Multimedia Environmental Models - Chapter 2 docx

Concentration The preferred unit is the mole per cubic metre mol/m3 or the gram per cubicmetre g/m3.. It may be convenient to lump all biota togetheras one phase or consider them as two

McKay, Donald "Some Basic Concepts"

Multimedia Environmental Models

Edited by Donald McKay

Boca Raton: CRC Press LLC, 2001

be unchanging with time The art of successful environmental modeling lies in theselection of the best, or “least-worst,” set of assumptions that yields a model that isnot so complex as to be excessively difficult to understand yet is sufficiently detailed

to be useful and faithful to reality The excessively simple model may be misleading.The excessively detailed model is unlikely to be useful, trusted, or even understand-able The aim is to suppress the less necessary detail in favor of the importantprocesses that control chemical fate

In this chapter, several concepts are introduced that are used when we seek tocompile quantitative descriptions of chemical behavior in the environment But first,

it is essential to define the system of units and dimensions that forms the foundation

of all calculations

2.2 UNITS

The introduction of the “SI” or “Système International d’Unités” or InternationalSystem of Units in 1960 has greatly simplified scientific calculations and communi-cation With few exceptions, we adopt the SI system The system is particularlyconvenient, because it is “coherent” in that the basic units combine one-to-one to

give the derived units directly with no conversion factors For example, energy(joules) is variously the product of force (newtons) and distance (metres), or pressure(pascals) and volume (cubic metres), or power (watts) and time (seconds) Thus, thefoot-pound, the litre-atmosphere, and the kilowatt-hour become obsolete in favor ofthe single joule Some key aspects of the SI system are discussed below Conversiontables from obsolete or obsolescent unit systems are available in scientific handbooks.

Length (metre, m)This base unit is defined as the specified number of wavelengths of a kryptonlight emission

Area

Square metre (m2) Occasionally, the hectare (ha) (an area 100 ¥ 100 m or 104 m2)

or the square kilometre (km2) is used For example, pesticide dosages to soils areoften given in kg/ha

Volume (cubic metre, m3)The litre (L) (0.001 m3) is also used because of its convenience in analysis, but

it should be avoided in environmental calculations In the United States, the spellings

“meter” and “liter” are often used

Mass (kilogram, kg)Kilogram (kg) The base unit is the kilogram (kg), but it is often more convenient

to use the gram (g), especially for concentrations For large masses, the megagram(Mg) or the equivalent metric tonne (t) may be used

Amount (mole abbreviated to mol)This unit, which is of fundamental importance in environmental chemistry, isreally a number of constituent entities or particles such as atoms, ions, or molecules

It is the actual number of particles divided by Avogadro’s number (6.0 ¥ 1023),which is defined as the number of atoms in 12 g of the carbon-12 isotope Whenreactions occur, the amounts of substances reacting and forming are best expressed

in moles rather than mass, since atoms or molecules combine in simple metric ratios The need to involve atomic or molecular masses is thus avoided

stoichio-Molar Mass or Molecular Mass (or Weight) (g/mol)This is the mass of 1 mole of matter and is sometimes (wrongly) referred to asmolecular weight or molecular mass Strictly, the correct unit is kg/mol, but it isoften more convenient to use g/mol, which is obtained by adding the atomic masses(weights) Benzene (C6H6) is thus approximately 78 g/mol or 0.078 kg/mol

Time (second or hour, s or h)

The standard unit of a second (s) is inconveniently short when consideringenvironmental processes such as flows in large lakes when residence times may bemany years The use of hours, days, and years is thus acceptable We generally usehours as a compromise

