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I. INTRODUCTION

EARTH’S COMPARTMENTS (Figure I.1). The earth can be divided into 4 distinctcompartments, three of them are physical compartments, i .e., atmosphere (the gaseous

envelope surrounding the earth), hydrosphere (the water belt around the earth, liquidand frozen, fresh and salt y), and lithosphere (the soil and rock of the earth’s crust). The4th compartment that overlaps the three physical compartments is the biosphere, whichis the global sum of all ecosystems. The biosphere is the global ecological system thatintegrates all living beings and their relationships, including their interactions with theatmosphere, hydrosphere, and lithosphere.

Atmosphere (Figure I.2). Atmosphere is the gaseous envelope surrounding theearth. The atmosphere is composed of a series of five concentric layers: troposphere,stratosphere, mesosphere, thermosphere, and exosphere. The troposphere is theatmosphere from the earth’s surface to the stratosphere which is characterized by thepresence of clouds, turbulent winds, and decreasing temperature with increasing altitude(i .e., −6oC/km). The troposphere extends to a height of approximately 10 km. Thestratosphere extends from 10 to 45 km above the earth’s surface and contains a layer ofozone (O3) critical to life because it absorbs much of the sun’s damaging ultraviolet (UV )radiation. The absorption of UV radiation by the O3 layer heats the air and therefore,temperature increases with increasing altitude in the stratosphere. There is a steady

wind but no turbulence in the stratosphere. There is little water, and the temperaturein the lower stratosphere, which commercial jets fly, remains largely uniform (i .e., −45oCto −75oC). The mesosphere, the layer of atmosphere directly above the stratosphere,extends from 45 to 80 km above the earth’s surface. Temperature drops steadily in themesosphere to the lowest in the atmosphere, i .e., as low as −138oC. The thermosphere extends from 80 to 500 km above the earth’s surface and is very hot. Gases in thethermosphere absorb X rays and short-wavelength UV radiations. This absorptiondrives the few gas molecules present to great speeds, raising their temperature to

> 1,000oC. The aurora, a colorful display of lights in dark polar skies, is produced whencharged particles from the sun hit oxygen and nitrogen molecules in the thermosphere.The thermosphere is important in long-distance communication because it reflectsoutgoing radio waves back to the earth without the aid of satellites. The outmost layerof the atmosphere, the exosphere, begins about 500 km above the earth’s surface. Theexosphere continues to thin out until it converges with interplanetary space. Thecomposition of atmosphere in ppmv (m3/106 m3 air) is: N2 (nitrogen), 780,840(78.084%); O2 (oxygen), 209,460 (20.946%); Ar (argon), 9,340 (0.9340%); CO2 (carbondioxide), 383 (0.0383%); Ne (neon), 18.18 (0.001818%); He (helium), 5.24(0.000524%); CH4 (methane), 1.745 (0.0001745%); Kr (krypton), 1.14 (0.000114%); andH2 (hydrogen), 0.55 (0.000055%). Water vapor accounts for about 0.4% over fullatmosphere and 1 to 4% near the earth’s surface.

p A ( patm) = Atmospheric Pressure = 101 kPa = 101 kN/m2 [1 N (Newton) = 1 kg-m/s2] S E = Earth’s Surface Area = 4 πR2 = 5.1×1014 m2 ( R = the earth’s radius = 6,371 km)

M A = Mass of Atmosphere =81.9

)101.5)(10101( 143

g

S p E A = 5.25×1018 kg (g = 9.81 m/s2)

Mass in Troposphere (~ 10 km in elevation) = 78%



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average salinity of the earth's oceans is about 35 g salt/kg sea water (3.5%). The globalocean is a single, continuous body of water, which is divided into five sections separated

by the continents (Figure I.3): the Pacific (155,557,000 km2), Atlantic (76,762,000 km2),Indian (68,556,000 km2), southern or Antarctica (20,327,000 km2) and Arctic(14,056,000 km2) Oceans. The Pacific Ocean is the largest which covers one-third of theearth’s surface and contains more than half of the earth’s water.

Ocean Mass ( M O) = S Od O ρO = (0.7)(5.1×1014 m2)(3,800 m)(1,030 kg/m3) = 1.40×1021 kg

where S O is ocean’s surface area, d O is ocean’s average depth, and ρO is ocean’s averagemass density.

Figure I.3. The world’s oceans.

Lithosphere (Figure I.4). The lithosphere is the solid part of the earth and it has twoparts: the crust and the upper mantle. The crust (rocks, minerals, and soil), which has athickness from 5 to 70 km, is the earth’s outermost layer. The rocks and minerals aremade of just 8 elements: oxygen (46.6%), silicon (27.72%), aluminum (8.13%), iron(5.00%), calcium (3.63%), sodium (2.83%), potassium (2.70%), and magnesium(2.09%). Soil is formed rock that is slowly broken down, or fragmented, into smaller andsmaller particles by biological, chemical, and physical weathering processes in nature.The thickness of soil varies from a thin film on young lands, near North and South Poles



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and on slopes near the top of mountains, to more than 3 m on old lands, such as certainforests. Soil is composed of 4 distinct parts: mineral particles, which make up about 45%of a soil; organic matter (about 5 %); water (about 25%); and air (about 25%). Theaverage bulk density of soil is about 1,500 kg/m3.

