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1 Permeability and Theta K varies as a function of moisture content in the vadose zone Williams, 2002 Modified after Selker,

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Presentation on theme: "1 Permeability and Theta K varies as a function of moisture content in the vadose zone Williams, 2002 Modified after Selker,"— Presentation transcript:

1 1 Permeability and Theta K varies as a function of moisture content in the vadose zone Williams, 2002 http://www.its.uidaho.edu/AgE558 Modified after Selker, 2000 http://bioe.orst.edu/vzp/

2 2 Extending Darcy to Unsaturated Media 1907 Buckingham saw Darcy could describe unsaturated flow q = - K(  )  H[2.102] K(  ): a function of the moisture content. Conductivity is not a function of pressure: the geometry of the water filled pores is all that matters, which is dictated by  alone. To express K as a function of pressure you must employ the hysteretic functional relationship between  and H. 1907 Buckingham saw Darcy could describe unsaturated flow q = - K(  )  H[2.102] K(  ): a function of the moisture content. Conductivity is not a function of pressure: the geometry of the water filled pores is all that matters, which is dictated by  alone. To express K as a function of pressure you must employ the hysteretic functional relationship between  and H.

3 3 How does K vary with  ? It decreases precipitously! Three factors are responsible for this behavior: 1. Large pores empty first. These are the pores with least resistance to flow, since they have the largest diameters (recall the 1/r 4 dependence of hydraulic resistance in the in the Hagen-Poiseuille equation). 2. Flow paths increase in length. Instead of proceeding straight through a volume of porous medium, the flow must avoid all the empty pores, making the path more tortuous. 3. There is less cross-section of flow. In any given area normal to flow, all the fluid must pass through a smaller portion of this area; for a given aerial flux the pore velocity must be higher. It decreases precipitously! Three factors are responsible for this behavior: 1. Large pores empty first. These are the pores with least resistance to flow, since they have the largest diameters (recall the 1/r 4 dependence of hydraulic resistance in the in the Hagen-Poiseuille equation). 2. Flow paths increase in length. Instead of proceeding straight through a volume of porous medium, the flow must avoid all the empty pores, making the path more tortuous. 3. There is less cross-section of flow. In any given area normal to flow, all the fluid must pass through a smaller portion of this area; for a given aerial flux the pore velocity must be higher.

4 4 Permeability with theta... To get a feel for how fast K drops, let’s consider the effect of tortuosity alone (item 2 above).  Particle flowpath and velocity must be distinguished from the “Darcian” flow length and the “Darcian” velocity. To get a feel for how fast K drops, let’s consider the effect of tortuosity alone (item 2 above).  Particle flowpath and velocity must be distinguished from the “Darcian” flow length and the “Darcian” velocity. Difference between the true microscopic fluid flow path length and flow velocity in comparison with the “Darcian” values, which are based on a macroscopic picture of the system.

5 5 A few illustrative calculations… So writing the true pressure gradient that acts on the fluid we have L e /L ratio of the true path length to the Darcian length. Next write equation for capillary velocity, v f (Hagen- Poiseuille) Translating into the Darcian velocity, q, we find So writing the true pressure gradient that acts on the fluid we have L e /L ratio of the true path length to the Darcian length. Next write equation for capillary velocity, v f (Hagen- Poiseuille) Translating into the Darcian velocity, q, we find

6 6 Conclusions on tortuosity... Solving for q we find or, after comparing to Darcy’s law we see that where C is some constant.  K goes down with (tortuosity) 2 and (radius) 2  K also hit by big pores emptying first  Conductivity will drop precipitously as  decreases Solving for q we find or, after comparing to Darcy’s law we see that where C is some constant.  K goes down with (tortuosity) 2 and (radius) 2  K also hit by big pores emptying first  Conductivity will drop precipitously as  decreases Characteristics of K(  ) 1: at saturation K= K s. 2-3: K(  ) = 0: pendular water

7 7 Adding Conservation of Mass: Richards Equation  Need to add the constraint imposed by the conservation of mass (L.A. Richards 1931).  There are many ways to obtain this result (we’ll do a couple). Consider arbitrary volume of porous medium. Keep track of fluid going into and out of this volume.  Need to add the constraint imposed by the conservation of mass (L.A. Richards 1931).  There are many ways to obtain this result (we’ll do a couple). Consider arbitrary volume of porous medium. Keep track of fluid going into and out of this volume.

8 8 Volume is independent of time, bring the time derivative inside Apply the divergence theorem to the right side Combining [2.110] and [2.111] Volume is independent of time, bring the time derivative inside Apply the divergence theorem to the right side Combining [2.110] and [2.111]

9 9 Since [2.112] is true for any volume element that the integrand must be zero for all points Replace q using Darcy’s law to obtain Richards equation where H is the total potential. This is also referred to as the Fokker-Plank equation Since [2.112] is true for any volume element that the integrand must be zero for all points Replace q using Darcy’s law to obtain Richards equation where H is the total potential. This is also referred to as the Fokker-Plank equation

10 10 In terms of elevation and pressure Total potential is sum of gravity potential and pressure, H = h + z but So we obtain noticing that Total potential is sum of gravity potential and pressure, H = h + z but So we obtain noticing that

11 11 And finally... So Richards Equation may be written (drum roll please....) Before we can get anywhere we need the relationships, K(  ) and h{  }.  First order in time and second order in space; require  1 initial condition and  2 boundary conditions So Richards Equation may be written (drum roll please....) Before we can get anywhere we need the relationships, K(  ) and h{  }.  First order in time and second order in space; require  1 initial condition and  2 boundary conditions

