Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Second order, non-linear, non-homogeneous differential eq.

  1. Jun 21, 2017 #1
    I have a physics project at university, we designed an experiment to measure the effectiveness of Poiseuilles law in a 'quasi non-steady state'. Poiseuilles law, simply being the measurement of the flow rate of a fluid in a pipe, holding only under steady state though. So by quasi steady state I mean that there is some acceleration of the fluid, but only a small amount. I've attached a picture of the apparatus below. The general idea is that we use compressed air to pump the fluid up a measuring tube and use a high speed camera to measure the flow rate. By varying the radius of the measuring tube and the viscosity of the fluid we hope to find a range were poiseuilles law begins to breakdown.

    So we did some experiments and after a while, realised that Poiseuilles law only matched actual results in the limits of low pressure and small radii (sub 10mm and 1bar). Which is kind of counter intuitive... If Force = Pressure x Area then by holding pressure constant and increasing the area we would have expected the Force to increase, by means of a faster flow rate. This wasn't the case, instead, at larger radii, we saw slower flow rates....

    Our theory then, is that with increased radius we have increased volume and hence mass, so the inertial force begins to dominate and slows down the flow rate significantly and it deviates from our expected outcome. However, by deriving an equation to describe this, we ended up with a monstrosity and can't think of a way of solving it. This is were we would like some help. Below are some pictures showing the free body diagram of our revised model and the derivations of the equations that describe the motion of fluid in a pipe. If anyone has any suggestions on how to go about solving this equation I'd be most grateful. The idea is that a solution (if found), plotted on a graph might match our measured data. In the last equation we have pressure as a solution but we wish to find the solution to the function h(t)....

    Thanks

    [​IMG]

    [​IMG]

    [​IMG]
     
  2. jcsd
  3. Jun 26, 2017 #2
    Hi Ron:
    I have been curious to see how someone who knows more than I do would respond to solving for h(t). Since no one else has responded, I will offer my thoughts.

    The equation 7.10 seems incomplete to me. I think you need an equation of the form
    Ftotal = M(t) A(t).​
    At any given time there is a quantity of fluid in the tube and it's mass M(t) can be determined from your experimental data.
    Also, the rate of acceleration of that mass can also be determined from your experimental data. This will give you a known function (Ftotal(t) equaling the RHS of equation 7.10. The resulting equation can be put into the form:
    C2 h(t) h'(t) + C3 h(t) = C1 - Ftotal(t)​
    The 3 Cis appear to be constants.
    If this is correct, this equation can be readily solved numerically.

    ADDED 1
    I just noticed that M(t) = h(t) Z ρ, where Z is the cross-section area of the tube, and ρ is the density of the fluid. Also, A(t) = h''(t). Therefore Ftotal(t) has the form
    Ftotal(t) = C4 h(t) h''(t).​

    ADDED 2
    I also just noticed that the pressure is time dependent. So I am guessing my thoughts will not be too helpful, except possibly that the final resulting DE can be solved numerically.

    Hope this helps.

    Regards,
    Buzz
     
    Last edited: Jun 26, 2017
  4. Jun 27, 2017 #3
    The acceleration in time is not the only reason Poiseuille's equation does not apply here. The flow also develops in the axial direction due to the no-slip boundary condition at the pipe walls. As the flow begins to accelerate from rest, initially only the fluid near the wall is slowed down by friction. This is the boundary layer. Throughout most of the pipe there will be no radial velocity gradient and therefore no viscous shear stress. As the fluid flows in the axial direction this region where there is a non-zero radial velocity gradient (the boundary layer) will grow thicker as the viscous effects diffuse into the flow until eventually you get the parabolic velocity profile (in the radial direction) of fully developed pipe flow. So in your model the viscous stress term is incorrect. It seems to me that you are attempting to apply your force balance to all of the fluid in the pipe simultaneously. If you are interested in determining the velocity profile then you should apply the force balance to a small element of fluid because the momentum and stresses are continuously distributed throughout the flow. Using this process you will arrive at the Navier Stokes Equations which is where you should start your analysis.

    When you increase the pipe radius while maintaining everything else constant you should expect the velocity to decrease due to conservation of mass. As for your argument regarding Force = Pressure x Area, you should think in terms of a local differential element of fluid with a cross sectional area dA. When you increase the area of the pipe but pressure is constant the force on this differential element remains the same dF = Pressure x dA.

    As for the reason you are finding better agreement with Poiseuille's equation at low pressure and small pipe radii. As I mentioned above the boundary layer grows thicker as the fluid flows in the axial direction. The rate at which this boundary layer grows depends the rate at which momentum diffuses in the radial direction and is convected in the axial direction. Diffusion increases the boundary layer thickness while convection generally works to decrease the boundary layer thickness. These two processes have different length and time scales. The diffusion scales with the viscosity while the convection scales with the velocity. In some arbitrary amount of time the radial distance over which the fluid diffuses is independent of the the pressure you apply and the radius of your pipe. But during that time the axial distance that the fluid convects will depend on the pressure since that will change the velocity. So it comes down to a balance between these two effects. Lets say for a given pressure a fluid element takes time T to travel the length of the pipe. But during that time T there will also be diffusion in the radial direction causing the boundary layer to grow. If the diffusion length scale during time T is large compared to the pipe radius then the flow will quickly become fully developed and Poiseuille's law applies after the initial development region. If the diffusion length scale is very small during this time then you will not have the correct velocity profile, the flow is still developing in the axial direction. So for low pressures and small radii you should see that Poiseuille's law applies over a portion of your pipe.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Second order, non-linear, non-homogeneous differential eq.
  1. Non linear dynamics (Replies: 1)

Loading...