[MUSIC] So, let us explore together a bit further the domain of small reduced velocities. As you know, this corresponds to cases where I can neglect the proper free velocity. We are still going to assume that the displacements of the solid are small, so that I can use all the much simpler linearized equations. Let us recall them. Here they are, on the fluid side, on the solid side, and at the interface. You recognize in the fluid force the part that is due to the motion of the solid in a pressure gradient. That gave us the fluid induced stiffness terms. Then, the other part corresponds to the effect of the pressure and velocity in the fluid caused by the motion of the solid. This is the feedback term. Let us focus on this feedback term. In the limit of very high Stokes number, the effect of the fluid viscosity could be neglected. And we showed that the fluid loading was just an added mass effect. Let us go now at the other limit, a very small stokes number, for instance, vibrations inside a bearing. There, the size of the fluid domain is small, say ten microns. The viscosity of oil is a thousand times more than water. For vibrations at a frequency of ten hertz you would have a Stokes number of 10^-6 This is very small, and you can imagine that the Stokes number may be even smaller when the size is smaller, the viscosity is larger, and the motion is slower. A very small Stokes number means that the viscous term in the momentum balance equation dominates over the acceleration term. We can neglect the du over dt. We have to rescale the pressure by using p times Stokes in place of p, otherwise your problem is ill-posed. This results in a simplified momentum balanced equation, zero equals minus gradient of p plus delta u. The mass balance is unchanged, div of u equals 0. Consistently, at interface, we keep the full kinematic condition, u equals q dot Phi. And the dynamic condition, where I have again removed the stiffness terms, contains the sum over the interface of the loading by pressure plus the loading by viscosity. Can we solve this in general case? Certainly. [MUSIC] In fact, we can use exactly the same technique done for the limit of very high Stokes number. At the interface, the fluid velocity takes the form of a function of time, q dot, times a function of space, Phi. This comes from the single mode description of the motion of the solid. We may then look for fluid velocity that would have the same form, not only on the interface but everywhere in the fluid. Of course, the time dependence would be q dot, the space dependence is unknown, and we note it phi u. In the same idea, because pressure is now proportional to the velocity u, let us look also for pressure field p of x and t as q dot times a function of space, say phi p of x. These are not the same phi u and phi p as for the added mass case, of course. Then, the mass balance implies that div of phi u is equal to zero. And in the momentum balance, the q dots cancel out, and we obtain minus grad of phi p, plus delta of phi u, equals 0. At the interface, the kinematic condition is simply phi u equals phi. Let us look at the dynamic condition. Here is the fluid loading we are looking for. Using our new single mode description of the pressure and velocity, it takes the form of minus q dot times a ratio of M over Stokes, times the sum over the whole interface of terms involving Phi, Phi p, Phi u, and a normal n. All these are time independent, so the quantity inside the bracket is just a constant. This means that the force induced by the fluid on the solid is a damping force, proportional to the velocity of the solid, q dot. The coefficient, CA, is the added damping. And again, the fluid loading on the solid at the time t only depends on the velocity at the same time. This is an instantaneous feedback. We just found for the limit of a very small Stokes number, a situation somewhat parallel to the limit of very large Stokes numbers. For large Stokes numbers, the effect of viscosity in dynamics of the fluid can be neglected. The motion of the fluid is not influenced by viscous diffusion, which is very slow. This results in an instantaneous intertial response of the fluid, and a consequence an instantaneous added mass effect. Conversely, for small Stokes numbers, the effect of the viscosity is dominant. The motion of the fluid is now governed by viscous diffusion which is very fast. This results in an instantaneous viscous response of the fluid, and as a consequence, an instantaneous added damping effect. In the first case, to compute the added mass, we just had to solve a Laplace problem on Phi p. In the second case, it is a Stokes problem, a bit more complicated but still it only has to be solved once. What are the effect of dimensionless numbers? The added mass and the added damping are both proportional to the mass number. The added damping decreases with the Stokes number. If we consider now the dimensional quantities, you remember that we had an added mass that was proportional to the fluid density Rho. Now the added damping is proportional to the fluid viscosity Mu, as you would expect. Using these very simple solutions for the fluid force acting on a solid, we know how the dynamics of the solid is going to be affected. At high Stokes number, as I said before, coupling with a fluid is just an increase of the mass. The oscillator equation for q is now (1 + m A) q double dot + q = 0. The frequency of the free oscillation would be lower in the presence of fluid. At low Stokes number, we now have a damped oscillator equation for q. Any free oscillation will be damped. The added damping caused by the viscous effects in the fluid is not a surprise. In practice, it is only significant when the Stokes number is quite small. This means very small systems or very viscous fluid, for instance. In industrial application this is the dominant damping for systems with a very confined fluid domain, such as fluid bearings for instance. You also find this effect in biomechanics when considering very small systems. And it is also important when dealing with a very viscous fluid, such as those found in food processing. Yes, there are people doing research on fluid-solid interactions in cheese making at the industrial level. Of course, the next question is what happens when the Stokes number is neither large nor small. Are we going to have added mass plus added damping? Next you will learn how to address this case. [MUSIC]