Introduction to non-newtonian fluid mechanics. Josef Málek

Transcription

1 Introduction to non-newtonian fluid mechanics Josef Málek October 212

2 In this lecture we will try to answer the following questions. Questions Q1. What is mechanics? Q2. What is a fluid? Q3. What is a Newtonian fluid? Q4. What are non-newtonian fluids? Q5. Why mechanics of non-newtonian fluids is important? Q6. What are the effects that can not be described by classical models? Q7. What will be the content of this course? Q8. Is J.M. appropriate lecturer? Q9. What are the unique features of the course? Q1. Are there other approaches to describe such strange features of materials? Classical mechanics Classical mechanics describes the evolution of mass particles which follow three Newton laws: Law of inertia If the external force F = then object in a state of uniform motion tends to remain in that state of motion. Law of force If there is an external force, it is equal to the time change of the momentum mv d mv) = F. dt Law of action and reaction For every action there is an equal and opposite reaction. We show four examples of one-dimensional linear spring. 1. Linear spring The displacement of the spring y satisfies ODE m d2 y dt 2 = F E + F S, where F S is the spring force and F E is the external force. If the spring force F S = ky and F E is gravitational force, then m d2 y dt 2 The initial-value problem has to be solved = mg ky. d 2 y dt 2 + k m y = y) = y dy dt ) = y 1. 1

3 2. Spring with damping force We can add the damping force for the spring F d = b dy dt d 2 y dt 2 + b dy m dt + k m y =. 3. Generalized non-linear spring Spring force is a nonlinear function of y : F S = F S y), for example Duffing spring F S y) = αy + βy 3. The well-known Duffing oscillator with the dumping satisfies d 2 y dt 2 + b dy m dt 1 m αy + βy3 ) =. 4. Further generalizations implicit relation Finally, we can have a fully implicit relation for the between the spring force F S and the displacement y, and between the damping force F D and the velocity dy dt Continuum mechanics f F S, y) = g F D, dy ) =. dt Instead of describing only a mass point we would like to describe whole body. To do this we describe it as a continuum. Let B is a body, K B) a configuration in the initial state and K t B) the configuration at time t see Figure). The we define a motion χ which maps from K B) to K t B) : x = χx, t). Further we define the following kinematical quantities Velocity Deformation gradient vx, t) = Left Cauchy-Green tensor χx, t) t F = = vx, t) := vχ 1 x, t), t) χx, t) X B = FF T = x X Right Cauchy-Green tensor C = F T F. 2

4 Balance equations Balance of mass This balance is derived in the following way: ϱ density, v fluid velocity) mp t ) = ϱ dx P t d dt mp t) = P t K t B) = d dt mp t) = ϱx, t) dx = P t P ϱ = P t + ϱ ) χ k χ k t + ϱdiv v det X χx, t) dx ) ϱ = t + xϱv + ϱdiv v dx = P t d dt ϱχx, t), t) det XχX, t) dx P t ϱ + div vϱ) dx t ϱ,t + div ϱv) = Balance of linear momentum ϱv),t + div ϱv v) div T = ϱb, b volume force, T Cauchy stress tensor. Balance of angular momentum T = T T Balance of energy ϱe) t E energy, q thermal flux, r thermal source. + div ϱev) + div q div Tv) = ϱb v + ϱr, Together with these balance equations we need to add some information to be able to solve the problem: 1. Constitutive equations characterize the response of class of materials at certain class of processes. So far we have 8 equations and 14 unknowns still necessary to add some constitutive equations. 2. Boundary conditions 3. Initial conditions Comments: Balance equations are formulated on averages similar to calculus of variation) it suggests to work with weak formulation. The last formulation is not equivalent in a weak sense only classical sense) 3

5 What is a Newtonian fluid? We have to distinguish between compressible and incompressible fluid. Constitutive equation for the Cauchy stress tensor T for compressible Newtonian fluid where A d := A 1 3 tr A) I. Incompressibility T = pϱ)i + 2µϱ)Dv) + λϱ)div v) I 1) = pϱ)i + 2µϱ)D d v) + 2µϱ) + 3λϱ) div v) I 3 Definition 1. Volume of any chosen subset at initial time t = ) remains constant during the motion. = d dt P t = d dx = div v dx = div v = dt P t P t in solid mechanics we can write det F = 1 det F dx = P Consequences of incompressibility, balance of mass + incompressibility P t dx { ϱ t + v ϱ + ϱdiv v = div v = = { ϱ t + v ϱ = div v = Non-homogeneous incompressible Newtonian fluid Homogeneous incompressible Newtonian fluid T = mi + 2µϱ)D d v), µϱ) >. 2) T = mi + 2µ D d v), µ >. 3) The mean normal stress m = tr T)/3 is the unknown. The pressure p in the compressible model is in a similar situation as m in the incompressible model, but in the compressible model the pressure has to be constitutively defined, and it is not the unknown. What is non-newtonian fluid? A fluid is non-newtonian if its response to mechanical loading can not be describe either 1), 2) or 3). Concept of viscosity Newton 1687): "The resistance arising from the want of lubricity in parts of the fluid, other things being equal, is proportional to the velocity with which the parts are separated from one another." 4

6 Definition 2 Viscosity). Viscosity is the coefficient proportionality between the shear stress and the shear-rate Generalized viscosity T xy = F z A vh + y) vy) h v y) = 2D xy µ g κ) := T xyκ), where κ = v κ vy) Simple shear flow: vx, y, z) = D = 1 v v. 2 bt xy = µv y). Experimental data show that the viscosity depends on the shear-rate, pressure, etc. One can understand the word proportional in the sense of arbitrary denpendence. Then we can understand Newton s statement in the sense of implicit relation GT xy, D xy,... ) =. Physical dimension of the dynamic viscosity and kinematic viscosity [µ] = N m kg m s 2 m 2 = m s 1 m 2 = kg m 1 s 1 =: 1 cp, s Kinematic viscosity of some fluids Fluid Viscosity [cp] Air at 18 C).2638 Benzene.5 Water at 18) 1 Olive oil at 2) 84 Motor oil SAE 5 54 Honey 2 3 Ketchup 5 7 Peanut butter Tar Earth lower mantle [ ] µ = m 2 s 1. ϱ 5

7 Non-Newtonian phenomena and model for non-newtonian fluids Definition 3. The fluid is non-newtonian if and only if its behavior can not be described by the model of Navier-Stokes fluid compressible incompressible T = pϱ, θ)i + 2µϱ, θ)d + λϱ, θ)div v)i T = pi + 2µϱ, θ)d. This definition is not useful. We describe the non-newtonian phenomena. Non-Newtonian phenomena 1) Shear thinning, shear thickening property 2) Pressure thickening property 3) Presence of activation/deactivation criteria connected with the stress or with the shear rate) 4) Presence of non-zero normal stress differences in a simple shear flow 5) Stress relaxation 6) Non-linear creep Consider a fluid velocity v = uy),, ) with the stress tensor in the form T = pi + S. We are interested in the shear rate D 12 and the shear stress S 12. The shear rate D 12 is in the literature denoted for example as κ neboor γ, shear stress S 12 is denoted as σ or τ. Ad 1) Shear thinning, shear thickening property We are interested in the dependence of S 12 on D 12. Let us define a generealized viscosity µ g as µ g D 12 ) := S 12 D 12 ). D 12 Newtonian fluid For the Newtonian fluid it holds S 12 = µd 12 and D 12 = u y), so S 12 is a linear function of D 12 with the proportionality coefficient µ see Figure 1). In the case of Newtonian fluid µ g = µ = konst., the graph µ g vs. γ is in the Figure 2. Non-Newtonian fluid Shear thickening property dilatant fluids) means that S 12 is a superlinear function of D 12 Figure 3). The generalized viscosity µ g is a increasing function, in the Figure 4 is the example where the generalized viscosity is degenerate at the beginning, usually the generealized visosity is positive in zero. Shear thinning property pseudoplastic fluids) means that S 12 is a sublinear function of D 12 Figure 5). The generalized viscosity µ g is a decreasing function, in the Figure 6 is the example where the generalized viscosity is singular at the beginning, usually the generealized visosity is finite in zero. Generally we can say that there is no reason that S 12 is monotone function of D It does not have to be even a function. 6

