robertson nonnewtoniannotes

7/31/2019 Robertson NonNewtonianNotes

1/85

Lecture Notes on Non-Newtonian Fluids

Part I: Inelastic Fluids

Professor A. M. RobertsonDepartment of Mechanical EngineeringUniversity of Pittsburgh

Fall 2005

1


2/85

2


3/85

Contents

1 Background Material on Continuum Mechanics 5

1.1 Description of Motion of Material Points in a Body . . . . . . . . . . . . . . . . . . . 51.2 Referential and Spatial Descriptions . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3 Velocity and Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Material Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.5 Deformation Gradient of the Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.6 Deformation Measures. Strain Measures . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.6.1 Right Cauchy Green Deformation Tensor . . . . . . . . . . . . . . . . . . . . 101.6.2 Relative Strain Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.7 Velocity Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.7.1 Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.8 Special Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.8.1 Rigid Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.8.2 Isochoric Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.8.3 Irrotational Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.8.4 Simple Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.8.5 Uniaxial Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.9 Governing Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.9.1 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.9.2 Balance of Linear Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.9.3 Balance of Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . 201.9.4 Mechanical Energy Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.10 Invariance Under Superposed Rigid Body Motions . . . . . . . . . . . . . . . . . . . 221.10.1 Relationship between kinematic variables in the original and superposed mo-

tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.10.2 Requirement: Invariance under a superposed rigid body motion . . . . . . . . 28

2 Non-Linear Viscous Fluids 312.1 Constitutive Assumption T = T(, L) . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.1.1 Incompressible Inelastic Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . 332.2 Reiner-Rivlin Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.2.1 Thermodynamic Restrictions on 1 and 2 for Reiner-Rivlin Fluids . . . . . . 352.2.2 Restrictions on 1, 2 based on behavior of real fluids in simple shear . . . . 36

3


4/85

2.3 Generalized Newtonian Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.3.1 Power-Law Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.3.2 Prandtl-Eyring Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.3.3 Powell-Eyring Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.3.4 Cross Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.3.5 Yasuda Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.3.6 Bingham Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.3.7 Modified Bingham Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.4 Steady, Fully Developed Flow of a Generalized Newtonian Fluid in a Straight Pipe . 452.4.1 Power-Law Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.4.2 Bingham Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.4.3 Q as a function of and (S) for unspecified generalized Newtonian fluid . . 482.4.4 Determination of (S) from measured values of driving pressure drop and

flow rate in a straight pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.5 Cone and Plate Flow for a Generalized Newtonian Fluid . . . . . . . . . . . . . . . . 502.5.1 Small Angle Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 542.5.2 Finite Cone Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

A The Relationship between Coordinate Free Notation and Indicial Notation 61

B Isotropic Tensors 63

C Isotropic Tensor Functions 67C.1 Scalar valued isotropic tensor functions . . . . . . . . . . . . . . . . . . . . . . . . . . 67C.2 Symmetric Isotropic Tensor Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 68

D Material Line Elements, Area Elements, Volume Elements 69

E Rectangular Coordinates 71

F Cylindrical Coordinates 75

G Spherical Coordinates 79

4


5/85

Chapter 1

Background Material on Continuum

Mechanics

1.1 Description of Motion of Material Points in a Body

We will first consider a physical body, for example a body of fluid, which we identify by the symbolB. We use the symbol Y to identify an arbitrary material particle in the body. The motion ofarbitrary material particles Y can then be described through a function,

x = (Y, t) (1.1)

where x is the vector identifying the spatial position at time t of the material particle identified

by Y. In continuum mechanics, we are interested in describing the position in time of an uncount-able number of points, which cannot be identified using letters or numbers. With this in mind,we identify material points in the body by their position in some chosen reference configuration.Namely, we assume the body is smooth and can be embedded in a three dimensional Euclideanspace. The mapping of each material particle of the body to Euclidean 3-space at an arbitrary timet will be called the configuration ofB at time t, denoted by (t). This mapping is assumed to beone-to-one, invertible and differentiable as many times as necessary for all time. The configuration,0, is a special configuration, chosen for the purpose of identifying material particles. For example,0 could be the configuration of the body at time zero.

We denote the region occupied by the entire body in this configuration as R0 with closedboundary R0. An arbitrary material region within B in the configuration 0 will be denoted as

V0 (

V0

R0) with boundary

V0, Fig. 1.1. We denote the corresponding volume in the current

configuration (t) as V(V R) with boundary V.An arbitrary material point identified by Y may also be identified by its position, denoted by

X, in the reference configuration 0. It is not necessary that the configuration of the body everhave coincided with the reference configuration. The motion of material points in the body B maythen be described through the relationship,

x = 0

(X, t). (1.2)

5


6/85

Figure 1.1: Schematic of notation used to identify regions of an arbitrary body B in the referenceand current configurations .

Figure 1.2: Schematic of position vector in reference and current configurations for a fixed materialpoint of an arbitrary body B.

6


7/85

The vector x is the position at time t of an arbitrary material particle will then be identified byits position X in reference configuration . We will assume that the transformation (1.2) possesses

an inverse,X = 1

0(x, t). (1.3)

For our current purposes it suffices to define one reference configuration for the body, so in furtherdiscussions we will drop the subscript 0 and it will be understood that the functions (X, t) and1(x, t) depend on the choice of reference configuration. In the discussion of viscoelastic fluidswith fading memory, we will use the current configuration as the reference configuration. Therelevant kinematics will be discussed later in the text.

1.2 Referential and Spatial DescriptionsField variables such as density, , may be written as a function ofX and t,

= (X, t). (1.4)

The function, (1.4) is called the referential or Lagrangian description of the density field. Whereas,(1.4) is sufficient to describe the density of all material points in the body, it is frequently of interestin classical fluid mechanics to have knowledge of field variables as a function of spatial positions xand time,

= (x, t). (1.5)

The function in (1.5) is called the spatial or Eulerian description of density and is independent ofinformation about the position of individual material particles. Rather, the function given in (1.5)gives the value of the density at a specified position in space at a given time t. The representations(1.4) and (1.5) are related through,

= (x, t) = ((X, t), t) = (X, t). (1.6)

1.3 Velocity and Acceleration

The velocity of a material particle is defined as

v =(X, t)

t(1.7)

and the acceleration,

a =2(X, t)

t2. (1.8)

7


8/85

Figure 1.3: Relationship between Eulerian and Lagrangian representation of density

As for the density field, the relationship between the Eulerian representation and Lagrangian rep-resentation of the velocity field can be obtained using (1.2),

v = v(x, t) = v((X, t), t) = v(X, t). (1.9)

1.4 Material Derivative

The material derivative of a field variable such as density, is defined as the partial derivative of thefunction with respect to time holding the material point fixed,

D

Dt=

(X, t)

t. (1.10)

Frequently, in fluid mechanics, the spatial description of the field variables is of interest rather thanthe material description. Using the chain rule, the material derivative can be written for the spatialformulation of a field variable. For example, in the case of the density field,

DDt

= (x, t)t

+ vi(x, t)xi

. (1.11)

Using (1.11), we may obtain the acceleration (1.8) from the spatial representation of the velocityfield,

ai =vi(x, t)

t+ vj

vi(x, t)

xj. (1.12)

8


9/85

1.5 Deformation Gradient of the Motion

The deformation gradient or displacement gradient of the motion relative to the reference configu-ration 0 is denoted by F. It is a second order tensor defined by,

F =(X, t)

X. (1.13)

The components ofF are then,

FiA =i(X, t)

XA. (1.14)

When the deformation gradient is given as a second order tensor field defined on

R, then the

deformation field can be determined from (1.14) which is a system of nine partial differentialequations for three unknowns xi. In order that there exists a solution, the integrability condition,

FiAXB

=FiBXA

(1.15)

must be satisfied. The deformation can then be determined to within an arbitrary constant, corre-sponding to an arbitrary rotation,

xi =

FiAdXA. (1.16)

To understand the physical significance of F it is helpful to consider the transformation of twomaterial points during the motion (1.2), which occupy positions Xo and X in the reference config-uration and positions xo and x in the current configuration, respectively. Using a Taylors seriesexpansion, we see that,

xi = xoi +i(X, t)

XA|X

o

(XA XoA) + higher order terms. (1.17)

or in index free notation,

x = xo + F(Xo, t)

(X

Xo) + higher order terms. (1.18)

In which the distance between X and Xo goes to zero,

dx = F dX, dxi = FiAdXA, (1.19)

where dx and dX are the infinitesimal vectors (x xo) and (X Xo), respectively in the limit ofinfinitesimal distance between points

9


10/85

1.6 Deformation Measures. Strain Measures

1.6.1 Right Cauchy Green Deformation TensorWe now consider the change in magnitude dx during the deformation. Well denote the magnitudeof dx and dX as ds and dS, respectively. Let m and M be the unit vectors tangent to dx and dX,respectively. The square of the magnitude of the infinitesimal vector dx is related to the square ofthe magnitude of dX through,

ds2 = |dx|2 = dx dx = dxidxi= FiAFiB MA MB dS

2

= CAB(X, t) MA MB dS2,

(1.20)

where CAB are the components of the right Cauchy-Green deformation tensor:

C FT F and CAB FiA FiB . (1.21)

Note that C is symmetric.

