Notes by: Andy Thaler Christel Hohenegger A simple model for ketchup-like liquid, its numerical challenges and limitations April 7, 2011 Many complex fluids are shear-thinning. Such a fluid has a shear rate-viscosity graph similar to the one below. Ketchup is not shear thinning, so its viscosity is independent of the shear rate. We will also need constitutive equations, which give the relation between stress (force per area) and strain (displacement). For example, in a Newtonian (purely viscous) fluid, we have σ = pi + µ ( v + v T ) σ = λtr (ɛ) + 2µɛ, where µ is the viscosity and ɛ = 1 2 ( u + u T ). Question: Can we derive a visco-elastic model? Answer: 1
1. Elastic network (like springs) Newtonian solvent. (Here the symbol refers to coupling). The problem in this case is determining how to combine the two pieces. 2. Suspension: the fluid has long polymer chains that sit in the liquid. (a) Microscale: Find an object describing your polymer. (b) Macroscale: Continuum. We also need to be able to find closure between the microscale and macroscale, which is hard. For example, we could be precise at the microscale, but this leads to a complicated model; on the other hand, we could also be less precise at the microscale and therefore more accurate at the macroscale. 1 Stokes-Oldroyd B Model James Oldroyd (1921-1982) was a British mathematician and rheologist (a person who studies the flow of matter ). 1.1 Stokes Equations The first part of this model is Stokes equations (the Navier-Stokes equations in the case where inertia is negligible, that is, the Reynolds number is zero). Re = Reynolds number = inertial forces viscous forces A table of some typical Reynolds numbers is given below: Bacteria 1 mm sphere in air Fish Car on Highway M. Phelps Plane Re 10 5 1 10 3 10 6 4 10 6 10 8 In particular, we assume that Re = 0 in the first two cases, as these values (10 5 and 1) are relatively small even compared to the Reynolds number of a fish. Assuming the Reynolds number 2
is zero leads us to Stokes equations: p + v = 0 v = 0 There is no time in these equations, since the Reynolds number is 0. This means that the process is fully reversible, which leads to the Scallop Theorem. The intuitive idea of this theorem is as follows: as a scallop opens its shell, it will be propelled in one direction. As it closes its shell, it will move in the opposite direction, returning to its original position. The idea is the same for the Stokes equations if we run the process backward in time, the solution will return to its original state. 1.2 Oldroyd B Dumbbell Model Consider a polymer, as shown in the figure below. We assume that the polymer is stiff, and that there are N links. We also assume that the length of each link is b, and that b is fixed. Although this last assumption is not very realistic, we could replace each link with a linear spring a statistical physics argument says that either assumption (that b is fixed or that each link is a linear spring) works fine. In particular, Φ (R, N) has a Gaussian pdf. Can we really get rid of the internal part and just keep track of R? Let Ω be the number of internal configurations for which R is fixed. Then the entropy is given by S = k B ln Ω, where k B is the Boltzmann constant. If we know S, and there is no enthalpy and no internal energy, then the free energy, E, is given by E = T S, 3
where T is the temperature. This implies that F S = E R = 2k BT β 2 R, where β = 3. Thus the force is linear in R and depends only on R, so we can replace the 2Nb2 whole thing with a linear spring. The equation for the total forces is then F S + F B + F D = 0 (no acceleration in system), where F B represents the Brownian motion forces and F D represents the drag forces. A few pages later, we get an equation for Ṙ, which we need to close the system. For closure, we have the Kirkwood formula, namely σ P = ν F S R T, where F S R T is a statistical average. This leads to the evolution equation: Dσ Dt ( x uσ P + σ P u T ) = 1 (σ GI), }{{} τ upper convective derivative where Dσ Dt is a material derivative. If we take a time derivative of σp, we obtain an equation for Ṙ. The Stokes-Oldroyd B Model is therefore x p + u = β x σ Wi σ = (σ I) x u = 0, where x σ is the force due to the linear spring and Wi is the Weissenberg number. We note that time is back!! Thus the system is not time reversible, so it is hard to simulate. The basic idea is as follows: (1) 1. Given σ 0, solve the Stokes part with prescribed boundary conditions using Fourier. 2. Now that we know u, solve the Oldroyd B part using the Adams-Bashforth II method, but be careful computing x u. 3. Iterate. 4
1.3 Problems First, we wanted to model ketchup, but ketchup has no shear thinning. Also, the solution blows up, as there is no restriction on how long the spring can get we want only finite extension. To correct this, we use the finite extension nonlinear spring (FENE) model, but there is no closure here. The FENE-p model, however, has closure where additional assumptions are needed in order to approximate closure. 5