CH.9. CONSTITUTIVE EQUATIONS IN FLUIDS. Multimedia Course on Continuum Mechanics

CH.9. CONSTITUTIVE EQUATIONS IN FLUIDS Multimedia Course on Continuum Mechanics

Overview Introduction Fluid Mechanics What is a Fluid? Pressure and Pascal s Law Constitutive Equations in Fluids Fluid Models Newtonian Fluids Constitutive Equations of Newtonian Fluids Relationship between Thermodynamic and Mean Pressures Components of the Constitutive Equation Stress, Dissipative and Recoverable Power Dissipative and Recoverable Powers Thermodynamic Considerations Limitations in the Viscosity Values Lecture 1 Lecture 3 Lecture 2 Lecture 4 Lecture 5 Lecture 6 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 = p 1 σ ij 0 { } = p0 δ i, j 1, 2,3 ij Where p 0 is the hydrostatic pressure. 8

Pressure Concepts Hydrostatic pressure, fluid in equilibrium. p 0 : normal compressive stress exerted on a Mean pressure, p : minus the mean stress. 1 p = σm = Tr σ 3 ( ) REMARK Tr ( σ) is an invariant, thus, so are and p. σ m 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 ( ρ, p, θ ) = 0 In a fluid at rest, p = p = p 0 9

Pressure Concepts Barotropic fluid: pressure depends only on density. ( ρ,p) = 0 = ( ρ) F p f Incompressible fluid: particular case of a barotropic fluid in which density is constant. ( ) ( ) F ρ,p, θ 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: ρ+ ρ v = 0 Conservation of Mass. Continuity Equation. 1 eqn. σ + ρb= ρv Linear Momentum Balance. Cauchy s Motion Equation. 3 eqns. 8 PDE + 2 restrictions T σ = σ ρu = σ : d+ ρr q Angular Momentum Balance. Symmetry of Cauchy Stress Tensor. Energy Balance. First Law of Thermodynamics. 3 eqns. 1 eqn. ( u θ s ) σ :d 0 ρ + 1 q θ 0 2 ρθ Clausius-Planck Inequality. Heat flux Inequality. Second Law of Thermodynamics. 2 restrictions 19 scalar unknowns: ρ, v, σ, u, q, θ, s. 12

Reminder Constitutive Eqns. Constitutive equations of the thermo-mechanical problem: (, θ, ) σ = σ v ζ Thermo-Mechanical Constitutive Equations. 6 eqns. (19+p) PDE + (19+p) unknowns Entropy s = s( v, θ, ζ ) Constitutive Equation. 1 eqn. ( θ) K q= q = θ Thermal Constitutive Equation. Fourier s Law of Conduction. 3 eqns. i (,,, ) u = f ρ v θ ζ ( ρθ,, ζ ) = 0 { 1, 2,..., } F i p Caloric Kinetic State Equations. (1+p) eqns. set of new thermodynamic variables: ζ = ζ1, ζ 2,..., 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. 14 In fluid mechanics, these are grouped into: Thermo-mechanical constitutive equations (, ρθ, ) ( d ) σ = p1 + f d σij = pδij + f ij, ρθ, i, j 1, 2, 3 Entropy constitutive equation (,, ) s = s d ρθ Fourier s Law q = k θ θ qi = k i, j 1, 2,3 xi Caloric equation of state ( ρθ) u = g, Kinetic equation of state F ( ρ, p, θ ) = 0 REMARK d v = ( ) s v

Viscous Fluid Models General form of the thermo-mechanical constitutive equations: σ = p1 + f d ij ij ij (, ρθ, ) f ( d,, ) i, j { 1, 2, 3} σ = pδ + ρθ Depending on the nature of f d, ρθ,, fluids are classified into : 1. Perfect fluid: ( ) ( ) f d, 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 σ where is the 4 th -order constant (viscous) constitutive tensor. Assuming: an isotropic medium the stress tensor is symmetrical Substitution of C into the constitutive equation gives: ( d) σ = p1+ λ Tr 1+ 2µ d { } σ = pδ + λd δ + 2 µ d i, j 1, 2,3 ij ij ll ij ij { } = pδ + C d i, j 1, 2,3 ij ij ijkl kl C C = λ 1 1+ 2µ I C ( ) = λδ δ + µ δ δ + δ δ ijkl ij kl ik jl il jk REMARK λ and µ { } i, jkl,, 1, 2, 3 are not necessarily constant. Both are a function of and. ρ θ 17

