Heavy Metals Release in Soils - Chapter 5 ppt

It is our intention in this chapter to do four things: 1 we will introducethe technique of phase plane analysis; 2 we will use this technique to address twoproblems from soil science, na

While considerable mathematical sophistication may be found in the soils ature, there has not been much use made of the qualitative theory of differentialequations It is our intention in this chapter to do four things: (1) we will introducethe technique of phase plane analysis; (2) we will use this technique to address twoproblems from soil science, namely, the time it takes to reach equilibrium in asorption problem and determining whether or not sorption is the only chemicalprocess occurring in our experiments; (3) we will use a more qualitative approach

liter-to differential equations liter-to develop a new model for multilayer sorption; and (4) wewill consider hysteresis in desorption and develop and analyze an elementary modelusing phase plane analysis

An elementary introduction to the mathematical techniques used here may befound in Blanchard et al (1998) and Borrelli and Coleman (1998) A more sophis-ticated discussion may be found in Coddington and Levinson (1955) and Hirsch andSmale (1974) Other papers and abstracts where this sort of analysis is used in soilwork are Kingery et al (1998) and Oppenheimer et al (in press) We use variousnumerical techniques in generating approximate solutions to systems of differentialequations throughout this chapter; Burden and Faires (1997) will contain the neededbackground Finally, we wrote our computer codes in the Matlab programminglanguage (Math Works, Inc 1992)

L1531Ch05Frame Page 109 Monday, May 7, 2001 2:33 PM

110 HEAVY METALS RELEASE IN SOILS

SINGLE DIFFERENTIAL EQUATIONS: PHASE LINE ANALYSIS

We want to use derivatives to describe physical phenomena, so we begin with

a simple example that leads to a single ordinary differential equation

Let us consider a chemical compound that is disappearing spontaneously from

an aqueous solution with probability P > 0 in any given second We assume we have

a whole macroscopic sample with mass M grams The mass of the sample will beproportional to the number of particles in the sample Let us also assume we canmeasure the mass at time t seconds to get the mass measurement of M(t). We expectthat in a given second, the rate at which the mass is decaying is given by pM(t) withunits of gs–1 where p = P× grams per particle of the substance Now we can alsowrite the rate of change in the mass as dM/dt Thus we end up with a differentialequation:

(5.1)

where the minus sign indicates disappearance

Before attempting a solution to Equation 5.1 we can get information on theprocess by looking at a phase line for this equation The phase line is simply anumber line provided with arrows to indicate in what direction a point will move(Figure 5.1)

The point M = 0 is called a stationary or equilibrium point If M = 0, there will be

no change over time Notice, that M = 0 is the value for M which makes dM/dt = 0.For M > 0, the arrow points down, because when we have a positive mass of material,the mass will decrease over time The up arrow for M < 0 is a mathematical artifactbecause we cannot have negative mass Mathematically, it states if there were such

a thing as negative mass, it would tend to become less negative over time Bymathematical artifact here, we mean a fact about the mathematical model that doesnot relate to the physical system we are trying to model

Notice that all of the arrows point toward M = 0; this means that 0 is an attractingequilibrium point or sink

We can approach this problem analytically and obtain

(5.2)

where M0 = M(0) Notice that the analytic solution does exactly what the phase linediagram says it should; go to M = 0 If we could not find an analytic solution, wecould do a numerical approximation, although we might lose our understanding ofthe long-term behavior of the system

In another numerical example, let us assume we are adding mass at the rate of

a gs –1 Our rate of change is now the loss of pM(t) plus a gs –1 This gives us adifferential equation of:

PHASE PLANE ANALYSIS AND DYNAMICAL SYSTEM APPROACHES 111

What are our equilibrium points? We obtain these by solving the problem dM/dt

= 0, or –pM + a = 0, or M = a/p When, M > a/p, we will get dM/dt < 0, and when

M < a/p, we will get dM/dt > 0 as is reflected in the phase line (Figure 5.2).Here we have no physical problems with M < a/p and we see that no matterwhat mass of material we start with, in the long run we tend toward an equilibrium

of M = a/p We note that we can draw the phase line in equilibrium mass of thismanner because the “right-hand side” of Equation 5.3 does not depend explicitly on

t, i.e., the equation is autonomous

It happens that we can also find an analytic solution to the above problem using

(5.4)but we have a great deal of information using only the phase line

Figure 5.1 Phase line diagram for Equation 5.1.