8 σ µ g σ = µ γ µ γ γ Figure 1: Newtonian fluid, shear stress / shear rate) Figure 2: Newtonian fluid, generalized viscosity / shear rate) σ µ g generalized viscosity degenerates γ γ Figure 3: Shear thickening, shear stress / shear rate) Figure 4: Shear thickening, generealized viscosity / shear rate) σ µ g generalized viscosity is singular γ γ Figure 5: Shear thinning, shear stress / shear rate) Figure 6: Shear thinning, generalized viscosity / shear rate) The best known models are power-law fluids where the stress tensor look like T = pi + 2µ Dv) r 2 Dv). }{{} µ g D )= µ g D 2 ) For r = 2 3) we speak about the Newtonian fluid, for r > 2 1) shear thickening fluid and for 1 < r < 2 2) shear thinning fluid, see Figures 7 and 8. Shear thinning fluid is for example the blood. Shear thickenning fluid are for example some painting colors, sauces dressings), ketchups. Viscosity is declared as a quality parameter for some industrial products. 7

9 Figure 7: Shear stress / shear rate. Figure 8: Generalized viscosity / shear rate. Ad 2) Pressure thickening property The generalized viscosity is not constant as in the case of Newtonian fluids 2 see Figure 9) but it is a function of pressure p see Figure 1). Experimental data say that the viscosity is an incresing function of the pressure, not decreasing. µ g µ g µ µ p p Figure 9: Newtonian fluid Figure 1: Viscosity dependent on pressure Around the year 189 Barus proposed a model µp) = µ expαp), α >. The Nobel prize was given for the experimental validation of the pressure dependence. These models are used for example in journal bearings where the pressure differences are very high. Compressible vs. incompressible fluid For the compressible fluid the pressure p = pϱ) is a function of the density ϱ. Let us suppose that we can invert this relation ϱ = ϱp), then the formulation is equivalent with T = pϱ)i + 2µϱ)D + λϱ)div v)i T = pi + 2 µp)d + λp)div v)i, which is the model with the viscosity dependent on the pressure. Een if the material is compressible but not extremely), the incompressible pressure dependence model is used. Such models are difficult to treat both analytically and numerically. 2 Already Stokes in his work wrote about the dependence of the viscosity on the pressure. For the simplicity he further supposed the viscosity to be constant. 8

10 Ad 3) Presence of activation/deactivation criteria The fluid starts to flow when it reaches the critical value of the stress τ called the yield stress. in the case that the following dependence of the stress on the shear rate is linear, respectively nonlinear, we call the fluid Bingham fluid, respectively Herschel-Bulkley fluid 3. The standard formulation is the following: D = S τ, T = pi + 2τ D D + µ g D 2 )D S > τ. If µ g is constant, it is the Bingham fluid, if not, it is Herschel-Bulkley fluid see Figure 11). Further example is the fluid where the response is connected with the chemical processes. At some value D 12 the fluid locks and does not flow, see Figure 12. σ σ locking τ τ γ γ Figure 11: Herschel-Bulkley Figure 12: Locking Except these explicit relations between the shear stress S 12 and shear rate D 12 we can consider a general implicit relation in the form g S 12, D 12 ) =. We do not have to just compute the tensor S from the knowledge of D and insert it into the balance of linear momentum, it is also possible to define a tensor function G that maps GD, S) = 2µ D 2 ) τ + S τ ) +) D S τ ) + S R d d sym R d d sym R d d sym and this equation add to the system of balance equations. The advantage is that G is continuous and no special tools has to be used. The disadvantage is that the system we are solving is bigger. Ad 4) Presence of non-zero normal stress differences in a simple shear flow In the three dimensional space we define three viscometric functions: µ viscosity N 1 := T 11 T 22 first normal stress difference N 2 := T 22 T 33 second normal stress difference In the simple shear flow we suppose a fluid velocity in the form v = uy),, ). 3 We call it fluid although the definition of the fluid says that the fluid can not sustain the shear stres. 9

11 Newtonian fluid Let us compute the Cauchy stress T D = 1 u y) p µu y) u y), T = pi + 2µD = µu y) p. 2 p We can see that N 1 = N 2 =, and so there are no normal stress differences in the Newtonian. Non-Newtonian fluid Non-Newtonian fluids like for example the amyl dissolved in the water, exhibit the presence of non-zero stress differences in the simple shear flow. If we press the fluid in one direction, it reacts in other direction, usually perpendicular. These effects are connected with this property: Rod climbing Die swell Delayed die swell Let us denote D the diameter of the pipe and D E the biggest diameter of the fluid after flowing out from the pipe. For the Newtonian fluids it was experimentally found that D E D = For non-newtonian fluids is this ratio equal even to four and it holds D E D = [ N1 2S ) ) ] 2 1/6, where S is the shear stress and a subscript w means that it is a value on the wall. Flow through the sloping channel In the case of Newtonian fluid the surface of the fluid is smooth free boundary), in the case of non-newtonian fluid the surface is not smooth, there are bumps. Inverted secondary flow Consider a container with a fluid, the fluid is covered with no gap between the fluid and the cover. Now we start to rotate with the cover. The secondary flow creates in the fluid, this secondary flow can be seen in the cross-section. In the case of non-newtonian fluid is the direction of the secondary flor oposite than in the case of Newtonian fluid and the direction depends on the size of N 2. Ad 5) Stress relaxation Stress relaxation test is the following see Figure 13): At time t = we deform the material with constant relative prolongation ε and then we study the shear stress σ. Consider two basic materials: linear elastic spring and linear dashpot two coaxial cylinders with almost same radii, between these cylinders is a Newtonian fluid with the viscosity µ). Linear spring is described with Hooke law σ = Eε, where E is the elastic modulus. Linear dashpot is described by the relation σ = µ ε, where µ is the viscosity. The result of the stress relaxation test is in the case of linear spring in the Figure 14, in the case of linear dashpot in the Figure 15. In the case of dashpot is the stress singular in zero. Most of materials are the combination of these two basic materials, their response is depicted in the Figure 16 w 1

12 ε σ ε??? t t Figure 13: Stress relaxation test σ σ Eε t t Figure 14: Linear spring Figure 15: Linear dashpot Figure 16: Viscoelastic solid Viscoelastic fluid Ad 6) Non-linear creep Non-linear creep test is the following see Figure 17): At time t = we deform the material with constant stress σ, at time t = t we turn off the stress and study the relative prolongation ε. The result of the creep test is in the case of linear spring in Figure 18, in the case of linear dashpot in Figure 19. Again, most of materials are the combination of two basic materials, in Figure 2 is the resultfor the material similar to viscoelastic solid a viscoelastic fluid. Except the algebraic models there exist also rate-type models, integral models and stochastic models. A detailed overview of non-newtonian models is in the following section. 11