Soil Mass ( M S ) = S S d S ρ S = (0.255)(5.1×1014 m2)(2 m)(1,500 kg/m3) = 3.90×1017 kg

where S S is earth’s land mass surface area, d S is soil’s average depth, and ρ S is soil’s bulkdensity.

Figure I.4. The earth’s lithosphere.

Biosphere. The biosphere is the global sum of all ecosystems (i .e., the zone of life onearth). From the broadest bio-physiological point of view, the biosphere is the globalecological system that integrates all living beings and their relationships, including theirinteractions with the atmosphere, hydrosphere, and lithosphere. The biosphere is

postulated to have evolved, beginning through a process of biogenesis or biopoesis, atleast some 3.5 billion years ago. Every part of the planet, from the polar ice caps to theequator, support life of some kind. Recent advances in microbiology have demonstratedthat microbes live deep beneath the earth’s terrestrial surface, and that the total mass ofmicrobial life is so called uninhabitable zones, may exceed all animal and plant life onthe surface. The actual thickness of biosphere on earth is difficult to measure. Birdstypically fly at altitudes of 650 to 1,800 m, and fish that live deep underwater can befound down to −8,372 m in the Puerto Rico Trench.

There are more extreme examples for life on the planet: Rüppell’s vulture has been foundat altitudes of 11,300 m; bar-headed geese migrate at altitudes of at least 8,300 m (overMount Everest); yaks live at elevations from 3,200 to 5,400 m above the seas level;

mountain goats live up to 3,050 m. Herbivorous animals at these elevations depend onlichens, grasses, and herbs. Microscopic organisms live at such extremes that, takingthem into consideration puts the thickness of the biosphere much greater. Culturablemicrobes have been found in the earth’s upper atmosphere as high as 41 km. It isunlikely, however, that microbes are active at such altitudes, where temperatures and airpressures are extremely low and UV radiation is very high. These microbes are morelikely brought into the upper atmosphere by winds or possibly volcanic eruptions.Barophilic marine microbes have been found at more than 10 km depth in the MarianasTrench. Microbes are not limited to the air, water, or the earth’s surface. Culturable



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thermophilic microbes have been extracted from cores drilled more than 5 km into theearth’s crust in Sweden, from rocks with temperatures ranging from 65 to 75oC.Temperature increases rapidly with increasing depth into the earth’s crust. The rate at

which the temperature increases depends on many factors, including type of crust(continental vs. oceanic), rock type, geographic location, etc. The upper known limit isabout 112oC ( Methanopyruskandleri Strain 116), and it is likely that the limit of life in

the deep biosphere is controlled by temperature rather than absolute depth.

The earth’s biosphere is divided into a number of biomes, inhabited by broadly similarflora and fauna. On land, biomes are separated primarily by latitude. Terrestrial biomeslying ithin the Arctic and Antarctic Circles are relatively barren of plant and animal life,

while most of the more populous biomes lie near the equator.

INVENTORY OF WATER AT THE EARTH’S SURFA CE (Table I.1). Although75% of the earth’s surface is covered with water, substantially less than 1% is available forhumans. Most water is salty, frozen, or inaccessible in the soil and atmosphere. Thedefinition of freshwater is water containing less than 1,000 mg/L of dissolved solids,most often salt. The U. S. Geological Survey (USGS) reports that about 1.54×109 m3 of

water per day were withdrawn for use in the U.S. during 2000.

Table I.1. Water Inventory at the Earth’s Surface. Reservoir Volume (106 km3) % of the Total

Oceans 1,370 97.25Ice Caps & Glaciers 29 2.05

Deep groundwater (750 – 4,000 m) 5.3 0.38Shallow Groundwater (< 750 m) 4.2 0.30

Lakes 0.125 0.01Soil Moisture 0.065 0.005

Atmosphere (measured as liquid water) 0.013 0.001Rivers 0.0017 0.0001

Biosphere 0.0006 0.00004TOTAL 1,408.7 100

Great Lakes (Figure I.5 & Table I.2). The total volume of water in Great Lakes isabout 24,260 km3 and the total water mass is therefore 2.46×1016 kg [i .e., (24,260 km3)(109 m3/km3)(1,000 kg/m3)]. The Great Lakes contain about 95% of total freshwatermass in the U.S. and 20% of world’s total freshwater mass. The effects the Great Lakeshave on weather in the region are known as the lake effects. In winter, the moisturepicked up by the prevailing winds from the west can produce very heavy snowfall,especially along lakeshores to the east such as Michigan, Ohio, Pennsylvania, Ontario,and New York.

The lakes also moderate seasonal temperatures somewhat, by absorbing

heat and cooling the air in summer, then slowly radiating that heat in autumn. Thistemperature buffering produces areas known as "fruit belts", where fruit typically grownfarther south can commercially be produced. The eastern shore of Lake Michigan, thesouthern shore of Lake Erie, and the Niagara Peninsula between Lake Erie and LakeOntario are homes to many wineries. Lake effects also cause the occurrence of fog overmedium-sized areas, particularly along the shorelines of the lakes. The Great Lakes havealso been observed to help strengthen storms, such as Hurricane Hazel in 1954, and afrontal system in 2007 that spawned a few tornadoes in Michigan and Ohio.



