Polymeric Fluids Lecture

8/11/2019 Polymeric Fluids Lecture

1/26

Polymeric Fluids

GM05 Part 1

Dr. Helen J. Wilson

9.1.2006 19.1.2006


2/26


3/26


4/26


5/26


6/26


7/26


8/26


9/26


10/26

3 Linear viscoelasticity

A linear viscoelastic uid is a uid which has a linear relationship between its strainhistory and its current value of stress:

(t) = t

G(t t ) (t ) dt

The function G(t) is the relaxation modulus of the uid. Because a uid can neverremember times in the future, G(t) = 0 if t < 0.

Physically, you would also expect that more recent strains would be more important thanthose from longer ago, so in t > 0, G(t) should be a decreasing function. There arentreally any other constraints on G.

A few often-used forms for G(t) where t 0 are:

Single exponential G0 exp[ t/ ]Multi-mode exponential G1 exp[ t/ 1 ] + G2 exp [ t/ 2 ] +

Viscous uid (t)Linearly elastic solid G0

Lets just check the last two. For the viscous form we have

(t) = t

G(t t ) (t ) dt = t

(t t ) (t ) dt = (t)

as we would expect for a Newtonian viscous uid. For the elastic solid we have

(t) = t

G(t t ) (t ) dt = t

G0 (t ) dt = G0 t

(t ) dt .

The integral on the right is the total strain (or shear) the material has undergone: sothis, too, gives the form we expect.

3.1 Creep

Given a sample of material, how would you go about modelling it? Even if you start byassuming it is a linear material (and they all are for small enough strains), how wouldyou calculate G(t)?

One way would be to carry out a step strain experiment: for shear ow this meansyou would set up your material between two plates and leave it to settle, so it loses thememory of the ow that put it there. Once it has had time to relax, you shear it throughone shear unit (i.e. until the top plate has moved a distance the same as the distancebetween the plates). Then stop dead. Measure the force required to maintain the platesin position as time passes. The results will look something like this:

11


11/26

Shear rate

Time

Stress

Time

What is the real relationship between the stress function (t) and the relaxation modulusG(t) in this case?

Suppose the length of time the material is sheared for is T . Then the shear rate duringthat time must be 1 /T (to get a total shear of 1):

(t ) =0 t < T 1/T T t 00 t > 0

Then the stress function becomes

(t) = t

G(t t ) (t ) dt =

1T

0

T G(t t ) dt .

Now suppose that we do the shear very fast so that T is very small. Then

0

T G(t t ) dt T G(t)

and so the stress is(t) G(t).

Thus the relaxation modulus is actually the response of the system to an instantaneous

unit shear.

3.2 Storage and Loss Moduli

An step shear is very difficult to achieve in practice. Real rheologists, working in industry,are far more likely to carry out an oscillatory shear experiment.

The material sample is placed in a Couette device , which is essentially a pair of con-centric cylinders, one of which is xed. The material is put in the narrow gap betweenthe cylinders, and the free cylinder is rotated. The ow set up between the two cylinders

12


12/26

is almost simple shear ow (as long as the gap is narrow relative to the radius of thecylinders, and no instabilities occur).

The rotation is set up to be a simple harmonic motion, with shear displacement:

(t) = sin(t)

where is the shear, the frequency and the amplitude. In reality must be keptsmall to ensure the material is kept within its linear regime; but if we are dealing withan ideal linear viscoelastic material, can be any size. Note that the shear rate of theuid, (t), will be

(t) = cos (t).

Suppose this motion started a long time ago ( t ). Then at time t, the stress prolefor a general linear viscoelastic material becomes

(t) = t

G(t t ) (t ) dt =

t

G(t t ) cos (t ) dt

We can transform this integral by changing variables using s = t t :

(t) = t

G(t t ) cos (t ) dt =

0G(s)cos((t s)) ds

and now if we write cos ((t s)) = [exp[i(t s)]] we get:

(t) =

0

G(s) [exp [i(t s)]] ds =

0

G(s)exp[i(t s)] ds

= exp[it ]

0G(s)exp[ is] ds

and the integral on the right is now a one-sided Fourier transform. Since it has nodependence on t, it is just a complex number. By convention we dene the complexshear modulus , G, as:

G = i

0G(s)exp[ is] ds,

and its real and imaginary parts:

G = G + iG

are called the storage modulus G and the loss modulus G .