13 σ ε σ??? t t t Figure 17: Non-linear creep test ε ε σ E σ µ t t t t Figure 18: Linear spring Figure 19: Linear dashpot Figure 2: Viscoelastic solid Viscoelastic fluid Non-Newtonian models Models with variable viscosity General form: T = pi + 2µD, T)D }{{} S Particular models mainly developed by chemical engineers. 4) Shear dependent viscosity Ostwald de Waele power law [24], [12] µd) = µ D n 1 5) 12

14 Fits experimental data for: polymer latex spheres ball point pen ink, molten chocolate, aqueous dispersion of Carreau Carreau Yasuda [8], [36] µd) = µ + µ µ 1 + α D 2 ) n 2 µd) = µ + µ µ ) 1 + α D a ) n 1 a 7) Fits experimental data for: molten polystyrene Eyring [14], [29] 6) µd) = µ + µ µ ) µd) = µ + µ 1 arcsinh α 1 D ) α 1 D arcsinh α D ) α D + µ 2 arcsinh α 2 D ) α 2 D 8) 9) Fits experimental data for: napalm coprecipitated aluminum salts of naphthenic and palmitic acids; jellied gasoline), 1% nitrocelulose in 99% butyl acetate Cross [11] µd) = µ + µ µ 1 + α D n 1) Fits experimental data for: aqueous polyvinyl acetate dispersion, aqueous limestone suspension Sisko [35] Fits experimental data for: lubricating greases Models with pressure dependent viscosity Barus [1] Fits experimental data for: mineral oils 4, organic liquids 5 Models with stress dependent viscosity Ellis [21] µd) = µ + α D n 1 11) µt) = µ ref exp β p p ref ) 12) µt) = µ 1 + α T δ n 1 13) Fits experimental data for:.6% w/w carboxymethyl cellulose CMC) solution in water, polyvynil chloride) 6 4 [2] 5 [6] 6 [33] 13

15 Glen [16] µt) = α T δ n 1 14) Fits experimental data for: ice Seely [34] ) µt) = µ + µ µ ) exp Tδ τ Fits experimental data for: polybutadiene solutions 15) Blatter [25], [5] Fits experimental data for: ice µt) = A T δ 2 + τ 2) n ) Models with discontinuous rheology Bingham Herschel Bulkley [4], [17] T δ > τ if and only if T δ = τ D D + 2µ D )D T δ τ if and only if D = 17) Fits experimental data for: paints, toothpaste, mango jam 7 Differential type models Rivlin Ericksen fluids Rivlin Ericksen [31], [32] General form: where T = pi + fa 1 A 2 A 3... ) 18) A 1 = 2D 19a) A n = da n 1 dt + A n 1 L + L T A n 1 19b) where d dt denotes the usual Lagrangean time derivative and L is the velocity gradient. Criminale Ericksen Filbey [1] T = pi + α 1 A 1 + α 2 A 2 + α 3 A 2 1 2) Fits experimental data for: polymer melts explains mormal stress differences) Reiner Rivlin [3] T = pi + 2µD + µ 1 D 2 21) Fits experimental data for: N/A 7 [2] 14

16 Rate type models Maxwell, Oldroyd, Burgers Maxwell [22] T = pi + S S + λ 1 S = 2µD 22a) 22b) M := dm dt LM MLT 23) Fits experimental data for: N/A Oldroyd-B [23] T = πi + S S + λ S = η 1 A 1 + η 2 A 1 24a) 24b) Fits experimental data for: N/A Oldroyd 8-constants [23] T = πi + S 25a) λ 3 S + λ 1S + 2 DS + SD) + λ 5 2 tr S) D + λ 6 S : D) I 25b) 2 = µ D + λ 2D + λ4 D 2 + λ ) 7 D : D) I 25c) 2 Fits experimental data for: N/A Burgers [7] T = πi + S S + λ 1 S + λ2 S = η1 A 1 + η 2 A 1 26a) 26b) Fits experimental data for: N/A Giesekus [15] T = πi + S 27a) Fits experimental data for: N/A S + λ S αλ 2 µ S2 = µd 27b) 15

17 Phan-Thien Tanner [26], [27] Y S + λ S + λξ 2 Fits experimental data for: N/A Johnson Segalman [19] T = πi + S 28a) DS + SD) = µd 28b) Y = exp ε λµ ) tr S 28c) T = pi + S 29a) S = 2µD + S 29b) ) ds S + λ dt + S D ad) + D ad) T S = 2ηD 29c) Fits experimental data for: spurt Johnson Tevaarwerk [18] T = pi + S S + α sinh S s = 2µD 3a) 3b) Fits experimental data for: lubricants Integral type models Kaye Bernstein Kearsley Zapas [3], [9] T = t ξ= Fits experimental data for: polyisobutylene, vulcanised rubber W I C + W II C 1 dξ 31) 16

18 Entropy and Newtonian fluids Definition 4. There exists a specific density of entropy η further just entropy) which is a function of state variables, η = ηy, y 1,... ), where y = e is a density of internal energy further just energy), and it holds: i) η e > e = ẽη, y 1, y 2,... ), denote a temperature θ := ẽ η Ω ii) η +, if θ + iii) St) = ϱη)t, x) dx S max for t +, if Ω is mechanically and thermally energetically) isolated For Newtonian fluids it holds Balance equations: Differentiate e, multiply by ϱ, η = ηe, ϱ) e = ẽη, ϱ). ϱ = ϱ div v, ϱ v = div T + ϱb, ϱė = divtv + q) + ϱb v. ϱė = ϱ ẽ ẽ η + ϱ η ϱ ϱ and insert it into the balance of linear momentum, ϱb v + divtv + q) div T) v ϱb v = ϱ ẽ ẽ η ϱ2 div v. η ϱ Now denote and obtain θ = ẽ η, ẽ p = ϱ2 ϱ Using the deviatoric parts we get ϱθ η = T v + div q + p div v symetrie T = T D + div q + p div v. ϱθ η = T d D d tr T)1 div v) I I 3 }{{} + div q + p div v 3 ) 1 = T d D d + tr T) + p div v + div q. 3 Dividing by θ we get q ) ϱ η div = 1 [ ) 1 T d D d + tr T) + p div v + q θ ]. θ θ 3 θ }{{} ξ 17

19 ξ is called rate of entropy production. The second law of thermodynamics says that ξ, and so ζ := ξ q ) θ = ϱ η div. θ If we denote m = tr T)/3 the mean normal stress, then it holds ) 1 ξ = T d D d + tr T) + p div v + q θ = { T d, m + p, q } { D d, div v, θ } 3 θ }{{} θ termodynamical fluxes }{{} termodynamical afinities What is ξ for Navier-Stokes equations? Compressible Navier-Stokes-Fourier equations 32) T = pϱ, η)i + 2µϱ, η)d + λϱ, η)div v)i, q = Kϱ, η) θ. It holds and Insert it into 32). m + p = 2µ + 3λ 3 T d = 2µD d. div v Option 1 ξ = 2µ D d 2 2µ + 3λ + div v) 2 + K θ 2. 3 θ If µ, 2µ + 3λ a K, then ξ. Option 2 ξ = 1 2µ Td µ + 3λ m + p)2 + 1 q 2 K θ. If µ >, 2µ + 3λ > a K >, then ξ >. We ask: When is ξ =? Option 1 We do not want any restriction for the velocity or temperature, that is why µ = 2µ + 3λ = K =. Option 2 It is possible only if T d =, m = p, q =. Incompressible Navier-Stokes-Fourier equations Option 1 If µ a K, then ξ. Option 2 If µ > a K >, then ξ >. ξ = 2µ D d 2 + K θ 2. θ ξ = 1 2µ Td q 2 K θ. 18

