Multimedia Environmental Models - Chapter 6 pot

©2001 CRC Press LLCCHAPTER 6 Advection and Reactions In Level I calculations, it is assumed that the chemical is conserved; i.e., it isneither destroyed by reactions nor conveyed out of

McKay, Donald "Advection and Reactions"

Multimedia Environmental Models

Edited by Donald McKay

Boca Raton: CRC Press LLC,2001

©2001 CRC Press LLC

CHAPTER 6 Advection and Reactions

In Level I calculations, it is assumed that the chemical is conserved; i.e., it isneither destroyed by reactions nor conveyed out of the evaluative environment byflows in phases such as air and water These assumptions can be quite misleadingwhen determining of the impact of a given discharge or emission of chemical.First, if a chemical, such as glucose, is reactive and survives for only 10 hours

as a result of its susceptibility to rapid biodegradation, it must pose less of a threatthan PCBs, which may survive for over 10 years But the Level I calculation treatsthem identically Second, some chemical may leave the area of discharge rapidly as

a result of evaporation into air, to be removed by advection in winds The ination problem is solved locally, but only by shifting it to another location It isimportant to know if this will occur Indeed, recently, considerable attention is beingpaid to substances that are susceptible to long-range transport Third, it is possiblethat, in a given region, local contamination is largely a result of inflow of chemicalfrom upwind or upstream regions Local efforts to reduce contamination by control-ling local sources may therefore be frustrated, because most of the chemical isinadvertently imported This problem is at the heart of the Canada–U.S., and Scan-dinavia–Germany–U.K squabbles over acid precipitation It is also a concern inrelatively pristine areas such as the Arctic and Antarctic, where residents have little

contam-or no control over the contamination of their environments

In this chapter, we address these issues and devise methods of calculating theeffect of advective inflow and outflow and degrading reactions on local chemicalfate and subsequent exposures It must be emphasized that, once a chemical isdegraded, this does not necessarily solve the problem Toxicologists rarely miss anopportunity to point out reactions, such as mercury methylation or benzo(a)pyreneoxidation, in which the product of the reaction is more harmful than the parentcompound For our immediate purposes, we will be content to treat only the parentcompound Assessment of degradation products is best done separately by havingthe degradation product of one chemical treated as formation of another

A key concept in this discussion that was introduced earlier, and is variouslytermed persistence, lifetime, residence time, or detention time of the chemical.

In a steady-state system, as shown in Figure 6.1a, if chemical is introduced at arate of E mol/h, then the rate of removal must also be E mol/h Otherwise, netaccumulation or depletion will occur If the amount in the system is M mol, then,

on average, the amount of time each molecule spends in the steady-state system will

be M/E hours This time, t, is a residence time and is also called a detention time

or persistence Clearly, if a chemical persists longer, there will be more of it in thesystem The key equation is

t = M/E or M = t E

This concept is routinely applied to retention time in lakes If a lake has a volume

of 100,000 m3, and if it receives an inflow of 1000 m3 per day, then the retentiontime is 100,000/1000 or 100 days A mean retention time of 100 days does not imply

Figure 6.1 Diagram of a steady-state evaluative environment subject to (a) advective flow, (b)

degrading reactions, (c) both, and (d) the time course to steady state.

that all water will spend 100 days in the lake Some will bypass in only 10 days,and some will persist in backwaters for 1000 days but, on average, the residencetime will be 100 days.