This gives

(t) = [exp [it]( iG )] = [(cos (t) + i sin(t))( G iG )]

= [G sin(t) + G cos(t)] = G (t) + G

(t).

13


13/26

Now a purely viscous uid would give a response

(t) = (t) = cos (t)

and a purely elastic solid would give

(t) = G0 (t) = G0 sin(t).

We can see that if G = 0 then G takes the place of the ordinary elastic shear modulusG0 : hence it is called the storage modulus, because it measures the materials ability tostore elastic energy. Similarly, the modulus G is related to the viscosity or dissipation of energy: in other words, the energy which is lost.

Since the r ole of the usual Newtonian viscosity is taken by G / , it is also common todene

= G

as the effective viscosity; however, the storage and loss moduli G and G are the mostcommon measures of linear rheology.

Let us check a few values:

G(t) = (t) G() = iG(t) = G0 G() = G0 NB: By observation not integration

G(t) = G0 exp [ t/ ] G() = iG0 (1 + i )

= G0 (2 2 + i )

1 + 2 2

Let us look more at the single-relaxation-time model. At slow speeds 1 we have

G iG 0

so the material looks like a viscous uid with viscosity G 0 . At high speeds 1, wehave insteadG G0

and the material looks like an elastic solid with modulus G 0 .

If we plot the two moduli, G and G against the graph looks like this:

G

G

Frequency,

14


14/26

The point where the two graphs cross is given by:

G = GG0 2 2

1 + 2 2 =

G0 1 + 2 2

= 1.

A real uid has more complex dynamics:

Typical dynamic data for an elastic liquid

Frequency, but a rst engineering approximation often used is to take the value of c at which thetwo curves cross, and use = 1c as the relaxation time of the uid. Alternatively, aseries of relaxation modes is used to t the curves as well as possible.

15


15/26

16


16/26

4 Microscopic dynamics

In this section we will look at the rst model that people came up with when they startedto model polymers from the microscopic level. Its called the Oldroyd B model. Wewill spend one double lecture to derive it and in the last lecture, look at its macroscopic(top-down) properties:

Linear rheology

Stress in shear

Stress in extensional ow

4.1 The stress tensor

Remember we had the three Navier-Stokes equations (in incompressible form): u = 0

This is mass conservation: integrated over a volume of space, it tells us that uid ow inmatches uid ow out, so uid does not accumulate in one place.

ut

+ u u =

= pI + ( u + ( u) )

In the last two equations, is the stress tensor . The physical meaning of this iscalculated from the force the force exerted across a small surface element with area S and unit normal n:

n

The force acting on the top surface of the tiny pillbox is n S .

4.2 Beads in a viscous uid

Probably the simplest object you could put into a ow is a solid spherical bead. Letslook briey at the motion of a microscopic solid sphere in a Newtonian uid.

There is really only one parameter in Newtonian uid ow (once youve made everythingdimensionless): the Reynolds number , Re, which gives the ratio of inertial to viscousforces. If the scale of the ow is very small, then the Reynolds number is very small andwe can assume it is zero. Then the governing equations are the Stokes equations :

u = 0

2

u =

p.17


17/26

There isnt time in this course to go into the details of these ows: well just accept onekey fact. In a uid which is otherwise at rest, a sphere of radius a, moving with constantvelocity U , experiences a drag force of

F = 6aU.

In the same way, a sphere moving with velocity U relative to the uid around it experi-ences the same drag force.

4.3 Microscopic model

4.3.1 Bead-spring dumbbell

The dynamics of a single sphere in a viscous uid are not very interesting. Interactionsbetween them can be. . . but thats a whole other research area.

