[MUSIC] So, we have found plenty of simple models for the coupling between a moving solid and a still fluid, as soon as the reduced velocity is small enough and the amplitude of the motion is also small. We showed that the force acting on the solid was made of an added stiffness part and depending on the value of the Stokes number, an added mass or an added damping part. These two extreme cases, added mass and added damping were obtained when the effect of viscosity was negligible or conversely dominant. The question now is of course what happens for intermediate Stokes number? The approach we used was based on the analysis of the momentum balance equation. You remember that using the approximation of either very small, or very large Stokes number, we could drop some of the terms in the momentum balance equation. Then we could derive the added mass and the added damping by using a very simple argument; As the solid velocity at the boundary was given by a simple approximation we built a single mode approximation for the velocity in the whole fluid domain. That allowed to cancel out all time dependence in the equations, and eventually to arrive at a force proportional to q dot, or q double dot. Can we use the same argument for any value of the Stokes number? Let us see. Unfortunately, no, we cannot. Why? Because in the momentum balance equation, we must keep the two terms involving the velocity, the inertial term and the viscous term. None of them can be neglected. So, any form of the velocity in the form of a function of space and function of time is going to mix in that equation the function of time and its time derivative. What does that mean in terms of physics? It means that there is a time scale inside the equations for the fluid domain. When you have an equation like dy / dt + 1 / Tau x Y = 0, you know that the solution is Y = e to the -t /Tau. The dynamics of y then includes the time scale tau. It is similar here, we have du over dt on one side, and one over Stokes delta u on the other side. If there is a time scale in the fluid dynamics there is no way that the fluid response can be instantaneous. We had instantaneous response for low and high Stokes number, because we have instantaneous viscous diffusion and instantaneous inertial response respectively. In between, there is competition between these two effects. So what does the force look like? Oh, no not him. [MUSIC] To try and answer this, let us consider the idealized problem of what is called the impulse response. Imagine that the motion of the solid is an impulsive start. Zero velocity then unit velocity, constant, the Heaviside side step function H of t. This is going to cause some fluid motion of velocity U_I and the pressure P_I, I as impulse. By integrating at the interface, these velocities and pressure will give me the force acting on the solid the soluble which I called f_l. I don't know how this force evolves in time but certainly it does. As this is a response of the system to an impulse displacement of the solid I call it the impulse force. Imagine now any arbitrary motion of the solid defined by q dot of t as sketched here. I can write this q dot as the sum of successive impulses of magnitude q double dot times the Heaviside step function. Because all the equations governing the dynamics of the fluid are linear, the response of the fluid and as a consequence, the force, is going to be the sum of successive impulse responses to the change of velocities the acceleration. This takes the form of a convolution product between the acceleration of the solid, and the impulse force. [MUSIC] Schematically, here is what happens. At the time t, the force exerted by the fluid on a solid, f(t), results from all fluid motion started earlier as a consequence of previous changes in velocity of the solid. The force f of t keeps the memory of the past. Why? Because the impulse response was not instantaneous, the fluid having its own time scale. To summarize, any motion of the solid is going to start a progressive diffusion of momentum in the fluid, and progressively the feedback force will take this into account. What is going to happen? Imagine a motion defined by the velocity q dot sketched here. At low Stokes number, the force will have the same shape being proportional to q dot. When the motion stops, the force stops. Conversionly at very high Stokes number, the force is proportional to acceleration but it will also stop when the motion stops. In between, because of the delay in the response, the force will continue to exist even after the solid has stopped its motion. This is a new very important result. Between these cases of added damping and added mass, the fluid feedback force will depend on the history of loading Not an added damping or added mass, but something like an added internal clock. [MUSIC] Actually, we can have exact solutions for all these when the geometry is simple enough. Imagine an infinite plate bounded on one side. by a semi-infinite fluid domain, the plate moves along the x axis only, tangentially to the fluid. In that case, the problem is independent of x, and the velocity only depends on y and t. The pressure is uniform and the velocity satisfies simply Du over dt equals 1 over Stokes, d square u over d y squared. This is a very classical formula of diffusion in one direction that you'll find in all domains of physics. The impulse response corresponding to q dot equals H of t is well known. It takes a self-similar form based on what is called the error function that I recall here. This means that the velocity in the fluid adapts as follows. At an early stage of the motion, only the fluid very close to the plate is entrained. As time goes, thanks to the viscous forces, more and more fluid is entrained. This is a classical diffusion of momentum by viscosity. The corresponding force, the impulse force, is a consequence of the viscous shear at the interface. Per unit area of the plate it can be computed using the general formula, which simplifies a lot here, and we have the impulse force equals to minus M over square root of Pi Stokes t. This means that an impulse start of the plate results in a feedback force that starts at infinite and decreases progressively. I could expect that. A very strong feedback as I try to shear the fluid at rest. Then as the fluid layers progressively start moving less and less force needed. As a consequence, the force, in the general case, reads as the convolution between the acceleration and that force decreasing in time. As you can see, the larger the Stokes number, the shorter the memory. To summarize, the effect of viscosity is all in the Stokes number. If the stokes number is very large, neglect viscosity and consider only added mass. If it is very small, neglect inertia and consider just added damping . In between, the competition between inertia and viscous diffusion will result in a memory effect that will give you a force depending on the history of motion of the solid. In practice, if you consider vibrations of a solid in a still fluid, here is what will happen. At large Stokes number the fluid force will be in phase with the acceleration, at small with the velocity, and in between the phase will be in between. Actually, by measuring the phase between the displacements and the force we can estimate the Stokes number and there by the fluid viscosity. This a technique used in food engineering, for instance, to measure the state of cooking of cheese. As you could see, a lot could be said in the field of coupling a small motion of solid
with a still fluid. We have obtained added stiffness, added mass, added damping and even memory effects For a given case the equations to solve are rather simple. This means that a lot of solutions are available in books and papers for geometries of interest. But the most important is, as always, to understand what is behind all this using the appropriate dimensional numbers. Actually, there are other cases where the dynamics of the fluid is governed by an internal time scale. One very important in practice is that of sloshing. Next, you will learn how to take that into account. [MUSIC]