The reason that this concept is so important is that chemicals exhibit variablelifetimes, ranging from hours to decades As a result, the amount of chemical present

in the environment, i.e., the inventory of chemical, varies greatly between chemicals

We tend to be most concerned about persistent and toxic chemicals, because tively small emission rates (E) can result in large amounts (M) present in theenvironment This translates into high concentrations and possibly severe adverseeffects A further consideration is that chemicals that survive for prolonged periods

rela-in the environment have the opportunity to make long and often tortuous journeys

If applied to soil, they may evaporate, migrate onto atmospheric particles, deposit

on vegetation, be eaten by cows, be transferred to milk, and then consumed byhumans Chemicals may migrate up the food chain from water to plankton to fish

to eagles, seals, and bears Short-lived chemicals rarely survive long enough toundertake such adventures (or misadventures)

This lengthy justification leads to the conclusion that, if we are going to discharge

a chemical into the environment, it is prudent to know

1 how long the chemical will survive, i.e., t, and

2 what causes its removal or “death”

This latter knowledge is useful, because it is likely that situations will occur inwhich a common removal mechanism does not apply For example, a chemical may

be potentially subject to rapid photolysis, but this is not of much relevance in long,dark arctic winters or in deep, murky sediments

In the process of quantifying this effect, we will introduce rate constants, tive flow rates and, ultimately, using the fugacity concept, quantities called D values,which prove to be immensely convenient Indeed, armed with Z values and D values,the environmental scientist has a powerful set of tools for calculation and interpre-tation

advec-It transpires that there are two primary mechanisms by which a chemical isremoved from our environment: advection and reaction, which we discuss individ-ually and then in combination

Thus, if there is river flow of 1000 m3/h (G) from A to B of water containing 0.3mol/m3 (C) of chemical, then the corresponding flow of chemical is 300 mol/h (N).Turning to the evaluative environment, it is apparent that the primary candidateadvective phases are air and water If, for example, there was air flow into the 1square kilometre evaluative environment at 109 m3/h, and the volume of the air inthe evaluative environment is 6 ¥ 109 m3, then the residence time will be 6 hours,

or 0.25 days Likewise, the flow of 100 m3/h of water into 70,000 m3 of water results

in a residence time of 700 hours, or 29 days It is easier to remember residencetimes than flow rates; therefore, we usually set a residence time and from it deducethe corresponding flow rate

Burial of bottom sediments can also be regarded as an advective loss, as canleaching of water from soils to groundwater Advection of freons from the tropo-sphere to the stratosphere is also of concern in that it contributes to ozone depletion

6.2.1 Level II Advection Algebra Using Partition Coefficients

If we decree that our evaluative environment is at steady state, then air and waterinflows must equal outflows; therefore, these inflow rates, designated G m3/h, mustalso be outflow rates If the concentrations of chemical in the phase of the evaluativeenvironment is C mol/m3, then the outflow rate will be G C mol/h This concept isoften termed the continuously stirred tank reactor, or CSTR, assumption The basicconcept is that, if a volume of phase, for example air, is well stirred, then, if some

of that phase is removed, that air must have a concentration equal to that of thephase as a whole If chemical is introduced to the phase at a different concentration,

it experiences an immediate change in concentration to that of the well mixed, orCSTR, value The concentration experienced by the chemical then remains constantuntil the chemical is removed The key point is that the outflow concentration equalsthe prevailing concentration This concept greatly simplifies the algebra of steady-state systems Essentially, we treat air, water, and other phases as being well mixedCSTRs in which the outflow concentration equals the prevailing concentration Wecan now consider an evaluative environment in which there is inflow and outflow

of chemical in air and water It is convenient at this stage to ignore the particles inthe water, fish, and aerosols, and assume that the material flowing into the evaluativeenvironment is pure air and pure water Since the steady-state condition applies, asshown in Figure 6.1a, the inflow and outflow rates are equal, and a mass balancecan be assembled The total influx of chemical is at a rate GACBA in air, and GWCBW

in water, these concentrations being the “background” values There may also beemissions into the evaluative environment at a rate E The total influx I is thus

I = E + GACBA + GWCBW mol/hNow, the concentrations within the environment adjust instantly to values CAand CW in air and water Thus, the outflow rates must be GACA and GWCW Theseoutflow concentrations could be constrained by equilibrium considerations; forexample, they may be related through partition coefficients or through Z values to

a common fugacity

This enables us to conceive of, and define, our first Level II calculation in which

we assume equilibrium and steady state to apply, inputs by emission and advectionare balanced exactly by advective emissions, and equilibrium exists throughout theevaluative environment All the phases are behaving like individual CSTRs