20 Remark Gibbs equation θ = e η, θ dη = de + p d e p = ϱ2 ϱ. ) 1, ϱ from this equation the relation for the pressure can be derived. This equation holds for compressible gas, there is no reason why it should hold in general for more complicated materials. Question: Is there a way how from the knowledge for constitutive relations for η and ξ get the constitutive relations for T and q? Principle of maximimzation of the rate of entropy production Our aim is to derive Navier-Stokes equations. It holds we choose a constitutive relation for ξ ξ = T d D d + m + p) div v + q θ, 33) θ ξ = ξd d, div v, θ) = 2µ D d 2 + 2µ + 3λ 3 div v) 2 + K θ 2. θ There is many possibilities when ξ = ξ, we choose the one when ξ is maximal. That is why we maximize it over D d, div v, θ such that 33) holds max ξd d, div v, θ). 33)+D d,div v, θ This heuristic principle is called principle of maximimzation of the rate of entropy production. The maximization is done using Lagrangeo multipiers. We define the Lagrange function L = ξd d, div v, θ) + λ ξd d, div v, θ) T d D d + m + p) div v + q θ )), θ and we are looking for its extreme, i.e. L D d =, L div v =, L θ =. 19

21 We obtain 1 + λ) ξ D d = λtd, 1 + λ) ξ = λm + p), div v 1 + λ) ξ θ = λq θ. 34) Now we eliminate λ in this way. We multiply the first equation by D d, the second by div v and the third by θ, 1 + λ λ = ξ ξ D D d + ξ ξ d div v div v + θ θ = 1 2. If we insert this result into 34) and do the differentiation, we obtain Navier-Stokes equations T d = 2µD d, 2µ + 3λ m + p = 3 q θ = K θ θ. div v, Maximization ξ for incompressible Navier-Stokes equations For isothermal incompressible materials it holds tr D = div v =, θ = θ = const. Constraint for ξ is First, let ξ is given by ξ = T D. 35) ξ = 2µ D 2. We use the principle of maximization of the rate of entropy production We define the Lagrange function and maximize it with respect to D, Multiply this equation by D and get max ξd). 35)+D+tr D= L = ξd) + λ 1 ξd) T D) ) + λ 2 tr D 1 + λ 1 ) ξ D = λ 1T λ 2 I. 1 + λ 1 λ 1 further take the trace of this equations and get = T D ξ D D = 1 2, 3λ 2 = λ 1 tr T λ 2 = 1 tr T. λ 1 3 2

22 We have T = 2µD + 1 tr T)I. 3 Incompressible Navier-Stokes equations can be also derived in other way. Let and we are maximizing ξ We obtain ξ = 1 2µ Td 2 q = ) 36) max 36)+T d ξt d ). D d = D = 1 2µ Td = 1 T 13 ) 2µ tr T)I, which is less usual, but equivalent formulation of incompressible Navier-Stokes equations. Integral form of entropy production Integrate the balance of energy b =, r =, v n = ) over Ω ) d ϱ e + v 2 dx = div Tv + q) dx = Tv + q) n ds = dt 2 Ω In order to have conservation of energy it suffices to assume pointwise Ω q n = Tv n = Tn v = Tn) τ + Tn) n ) v τ + v n)n) = Tn) τ v τ For the total entropy S = ϱη dx we also have Ω d dt St) = 1 [ T δ : D δ +... ] q n θ θ ω d dt St) = T δ, D δ) L 2 Ω) + m + p, div v) L 2 Ω) + The simplest proportional relation with Ω T d = 2µD d, guarantee the second law of thermodynamics. In general we have 2µ + 3λ m + p = div v, 3 Q θ = K θ θ Tn) τ = γv τ Ω µ, 2µ + 3λ, K, γ G 1 T d, D d) =, G 2 m + p, div v) =, Q G 3 θ, θ ) =, θ G 4 Tn) τ, v τ ) =. q, θ ) + Tn) τ, v τ ) θ L 2 Ω) L 2 Ω) If D δ = or div v = or θ = or v τ = we get constrained material rigid), resp. incompressible, resp. isothermal process, resp. no-slip boundary condition. No dissipation if T δ = and q = and m = p and Tn) τ =. 21

23 Remark: If you have dissipation you have energy estimates Cauchy model for finite elasticity Kinematical quantities F K = χ X, Ḟ K = LF K B K = F K F T K, where B K is the left Cauchy-Green tensor, we can compute its time derivative Let us have the entropy For example the neo-hookean material fullfils Insert e into the balance of energy We get Ḃ K = LB K + B K L T a tr ḂK = 2D B K. η = ηe, B K ) e = ẽη, B K ), ẽ = e η) + µ 2 tr B K 3). ϱθ η = divtv + q) div T v µb K D = T D µb K D + div q = T µb K ) D + div q. q ) ϱ η + div = 1 [ T µb K ) D + q θ ]. θ θ θ }{{} produkce entropie Elastic material does not disipate, that is why the rate of entropy production has to be zero, and so T = µb K, and for the incompressible material we have Korteweg fluids T = ΦI + µb K. We will use the principle of maximization of the rate of entropy production to derive the model derived in 191 by Dutch mathematician Korteweg. This model is used for describing capilar effects, multiphase and granulated materials. The model is formulated by the following equations ϱ = ϱ div v, ϱ v = div T, T = pϱ)i + 2µϱ)Dv) + λϱ)div v)i + K, K = κ 2 ϱ2 ) ϱ 2 )I κ ϱ ϱ). 22

24 In the paper [13] Dunn and Serrin showed that this model is thermodynamically compatible. We know that if we suppose this constitutive relation η = ηe, ϱ), then for compressible Navier-Stokes equations we obtain q ) ϱ η + div = 1 [ T d D d + m + p) div v + q θ ] θ θ θ = 1 [ T d dis D d + t dis div v + q ] dis θ. θ θ We introduce disipative terms with subscript dis) and write the constitutive relation for the rate of entropy production ξ = ξt d dis, t dis, q dis ) = 1 2µ Td dis µ + 3λ t2 dis + 1 K q dis 2 and using the principle of maximization of the rate of entropy production we get Navier- Stokes-Fourier system. For the Korteweg model we consider the entropy dependent on the gradient of density ϱ η = ηe, ϱ, ϱ) e = ẽη, ϱ, ϱ). Differentiate ϱė = ϱ ẽ ẽ ẽ η + ϱ ϱ + ϱ ϱ, η ϱ ϱ we know that ϱ = ϱ div v, and need to compute ϱ, Let us denote therefore we have ϱθ η = ϱė ϱ v v p ϱ ϱ ϱ ẽ ϱ = ϱ ϱ = ϱ div v) v) ϱ. θ = ẽ ẽ, p = ϱ2 η ϱ, = divtv + q) div T v + p div v + ϱ ẽ ϱ = T v+ ϱ ẽ ) ϱ ϱ v+div q+p div v+div ẽ ϱ div v) + ϱ ϱ 2 div v ẽ ϱ The principle of objectivity says that e is a function of the size of ϱ, this implies that is a symmetric tensor. We obtain T + ϱ ẽ ) ϱ ϱ ) ) d ϱθ η = ẽ = T d + ϱ ϱ ϱ }{{} T d dis e = ẽη, ϱ, ϱ ), ẽ ϱ ϱ D + p ϱ div ϱ ẽ ϱ D d + ) ϱ ϱ v = ) ϱdiv v) div ϱ ẽ ϱ )) div v + div q + ϱ 2 div v) ẽ ϱ ). ) = p ϱ div ϱ ẽ ) + m + ϱ ) ẽ ϱ 3 ϱ ϱ div v+div q + ϱ 2 div v) ẽ ). ϱ }{{}}{{} t dis q dis 23

