Ch.9. Constitutive Equations in Fluids

CH.9. CONSTITUTIVE EQUATIONS IN FLUIDS

Multimedia Course on Continuum Mechanics Overview

 Introduction  Fluid Mechanics Lecture 1  What is a Fluid?  Pressure and Pascal´s Law Lecture 3  Constitutive Equations in Fluids Lecture 2  Fluid Models  Newtonian Fluids  Constitutive Equations of Newtonian Fluids Lecture 4  Relationship between Thermodynamic and Mean Pressures  Components of the Constitutive Equation Lecture 5  Stress, Dissipative and Recoverable Power  Dissipative and Recoverable Powers Lecture 6  Thermodynamic Considerations  Limitations in the Viscosity Values

2 9.1 Introduction

Ch.9. Constitutive Equations in Fluids

3 What is a fluid?

Fluids can be classified into:

 Ideal (inviscid) fluids:  Also named perfect fluid.  Only resists normal, compressive stresses (pressure).  No resistance is encountered as the fluid moves.

 Real (viscous) fluids:  Viscous in nature and can be subjected to low levels of shear stress.  Certain amount of resistance is always offered by these fluids as they move.

5 9.2 Pressure and Pascal’s Law

Ch.9. Constitutive Equations in Fluids

6 Pascal´s Law

 Pascal’s Law: In a confined fluid at rest, pressure acts equally in all directions at a given point.

7 Consequences of Pascal´s Law

 In fluid at rest:  there are no shear stresses  only normal forces due to pressure are present.

 The stress in a fluid at rest is isotropic and must be of the form:

σ = − p01

σδij =−∈p0 ij ij,{} 1, 2, 3

Where p 0 is the hydrostatic pressure.

8 Pressure Concepts

 Hydrostatic pressure, p 0 : normal compressive stress exerted on a fluid in equilibrium.

 Mean pressure, p : minus the mean stress. REMARK 1 p = −σ= − Tr () σ Tr () σ is an invariant, m 3 thus, so are σ m and p .

 Thermodynamic pressure, p : Pressure variable used in the constitutive equations . It is related to density and temperature through the kinetic equation of state. REMARK F ()ρθ,p,= 0 In a fluid at rest, p0 = pp =

9 Pressure Concepts

 Barotropic fluid: pressure depends only on density.

F(ρρ,p) = 0 pf= ( )

 Incompressible fluid: particular case of a barotropic fluid in which density is constant.

F(ρθ,p, ) ≡ F( ρ) = ρ −= k 0 ρ==k const.

10 9.3 Constitutive Equations

Ch.9. Constitutive Equations in Fluids

11 Reminder – Governing Eqns.

 Governing equations of the thermo-mechanical problem: Conservation of Mass. ρρ+ ∇⋅ = 1 eqn.  v 0 Continuity Equation.

Linear Momentum Balance. ∇⋅σ +ρρbv =  3 eqns. Cauchy’s Motion Equation. Angular Momentum Balance. Symmetry 8 PDE + T 3 eqns. σσ= of Cauchy Stress Tensor. 2 restrictions Energy Balance. First Law of ρρur =σ :dq + −∇⋅ 1 eqn. Thermodynamics.

−−+ρθ()usσ :d ≥0 Clausius-Planck Inequality. Second Law of 1 2 restrictions −q ⋅≥∇θ 0 Heat flux Thermodynamics. ρθ 2 Inequality.

 19 scalar unknowns: ρ , v , σ , u , q , θ , s .

12 Reminder – Constitutive Eqns.

 Constitutive equations of the thermo-mechanical problem: Thermo-Mechanical 6 eqns. σσ= ()v,,θ ζ Constitutive Equations.