Of course, starting with a clean environment and introducing these inflows, itwould take the system some time to reach steady-state conditions, as shown in Figure6.1d At this stage, we are not concerned with how long it takes to reach a steadystate, but only the conditions that ultimately apply at steady state We can thereforedevelop the following equations, using partition coefficients and later fugacities

I = E + GACBA + GWCBW = GACA + GWCWBut

CA = KAWCWTherefore,

I = CW[GAKAW + GW] and CW = I/[GAKAW + GW]Other concentrations, amounts (m), and the total amount (M) can be deducedfrom CW The extension to multiple compartment systems is obvious For example,

if soil is included, the concentration in soil will be in equilibrium with both CA and

CW

6.2.2 Level II Advection Algebra Using Fugacity

We assume a constant fugacity f to apply within the environment and to theoutflowing media, thus,

0.1, and soil 1.0 There is also an emission of 4 mol/h Calculate the fugacity

concentrations, persistence amounts and outflow rates

In this example, the total amount of material in the system, M, is 450 mol The

inflow rate is 15 mol/h, thus the residence time or the persistence of the chemical

is 30 hours This proves to be a very useful time Note that the air residence time

is 10 hours, and the water residence time is 100 hours; thus, the overall residence

time of the chemical is a weighted average, influenced by the extent to which the

chemical partitions into the various phases The soil has no effect on the fugacity

or the outflow rates, but it acts as a “reservoir” to influence the total amount present

M and therefore the residence time or persistence

6.2.3 D values

The group G Z, and other groups like it, appear so frequently in later calculations

that it is convenient to designate them as D values,

i.e.,

G Z = D mol/Pa h The rate, N mol/h, then equals D f These D values are transport parameters, with

units of mol/Pa h When multiplied by a fugacity, they give rates of transport They

are thus similar in principle to rate constants, which, when multiplied by a mass of

chemical, give a rate of reaction Fast processes have large D values We can write

the fugacity equation for the evaluative environment in more compact form, as shown

below:

f = I/(DAA + DAW) = I/SDAiwhere DAA = GAZA, DAW = GWZW, and the first subscript A refers to advection

Recalculating Example 6.1,

DAA = 0.4 and DAW = 0.1 and SDAi = 0.5Therefore,

f = 15/0.5 = 30 and the rates of output, Df, are 12 and 3 mol/h, totaling 15 mol/h as before.

It is apparent that the air D value is larger and most significant D values can be

added when they are multiplied by a common fugacity Therefore, it becomes

obvious which D value, and hence which process, is most important We can arrive

at the same conclusion using partition coefficients, but the algebra is less elegant

Note that how the chemical enters the environment is unimportant, all sources

being combined or lumped in I, the overall input This is because, once in the

environment, the chemical immediately achieves an equilibrium distribution, and it

“forgets” its origin

6.2.4 Advective Processes

In an evaluative environment, there are several advective flows that convey

chemical to and from the environment, namely,

1 inflow and outflow of air

2 inflow and outflow of water

3 inflow and outflow of aerosol particles present in air

4 inflow and outflow of particles and biota present in water

5 transport of air from the troposphere to the stratosphere, i.e., vertical movement

of air out of the environment

6 sediment burial, i.e., sediment being conveyed out of the well mixed layer to depths

sufficient that it is essentially inaccessible

7 flow of water from surface soils to groundwater (recharge)

It also transpires that there are several advective processes which can apply to

chemical movement within the evaluative environment Notable are rainfall, water

runoff from soil, sedimentation, and food consumption, but we delay their treatment

until later

In situations 1 through 4, there is no difficulty in deducing the rate as GC or Df,

where G is the flowrate of the phase in question, C is the concentration of chemical

in that phase, and the Z value applies to the chemical in the phase in which it is

dissolved or sorbed

For example, aerosol may be transported to an evaluative world in association

with the inflow of 1012 m3/h of air If the aerosol concentration is 10–11 volume

fraction, then the flowrate of aerosol GQ is 10 m3/h The relevant concentration of

chemical is that in the aerosol, not in the air, and is normally quite high, for example,

