the Reynolds number R, and is obtained by solving the differential equation (Eqn. 3) subject to the boundary conditions ##f(1) = f(-1) =0## in conjunction with the condition that $$\int_{-1}^{+1}{f(\eta) d\eta} = 2\tag{6}$$This latter condition is required to determine all the unknown constants and to guarantee that the velocity averaged over z is equal to ##\frac{Q}{4\pi rh}##.
    Last edited: Oct 26, 2016
  10. Oct 26, 2016 #9
    Thanks so much! I followed most of that, the only thing that I'm confused over is the constant C. So you're saying that you solved Eqn. 3 analytically and that that's how you found the value of C? Thanks again.
  11. Oct 26, 2016 #10


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    I haven't quite figured out what they did in the Wikipedia article yet, but I don't think you guys are on the right track if you still want to match that solution, to be honest. I've gotten close to what they had with the exception of the fact that I haven't yet removed the ##r## dependence. It seems pretty clear to me that they are citing a similarity solution, and I just haven't yet found the correct similarity variable. You can remove the pressure by assuming that the radial pressure gradient is constant and known a priori and use it as part of the process of making the equations dimensionless. It then works out that the pressure term on the RHS of the momentum equation is non-dimensionalized to unity if you specify the reference velocity by which you scale ##v_r## using the pressure gradient rather than just picking some arbitrary velocity. This makes sense because there is no characteristic velocity to use to make the equation dimensionless. The pressure gradient is actually what specifies the flow field.

    So far, by defining a pressure velocity of
    [tex]U^*_p = \sqrt{\dfrac{h^*}{\rho^*}\dfrac{dp^*}{dr^*}}[/tex]
    and using it to scale the equation, I've come up with
    [tex]f^{\prime\prime} + \dfrac{R}{r^2}f^2 = Rr,[/tex]
    which is close but not quite right. I probably need to incorporate #r^*# in my velocity parameter most likely but haven't quite decided how I want to go about it.
    Last edited: Oct 26, 2016
  12. Oct 26, 2016 #11
    Oh, and as an aside while we all try to work things out ; it turns out that the ODE F" + R*F^2 = -C for F(-1) = 0 and F(1) = 0 has existence problems. Numerically solving it is turning out to be problematic.
  13. Oct 26, 2016 #12


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    I am not entirely convinced that the solution is even self-similar and it's hard to say for sure without spending more time with this or else Wikipedia would have to cite its source. The problem doesn't show up in any of the textbooks I have on the topic.
  14. Oct 26, 2016 #13
    Let me try to clarify what I am saying with the following points:

    1. In the limit of low Reynolds number, the radial fluid velocity described by Eqn. 1 of my post #8 provides an exact solution; this solution involves streamlines that are parallel to the two disks. However, for finite Re, as both Boneh3ad and I have pointed out, the presence of the r2 in the denominator of the inertial term in the differential equation guarantees that the form of the velocity given by Eqn. 1 is not an exact solution, and the streamlines in the finite Re case will not be parallel to the two disks.

    2. In order to get an approximate solution to the problem with finite Re, we can integrate the equations with respect to r so that the r variation in the inertial term is averaged out. This leads to Eqns. 3-5 in my analysis.

    3. In post #11, Chuckstabler indicates that numerically solving Eqns. 3-5 is problematic, and that a solution may not exist. In my judgment, this is not correct. I guarantee that I can integrate these equations to get a solution. Certainly for the case of R = 0, this can be done, and a solution does exist. @Chuckstabler, have you tried to integrate this case? I would use the shooting method to integrate from -1 to +1, iterating on the values for f'(-1) and C using Newton's method (numerically). I also feel that something can be done with these equations analytically, by expanding in f and C in powers of the Reynolds number R: ##f=f^{(0)}(\eta)+Rf^{(1)}(\eta)+...## and ##C=C^{(0)}+RC^{(1)}...##, where ##f^{(0)}## and ##C^{(0)}## correspond to the solution for R = 0.

    The solution for R = 0 is ##f^{(0)}=\frac{3}{2}(1-\eta^2)## and ##C^{(0)}=3##
    Last edited: Oct 26, 2016
  15. Oct 26, 2016 #14
    Hey. I haven't yet, I'm still at school and my brains fried. I'll post an answer tomorrow morning when my brain is no longer dead.

    As a side note; I did find a problem which is amenable to an analytical solution without integrating the radial variation out. It's the problem of a 2d flow using polar coordinates, which models a converging flow. It separates nicely. Anyways, I'll post an update tomorrow.
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