Physical Significance of Diagonal elements of C

Consider an infinitesimal element represented by dX in the reference configuration and dx in thecurrent configuration. If we now consider the special case where dX = dSe1, then it follows from(1.20), that

ds2

dS2= C11. (1.22)

The ratio ds2/dS2 is typically called the stretch ratio and denoted by . We can describe the result(1.22) in words as,

C11 is equal to the stretch ratio squared of an infinitesimal material elementwhich was aligned with the e1 axis in the reference configuration.

The other diagonal elements can be interpreted in a similar way.

Physical significance of off-diagonal elements of C

Now consider two infinitesimal material elements corresponding to dX(1) = dS(1)M(1) and dX(2) =dS(2)M(2), where M(1) and M(2) are unit vectors. In the current configuration, these infinitesimalelements are denoted as dx(1) = ds(1)m(1) and dx(2) = ds(2)m(2), respectively. The angle betweenthese same material elements in the current configuration will be denoted as . Therefore,

cos =dx(1) dx(2)

ds(1)ds(2)=

FiAFiBM(1)A M

(2)B dS

(1)dS(2)

ds(1)ds(2)=

CABM(1)A M

(2)B

(1) (2). (1.23)

10


11/85

Now consider the special case, where the infinitesimal elements are perpendicular in the referenceconfiguration, for example, M(1) = e1 and M

(2) = e2, in which case, (1.23) reduces to,

cos =C12

C11C22. (1.24)

We can describe in words, the physical significance of

C12 provides information about the angle between two material elementswhich were parallel to e1 and e2 in the reference configuration.

The other off-diagonal elements can be interpreted in a similar way. For example, if the anglebetween the elements remains unchanged (still 90o), then C12 = 0. If the angle decreases from 90

o,

then C12 will be a positive number. If the (90o, 270o), then C12 will be negative.

Jacobian of the transformation

In future discussions, we denote the the Jacobian of the transformation (1.2) by J. Namely, theJacobian is equivalent to the determinant of F,

J det i(X, t)XA

= det(F) . (1.25)

Since (X, t) is assumed to have an inverse,

0 < J < . (1.26)

Recall from continuum mechanics that the relationship between the infinitesimal volume occupiedby material points in the current configuration dv is related to the infinitesimal volume occupiedby the same material points in the reference configuration dV is,

dv = J dV. (1.27)

It will be useful in the discussion of the governing equations to note the following result for the

material derivative of the Jacobian of the transformation (1.2),

DJ

Dt= Jdivv, (1.28)

where v is the velocity vector.

11


12/85

1.6.2 Relative Strain Tensor

When the motion is rigid, C is equal to the identity tensor. It is useful to define a relative measureof strain which vanishes for rigid motions:

E 12

(C I), (1.29)

where E, is often referred to as the Lagrangian strain. A second measure of relative strain, e,defined as

e 12

(I B1) (1.30)

is referred to as the Eulerian strain where B is the left Cauchy-Green tensor

B F FT

and Bij FiA FjA . (1.31)

1.7 Velocity Gradient

We denote L as the gradient of the spatial form of the velocity vector, so that the components ofL with respect to rectangular coordinates are 1

Lij =vi(x, t)

xj. (1.32)

Recall that any second order tensor can be decomposed into the sum of a symmetric and skew

symmetric second order tensor. We can represent L in this way,

Lij =1

2

vixj

+vjxi

+

1

2

vixj

vjxi

. (1.33)

We define the rate of deformation tensor, D, as the symmetric part of L and the vorticity tensor,W, as the skew-symmetric part of L,

Dij =1

2

vixj

+vjxi

, Wij =

1

2

vixj

vjxi

. (1.34)

Physical Significance of Diagonal elements of DThe physical significance of D can be studied by considering the rate of change in magnitude of aninfinitesimal material element dx of length ds. We first consider the rate of chance of ds2 which isequal to

D(ds2)

Dt= 2 dxi

DdxiDt

. (1.35)

1Note that in some books, the gradient of a second order tensor is defined as the transpose of that used here.

12


13/85

The rate of change in the infinitesimal material element dx is,

DdxiDt

= DDt

FiAdXA

=FiA

tdXA

=2i(X, t)

XA tdXA

=vi

XAdXA

=vixj

FjA dXA

= Lijdxj

(1.36)

and so from (1.35) and (1.36) we find,

D(ds2)

Dt= 2 Dijdxidxj. (1.37)

Thus,

Dds

Dt=

Dijdxidxjds

. (1.38)

We see from (1.38) that the rate of change of magnitude of an infinitesimal element which at timet is parallel to the e1 axis is

Dds

Dt= D11ds. (1.39)

D11 is the rate of change of ds divided by ds of a material element which attime t is aligned with the e1 axis.

The other diagonal elements can be interpreted in a similar way.

Physical Significance of the off diagonal elements of D

Now consider two infinitesimal material elements x and y which intersect at angle with lengths|dx| and |dy| respectively. Then,

cos =dx dy|dx| |dy| (1.40)

13


14/85

and therefore,

DDt

cos = DDt

dx dy|dx| |dy|

=D

Dt

dx dy 1|dx| |dy| dx dy|dx| |dy| DDt |dx| |dy|

=Dijdxidyj|dx| |dx|

dx dy|dx| |dy|

D

Dt

|dx| |dy|(1.41)

In particular, consider (1.41) for two infinitesimal material elements, dx, dy which at time t areparallel to the base vectors e1 and e2. We see that the rate of change of the angle between thesetwo vectors at time t is

DDt

= D12. (1.42)

Therefore,

D12(x, t) is the rate of change of angle between the two infinitesimal vectors whichat time t are located at position x and parallel to base vectors, e1 and e2.

The other off-diagonal elements can be interpreted in a similar way. Notice that this interpretationof the components of D does not require knowledge of the behavior of specific material elements.Rather D(x, t) is related to the rate of change of material elements which at time t are located atposition x.

1.7.1 Vorticity

Recall the definition of the vorticity, is the curl of the velocity vector,

v, (1.43)

or in indicial notation 2,

i ijk vkxj

. (1.44)

The components of the vorticity vector are related to the components of the vorticity tensor through,

i = ijk Wjk , (1.45)and

Wij = 12

ijk k. (1.46)

2Note that alternate definitions of vorticity are sometimes used. For example, sometimes the vorticity is definedas the negative of that given in (1.43). In other cases, the vorticity is taken to be twice that in (1.43).

14


15/85

1.8 Special Motions

1.8.1 Rigid MotionsA rigid motion is one in which the distance between points remains constant. Therefore, for rigidmotions, the material derivative of ds is zero for all points in the body for all time during the rigidmotion. We see from (1.38) that for rigid motions, D must be identically zero at all points in thebody for all time. Alternatively, we see from (1.38) that motions in which D is zero at all pointsin the body for all time are rigid motions.

One can show that the most general rigid motion can be written as,

x = xo(t) + Q(t) X, (1.47)where Q is a proper orthogonal second order tensor. Namely,

Q QT = I, and therefore, QT Q = I. (1.48)In indicial notation,

QijQkj = ik, and therefore, QjiQjk = ik. (1.49)

In preparation for considering the form of the velocity field for a rigid body motion, we will firstconsider the time derivative of Q. If we take the derivative with respect to time of equation, (1.48),we find,

dQ

dt QT + Q dQ

T

dt= 0, (1.50)

and so,

dQdt

QT = ( dQdt

QT)T (1.51)

We see from (1.51) that the quantity dQ/dt QT is a skew symmetric tensor, which we denote as, so

dQ

dt= Q. (1.52)

Since is skew symmetric there exists a vector which we will denote as c, where,

ij = ijkck. (1.53)

The velocity field for a rigid motion can be obtained by taking the material derivative of (1.47),

v = vo(t) + c (x xo), (1.54)where vo is the time derivative of xo. If we take the curl of (1.54), we find that,

= 2 c. (1.55)

Therefore, the velocity field for the most general rigid body motion can be written as,

v = vo(t) +1

2 (x xo). (1.56)

15


16/85

1.8.2 Isochoric Motions

Isochoric motions are those in which the volume occupied by fixed material particles is unchangedduring the motion. A material does not have to be incompressible to undergo isochoric motions.As will be discussed in the next section, all motions undergone by incompressible fluids must beisochoric. We see from (1.27) that if a motion is isochoric then the value of J is one throughoutthe motion. In this case, we have from, (1.28) that the divergence ofv is equal to zero,

v = 0 (1.57)and hence the trace of the rate of deformation tensor is zero.

1.8.3 Irrotational Motions

Irrotational motions are those for which the vorticity is zero. From (1.46), the vorticity tensor Wis zero for irrotational motions.