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Table I.2. Morphological Data of the Great Lakes.Lake Area (km2) d max1 (m) d mean (m) Volume (km3) Shoreline (km) θ2 (years)

Superior 83,300 397 145 12,000 3,000 190Huron 59,510 223 76 4,600 2,700 40

Michigan 57,850 265 99 5,760 2,210 37

Erie 28,280 60 21 540 1,200 3Ontario 18,760 225 97 1,720 1,380 81 d max is the maximum depth and d mean is the mean depth.2 θ is the mean residence time, which is the average amount of time that a water molecule

resides in the system. θ =

V

V , where V is the volume of the surface water body and

V is

the total water volumetric inflow to the surface water body.

Figure I.5. The Great Lakes.

The World’s Largest Lakes (surface area).

1. Caspian Sea (Asia), 371,000 km2. Some consider the Caspian Sea as a full-fledgeseas, but it is surrounded by land and thus meets the generally accepteddefinition of a lake.

2. Lake Superior (North America), 82,100 km2.

3. Lake Victoria (Africa), 68,800 km2.4. Lake Huron (North America), 59,600 km2.5. Lake Michigan (North America), 57,800 km2.6. Lake Tanganyika (Africa), 32,900 km2.7. Baikal (Asia), 30,500 km2.8. Great Bear Lake (North America), 31,328 km2.9. Lake Malawi (Africa), 30,044 km2.10. Great Slave Lake (North America), 28,568 km2.



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boundary. The depth of the water below the surface of the land ranges from almost 122m in parts of the north to between 30 to 61 m throughout much of the south. Present-day recharge of the aquifer with fresh water occurs at an exceedingly slow rate,suggesting that much of the water in its pore spaces is paleowater, dating back to themost recent ice age and probably earlier.

Figure I.6. The Ogallala Aquifer.

WATER POVERTY INDEX (WPI ). A holistic water poverty index (WPI ),developed by Britain’s Center for Ecology and Hydrology and the World Water Council,grades countries according to their water resource ( R), access ( A), use (U ), capacity (C ),and environment ( E ), as shown in (I.1).

1

ecuar

ecuar

wwwww

E wC wU w Aw RwWPI (I.1)



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WPI was created as an interdisciplinary indicator to assess water stress and scarcity,linking physical estimates of water availability with the socioeconomic drivers of poverty. WPI is the weighted sum of the above 5 components, each having a value ranging from 0to 100. wr,…, we are weighing factors assigned and each has a value that ranges from 0to 1. Based on the WPI rankings summarized in Table I.3 Finland is rated as the topscorer, with plenty of water wisely used; Haiti comes bottom. On the whole, poor

countries do worse than rich ones, with some exceptions: Guyana’s score, for example,is raised by good access to safe water, and Japan is marked down by problems withpollution. U.S.’s standing is lowered by inefficient use of water in fields, factories, and homes (e.g., 3.4% of the rainfall is used for irrigational purposes and 3.6% of the rainfallis used for industrial purposes).

Table I.3. WPI Rankings. Country WPI Country WPIFinland 79.9 Argentina 62.8

Suriname 78.6 Poland 61.3Iceland 74.4 Mexico 59.2Norway 73.8 Brazil 58.8

Guyana 72.8 U.S. 58.5 Austria 72.2 India 58.2Ireland 71.9 Saudi Arabia 58.0Chile 70.5 Singapore 57.2

Canada 70.4 China 56.6Thailand 67.7 South Africa 55.7Sweden 67.2 Israel 55.6

Indonesia 66.2 Morocco 52.3Czech Republic 64.5 Cambodia 44.9

Pakistan 64.5 Uganda 43.3Japan 64.2 Niger 36.0Russia 63.4 Haiti 32.7

HYDROLOGIC CYCLE (Figure I.7). The hydrologic cycle depicts the continuouscirculation of water from the ocean to the atmosphere, to the land, and back to the ocean.Most of the freshwater formed by evaporation/evapotranspiration returns directly to thesea (i .e., 75% of 389,500 km3/year globally). Small portion of the precipitation that fallson the land is used in agriculture, in industry, and for municipal purposes. In the UnitedStates only 0.6% of the rainfall is used for municipal purposes, and of that only 5% isused for food preparation or drinking. By comparison, 3.4% of the rainfall is used forirrigational purposes and 3.6% of the rainfall is used for industrial purposes. Thefollowing are some useful terminologies that are widely used in the description of thehydrologic cycle.

Precipitation: Any form of moisture condensing in the air and depositing on the

ground.Evaporation: Molecules leaving the liquid state and entering the vapor or gaseous stateas, e.g., water evaporates to form water vapor.

Surface Runoff: The portion of precipitation which runs off the surface as opposed toeither percolating through the subsurface soil or soaking into the soil.

Groundwater Infiltration: The process of water percolating through the subsurface



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soil and entering the groundwater aquifers. An aquifer is an underground layer ofporous rock, sand, or other materials that allows the movement of water.

Figure I.7. The hydrologic cycle.

Some useful information on the global and U.S. hydrologic cycles is summarized asfollows:

I. Global Hydrologic Cycle:

1. Precipitation on the land: 110,000 km3/year.2. Evaporation from soil/streams/rivers/lakes and evapotranspiration from

vegetation: 71,000 km3/year.3. Precipitation to the ocean: 385,000 km3/year.4. Evaporation from the ocean: 425,000 km3/year.5. Runoff to the ocean: 40,000 km3/year.6. Water in the atmosphere: 13,000 km3.7. Water in the ocean: 1,350,000,000 km3;8. Water in groundwater aquifers: 15,300,000 km3.9. Movement of moist air in the atmosphere: 40,000 km3/year.