12 12 The diffusion form of R’s eq. Can put in more familiar form by introducing the soil diffusivity. Note that Can put in more familiar form by introducing the soil diffusivity. Note that

13 13 The diffusion form of R’s eq. D(  ) gives us a diffusion equation in   Favorite trick in solving diffusion problems is to assume that D is constant over space and pull it outside the derivative: it will find no place here!  D(  ) strongly non-linear function of q and varies drastically as a function of space.  This makes Richards equation quite an interesting challenge in terms of finding tidy analytical solutions, and even makes numerical modelers wince a bit due to the very rapid changes in both D and  D(  ) gives us a diffusion equation in   Favorite trick in solving diffusion problems is to assume that D is constant over space and pull it outside the derivative: it will find no place here!  D(  ) strongly non-linear function of q and varies drastically as a function of space.  This makes Richards equation quite an interesting challenge in terms of finding tidy analytical solutions, and even makes numerical modelers wince a bit due to the very rapid changes in both D and 

14 14 Gas Flow in Porous Media Might expect movement of gases is a simple extension of liquids  use gas density & viscosity with intrinsic permeability to get K  use Darcy's law. It isn’t that simple because of the shear forces. Recall the "no-slip" boundary condition  Idea: collisions between molecules in liquid so frequent that near a fixed surface the closest molecules will be constantly loosing all of their wall-parallel energy, rendering them motionless from the macroscopic perspective.  Requisite: mean free path of travel short compared to the aperture through which the liquid is moving. Might expect movement of gases is a simple extension of liquids  use gas density & viscosity with intrinsic permeability to get K  use Darcy's law. It isn’t that simple because of the shear forces. Recall the "no-slip" boundary condition  Idea: collisions between molecules in liquid so frequent that near a fixed surface the closest molecules will be constantly loosing all of their wall-parallel energy, rendering them motionless from the macroscopic perspective.  Requisite: mean free path of travel short compared to the aperture through which the liquid is moving.

15 15 Gas flow in porous media Gases  mean free length of travel on the order of medium pore size.  No-slip condition does not apply  No-slip results for liquids inapplicable. Taken together, results in "Klinkenberg effect" (experimental/theoretical 1941 paper by L.J. Klinkenberg). Gas permeability,  g, a function of gas pressure (dictates mean free path length) Gases  mean free length of travel on the order of medium pore size.  No-slip condition does not apply  No-slip results for liquids inapplicable. Taken together, results in "Klinkenberg effect" (experimental/theoretical 1941 paper by L.J. Klinkenberg). Gas permeability,  g, a function of gas pressure (dictates mean free path length)

16 16 Parameters  intrinsic permeability to liquid flow rcharacteristic radius apertures of the medium mean free path length of the gas molecules cproportionality factor between the mean free path for the free gas compared to the mean free path for gas which just collided with the capillary wall (c is just slightly less than 1, Klinkenberg, 1941). Coarse media: capillaries larger than the mean free path length, the intrinsic permeability for liquids is recovered Fine media: gas permeability exceeds liquid permeability. Parameters  intrinsic permeability to liquid flow rcharacteristic radius apertures of the medium mean free path length of the gas molecules cproportionality factor between the mean free path for the free gas compared to the mean free path for gas which just collided with the capillary wall (c is just slightly less than 1, Klinkenberg, 1941). Coarse media: capillaries larger than the mean free path length, the intrinsic permeability for liquids is recovered Fine media: gas permeability exceeds liquid permeability.

17 17 Final notes on gas permeability  g /  measure of size of conducting pathways,  can be used as a diagnostic parameter (Reeve, 1953). As P  the mean free path length  r  limit for  g  10   Klinkenberg's found max  g  5 - 20 times  Practical purposes correction for vicinity of 1 bar 20 to 80% over the liquid permeability, depending upon the medium (Klinkenberg, 1941). Methods of measurement of gas permeability: Corey (1986), New methods: Moore and Attenborough, 1992.  g depends on the liquid content, analogous to hydraulic conductivity with moisture content (see Corey, 1986).  g /  measure of size of conducting pathways,  can be used as a diagnostic parameter (Reeve, 1953). As P  the mean free path length  r  limit for  g  10   Klinkenberg's found max  g  5 - 20 times  Practical purposes correction for vicinity of 1 bar 20 to 80% over the liquid permeability, depending upon the medium (Klinkenberg, 1941). Methods of measurement of gas permeability: Corey (1986), New methods: Moore and Attenborough, 1992.  g depends on the liquid content, analogous to hydraulic conductivity with moisture content (see Corey, 1986).

18 18 Expressions for Conductivity & Retention  To solve Richards equation we need the mathematical relationships between pressure, moisture content and conductivity.  Conductivity is a non-hysteretic function of moisture content  Moisture content and pore pressure are related through a hysteretic functional (often hysteretic ignored; gives expressions of conductivity in terms of matric potential).  For analytical solution, must use analytical expressions.  For numerical solutions may use constructed as interpolations between successive laboratory data.  To solve Richards equation we need the mathematical relationships between pressure, moisture content and conductivity.  Conductivity is a non-hysteretic function of moisture content  Moisture content and pore pressure are related through a hysteretic functional (often hysteretic ignored; gives expressions of conductivity in terms of matric potential).  For analytical solution, must use analytical expressions.  For numerical solutions may use constructed as interpolations between successive laboratory data.

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