Entropy ss= ()v,,θ ζ Constitutive Equation. 1 eqn. (19+p) PDE + (19+p) unknowns Thermal Constitutive Equation. Fourier’s qq=()θθ =−∇K Law of Conduction. 3 eqns.

uf= ()ρθ,,,v ζ Caloric State Equations. (1+p) eqns. Kinetic Fi (){}ρθ, ,ζ = 0 ip ∈ 1,2,...,

set of new thermodynamic

variables: ζ = {} ζζ 12 , ,..., ζ p .  The mechanical and thermal problem can be uncoupled if the temperature distribution is known a priori or does not intervene in the constitutive eqns. and if the constitutive eqns. involved do not introduce new thermodynamic variables.

13 Constitutive Equations

 Constitutive equations  Together with the remaining governing equations, they are used to solve the thermo/mechanical problem.

 In fluid mechanics, these are grouped into: Thermo-mechanical constitutive equations Caloric equation of state σ =−+p1 fd(),,ρθ u = g,()ρθ  σ=−+ δ ρθ ∈  ijp ijf ij ()d , ,ij , 1, 2, 3 Entropy constitutive equation Kinetic equation of state ss= ()d,,ρθ F ()ρθ,p,= 0 q =−⋅k ∇θ  REMARK Fourier’s Law  ∂θ s =−∈ dv() = ∇ v qi k ij, 1, 2, 3  ∂xi 14 Viscous Fluid Models

 General form of the thermo-mechanical constitutive equations:

σ =−+p1 fd(),,ρθ

σij=−+p δ ijf ij (){}d , ρθ ,ij , ∈ 1, 2, 3

 Depending on the nature of fd () ,, ρθ , fluids are classified into : 1. Perfect fluid: fd(),,ρθ =⇒=− 0 σ p1 2. Newtonian fluid: f is a linear function of the strain rate 3. Stokesian fluid: f is a non-linear function of its arguments

15 9.4. Newtonian Fluids

Ch.9. Constitutive Equations in Fluids

16 Constitutive Equations of Newtonian Fluids

 Mechanic constitutive equations: σ =−+p1 C :d

σδij=−+p ijC ijkl d kl ij, ∈{} 1, 2, 3 where C is the 4th-order constant (viscous) constitutive tensor.

C =λµ11 ⊗+2 I  Assuming: =++λδ δ µ δ δ δ δ  an isotropic medium Cijkl ij kl( ik jl il jk )  the stress tensor is symmetrical i, jkl , ,∈{} 1, 2, 3  Substitution of C into the constitutive equation gives: σ =−+p11λµ Tr ()dd +2 REMARK λµ σ=−++p δ λδ d2 µ d ij , ∈{} 1, 2, 3 and are not necessarily constant. ij ij ll ij ij Both are a function of ρθ and .

17 Relationship between Thermodynamic and Mean Pressures

 Taking the mechanic constitutive equation,

σij=−++p δ ij λδ d ll ij2 µ d ij ij , ∈{} 1, 2, 3

 Setting i=j, summing over the repeated index, and noting that

δ ii = 3 , we obtain 1 σ=−++3p() 32 λµ dp =− 3 = − σ ii ll ()p ii −3p Tr()d 3

bulk viscosity κ 2 2 p=++ p()λµ Tr()dd =+ p κ Tr () κλ= + µ 3 3

18 Relationship between Thermodynamic and Mean Pressures

 Considering the continuity equation, ddρρ1 +ρ ∇⋅vv =0 ∇⋅ =− dt ρ dt  And the relationship p= p +κ Tr (d) ∂vi Tr(d) = dii = = ∇ ⋅ v ∂xi

REMARK κρd For a fluid at rest, v =0 ppp = = pp= +κ ∇⋅v = p − 0 ρ dρ dt For an incompressible fluid, =0 pp = dt 2 For a fluid with , κ  = 0 λµ=−=pp 3 Stokes' condition

19 9.5 Components of the Constitutive Equations Ch.9. Constitutive Equations in Fluids

20 Components of the Constitutive Equation

 Given the Cauchy stress tensor, the following may be defined: σσ= + σ′ σ =−+p11λµ Tr ()dd +2 sph = − p1

 SPHERICAL PART – mean pressure p= p −κκ ∇⋅vd = p − Tr ()

