I. Borsi. EMS SCHOOL ON INDUSTRIAL MATHEMATICS Bedlewo, October 11 18, 2010

Transcription

1 : an : an (Joint work with A. Fasano) Dipartimento di Matematica U. Dini, Università di Firenze (Italy) borsi@math.unifi.it porous EMS SCHOOL ON INDUSTRIAL MATHEMATICS Bedlewo, October 11 18, 2010

2 Outlook porous : an porous

3 Porous materials The structure A porous medium: any medium in which are present a matrix and a void space. In a porous medium the water (or any other fluid) flows through a very complex network of pores and capillaries. The latter form the void space of the medium. The flow The boundaries of this flow are the microscopic interfaces between solid and fluid(s). the complete study of the hydrodynamics at the micro-scale is very involved! : an porous

4 Continuum approach : an porous To avoid such a difficulty, we may address the problem from a macro-scale. At this scale the quantities involved in the problem can be measured. Continuum approach: the real medium it is substituted by an artificial model, in which any phase (liquid and air, for instance) is view as a continuum filling all the medium.

5 Continuum approach: how can we define a quantity? For instance: we want to know the density ρ of the medium in a point x. Imagine to expand an ideal sphere (whose center is x) At every step, measure the value of the density averaged on the sphere Increasing the radius: we get different values (random structure of the medium) : an porous There is an interval of the sphere radius in which the value of ρ becomes stable!

6 Continuum approach (ctd.) : an porous We can fix a radius R belonging to this interval We call REV (Representative Element Volume) the sphere whose radius is R. Conclusion: Any quantity defined at the macroscopic scale in a point x has to be thought as a quantity averaged on the REV centered in x.

7 Darcy s law Definitions: volume of the void space Porosity: φ = volume of the REV Saturation (assume only one fluid flowing through the medium): S = volume of fluid volume of the void space, The medium is saturated (namely all the void space is occupied by the fluid), if S = 1. Specific discharge, q: the flux of fluid per unit area of the surface, (dims. LT 1 ). : an porous

8 Darcy s law (ctd.) Darcy s experiment To determine the law linking q to the water pressure, the French engineer Henry Darcy, studying the system of water fountains of its city, Dijon, built up a particular experiment, reported in its famous book Les Fontaines Publiques de la Ville de Dijon (1856). : an porous

9 Darcy s law (ctd.) Darcy measured the relationship between q and the hydraulic head h, defined as h = z + p w /ρg, where z is the quote at which the water is raised, p w is the water pressure and g is the gravity acceleration. He found a proportionality relation between discharge Q (volume of water per unit time) and increment of the hydraulic head with respect to the quote, i.e. : an porous Q = K A h 1 h 2 L where A is the area of the column section and L its length. The coefficient K is called hydraulic conductivity.

10 Generalization of Darcy s law In a differential form: q = K h z = K z ( ) pw ρ g + z If density and/or the viscosity are not constant: q = k µ z (p w + ρgz) where k is the medium permeability and µ is the fluid viscosity. We have: K = k ρ g, [k] = L2 µ : an porous k : property of the medium; ρ, µ: properties of the fluid Generalized form of Darcy s law: q = k ( p ρg), µ where p is the pressure and g is the gravity vector.

11 Definitions and notation Three phases (unsaturated ): a = air (inert gas) v = vapour w = water = g = v + a = gas mixture = vapour + air φ porosity S α = saturation of phase α = w, g ρ α = density of phase α = w, g ρ i = density of species i (w.r.t. the gas volume),i = v, a = S g = S v + S a ; S w + S g = 1 : an porous

12 Definitions (ctd.) : an porous p w, p v = water and vapour pressure P = (total) gas pressure (P p v ) = air pressure p c = P p w = capillary pressure of the liquid phase (see later on) p w = P p c

13 Mass balance : an porous t (ρ w φs w ) + J w = Γ (1) t (ρ v φs g ) + J v = + Γ (2) t (ρ aφs g ) + J a = 0 (3) where Γ is the evaporation rate and J α is the total mass flux of phase α.