II. U.S. Hydrologic Cycle:

1. Precipitation (100%): 5.867×1012 m3/year.

2. Evaporation (30%): 2.139×1012 m3/year.3. Evapotranspiration (Transpiration) (40%): 2.347×1012 m3/year.4. Surface Runoff (20%): 1.173×1012 m3/year.5. Groundwater Infiltration (10%): 0.587×1012 m3/year.



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asthma). PM 10 and PM 2.5 refer to airborne particles with diameters < 10 and 2.5 μm,respectively (1 μm = 10-6 m). Particulate matter is a mixed primary and secondarypollutant. Coarse particles are mainly primary pollutants, with dominant sources thatinclude fugitive dust from roads, construction, and agricultural emissions. Winderosion, especially during dry seasons, is another important source. Industrialemissions, transportation sources (e.g., diesel soot), and residential wood smoke are

major sources of fine particles.

Lead (Pb). Lead is regulated both as a criteria pollutant and as a hazardous airpollutant. Lead is strictly a primary pollutant. In the atmosphere, lead is associated withsuspended particulate matter, mainly in the fine mode. Lead in fine particles has a longatmospheric lifetime and deposits broadly over the globe. The removal of lead fromgasoline has largely eliminated atmospheric lead problems in urban areas of the U.S.

UNITS OF MEASUREMENTS. For most water pollution work, it is customary toexpress pollutant concentrations in mg/L or mg/m3. Since the density of water isroughly 1 kg/L under normal conditions, then 1 mg/L = 1 g/m3 = 1 mg pollutant/kg water= 1 mg/106 mg = 1 parts per million ( ppm). In addition, 1 μg/L = 1 mg/m3 = 1 μg

pollutant/kg water = 1 mg/109 mg = 1 parts per billion ( ppb). For most air pollution work, on the other hand, it is customary to express pollutant concentrations in volumetric terms. For instance, the concentration of a gaseous pollutant in parts permillion is the volume of pollutant per million volumes of the air mixture ( ppm or

ppmv). Concentrations are sometimes expressed as mass pollutant per air volume, suchas μg/m3 or mg/m3.

The relationship between ppmv and mg pollutant/m3 air, which depends on thepressure, temperature, and molecular weight ( MW ) of the pollutant, can be establishedusing the ideal gas law, i .e., PV = nRT , where P is the absolute pressure, atm; V is the

volume, m3; n is the moles gas, moles; R is the universal gas constant, 0.082056 L-atm/mol-oK; and T is the absolute temperature, oK. Note that 1 atm = 101.325 kPa =

101,325

Pa = 101.325 bar.

Since the specific volume of an ideal gas (i .e.,n

V V

) at 1 atm and 273.15oK (0oC) is

22.4×10-3 m3/mole, then at P and T the specific volume of the same ideal gas is

calculated as)15.273(

))(10414.22)(1( 3

P

T , m3/mole, i .e.,

S

S S

S

S S

PT

T V P V

T

V P R

T

V P

.

If the gaseous pollutant behaves like an ideal gas, then its specific volume at P and T can

be expressed as ,

))(10)(15.273)((

))(10414.22)(1(3

3

MW P

T m3 pollutant/mg pollutant; or

,))(414.22)(1(

)10)()(15.273)((

),(

1 6

T

MW P

T P V

mg pollutant/m3 pollutant. Therefore,

),(

1)10)(( 6

T P V

ppmv yields mg pollutant/m3 air, i .e.,



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))(414.22)(1(

))(15.273)()((3 T

MW P ppmv

m

mg

AIR

POLLUTANT (I.2)

SOLID WASTES. Solid wastes are wastes that are in either liquid or gaseous forms,

e.g., food scraps, packaging materials, yard trimmings, miscellaneous inorganic wastes.Municipal solid wastes ( MSW ) are the solid wastes produced from residential,commercial, institutional, and industrial sources, but they do not include such wastes asconstruction wastes, automobile bodies, municipal wastewater sludge, combustion ash,and serving food. Major sources of solid wastes in the U.S. are: mining, 75%; agriculture,12%; industry, 9.0%; municipalities, 3.0%; and municipal wastewater sludge, 1.0%. U.S.EPA estimates that the U.S. generated 246 million tons of MSW in 2005, which is about2.2 kg/capita-day. Paper and paperboard products accounted for about 34.2% of totalMSW weight. Another 37% is roughly equally divided among yard trimmings (13.1%),food scraps (11.9%), and plastics (11.8%).

Solid Waste Management Strategies. New concepts about the management of

solid wastes can be applies depending on the countries or regions. Some of the mostgeneral, widely used concepts include:

Waste Hierarchy (Figure I.8). The waste hierarchy refers to the "3 Rs" (i .e., reduce,reuse, and recycle), which classify waste management strategies according to theirdesirability in terms of waste minimization. The waste hierarchy remains thecornerstone of most waste minimization strategies. The aim of the waste hierarchy is toextract the maximum practical benefits from products and to generate the minimumamount of waste. 3 Rs along with composting and energy recovery are all examples ofresource conservation. Resource conservation reduces greenhouse gas (GHG ) emissionsfrom common solid waste management pathways, including: (1) the incineration of solid

wastes (emissions of CO2 and nitrous oxide, a GHG that is 310 times as potent as CO2),

(2) the transportation of solid wastes to disposal sites (GHG emissions from thecombustion of the fuel used in the equipment), and (3) landfills (wastes in landfillsdecompose anaerobically and produce CH4, a GHG that is 21 times as potent as CO2).