 DEVIATORIC PART −+p11λµ Tr ()dd +2 =−+p 1σ′ σ=′ ()()p−+ p11λµ Tr dd +2

p= p −κ Tr ()d 2 2 κλ= + µ σ=′ −+()λµTr ()()d11 + λTr dd +2 µ 3 3 1 deviatoric part of the rate of σ=2′′µµ()dd−=Tr ()1 2 d strain tensor   3    =d′ 21 Components of the Constitutive Equation

 Given the Cauchy stress tensor, the following may be defined:  SPHERICAL PART – mean pressure p

p= p −κκ ∇⋅vd = p − Tr ( ) p −κ

 DEVIATORIC PART – deviator stress tensor σ ij′ Tr (d)

σ′′= 2µ d 2µ  The stress tensor is then dij′ 1 σ=Tr ( σσ)11 +=−+′′p σ 3 REMARK = −3p from the definition Note that κ is not a of mean pressure function of d, while µµ = ( d ) . 22 9.6 Stress, Dissipative and Recoverable Powers Ch.9. Constitutive Equations in Fluids

23 Reminder – Stress Power

 Mechanical Energy Balance:

d 1 P() t=ρρb ⋅ v dV +⋅ t v dS =v2dV +σ :d dV e ∫∫VV∂ dt ∫∫2 VVt ≡ V

external mechanical power kinetic energy stress power entering the medium d Pt() =() t + P e dt K σ REMARK The stress power is the mechanical power entering the system which is not spent in changing the kinetic energy. It can be interpreted as the work per unit of time done by the stress in the deformation process of the medium. A rigid solid will have zero stress power.

24 Dissipative and Recoverable Powers

 Stress Power = σ :d dV ∫ 1 V d=Tr() d1 + d′ 3 σσ=−+p1 ′ 1 σσ::d=−+( p11′′)  Tr (dd) + = 3 ′ = 3 =Tr (d ) = 0 11 =−pTr (d) 11:: +−σσ′′ ddp 1 : ′ + Tr ( d) ′ :1 = 33=Tr (σ′) = 0 =−+pTr (dd) σ′′:

σ′′= 2µd σ :d=−+pTr( d) κµ Tr 2 ( d) +2: dd′′ p= p −κ Tr (d) RECOVERABLE STRESS DISSIPATIVE STRESS POWER, . POWER, . WR 2WD

25 Dissipative and Recoverable Parts of the Cauchy Stress Tensor

 Associated to the concepts of recoverable and dissipative powers, the Cauchy stress tensor is split into:

σ =−+p11λµ Tr ()dd +2 RECOVERABLE DISSIPATIVE PART, . PART, . σ R σ D

 And the recoverable and dissipative powers are rewritten as:

WRR=−=−= pTr()d p1 :dσ :d 2 2WDD=κµ Tr ()d +=2σ d′′: d :d REMARK For an incompressible fluid,

W0R =−=pTr ()d

26 Thermodynamic considerations

 Specific recoverable stress power is an exact differential, 11 dG W =σ :d = → (exact differential) ρρRRdt Then, the recoverable stress work per unit mass in a closed cycle is zero: BA≡≡11BA BA ≡ =σ = = −= ∫∫WR dt R:d dt ∫ dG GBA≡ G A 0 AAρρ A

 This justifies the denomination “recoverable stress power”.

28 Thermodynamic Considerations

 According to the 2nd Law of Thermodynamics, the dissipative power is necessarily non-negative,

2 2WDD≥ 0 2W =κµ Tr (d) +=2 dd′′: 0 d = 0

In a closed cycle, the work done by the dissipative stress per unit mass will, in general, be different to zero: BB≡ 1 σ > ∫ D :d dt 0 A ρ

2WD > 0

 This justifies the denomination “dissipative power”.