In order to model a polymer, we need something which can in some way remember theows it has been through. We do this by creating something which can change its physicalconguration as it responds to ow: then when the ow stops, its conguration is theway it carries the information about the ow history. This is exactly what real polymersdo: but our model will be much simpler than real polymers.

We take two spherical beads, and connect them together with a linear spring. This isthe simplest possible deformable object. The spring is intended to represent the entropictendency of the polymer coil to end up in a random, relaxed state. So if you stretch outthe coil, it will tend to recoil: the spring tends to bring the beads back together.

Now there are three effects acting on each bead:

The spring force

A drag force if it is moving relative to the uid around it

Brownian motion

We will build up the behaviour of the whole dumbbell bit by bit.

4.3.2 Separate beads in deterministic motion

Let us just look at one bead of radius a. Well choose the radius to represent the typicalsize of a relaxed polymer coil. Now say its instantaneously at a position x; and supposethe other bead is instantaneously at x + r . Well look at what it does under the actionof a uid ow and the spring force and nothing else.

Because it is very small, it has no inertia, and Newtons second law becomes a forcebalance: total force on our bead is zero.

The spring force pulling it towards the other bead is r . is just a spring constant.Because it comes from thermal forces, we know its size:

= 3kT

a2

18


18/26

which also depends on the typical size of the relaxed polymer coil.

If the bead at x is moving with velocity U 1 , and the uid around it has velocity u(x),the drag force on the bead will be

6a[u(x) U 1 ].

The total force on the bead (which must be zero) is

3kT a2

r + 6 a[u(x) U 1 ] = 0

so we can work out the velocity of the bead:

U 1 = kT

2a3 r + u(x).

The other bead is positioned at x + r and the spring force on it is r so its velocity is

U 2 = kT 2a 3

r + u(x + r).

In the absence of Brownian motion, these two velocities tell us how the bead positionsevolve:

dxdt

= kT 2a 3

r + u(x)

d(x + r)dt

= kT 2a 3

r + u(x + r)

Finally we subtract the two to get the evolution of r :

drdt

= kT a 3

r + u(x + r) u(x)

Since r is microscopically small relative to the scale over which u changes, we can use aTaylor series for u(x + r):drdt

= kT a3

r + r u(x)

which is correct up to terms of order r.

The constant 2 kT/a 3 is called 1/ ; it represents the inverse relaxation time of thedumbbell following distortion caused by the ow.

dr

dt =

1

2 r + r u.

19


19/26

4.3.3 Stress in the uid

Now suppose we have a suspension containing many of these dumbbells. What will theextra stress they contribute be? Of course, the uid they are suspended in, which hasits own viscosity , will continue to contribute a Newtonian stress; were just looking forthe polymer extra stress , p.Again, we look at the tiny surface element with area S and unit normal n. We areinterested in dumbbells that cross the surface.

n

The force associated with a single dumbbell that crosses the surface is r ; how many willcross the surface?

Clearly a long dumbbell is more likely to cross the surface than a short one; and adumbbell which is lined up with n is more likely to cross the surface than one whichis perpendicular to it. If there are m dumbbells per unit volume, then we expect thenumber crossing our surface element to be

mr n S.

Then the extra stress exerted by the dumbbells is

p n S = r (mr n S ) so p = mrr = 3am2

rr.

Again, we introduce a name for the new constant: this time, G = 3am/ 2 so that

p = Grr.

4.3.4 Adding Brownian motion

We have worked out how a dumbbell will evolve if its motion is deterministic, and the

stress it will cause. But to complete the model, we need to add Brownian motion of thetwo beads.

We simply add a standard 3D Brownian motion (i.e. each of the x, y, z components is aBrownian motion) to the evolution of the vector r:

dr = 12

r + r u(x) dt + 1 / 2 dB t

We have chosen 1 / 2 as the coefficient for two reasons: it has the correct dimension; butalso because (as we will see later) this leads to a dumbbell with no ow having a natural

length of 1.20


20/26

Because we eventually want to work out the extra stress, we then dene the conforma-tion tensor

A(x, t ) = IE [ r (x, t )r (x, t )]

which will give us

p

= GA.Now we use the rst-step method (see GM01 part 1). As we take a time-step dt, theposition x of our dumbbell will move to

x + dx = x + u(x)dt.

