Saturated flow

 

Now that we have looked at solids and fluids, we start to look at the behavior of a single fluid in a solid (a porous medium).  We will start with the issue of single-phase flow.  For a soil, that could be saturated flow of water, or air-dry flow of air.

 

The good news is that the basic relationship for saturated flow is really easy:

            Flow is proportional to pressure gradient:  q = -K dP/dL

(Makes sense, doesn’t it?!)  This relationship is called Dary’s Law, after the French civil engineer Henry Darcy of Dijon, who worked it out empirically around 1856.  Analogy with Ohm’s law, Hooke’s law, Fourier’s law, and Fick’s law.  Note that conductivity = 1/resistivity.

 

The issues involved in really understanding this are:

1)      properly defining the terms

2)      measuring the proportionality constant K (“hydraulic conductivity”), and

3)      Determining under what conditions this equation actually holds.

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What is a fluid, and how does it flow?  We established earlier that a Newtonian fluid has resistance to compression but not to shear – deformation is proportional to shear, and the resistance is expressed as a viscosity.  For the simple situation of a fluid between two plates when one plate moves relative to the other,

           

where t is shearing stress, eta is viscosity, and du/dy is the velocity gradient.  Notice the similarity to equations of elastic deformation.

 

Poiseuille’s law (derivation based on Hillel):

Suppose we have a pipe of radius R (diameter D=2R), and we push fluid through it.  The fluid pressure acts on the whole face equally, so

           

and similarly for any concentric cylindrical sub-set within the pipe.

 

Looking at such a concentric cylinder of radius y, if flow is steady, force mush be equal to viscous resistance.  This resistance acts at the edge of the imaginary cylinder, so

           

Substituting in the earlier definition of t, we have           

           

Integrating gives

           

Given that when y=R, u=0 (no-slip condition at the wall), we have c=R2/4, so we can substitute this in, getting

           

Showing that we have a parabolic velocity distribution.

The maximum velocity is at the center, and is proportional to R2:

           

Likewise the mean velocity is

           

Solving for total flux (discharge), we have

           

As you’d expect, Q is proportional to the pressure gradient and inversely proportional to the viscosity.  The R4, though, is the surprise: a huge influence of radius. 

 

But don’t get carried away: 2 powers are for the mean velocity, and 2 for the increase in cross-sectional area.  If you double the size you take up more area, so you get fewer pores.  So for a given cross-sectional area of soil, the discharge is proportional to the velocity, or R2.

 

This raises the question, is there a relationship between the R of a soil (say, mean pore size) and the K (hydraulic conductivity)?  Specifically, K ~ R2?  As it happens, there is some justification for this, which we’ll see next lecture.

 

Suppose now we have two plates that are not moving, and the fluid moves between them.  Local cubic law: in this geometry, the velocity is not a quadratic but a cubic, and

vmax = R3

 

But most pores aren’t round tubes or infinitely long slits (what shape are they, and how would we know?).  We can generalize this equation to get a sort of effective “hydraulic radius”:

           

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How would we measure the K in Darcy’s law?

First we have to understand more about how we define dP and dL.  Consider a swimming pool.  The water at the bottom is under more pressure than the water at the top, but the water doesn’t flow upwards.  In other words, there is no net gradient causing flow, so the pressure difference must be compensated by something else.  Specifically, gravity.  More generally than pressure, we can call the driving force potential, and say

Total potential = pressure potential + gravitational potential