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 () + 2µ ,
where µ is the viscosity and =
1
∇u + ∇uT .
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Question: Can we derive a visco-elastic model?
Answer:
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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.
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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
Re
10−5
1 mm sphere in air
Fish
Car on Highway
1
103
106
M. Phelps
4×
106
Plane
108
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 = kB ln Ω ,
where kB 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 ,
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where T is the temperature. This implies that
FS =
∂E
= 2kB T β 2 R ,
∂R
3
. Thus the force is linear in R and depends only on R, so we can replace the
2N b2
whole thing with a linear spring. The equation for the total forces is then
where β =
FS + FB + FD = 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 = νhF S RT i ,
where hF S RT i is a statistical average. This leads to the evolution equation:
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Dσ
− ∇x uσ P + σ P ∇uT = − (σ − GI) ,
τ
|Dt
{z
}
upper convective derivative
where
Dσ
Dt
is a material derivative. If we take a time derivative of σ P , we obtain an equation for Ṙ.
The Stokes-Oldroyd B Model is therefore


− ∇x p + ∆u = −β∇x · σ



∇
Wi σ = − (σ − I)



∇ · u = 0,
(1)
x
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. Given σ0 , 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.
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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.
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