1.8.4 Simple Shear

A flow field that is of great significance in fluid mechanics is simple shear flow, sometimes calledlineal Couette flow. In this flow, the velocity field is steady, fully developed and uni-directional.The magnitude of the velocity depends linearly on the spatial component with axis perpendicularto the direction of flow. For example, if rectangular coordinates xi are chosen such that the flowdirections is parallel to the x1 axis, then the orientation and origen of the coordinate system canbe chosen such that the velocity field can be written as,

v = x2 e1. (1.58)

Clearly simple shear is an example of an isochoric motion. It is possible to show (see later chapters)that simple shear flow can be generated between parallel plates for a wide class of fluids called simplefluids (see, e.g. [8]). As such, it has great value in the design of rheometers. For example, considersimple shear flow between two parallel plates separated by a distance h, when the upper plate ismoving with speed U in the e1 direction and the origin of the coordinate system coincides witha point on the bottom plate, Figure 1.4. Using the no-slip boundary conditions at the top andbottom surface it can be shown that = U/h. Using the result (1.7) and the initial conditionx = X at t = 0, we can show that the motion corresponding to (1.58) is,

x1 = X1 + X2 t, x2 = X1, x3 = X3. (1.59)

It is helpful for discussions in later chapters, to calculate several kinematic variables for simple shear.In particular, using the results (1.58) and (1.59) for simple shear, it follows that the correspondingvalues of D, W are,

[D] =

2

0 1 01 0 0

0 0 0

, [W] =

2

0 1 01 0 0

0 0 0

. (1.60)

16


17/85

Figure 1.4: Schematic of velocity field in simple shear flow.

while F , C and E are

[F] =

1 t 00 1 0

0 0 1

, [C] =

1 t 0t 1 + 2t2 0

0 0 1

, [E] = 1

2

0 t 0t 2t2 0

0 0 0

.(1.61)

1.8.5 Uniaxial Extension

Another relatively simple flow field which is of particular interest for viscoelastic fluids is uniaxialextension. The Eulerian representation for the velocity field for this motion is,

v1 = x1, v2 = 2

x2, v3 = 2

x3. (1.62)

As can be seen from (1.62), uniaxial extension is an isochoric motion. Using the result (1.7) and theinitial condition x = X at t = 0, the position of material particles in this motion can be describedthrough,

x1 = X1 et, x2 = X2 e

t/2, x3 = X3 et/2. (1.63)

Using the results (1.62) and (1.63), the corresponding values of D, W are,

[D] = 1 0 0

1 1/2 00 0 1/2

, [W] =

2 0 0 0

0 0 00 0 0

. (1.64)

while F , C and E are

[F] =

et 0 00 et/2 0

0 0 et/2

, [C] =

e2t 0 00 et 0

0 0 e

, (1.65)

17


18/85

and

[E] =

1

2 e

2t 1 0 00 e

t

1 00 0 e 1

. (1.66)

1.9 Governing Equations

1.9.1 Conservation of Mass

The mass M of a fixed subset of material particles of the body occupying a region V(t) ofR(t) is

M = V(t)

dv, (1.67)

where is the mass density of the fluid. The volume V(t) is a material volume of the fluid: thevolume occupied by a fixed set of material particles in the body, that may change in time. Theprinciple of conservation of mass is the postulate that the mass of this fixed set of material particlesdoes not change in time,

d

dt

V(t)

dv = 0. (1.68)

Making use of the Transport Theorem, we can write the principle of conservation of mass withrespect to the spatial (Eulerian) representation of the field variables,

0 =

V(t)

t+ v + v

dv, (1.69)

where v is the velocity vector. Making suitable assumptions about continuity of the field variablesand making use of the arbitrariness of the part of the body V(t), we obtain the local form of (1.69),

t+ v () + v = 0 (1.70)

or in indicial notation,

0 =

(x, t)

t + vi(x, t)

xi +

vi(x, t)

xi . (1.71)

Implications of Conservation of Mass for Incompressible FluidsThe motion of an incompressible fluid is constrained to be isochoric, in which case the divergenceof the velocity vector must be zero for all motions. In this case, the conservation of mass reducesto,

t+ v () = 0. (1.72)

18


19/85

From (1.72) we see that for an incompressible fluid, the material derivative of the density is alwayszero,

DDt

= 0. (1.73)

Note that the even thought the material derivative of the density it is zero, it is not necessarythat the density be constant in space. For example, a stratified fluid with density distribution, = 0 + y, where is a constant, can experience simple shear, Section 1.8.4. It is easily seenthat this motion is isochoric, a necessary condition for an incompressible fluid to experience thismotion. In addition,

D

Dt=

t+

xivi = 0 (1.74)

for the stratified fluid and given motion and therefore conservation of mass is satisfied.

1.9.2 Balance of Linear Momentum

The postulate of balance of linear momentum is the statement that the rate of change of linearmomentum of a fixed mass of the body is equal to the sum of the forces acting on the body. Theseforces can show up as body forces, or as forces due to stress vectors acting on the surface of thebody,

d

dt

V(t)

vdv =

V(t)

b d v +

dV(t)

t da (1.75)

where b is the body force per unit mass, t(x,n,t) is the surface force acting on the body in thecurrent configuration per unit area of Pand n is the unit normal to the surface Pat x at timet. Recall that the stress vector is in general a function of the surface unit normal n. The first and

second integrals on the right hand side of (1.75) represent the contributions due to body forcesand to surface forces respectively. Note that stress vector depends on position, time and the unitnormal to the surface at x. The first and second integrals on the right hand side of (1.75) representthe contributions due to body forces and to surface forces respectively. Recall, Cauchys Lemmawhich follows from suitable smoothness assumptions on the function t(x,n,t) and the balance oflinear momentum

t(x,n,t) = t(x, n, t). (1.76)In addition, the existence of a second order tensor, T can be shown, where,

t = t(x,n,t) = T(x, t)n. (1.77)

The second order tensor T is called the Cauchy stress tensor. Significantly, T is independent of then.

Making use of the divergence theorem as well as the relationship between the stress tensor andthe stress vector, (1.75) can be written as,

d

dt

V(t)

vdv =

V(t)

(b + T n) dv. (1.78)

19


20/85

The local form of the equation of linear momentum can be obtained form (1.78) by using thetransport theorem and making suitable assumptions about the continuity of the field variables,

v

t+ (v) v

= T + b, (1.79)

or, using index notation,

vit

+ vjvixj

=

Tijxj

+ bi. (1.80)

It is often convenient to represent T as the sum of a deviatoric and spherical part.

T = + t I, (1.81)

where,

ii = 0, and t = 13

Tkk. (1.82)

The tensor is often referred to as the deviatoric part ofT and tI as the spherical part. When theCauchy stress tensor is decomposed in this way, t is often called the pressure and denoted by p.Using the decomposition, (1.81), the balance of linear momentum can be written as,

v

t+ (v) v

= p + + b. (1.83)

Alternatively,

vi

t

+ vjvi

xj =

p

xi

+ij

xj

+ bi. (1.84)

For compressible fluids, p is a thermodynamic pressure. An equation of state relating pressure toother thermodynamic variables such as mass density and temperature will be necessary. For in-compressible fluids, p is a mechanical pressure arising from the constraint of incompressibility. Noequation of state is necessary, rather, p will be determined as part of the solution to the governingequations and boundary conditions.

1.9.3 Balance of Angular Momentum

In the absence of body couples, the integral form of the balance of angular momentum is

d

dt

V(t)

x v dv =V(t)

x b dv +

dV(t)x tda. (1.85)

Making use of the balance of linear momentum, it can be shown that (1.85) reduces to the require-ment that the Cauchy stress tensor must be symmetric,

Tij = Tji. (1.86)

20


21/85

1.9.4 Mechanical Energy Equation

It is sometimes useful to consider the Mechanical Energy Equation which is not derived from theequation of balance of energy, rather it is obtained from the equation of linear momentum makinguse of the equation of conservation of mass. If we take the inner product of the velocity vector andthe equation of linear momentum (1.80) we obtain,

1

2

D

Dt(vivi) =

Tijxj

vi + bivi. (1.87)

It is sometimes useful to have an integral form of this equation. This can be obtained by integrating(1.87) over a fixed part of the fluid (same material particles) occupying region V(t) with surfaceV(t), to obtain,

12

ddt

V(t)

v vdv = V(t)

t vda V(t)

tr(T DT)dv + V(t)

b vdv. (1.88)

where we have made use of the divergence theorem and the conservation of mass. For futurereference, we introduce the following notation for terms in (1.88):

K 12

V(t)

v vdv = kinetic energy in V(t)

Rc V(t)

t vda = rate of work done by surface forces on the boundary V(t)

Rb V(t)

b vdv = rate of work done on the material volume V(t) by body forces

(1.89)

The scalar, tr(T DT) = TijDij is the rate of work by stresses per unit volume of the body andis called the stress power. If we consider a part of the body V(t) in the current configuration, wecan hypothesize the existence of a scalar called the specific internal energy, u = u(x, t) (internalenergy per unit mass). The internal energy for the part V(t) of the body, will then be

V(t)

udv. (1.90)

Recall that the kinetic energy of the part V(t)of the body isV(t)

1

2v vdv. (1.91)

Heat may enter the body through the surface dV(t) of the body with outward unit normal n. Itcan be shown, that this heat flux can be represented as the scalar product of a vector q and thenormal to the surface n. Where q n positive is associated with heat leaving the surface and q n

21


22/85

negative is associated with heat entering the surface. In addition, heat may enter the body as aspecific heat supply, r = r(x, t): the heat entering the body per unit mass per unit time. Therefore

the rate of heat entering the part V(t) of the body is

dV(t)q nda +

V(t)

rdv. (1.92)

The balance of energy is a statement that the rate of increase in internal energy and kinetic energyin the part V(t) of the body is equal to the rate of work by body forces and contact forces plusenergies due to heat entering the body per unit time. We can write this statement as

d

dt

V(t)

(u +1

2v v)dv =

dV(t)

t vda +V(t)

b vdv

dV(t)q nda +

V(t)

rdv (1.93)

where the first integral on the left hand side is the rate of work by contact forces, the secondintegral is the rate of work by body forces. Making suitable assumptions about continuity of thefield variables, we can obtain the local form of (1.93),

Du

Dt+ v Dv

Dt

= T : D + v ( T) + v b q + r. (1.94)

Using results from the mechanical energy equation, we can rewrite (1.94) as

Du

Dt= T : D q + r, (1.95)

or in indicial form

Du

Dt= TijDji qi

xi+ r. (1.96)

1.10 Invariance Under Superposed Rigid Body Motions

A variety of criteria have been used to assess whether a proposed constitutive equation is rea-sonable. In addition to assessing how well a constitutive equation models the behavior of a realfluid for a chosen flow, criteria such as invariance of the constitutive equation under a superposedrigid motion (or, alternatively, material frame indifference3) [30, 14]; thermodynamic considera-tions [28, 29, 18]; stability of the rest state and stability of some steady flows [10, 17, 16, 13]; andwell-posedeness of classes of initial value problems [6, 15, 12] have been used. For example, basedon thermodynamic restrictions, we require that the viscosity of a Newtonian fluid be non-negative.