100 mol/m3 Therefore, the rate of chemical input in the aerosol is 1000 mol/h This

can be calculated using the D and f route as follows, giving the same result

If ZQ = 108, then

f = CQ/ZQ = 100/108 = 10–6 Pa

DAQ = GQZQ = 10 ¥ 108 = 109

N = Df = 109¥ 10–6 = 1000 mol/hTreatment of transport to the stratosphere is somewhat more difficult We canconceive of parcels of air that migrate from the troposphere to the stratosphere at

an average, continuous rate, G m3/h, being replaced by clean stratospheric air thatmigrates downward at the same rate We can thus calculate the D value As discussed

by Neely and Mackay (1982), this rate should correspond to a residence time of thetroposphere of about 60 years, i.e., G is V/t Thus, if V is 6 ¥ 109 and t is 5.25 ¥

105 h, G is 11400 m3/h This rate is very slow and is usually insignificant, but thereare situations in which it is important

We may be interested in calculating the amount of chemical that actually reachesthe stratosphere, for example, freons that catalyze the decomposition of ozone Thisslow rate is thus important from the viewpoint of the receiving stratospheric phase,but is not an important loss from the delivering, or tropospheric, phase Second, if

a chemical is very stable and is only slowly removed from the atmosphere by reaction

or deposition processes, then transfer to the troposphere may be a significant anism of removal Certain volatile halogenated hydrocarbons tend to be in this class

mech-If we emit a chemical into the evaluative world at a steady rate by emissions andallow for no removal mechanisms whatsoever, its concentrations will continue tobuild up indefinitely Such situations are likely to arise if we view the evaluativeworld as merely a scaled-down version of the entire global environment There iscertainly advective flow of chemical from, for example, the United States to Canada,but there is no advective flow of chemical out of the entire global atmosphericenvironment, except for the small amounts that transfer to the stratosphere Whetheradvection is included depends upon the system being simulated In general, thesmaller the system, the shorter the advection residence time, and the more importantadvection becomes

Sediment burial is the process by which chemical is conveyed from the activemixed layer of accessible sediment into inaccessible buried layers As was discussedearlier, this is a rather naive picture of a complex process, but at least it is a startingpoint for calculations The reality is that the mixed surface sediment layer is rising,eventually filling the lake Typical burial rates are 1 mm/year, the material beingburied being typically 25% solids, 75% water But as it “moves” to greater depths,water becomes squeezed out Mathematically, the D value consists of two terms,the burial rate of solids and that of water

For example, if a lake has an area of 107 m2 and has a burial rate of 1 mm/year,the total rate of burial is 10,000 m3/year or 1.14 m3/h, consisting of perhaps 25%solids, i.e., 0.29 m3/h of solids (GS) and 0.85 m3/h of water (GW) The rate of loss

of chemical is then

GSCS + GWCW = GSZSf + GWZWf = f(DAS + DAW)Usually, there is a large solid to pore water partition coefficient; therefore, CSgreatly exceeds CW or, alternatively, ZS is very much greater than ZW, and the term

DAS dominates A residence time of solids in the mixed layer can be calculated asthe volume of solids in the mixed layer divided by GS For example, if the depth ofthe mixed layer is 3 cm, and the solids concentration is 25%, then the volume ofsolids is 75,000 m3 and the residence time is 260,000 hours, or 30 years Theresidence time of water is probably longer, because the water content is likely to behigher in the active sediment than in the buried sediment In reality, the water wouldexchange diffusively with the overlaying water during that time period.