Relationship between Thermodynamic and Mean Pressures Taking the mechanic constitutive equation, Setting i=j, summing over the repeated index, and noting that δ ii = 3, we obtain 1 σ = 3p+ 3λ + 2µ d = 3p ( p = σ ) ii ( ) ll 3 p Tr( d) { } σ = pδ + λd δ + 2 µ d i, j 1, 2,3 ij ij ll ij ij 3 ii bulk viscosity κ 2 p = p + ( λ + µ ) Tr ( d) = p + κtr ( d 2 ) κ = λ + µ 3 3 18

Relationship between Thermodynamic and Mean Pressures Considering the continuity equation, dρ 1 dρ + ρ v= 0 v= dt ρ dt And the relationship Tr v x i ( d) = d = = v ii REMARK For a fluid at rest, v = 0 p= p = p 0 d For an incompressible fluid, ρ = 0 p = p dt For a fluid with, κ = 0 2 λ = µ p = p Stokes' 3 condition i p = p + κ Tr ( d) κ d ρ p = p+ κ v = p ρ dt 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: σ = p1+ λ Tr ( d) 1+ 2µ d σ = σ sph + σ = p1 SPHERICAL PART mean pressure p = p κ v = p κtr ( d) 21 DEVIATORIC PART 1 λ ( d) 1 2µ d 1 σ σ= ( p p) 1+ Tr ( d) 1+ 2 p + Tr + = p + 2 σ= ( λ + µ ) Tr ( d) 1+ λ Tr ( d) 1+ 2µ d 3 1 σ=2 µ ( d ( ) ) 3 Tr d 1 = 2 µ d = d p = p κ Tr ( d) λ µ d 2 κ = λ + µ 3 deviatoric part of the rate of strain tensor

Components of the Constitutive Equation Given the Cauchy stress tensor, the following may be defined: SPHERICAL PART mean pressure p p = p κ v = p κ Tr ( d) p κ DEVIATORIC PART deviator stress tensor σ ij Tr ( d) 22 σ = 2µ d The stress tensor is then 1 σ = ( ) 3 Tr σ 1+ σ = p 1+ σ = 3p from the definition of mean pressure 2µ d ij REMARK κ Note that is not a function of d, while. µ = µ ( d)

9.6 Stress, Dissipative and Recoverable Powers Ch.9. Constitutive Equations in Fluids 23

Reminder Stress Power Mechanical Energy Balance: d 1 2 Pe ( t) = ρb vdv + ds ρ v dv dv V t v = + V dt 2 σ :d V V V t external mechanical power entering the medium d Pe ( t) = K ( t) + P dt kinetic energy σ stress power 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 ( d) = σ :d dv 1 V d= ( ) 3 Tr d1+ d σ = p 1 + σ 1 σ: d= ( p1+ σ ) : Tr ( d) 1+ d = 3 = 3 = Tr ( d ) = 0 1 1 = ptr ( d) 11 : + σ : d p1: d + Tr ( d) σ : 1= 3 3 = Tr σ = = ptr + σ : d σ = 2µ d 2 σ ptr ( ) Tr ( ) p = p κtr ( d) ( ) 0 : d= d + κ d + 2 µ d : d RECOVERABLE STRESS POWER,. W R DISSIPATIVE STRESS POWER,. 2W D 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 = p1+ λ Tr ( d) 1+ 2µ d RECOVERABLE PART,. σ R DISSIPATIVE PART,. σ D And the recoverable and dissipative powers are rewritten as: R ( ) 2 ( ) W = ptr d = p1 :d = σ :d 2 W = κ Tr d + 2µ d : d = σ :d D R D REMARK For an incompressible fluid, W = ptr d = 0 R ( ) 26

Thermodynamic considerations Specific recoverable stress power is an exact differential, 1 1 dg W R R :d ρ = σ ρ = dt (exact differential) Then, the recoverable stress work per unit mass in a closed cycle is zero: B A B A B A 1 1 W Rdt = σ R:d dt = dg = GB A GA = 0 ρ ρ A A A This justifies the denomination recoverable stress power. 28

Thermodynamic Considerations According to the 2 nd Law of Thermodynamics, the dissipative power is necessarily non-negative, ( ) = κ d + 2µ d d = d= 2 2WD 0 2W D Tr : 0 0 In a closed cycle, the work done by the dissipative stress per unit mass will, in general, be different to zero: B B 1 σ D :d dt ρ A 2WD > 0 > 0 This justifies the denomination dissipative power. 29