Figure 5.2 Phase line diagram for Equation 5.3.

pt o pt

L1531Ch05Frame Page 111 Friday, May 11, 2001 9:03 AM

112 HEAVY METALS RELEASE IN SOILS

It is worthwhile to consider how an experiment’s data would normally be sented For this purpose, we will take a = 2 and p = 1 In Figure 5.3, the graph is aplot of mass vs time with a starting mass of M o= 3.5

pre-In Figure 5.4, the graph is a plot of mass vs time with a starting mass of M0 =0.5 for Equation 5.4 Looking at the two figures, we can see what is going on interms of an approach to an equilibrium, but not as easily as when we use the phaseline

SYSTEMS OF TWO DIFFERENTIAL EQUATIONS AND PHASE PLANES

As in the case of modeling with a single differential equation, we shall discussthe techniques we wish to introduce in the context of a concrete example In fact,our examples will reflect actual situations we have encountered in our research

Figure 5.3 A plot of mass vs time with a starting mass of M0 = 3.5 for Equation 5.4.

Figure 5.4 A plot of mass vs time with a starting mass of M0 = 0.5 for Equation 5.4 L1531Ch05Frame Page 112 Monday, May 7, 2001 2:33 PM

PHASE PLANE ANALYSIS AND DYNAMICAL SYSTEM APPROACHES 113

We consider an agitated vessel containing soil with a of total mass Mg and a

solution of total volume VmL We assume that there is a heavy metal that is dissolved

in the solution that can be sorbed to the soil The concentration of the metal in the

solution at time t will be given by c(t) µgmL –1 and the concentration of the metal

sorbed to the soil will be given by q(t)µg g–1

Recall that there exists an equilibrium relationship between the solution

concen-tration and the sorbed concenconcen-tration That is, for any given solution concenconcen-tration

c0 there is a sorbed concentration q0 such that if the solution concentration is c0 and

the sorbed concentration is q0 the concentrations will not change over time This

defines a functional relationship where one inputs the solution concentration and

the output is the equilibrium sorbed concentration The function, defined at a fixed

temperature, is called the sorption isotherm and will be denoted by f Some typical

examples are the Henry isotherm:

(5.5)the Langmuir isotherm:

for our examples, where the constants have been chosen for convenience This is

plotted in Figure 5.5 (We note that we are ignoring the possibility of hysteresis

effects that would allow for multiple equilibrium sorbed concentrations and possible

differing sorption and desorption behaviors We will discuss this in the section titled

Desorption–Sorption Modeling.)

It is important to realize that understanding the equilibrium behavior is not

sufficient for problems involving transport, such as site remediation and the study

of agricultural wastes Therefore, we must give a model for how c and q change in

time We will assume that the system will tend toward a condition of equilibrium

and the rate of change will be proportional to the distance from equilibrium That

is to say, the rate at which the sorbed concentration is changing at time t will be

proportional to f(c(t)) – q(t) Notice that if the current sorbed concentration is smaller

q0= γc0

c

0 0 0

1

=+ β

L1531Ch05Frame Page 113 Monday, May 7, 2001 2:33 PM

114 HEAVY METALS RELEASE IN SOILS

than the equilibrium sorbed concentration for the current solution concentration,

f(c(t)), the rate of change will be positive On the other hand, if the current sorbed

concentration is larger than the equilibrium sorbed concentration for the current

solution concentration, the rate of change will be negative This leads to the equation

(5.8)

where the rate constant rq is the constant of proportionality with units of s –1 Similarly,

we obtain an equation for the rate of change in the solution concentration

(5.9)

where r c has units of g mL–1 s–1 This yields a two-by-two system of ordinary

differential equations

(5.10)

Again we note that the “right-hand sides” of the equations have no explicit

dependence on t That is, t does not appear except as an argument of c and q, and

the equations are autonomous

We will now discuss some ways of analyzing this system without solving it or

considering experimental data We will then apply this analysis to two data sets Let

us again consider the graph of the isotherm, but now we will view it as a phase

plane which we can use to understand what the dynamical behavior of the system

PHASE PLANE ANALYSIS AND DYNAMICAL SYSTEM APPROACHES 115

As we see in Figures 5.6 and 5.7, the isotherm divides the first quadrant of theplane into two distinct regions, I and II Each point on the plane represents a possiblestate of the system; the horizontal coordinate giving the solution concentration and

the vertical coordinate giving the sorbed concentration In region I we have f(c) >

q and thus from Equation 5.10 we have

(5.11)

In a similar fashion, for region II we have f (c) < q and thus

(5.12)

This means that if, at a given time t the state of our system places it in region

I, the solution concentration will be decreasing and the sorbed concentration will

Figure 5.6 Langmuir isotherm as a phase plane with two distinct states, I and II.