As discussed in Chapter 5, there are occasions in which it is convenient tocalculate a “bulk” Z value for a medium containing a dispersed phase such as anaerosol This can be used to calculate a “bulk” Z value, thus expressing two lossprocesses as one D is then GZ where G is the total flow and Z is the bulk value

consider reaction to have occurred In the literature, the word reaction is occasionally,

and wrongly, applied to these processes, especially to sorption

We have two tasks The first is to assemble the necessary mathematical work for treating reaction rates using rate constants, and the second is to devisemethods of obtaining information on values of these rate constants

frame-6.3.1 Reaction Rate Expressions

We prefer, when possible, to use a simple first-order kinetic expression for allreactions The basic rate equation is

rate N = VCk = Mk mol/hwhere V is the volume of the phase (m3), C is the concentration of the chemical(mol/m3), M is the amount of chemical, and k is the first-order rate constant withunits of reciprocal time The group VCk thus has units of mol/h

The classical application of this equation is to radioactive decay, which is usuallyexpressed in the forms

dM/dt = –kM or dC/dt = –CkThe use of C instead of M implies that V does not change with time

Integrating from an initial condition of CO at zero time gives the followingequations:

ln(C/CO) = –kt or C = CO exp(–kt)Rate constants have units of frequency or reciprocal time and are therefore noteasily grasped or remembered A favorite trick question of examiners is to ask astudent to convert a rate constant of 24 h–1 into reciprocal days The correct answer

is 576 days–1, so beware of this conversion! It is more convenient to store andremember half-lives, i.e., the time, t1/2, which is the time required for C to decrease

to half of CO This can be related to the rate constant as follows

When C = 0.5 CO, then t = t1/2

ln (0.5) = –kt1/2, therefore, t1/2 = 0.693/kFor example, an isotope with a half-life of 10 hours has a rate constant, k, of0.0693 h–1

6.3.2 Non-First-Order Kinetics

Unfortunately, there are many situations in which the real reaction rate is not afirst-order reaction Second-order rate reactions occur when the reaction rate isdependent on the concentration of two chemicals or reactants For example, if

A + B Æ D + Ethen the rate of the reaction is dependent on the concentration of both A and B.Therefore, the reaction rate is as follows:

N = Vk CA CBReactant “B” is often another chemical, but it could be another environmentalreactant such as a microbial population or solar radiation intensity Third-orderreaction rates, when the rate of reaction is dependent on the concentration of threereactants (N = Vk CA CB CC), are very rare and are unlikely to occur under environ-mental conditions

We can often circumvent these complex reaction rate equations by expressingthem in terms of a pseudo first-order rate reaction The primary assumption is thatthe concentration of reactant “B” is effectively constant and will not change appre-ciably as the reaction proceeds Thus, the constant k and concentration of reactant

“B” can be lumped into a new rate constant, kP, and the second-order reactionbecomes a pseudo first-order reaction Therefore,

N = Vk CACB

kP = k CBTherefore,

N = VkP CAwhich has the form of a simple first-order reaction Examples of pseudo first-orderreactions include photolysis reactions where reactant “B” is the solar radiationintensity (I, in photons/s) or microbial degradations processes where “B” is thepopulations of microorganisms Reactions between two chemicals can also be con-sidered a pseudo first-order reaction when CA << CB, so the concentration of B doesnot change as the reaction proceeds

Second-order rate expressions also arise when a chemical reacts with itself,giving rise to a messy quadratic equation Thus, if A + A Æ D, the rate equation is

Our strategy is to use every reasonable excuse to force first-order kinetics onsystems by lumping parameters in k The dividends that arise are worth the effort,because subsequent calculations are much easier

Perhaps most worrisome are situations in which we treat the kinetics of microbialdegradation of chemicals It is possible that, at very low concentrations, there is aslower or even no reaction, because the required enzyme systems are not “turnedon.” At very high concentrations, the enzyme may be saturated; therefore, the rate

of degradation ceases to be controlled by the availability of the chemical and becomescontrolled by the availability of enzyme In other cases, the rate of conversion may

be influenced by the toxicity of the chemical to the organism or by the presence of

co-metabolites, chemicals that the enzyme recognizes as being similar to that of the

chemical of interest Microbiologists have no difficulty conceiving of a multitude

of situations in which chemical kinetics become very complicated and very difficult

to predict and express They seem to obtain a certain perverse delight in findingthese situations