14 Expression of fluxes Vapour: : an J v = ρ v q }{{} v Darcy s flux due to pressure gradient We have (gravity is neglected): where k sat, saturated permeability + j }{{} v flux of (binary) diffusion (4) q v = k satk g P (5) µ g ( ( pv )) j v = D (6) P k g = k g (S g ), relative permeability µ g, viscosity porous

15 Expression of fluxes (ctd.) : an porous Similarly, for air: J a = k sat k g ρ a P µ g }{{} Flux due to pressure ( ( )) P pv D P }{{} binary diffusion (7)

16 Expression of fluxes (ctd.) : an porous Water: J w = ρ w q w = ρ w k sat k w µ w [(P p c ) ρ w g] (8) Here the gravity may play a significant role, so that we include it in the Darcy s flux.

17 Constitutive eq.s Capillary pressure p c = p c (S w, T ). For instance, Leverett function: φ p c (S w, T ) = σ(t )f (S w ), k sat where: S w,r, irreducible water saturation, T, temperature. σ(t ) = σ 0 βt, surface tension. f (S w ) = e ( 40(1 Sw )) (1 S w ) S w S w,r....or many other forms (vangenuchten, Brooks & Corey, etc.) : an porous

18 Constitutive eq.s (ctd.) Relative permeability: Example (Verma et al., 1985): k w = k w (S w ) k g = k g (S w ). ( ) k w = (S eff ) 3 Sw S 3 w,r =, for S w S w,r 1 S w,r k g = a + bs eff + cs 2 eff, a, b, c = const. : an porous or many other forms (e.g. Mualem, Corey, Grant). Liquid viscosity & density (only in of high temperature and/or pressure gradient): µ w = µ w (T ), ρ w = ρ w (p w, T ).

19 Constitutive eq.s (ctd.) : an Moreover, the perfect gas law for vapour and air ρ v = p v M v RT ρ a = (P p v ) M a RT with M v and R constants. (9) (10) porous

20 More on water flux... : an porous In particular, exploiting the expression for the capillary pressure in the Darcian water flux q w, we have J w = k sat k w ρ w (P ρ w g) µ } w {{} Darcy s flux D s S w D T T }{{} Capillary flux due to saturation & temperature

21 Energy balance : an where: ( ρ s (1 φ)h s + t α ρ α φs α ) + α h α J α = (λ mix T ) + (h v h w ) Γ, (11) subscript ( ) s refers to the solid matrix h α, enthalpy of phase α λ mix = λ s (1 φ) + φ α λ αs α, thermal conductivity of the mixture. porous

22 Closure to the system Exploiting the expression of fluxes and the constitutive laws in mass and energy balance, we have: 4 eq.s (1), (2), (3), (11) 5 unknowns P, p v, S w, T, Γ -1 2 options for closure: The description of evaporation can be based on either 1. Equilibrium 2. Non-equilibrium condition : an porous

23 Closure to the system (ctd.) Equilibrium: we assume there is a water-vapour equilibrium relation: p v = F (S w, T ) (12) and sum mass balance of water and vapour [(1) + (2)], so that Γ disappears! For instance, Kelvin s equation: p v = pv sat (T ) exp { Mv p c (S w ) ρ w RT }, (13) : an porous which is derived from Clapeyron s eq., assuming a local thermodynamic equilibrium.