29 Limitations in the Viscosity Values

 The thermodynamic restriction, 2 2WD =κµ Tr ()d +≥2 dd′′: 0 introduces limitations in the values of the viscosity parameters κλ , and µ :

1. For a purely spherical deformation rate tensor:

Tr ()d ≠ 0 2 20W=κ Tr 2 ()d ≥ κλ=+≥ µ0 d′ = 0 D 3 2. For a purely deviatoric deformation rate tensor: Tr ()d = 0 2W = 2:µµdd′ ′ = 2dd ′′ ≥ 0 ≥µ 0 d′ ≠ 0 D ij ij > 0

30 Chapter 9 Constitutive Equations in Fluids

9.1 Concept of Pressure Several concepts of pressure are used in continuum mechanics (hydrostatic pressure, mean pressure and thermodynamic pressure) which, in general, do not coincide. 9.1.1 Hydrostatic Pressure

Definition 9.1. Pascal’s law In a confined fluid at rest, the stress state on any plane containing a given point is the same and is characterized by a compressive normal stress. Theory and Problems In accordance with Pascal’s law, the stress state of a fluid at rest is characterized by a stress tensor of the type Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar σ = −p 1 0 , (9.1) σij = −p0 δij i, j ∈{1,2,3} where p0 is denoted as hydrostatic pressure (see Figure 9.1).

Deﬁnition 9.2. The hydrostatic pressure is the compressive normal stress, constant on any plane, that acts on a ﬂuid at rest.

439 440 CHAPTER 9. CONSTITUTIVE EQUATIONS IN FLUIDS

Figure 9.1: Stress state of a ﬂuid at rest.

Figure 9.2: Mohr’s circle of the stress tensor of a ﬂuid at rest.

Remark 9.1. The stress tensor of a ﬂuid at rest is a spherical tensor and its representation in the Mohr’s plane is a point (see Figure 9.2). Consequently, any direction is a principal stress direction and the stress state is constituted by the state deﬁned in Section 4.8 of Chap- ter 4 as hydrostatic stress state.

Theory and Problems 9.1.2 Mean Pressure Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

Definition 9.3. The mean stress σm is defined as 1 1 σ = Tr (σ )= σ . m 3 3 ii The mean pressure p¯ is defined as minus the mean stress,

def 1 1 p¯ = mean pressure = −σ = − Tr (σ )=− σ . m 3 3 ii

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Concept of Pressure 441

Remark 9.2. In a ﬂuid at rest, the mean pressurep ¯ coincides with the hydrostatic pressure p0, 1 σ = −p 1 =⇒ σ = (−3p )=−p =⇒ p¯ = p . 0 m 3 0 0 0 Generally, in a ﬂuid in motion the mean pressure and the hydrostatic pressure do not coincide.

Remark 9.3. The trace of the Cauchy stress tensor is a stress invariant. Consequently, the mean stress and the mean pressure are also stress invariants and, therefore, their values do not depend on the Cartesian coordinate system used.

9.1.3 Thermodynamic Pressure. Kinetic Equation of State A new thermodynamic pressure variable, named thermodynamic pressure and denoted as p, intervenes in the constitutive equations of ﬂuids or gases.

Deﬁnition 9.4. The thermodynamic pressure is the pressure variable that intervenes in the constitutive equations of ﬂuids and gases, and is related to the density ρ and the absolute temperature θ by means of the kinetic equationTheory of state, F and(p,ρ,θ Problems)=0. Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar Example 9.1 The ideal gas law is a typical example of kinetic equation of state: F (p,ρ,θ) ≡ p − ρRθ = 0 =⇒ p = ρRθ , where p is the thermodynamic pressure and R is the universal gas constant.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 442 CHAPTER 9. CONSTITUTIVE EQUATIONS IN FLUIDS

Remark 9.4. In a ﬂuid at rest, the hydrostatic pressure p0, the mean pressurep ¯ and the thermodynamic pressure p coincide.

Fluid at rest : p0 = p¯ = p Generally, in a ﬂuid in motion the hydrostatic pressure, the mean pressure and the thermodynamic pressure do not coincide.