25 We get the equation ϱ η div qdis θ ) = 1 θ ξ =, pokud T d dis =, t dis =, q dis =, pak [ T d dis D d + t dis div v + q ] dis θ. θ ) d ẽ T = T d + mi = ϱ ϱ ϱ p... )I ϱ ) ẽ 3 ϱ ϱ I + ϱ div ϱ ẽ ) I ϱ = pe, ϱ, ϱ)i + ϱ div ϱ ẽ ) ) ẽ I ϱ ϱ ϱ ϱ. The special choice of internal energy e = ẽη, ϱ, ϱ) = e η, ϱ) + β 2ϱ ϱ 2 ϱ ẽ ϱ = β ϱ gives the Korteweg model T = p... )I + ϱβ div ϱi β ϱ ϱ) = pi + β ϱ 2 ) ϱ 2) I β ϱ ϱ). 2 From the choice of constitutive equations for the rate of entropy production and the choice 8 ξ = ξt d dis, t dis, q dis ) = 1 2µ Td dis µ + 3λ t2 dis + 1 K q dis 2 T d dis = 2µD d, 2µ + 3λ t dis = div v, 3 q dis = K θ we see that ξ and the second law of thermodynamics is satisfied. Korteweg model Then we get the q = K θ ϱ 2 div v ẽ = K θ βϱdiv v) ϱ, ϱ T = pi + 2µD + λdiv v)i + β ϱ 2 ) ϱ 2) I β ϱ ϱ). 2 Viscoelastic materials Our aim is to model materials capable of describing non-newtonian phenomena like nonlinear stress relaxation, nonlinear creep or the normal stress differences in simple shear flow. Such materials are for example geomaterials, biological materials, polymers and other chemical products like food for example. Repeat what is the stress relaxation test and creep test. Stress relaxation test Stress relaxation test is the following see Figure 21): At time t = we deform the material with constant relative prolongation ε and then we study the shear stress σ. For linear spring it holds σ = Eε, 24

26 ε σ ε??? t t Figure 21: Stress relaxation test σ σ Eε t t Figure 22: Linear spring Figure 23: Linear dashpot for the linear dashpot it holds σ = µ ε. Let us define the stress relaxation function Real materials exhibit the following bahavior Gt) = σt) ε. Figure 24: Viscoelastic solid Viscoelastic fluid Creep test Creep test is the following see Figure 25): At time t = we deform the material with constant stress σ, at time t = t we turn off the stress and study the relative prolongation ε. 8 It is possible to use the principle of maximization of the rate of entropy production, but for the simplicity we can guess it, because the rate of entropy production is quadratic. 25

27 σ ε σ??? t t t Figure 25: Creep test ε ε σ E σ µ t t t t Figure 26: Linear spring Figure 27: Linear dashpot Let us define the creep test function Real materials exhibit the following bahavior Jt) = εt) σ. Figure 28: Viscoelastic solid Viscoelastic fluid The result of stress relaxation test is in the case of linear spring in Figure 26, in the case of linear dashpot in Figure 27. Creation of simple models by combining linear springs and linear dashpots Maxwell element We get the Maxwell element if we connect the linear spring and linear dashpot in series. Denote F resp. F S resp. F D the force in the whole element resp. in the 26

28 Spring Dashpot l Figure 29: Maxwell element spring resp. in the dashpot, and resp. S resp. D o the prolongation of the whole element resp. of the spring resp. of the dashpot. Then for the linear spring holds for the linear dashpot F S = E S, σ S = Eε S, F D = µ D, σ D = µ ε D. For the Maxwell element we do the computation in details. First we derive the constitutive relation. In the series connection is the stress in the spring and dashpot the same, i.e. σ D = σ S, the whole prolongation is the sum of the prolongation of the spring and the dashpot, i.e. = S + D. We substitute from the constitutive relation for the linear spring and the linear dashpot = S + D = F S E + F D µ, and get the constitutive equation fot the Maxwell element From the initial condition we obtain E ε = σ + E µ σ p 1 σ + p σ = q 1 ε. p 1 σ+) = q 1 ε+). Now copmute how the Maxwell element behaves in the creep test. Let σt) = σ, then and the creep function is equal to εt) = σ q 1 p 1 + p t) Jt) = εt) σ = 1 q 1 p 1 + p t). The response is linear it is not a satisfacory result. Further compute the stress relaxation test for Maxwell element. Let εt) = ε Ht), where Ht) is the Heavisideova function and the stress relaxation function is equal to σt) = q 1 ε e p p 1 p 1 t This response looks reasonably. Gt) = σt) = q 1 e p ε p 1 p 1 t. 27

29 Now compute how the stress σ depends on th deformation ε σt)e ) p p t 1 = q 1 εe p p t 1 p 1 σt) = σ+)e p p t 1 + q t 1 εt)e p p t τ) 1 dτ p 1 initial cond. = q 1 ε+)e p p t 1 + q 1 p 1 = ε+)gt) + q 1 p 1 p 1 t t εt)e p p 1 t τ) dτ εt)gt τ) dτ. In the higher dimension this result corresponds to the fact that we obtained the integral relation between T and D. Now compute how ε depends on σ. t t q 1 εt) = q 1 ε+) + p 1 στ) dτ + p στ) dτ Using initial condition we obtain p1 εt) = + p ) t q 1 q 1 t = q 1 ε+) + στ)p 1 + p t τ)) dτ + p σ+)t. = Jt)σ+) + t σ+) + t στ)jt τ) dτ. p1 στ) + p ) t τ) dτ q 1 q 1 We obtained the rate-type model and two integral type models, all equivalent. All models hold in one dimensional space, but it is not clear how to generalize them into three dimensional space. 1. Linear dashpot is not an approximative theory, but the linear spring is. 2. What to do if the dashpot or the spring depend on the deformation in nonlinear way? 3. Partial time derivative is neither material time derivative nor objective time derivative, the generalizations are not unique. 4. Many nonlinear 3D models can reduce in 1D to the same equation. So many models can be used to capture 1D experimental data. Kelvin-Voigt element We get the Kelvin-Voigt element if we paralelly connect the spring and the dashpot. The prolongations are the same for the spring and the dashpot. The total force is equal to the sum of the forces in spring and dashpot, so F = F S + F D, = S = D. Using constitutive relations for the linear spring and the linear dashpot we get The stress relaxation function is equal to p σ = q ε + q 1 ε, ε+) =. Gt) = q p + q 1 p δt), 28

