ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 3 ppt

The terms mass, moles, or such as mass or moles of material per volume of medium resulting in mass or molar concentration; moles of material per mole of medium resulting in mole fraction

CHAPTER 3 Environmental Modeling 3.1 INTRODUCTION

Interest in the field of environmental monitoring and quantitative assessment of environmental problems is growing For some years now, the results of environmental models and assessment analyses have been influencing environmental regulation and policies These results are widely cited by politicians in forecasting consequences of greenhouse gas emissions like carbon dioxide (CO2) and in advocating dramatic reductions of energy consumption at local, state, national, and international levels For this reason and because environmental modeling is often based on extreme conceptual and numerical intricacy and uncertain validity, environmental modeling has become one

of the most controversial topics of applied mathematics

Having said this, environmental modeling continues to be widely used in environmental engi-neering, with its growth only limited by the imagination of the modelers Environmental engineering problem-solving techniques incorporating modeling are widely used in watershed mapping; surface water information; flood hazard mapping; climate modeling; groundwater modeling; and others Keep in mind that the end product produced on any modeling system will be, at least in part, a reflection of the modeling system — sometimes more than a reflection of actual conditions

In this chapter, we do not provide a complete treatment of environmental modeling (For the reader who desires such a treatment, we highly recommend Nirmalakhandan’s Modeling Tools for

modeled after his work.) We present an overview of quantitative operations implicit to environmental modeling processes

3.2 MEDIA MATERIAL CONTENT

Media material content is a measure of the material contained in a bulk medium, quantified by the ratio of the amount of material present to the amount of the medium The terms mass, moles, or

such as mass or moles of material per volume of medium (resulting in mass or molar concentration); moles of material per mole of medium (resulting in mole fraction); and volume of material per volume of medium (resulting in volume fraction)

When dealing with mixtures of materials and media, the use of different forms of measures in the ratio to quantify material content may become confusing With mixtures, the ratio can be expressed in concentration units The concentration of a chemical substance (liquid, gaseous, or solid) expresses the amount of substance present in a mixture Concentration can be expressed in many different ways

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74 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

Chemists use the term solute to describe the substance of interest and the term solvent to describe the material in which the solute is dissolved For example, in a can of soft drink (a solution

of sugar in carbonated water), approximately 12 tablespoons of sugar (the solute) are dissolved in the carbonated water (the solvent) In general, the component present in the greatest amount is termed the solvent

Some of the more common concentration units are:

• Mass per unit volume Some concentrations are expressed in milligrams per milliliter (mg/mL) or milligrams per cubic centimeter (mg/cm 3 ) Note that “1 mL = 1 cm 3 ” is sometimes denoted as a

“cc.” Mass per unit volume is handy when discussing how soluble a material is in water or a particular solvent For example, “the solubility of substance X is 4 grams per liter.”

• Percent by mass Also called weight percent or percent by weight, this is the mass of the solute divided by the total mass of the solution and multiplied by 100%:

(3.1)

The mass of the solution is equal to the mass of the solute plus the mass of the solvent For example,

a solution consisting of 30 g of sodium chloride and 70 g of water would be 30% sodium chloride

by mass: (30 g NaCl)/(30 g NaCl + 70 g water) × 100% = 30% To avoid confusion over whether a solution is percent by weight or percent by volume, the symbol “w/w” (for weight to weight) is often used after the concentration: “10% potassium iodide solution in water (w/w).”

• Percent by volume Also called volume percent or percent by volume, this is typically only used for mixtures of liquids and is the volume of the solute divided by the sum of the volumes of the other components, multiplied by 100% If we mix 30 mL of ethanol and 70 mL of water, the percent ethanol by volume will be 30%; however, the total volume of the solution will NOT be

100 mL (although it will be close) because ethanol and water molecules interact differently with each other than they do with themselves To avoid confusion over whether we have a percent by weight or percent by volume solution, we could label this as “30% ethanol in water (v/v)” where v/v stands for “volume to volume.”