Remark 9.5. A barotropic fluid is defined by a kinetic equation of state in which the temperature does not intervene. Barotropic fluid : F (p,ρ)=0 =⇒ p = f (ρ)=⇒ ρ = g(p)

Remark 9.6. An incompressible ﬂuid is a particular case of barotropic ﬂuid in which density is constant (ρ (x,t)=k = const.). In this case, the kinetic equation of state can be written as

F (p,ρ,θ) ≡ ρ − k = 0 and does not depend on the pressure or the temperature.

Theory and Problems

9.2 ConstitutiveContinuum Equations Mechanics in Fluid for Mechanics Engineers © X. Oliver and C. Agelet de Saracibar Here, the set of equations, generically named constitutive equations, that must be added to the balance equations to formulate a problem in ﬂuid mechanics (see Section 5.13 in Chapter 5) is considered. These equations can be grouped as follows: a) Thermo-mechanical constitutive equation This equation expresses the Cauchy stress tensor in terms of the other thermodynamic variables, typically the thermodynamic pressure p, the strain rate tensor d (which can be considered an implicit function of the velocity, d(v)=∇Sv), the density ρ and the absolute temperature θ.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Constitutive Equation in Viscous Fluids 443

Thermo-mechanical σ = −p1 + f(d,ρ,θ) 6 equations (9.2) constitutive equation:

b) Entropy constitutive equation An algebraic equation that provides the speciﬁc entropy s in terms of the strain rate tensor, the density and the absolute temperature.

Entropy s = s(d,ρ,θ) 1 equation (9.3) constitutive equation:

c) Thermodynamic constitutive equations or equations of state These are typically the caloric equation of state, which deﬁnes the speciﬁc internal energy u, and the kinetic equation of state, which provides an equation for the thermodynamic pressure. Caloric equation of u = g(ρ,θ) state: 2 equations (9.4) Kinetic equation of F (ρ, p,θ)=0 state:

d) Thermal constitutive equations The most common one is Fourier’s law, which defines the heat flux by conduction q as ⎧ ⎨ q = −k · ∇θ Fourier’s ∂θ ⎩ 3 equations (9.5) law: Theoryqi = kij andi ∈{1 Problems,2,3} ∂x j whereContinuumk is the (symmetrical Mechanics second-order) for tensor Engineers of thermal conductivity, which is a property of© the X. fluid. Oliver For and the isotropic C. Agelet case, de the Saracibar thermal conductivity tensor is a spherical tensor k = k1 and depends on the scalar parameter k, which is the thermal conductivity of the fluid.

9.3 Constitutive Equation in Viscous Fluids The general form of the thermo-mechanical constitutive equation (see (9.2)) for a viscous ﬂuid is

σ = −p 1 + f(d,ρ,θ) , (9.6) σij = −p δij+ fij(d,ρ,θ) i, j ∈{1,2,3}

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 444 CHAPTER 9. CONSTITUTIVE EQUATIONS IN FLUIDS where f is a symmetrical tensor function. According to the character of the function f, the following models of fluids are defined: a) Stokesian or Stokes fluid: the function f is a non-linear function of its arguments. b) Newtonian fluid: the function f is a linear function of its arguments. c) Perfect fluid: the function f is null. In this case, the mechanical constitutive equation is σ = −p1. In the rest of this chapter, only the cases of Newtonian and perfect fluids will be considered.

Remark 9.7. The perfect ﬂuid hypothesis is frequently used in hy- draulic engineering, where the ﬂuid under consideration is water.