30 Spring Dashpot l Figure 3: Kelvin-Voigtů element this is disadvantage for Kelvin-Voigt. The creep function is Jt) = p ) 1 e q q1 t, q which is the advantage of Kelvin-Voigt element Both derived models were onedimensional, we show teh modification into the higher dimension. Suppose that the fluid is incompressible, tr D =. Consider the stress tensor in the form T = pi + S and say that S satisfies the same equation as derived earlier, let us try S + λṡ = 2µD. Suppose we have two coordinate systems starred and unstarred, the starred one is related to the unstarred by x = Qx x ) + c, where Q is orthogonal. We want the tensor S to transform as S = QSQ T. In the starred system the material satisfies the equation S + λṡ = 2µD, which is equivalent to If it holded QSQ T + λ d dt QSQ T + λ d dt QSQ T ) = 2µQDQ T. QSQ T ) ) = Q S + λṡ Q T, everything would be OK, but this does not hold because d QSQ T ) = dt QSQ T + QṠQT + QS Q T and the sum of the first and last term can never be equal zero. So, our try is bad Ṡ is not objective. Let us denote the objective derivative of S by S satisfying S = Q S Q T. 29

31 Define now S= ds dt LS SLT + tr L)S and verify that this derivative is objective. We want ) S = Q S ds Q T = Q dt LS SLT + tr L)S Q T. Compute. ds dt L S S L T ) + tr L )S = QSQ T = = d QSQ T ) L QSQ T QSQ T L T ) + tr L )QSQ T. dt We need to know what is L. After some algebraic manipulation it can be computed that L = QQ T + QLQ T. Inserting this into the previous calculation we get d QSQ T ) L QSQ T QSQ T L T ) + tr L )QSQ T = dt = QSQ T + QṠQT + QS Q T QQ T + QLQ T) QSQ T QSQ T QQ T + QLQ T) + tr QQ T + QLQ T) QSQ T = = QṠQT QLSQ T QSL T Q T + tr QQ T + QLQ T) QSQ T = because the trace is cyclic and = QQ T + Q Q T. = Q Ṡ LS SL T ) Q T + Qtr L)SQ T, The derivative is objective even if there is not the last term with the trace S= ds dt LS SLT. This derivative is called Oldroyd upper convected derivative. If we use it in the Maxwell element we obtain the Maxwell model T = pi + S, S + λ S= 2µD. Incompressible rate-type fluid models derivation of thermodynamically compatible models Motivaation and the aim Combining two linear dashpots and one linear spring we obtain the relation between the shear stress and shear rate p 1 T xy + p T xy = q 1 Ḋ xy + q D xy. 3

32 In the previous section we explained why it is not trivial to generalize this model into three dimensional space. A similar model is a Yeleswerapu model for blood: div v =, ϱ v = div T + ϱb, T = pi + S, S + λ 1 Ṡ ) LS SL T = µ D )D + λ 2 Ḋ LD DLT ), [ ] 1 + ln1 + Λ D ) µ D ) = µ + µ µ ). 1 + Λ D But is this model thermodynamically compatible satisfying the second law of thermodynamics? First we derive three dimensional Kelvin-Voigt model. Kelvin-Voigt model Suppose the entropy in the form η = ηe, ϱ, B kr ) e = ẽη, ϱ, B kr ) = êη, ϱ, tr B kr ) and define kinematical quantities. First the deformation gradient left and right Cauchy-Green tensor F kr = χ k R X Compute B kr = F kr F T k R, C kr = F T k R F kr. ê ê ϱė ϱ v v = ϱė = ϱ η + ϱ η ϱ ϱ + ϱ ê tr B kr ) I Ḃk R. We know that Ḟk R = LF kr which implies Continue computing Ḃ kr = LB kr + B kr L T I Ḃk R = 2B kr D. ê ϱθ η div q = T D + p div v 2ϱ tr B kr ) B k R D = ) d ê = T 2ϱ tr B kr ) B k R D d + m 2 ) 3 ϱ ê tr B kr ) tr B k R + p div v. Divide by temperature θ and get q ) ϱ η div = 1 [ T d dis D d + t dis div v + q θ ]. θ θ θ }{{} ξ The incompressibility and constant temperature gives ) d ξ = T d dis D d ê = T 2ϱ tr B kr ) B k R D d. Further for a general ξ = ξt dis, D) choose the constitutive relation ξ = 1 2ν T dis 2, 31

33 or ξ = 2ν D 2. In the first case we maximize ξ with respect to T dis with no other constrains, in the second case we maximize with respect to D with the incompressibility condition tr D =. We obtain If we use neo-hookean free energy ê T = mi + 2νD + 2ϱ tr B kr ) Bd k R. ϱψ = µ 2 tr B k R 3), the it holds and we obtain Kelvin-Voigt model ê tr B kr ) = µ 2 I T = mi + 2νD + µb d k R. If the rate of entropy production is zero, then and we obtain we get T d dis =, p + m + µ 1 3 tr B k R =, q = T = T d + mi = T d dis + µb d k R + mi = pi + µb kr. If we do not require the incompressibility and use the relations as in Navier-Stokes-Fourier T d dis = 2νD d, p + m + µ 1 3 tr B 2ν + 3λ k R = div v, 3 q = k θ, T = pi + 2νD + λdiv v)i + µb kr, q = k θ. Viscoelastic model K.R. Rajagopal, A.R. Srinivasa [28]) The body made from the viscoelastic material is at time t in configuration κt), this configuration is called the current configuration. We will compare this configuration with the configuration κr) which is a configuration of the system in rest at the beginning, so called the reference configuration. We define now the natural configuration κpt)). It is a configuration of the system corresponding to the current configuration κt) the body would go to on the sudden removal of external stimuli. Then the only difference between the natural and current configuration is purely elastic deformation. By defining the natural configuration we split the whole deformation into purely elastic part and the dissipative part. We motivate the notion of the natural configuration using a simple example. Suppose we have a simple linear viscoelastic material where in every material point is Maxwell element. The spring represent the elastic part and the dashpot the viscous part In Figure 31a) jthe material is at the beginning in rest, relaxed. The spring and the dashpot are not stretched. 32

34 a) b) c) Figure 31: Motivational example for the natural configuration The system is in the reference configuration κr). Then we stretch it Figure 31b)) and the system goes to the current configuration κt). Now we let the system relax, the spring shrinks to its reference length but the dashpot remains stretched. Figure 31c)). The system is in the natural configuration κpt)). Because during the relaxation the deformation of the dashpot does not change, we can think that the whole free energy of the system is hidden only in the deformation of the natural configuration. konfigurace. F κr) κt) G F pt) κpt)) We define some kinematical quantities. Figure 32 denoted by F) Figure 32: Přirozená konfigurace F kr First we define the deformation gradient in = χ that maps infinitesimal element from κr) to κt). We define standard left and right Cauchy- Green tensors. B kr = F kr F T k R, C kr = F T k R F kr, Further we define F kpt) as a mapping of the infinitesimal element from κpt)) to κt) in Figure. 32 denoted by F pt) ) and G that maps from κr) to κpt)). These two new variables are related through the deformation gradient G = F 1 k pt) F kr. 33