So when we take our time-step, we get

A(x + u(x)dt,t + dt) = IE [( r + dr)(r + dr)]= IE [ r r + r dr + dr r + dr dr ]

A(x + u(x)dt,t + dt) = IE [ r r ] + IE [r dr ] + IE [dr r ] + IE [dr dr ]

Now lets look at these four terms separately. As usual, we only keep terms up to orderdt.

IE [r r ] = A(x, t )

IE [r dr ] = IE r 12

r + r u(x) dt + 1 / 2 dB t

= IE r 12

r + r u(x) dt

= 12

Adt + A udt

IE [dr r ] = IE 12

r + r u(x) dt + 1 / 2 dB t r

= IE 12

r + r u(x) dt r

= 12

Adt + ( u) Adt

IE [drdr ] = IE r

2

+ r u dt + 1 / 2 dB t r

2

+ r u dt + 1 / 2 dB t

= IE 1 / 2 dB t 1 / 2 dB t

= 1 Idt

We also need to look at the left-hand side term.

A(x + u(x)dt,t + dt) = A(x + u(x)dt,t + dt) A(x, t + dt) + A(x, t + dt)= ( udt )A + A(x, t + dt)= ( udt )A + A(x, t + dt) A(x, t ) + A(x, t )

= ( u )Adt +A

t dt + A(x, t )

21


21/26

Putting these all together gives:

At

+ ( u )A dt = 12

Adt + A udt 12

Adt + ( u) Adt + 1 Idt

At

+ ( u )A A u ( u) A = 1

A I

and the polymer extra stress is p = GA.

The terms on the left-hand side of the A equation are called the co-deformational timederivative because they represent what happens to a line element that is deforming withthe ow.

We have now derived the Oldroyd-B equations :

u = 0

ut

+ u u =

= pI + u + ( u) + GA

At

+ ( u )A A u ( u) A = 1

A I

In the nal lecture we will see how the uid corresponding to these equations behaves.

22


22/26

5 The Oldroyd-B uid

Last time we started from a microscopic dumbbell with a linear entropic spring, andderived the Oldroyd-B equations:

u = 0 (1)

ut

+ u u = (2)

= pI + u + ( u) + GA (3)

At

+ ( u )A A u ( u) A = 1

A I (4)

Note that since A = IE[r r ], it must be symmetric.

5.1 Shear ow

Suppose we make our uid carry out an unsteady shear ow:

u = ( (t)y, 0, 0).

If the forcing all depends on y and t only, we expect all the physical variables only todepend on y and t. The mass conservation equation (1) is satised. The momentumequation (2) becomes

ut

= .

Now the tensor u is u = 0 0 (t) 0

so (3) gives

= p (t) (t) p + GA

and (4) becomes

t

Axx AxyAxy Ayy

A xy 0A yy 0

A xy A yy0 0 =

1

Axx 1 AxyAxy Ayy 1

5.1.1 Steady shear ow

Now set as a constant. This means that all the variables should be independent of t:

= 0.

= p

p + GA

23


23/26

A xy 0A yy 0

A xy A yy0 0 =

1

Axx 1 AxyAxy Ayy 1

Lets look at the last equation, for the components of A. It gives three scalar equations:

A xy A xy = 1

(Axx 1)

A yy = 1 Axy

0 = Ayy 1.

Solving from the bottom up gives

Ayy = 1 Axy = Axx = 1 + 2 2 2.

The total stress becomes

= p + G + 2G 2 2 ( + G )

( + G ) p + G .

The presence of the polymers has made two changes to the Newtonian stress:

The viscosity is increased from to + G

There is a difference between the two diagonal stress components.

This difference is called the rst normal stress difference

N 1 = xx yy = 2G 2 2.