9.4 Constitutive Equation in Newtonian Fluids The mechanical constitutive equation1 for a Newtonian ﬂuid is

σ = −p 1 +C : d , (9.7) σij = −p δij+ Cijkldkl i, j ∈{1,2,3} where C is a constant fourth-order (viscosity) constitutive tensor. A linear dependency of the stress tensor σ on the strain rate tensor d is obtained as a result of (9.7). For an isotropic NewtonianTheory ﬂuid and, the constitutive Problems tensor C is an isotropic fourth-order tensor. ContinuumC = λ1 ⊗ 1 + 2μI Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar (9.8) Cijkl = λδijδkl + μ δikδ jl + δilδ jk i, j,k,l ∈{1,2,3}

Replacing (9.8) in the mechanical constitutive equation (9.7) yields σ = −p 1 +(λ1 ⊗ 1 + 2μI) : d = −p 1 + λ Tr (d)1 + 2μ d , (9.9) which corresponds to the constitutive equation of an isotropic Newtonian ﬂuid.

1 Note that the thermal dependencies of the constitutive equation are not considered here and, thus, the name mechanical constitutive equations.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Constitutive Equation in Newtonian Fluids 445

Constit. eqn. of σ = −p 1 + λ Tr (d)1 + 2μ d (9.10) an isotropic σ = − δ + λ δ + μ , ∈{ , , } Newtonian ﬂuid ij p ij dll ij 2 dij i j 1 2 3

Remark 9.8. Note the parallelism that can be established between the constitutive equation of a Newtonian ﬂuid and that of a linear elastic solid (see Chapter 6):

Newtonian ﬂuid Linear elastic solid σ = −p 1 +C : d σ = C : ε

σij = −p δij+ Cijkldkl σij = Cijklεkl

Remark 9.9. The parameters λ and μ physically correspond to the viscosities, which are understood as material properties. In the most general case, they may not be constant and can depend on other thermodynamic variables, λ = λ (ρ,θ) and μ = μ (ρ,θ) .

A typical example is the dependency of the viscosity on the temper- −α(θ−θ ) ature in the form μ (θ)=μ0 e 0 , which establishes that the ﬂuid’s viscosity decreases as temperature increases (see Figure 9.3). Theory and Problems

Figure 9.3: Possible dependency of the viscosity μ on the absolute temperature θ.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 446 CHAPTER 9. CONSTITUTIVE EQUATIONS IN FLUIDS

9.4.1 Relation between the Thermodynamic and Mean Pressures In general, the thermodynamic pressure p and the mean pressurep ¯ inaNew- tonian ﬂuid in motion will be different but are related to each other. From the (mechanical) constitutive equation of a Newtonian ﬂuid (9.10),

σ = −p 1 + λ Tr (d)1 + 2μ d =⇒

(σ ) = − ( )+λ ( ) ( )+ μ ( )=− +( λ + μ) ( )=⇒ Tr p Tr 1 Tr d Tr 1 2 Tr d 3p 3 2 Tr d − 3¯p 2 p = p¯ + λ + μ Tr (d)=p¯ + K Tr (d) 3 K (9.11) where K is denoted as bulk viscosity.

2 Bulk viscosity : K = λ + μ (9.12) 3

Using the mass continuity equation (5.24), results in dρ 1 dρ + ρ∇ · v = 0 =⇒ ∇ · v = − (9.13) dt ρ dt Then, considering the relation ∂ ( )= = vi = ∇ · Tr d dii ∂ v (9.14) Theory andxi Problems and replacing in (9.11), yields Continuum Mechanics for Engineers © X. Oliver and C. AgeletK dρ de Saracibar p = p¯ + K∇ · v = p¯ − (9.15) ρ dt which relates the mean and thermodynamic pressures.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Constitutive Equation in Newtonian Fluids 447

Remark 9.10. In accordance with (9.15), the thermodynamic pressure and the mean pressure in a Newtonian ﬂuid will coincide in the following cases:

• Fluid at rest: v = 0 =⇒ p = p¯ = p0 dρ • Incompressible ﬂuid: = 0 =⇒ p = p¯ dt • Fluid with null bulk viscosity K (Stokes’ condition2): 2 K = 0 =⇒ λ = − μ =⇒ p = p¯ 3

9.4.2 Constitutive Equation in Spherical and Deviatoric Components Spherical part From (9.15), the following relation is deduced.

p¯ = p − K ∇ · v = p − K Tr (d) (9.16)

