2. Modeling of shrinkage during first drying period

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1 2. Modeling of shrinkage during first drying period In this chapter we propose and develop a mathematical model of to describe nonuniform shrinkage of porous medium during drying starting with several assumptions which will help us to simplify the problem and get reasonable results. 2.1 Basic ideas The fundamental principle of the model is that the drying is in the first period and will remain so through the entire drying process. As it is discussed in the first chapter this means that all the pores are completely filled with liquid and thus, the surface is saturated. So when drying goes on, the amount of liquid in the porous media will decrease and since the whole media is only filled with liquid, there needs to be a reduction in the volume of the media to compensate the volume reduction caused by the evaporation of the liquid inside the media. The porous media is modeled here. The void space of porous media is approximated by a network of cylindrical pores (pores), which are connected together at nodes. The solid phase is approximated by a particle network. Particles are assumed to be spheres of uniform size and so, all pores have the same size initially. The liquid flow in the pores is modeled same as the flow in a pipe (which is explained in detail in the next part). The moment drying process starts, liquid inside the pores flows towards the drying surface. The liquid flow in pores is always in the direction of negative pressure gradient, which means liquid flows from points with higher pressure to the points with lower pressure. The pressure gradient in the porous media is expected to increase from the top towards the network's bottom (in case evaporation is only from the top) Liquid flow in the pores To model liquid flow in each pore the Hagen-Poiseuille equation is used. This equation describes a slow viscous incompressible flow through a constant circular cross-section. One of its uses is in the experimental measurement of viscosity, by measuring the pressure drop and volumetric flow rate through a tube of known length and diameter. The law in fluid dynamic notation can be written as: P = 8μLV πr 4 (2.1)

2 Where P is the pressure drop along the pore a), μ the liquid viscosity a.s), V the volumetric flow rate in the pore ( m3 ), and L and r are length and radius of the pores (m) s respectively. If the liquid pressure inside a pore ) and that at the inlet and outlet are known, the liquid flow rate can easily be calculated using Hagen-Poiseuille equation as follows: V in = πr4 8μL in P) (2.2) V out = πr4 8μL P out) If the outgoing flow rate is more than incoming flow rate, the amount of liquid in the pore decreases which may lead to reduction in pore volume (in our case shrinkage). In order to calculate the flow rates, the liquid pressures need to be calculated. Calculation of liquid pressure is described in the next part Pressure of each pore In order to calculate the liquid pressure, it is needed to solve multiple equations and inspect the evolution of menisci inside the system. Since the shrinkage in the first drying period is to be modeled here, it is not possible build the model around menisci and its effect on pressure drop in the system; so one needs to come up with another idea on acquiring the liquid pressures. In the first drying period pores are fully filled with liquid. So if there is a reduction in the amount of the liquid in the pores, it will cause the pressure to decrease in the system; however, there is a pressure drop in the pores on the surface of the medium which is caused due to formation of menisci in capillaries. This reduction in liquid pressure may induce a stress on the solid phase and moves the solid particles towards each other. This stress is expected to cause the solid phase to shrink so as to compensate the reduction in liquid volume and the pores remain fully wet. The shrinkage in the solid phase may cause an overpressure in the body. This overpressure is expected to be equal to the pressure reduction caused by the reduction of the liquid volume so that it can be compensated (a system in equilibrium cannot have pressure lower than surrounding atmosphere). So we came up with the idea of calculating the solid overpressure using the geometry of the system, for each pore and use the negative of this amount to calculate the liquid pressure.

3 The solid particles are assumed to be spheres with radius R. 4 particles are placed next to each other as shown in Figure 2.1, and a pore with radius r lies within them. Figure 2.1: Schematic representation of solid particles and a pore geometry When the liquid volume drops in the pore and the pressure decreases, the pores are expected to shrink and the solid particles overlap with each other. This phenomenon causes the overpressure in solid phase which can be calculated using the linear contact model which states that the force between two particles is proportional to the product of the particle overlap u and stiffness k: F = k. u (2.3) Using the Phythagoras theorem (Figure 2.1), the distance between the center of two neighboring particles and therefore, the overlap ( u), can be calculated: d = 2(r + R) (2.4)

4 u = 2R d = 2R 2(r + R). (2.5) Equation (2.5) is replaced into Equation (2.3) and simplifying gives: F = ((2 2)R 2r) k. (2.6) In order to calculate the pressure caused by this force, the area (A) on which this force acts, has to be known. Again it can be calculated using the geometry: A = L 2(r + R). (2.7) And thus the pressure is: P = F A = ((2 2)R 2r)k L 2(r+R). (2-8) Figure 2.2 shows the liquid pressure versus the pore radius, which is plotted using Equation (2.8). Figure 2.2: pressure dependency on pore radius The figure above is for the particles with radius of 1 µm, and stiffness of 1 N/m and pore length of 2 µm. As can be seen at the pore radius of 0.42 µm the overpressure is zero. This