Recycling typically requires less energy than producing goods from virgin materials.Energy savings from recycling translates into GHG savings because fewer fossil fuels arecombusted to produce energy. Source reduction and recycling can also increase carbonstorage in forests. By preventing or reducing harvests of raw materials, preserved treescontinue to remove CO2 from the atmosphere. Composting produces a useful productfrom organic waste that otherwise would have been landfilled, therefore compostinghelps prevent methane emissions from and leachate formulation in the landfills. Energyrecovery at incineration facilities and landfills can conserve resources by offsetting fossil

Recycling typically requires less energy than producing goods from virgin materials.Energy savings from recycling translates into GHG savings because fewer fossil fuels arecombusted to produce energy. Source reduction and recycling can also increase carbon



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Figure I.8. The solid waste management hierarchy.

storage in forests. By preventing or reducing harvests of raw materials, preserved treescontinue to remove CO2 from the atmosphere. Composting produces a useful productfrom organic waste that otherwise would have been landfilled, therefore compostinghelps prevent methane emissions from and leachate formulation in the landfills. Energyrecovery at incineration facilities and landfills can conserve resources by offsetting fossilfuels used for energy. Energy recovery is often associated with electricity generationsuch as landfill methane capture, although it can also offset fossil fuels used at industrialsites, resulting in fewer GHG emissions. In terms of climate benefits, reducing andreusing materials offers the best approaches to reduce GHG s. Because no wastes aregenerated, source reduction and reuse avoid all emissions associated with recycling,

composting, combusting, or landfilling the materials.

Extended Producer Responsibility ( EPR). EPR is a strategy designed to promote theintegration of all costs associated with products throughout their life cycle (includingend-of-life disposal costs) into the market price of the product. EPR is meant to imposeaccountability over the entire lifecycle of products and packaging introduced to themarket. This means that firms which manufacture, import, and/or sell products arerequired to be responsible for the products after their useful life as well as duringmanufacture.

Polluter Pay Principle. This is the principle that requires the polluting party to pay forthe impacts caused to the environment. With respect to solid waste management, this

generally refers to the requirement for a waste generator to pay for appropriate disposalof the wastes. Table I.4 summaries the net environmental impacts of recycling,incineration, and landfilling of 1 kg MSW . Negative values results from displacement ofenergy and processing of raw materials when incineration or recycling are used. MSWlandfills generate CH4 as a by-product because of the anaerobic biodegradation oforganic components in MSW .



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Table I.3. Recycling, Incineration, and Landfilling of MSW .Recycling Incineration Landfilling

Solid Waste (kg/kg) −0.416 +0.233 +1.000CO2 (kg/kg) −1.247 +0.490 +0.159CH4 (kg/kg) 0 0 +0.062

CO (kg/kg) −0.013 +4×10-4 +5×10-4

NOX (kg/kg) −4.7×10-3 −1×10-4 +5×10-4 SO2 (kg/kg) −5.5×10-3 −2.9×10-3 +1×10-4

Combustion of MSW in Various Countries (1996).

Country %Japan 67.0

Switzerland 64.0Sweden 43.0France 40.0

Germany 37.0Netherlands 33.0

U.S. 18.0

Italy 14.0U.K. 7.0

Spain 4.0

Example I.1. If the current recycling rate for aluminum can is 63%, find the primaryenergy required to produce the aluminum in a 355 mL (12 oz) aluminum can having amass of 16 g. (1) How much energy is saved when a can is recycled instead of beingthrown away? (2) What is the equivalent amount of gasoline wasted when one can isthrown away? Gasoline has an energy content of about 35,000 kJ/L.

Source Electricity(kWh/kg)1

Fossil Fuel(kJ/kg)

Primary Energy(kJ/kg)2

CO2 Emissions3 (kg/kg)

Bauxite 15 60,000 235,0004 13.1

Recycled 0.08 4,000 5,1505 0.481 1 kWh (kW-hr) = (1 kJ/s)(3,600) = 3,600 kJ. 2 The equivalent heat energy an average thermalpower plant (e.g., a 33-% efficient coal-burning power plant) would need to generate 1 kW-hr(kWh) electricity (~ 11,700 kJ/kWh for the U.S.). 3 Half hydroelectric, half coal-fired powerplants. 4 15×11,700 + 60,000. 5 0.08×11,700 + 4,000.

1.1. The primary energy needed for the 37% of the can made from bauxite is:(0.016 kg/can)(0.37)(235,000 kJ/kg) = 1,391 kJ/can.

1.2. The primary energy needed to produce the rest from recycled aluminum is:(0.016 kg/can)(0.63)(5,150 kJ/kg) = 52 kJ/can. Therefore, total primary energy

for an average can with 63% recycled aluminum is 1,391 + 52 = 1,443 kJ/can.2.1. The primary energy needed to produce a new can from ore is:(0.016 kg/can)(235,000 kJ/kg) = 3,760 kJ/can.