Deviatoric part Using the decomposition of the stress tensor σ and the strain rate tensor d in its spherical and deviator components, and replacing in the constitutive equation (9.10), results in Theory and Problems 1 σ = Tr (σ )1 + σ = −p¯ 1 + σ = −p1 + λ Tr (d)1 + 2μ d =⇒ 3 − Continuum3¯p Mechanics for Engineers σ =(− ) +©λ X.( Oliver) + μ and= C.λ − AgeletK de( ) Saracibar+ μ =⇒ p¯ p 1 Tr d 1 2 d Tr d 1 2 d −K ( ) Tr d λ + 2 μ 3 2 1 σ = − μ Tr (d)1 + 2μd = 2μ d − Tr (d)1 =⇒ 3 3 d (9.17)

2 Stokes’ condition is assumed in certain cases because the results it provides match the experimental observations.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 448 CHAPTER 9. CONSTITUTIVE EQUATIONS IN FLUIDS

σ = 2μd (9.18) where (9.16) and (9.12) have been taken into account.

9.4.3 Stress Power, Recoverable Power and Dissipative Power Using again the decomposition of the stress and strain rate tensors in their spherical and deviatoric components yields

1 σ = −p¯ 1 + σ and d = Tr (d)1 + d , (9.19) 3 and replacing in the expression of the stress power density (stress power per unit of volume) σ : d, results in3

1 σ : d =(−p¯ 1 + σ ) : Tr (d)1 + d = 3 1 1 = − p¯ Tr (d) 1 : 1 +σ : d − p¯ 1 : d + Tr (d) σ : 1 = 3 3 (9.20) 3 Tr d = 0 Tr σ = 0 = −p¯ Tr (d)+σ : d .

Replacing (9.16) and (9.17)in(9.20) produces σ : d = − p − K Tr (d) Tr (d)+2μ d : d . (9.21) Theory and Problems

2 σ : d = −p Tr (d) + KTr (d)+2μ d : d = WR + 2WD Continuum Mechanics for Engineers (9.22) recoverable power© X. Oliverdissipative and C. power Agelet de Saracibar WR 2WD

Recoverable power density: WR = −p Tr (d) (9.23) 2 Dissipative power density: 2WD = KTr (d)+2μ d : d

3 The property that the trace of a deviator tensor is null is used here.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Constitutive Equation in Newtonian Fluids 449

Associated with the concepts of recoverable and dissipative powers, the recoverable and dissipative parts of the stress tensor, σ R and σ D, respectively, are deﬁned as σ = − +λ ( ) + μ =⇒ σ = σ + σ . p1 Tr d 1 2 d R D (9.24) σ R σ D Using the aforementioned notation, the recoverable, dissipative and total power densities can be rewritten as ⎧ ⎨ WR = −p Tr (d)=−p 1 : d = σ R : d , ⎩ = K 2 ( )+ μ = σ , 2WD Tr d 2 d : d D : d (9.25)

σ : d =(σ R + σ D) : d = σ R : d + σ D : d = WR + 2WD .

Remark 9.11. In an incompressible ﬂuid, the recoverable power is null. In effect, since the ﬂuid is incompressible, dρ/dt = 0 , and considering the mass continuity equation (5.24), 1 dρ ∇ · v = − = 0 = Tr (d)=⇒ W = −p Tr (d)=0 . ρ dt R

Remark 9.12. IntroducingTheory the and decomposition Problems of the stress power (9.25), the balance of mechanical energy (5.73) becomes dK dK ContinuumP = + σ : d MechanicsdV = + σ for: d dV Engineers+ σ : d dV e dt © X. Oliverdt and C.R Agelet de SaracibarD V V V

dK P = + W dV + 2W dV , e dt R D V V

which indicates that the mechanical power entering the ﬂuid Pe is invested in modifying the kinetic energy K and creating recoverable power and dissipative power.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 450 CHAPTER 9. CONSTITUTIVE EQUATIONS IN FLUIDS

9.4.4 Thermodynamic Considerations 1) It can be proven that, under general conditions, the speciﬁc recoverable power (recoverable power per unit of mass) is an exact differential 1 1 dG W = σ : d = . (9.26) ρ R ρ R dt In this case, the recoverable work per unit of mass performed in a closed cycle will be null (see Figure 9.4),

B≡A B≡A B≡A 1 = 1 σ = = − = , ρ WR dt ρ R : d dt dG GB≡A GA 0 (9.27) A A A which justiﬁes the denomination of WR as recoverable power.