In this section we focus on the assumption that a constitutive equation be invariant under asuperposed rigid body motion to be physically reasonable . In some works, authors apply the re-lated but different physical requirement that the form of the constitutive equation be independent

3Truesdell and Noll [30] include an interesting discussion of the history of these two principles.

22


23/85

of observer. This assumption is often referred to as the principle of material frame indifference.In both cases, an assumption is necessary to define how the stress vector (or stress tensor) in the

original and superposed motion (or between the two observers) are related. In the following subsec-tion, we discuss the kinematics associated with a general superposed motion without introducinginvariance assumptions. In the subsequent subsection, we provide a mathematical statement of therequirement that a constitutive equation be invariant under a superposed rigid body motion. Thediscussion of invariance under superposed rigid body motion follows that of Green and Naghdi, [14]and Naghdi (1972,pp. 484-486) [20]).

Superposed Rigid Body Motions

In this section, we consider two motions of a body: an arbitrary motion

x = (X, t) (1.97)

and a second motion which differs from the first by a superposed rigid body motion

x+ = +(X, t). (1.98)

Namely, the position of a material point which at time t is at position x in the first motion, is atposition x+ in the second motion, Figure 1.5. The spatial description of (1.98) is then,

x+ = +(x, t). (1.99)

Significantly, the motion (1.99) is not an arbitrary second motion. Since the second motion differs

Figure 1.5: Identification of points in the body in the current and superposed configurations.

from the first motion by a rigid body motion, the function +(x, t) must belong to a restricted class

23


24/85

of motions. In a rigid motion the distances between material points in the body are kept constant.In this section, we shown that the most general form for this function is,

x+ = +(x, t) = c(t) + Q(t) x, (1.100)

where Q(t) is a proper orthogonal second order tensor,

Q QT = I, det Q = +1. (1.101)

First consider, two material points which we can identify as P and Q. These points can also beidentified by their respective positions, X and Y, in the reference configuration. The positions ofpoints P and Q in the current configuration, denoted by x and y respectively, are

x = (X, t), y = (Y , t). (1.102)

The positions occupied by these materials points at time t in the superposed motion, (1.99), arethen,

x+ = +(x, t). y+ = +(y, t). (1.103)

By defintion, the distance between two arbitrary points in the superposed motion must be the sameas the distance between those in the motion, (1.97),

(x+ y+) (x+ y+) = (x y) (x y), (1.104)

namely,

(+(x, t) +(y, t)) (+(x, t) +(y, t) ) = (x y) (x y). (1.105)Equation (1.105) can be written in indicial notation as,

(+i (x, t) +i (y, t))(+i (x, t) +i (y, t) ) = (xi yi)(xi yi). (1.106)

Recalling that x and y are independent, we can differentiate (1.106) with respect to xj and thenwith respect to yk to obtain,

+i (x, t)

xj

+i (y, t)

yk= jk . (1.107)

Equation (1.107) can be written in coordinate free notation as,

+(x, t)

x

T

+(y, t)

y= I. (1.108)

Equivalently, +(x, t)

x

T=

+(y, t)

y

1. (1.109)

24


25/85

Equation (1.109) must hold for arbitrary choices of x and y in the body. Since the left side ofequation (1.109) is indpendent ofy and the right side is independent of x, the second order tensors

on the left and right hand side of (1.109) must be functions of t only, which we denote as thetranspose of Q(t),

Q(t) =

+(x, t)

x

, or Qij(t) =

+i (x, t)

xj. (1.110)

Since (1.110) must hold for all x in R,

Q(t) =

+(y, t)

x

(1.111)

and therefore from (1.108)

QT Q = I, and det Q = 1, (1.112)

or

QliQlj = ij . (1.113)

We see from (1.112) that Q is an orthogonal tensor. Each superposed motion, (1.99), must includethe particular case in which +(x, t) = x. It can be seen from (1.110), that for this case, Q = Iand det Q = 1. Since the motions under consideration are continuous, we must always have,

det Q = 1. (1.114)

The equation (1.110), may be integrated with respect to x to yield,

x+ = +(x, t) = c(t) + Q(t) x, or x+i = +i (x, t) = ci(t) + Qik(t)xk, (1.115)

where c is a vector function of time. Therefore, the most general superposed rigid body motion isdescribed by (1.100). From (1.115), we also have

x = QT (x+ c), xk = Qlk(x+l cl). (1.116)

So far, we have shown that necessarily a superposed rigid body motion is represented by (1.100).We now show that it is also sufficient, namely, that all motions of the form (1.100) preserve distance

between points.

|x+ y+|2 = (x+ y+) (x+ y+) = Q (x y) Q (x y)= (x y) QT Q (x y)= (x y) (x y)= |x y|2.

(1.117)

25


26/85

1.10.1 Relationship between kinematic variables in the original and superposed

motions

We now consider other kinematical quantities under the superposed rigid body motion, (1.100),Figure 1.6. The velocity vector v, transforms as,

v+ =+(X, t)

t=

dc(t)

dt+

dQ

dt x + Q v. (1.118)

We define as

=dQ

dt QT. (1.119)

Recall, that Q QT = I, sodQdt

QT + Q dQTdt

= 0 (1.120)

and hence, using (1.119) and (1.120), we find,

+ T = 0 (1.121)

and therefore, is skew-symmetric. Using the definition of , (1.119), we can write the expressionfor v+ as,

v+ =dc(t)

dt+ Q x + Q v, (1.122)

and

v+i =dci(t)

dt+ ilQlmxm + Qilvl. (1.123)

For future use, we now derive the relationship between L+ and L,

L+ij =v+ix+j

=v+ixk

xk

x+j

=

ilQlmmk + Qil

vlxk

xk

x+j

= [ilQlk + QilLlk ] Qjk

= ij + QilLlkQjk

(1.124)

or in coordinate free notation,

L+ = + Q L QT. (1.125)Using (1.125), it can be shown that

26


27/85

Figure 1.6: Representative kinematic variables for the original motion (F , D) and the superposedmotion (F+, D+) as well as the relationship between these kinematic variables.

D+ = Q D QT,W+ = + Q W QT. (1.126)

Recalling (1.14) and (1.98), it is clear that

F+ =

+

(X, t)X

, or F+iA = +

i (X, t)XA

, (1.127)

Using (1.127) and (1.100),

F+iA =+i (x, t)

xj

xjXA

,

= Qij FjA

(1.128)

and therefore,

F+ = Q F . (1.129)Using (1.129) it can also be shown that,

J+ = J. (1.130)

Additionally, the density and the normal to a surface n, can be shown to transform as,

+ =

n+ = Q n. (1.131)

27


28/85

A scaler, vector or second order tensor which transform as (1.131)1, (1.131)2 and (1.126)1, respec-tively are called objective.

1.10.2 Requirement: Invariance under a superposed rigid body motion

In this section, we consider the mechanical behavior of a fixed material undergoing two differentmotions: x = (X, t) and a second motion which is related to that at any arbitrary time t differsfrom the first by only a superposed rigid body motion. Since the material is the same for bothmotions, the form of the constitutive equation will remain the same. For example, suppose thematerial is linear in the symmetric part of the velocity gradient,

T = I + D. (1.132)

Then for the superposed rigid body motion,

T+ = I + D+. (1.133)

More generally, if T = T(D), then

T+ = T(D+). (1.134)

Note that the functional form does not change.In this section, we are going to impose a requirement on the relationship between t and t+ and

hence on T and T+. In particular, we might expect that the magnitude of the stress vector isunchanged by a superposed rigid body motion ( t+ and t to have the same magnitude) and that t+

and t to have the same orientation relative to n+ and n,respectively. Based on these expectations,we introduce the following assumption

t+ = Q t. (1.135)Using (1.135) and (1.131), it can be shown that,

|t+| = |t|, t+ n+ = t n. (1.136)

The next step is to determine the relationship between T and T+. Recall that,

t = t(x, t; n) = T(x, t) n. (1.137)

Therefore,t+ = t+(x+, t; n+) = T+(x+, t) n+. (1.138)

In addition, recall the relationship between the outward unit normal n to a material surface inthe current configuration and the corresponding outward unit normal in the superposed motion,n+ = Q n. Therefore,

t+ = T+ n+ = T+ Q n. t+i = T+ij n+j = = T+ij Qjknk. (1.139)

28


29/85

Using the assumption (1.135) and (1.137), it then follows that,

t+

= Q t = Q T n, and t+

i = Qijtj = QijTjknk. (1.140)

Combining (1.139) and (1.140), we have that

T+ Q Q T n = 0, and T+ij Qjk QijTjk nk = 0. (1.141)

Since (1.141) must hold for all n and the expression in brackets is independent ofn, we can concludethat the expression in brackets is equal to zero. Hence,

0 = T+ Q Q T= T+

Q

QT

Q

T

QT

= T+ Q T QTor

0 = T+ij Qjk QijTjk= T+ij QjkQlk

QijTjkQlk

= T+il QijTjkQlk(1.142)

and therefore,

T+ = Q T QT, and T+il = QijTjkQlk . (1.143)

29


30/85

30


31/85

Chapter 2

Non-Linear Viscous Fluids

In this chapter, we start with the constitutive assumption that the stress tensor depends on thefluid mass density and the velocity gradient L.