35 Define leftand right Cauchy-Green tensor corresponding to κpt)) B kpt) = F kpt) F T k pt), C kpt) = F T k pt) F kpt) and the rate of the deformation of κpt)) in the analogy with the usual relation between the velocity gradient and deformation gradient), resp. its symmetric part L kpt) = ĠG 1, D kpt) = L k pt) + L T k pt). 2 The material time derivative of B kpt) is equal to Ḃ kpt) = Ḟk pt) F T k pt) + F kpt) Ḟ T k pt) = where we used Ḟ kr G 1 F T k pt) + F kr G 1 F T k pt) + F kpt) G T F T k R + F kpt) G T Ḟ T k R = A 1 = A 1 ȦA 1. LB kpt) + B kpt) L T 2F kpt) D kpt) F T k pt), And we naturally obtained the Oldroyd upper convected time derivative Bk pt)) = Ḃk pt) LB kpt) B kpt) L T = 2F kpt) D kpt) F T k pt). Note that I = 2D. It holds that tr Ḃk pt) = 2B kpt) D 2F kpt) D kpt) F kpt). Choose the entropy in the form η = ηe, ϱ, B kpt) ), we can invert it and use the internal energy e = ẽη, ϱ, B kpt) ), concretely choose e = e η, ϱ) + µ 2ϱ and as in the case for Kelvin-Voigt we obtain tr Bkpt) 3) ξ = T µb kpt) ) D + p + m) div v + q θ θ + µf kpt) D kpt) F kpt). For simplicity q = and div v = and like in [28], we suppose that the elastic response is also incompressible, i.e. det B kpt) = 1. From this and incompressibility F kr results that det G = 1. Differentiating det G gives tr D kpt) =. For the rate of entropy production holds we the zero traces of D and D kpt) ) ξ = T µb kpt) ) d D + µc d k pt) D kpt) = T d dis D + µc d k pt) D kpt). We choose the following constitutive relation for the rate of entropy production ξ = ξd, D kpt), C kpt) ) = 2ν D 2 + 2ν 1 D kpt) C kpt) D kpt). We observe that D kpt) B kpt) D kpt) = D kpt) F kpt) 2, and the second law of thermodynamics is satisfied. We define the Lagrange function LD, D kpt) ) = ξ... ) + λ 1 ξ... ) T d dis D µc d k pt) D kpt) ) + λ 2 tr D + λ 3 tr D kpt). and maximize. The necessary conditions are L D =, L D kpt) =. 34

36 We substite for L 1 + λ 1 λ λ 1 λ 1 ξ D = Td dis + λ 2 λ 1 I, 37) ξ D kpt) = µc d k pt) + λ 3 λ 1 I. 38) First we do the differentiation ξ D = 4ν D, ξ D kpt) = 4ν 1 C kpt) D kpt). and eliminate the Lagrange multipliers. To do this we add 37) D + 38) D kpt), and use the constrains of incompressibility 1 + λ 1 λ 1 = Td dis D + µcd k D pt) k pt) ξ D D + ξ = D kpt) D kpt) ξ 2 ξ = 1 2. We take the trace of 37) and obtain If we take the trace of 38), we obtain = tr T d dis 3 λ 2 λ 1 λ 2 λ 1 =. We get λ 3 λ 1 = 2 3 ν 1D kpt) C kpt). From the equations 39) it implies that T d dis = 2ν D, 39) 2ν 1 C kpt) D kpt) 1 ) 3 D k pt) C kpt) )I = µc d k pt). 4) Before the further calculation we remind that T = pi + 2ν D + µb d k pt). Ḃ kpt) = LB kpt) + B kpt) L T 2F kpt) D kpt) F T k pt), and compute the Oldroyd upper convected time derivative of the deviatoric part of B kpt). B d k pt) = B kpt) 1 ) 3 tr B k pt) )I = Bk pt) 1 3 tr Ḃk pt) )I 1 3 tr B k pt) ) I= = 2F kpt) D kpt) F T k pt) 1 3 trlb k pt) + L T B kpt) )I trft k pt) F kpt) D kpt) )I tr B k pt) )D = = 2F kpt) D kpt) F T k pt) 2 3 D B k pt) )I C k pt) D kpt) )I tr B k pt) )D. Multiply 4) from the left by F T k pt) and from the right by F T k pt), we obtain 2ν 1 F kpt) D kpt) F T k pt) 1 ) 3 C k pt) D kpt) )I = µc d k pt). 35

37 From the relations for the Oldroyd derivative of the deviatoric part of B kpt) we get ) 2ν 1 1 B d k 2 pt) 1 3 D Bd k pt) )I tr B k pt) )D = µb d k pt). Note that the trace both sides of the equation is zero, in the three dimensional space it is five scalar equations for six unknowns. We complete the system of equations with the equation for incompressibility of the elastic response det B kpt) + 1 ) 3 tr B k pt) )I = 1. We obtain the set of equations for nonlinear viscoelastic rate-type fluid model div v =, ϱ v = div T, T = pi + 2ν D + µb d k pt), ) 2ν 1 1 B d k 2 pt) 1 3 D Bd k pt) )I tr B k pt) )D = µb d k pt), 41) det B kpt) + 1 ) 3 tr B k pt) )I = 1. 42) Linearization of the viscoelastic model By linearization of the elastic response we show that this model reduces to Oldroyd-B model. Suppose that the left Cauchy-Green tensor is in the form B kpt) = I + A, where A = ε, < ε 1. By linearization we understand that after substitution into the nonlinear viscoelastic model we will neglect all term of the order Oε 2 ) and higher. We use the last equation for incompressibility 42) and do the Taylor expansion of determinant. We obtain 1 = det ) 1 B kpt) = deti + A) = 1 + tra) + tr A) 2 tr A 2)) + Oε 3 ) 2 tra) = 1 tr A 2 ) tr A) 2) + Oε 3 ). } 2 {{} Oε 2 ) From this it implies that tr B kpt) = we obtain the relation for B d k pt), tr A 2 ) tr A) 2) + Oε 3 ), B d k pt) = B kpt) 1 3 tr B k pt) )I = A 1 6 Further we need to compute B d k pt). tr A 2 ) tr A) 2) I + Oε 3 )I. B d k pt) = Ḃd k pt) LB d k pt) B d k pt) L T = A 1 A tr A) tr 3Ȧ }{{} Ȧ)) I + Oε 2 ). Oε 2 ) Inserting all into 41) we get µa + ν 1 A= 2ν1 D 1 3 ν 1 [ ) ] 2D Ȧ A I + Oε 2 ). 43) 36

38 This model is ver similar to Oldroyd-B model, the difference is in the last term. We show that this term is of the order Oε 2 ), and so we can neglect it. Take the scala product of 43) with tensor A, we get µ A 2 }{{} Oε 2 ) +ν 1 Ȧ A ν 1 LA A + AL T A) = 2ν 1 D A) 1 [ ) ] }{{} 3 ν 1 2D Ȧ A Oε 2 ) From this it implies that 9 ) 2D Ȧ A = Oε 2 ). We showed that the nonlinear viscoelatic model linearizes to Oldroyd-B model. div v =, ϱ v = div T, T = pi + 2ν D + µa, µa + ν 1 A = 2ν1 D, tr A }{{} Oε 2 ) and so derived thermodynamically compatible nonlinear viscoelastic model is in some sense consistent with the well-known viscoelastic model. Nonlinear two-component viscoelastic model Let consider the material consisting of two components, to each component corresponds one natural configuration k p1t) a k p2t) see Figure 33)). We choose the entropy η = F. κr) κt) G 1 F kp1 t) κp 1 t)) F kp2 t) G 2 κp 2 t)) Figure 33: Two natural configurations ηe, ϱ, B, B kp1 t) k ), we can invert it and use the internal energy e = ẽη, ϱ, B p2 t) k, B p1 t) k ) p2 t) and concretely we choose e = e η, ϱ) + µ ) 1 tr B kp1t) 2ϱ 3 + µ ) 2 tr B kp2t) 2ϱ 3, 9 Note that from this we can read which is expected. Ȧ = 2D + Oε), 37