Concentration

The preferred unit is the mole per cubic metre (mol/m3) or the gram per cubicmetre (g/m3) Most analytical data are reported in amount or mass per litre (L),because a litre is a convenient volume for the analytical chemist to handle andmeasure Complications arise if the litre is used in environmental calculations,because it is not coherent with area or length The common mg/L, which is oftenambiguously termed the “part per million,” is equivalent to g/m3 In some circum-stances, the use of mass fraction, volume fraction, or mole fraction as concentrations

is desirable

It is acceptable, and common, to report concentrations in units such as mol/L ormg/L but, prior to any calculation, they should be converted to a coherent unit ofamount of substance per cubic metre

Concentrations such as parts per thousand (ppt), parts per million (ppm), partsper billion (ppb), and parts per trillion (also ppt) should not be used There can beconfusion between parts per thousand and per trillion The billion is 109 in NorthAmerica and 1012 in Europe The air ppm is usually on a volume/volume basis,whereas the water ppm is usually on a mass/volume basis The mixing ratio usedfor air is the ratio of numbers of molecules or volumes and is often given in ppm.Concentrations must be presented with no possible ambiguity

Density (kg/m3)

This has identical units to mass concentrations, but the use of kg/m3 is preferred,water having a density of 1000 kg/m3 and air a density of approximately 1.2 kg/m3

Force (newton, N)

The newton is the force that causes a mass of 1 kg to accelerate at 1 m/s2 It is

105 dynes and is approximately the gravitational force operating on a mass of 102 g

at the Earth’s surface

Pressure (pascal, Pa)

The pascal or newton per square metre (N/m2) is inconveniently small, since itcorresponds to only 102 grams force over one square metre, but it is the standardunit, and it is used here The atmosphere (atm) is 101325 Pa or 101.325 kPa Thetorr or mm of mercury (mmHg) is 133 Pa and, although still widely used, should

be regarded as obsolescent

Energy (joule, J)

The joule, which is one N-m or Pa-m3, is also a small quantity It replaces theobsolete units of calorie (which is 4.184 J) and Btu (1055 J)

Temperature (K)

The kelvin is preferred, although environmental temperatures may be expressed

in degrees Celsius, °C, and not centigrade, where 0°C is 273.15 K There is nodegree symbol prior to K

corre-Gas Constant (R)

This constant, which derives from the gas law, is 8.314 J/mol K or Pa-m3/mol K

An advantage of the SI system is that R values in diverse units such as cal/mol Kand cm3·atm/mol K become obsolete and a single universal value now applies

2.2.1 Prefices

The following prefices are used:

Note that these prefices precede the unit It is inadvisable to include more than oneprefix in a unit, e.g., ng/mg, although mg/kg may be acceptable, because the baseunit of mass is the kg The equivalent µg/g is clearer The use of expressions such

as an aerial pesticide spray rate of 900 g/km2 can be ambiguous, since a kilo(metre2)

is not equal to a square kilometre, i.e., a (km)2 The former style is not permissible

Expressing the rate as 9 g/ha or 0.9 mg/m2 removes all ambiguity The prefices deka,hecto, deci, and centi are restricted to lengths, areas, and volumes A common (anddisastrous) mistake is to confuse milli, micro, and nano.

We use the convention J/mol-K meaning J mol–1 K–1 Strictly, J/(mol-K) is correctbut, in the interests of brevity, the parentheses are omitted

2.2.2 Dimensional Strategy and Consistency

When undertaking calculations of environmental fate, it is highly desirable toadopt the practice of first converting all the supplied input data, in its diversity ofunits, into the SI units described above and eliminate the prefices, e.g., 10 kPa shouldbecome 104 Pa Calculations should be done using only these SI units If necessary,the final results can then be converted to other units for the convenience of the user.When assembling quantities in expressions or equations, it is critically importantthat the dimensions be correct and consistent It is always advisable to write downthe units on each side of the equation, cancel where appropriate, and check thatterms that add or subtract have identical units For example, a lake may have aninflow or reaction rate of a chemical expressed as follows:

It is useful to view the environment as consisting of a number of connected

phases or compartments Examples are the atmosphere, terrestrial soil, a lake, thebottom sediment under the lake, suspended sediment in the lake, and biota in soil

or water The phase may be continuous (e.g., water) or consist of a number ofparticles that are not in contact, but all of which reside in one phase [e.g., atmospheric

particles (aerosols), or biota in water] In some cases, the phases may be similarchemically but different physically, e.g., the troposphere or lower atmosphere, andthe stratosphere or upper atmosphere It may be convenient to lump all biota together

as one phase or consider them as two or more classes each with a separate phase.Some compartments are in contact, thus a chemical may migrate between them (e.g.,air and water), while others are not in contact, thus direct transfer is impossible (e.g.,air and bottom sediment) Some phases are accessible in a short time to migratingchemicals (e.g., surface waters), but others are only accessible slowly (e.g., deeplake or ocean waters), or effectively not at all (e.g., deep soil or rock)

Some confusion is possible when expressing concentrations for mixed phasessuch as water containing suspended solids (SS) An analysis may give a total or bulkconcentration expressed as amount of chemical per m3 of mixed water and particles.Alternatively, the water may be filtered to give the concentration or amount ofchemical that is dissolved in water per m3 of water The difference between these

is the amount of chemical in the SS phase per m3 of water This is different fromthe concentration in the SS phase expressed as amount of chemical per m3 of particle.Concentrations in soils, sediments, and biota can be expressed on a dry or wet weightbasis Occasionally, concentrations in biota are expressed on a lipid or fat contentbasis Concentrations must be expressed unambiguously

2.3.1 Homogeneity and Heterogeneity

A key modeling concept is that of phase homogeneity and heterogeneity Wellmixed phases such as shallow pond waters tend to be homogeneous, and gradients

in chemical concentration or temperature are negligible Poorly mixed phases such

as soils and bottom sediments are usually heterogeneous, and concentrations varywith depth Situations in which chemical concentrations are heterogeneous aredifficult to describe mathematically, thus there is a compelling incentive to assumehomogeneity wherever possible A sediment in which a chemical is present at aconcentration of 1 g/m3 at the surface, dropping linearly to zero at a depth of 10 cm,can be described approximately as a well mixed phase with a concentration of 1 g/m3

and 5 cm deep, or 0.5 g/m3 and 10 cm deep In all three cases, the amount of chemicalpresent is the same, namely 0.05 g per square metre of sediment horizontal area.Even if a phase is not homogeneous, it may be nearly homogeneous in one ortwo of the three dimensions For example, lakes may be well mixed horizontallybut not vertically, thus it is possible to describe concentrations as varying only inone dimension (the vertical) A wide, shallow river may be well mixed verticallybut not horizontally in the cross-flow or down-flow directions

2.3.2 Steady- and Unsteady-State Conditions

If conditions change relatively slowly with time, there is an incentive to assume

“steady-state” behavior, i.e., that properties are independent of time A severe ematical penalty is incurred when time dependence has to be characterized, and

math-“unsteady-state,” dynamic, or time-varying conditions apply We discuss this issue

in more detail later

2.3.3 Summary

In summary, our simplest view of the environment is that of a small number ofphases, each of which is homogeneous or well mixed and unchanging with time.When this is inadequate, the number of phases may be increased; heterogeneity may

be permitted in one, two, or three dimensions; and variation with time may beincluded The modeler’s philosophy should be to concede each increase in complex-ity reluctantly, and only when necessary Each concession results in more mathe-matical complexity and the need for more data in the form of kinetic or equilibriumparameters The model becomes more difficult to understand and thus less likely to

be used, especially by others This is not a new idea William of Occam expressedthe same sentiment about 650 years ago, when he formulated his principle ofparsimony or “Occam’s Razor,” stating

Essentia non sunt multiplicanda praeter necessitatem

which can be translated as, “What can be done with fewer (assumptions) is done invain with more,” or more colloquially, “Don’t make models more complicated than

is necessary.”