T = T(, L). (2.1)

Inherent in the form of the stress tensor, (2.1) is the assumption that the current state of stressdepends only on the velocity gradient at the current time and not on any previous deformation thefluid might have undergone. Using invariance of the stress tensor under a superposed rigid bodymotion and a representation theorem for symmetric isotropic tensor functions, it is shown that themost general form of (2.1), which does not violate invariance requirements is

T = 0I + 1D + 2D2 (2.2)

where in general 0, 1, 2 depend on and the three principal invariants of D.1

Similarly, we can start by considering incompressible fluids for which the extra stress tensoris a function of the velocity gradient. After imposing invariance requirements and using the samerepresentation theorem, we can deduce that the most general form of the Cauchy stress tensor thatsatisfies invariance can be written as,

T = I + 1D + 2D2 (2.3)

where is a Lagrange multiplier arising from the incompressibility constraint and 1 and 2 dependonly on the second and third principal invariants of D. Such incompressible fluids are typicallycalled Reiner-Rivlin fluids. The Navier-Stokes fluid is a special Reiner-Rivlin fluid with 2 equalto zero and 1 constant. The behavior of Reiner-Rivlin fluids with non-zero values of 2 in simple

shear does not match experimental results on real fluids [1]. In addition, the dependence on thevalue of IIID is often considered negligible [1]. For this reason, attention is typically confined toa special class of Reiner-Rivlin fluids called generalized Newtonian fluids. These are Reiner-Rivlinfluids with 2 zero and 1 dependent only on the second principal invariant of D. In the latter partof this chapter, specific generalized Newtonian fluids are considered and the behavior is contrastedand compared with experimental results.

1Recall the principal invariants of a second order tensor A are IA = tr A, IIA = 1/2((tr A)2 tr (A2)), IIIA =

detA.

31


32/85

2.1 Constitutive Assumption T = T(, L)

In this section, we start with the constitutive assumption that the stress tensor depends on thefluid mass density and the velocity gradient L. Namely, we assume, T = T(, L). We then obtainthe most general function which satisfies invariance requirements. Using the relationship

L = W + D (2.4)

we can rewrite the constitutive assumption as

T = T(,D,W). (2.5)

If the constitutive equation (2.5) is a valid one, then it must hold for all motions, in particular, it

must hold for a superposed rigid motion,

T+ = T(+, D+, W+). (2.6)

Recalling that,

T+ = Q T QT, (2.7)

and using (2.6)

T(+, D+, W+) = Q T(,D,W) QT. (2.8)

Recalling,

+ = , D+ = Q D QT, W+ = Q W QT + , (2.9)

we can write (2.8) as

T(+, Q D QT, Q W QT + ) = Q T(,D,W) QT. (2.10)

Now consider a superposed motion, for which at some arbitrary time t, Q = I, and = 0. In thiscase, (2.10) becomes,

T( ,D,W+ ) = T(,D,W), (2.11)

for arbitrary choices of the skew-symmetric tensor . It then follows that the response function Tcannot depend on W, and so (2.5) is reduced to

T = T(, D), (2.12)

32


33/85

and must satisfy

T(, Q D QT) = Q T(, D) QT, (2.13)

for all proper orthogonal Q. Then, from (2.13), we can conclude that T(, D) must be a symmetricisotropic tensor function of D, (see, for example, Appendix C). A representation theorem fortensors (Appendix C), states that the most general second order isotropic tensor function T(, D)for symmetric second order tensor D, has the form

T = 0 I + 1D + 2D2, (2.14)

where in general 0, 1, 2 are functions of the three principal invariants of D as well as the massdensity .

2.1.1 Incompressible Inelastic Fluids

In the case, where the fluid is incompressible, the density of a fixed material point is constant. Thestress tensor is therefore independent of the density ,

T = T(D). (2.15)

The Cauchy stress tensor is determined by D only to within a stress N which does no work inany motion satisfying the constraint of incompressibility. We can rewrite (2.15) as the sum of theindeterminant stress N and an extra stress , (see, for example, Truesdell [31], pp. 70-73),

T = N + (D). (2.16)

As discussed previously in Section 1.9, the expression for the rate at which stresses do work perunit volume, P, is

P = tr(T DT). (2.17)Recall, the scalar quantity tr(A BT) = AijBij satisfies the requirements for an inner product oftwo arbitrary second order tensors, A, B. We will use the notation (A, B) to denote this innerproduct. Namely,

(A, B) tr(A BT). (2.18)Using this notation, the condition of incompressibility can be written as (D, I) = 0. Since the

indeterminate stress does no work on the constrained motions,

0 = (N, D) for all motions in which(D, I) = 0. (2.19)

We can state the conditions on N given in (2.19) as, N must be orthogonal to all D which areorthogonal to I. It therefore follows, that N must be a scalar multiple of I,

N = I, (2.20)

33


34/85

where is a Lagrange multiplier. Therefore the Cauchy Stress, for an incompressible fluid takesthe form,

T = I + (D). (2.21)

The most general form for (D) can be deduced in essentially the same way we deduced theform of the stress tensor for a compressible fluid. For an incompressible fluid, the most generalrepresentation for the extra stress, (D) is

= 0 I + 1 D + 2 D2. (2.22)

The first principal invariant of D is 0 for isochoric motions, so 0, 1, 2 are functions of only IIDand IIID. Without loss in generality, we can absorb 0 into the Lagrange multiplier and write(2.22) as

Reiner Rivlin Fluids

T = I + 1 D + 2 D2

where 1 = 1(IID, IIID) and 2 = 2(IID, IIID)

is a Lagrange multiplier arising from the incompressibilityconstraint.

(2.23)

It remains to specify the two material functions 1(IID

, IIID

) and 2(IID

, IIID

). The Lagrangemultiplier will be determined from solving the governing equations subject to the boundary con-ditions. Incompressible fluids for which the stress tensor is related to the velocity gradient through(2.23) are called Reiner-Rivlin fluids.

2.2 Reiner-Rivlin Fluids

As just stated, Reiner-Rivlin fluids are incompressible fluids with constitutive equation,

T = I + 1 D + 2 D2, (2.24)

where 1(IID, IIID) and 2(IID, IIID). We have not yet considered specific forms for the material

functions, 1(IID, IIID) or 2(IID, IIID). In this section, we discuss some restrictions on thesematerial functions which can be motivated by thermodynamic considerations. In addition, we willcompare the behavior of the Reiner-Rivlin fluid in simple shear with the behavior of real fluids tomotivate restricting our attention in the remainder of this chapter to a special case of Reiner-Rivlinfluids with 2 equal to zero and 1 a function only of IID. Fluids in this subclass of Reiner-Rivlinfluids are called generalized Newtonian fluids.

34


35/85

2.2.1 Thermodynamic Restrictions on 1 and 2 for Reiner-Rivlin Fluids

From the Second Law of Thermodynamics we require that the stress power be non-negative,

T : D 0. (2.25)

This requirement can be used to deduce restrictions on the choices of the material functions 1, 2,[28, 29, 7, 18]. For a Reiner-Rivlin fluid,

T : D tr (T D) = tr D + 1 D2 + 2 D3 ,= tr

D + 1 D

2 + 2 (IIDD + IIIDI)

,

= 1 tr (D2) + 32IIID,

= 1 tr (D2

) + 32detD,

(2.26)

where we have made use of the Cayley-Hamilton Theorem and the condition of incompressibility.Therefore from thermodynamics, we require,

0 1 tr (D2) + 32detD. (2.27)

The first term on the right hand side of (2.27) is non-negative while the second term may bepositive, negative or zero. In order to deduce further information from (2.27), we consider somespecial cases of Reiner-Rivlin fluids.

Case I: 2 = 0, 1 not necessarily constant.For the special case of a Reiner-Rivlin fluid with 2 equal to zero, we see from (2.27) that necessarily1 0.Case II: 1, 2 both constant.We now consider the particular case where 1 and 2 are both constants. In this case, 1 and 2will be independent of the invariants of D, so any conclusions we make based on specific choices ofD will hold for all D.