As a stress acting along the ow lines (xx-direction) it has the opposite sign to pressureso it acts like a tension: the streamlines are like stretched rubber bands. It is the drivingforce behind the rod-climbing experiment:

24


24/26

This picture is from Boger & Walters book Rheological Phenomena in Focus. Therod in the middle is rotated, causing a shear ow round the outside. The streamlines arecircular, so their tension causes the uid to move to the middle and the only place itcan go is up the rod.

5.1.2 Linear rheology

Lets return to the time-dependent shear ow:

u = ( (t)y, 0, 0).

ut

= ; = p (t) (t) p + GA

and t

Axx AxyAxy Ayy

A xy 0A yy 0

A xy A yy0 0 =

1

Axx 1 AxyAxy Ayy 1

For the standard linear rheology experiment, we set

(t) = cos (t).

Well assume the motion starts at t = 0. Before that there is no ow so the uid wasrelaxed and A = I .

Lets look at the evolution of A. To get the rheology, we will only need Axy , so well onlyuse the Axy and Ayy equations.

Axyt A yy = 1 Axy

Ayyt

= 1 (Ayy 1).

The second of these is satsed by Ayy = 1, so the rst gives

Axyt

cos (t) = 1 Axy

Axyt

+ 1 Axy = cos (t)

t Axy e

(t/ )

= cos (t)e(t/ )

This is one of those tricky integrals that works iteratively by parts:

I = t

0 cos (t )e(t / ) dt

= [ cos (t )e(t / )]t0 + t

02 sin (t )e(t / ) dt

= [ cos (t )e(t / )]t0 + [2 2 sin(t )e(t / )]t0

t

03 2 cos(t )e(t / ) dt

= cos(t)e(t/ )

+ 2

2

sin(t)e(t/ )

2

2

I 25


25/26

Finally the expression for Axy is

Axy = 1

1 + 2 2 cos(t) + 2 2 sin(t) e (t/ )

For long times, the e (t/ )

term becomes very small so we will neglect it. The total shearstress xy then becomes

xy = + G

1 + 2 2 (t) +

G2 2

1 + 2 2 (t)

Thus the linear rheology functions for the Oldroyd-B uid are

G = G2 2

1 + 2 2 G = +

G 1 + 2 2

.

This is just like the single exponential relaxation uid if we set = 0; that would give the

Upper Convected Maxwell model. With a nonzero viscosity, although the relaxationtime is still , the storage and loss modulus no longer cross at = 1.

5.2 Extensional ow

Finally, we look at a steady 2D extensional ow:

u = ( x, y).

Again, this satises mass conservation. This time we have

u = 00 .

The stress is =

p + 2 00 p 2 + GA

and the evolution of A (independent of time and position) becomes

Axx AxyAxy Ayy

Axx Axy Axy Ayy

= 1

A I

soAxx =

1(1 2 )

Axy = 0 Ayy = 1

(1 + 2 )The total stress is

= p + 2 + G/ (1 2 ) 0

0 p 2 + G/ (1 + 2 ) .

Since in a Newtonian uid we have

= p + 2 0

0 p 2 26


26/26

we can dene an extensional viscosity as

ext = xx yy

4 .

The Oldroyd-B uid has extensional viscosity

ext = + G

(1 42 2)

This strain-hardens (gets thicker with increasing speed) for low strain rates but for higherstrain rates disaster strikes:

The viscosity diverges at a strain rate of = 1/ 2 and for strain rates slightly larger, theviscosity value is negative!

The moral of this: a linear spring is ne for shear ows, where the stretch is fairlymoderate; for stretching ows, a linear spring can stretch indenitely and give inniteforces. The standard workaround at this point is to use a nonlinear spring law (FENEmodel, nie extensibility nonlinear elasticity) which brings with it its own complications.

There is no single right answer to polymer modelling but hopefully you now have anidea about how to start!

27

Polymeric Fluids Lecture

Documents

Transcript of Polymeric Fluids Lecture