Department of Mathematics

IDEA: Internet Differential Equations Activities

Landfills and Pollutants

For most of us, when we throw something into the trash, we forget about its existence. For environmental engineers and cities across the United States, however, this is when the problem begins. It takes considerable resources to plan and create containment systems for the waste we produce. Some chemicals, if introduced into our aquifers, can contaminate literally billions of liters of our drinking water. According to Johnson, et al. (1), nineteen kilograms of benzene can contaminate 3.8 billion liters of water at the EPA standard.

Recycling, burning,and burial in landfills are some of the ways used to dispose of waste. When engineers design a system to store hazardous or toxic materials they have to take into account the type of building materials and how the waste interacts with it. One method is to use local clay in compacted layers, another is to use a synthetic geomembrane, due to its low permeability.

Engineers and city planners can use differential equations to determine how far and how fast contaminants will spread. Near a coal mining region, for example, we may need to dispose of naphthalene. We can determine how thick the liner must be to contain the leachate from the waste for two hundred years, one thousand years, or more. Understanding diffusion allows us to model these physical processes.


Deriving the Steady State Diffusion Model

The complete steady state Diffusion Model, with boundary conditions, is given by:

(1) -D(A(x)*c'(x))' + (A(x)*V(x)*c(x))' = 0

c(0)=c0 and c(L)=cL

How was this equation determined? First of all let's look at its component parts. -D(A(x)c'(x))' models the rate at which some material moves through a unit area due strictly to diffusion. The D stands for the material's diffusion coefficient, A(x) for the unit area, and C (x) for the concentration gradient. The concentration gradient is the rate of the change in concentration from one point to another, and hence we find the concentration gradient by taking the derivative of the concentration function, C(x).

When diffusive action is aided by physical flow then C(x), multiplied by the flow's velocity, V(x), through a unit area, is added to the diffusion rate. This physical flow is also known as convection. This is where (A(x)*V(x)*c(x))' comes from.

At equilibrium the total diffusive flow through a unit volume is constant. Due to the Law of Conservation of Mass, the rate of mass in to any volume element is equal to the rate of mass out of that same element. Therefore equation (2) says that the flow in at a point x is equal to the flow out at a point .

(2) -A(x)Dc'(x) + A(x)V(x)c(x) = -A()Dc'() + A()V()c()

Problem 1: Collect like terms of equation (2) and divide by  . If we take the limit as  goes to zero what equation should we expect to see?


Simple Diffusion Downward

A barrel of radius .75 m sits atop a clay liner of 1 m depth. Benzene is leaking along the entire bottom of the barrel. Find the equilibrium concentration of benzene between the top of the clay liner and the bottom of the clay liner. The concentration of benzene leaking from the barrel is .27 g/cm3 , while at the top of the drainage blanket it is equal to cL. Notice that in this situation the movement of the benzene compound is due to diffusion alone. What does this mean for the term containing velocity?

Problem 2:

  1. Using what you learned in the section above determine if the second order differential equation c''(x) = 0 models the process given.
  2. Solve for the general solution of this ODE.
  3. Using the data provided above determine what the constants of integration must be and rewrite the ODE with them.
  4. If the concentration at 1 meter below the barrel is greater than .27 g/ cm3 what physical reality is the equation describing? What if the concentration is equal to .27 g/ cm3 ? Or less than .27 g/ cm3 ? Graph each case and describe what the equation says the benzene is doing.

Diffusion through the Drainage Blanket

In many landfills that use several clay liners to prevent the spread of pollutants, a layer of sand or gravel is placed beneath the first clay liner to allow the contaminant to be drained or pumped off to waste sumps. This layer is known as a "drainage blanket" and can act as a warning sign that pollution has broken through the first clay layer and is approaching the second.

In our model above, benzene has diffused downward and is entering the drainage blanket. Assume a concentration of .15 g/ cm3 , which is slowly diffusing towards the waste sump, through an area roughly square in shape, each side approximately 1.75 cm. We wish to find an equilibrium concentration that describes this. However, as the benzene is diffusing outward, water has accidentally been allowed to enter the drainage blanket and is moving in the same direction as the benzene at a rate of .05 meters per hour. The waste sump is 8 meters away and the benzene concentration being recorded by sensors there is at .03 g/ cm3 .