Figure 9.4: Closed cycle.

2) The second law of thermodynamicsTheory and allows Problems proving that the dissipative power 2WD in (9.25) is always non-negative, Continuum2W ≥ 0; Mechanics 2W = 0 for⇐⇒ Engineersd = 0 (9.28) D © X. Oliver andD C. Agelet de Saracibar and, therefore, in a closed cycle the work performed per unit of mass by the dissipative stresses will, in general, not be null,

B 1 σ : d dt > 0 . (9.29) ρ D A 2WD > 0

This justiﬁes the denomination of 2WD as (non-recoverable) dissipative power. The dissipative power is responsible for the dissipation (or loss of energy) phe- nomenon in ﬂuids.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Constitutive Equation in Newtonian Fluids 451

Example 9.2 – Explain why an incompressible Newtonian ﬂuid in motion that is not provided with external power (work per unit of time) tends to reduce its velocity to a complete stop.

Solution The recoverable power in an incompressible ﬂuid is null (see Remark 9.11). In addition, the dissipative power 2WD is known to be always non-negative (see (9.28)). Finally, applying the balance of mechanical energy (see Re- mark 9.12) results in dK 0 = P = + W dV + 2W dV =⇒ e dt R D V = 0 V dK d 1 = ρv2dV = − 2 W dV < 0 dt dt 2 D V V > 0 and, therefore, the ﬂuid looses (dissipates) kinetic energy and the velocity of its particles decreases.

9.4.5 Limitations in the Viscosity Values

Due to thermodynamic considerations, the dissipative power 2WD in (9.25) has been seen to always be non-negative,

2 . 2WD = KTr (d)+2μ d : d ≥ 0 (9.30)

This thermodynamic restrictionTheory introduces and limitations Problems in the admissible values of the viscosity parameters K, λ and μ of the fluid. In effect, given a certain fluid, theContinuum aforementioned restriction Mechanics must be verifiedfor Engineers for all motions (that is, for all velocity fields v)© that X. the Oliver fluid may and possibly C. Agelet have. de Therefore, Saracibar it must be verified for any arbitrary value of the strain rate tensor d = ∇S (v). Consider, in particular, the following cases: a) The strain rate tensor d is a spherical tensor. In this case, from (9.30) results 2 Tr (d) = 0; d = 0 =⇒ 2WD = KTr (d) ≥ 0 =⇒ 2 (9.31) K = λ + μ ≥ 0 3 such that only the non-negative values of the bulk viscosity K are feasible.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 452 CHAPTER 9. CONSTITUTIVE EQUATIONS IN FLUIDS

b) The strain rate tensor d is a deviatoric tensor. This type of ﬂow is schematically represented in Figure 9.5. In this case, from (9.30) results ( )= = =⇒ = μ = μ ≥ =⇒ Tr d 0; d 0 2WD 2 d : d 2 d ij : d ij 0 > 0 (9.32) μ ≥ 0

⎡ ⎤ ⎡ ⎤ 1 ∂vx v (y) ⎢ 0 0 ⎥ ⎢ x ⎥ ⎢ 2 ∂y ⎥ ( , )=⎢ ⎥ = ⎢ 1 ∂vx ⎥ = v x y ⎣ 0 ⎦ ; d ⎢ 00⎥ d ⎣ 2 ∂y ⎦ 0 000

Figure 9.5: Flow characterized by a deviatoric strain rate tensor. Theory and Problems

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961