2.2. The primary energy needed to produce a new can from recycled aluminum is:(0.016 kg/can)(5,150 kJ/kg) = 82 kJ/can. Therefore, the energy saved byrecycling is: 3,760 − 82 = 3,678 kJ/can.

3. The equivalent amount of gasoline thrown away when a can is not recycled is:



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000,35

678,3 = 0.105 L = 105 mL.

Mineral Ore Requirements. The magnitude of the earth- and ore-moving tasksneeded to provide the minerals society demands is truly monumental. Digging holes,

removing the ores, and piling up the leftover residues create enormous aesthetic,environmental, economic, and energy problems.

Mineral Average Grade (%) Ore (Residue) (kg/100 kg Product)

Aluminum 23.00 435 (335)

Copper 0.91 10,900 (10,890)

Iron 40.00 250 (150)

Lead 2.50 4,000 (3,900)

Nickel 2.50 4,000 (3,900)

Generation and Recovery of Materials in MSW in the U.S. (1993).

Material Generated (million tones) Recovered (million tones)Paper/Paperboard 70.6 24.0

Glass 12.4 2.7Metals

◙ Ferrous Metals◙ Aluminum◙ Non-Ferrous Metals

11.72.71.1

3.11.00.7

Plastics 17.5 0.6Rubber & Leather 5.6 0.4

Textiles 5.5 0.6

Wood 12.4 1.2Other Materials 3.0 0.6Food Wastes 12.5 Nil

Yard Trimmings 29.8 5.9Miscellaneous Inorganic

Wastes2.8 Nil

Total MSW 187.7 40.8

Figure I.9 illustrates the energy values of various constituents found in solid wastestreams. Figure I.10 illustrates plastics that can be recycled.

HAZARDOUS WASTES. A waste is hazardous if (1) it is, or contains, a listed waste; or (2) it demonstrates any of the Following 4 characteristics: ignitability,corrosivity, reactivity, and toxicity; or (3) it is otherwise capable of causingenvironmental or health damage if disposed of improperly.

Ignitability. Ignitable wastes can create fires under certain conditions, arespontaneously combustible, or have a flash point less than 60°C (140°F). Examplesinclude waste oils and used solvents.



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Figure I.9. Energy values of various constituents in solid waste streams.

Figure I.10. Recyclable plastics chart.

Corrosivity . Corrosive wastes are acids or bases (pH < 2, or > 12.5) that are capable ofcorroding metal containers, such as storage tanks, drums, and barrels. Battery acid is anexample.



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Reactivity. Reactive wastes are unstable under "normal" conditions. They can causeexplosions, toxic fumes, gases, or vapors when heated, compressed, or mixed with water.Examples include lithium-sulfur batteries and explosives.

Toxicity. Toxic wastes are those containing concentrations of certain substances inexcess of regulatory thresholds which are expected to cause injury or illness to human

health or the environment. There are generally three types of toxic entities; chemical, biological, and physical. Chemical toxicants include inorganic substances such as lead;mercury; hydrofluoric acid; and chlorine gas, organic compounds such as methylalcohol; most medications; and poisons from living things. Biological toxicants include

bacteria and viruses that can induce disease in living organisms. Biological toxicity can be difficult to measure because the "threshold dose" may be a single organism.Theoretically one virus, bacterium, or worm can reproduce to cause a serious infection.However, in a host with an intact immune system the inherent toxicity of the organism is

balanced by the host's ability to fight back; the effective toxicity is then a combination of both parts of the relationship. A similar situation is also present with other types of toxicagents. Physical toxicants are substances that, due to their physical nature, interfere

with biological processes. Examples include coal dust and asbestos fibers, both of which

can ultimately be fatal if inhaled.

Examples of hazardous wastes include: (1) spent halogenated solvents used fordegreasing, e.g., trichloroethylene (TCE) and methylene chloride; (2) spent non-halogenated solvents such as xylene, acetone, and ethylbenzene; (3) wastewatertreatment sludges from electroplating operations; (4) dewatered air pollution controlscrubber sludges from coke ovens and blast furnaces; (5) sludges generated during theproduction of various chromium compounds; and (6) API (American PetroleumInstitute) separator sludges from petroleum refineries.

Cradle-to-Grave Hazardous Waste Management (Figure I.10). The EPA’scradle-to-grave hazardous waste management system is an attempt to track hazardous

waste from its generation point (the cradle) to its ultimate disposal point (the grave).The system requires generators to attach a manifest (itemized list describing thecontents) form to their hazardous waste shipments. This procedure is designed toensure that wastes are directed to, and actually reach, a permanent disposal site. Underthe cradle-to-grave concept, a generator of hazardous wastes can no longer avoid theliability by contracting with a third party to dispose of the wastes. Even if it can beshown that the wastes were mishandled through the actions of a third party, the originalgenerator will remain liable for improper disposal. This compels generators to exercisecare in the selection of the disposal companies they utilize.

Example I.2 (Site audit of hazardous waste inventories). The state agency hasaccused your waste transfer facility of accidental PCB (polychlorinated biphenyl) spillage

over the past year that has contaminated the upper 5 cm of the soils in the facility (~ 140,000kg of soil) with PCBs > the regulatory action level of 50 ppm. Since measurements of soilsamples for PCBs will cost $60,000, you instead decide to perform a waste audit on PCBinventories at the facility over the past year. Can you refute the agency’s claims? BecausePCBs have low volatility and water solubility, volatilization and leaching losses are negligible.The facility generates a wastewater stream with a mean flow of 890,000 L/day and a meanPCB concentration of 0.002 mg/L.