Saturation kinetics is usually treated by the Michaelis–Menten equation, whichcan be derived from first principles or, more simply, by writing down the basic first-order equation and multiplying the rate expression by the group shown below

Basic expression N = VCk

Combined expression N = VCCMk/(C + CM)

When C is small compared to CM, the rate reduces to VCk When C is large compared

to CM, it reduces to VCMk, which is independent of C, is constant, and corresponds

to the maximum, or zero-order rate The concentration, CM, therefore corresponds

to the concentration that gives the maximum rate using the basic expression When

C equals CM, the rate is half the maximum value This can be (and usually is)expressed in terms of other rate constants for describing the kinetics of the associ-ation of the chemical with the enzyme

The rate expression is usually written in biochemistry texts in the form

N/V = C vM/(C + kM)

where vM is a maximum rate or velocity equivalent to kCM, and kM is equivalent to

CM and is viewed as a ratio of rate constants A somewhat similar expression, theMonod equation, is used to describe cell growth

If kinetics are not of the first order, it may be necessary to write the appropriateequations and accept the increased difficulty of solution A somewhat cunning butunethical alternative is to guess the concentration, calculate the rate N using thenon-first-order expression, then calculate the pseudo first-order rate constant in theexpression For example, if a reaction is second order and C is expected to be about

2 mol/m3, V is 100 m3, and the second-order rate constant, k2, is 0.01 m3/mol·h,then N equals 4 mol/h We can set this equal to VCk; then, k is 0.02 h–1 Essentially,

we have lumped Ck2 as a first-order rate constant This approach must be used, ofcourse, with extreme caution, because k depends on C

6.3.3 Additivity of Rate Constants

A major advantage of forcing first-order kinetics on all reactions is that, if achemical is susceptible to several reactions in the same phase, with rate constants

kA, kB, kC, etc., then the total rate constant for reaction is (kA + kB + kC), i.e., therate constants are simply added Another favorite trick of perverse examiners is toinform a student that a chemical reacts by one mechanism with a half-life of 10hours, and by another mechanism with a half-life of 20 hours, and asks for the totalhalf-life The correct answer is 6.7 hours, not 30 hours Half-lives are summed asreciprocals, not directly

6.3.4 Level II Reaction Algebra Using Partition Coefficients

We can now perform certain calculations describing the behavior of chemicals

in evaluative environments The simplest is a Level II equilibrium steady-state

reaction situation in which there is no advection, and there is a constant inflow ofchemical in the form of an emission, as depicted in Figure 6.1b When a steady state

is reached, there must be an equivalent loss in the form of reactions Starting from

a clean environment, the concentrations would build up until they reach a level suchthat the rates of degradation or loss equal the total rate of input We further assumethat the phases are in equilibrium, i.e., transfer between them is very rapid As aresult, the concentrations are related through partition coefficients, or a commonfugacity applies The equations are as follows:

E = V1C1k1 + V2C2k2 etc = SViCiki

Using partition coefficients,

E = SViCwKiwki = CwSViKiwki

from which Cw can be deduced, followed by other concentrations, amounts, rates

of reaction, and the persistence In the general expression, KWW, the water-waterpartition coefficient is unity

Worked Example 6.2

The evaluative environment in Example 6.1 is subject to emission of 10 mol/h

of chemical, but no advection The reaction half-lives are air, 69.3 hours; water, 6.93hours; and soil, 693 hours Calculate the concentrations Recall that KAW = 0.004and KSW = 10.

The rate constants are 0.693/half-lives or air, 0.01; water, 0.1; soil, 0.001; h–1

E = VACAkA + VWCWkW + VSCSkS

= CW(VAKAWkA + VWkW + VSKSWkS)

= CW(0.4 + 10 + 0.01) = CW(10.41) = 10Therefore,

CW = 0.9606 mol/m3, CA = 0.0038, CS = 9.606The rates of reaction then are