5 is exactly at the beginning of the drying where there is no overlap in the solid phase. And as the pore radius decreases the overpressure increase Liquid evaporation At the surface of the body, liquid is evaporated and the evaporation rate is calculated by equation (1.1): M v = β g ρa(y Y ) = β gam v R T v P v, ) Where β g is gas-side mass transfer coefficient, ρ v the water vapor density, A evaporating surface, Y moisture content in air, M v molecular weight of water, R universal gas constant, T the temperature and P v and P v, vapor partial pressure in air at the surface and in the bulk respectively. Density is transformed to the other parameters using the ideal gas law. There are several ways of calculating mass transfer coefficient β g. One of the easiest ways is to use Sherwood number and the empirical equations to calculate the Sherwood number. When Sherwood number is known, the mass transfer coefficient can easily be calculated: β g = Shδ L. (2.9) Where δ is the mass diffusivity between 2 known gasses (water vapor and air) and L is the characteristic length which depends on geometry and dynamic model (e.g. direction of flow onto a surface). As pores tend to shrink during the drying process the area available for evaporation also decreases, and thus the evaporation rate decreases during the whole drying process. 2.2 shrinkage of a system of vertical pores The first model system is a combination of 20 vertical pores attached together and the evaporation is only from top pore. Such system can be constructed using spherical particles as shown in Figure 2.3. In such system, pores are assumed to be cylinders and the dimensions are calculated using the particles size. There is an inlet flow of water to each pore and an outlet flow from each. The only cylinder which does not have any inlet flow is the bottom pore (Figure 2-4).

6 Figure 2.3: 3D illustration of pores and particles water. As it can be seen in Figure 2.3 the space inside the pores are completely filled with

7 Figure 2-4: Schematic illustration of the pores As it is shown in Figure 2-4 the pores in the top part of the system are expected to shrink faster and more than the pores in the bottom of the system (since the evaporation is only from top) Governing differential equations In order to model the whole system it is needed to solve a differential equation for each throat. Since each middle pore is linked to the one on top and the one beneath, differential equations are coupled together and need to be solved simultaneously. So there is a system of ordinary differential equations and the boundary conditions as shown in Figure 2.4 are no inlet for the bottom pore and evaporation from the top pore.

8 Each pore is treated as a separate cylinder that shrinks independently and the only coupling between the equations is due to change of the inlet and outlet pressure of each pore when the neighboring pores change in dimensions. Figure 2.5 shows the control volume and the pressures acting on each pore. Figure 2.5: control volume used to obtain the differential equations The ordinary differential equations are derived using mass balance at pore. The differential equations for the middle pores are as follows: dv i = r 4 i 1 8μL w,i 1 P w,i ) i 1 inlet r i 4 8μL i w,i P w,i+1 ) outlet i=2 n-1 (2.10) For the top and bottom pores the flow equations are as follows: dv 4 1 = r 1 8μL w,1 P w,2 ) (bottom) (2.11) 1 dv n = r 4 n 1 8μL w,n 1 P w,n ) M v n 1 ρ (top) (2.12)

9 In order to solve this system of equations it is needed to get a relationship between the rate of change in volume ( dv i ) and radius and length of each pore. The relationship for a cylinder's volume can be written as V = πr 2 L. The time derivative of this relationship by time gives us the following equation: dv i = r i 2 dl i + 2L d(r 2 i ) 2 i = r dl i + 2L dr i i ir i (2.13) By inserting Equation (2.13) into Equations (2.10) the ODEs are not yet solvable. Since there are two variables (L and r) evolving by time: possible. r i 2 dl i + 2L ir i dr i = r 4 i 1 4 8μL w,i 1 P w,i ) r i i 1 8μL w,i P w,i+1 ) i=2 n-1 (2.14) i So there needs to be further simplifications to make solving of this system of ODE Shrinkage only in radial direction The first simplifying assumption we used for the model was to consider the shrinkage only in radial direction for each pore, so the term dl i in equation (2.14) becomes zero. This way solving the ODEs is made possible and the governing differential equation (2.14) changes to: 2L i r i dr i = r 4 i 1 4 8μL w,i 1 P w,i ) r i i 1 8μL w,i P w,i+1 ) i=2 n-1 (2.15) i It is obvious that if the length of the pores will not change, then since all of the lengths are equal in the beginning we don t need to indicate any indices for L in the governing differential equation: 4 r i 1 dr i = [ 16μL 2 w,i 1 P w,i ) 16μL 2 w,i P w,i+1 )] r 4 i r i i=2 n-1 (2.16) The top and bottom pores' differential equations will also change: 3 dr 1 = r 1 16μL 2 w,1 P w,2 ) (2.17) dr n = 4 (r n 1 8μL w,n 1 P w,n ) M v ρ ) 1 2r n L (2.18)

10 M v can be calculated using the mass transfer equation (1-1) Solution approach This system of ordinary differential equation is solved numerically. The sequence used to solve this system is as follows: 1. Initial values for pore radius and length are calculated from the geometry of the system using particle radius. 2. Using these values to calculate pressure inside the pores (which is zero at the beginning because there is no overpressure). 3. The evaporation rate is corrected for the new top pore radius at the top. 4. Solving the system of ODE to get the new values for pore radius. 5. If the end of the drying process has not yet reached go to step Stop the simulation.

Shell Balances in Fluid Mechanics

Shell Balances in Fluid Mechanics R. Shankar Subramanian Department of Chemical and Biomolecular Engineering Clarkson University When fluid flow occurs in a single direction everywhere in a system, shell