Limitations in the Viscosity Values The thermodynamic restriction, 2 κ Tr d 2µ d d ( ) 2W = + : 0 D introduces limitations in the values of the viscosity parameters κ, λ and µ : 1. For a purely spherical deformation rate tensor: Tr ( d) 0 d = 0 ( d) = κ 2 2WD Tr 0 2 κ = λ + µ 3 0 2. For a purely deviatoric deformation rate tensor: Tr ( d) = 0 d 0 2WD = 2 µ d : d = 2µ dd ij ij 0 µ 0 > 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. 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 Theory and Problems X. Oliver and C. Agelet de Saracibar σ = p 0 1 σ ij = p 0 δ ij i, j {1,2,3}, (9.1) where p 0 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 Chapter 4 as hydrostatic stress state. 9.1.2 Mean Pressure Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Definition 9.3. The mean stress σ m is defined as σ m = 1 3 Tr (σ)=1 3 σ ii. The mean pressure p is defined as minus the mean stress, p def = mean pressure = σ m = 1 3 Tr (σ)= 1 3 σ 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 pressure p coincides with the hydrostatic pressure p 0, σ = p 0 1 = σ m = 1 3 ( 3p 0)= p 0 = p = p 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 equation of state, F (p,ρ,θ)=0. Continuum Mechanics for Engineers Theory and Problems 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 p 0, the mean pressure p and the thermodynamic pressure p coincide. Fluid at rest : p 0 = 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. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 9.2 Constitutive Equations in Fluid Mechanics 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)= S v), 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 constitutive equation: σ = p1 + f(d,ρ,θ) 6 equations (9.2) 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 constitutive equation: s = s(d,ρ,θ) 1 equation (9.3) 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 state: Kinetic equation of state: d) Thermal constitutive equations u = g(ρ,θ) F (ρ, p,θ)=0 2 equations (9.4) 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: q i = k ij i {1,2,3} x j Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar where k is the (symmetrical second-order) tensor of thermal conductivity, which is a property of the fluid. For the isotropic case, the thermal conductivity tensor is a spherical tensor k = k 1 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) σ ij = p δ ij + f ij (d,ρ,θ) i, j {1,2,3}, (9.6) 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 hydraulic engineering, where the fluid under consideration is water. 9.4 Constitutive Equation in Newtonian Fluids The mechanical constitutive equation 1 for a Newtonian fluid is σ = p 1 + C : d σ ij = p δ ij + C ijkl d kl i, j {1,2,3}, (9.7) 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 Newtonian fluid, the constitutive tensor C is an isotropic fourth-order tensor. { C = λ1 1 + 2μI C ijkl = λδ ij δ kl + μ ( ) (9.8) δ ik δ jl + δ il δ jk Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 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 an isotropic Newtonian fluid σ ij = p δ ij + λ d ll δ ij + 2μ d ij i, j {1,2,3} (9.10) 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 { σ = p 1 + C : d σ ij = p δ ij + C ijkl d kl Linear elastic solid { σ = C : ε σ ij = C ijkl ε kl 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 temperature in the form μ (θ) =μ 0 e α(θ θ 0), which establishes that the fluid s viscosity decreases as temperature increases (see Figure 9.3). Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 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 pressure p in a Newtonian 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 = Tr (σ) = p Tr (1)+λ Tr (d)tr (1)+2μTr (d)= 3p +(3λ + 2μ)Tr (d) = }{{} 3 p p = p + (λ + 23 ) μ Tr (d)= p + K Tr (d) }{{} K (9.11) where K is denoted as bulk viscosity. Bulk viscosity : K = λ + 2 3 μ (9.12) Using the mass continuity equation (5.24), results in Then, considering the relation and replacing in (9.11), yields dρ dt + ρ v = 0 = v = 1 dρ ρ dt (9.13) Tr (d)=d ii = v i = v (9.14) x i Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar p = p + K v = p K ρ dρ dt (9.15) 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 = p 0 dρ Incompressible fluid: = 0 = p = p dt Fluid with null bulk viscosity K (Stokes condition 2 ): K = 0 = λ = 2 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 σ = 1 Tr (σ) 1 + σ = p1 + σ = p1 + λ Tr (d)1 + 2μ d 3 }{{} = σ = 3 p ( p p) 1 + λ Tr (d)1 + 2μd = ( ) λ K Tr (d)1 + 2μd }{{}}{{} K Tr (d) = Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar λ + 2 3 μ σ = 2 ( 3 μ Tr (d)1 + 2μd = 2μ d 1 ) 3 Tr (d)1 } {{ } 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 where (9.16) and (9.12) have been taken into account. σ = 2μd (9.18) 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 σ = p1 + σ and d = 1 3 Tr (d)1 + d, (9.19) and replacing in the expression of the stress power density (stress power per unit of volume) σ : d, results in 3 ( ) 1 σ : d =( p1 + σ ) : Tr (d)1 + d = 3 = 1 p Tr (d) 1 : 1 +σ : d p 1 : d + 1 3 }{{}}{{} 3 Tr ( d ) 3 Tr (d) σ : 1 = }{{} = 0 Tr ( σ ) (9.20) = 0 = p Tr (d)+σ : d. Replacing (9.16) and (9.17) in(9.20) produces ( ) σ : d = p K Tr (d) Tr (d)+2μd : d. (9.21) σ : d = p Tr (d) + KTr 2 (d)+2μd : d = W R + 2W D }{{}}{{} (9.22) recoverable power dissipative power W R 2W D Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Recoverable power density: W R = p Tr (d) (9.23) Dissipative power density: 2W D = KTr 2 (d)+2μd : 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 = p1 +λ Tr (d)1 + 2μ d = σ = σ R + σ D. (9.24) }{{}}{{} σ R σ D Using the aforementioned notation, the recoverable, dissipative and total power densities can be rewritten as W R = p Tr (d)= p1: d = σ R : d, 2W D = KTr 2 (d)+2μd : d = σ D : d, σ : d =(σ R + σ D ) : d = σ R : d + σ D : d = W R + 2W D. 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), v = 1 dρ ρ dt = 0 = Tr (d) = W R = p Tr (d)=0. Remark 9.12. Introducing the decomposition of the stress power (9.25), the balance of mechanical energy (5.73) becomes P e = dk + σ : d dv = dk + σ R : d dv + σ D : d dv dt dt (9.25) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar V V V P e = dk dt + V W R dv + V 2W D dv, which indicates that the mechanical power entering the fluid P e 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 ρ W R = 1 ρ σ R : d = dg. (9.26) 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 A 1 ρ W R dt = B A A 1 ρ σ R : d dt = B A which justifies the denomination of W R as recoverable power. Figure 9.4: Closed cycle. A dg = G B A G A = 0, (9.27) 2) The second law of thermodynamics allows proving that the dissipative power 2W D in (9.25) is always non-negative, Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 2W D 0 ; 2W D = 0 d = 0 (9.28) and, therefore, in a closed cycle the work performed per unit of mass by the dissipative stresses will, in general, not be null, B A 1 ρ σ D : d }{{} dt > 0. (9.29) 2W D > 0 This justifies the denomination of 2W D as (non-recoverable) dissipative power. The dissipative power is responsible for the dissipation (or loss of energy) phenomenon 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 2W D is known to be always non-negative (see (9.28)). Finally, applying the balance of mechanical energy (see Remark 9.12) results in 0 = P e = dk + W R dv + 2W D dv = dt }{{} V = 0 V dk = d 1 dt dt 2 ρv2 dv = 2 W D dv < 0 }{{} 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 2W D in (9.25) has been seen to always be non-negative, 2W D = KTr 2 (d)+2μ d : d 0. (9.30) This thermodynamic restriction introduces limitations in the admissible values of the viscosity parameters K, λ and μ of the fluid. In effect, given a certain fluid, the aforementioned restriction must be verified for all motions (that is, Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar for all velocity fields v) that the fluid may possibly have. Therefore, 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 Tr (d) 0; d = 0 = 2W D = KTr 2 (d) 0 = K = λ + 2 3 μ 0 (9.31) 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 Tr (d)=0; d 0 = 2W D = 2μ d : d = 2μ d ij : d ij 0 = }{{} > 0 (9.32) μ 0 1 v x v x (y) 0 0 v(x,y)= 0 ; d = 2 y 1 v x 0 0 2 y = d 0 0 0 0 Figure 9.5: Flow characterized by a deviatoric strain rate tensor. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg.2.2.25821.20961