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### Transcript of NONLINEAR VISCOSITY Linear, isotropic, ... NONLINEAR VISCOSITY Linear, isotropic,incompressible,...

• Plasticity http://imechanica.org/node/17162 Z. Suo

October 13, 2014 Nonlinear viscosity 1

NONLINEAR VISCOSITY Linear, isotropic, incompressible, viscous fluid. A fluid deforms

in a homogeneous state, with stress σ ij

and rate of deformation D ij

. The mean

stress is σ m =σ

kk /3 , and the deviatoric stress is s

ij =σ

ij −σ

m δ ij

. By a linear,

isotropic, incompressible, viscous fluid we mean a model specified by

D kk =0 ,

s ij =2ηD

ij ,

where η is the viscosity. The model assumes that η is independent of the rate of deformation, so that the deviatoric stress is linear in the rate of deformation. The expression

D kk =0 ensures incompressibility, and represents a single equation:

D 11 +D

22 +D

33 =0 .

The expression s ij =2ηD

ij ensures isotropy, and represents six equations:

σ 12 =2ηD

12 ,

σ 23 =2ηD

23 ,

σ 31 =2ηD

31 ,

σ 11 − σ 11 +σ

22 +σ

33

3 =2ηD

11 ,

σ 22 − σ 22 +σ

33 +σ

11

3 =2ηD

22 ,

σ 33 − σ 33 +σ

11 +σ

22

3 =2ηD

33 .

The sum of the last three equations gives an identity, 0 = 0. Thus, the last three equations consist of only two independent equations.

The two types of relations, D kk =0 and s

ij =2ηD

ij , consist of a total of six

independent linear equations between the twelve components of stress and rate of deformation. Thermodynamic state and thermodynamic property. A fluid can be in many thermodynamic states, and has many thermodynamic properties. Examples of thermodynamic properties include temperature, pressure, volume, energy, entropy, smell, color, electric field, and polarization. When the fluid is in a particular thermodynamic state, the value of every thermodynamic property is fixed. We specify a thermodynamic state of a fluid by values of its thermodynamic properties.

• Plasticity http://imechanica.org/node/17162 Z. Suo

October 13, 2014 Nonlinear viscosity 2

How many thermodynamic properties do we need to differentiate thermodynamic states of a fluid? The answer is two. How do we know? The answer depends on the evidence from our experience and on the limit of our attention. One property is too few: if we hold the value of the temperature we can still change the thermodynamic state of the fluid by changing the pressure. Three properties are too many: once the temperature and pressure are fixed, the thermodynamic state of the fluid is fixed if we neglect minor effects of, say, the electric field and magnetic field. As a model, we specify the thermodynamic state using two thermodynamic properties, for example, temperature and pressure. Once the values of the two thermodynamic properties are fixed, the fluid is in a fixed thermodynamic state, and values of all other thermodynamic properties are fixed. In this model, the deviatoric stress is not a thermodynamic property, and does not affect thermodynamic state of the fluid. Viscosity is a thermodynamic property. The viscosity of a fluid is a

function of temperature and pressure, η T ,p( ) . The experimental data are sometimes fit to an expression

η T ,p( ) =η0 exp q+V

a p

kT

!

" ##

\$

% && .

Here kT is the temperature in the unit of energy; the pre-factor η 0

, the activation

energy q and the activation volume V a

are parameters used to fit the

experimental data. The model also requires that the mean stress equal the thermodynamic pressure:

σ 11 +σ

22 +σ

33

3 = p .

The pressure-dependent viscosity makes the model of viscosity, s ij =2ηD

ij ,

nonlinear. Is this nonlinear effect significant? At room temperature,

kT = 1.38×10−23J/K( ) 300K( ) = 4×10−21 J . Assume that the activation volume is comparable to the volume per molecule, V

a ≈ 3×10−29m3 , and assume a value of

pressure p = 106Pa . We find that V a p > kT , and the viscosity is sensitive to temperature. At the atmospheric pressure, the viscosity of water is around 10-3

Pa.s at room temperature, 1.8×10−3 Pa.s at the freezing point, and 0.28×10−3 Pa.s