• Molarity This is the number of moles of solute dissolved in one liter of solution For example, if

we have 90 g of glucose (molar mass = 180 g/mol), this is (90 g)/(180 g/mol) = 0.50 mol of glucose If we place this in a flask and add water until the total volume = 1 L, we would have a 0.5 molar solution Molarity is usually denoted with a capital, italicized “M” — a 0.50-M solution Recognize that molarity is moles of solute per liter of solution, not per liter of solvent Also recognize that molarity changes slightly with temperature because the volume of a solution changes with temperature.

• Molality (m, used for calculations of colligative properties) Molality is the number of moles of solute dissolved in 1 kg of solvent Notice two key differences between molarity and molality: molality uses mass rather than volume, and solvent instead of solution.

(3.2)

Unlike molarity, molality is independent of temperature because mass does not change with tem-perature If we place 90 g of glucose (0.50 mol) in a flask, then add 1 kg of water, we have a 0.50 molal solution Molality is usually denoted with a small, italicized “m” — a 0.50-m solution.

• Parts per million (ppm) Parts per million works like percent by mass, but is more convenient when only a small amount of solute is present It is defined as the mass of the component in solution, divided by the total mass of the solution, multiplied by 10 6 (1 million):

Percent by mass Mass of component

Mass of s

=

o

Kilograms of solu

=

ttion L1681_book.fm Page 74 Tuesday, October 5, 2004 10:51 AM

ENVIRONMENTAL MODELING 75

(3.3)

A solution with a concentration of 1 ppm has 1 g of substance for every million grams of solution Because the density of water is 1 g/mL and we are adding such a tiny amount of solute, the density

of a solution at such a low concentration is approximately 1 g/mL Therefore, in general, 1 ppm im-plies 1 mg of solute per liter of solution Finally, recognize that 1% = 10,000 ppm Therefore, some-thing that has a concentration of 300 ppm could also be said to have a concentration of (300 ppm)/(10,000 ppm/percent) = 0.03% percent by mass.

• Parts per billion (ppb) This works like parts per million, but we multiply by 1 billion (10 9 ) (Caution: the word billion has different meanings in different countries.) A solution with 1 ppb of solute has 1 µg (10 –6 ) of material per liter.

• Parts per trillion (PPT) Again, this works like parts per million and parts per billion, except that

we multiply by 1 trillion (10 12 ) Few, if any, solutes are harmful at concentrations as low as 1 ppt.

The following notation and examples can help in standardizing these different forms; subscripts for components are i = 1, 2, 3, … N; and subscripts for phases are g = gas; a = air; l = liquid; w = water; and s = solids and soil

3.2.1 Material Content: Liquid Phases

Mass concentration, molar concentration, or mole fraction can be used to quantify material content

in liquid phases

(3.4)

(3.5)

Because moles of material = mass/molecular weight (MW), mass concentrations, p i,w, are related as:

(3.6)

For molarity M, [X] is molar concentration of “X.”

Mole fraction, X, of a single chemical in water can be expressed as follows:

(3.7)

For dilute solutions, the moles of chemical in the denominator of the preceding can be ignored in comparison to the moles of water, n w, and can be approximated by:

(3.8)

Parts per million Mass of component

Mass o

=

ff solution (1,000,000)

Mass conc of component in wateri = pi, w MMass of material,

Volume of water

i

Mass conc of component in wateri = Ci, w MMoles of material,

Volume of water

i

MW

i, w

i, w

i

Total moles of sol (Moles of chemical ++ Moles of water)

Moles of water

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76 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

If X is less than 0.02, an aqueous solution can be considered dilute On a mass basis, similar expressions can be formulated to yield mass fractions Mass fractions can also be expressed as a percentage or as other ratios, such as parts per million or parts per billion

The mole fraction of a component in a solution is simply the number of moles of that component divided by the total moles of all the components We use the mole fraction because the sum of the individual fractions should equal 1 This constraint can reduce the number of variables when modeling mixtures of chemicals Mole fractions are strictly additive; the sum of the mole fractions

of all components is equal to one Mole fraction, X i, of component i in an N-component mixture

is defined as follows:

(3.9)

(3.10)

For dilute solutions of multiple chemicals (as in the case of single chemical systems), mole fraction X iof component i in an N-component mixture can be approximated by:

(3.11)

Note that the preceding ratio is known as an intensive property because it is independent of the system and the mass of the sample An intensive property is any property that can exist at a point

in space Temperature, pressure, and density are good examples An extensive property is any property that depends on the size (or extent) of the system under consideration — volume, for example If we double the length of all edges of a solid cube, the volume increases by a factor of eight Mass is another extensive property The same cube undergoes an eightfold increase mass when the length of the edges is doubled

Note: The material content in solid and gas phases is different from those in liquid phases For example, the material content in solid phases is often quantified by a ratio of masses and expressed

as parts per million or parts per billion The material content in gas phases is often quantified by

a ratio of moles or volumes and expressed as parts per million or parts per billion Reporting gas phase concentrations at standard temperature and pressure (STP — 0°C and 769 mm Hg or 273 K and 1 atm) is the preferred form

Example 3.1

A certain chemical has a molecular weight of 80 Derive the conversion factors to quantify the following:

1 1 ppm (volume/volume) of the chemical in air in molar and mass concentration form

2 1 ppm (mass ratio) of the chemical in water in mass and molar concentration form

3 1 ppm (mass ratio) of the chemical in soil in mass ratio form





∑  +

i

N

w 1

The sum of all the mole fractions = 

∑Xw

N

1

  =1

nw

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ENVIRONMENTAL MODELING 77

Solution:

1 Gas phase The volume ratio of 1 ppm can be converted to the mole or mass concentration form using the assumption of ideal gas, with a molar volume of 22.4 L/g mol at STP conditions (273

K and 1.0 atm).

The general relationship is 1 ppm = (MW/22.4) mg/m 3

2 Water phase The mass ratio of 1 ppm can be converted to mole or mass concentration form using the density of water, which is 1 g/cm 3 at 4°C and 1 atm.

3 Soil phase The conversion is direct

Example 3.2

Analysis of a water sample from a pond gave the following results: volume of sample = 2 L; concentration of suspended solids in the sample = 15 mg/L; concentration of dissolved chemical = 0.01 mol/L; and concentration of the chemical adsorbed onto the suspended solids = 400 µg/g solids If the molecular weight of the chemical is 125, determine the total mass of the chemical in the sample

1,000,000 m of air

v

3

3

=

1,000,000 m of air

v

3

3

22.4 L

1000 L m

mol

3















≡4 46 10× −

5

m3





− mol m

80 g gmol

g m

m

µg L

3≡

1,000,000 g water

=

1,000,000 g water

g

cm3













100 cm m

g m

mg L

3 3

g m

mol

80 g

mol m





1,000,000 g soil

=

1,000,000 g soil

1000

kg

1000 mg g

mg kg















= 1 L1681_book.fm Page 77 Tuesday, October 5, 2004 10:51 AM

78 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

Solution:

Dissolved concentration = molar concentration × MW

Dissolved mass in sample = dissolved concentration × volume

Mass of solids in sample = concentration of solids × volume

Adsorbed mass in sample = adsorbed concentration × mass of solids

Thus, total mass of chemical in the sample = 0.25 g + 0.00020 g = 0.25020 g

3.3 PHASE EQUILIBRIUM AND STEADY STATE

The concept of phase equilibrium (balance of forces) is an important one in environmental modeling

In the case of mechanical equilibrium, consider the following example A cup sitting on a table top remains at rest because the downward force exerted by the Earth’s gravity action on the cup’s mass (this is what is meant by the “weight” of the cup) is exactly balanced by the repulsive force between atoms that prevents two objects from simultaneously occupying the same space, acting in this case between the table surface and the cup If one picks up the cup and raises it above the tabletop, the additional upward force exerted by the arm destroys the state of equilibrium as the cup moves upward If one wishes to hold the cup at rest above the table, it is necessary to adjust the upward force to balance the weight of the cup exactly, thus restoring equilibrium