When describing a volume of the environment, it is obviously essential to defineits limits in space This may simply be the boundaries of water in a pond or the airover a city to a height of 1000 m The volume is presumably defined exactly, as arethe areas in contact with adjoining phases Having established this control “envelope”

or “volume” or “parcel,” we can write equations describing the processes by which

a mass of chemical enters and leaves this envelope

The fundamental and now axiomatic law of conservation of mass, which wasfirst stated clearly by Antoine Lavoisier, provides the basis for all mass balanceequations Rarely do we encounter situations in which nuclear processes violate thislaw Mass balance equations are so important as foundations of all environmentalcalculations that it is essential to define them unambiguously Three types can beformulated and are illustrated below We do not treat energy balances, but they areset up similarly

2.4.1 Closed System, Steady-State Equations

This is the simplest class of equation It describes how a given mass of chemicalwill partition between various phases of fixed volume The basic equation simplyexpresses the obvious statement that the total amount of chemical present equals thesum of the amounts in each phase, each of these amounts usually being a product of

a concentration and a volume The system is closed or “sealed” in that no entry orexit of chemical is permitted In environmental calculations, the concentrations areusually so low that the presence of the chemical does not affect the phase volumes

Worked Example 2.1

A three-phase system consists of air (100 m3), water (60 m3), and sediment (3

m3) To this is added 2 mol of a hydrocarbon such as benzene The phase volumes

are not affected by this addition, because the volume of hydrocarbon is small

Subscripting air, water, and sediment symbols with A, W, and S, respectively, and

designating volume as V (m3) and concentration as C (mol/m3), we can write the

mass balance equation

total amount = sum of amounts in each phase mol

2 = VACA + VWCW + VSCS = 100 CA + 60 CW + 3 CS mol

To proceed further, we must have information about the relationships between CA,

CW, and CS This could take the form of phase equilibrium equations such as

CA/CW = 0.4 and CS/CW = 100These ratios are usually referred to as partition coefficients or distribution coef-

ficients and are designated KAW and KSW, respectively We discuss them in more

The amounts in each phase (mi) mol are the VC products as follows:

This simple algebraic procedure has established the concentrations and amounts in

each phase using a closed system, steady-state, mass balance equation and

equilib-rium relationships The essential concept is that the total amount of chemical present

mW = VWCW = 0.30 mol (15%)

mA = VACA = 0.20 mol (10%)

mS = VSCS = 1.50 mol (75%)Total 2.00 mol

must equal the sum of the individual amounts in each compartment We later refer

to this as a Level I calculation It is useful because it is not always obvious where

concentrations are high, as distinct from amounts.

Example 2.2

In this example, 0.04 mol of a pesticide of molar mass 200 g/mol is applied to

a closed system consisting of 20 m3 of water, 90 m3 of air, 1 m3 of sediment, and

2 L of biota (fish) If the concentration ratios are air/water 0.1, sediment/water 50,and biota/water 500, what are the concentrations and amounts in each phase in bothgram and mole units?

Answer

In this case, 8.3% is present in each of SS and biota and 83% in water with aconcentration in water of 39.8 µg/m3

2.4.2 Open System, Steady-State Equations

In this class of mass balance equation, we introduce the possibility of thechemical flowing into and out of the system and possibly reacting or being formed.The conditions within the system do not change with time, i.e., its condition looksthe same now as in the past and in the future The basic mass balance assertion isthat the total rate of input equals the total rate of output, these rates being expressed

in moles or grams per unit time Whereas the basic unit in the closed system balancewas mol or g, it is now mol/h or g/h

Worked Example 2.4

A 104 m3 thoroughly mixed pond has a water inflow and outflow of 5 m3/h Theinflow water contains 0.01 mol/m3 of chemical Chemical is also discharged directlyinto the pond at a rate of 0.1 mol/h There is no reaction, volatilization, or otherlosses of the chemical; it all leaves in the outflow water