First, consider the following choice for the rectangular components of D,

[D] = 2

0 1 01 0 00 0 0

. (2.28)

For the choice of D defined in (2.28), tr (D2) = 2/2 and det D = 0. Therefore the inequality(2.27) takes the form

0 2

21, (2.29)

35


36/85

and therefore for this case 1 must be greater than or equal to zero. Since 1 is a constant, thisresult holds not only for the choice (2.28) for D but for all D. Now, consider the following two

choices for the rectangular components of D,

[D] =

2 0 00 1 0

0 0 1

, [D] =

2 0 00 1 0

0 0 1

. (2.30)

For both cases, tr (D)2 = 62. For the first case, det D = 23 while for the second case det D =23. Therefore the inequality (2.27) must satisfy both of the following,

0 1 + 2,0

1

2.

(2.31)

Since 1 is always constant and is arbitrary, we see from that (2.31) that 2 must be zero cor-responding to an incompressible Newtonian fluid with kinematic viscosity equal to 1 /21. If 2 iszero, then from (2.27), 1 must be greater than zero. This is equivalent to the familiar requirementthat the kinematic viscosity be positive. Therefore, we can conclude that the only physically rea-sonable case with 1 and 2 both constant is the case of an incompressible Newtonian fluid.

2.2.2 Restrictions on 1, 2 based on behavior of real fluids in simple shear

Consider flow of an incompressible Reiner-Rivlin fluid between two infinite parallel plates which isdriven by the motion of the upper plate moving at a constant speed U, Fig. 1.4. We have selecteda rectangular coordinate system with the x1 axis parallel to the plates and x2 perpendicular to theplates. The location of the x1 axis is chosen such that x2 = 0 is coincident with the bottom plate.As discussed in Section 1.58, we look for solutions of the form,

v = v(x2) = x2 e1 (2.32)

where is a constant. Recall that for (2.32) to be a solution it must satisfy the boundary con-ditions for the problem, the incompressibility condition, the equation of linear momentum, andthe constitutive equation for the stress tensor. The boundary conditions for this problem are the

no-slip condition at the two plate surfaces,

v(x2 = 0) e1 = 0, v(x2 = h) e1 = U, (2.33)

and the condition of impermeability of both plates,

v(x2 = 0) e2 = 0, v(x2 = h) e2 = 0. (2.34)

36


37/85

We see that boundary conditions, (2.33) and (2.34) are met when = U/h. For a velocity fieldof the form, (2.32), the components of the rate of deformation tensor D are

[Dij] =

1

2(

vixj

+vjxi

)

=

2

0 1 0

1 0 0

0 0 0

ij

. (2.35)

It is clear from (2.35) that for the motion, (2.32), the deformation tensor D is a constant secondorder tensor. The components of D2 are,

D2

= [DijDjk ] =2

4

1 0 0

0 1 0

0 0 0

, (2.36)

and the three principal invariants of D are,

ID = tr D = 0,

IID =1

2

(tr D)2 (tr (D)2) = 2

4,

IIID = det D = 0.

(2.37)

Using (2.37), we see that the incompressibility condition

0 = vixi

= tr D (2.38)

is identically satisfied.The incompressible Reiner-Rivlin constitutive equation (2.24) can be split into the constraint

response and response that does work,

T = I + (D), where (D) = 1(IID, IIID)D + 2(IID, IIID)D2. (2.39)

Using the results (2.36) with (2.39), it follows that,

[] = = 1 2

0 1 0

1 0 0

0 0 0

+ 2 24

1 0 0

0 1 0

0 0 0

, (2.40)

Therefore, the only non-zero components of are,

12 = 21 = 1

211 = 22 = 2

2

4. (2.41)

37


38/85

where 1 and 2 are both even functions of ,

1 = 1(1

4 2

, 0), 2 = 2(1

4 2

, 0). (2.42)

It follows from (2.41) and (2.22), that the only non-zero components of T are,

T12 = T21 = 1

2, T11 = T22 = + 2

2

4, T33 = 0. (2.43)

We see from (2.43) that T12 is an odd function of while T11 and T22 are even functions of .It follows from (2.43) that to fully determine the solution for the Cauchy stress tensor, we need

to solve for , which will in general require consideration of the balance of linear momentum as wellas the boundary conditions. We will consider this issue at a later time. Here we focus attentionon the viscometric functions. As will be discussed in more detail in the following sections, the

following three functions, determine the behavior of a certain category of fluids in class of flowscalled viscometric flows:

() = T12N1() = T11 T22N2() = T22 T33

(2.44)

where , N1, N2 are referred to as the shear stress, first normal stress and second normal stress.Three material functions commonly used in the literature are defined relative to , N1, N2,

() = T12/

1() = N1/2 = (T11 T22)/2

1() = N2/

2

= (T22 T33)/2

(2.45)

where , 1, 2 are called the viscometric functions, referred to as the viscosity, first normal stresscoefficient and second normal stress coefficient. Most rheometers are based on viscometric flowsand are directed at measuring the viscometric functions.

It follows from (2.43) and (2.45), that for a Reiner-Rivlin fluid,

=12

, 1 = 0, 2 =1

42. (2.46)

and therefore,

1 = 2, 2 = 42. (2.47)

However, there is no evidence of real fluids exhibiting a zero value for 1 and a non-zero value for2 [1]. For this reason, attention is typically confined to Reiner-Rivlin fluids with 2 equal to zero.Therefore, no constitutive equation of the form T = T(D) will be suitable for describing real fluidswith nonzero normal stress coefficients. In summary, based on invariance requirements, thermo-dynamic considerations and the behavior of real fluids, the most general reasonable constitutiveequation of the form T = T(D) is,

T = I + 1(IID, IIID)D where 1 0. (2.48)

38


39/85

Recalling that the mechanical pressure is defined as p = tr(T)/3, it follows from (2.48), that

= p. (2.49)

In addition, in simple shear (and other viscometric flows), the quantity IIID is identically zeroand therefore the functional dependence of 1 on this variable cannot be determined from mostrheometers. In the remainder of this chapter, we will only consider fluids for which 1 is assumedindependent of IIID. As discussed in [1], page 54, there is some evidence that this may be reasonablefor real fluids. After imposing this restriction on (2.48), we obtain the equation for a generalizedNewtonian Fluid (incompressible),

T = pI + 1(IID)D where 1 0. (2.50)

If we rewrite this last results in terms of the viscometric functions we obtain,

T = pI + 2(IID)D where 0. (2.51)

Recalling that IID = 1/2tr(D2), it is clear that it will be convenient to replace IID with anon-negative quantity such as,

S = 4IID (2.52)

where the factor of two is introduced in order that S = 2 in simple shear. Using this new notation,

Incompressible, generalized Newtonian Fluid

T = pI + 2(S)D

where S 4IID, p is the mechanical pressure,From Thermodynamics: 0.

(2.53)

2.3 Generalized Newtonian Fluids

In this section we first consider some commonly used examples of generalized Newtonian fluids.We then consider the behavior of these models in several important flow fields: (i) steady, fullydeveloped flow in a straight pipe and (ii) steady cone and plate flow and (iii) steady flow betweenconcentric cylinders.

39


40/85

2.3.1 Power-Law Fluid

The Power-Law Fluid is an example of one of the simpler generalized Newtonian fluids. It isparticularly popular because of the number of exact solutions which can be obtained for this model.The Cauchy Stress tensor for a Power-Law fluid is defined as,

Tij = pij + 2 K|4IID|(n1)/2Dij (2.54)and therefore, (S) is defined as

(S) = K S(n1)/2 (2.55)

where n is called the power law index and K is called the consistency. When n is equal to one, theNavier-Stokes constitutive equation is recovered. When n is less than one, the constitutive equation

Power Law Fluids for n less than or equalto 1 (shear thinning or Newtonian)

Power Law Fluids for n greater than orequal to 1 (shear thickening or Newtonian)

is shear thinning while when n is greater than one the behavior is shear thickening. The viscosityin the limit of vanishing S and in the limit ofS tending to infinity are,

n < 1 : limS0

(S) = , limS

(S) = 0,

n > 1 : limS0

(S) = 0, limS

(S) = .(2.56)

The unboundedness of the viscosity function and the lack of a non-zero viscosity at zero shear ratedoes not match experimental results for real fluids and so limits the applicability of the Power-Lawmodel.

40


41/85

2.3.2 Prandtl-Eyring Model

An alternative to the Power-Law model is the Prandtl-Eyring Model which tends to a constantviscosity in the limit of going to zero. However, the viscosity function tends to zero as tendsto infinity,

(S) = 0sinh1(

S)

S(2.57)

where 0 and are material constants and

limS0

(S) = 0, limS

(S) = 0. (2.58)

2.3.3 Powell-Eyring Model

The Powell-Eyring Model is a three constant model which displays a non-zero bounded viscosityat both the upper and lower limits,

(S) = + (o ) sinh1(

S)

S(2.59)

where 0, and are material constants and

limS0

(S) = 0, limS

(S) = . (2.60)

It is convenient to write rewrite (2.59) as,

0 =

sinh1(S)S

(2.61)

2.3.4 Cross Model

The Cross Model [9] is a four constant model which also displays a non-zero bounded viscosity atboth the upper and lower limits

0 =

1

1 + (K2|S|)(1n)/2(2.62)

The notation used in (2.62) is chosen to fit the shear thinning case where n < 1,

limS0

(S) = 0, limS

(S) = . (2.63)

This model is also suited for shear thickening, though in this case, the notation should be modified,since, while for n > 1,

limS0

(S) = , limS

(S) = 0. (2.64)

41


42/85

Ellis Model

Frequently, the flow regimes of interest are such that it is not essential to specify a nonzero viscosityinfty. Setting to zero in (2.62) yields the Ellis model [?],

=0

1 + (K2|S|)(1n)/2. (2.65)

A major advantage of the Ellis model is that it is possible to obtain analytical solutions for somesimple flows (e.g. [3], [27]).