For more pertinent examples (chemical equilibrium, for example) consider the following Chemical equilibrium is a dynamic system in which chemical changes are taking place in such a way that no overall change occurs in the composition of the system In addition to partial ionization, equilibrium situations include simple reactions — for example, when the air in contact with a liquid

is saturated with the liquid’s vapor, meaning that the rate of evaporation is equal to the rate of condensation When a solution is saturated with a solute, the dissolving rate is just equal to the precipitation rate from solution In each of these cases, both processes continue The equality of rate creates the illusion of static conditions, and no reaction actually goes to completion

Equilibrium is best described by the principle of Le Chatelier, which sums up the effects of changes in any of the factors influencing the position of equilibrium It states that a system in equilibrium, when subjected to a stress resulting from a change in temperature, pressure, or





=

0.001Mol L

125 g gmol

g L

0 125





0.125 g

L (2L) 0 25. g









×







=

400 µg

g

10 µg6 0.000

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ENVIRONMENTAL MODELING 79

concentration, and causing the equilibrium to be upset, will adjust its position of equilibrium to relieve the stress and re-establish equilibrium.

What is the difference between steady state and equilibrium? Steady state implies no changes with the passage of time; similarly, equilibrium can also imply no change of state with passage of time In many situations, the system not only is at steady state, but also is at equilibrium However, this is not always the case Sometimes, when flow rates are steady but the phase contents, for example, are not being maintained at the “equilibrium values,” the system is at steady state but not

at equilibrium

3.4 MATH OPERATIONS AND LAWS OF EQUILIBRIUM

Earlier we observed that no chemical reaction goes to completion; the qualitative consequences of this insight go beyond the purpose of this text However, in this text we describe and use the basic quantitative aspects of equilibria The chemist usually starts with the chemistry of the reaction and fully uses chemical intuition before resorting to mathematical techniques That is, science should always precede mathematics in the study of physical phenomena Note, however, that most chemical problems do not need an exact, closed-form solution, and the direct application of mathematics to

a problem can lead to an impasse

Several basic math operations and fundamental laws from physical chemistry and thermody-namics serve as the tools, blueprints, and foundational structures of mathematical models They can be used and applied to environmental systems under certain conditions, serving to solve various problems Many laws serve as important links between the state of the system, chemical properties, and their behavior In the following sections, we review some of the basic math operations used

to solve basic equilibrium problems, as well as laws essential for modeling the fate and transport

of chemicals in natural and engineered environmental systems

3.4.1 Solving Equilibrium Problems

In the following math operations, we provide examples of the various forms of hydrogen combustion

to yield water to demonstrate the solution of equilibrium problems The example reactions are represented by the following equation Note: Consider the reaction at 1000.0 K where all constit-uents are in the gas phase and the equilibrium constant is 1.15 × 1010 atm–1:

and the equilibrium constant expression:

(3.12)

where concentrations are given as partial pressures in atmospheres Note that K is very large, and consequently the concentration of water is large and/or the concentration of at least one of the reactants is very small

Example 3.3

Consider a system at 1000.0 K in which 4.00 atm of oxygen is mixed with 0.500 atm of hydrogen and no water is initially present Note that oxygen is in excess and hydrogen is the

2 H (g)2 + O (g)2 = 2 H O(g)2

K = [H O] /[H ] [O ]2 2 22 2 L1681_book.fm Page 79 Tuesday, October 5, 2004 10:51 AM

80 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

limiting reagent Because the equilibrium constant is very large, virtually all the hydrogen is converted to water yielding [H2O] = 0.500 atm and [O2] = 4.000 – 0.5(0.500) = 3.750 atm The final concentration of hydrogen, a small number, is an unknown — the only unknown

Solution:

Using the equilibrium constant expression, we obtain:

from which we determine that [H2] = 2.41 × 10–6 atm Because this is a small number, our initial approximation is satisfactory

Example 3.4

Again, consider a system at 1000.0 K, where 0.250 atm of oxygen is mixed with 0.500 atm of hydrogen and 2000 atm of water

Solution:

Again, the equilibrium constant is very large and the concentration of least reactants must be reduced to a very small value

In this case, oxygen and hydrogen are present in a 1:2 ratio, the same ratio given by the stoichio-metric coefficients Neither reactant is in excess, and the equilibrium concentrations of both will

be very small values We have two unknowns, but they are related by stoichiometry Because neither product is in excess and one molecule of oxygen is consumed for two of hydrogen, the ratio [H2]/[O2] = 2/1 is preserved during the entire reaction and [H2] = 2[O2]