19

Figure I.10. Hazardous waste management (the cradle-to-grave approach).

◙ RCRA Disposal Manifests:

Manifest # Volume (L) PCB Concentration (mg/L) Mass (kg)

0124 1,890 22,100 41.8

0125 1,890 18,400 34.8

0126 1,890 9,550 18.0

0127 1,890 6,420 12.1

◙ Incoming Shipments of Transformer Oils:

Invoice # Volume (L) PCB Concentration (mg/L) Mass (kg)01137 208 42,000 8.75 (208×42,000×10-6)

01138 1,740 1,300 2.27

01139 795 22,000 17.5

01140 3,260 8,200 26.7

01141 1,440 38,000 54.6

1. Mass of PCBs delivered to the waste transfer facility: 8.75 + 2.27 +17.5 + 26.7 +54.6 = 109.8 kg.

2. Mass of PCBs leaving the waste transfer facility:

2.1. Industrial wastewater: (1,890,000 L/day)(0.002 mg/L)(365 days) = 1.38 kg.2.2. RCRA manifests: 41.8 + 34.8 + 18.0 + 12.1 = 106.7 kg.3. Total (Wastewater + RCRA): 108.1 kg.4. Total mass unaccounted for: 109.8 − 108.1 = 1.7 kg.5. Total mass that triggers regulatory action: (50 mg/kg)(140,000 kg) = 7.0 kg >

1.7 kg. Therefore, a case could be presented to the state regulatory agency thatthe losses do not provide sufficient PCB concentrations in the soil to exceed theregulatory action level.

U.S. EPA

Generator (cradle)

Transporter

Treatment/Storage/Disposal Facility (grave)



20

U.S. ENERGY USE. The U.S. is the largest energy consumer in terms of total use,using 1.05×1017 kJ (1 kJ = 1,000 J) in 2005. This is three times the consumption by theU.S. in 1950. The U.S. ranks 7th in energy consumption per-capita after Canada and anumber of small countries. The majority of this energy is derived from fossil fuels: in2005, it was estimated that 40% of the nation's energy came from petroleum, 23% fromcoal, and 23% from natural gas. Nuclear power supplied 8.4% and renewable energysupplied 7.3%, which was mainly from the hydropower. Energy consumption hasincreased at a faster rate than energy production over the last 50 years in the U.S., andthe difference is now largely met through imports. The per capita energy consumption inthe U.S. has been somewhat consistent from the 1970s to today (i .e., 3.27×108 kJ/capita-

year, from 1980 to 2006). Major reasons for this include the significant improvementsin manufacturing efficiencies, better energy conservation practices, and overseasproductions of many consumer goods purchased by the U.S. consumers. However, U.S."off-shoring" of manufacturing is sometimes exaggerated, because U.S. domestic

manufacturing has grown by 50% since 1980.

Industrial (37%) 1. Primary Metals (26.1%).2. Chemicals (19.3%).3. Petroleum and Coal (13.8%)4. Paper (7.9%).5. Stone and Glass (6.9%).6. Other Industry (20.6%).

Transportation (28%)1. Highway Vehicles (74%).2. Air (14%).3. Marine (7%).4. Rail (5%).

World Proven Conventional Natural Gas (NG) Reserves (can last ~ 60 years).

Former U.S.S.R. 44%Iran 15%

U.S.A. 6%Saudi Arabia 4%

Algeria 3%Other 28%

Shale Gas (Figures I.11 and I.12). Shale gas is the natural gas that is being trapped within shale formations. Shale gas has become an increasingly important source of

natural gas in the U.S. since the start of the 21 th century. In 2000 shale gas providedonly 1% of U.S. natural gas production; by 2010 it was over 20% and the U.S.government's Energy Information Administration predicts that by 2035 shale gas willaccount for about 46% of the natural gas supply in the U.S. Shale gas is one of a numberof unconventional sources of natural gas; others include coal-bed methane, tightsandstones, and methane hydrates.

PCBs



21

Figure I.11. The location of shale gas.

(1) Marcellus (141 tcf), Haynesville (66 tcf), Eagle Ford (50 tcf), Barnett-Woodford (27 tcf),and Woodford (24 tcf).

(2) Total for Continental U.S. Shale Gas Reserve: 542 tcf.(3) Annual U.S. NG Consumption (2011): 2.43 tcf.(4) 1 tcf = 1×1012 ft3 = 2.83×1010 m3.

Figure I.12. Shale gas deposits in the continental U.S.

Since shales have low permeability to allow significant fluid flow to a well bore, they arenot commercially viable sources of natural gas until the technology of hydraulicfracturing (fracking) matures after nearly 20 years of development. In general, thetechnology of fracking has two parts, vertical drilling and horizontal drilling. First, a wellis drilled down to the shale, and then the well is drilled horizontally up to 3,000 m within



22

the shale to create maximum borehole surface in contact with the shale. A fluid mix of water, sand, and chemicals is pumped down the well at the high pressure, creatingfissures in the shale that permit gas flow to the well. Pathways may be created during the

whole drilling process that allow gas or chemicals to enter the streams and groundwateraquifers. Contaminated water from fracking is often stored in surface ponds, which canoverflow or leak, polluting streams or groundwater. Wells are usually reinforced with

steel casing and sealed with concrete. But poor cementing can leave gaps that allowmethane or fracking chemicals to contaminate the groundwater aquifers. Frackingfissures might connect to natural ones, allowing pollutants to migrate. Whether they

would be able to move hundred meters upwards to reach shallow aquifers is unclear.