39 again we consider only isothermal processes and the incompressible material tr D = tr D = kp1 t) tr D kp2 =. After some calculation we obtain a relation for the rate of entropy production t) ξ = T µ 1 B kp1 t) µ 2B kp2 t) ) D + µ 1C d k p1 t) D k p1 t) + µ 2C d k p2 t) D k p2 t) = T d dis D + µ 1 C d k p1 t) D k p1 t) + µ 2C d k p2 t) D k p2 t). We choose the following constitutive relation for the rate of entropy production ξ = ξd, D kp1 t), D k p2 t), B k p1 t), B k p2 t) ) = 2ν D 2 + 2ν 1 D kp1 t) B k p1 t) D k p1 t) + 2ν 2 D kp2 t) B k p2 t) D k p2 t). Again we observe that the rate of entropy production is non-negative and so the second law of thermodynamics is satified. We define the Lagrange function LD, D kp1 t), D k p2 t) ) = ξ... )+λ 1 ξ... ) T d dis D µ 1 C d k p1 t) D k p1 t) µ 2C d k p2 t) D k p2 t) )+ Necessary conditions for the maximum are λ 2 tr D + λ 3 tr D kp1 t) + λ 4 tr D kp2 t). L D =, L D kp1 t) =, L D kp2 t) =. We substitute for L 1 + λ 1 λ λ 1 λ λ 1 λ 1 ξ D kp1 t) ξ D kp2 t) ξ D = Td dis + λ 2 λ 1 I, 44) = µ 1 C d k p1 t) + λ 3 λ 1 I, 45) = µ 1 C d k p2 t) + λ 4 λ 1 I. 46) Do the differentiation ξ D = 4ν D, ξ D kp1 t) = 4ν 1 C kp1 t) D k p1 t), ξ D kp2 t) = 4ν 2 C kp2 t) D k p2 t). and eliminate the Lagrange multipliers. To do this we add 44) D+45) D kp1 t) +46) D k p2 t), use the constrains of incompressibility and get 1 + λ 1 λ 1 = Td dis D + µ 1 Cd D kp1t) k + µ 2Cd D p1t) kp2t) k p2t) ξ D D + ξ D kp1 D + ξ kp1 t) D t) kp2 D kp2 t) t) = ξ 2 ξ = 1 2. Now take the trace of 44) and obtain = tr T d dis 3 λ 2 λ 1 λ 2 λ 1 =. By taking the traces of 45) and 46), we obtain λ 3 λ 1 = 2 3 ν 1D kp1 t) C k p1 t), λ 4 λ 1 = 2 3 ν 2D kp2 t) C k p2 t). 38

40 We get From 47) it implies that T d dis = 2ν D, 47) 2ν 1 C D kp1 t) k 1 ) p1 t) 3 D k C p1 t) k )I p1 = µ t) 1 C d k, 48) p1 t) 2ν 2 C D kp2 t) k 1 ) p2 t) 3 D k C p2 t) k )I p2 = µ t) 2 C d k. 49) p2 t) T = pi + 2ν D + µ 1 B d k p1 t) + µ 2B d k p2 t). As in the case of model with one natural configuration it holds B d k = 2F p1 t) k D p1 t) k p1 t) FT k 2 p1 t) 3 D B k )I + 2 p1 t) 3 C k D p1 t) k )I + 2 p1 t) 3 tr B k )D, p1 t) B d k p2 t) = 2F k p2 t) D k p2 t) FT k p2 t) 2 3 D B k p2 t) )I C k p2 t) D k p2 t) )I tr B k p2 t) )D Multiplying 48) from the left by F T k and from the right by FT p1 t) k p1, and 49) from the left t) by F T k and from the right by FT p2 t) k p2, we get t) 2ν 1 F D kp1 t) k p1 t) FT k 1 ) p1 t) 3 B k D p1 t) k )I p1 = µ t) 1 B d k, p1 t) 2ν 2 F D kp2 t) k p2 t) FT k 1 ) p2 t) 3 B k D p2 t) k )I p2 = µ t) 2 B d k. p2 t) From the relations for Oldroyd derivatives of B d k a Bd p1 t) k p2 we obtain t) 2ν 1 2ν B d k p1 t) 1 3 D Bd k p1 t) )I tr B k p1 t) )D ) = µ 1 B d k p1 t), B d k p2 t) 1 3 D Bd k p2 t) )I tr B k p2 t) )D ) = µ 2 B d k p2 t). The trace of both equations is zero, we have 2 5 scalar equations for 2 6 unknowns. We complete the system of equations with two equations for incompressibility of two elastic responses det B + 13 ) tr B )I kp1t) kp1t) = 1, det B + 13 ) tr B )I kp2t) kp2t) = 1. 39

41 We obtain the governing equations for the nonlinear two-component viscoelastic rate-type fluid model 2ν 1 2ν div v =, ϱ v = div T, B d k p1 t) 1 3 D Bd k p1 t) )I tr B k p1 t) )D ) = µ 1 B d k p1 t), B d k p2 t) 1 3 D Bd k p2 t) )I tr B k p2 t) )D ) = µ 2 B d k p2 t), det B + 13 ) tr B )I kp1t) kp1t) = 1, det B + 13 ) tr B )I kp2t) kp2t) = 1. T = pi + 2ν D + µ 1 B d k p1 t) + µ 2B d k p2 t), 4

42 Homework This is not full list of homework. It is just to complete the lecture notes. Homework 1. Under the assumption ξ = T d D d + m + p) div v + q θ, 5) θ derive Navier-Stokes equations by principle of maximization of entropy production max ξt d, m, q), 5)+T d,m,q where ξ is given by ξ = 1 2µ Td µ + 3λ m + p)2 + 1 q 2 K θ. Homework 2. Consider the following model ν ν T = pi + ν + D, 1 + Γ 2 D 2 ) 1 n 2 where ν, ν, Γ >. Explain which non-newtonian behaviour exhibits fluid described by that model for n R. Homework 3. Show that constitutive relation for Bingham fluid is equivalent to S τ D =, S > τ S = τ D D + 2µ D 2 )D 2µ D 2 ) τ + S τ ) +) D = S τ ) + S, where τ >, µ ) : R + R+ and x+ = max{, x}. Homework 4. Use the principle of maximization of entropy production ξ = ξt d, D d ) for T d A, where A := { T d R 3 3 sym, ξ = T d : D d}. In two cases for ξ given by: 1. ξ = T d r/r 1), 2. ξ = D d 2 r T d 2. Homework 5. Stokes fluid is given by constitutive relation T = pi + 2µD + αd 2, where α and µ are constant. Explain which non-newtonian behaviour exhibits Stokes fluid. Homework 6. Consider Korteweg s model in unidirectional velocity field v = uy),, ). Show that model exhibits normal stress differences T 11 T 22, T 22 T