3.4.2 Laws of Equilibrium

Some of the laws essential for modeling the fate and transport of chemicals in natural and engineered environmental system include:

• Ideal gas law

• Dalton’s law

• Raoults’ law

• Henry’s law

3.4.2.1 Ideal Gas Law

An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly elastic and have no intermolecular attractive forces One can visualize it as collections of perfectly

1.15 × 1010 = (0.500) /[H ] (2 22 3 750 )

[H O]2 =2.000+0.500=2.500 atm

1.15× 1010 = 2.500 /(2[O ]) [O ]2 2 2 2

[O ]2 = 5.14 × 10–4 atm and [H ]2 = 2[O ]2 = 031 × 10–3 atm L1681_book.fm Page 80 Tuesday, October 5, 2004 10:51 AM

ENVIRONMENTAL MODELING 81

hard spheres, which collide but otherwise do not interact with each other In such a gas, all the

internal energy is in the form of kinetic energy, and any change in the internal energy is accompanied

by a change in temperature An ideal gas can be characterized by three state variables: absolute

pressure (P); volume (V); and absolute temperature (T) The relationship between them may be

deduced from kinetic theory and is called the ideal gas law:

(3.13)

where:

n = number of moles

R = universal gas constant = 8.3145 J/mol K or 0.821 L-atm/de-mol

N = number of molecules

k = Boltzmann constant = 1.38066 × 10–23 J/K

= R/N A = where N A = Avogadros number = 6.0221 × 1023

Note: At standard temperature and pressure (STP), the volume of 1 mol of ideal gas is 22.4 L, a

volume called the molar volume of a gas

Example 3.5

Calculate the volume of 0.333 mol of gas at 300 K under a pressure of 0.950 atm

Solution:

Most gases in environmental systems can be assumed to obey this law The ideal gas law can

be viewed as arising from the kinetic pressure of gas molecules colliding with the walls of a

container in accordance with Newton’s laws However, a statistical element is also present in the

determination of the average kinetic energy of those molecules The temperature is taken to be

proportional to this average kinetic energy; this invokes the idea of kinetic temperature

3.4.2.2 Dalton’s Law

Dalton’s law states that the pressure of a mixture of gases is equal to the sum of the pressures of

all of the constituent gases alone Mathematically, this can be represented as:

(3.14)

where:

PTotal = total pressure

P1 = partial pressure and

(3.15)

P

PTotal = P1 + P2 …Pn

V

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82 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

where n j is the number of moles of component j in the mixture.

Note: Although Dalton’s law explains that the total pressure is equal to the sum of all of the

pressures of the parts, this is only absolutely true for ideal gases, although the error is small for real gases

Example 3.6

Problem:

The atmospheric pressure in a lab is 102.4 kPa The temperature of the water sample is 25°C, with pressure as 23.76 torr If we use a 250-mL beaker to collect hydrogen from the water sample, what is the pressure of the hydrogen, and the moles of hydrogen using the ideal gas law?

Solution:

Step 1 Make the following conversions: a torr is 1 mm Hg at standard temperature In kilopascals, that would be 3.17 (1 mm Hg = 7.5 kPa) Convert 250 mL to 0.250 L and 25°C to 298 L.

Step 2 Use Dalton’s law to find the hydrogen pressure

Step 3 Recall that the ideal gas law is:

where:

P is pressure

V is volume

n is moles

R is the ideal gas constant (0.821 L-atm/mol-K or 8.31 L-kPa/mol-K)

T is temperature

Therefore,

rearranged:

PTotal= PWater + PHydrogen

102.4 kPa= 3.17 kPa + PHydrogen

PHydrogen= 99.23 kPa or pp.2 kPa

99.2 kPa × 250 L = n × 8.31 L-kPa/mol × 2988 K

n = 99.2 kPa × 250 L/8.31 L-kPa/mol-K/298K

n = 0100 mol or 1.00 × 10–2 mol H