• Plasticity http://imechanica.org/node/17162 Z. Suo

October 13, 2014 Nonlinear viscosity 3

at the boiling point. We will usually consider isothermal flows, in which the temperature is held fixed. Thermodynamic inequality. A fluid is subject to an applied force, and is in thermal contact with a heat bath held at a fixed temperature. We assume that the fluid and the heat bath are in thermal equilibrium, so that the temperature of the fluid is fixed at that of the heat bath. We further assume that the fluid is incompressible. Once the values of the two thermodynamic properties—temperature and volume—are fixed, the fluid is in a fixed thermodynamic state, with all its thermodynamic properties fixed. In particular, the Helmholtz free energy of the fluid is constant, even when the fluid flows. The

potential energy of the applied force changes at the rate −σ ij D ij V , where V is the

volume of the fluid. For an incompressible fluid, D kk =0 , we can confirm an

identity, σ ij D ij = s

ij D ij

. The fluid and the applied force together constitute a

composite thermodynamic system at a fixed temperature. The Helmholtz free

energy of the composite system sums over its parts: 0−σ ij D ij V .

Thermodynamics requires that the Helmholtz free energy of the composite system should never increase:

s ij D ij ≥0 .

The decrease in the potential entirely dissipates into the heat bath. That is, in an isothermal flow, the incompressible, viscous fluid does not change its own Helmholtz free energy, but converts the potential energy of the applied force into energy in the heat bath. This thermodynamic condition applies when an incompressible, viscous fluid flows at a fixed temperature. The condition does not require linearity and isotropy.

For a linear, isotropic, incompressible, viscous fluid, s ij D ij =2ηD

ij D ij

.

Note that D ij D ij ≥0 for any rate of deformation. The model satisfies the

thermodynamic inequality for arbitrary rate of deformation if and only if the viscosity is non-negative: η ≥0 . For many materials, stress is nonlinear in rate of deformation. Metals creep at elevated temperatures, but the stress is often nonlinear in the rate of deformation (Frost and Ashby 1982). Nonlinear viscosity is also observed in many other materials, such as ice, ice cream, skin cream, toothpaste, and chocolate. We can test a material in shear, and measure the relation between the stress and the rate of deformation. We often fit the experimental data to a power law:

τ = A γ N ,

• Plasticity http://imechanica.org/node/17162 Z. Suo

October 13, 2014 Nonlinear viscosity 4

where A and N are parameters used to fit the experimentally measured curve between the stress and the rate of deformation. The power law is also written as

γ = τ A

!

" #

\$

% &

n

,

where n = 1/N. For example, a representative values for ice is n = 3 (Glen 1955). The creep of ice contributes to the dynamics of glaciers. We plot the experimental data as a curve in the plane with the axes of τ and γ . The curve is often monotonic. We represent the curve as a function:

τ = g γ( ) . The same relation can also be written in another form:

γ = h τ( ) . Nonlinear, isotropic, incompressible, viscous fluid. We have tested a material under shear, and measured the curve between the stress and the

rate of deformation, τ = g γ( ) . What can we do with the curve? We can compare the curve with that of another material. We can study the microscopic origins for the values of A and N (Ashby and Frost, 1982). We can even use the data to solve some boundary-value problems. But to solve boundary-value problems in general, we will need to have relation between stress and rate of deformation in arbitrary state of stress, not just shear.

Isotropy of the material implies that the relation τ = g γ( ) applies in all shearing directions:

σ 12 = g 2D

12( ) , σ

23 = g 2D

23( ) , σ

31 = g 2D

31( ) . The experimental data, τ = g γ( ) , in general do not allow us to predict the relation between a tensile stress and the rate of extension.

stress, τ

rate of deformation,

γ = h τ( )τ = g γ( )

γ

• Plasticity http://imechanica.org/node/17162 Z. Suo

October 13, 2014 Nonlinear viscosity 5

By viscosity here we mean that the deviatoric stress depends on the rate of deformation. In general we need to determine functions of five independent variables:

s ij = s

ij D 11 ,D 22 ,D 23 ,D 31 ,D 12( ) .

Here we have dropped the dependence on D 33

; due to incompressibility, D 33

is

not an independent quantity, D 33 = −D

11 −D

22 .

A brute-force method to determine these functions