2.3.5 Yasuda Model

Yasuda proposed a model similar to the Cross model but with an extra material constant a withwhich to fit the data [32],

0 =

1[1 + aS](1n)/a

. (2.66)

Several special cases of the Yasuda model are notable including the Carreau model.

Carreau Model

Ifa is set to 2 in (2.66), the four constant Carreau model is obtained [4]. It also displays a non-zerobounded viscosity at both upper and lower limits,

0 =

1

1 + 2 S

(1n)/2 . (2.67)

Comments regarding the limiting behavior for the Cross model hold for the Yasuda model aswell.

2.3.6 Bingham Fluids

There are many real fluids which if tested in simple shear, would appear not to flow at all untilthe magnitude of the applied shear stress surpassed a fixed finite value termed the yield stress.E. C. Bingham (1916) observed this type of behavior in paints. Many other real fluids show anapparent yield stress as well. One type of constitutive equation which is used to describe fluidswhich appear to display a yield stress is the Bingham model (see, e.g. [2] the review articleby Barnes, for a discussion of experimental evidence supporting the lack of existence of a true

yield stress in real materials.). In simple shear of a Bingham fluid, there is no motion unless themagnitude of the applied shear stress is above a threshold, Y

, in which case, the material behaveslike an incompressible Newtonian fluid, Fig. (2.1). It is a simple matter to provide a mathematicalstatement of this response,

12 < Y = = 0

12 Y = =1

(12 Y).

(2.68)

42


43/85

Figure 2.1: Shear stress 12 versus shear rate in simple shear for (i) Newtonian fluids (ii) Binghamfluids with yield stress Y .

When the flow is more complex than simple shear, a more general measure of the magnitudeof the shear stress is used. Namely, we cannot simply use 12 = Y as a criterion. Note thatthis expression is even incorrect for simple shear under a change of basis. The condition, mustbe rephrased in terms of an invariant of the stress tensor. For a Bingham fluid, the more generalmeasure of the shear stress is the second invariant of the extra stress tensor, II. The threedimensional representation of the Bingham model can be represented as (see, e.g. [11],[25]),

|II| < Y = Dij = 0|II| Y =

Dij =1

2

1 Y

|II|

ij

ij = 2

+ Y

2

|IID|

Dij

(2.69)

where Y

and are material constants. Recall that in simple shear, I, III are zero for a Binghamfluid. In simple shear, 2.69 can be used to obtain .

Early studies of this constitutive equation were carried out by [22, 23, 26, 25]. Other materi-

als which seem to be well modeled by the Bingham constitutive model include pastes, margarine,mayonnaise, ketchup. and concentrated suspensions of particles. In simple shear, the shear stressis constant throughout the flow field. In more complex flows, such as fully developed flow in astraight pipe or flow between two coaxial cylinders driven by the relative rotation of the cylinders,we will see that the shear stress is not constant through out the flow field, and as a result, theremay be regions where the yield criterion is reached ( and the fluid is flowing) while in other regionsthe value of II is below Y and the fluid does not flow (D = 0 in that region). Note that D = 0does not necessarily imply v = 0. A collection of analytical solutions for the Bingham model and

43


44/85

extensive reference list can be found in a review article by Bird, Dai and Yarusso [5].

2.3.7 Modified Bingham Model

The Bingham fluid equation is important because it appears to model some real fluids well andyet it is possible, in some cases, to obtain analytical solutions for the flow field, (for example,Poiseuille and Couette flow). However, this constitutive equation is in general challenging to modelnumerically due to the difficulty in tracking the yield surfaces in the flow field. Papanastasiou [24]modified the Bingham model (2.69) to eliminate the yield stress. Recall, that for Bingham fluids, atzero shear rate, the viscosity function (IID) abruptly changes from an infinite value to a constant(finite) viscosity . Papanastasiou defined the viscosity function as an increasing function of IID,as

(S) = +Y

1 enS/2

S, (2.70)

where n, , Y

are material constants. The value of n determines the rate of exponential increase

Figure 2.2: Shear stress versus shear rate for simple shear of a Bingham fluid and modified Bingham

fluid.

to the constant viscosity . For simple shear, it follows from (2.70),

12 =

+ Y

1 en

. (2.71)

In the limit ofn tending to infinity the shear stress for the Bingham fluid is obtained.

44


45/85

2.4 Steady, Fully Developed Flow of a Generalized Newtonian

Fluid in a Straight Pipe

In this section, we consider steady, fully developed flow in a straight pipe of circular cross section ofradius R driven by a pressure drop per unit length in the axial direction equal to . In particular,we look for velocity fields of the form,

v = w(r)ez, (2.72)

where we have used standard cylindrical coordinates in (2.72) such that the z-axis is aligned withthe pipe centerline. The volumetric flow rate, Q, for velocity fields of the form (2.72) is

Q = 2 R

0w(r)rdr. (2.73)

In many engineering applications, it is of interest to obtain a relationship between Q and p/z(denoted as -). If a closed form solution for the velocity field can be calculated for a givenviscosity function (S), then it is relatively straightforward to calculate the corresponding flow rateas a function of from (2.73). In the first part of this subsection, we consider the Power-Law,Cross and Bingham fluids for which it is possible to obtain closed form solutions for w(r). For mostGeneralized Newtonian fluids, it is not possible to obtain a closed form solution for the velocityfield for steady, fully developed flow in a straight pipe. In the second part of this subsection weconsider the more general case and obtain an equation relating Q as a function of and (S).In the last part of this subsection, we discuss the inverse problem of using measured relationshipsbetween driving pressure drop and volumetric flow rate to obtain the viscosity function (S) for a

generalized Newtonian fluid.

Velocity fields of the form (2.72) identically satisfy the incompressibility condition and thecomponents of D and D2 in cylindrical coordinates are,

[D] = 2

0 0 1

0 0 0

1 0 0

, D2 =

2

4

1 0 0

0 0 0

0 0 1

, (2.74)

where we have denoted dw(r)/dr by . It follows from (2.74) that the invariants of D are,

ID = 0, IID = 2/4, III D = 0. (2.75)

From these results and (2.53), we see that the only nonzero components of the extra stress tensorfor a Generalized Newtonian Fluid in simple shear are,

rz = zr = (S) where S 4 IID = 2. (2.76)

45


46/85

Note that and S depend at most on r. If we use the results (2.76) in the equation of linearmomentum in cylindrical coordinates we find,

0 = pr

0 = 1r

p

0 = pz

+1

r

(rrz)

r.

(2.77)

It follows directly from (2.77), that p is independent of both r and and p/z is a constant whichwe will denote as . Hence, the pressure is p = z + f(t). We can now integrate (2.77)3 withrespect to r to obtain, rz = r/2+ constant/r. In order that the shear stress be bounded at thecenter of the pipe, we set the constant to zero.

rz = r2

. (2.78)

Combining this with the result (2.74), we obtain

(2) = r

2. (2.79)

We cannot proceed further without specifying the viscosity function.

2.4.1 Power-Law Model

From (2.79) and (2.55), we obtain,

Kn =r

2(2.80)

and upon integration and use of the no-slip boundary condition at the wall, we have,

w(r) =

R

2K

1/n R1 + 1/n

1 ( r

R)1+1/n

. (2.81)

After substituting (2.83) in (2.73) and integrating, we obtain the volumetric flow rate as a functionof the axial components of the pressure gradient , the materials constants K and n and the piperadius, R.

Q =R

2K1/n R3

3 + 1/n . (2.82)

Using (2.82), we can write the axial velocity (2.83) as,

w(r) =Q

R23 + 1/n

1 + 1/n

1 ( r

R)1+1/n

. (2.83)

For the same flow rate, the velocity field in the shear thinning power-law model is flatter than forthe constant viscosity case, Fig. 2.4.1.

46


47/85

Figure 2.3: Nondimensional axial velocity profile, w(r)/(Q/R2, as a function of nondimensionalradius r/R for steady, fully developed flow of power-law fluids in a straight pipe. The profiles shownare for the same flow rate but with different values of n.

2.4.2 Bingham Fluid

Recall that for a Bingham Fluid, D is zero at all points where

|II| is less than the yield stress.

For the flow under consideration |II| is equal to the magnitude ofrz which is equal to r/2. It isclear from the form of the solution for rz that the magnitude of|II| will be largest at the largestradial position: r = R. When R/2 < Y, the yield criterion will not be met anywhere in the fluidand hence D will be zero throughout the fluid. From the boundary conditions, it then follows thatthe velocity is zero everywhere. If the driving pressure gradient increased so that R/2 > Y , therewill be a region r [0, 2Y/], where the yield criterion is not met, and r (2Y/,R] where it ismet. We will denote this critical radius as ry 2Y/. From (2.69),

Drz =

0 for r rY

r4

+Y

2for r r

Y

(2.84)

As previously discussed, for the velocity field under consideration, Drz is equal to 1/2 dw/dr andso, from (2.84), we can obtain the axial velocity as a function of the radial variable r,

w(r) =

C1 for r rY

r2

4+

Y

r

+ C2 for r rY

(2.85)

47


48/85

where C1 and C2 are constants of integration. If we require that the value of w(r) be zero at thepipe boundary and be continuous throughout the flow field (including r equal to r

Y), then

w(r) =

R2

4

1 rY

R

2for r r

Y

R2

4

1 r

2

R2

Y R

1 r

R

for r r

Y

(2.86)

The flow rate as a function of can be calculated directly from (2.73) and (2.86) and is left as anexercise.