(i) What is the concentration (C) in the outflow water? We designate this as anunknown C mol/m3.

total input rate = total output rate

5 m3/h ¥ 0.01 mol/m3 + 0.1 mol/h = 0.15 mol/h = 5 m3/h ¥ C mol/m3 = 5 C mol/hThus,

C = 0.03 mol/m3

The total inflow and outflow rates of chemical are 0.15 mol/h

(ii) If the chemical also reacts in a first-order manner such that the rate isVCk mol/h where V is the water volume, C is the chemical concentration in the wellmixed water of the pond, and k is a first-order rate constant of 10–3 h–1, what will

be the new concentration?

The output by reaction is VCk or 104¥ 10–3 C or 10 C mol/h, thus we rewritethe equation as:

0.05 + 0.1 = 5 C + 10 C = 15 C mol/hThus,

C = 0.01 mol/m3

The total input of 0.15 mol/h is thus equal to the total output of 0.15 mol/h, consisting

of 0.05 mol/h outflow and 0.10 mol/h reaction

An inherent assumption is that the prevailing concentration in the pond is stant and equal to the outflow concentration This is the “well mixed” or “continu-ously stirred tank” assumption It may not always apply, but it greatly simplifiescalculations when it does

con-The key step is to equate the sum of the input rates to the sum of the outputrates, ensuring that the units are equivalent in all the terms This often requires someunit-to-unit conversions

Worked Example 2.5

A lake of area (A) 106 m2 and depth 10 m (volume V 107 m3) receives an input

of 400 mol/day of chemical in an effluent discharge Chemical is also present in theinflow water of 104 m3/day at a concentration of 0.01 mol/m3 The chemical reactswith a first-order rate constant k of 10–3 h–1, and it volatilizes at a rate of (10–5 C)mol/m2s, where C is the water concentration and m2 refers to the air-water area Theoutflow is 8000 m3/day, there being some loss of water by evaporation Assumingthat the lake water is well mixed, calculate the concentration and all the inputs andoutputs in mol/day Use a time unit of days in this case

Discharge = 400 mol/day

Inflow = 104 m3/day ¥ 0.01 mol/m3 = 100 mol/day

Total input = 500 mol/day

Reaction rate = VCk = 107 m3¥ C mol/m3¥ 10–3 h–1¥ 24 h/day = 24 ¥ 104C

mol/dayVolatilization rate = 106 m2¥ 10–5 C mol/m2 s ¥ 3600 s/h ¥ 24 h/day = 86.4 ¥

104C mol/dayOutflow = 8000 m3/day ¥ C mol/m3 = 0.8 ¥ 104C mol/day

Thus,

500 = 24 ¥ 104C + 86.4 ¥ 104C + 0.8 ¥ 104C = 111.2 ¥ 104C

C = 4.5 ¥ 10–4 mol/m3

Reaction rate = 107.9 mol/day (i.e., 108 mol/day)

Volatilization rate = 388.5 mol/day (i.e., 390 mol/day)

Outflow = 3.6 mol/day

Total rate of loss = 500 mol/day = input rate

Until proficiency is gained in manipulating these multi-unit equations, it is best

to write out all quantities and units and check that the units are consistent Judgementshould be exercised when selecting the number of significant figures to be carriedthrough the calculation It is preferable to carry more than is needed, then go backand truncate Remember that environmental quantities are rarely known with betterthan 5% accuracy Avoid conveying an erroneous impression of accuracy by usingtoo many significant figures

Example 2.6

A building, 20 m wide ¥ 25 m long ¥ 5 m high is ventilated at a rate of 200 m3/h.The inflow air contains CO2 at a concentration of 0.6 g/m3 There is an internalsource of CO2 in the building of 500 g/h What is the mass of CO2 in the buildingand the exit CO2 concentration?

2500 kg/m3?