Take the coefficient of diffusion in sand to be 2 * 10-4 cm2/s.
[Hint: Assume the area is constant from x = 0 to x = 8 meters]

Problem 3:

  1. What are the boundary conditions?
  2. Show that the differential equation : -(2 * 10-4 )*c''(x) + (.001389)*c'(x) = 0 follows from the steady state Diffusion model for this problem.
  3. Solve for the general solution of the ODE of part b.
  4. Solve for the constants of integration of this ODE and rewrite the equation to include them.

Diffusion and Partial Differential Equations

Diffusion is a process where a solute in some solution moves from a high concentration to a lower concentration. This occurs along the solute's concentration gradient. Movement of the solute is not necessarily due to movement by the solution itself. We saw this when we derived our steady state Diffusion model of part one. The model is composed of two pieces. One piece, -D(A(x)*c'(x))', showed the rate due strictly to solute diffusion. The other, (A(x)*V(x)*c(x))' , added in the movement due to the solution containing the solute. Fick's second laws deal with the first piece of our model. It gives the concentration of a material at some point at any given time. This is useful if we want to understand how a concentration is changing over time before it reaches its steady state value.

At first glance there doesn't seem much to relate our steady state Diffusion model of equation (1) to (3). However, if we replace the zero in equation (1) with  ,we can begin to see where it comes from.
 

In fact, it is when  = 0 that we have reached equilibrium, since this is when there is no change in concentration over time.

Recall that in our model that A(x), as a unit area, is equal to one. Also, the concentration is now a function of not only position, but of time as well. This means that when we take the derivative of the concentration with respect to x we get a partial derivative with respect to x, and the same for when we derive c(x,t) with respect to time. So now our diffusion model looks like this:
 

Or equivalently:

   (6) 

which is what we wished to derive. Note that since we don't know the sign of the Diffusion coefficient, D, the minus sign in our equation does not change the equivalence of (3) and (6).


Another look at Diffusion and Partial Differential Equations

Fick's second law can be written as:

(7) 

where  Now assume C(x,t) = F(x)*G(t).

Problem 4:

  1. Compute Ct and Cxx.
  2. Substitute what you found in problem 4a into equation (7). Collect the G and D terms on one side of the equation and F terms on the other. What does the equation look like?
  3. Take the left-hand side of problem 4b and set it equal to -k. It should look like a familiar ODE. Solve this ODE symbolically. Do the same for the right hand side of your problem 4b equation. It also should be a familiar ODE. Again, solve it symbolically.
What you have just done is the beginning of solving a partial differential equation. For the rest of the steps you should see your textbook on separation of variables for PDE's.


Breakthrough Time

In clay lined landfills, pollutants will most probably escape their liner. This is one reason drainage blankets are sometimes built, so as to channel contaminants into waiting sumps. Using the solution to equation (3) we can determine the time it will take for some organic compounds, such as benzene, to reach a steady state value (or equilibrium concentration) across a control volume. Incidentally this equation should look somewhat familiar to what you did in the section above.

(8) 

[L = liner thickness, z = vertical distance compound travels, D = diffusion coefficient, t = breakthrough time in years, c = concentration, c0 = concentration at interface]

Problem 5:

Given  = .001 (breakthrough, for some organic compounds, is defined as the arrival of  = .001 ), liner thickness equal to 1 meter, and vertical distance compound travels as 1 meter, determine the breakthrough time for:

  1. Benzene with diffusion coefficient D = 2 * 10-7 cm2/s 
  2. Naphthalene with diffusion coefficient D = 2 * 10-8 cm2/s 
What if we increased the liner thickness and vertical distance values to 50m? What would the values be for questions a and b? Is this as expected?

[Note: In calculating the value of t we used a finite upper value for N. For example, you may wish to set the iterations to 3, 5, or 10. Using a computer algebra system, such as Mathematica, is also valid.]


LITERATURE CITED
(1)Johnson, et al.,"Diffusive Contaminant Transport in Natural Clay: A Field Example and Implications for Clay-Lined Waste Disposal Sites", Environ. Sci. Technol. 1989, 23, 340-349.

REFERENCES
T. Christensen, et al., Sanitary Landfilling: Process, Technology, and Environmental Impact, Academic Press Inc., San Diego, 1989

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