2.4.3 Q as a function of and (S) for unspecified generalized Newtonian fluid

For most viscosity functions it will be difficult, if not impossible to obtain a closed form solutionfor the velocity field. In this section, we consider the possibility of directly obtaining an expressionfor the flow rate as a function of without first obtaining an explicity expression for the velocityfield. While in most cases we will still need to use numerics to obtain the relationship, we will beable to avoid the intermediate step of first calculating the velocity field.

Beginning with (2.73) and the definition of and performing integration by parts twice, we canwrite the volumetric flow rate as

Q =

3

R3 w

w0

r()3 d

, (2.87)

where w denotes the value of at the pipe wall. Eq. (2.79) provides and expression for r as afunction of ,

r =2

(2). (2.88)

Using the relationship (2.88) in (2.87), we obtain

Q =

3

R3 w 8

3

w0

3 (2)3 d

. (2.89)

We can obtain w as a function of and R by solving the equation resulting from evaluating (2.88)at the wall,

w (

2

w) =

2 R. (2.90)

Once this value is obtained, (2.89) can be integrated to obtain volumetric flow rate as a functionof .

Exercise: Use (2.89) to obtain the flowrate as a function of for the Power-Law model.

Exercise: Is it possible to evaluate (2.89) in closed form for the Carreau model?

48


49/85

2.4.4 Determination of(S) from measured values of driving pressure drop andflow rate in a straight pipe

Capillary viscometers are commonly used to measure the viscosity function (S). These rheometersare designed to reproduce, with some level of experimental error, the velocity field for steady fullydeveloped flow in a pipe of constant radius v = w(r)ez. The general idea is that using results frombalance of linear momentum for this velocity field, and measurements of flow rate and pressure dropat different flow rates, we can back out and expression for (S). Note that we want this expressionto be for an unspecified generalized Newtonian fluid (in fact, we will show in later chapters that thecapillary viscometer can in principal be used to measure (S) for all simple fluids). The argumentsin this section, follow that of [1].

As will be clear at the end of this section, it is useful to rephrase the problem as one of obtaining as a function of a characteristic shear rate a and the shear stress at the wall, w. We will definea as,

a 4QR3

, (2.91)

which is the shear rate at the pipe wall if the fluid is Newtonian. Using measured values of Q it isa simple matter to obtain a. Using the definition of flow rate, integration by parts and (2.91), wecan obtain an expression for the characteristic shear rate

a =4

R

us +

1

R2

R0

r2 dr

, (2.92)

where the notation us denotes the slip velocity at the wall of the pipe. Here we focus attentionon fluids for which the slip velocity is assumed to be zero. However, there is evidence that this

assumption is not appropriate for some real fluids. See [1] for further discussion of using capillaryrheometers in cases where us is not negligible.

Recall, that for a generalized Newtonian fluid undergoing a velocity field v = vz(r)ez, thatnecessarily

rz = r2

, (2) = rz/, (2.93)

based on balance of linear momentum and the constitutive equation. Evaluating (2.93)1 at thewall,

w rz |r=R = 12

R. (2.94)

It follows that w can easily be calculated from measured values of . As mentioned above, we areinterested in obtaining an expression relating w and a. We can do so, by rewrite the integral in(2.92) over r as an integral over rz . It follows from (2.93)1 and (2.94) that

r = rzR/w. (2.95)

When (2.93)2 is invertible,

= (rz). (2.96)

49


50/85

After using (2.95) and (2.96) in (2.92), we find

a = 43w

w0

2rz (rz) drz. (2.97)

If we take the derivative of the expression, (2.97) with respect to w, we find,

dadw

= 124w

w0

2rz (rz) drz +4

w(w),

= 3 aw

+4 (w)

w.

(2.98)

Rearranging (2.98) and using the chain rule, we obtain

4wa

=

3 +

ln aln w

(2.99)

where w = (w). Defining,

n =ln wln a

(2.100)

we can rewrite (2.99) as,

w = a3n + 1

4n. (2.101)

Equation (2.101) is the Mooney-Rabinowitsch equation. Experiments can be run at different pres-sure drops to obtain a curve of a as a function of w (recall, (2.91) and (2.94)). If these resultsare plotted on a log scale, the value n can be obtained from the slope of the curve, which will ingeneral be a function of w. From the definition of the viscosity function, applied at the wall,

(w)2 = w/w. (2.102)

Using (2.101) with this last result, enables us to determine results for (S) for this flow. Astaritaand Marrucci provide the details of applying this result to a specific data set (page 73, [1]).

2.5 Cone and Plate Flow for a Generalized Newtonian Fluid

The cone and plate rheometer is one of the most commonly used instruments for measuring viscosityand normal stress coefficients. In particular, it can be used to measure the first normal stresscoefficient 1 in addition to the viscosity function

2. The implementation of the rheometer isbased on an idealized flow of fluid between a cone and plate, driven by the rotation of the cone aboutthe z-axis, Figure 2.5. In this section, we focus on the derivation of the fundamental relationships

2While the focus in this chapter is on generalized Newtonian fluids, this rheometer can also be used for viscoelasticfluids. Recall that 1 is zero for generalized Newtonian fluids.

50


51/85

for this rheometer including the implementation of the small angle approximation at the basis ofcone and plate rheometers. A detailed discussion of the assumptions involved in the use of this

rheometer can be found in [19], pages 205-217.It is convenient to use standard spherical coordinates for which the the solid walls of the rheome-

ter are coordinate surfaces. The angle between the cone and the plate is denoted by and themagnitude of the angular velocity by . The corresponding applied torque necessary to generatethis angular velocity as m = mzez . Using this notation, the boundary conditions on the velocityfield are,

v|=/2 = 0, and v|=(/2) = r cos e. (2.103)The boundary condition on the stress vector at the free surface, denoted by 1, is

t = patm n, on 1 (2.104)

where n is the normal to the free surface and patm denotes the atmospheric pressure. At this point,

Figure 2.4: Schematic of cone and plate geometry with spherical coordinates. Note: The angle is typically less that 6 degrees for cone and plate rheometers.

we have not specified the shape of the free surface, which in general should be determined as partof the solution. Motivated by the boundary conditions (2.103) and (2.104), we look for a velocityfield of the form,

v = v(r, ) e. (2.105)

Moreover, the boundary conditions will be met if v depends linearly on r, so we look for v of theparticular form,

v = r () e. (2.106)

51


52/85

From the assumed form of the velocity field, (2.106), it is straightforward to calculate severalkinematic variables (see, e.g. Appendix G),

[D] = 2

0 0 00 0 1

0 1 0

, D2 = 2

4

0 0 00 1 0

0 0 1

, (2.107)

where

sin dd

sin

. (2.108)

For this flow field, we can evaluate (2.52) to show S = 2. Therefore, the only non-zero componentsof the extra stress tensor for a Generalized Newtonian fluid are (2.53),

= =

(2) . (2.109)

Since is only a function of , is also only a function of . From (2.106) and (2.109) theequation of linear momentum reduces to,

r 2 = pr

,

2 r2 cot = p

,

0 = 1sin

p

+

1

sin2

( sin

2 ).

(2.110)

From (2.110) and because depends only on , we can conclude that p/ is a constant.Furthermore, in order that the boundary condition (2.104) be met for all , necessarily, this constantis zero and hence p is at most dependent on r and .

Neglect Inertial Effects

By differentiating (2.110)1 with respect to and (2.110)2 with respect to r we obtain,

(

p

r) = 2 r

d

d

r(

p

) = 2 r 2 cot .

(2.111)

It then follows that either = 0 or = A sin where A is a constant of integration. Neither of thesesolutions can satisfy both boundary conditions. Therefore, our assumed form of the velocity fieldis only meaningful in the limit of negligible inertial effects in which case, the first two componentsof the balance of linear momentum reduce to the condition that p is independent of r and andhence p is constant throughout the flow field. It then follows from the condition at the free surface(2.104), the fact that r = r are the only non-zero components of the extra stress, the freesurface must be a surface of constant radius and p = patm everywhere in the fluid.

52


53/85

Relationship between r and mz

.It follows from the third component of linear momentum (2.111)3, that

= C1sin2

, (2.112)

where C1 is a constant of integration. It is convenient to write this constant in terms of measurablequantities, in particular the torque to the fluid by the cone. m = mzez. Recall, that in general,the applied torque on a surface with outward unit normal n is related to the stress vector t actingon that surface through

m =

A

x t|1 da (2.113)

where x is the lever arm, da is the elemental area. At the fluid surface adjacent to the cone, isequal to /2 and

x = r cos (sin e + cos er),

da = r cos drd,n = et|1 = T n

= pe |1e

(2.114)

and therefore,

x t|1 = p r cos2 e |1 r cos ez, (2.115)where the base vector,

robertson nonnewtoniannotes

Documents