24 Closure to the system (ctd.) Non-equilibrium: we assume there is an equilibrium value for p v, and the evaporation rate is defined as p v,eq = F (S w, T ) (14) Γ = γ (p v,eq p v ) (15) This approach has to be used if the time to reach the equilibrium is much longer than the time scale of the evaporation process (e.g. fast drying process in food industry). In this Γ is a function of p v, so that we have 4 eq.s and 4 unknowns. : an porous

25 Boundary conditions on the external surface In general, outside the domain, the following conditions are known: Air pressure Temperature Relative humidity: H = p v p sat v (T ). (16) Assuming an equilibrium condition for the vapour pressure (Kelvin s eq.), (16) is related to S w, { } Mv p c (S w ) H = exp ρ w RT : an porous

26 Boundary conditions (ctd.) : an Under these assumptions, the conditions on the boundary are: 1. Mass flux: H n = ν (H H ext ). 2. Pressure: (P p v ) = P air ext 3. Temperature: T = T ext Remark: in condition 1, the coefficient ν represents the interface effect (permeability of the boundary). As ν high permeability: the effect vanishes and the condition is H = H ext porous As ν 0 low permeability: no flux condition

27 Analysing the physics of the process, one may assume: 1. Air (inert gas) is always at the same pressure. p a = (P p v ) = const. Only eq.s for water & vapour 2. Capillarity as primary mode, p c P P 0 The process is driven by the capillary gradient only equation for water. 3. Sharp interface A sharp interface divides the dry region (S w S w,r ) to the wet region. All the evaporation takes place at the interface We obtain a free boundary problem 4. Isothermal condition, when T const. : an porous

28 : an If the initial porosity (= before the drying) is quite large, the loss of water may cause a significant stress within the structure of the medium. Then, the porosity φ is no longer constant and the problem becomes very involved! In particular: In the mass balance, among the fluxes, we have to include the one describing the movement of the matrix (since Darcy s law refers to flow relative to the skeleton!) porous

29 Shrinkage (ctd.) The simplest way to model this process is the definition of a constitutive law between porosity ans water saturation, For instance: Linear: Nonlinear (cubic): φ = F(S w ) (17) ( ) φ = φ (0) S w (0) S w + c 1 S w (0). ( ) φ = φ (0) S (0) 3 w S w + c 2 S w (0). : an porous and many other forms (see Mayor, J. Food Eng., 2004)

30 The simpler problem in isothermal conditions (T = const.) : an porous This approach can be applied whenever the drying takes place in a natural environment (=not forced drying). Assume also Constant air pressure, p a = (P p v ) = const. Therefore: rescale p a to 0, so that P = p v Equilibrium condition (Kelvin s equation): p v = p v (S w ).

31 (ctd.) : an porous Under these conditions, the system simplifies to t [ρw φsw + ρv φ(1 Sw )] = j» k satk w ρ w µ w dpc ds w dpv ds w ρ w g «ff k satk g dp v + ρ v S w µ g ds w + EOS: ρ v = ρ v (p v ) ; p c = p c (S w ); k α = k α (S w ) Only 1 equation (in S w ), even if strongly nonlinear!

32 External B.C. We assume a condition on the relative humidity. Since T = const., saturated vapour pressure is a constant: pv sat. Therefore, the condition on the relative humidity becomes a condition on p v (and in turn on S w )! Remember: H = p v pv sat = 1 pv sat p v (S w ) Thus, the B.C. H = ν (H H ext ) reads as: [p v (S w )] = ν ( p v (S w ) H ext pv sat ) : an porous which is a Robin s condition for S w.

33 Some references For a complete on porous : J. Bear, Dynamics of fluids in porous, 1972 (or any other book by J. Bear). For a rigorous derivation of the eq.s (from micro- to macro-scale): S. Whitaker, Simultaneous heat, mass and momentum : a theory of drying, Advances in Heat Transfer, 13 (1972), A review on modelling the drying in food processes: A.K. Datta, Porous approaches to studying simultaneous heat and mass transfer in food processes. I: Problem formulations, Journal of Food Engineering, 80, Issue 1 (2007), A review on the in food processes: L. Mayor and A. M. Sereno, Modelling during convective drying of food materials: a review, Journal of Food Engineering, 61, Issue 3 (2004), : an porous