Figure 5.7 Langmuir isotherm with the motion of the state vector and the derivatives of rates

of solutes in the solution and solid states.

dc dt

dq dt

<0 and >0

dc dt

dq dt

>0 and <0

116 HEAVY METALS RELEASE IN SOILS

be increasing and the point that represents the state of the system will be moving

up and to the left, toward an equilibrium point on the isotherm Similarly, if, at a

given time t, the state of our system places it in region II, the solution concentration

will be increasing and the sorbed concentration will be decreasing and the point thatrepresents the state of the system will be moving down and to the right toward anequilibrium point on the isotherm This analysis tells us what behavior we can expect

from our experimental data if our model is correct and the experiment has been

done correctly

Using conservation of mass, we can get still more information about how our

experimental data can be expected to behave if our model is correct At t = 0 we will know the initial concentrations c(0) and q(0) Since no mass is removed from

the system and since mass is neither created nor destroyed, the total mass at any

given t should remain constant That is, the total initial mass of contaminant equal

to the mass in solution plus mass sorbed to the soil = c(0)V + q(0)M must be the same at any given t, or

The same sort of conservation of mass analysis forces a relationship between r q

and r c The rate at which the mass of metal disappears from the solution shouldequal the rate at which mass is sorbed onto the soil That is,

(5.15)

This implies that

(5.16)or

rate at which mass is removed from the solution

= rate at which mass sorbs onto soil = dq

dt

−r q t c( ( )− f c t( ) ( ) )V=r f c t q( ( ) ( ) −q t M( ) )

r r

M V c q

=

PHASE PLANE ANALYSIS AND DYNAMICAL SYSTEM APPROACHES 117

(It is a useful exercise for the reader to make sure that the units work outcorrectly.) This can be used as a check when we are seeking to find the rate constantsfrom experimental data

We identified the parameters α, β, and γ to obtain α = 1.057, β = 4.298, and γ

= 1.396 minimizing a sum of square error objective function, constructed usingexperimental data The fitted curve and the experimental values are compared inFigure 5.8

How long should an experiment measuring the dynamic process of sorption last?Recall that we are assuming the differential equation model (Equation 5.10) withthe Langmuir-Freundlich isotherm given above After two hours, we recorded ourdata points on the phase plane as seen in Figure 5.9 Clearly, two hours are notenough to reach equilibrium Merely plotting our data in phase space showed us

Figure 5.8 An example of Langmuir-Freundlich isotherm of a protein sorption on

calcium-saturated Wyoming smectite.

1

L1531Ch05Frame Page 117 Friday, May 11, 2001 9:04 AM

118 HEAVY METALS RELEASE IN SOILS

that we need to run our kinetic experiments longer Our results for 2880 minutesare shown in Figure 5.10

We can also plot the predicted path in phase space predicted by Equation 5.10

with r c = 0032 and r q = 0011 as shown in Figure 5.11 In this case, M/V = 2.9 and

r c /r q = 2.91

Figure 5.9 Langmuir-Freundlich isotherm and experimental kinetics during the first 120 min

of sorption of a protein on calcium-saturated Wyoming smectite.

Figure 5.10 Langmuir-Freundlich isotherm and experimental kinetics during 2880 min of

sorp-tion of a protein on calcium-saturated Wyoming smectite.

PHASE PLANE ANALYSIS AND DYNAMICAL SYSTEM APPROACHES 119

Our next example uses the same identified isotherm We ran some experimentswith somewhat higher initial solution concentrations Our results, plotted on thephase plane, are given below We note that the rate constants identified using thisdata set should be the same as those we identified earlier; however, they are not,

and we have r c = 0054 and r q = 0016 In this case, M/V = 2.9 and r c /r q = 3.38 Thisleads to a loss of conservation of mass in the theoretical results

In Figure 5.12 we see a problem Notice that the data points start at the bottom

of the quadrant and move through and above the isotherm Now the isothermrepresents a continuous set of equilibrium points to the set of differential equations,(5.10), we are using to model the experiment This means that our data passes through

an equilibrium point Mathematically, this is impossible Therefore, either there is

a problem with our data or there is a problem with our model It is possible that theexperiment to determine the equilibrium concentration was not run long enough toreach equilibrium, which would result in an incorrectly identified isotherm However,this isotherm worked well at the lower concentration Among the possible difficultiesfor our model are: (1) perhaps the isotherm is not accurate for high concentrations

of contaminant; (2) perhaps there are other mechanisms coming into play that onlyoccur at high concentrations such as precipitation (Oppenheimer et al., in press); or,multisite sorption In any case, the phase plane analysis of the data has shown usthat there is a problem with our model or our experiment

A NEW MULTILAYER SORPTION MODEL

The following new model for multilayer sorption is worked out in detail for atwo-layer model However, the same reasoning can be used for any fixed number

Figure 5.11 Langmuir-Freundlich isotherm, theoretical and experimental kinetics during

2880 min of sorption of a protein on calcium-saturated wyoming smectite.