World Proven Conventional Oil Reserves ( ~ 4 trillion barrels).

Saudi Arabia 18.4%Iraq 11.0%

United Arab Emirates (UAE) 10.8%Iran 10.6%

Kuwait 10.3%

Former U.S.S.R. 6.5% Venezuela 6.3%Mexico 6.2%U.S.A. 3.5%Other 16.4%

Oil Sands (Figure I.13). Oil sands, tar sands or, more technically, bituminous sands,are unconventional oils. The oil sands are loose sand or partially consolidatedsandstone containing naturally occurring mixtures of sand, clay, and water that aresaturated with bitumen (or tar). Natural bitumen deposits are reported in manycountries, but in particular are found in extremely large quantities in Canada. Otherlarge reserves are located in Kazakhstan and Russia. The estimated deposits in the U.S.could be as much as 2 trillion barrels. The estimates include deposits that have not yet

been discovered; proven reserves of bitumen contain approximately 100 billion barrels.Total natural bitumen reserves are estimated at 249.67 billion barrels (39.7×109 m3)globally, of which 176.8 billion barrels (28.1×109 m3), or 70.8%, are in Canada. U.S.consumed roughly 6.9 billion barrels of oil (18.8 million barrels per day) in 2011.

Making liquid fuels from oil sands requires energy for steam injection and refining. Thisprocess generates a 12 percent higher amount of greenhouse gases per barrel of finalproduct as the "production" of conventional oil. The oil sands must be extracted by stripmining or the oil made to flow into wells by in-situ techniques, which reduce the

viscosity by injecting steam, solvents, solvents, and/or hot air into the sands. Theseprocesses can use more water and require larger amounts of energy than conventional oilextraction, although many conventional oil fields also require large amounts of water

and energy to achieve good rates of production. After excavation, hot water and causticsoda (NaOH) is added to the sand, and the resulting slurry is piped to the extractionplant where it is agitated and the oil skimmed from the top. Provided that the waterchemistry is appropriate to allow bitumen to separate from sand and clay, thecombination of hot water and agitation releases bitumen from the oil sands, and allowssmall air bubbles to attach to the bitumen droplets. The bitumen froth floats to the topof separation vessels, and is further treated to remove residual water and fine solids.



23

About 2,000 kg of oil sands are required to produce one barrel of oil. Originally, roughly75% of the bitumen was recovered from the sand.

Figure I.13. Oil sands.

However, recent enhancements to this method include Tailings Oil Recovery (TOR) units which recover oil from the tailings, Diluent Recovery Units to recover naptha from thefroth, Inclined Plate Settlers (IPS) and disc centrifuges. These allow the extractionplants to recover well over 90% of the bitumen in the sand. After oil extraction, thespent sand and other materials are then returned to the mine, which is eventuallyreclaimed.

LAND USE. The United States has a total land area of nearly 2.3 billion acres. Majoruses in 2002 were forest-use land, 651 million acres (28.8 percent); grassland pastureand range land, 587 million acres (25.9 percent); cropland, 442 million acres (19.5percent); special uses (primarily parks and wildlife areas), 297 million acres (13.1percent); miscellaneous other uses, 228 million acres (10.1 percent); and urban land, 60million acres (2.6 percent). Soil, water, air, minerals, biota, and energy are resourcesessential to human. These resources are fundamentally dependent on land.

1. Land contains water the largest reservoir of fresh water, both surface water andgroundwater.

2. Land supports all terrestrial biota, both natural and domesticated. This biota

provides over 90% of food, all timber, and natural fibers that human consumes.3. By supporting natural biota and ecosystems, land provides natural servicesaffecting water and air quality.

4. Nearly all mineral resources are mined from the land.5. Coal, oil, natural gas, and uranium mined from the land. In addition, land

provides space for hydroelectric reservoirs, power plants, wind mills, and solarinstallations.

6. Land provides space for all human activities, homes, stores, schools, offices,factories, roadways, parking lots, airports, landfills, and other facilities that make



24

up cities and towns and connections between them. Also, land is used forrecreational activities and aesthetic pleasure. Even water-based activities requireland facilities such as beaches and marinas.

Overview of Land Use in the U.S. The U.S. has 2.3 billion acres of land. However,375 million acres are in Alaska and not suitable for agricultural production. The land

area of the lower 48 states is approximately 1.9 billion acres. To put things inperspective, keep in mind that California is 103 million acres, Montana 94 million acres,Oregon 60 million acres and Maine 20 million acres. Only 66 million acres in the U.S. (~3%) are considered developed lands, which is home to 75% of the population in thecountry. In general, urban lands are nearly useless for biodiversity preservation.Moreover, urbanized lands, once converted, usually do not shift to another use.

Rural residential land comprises nearly all sprawl and subdivisions along withfarmhouses scattered across the country. The total acreage for rural residential is 73million acres. Of this total, 44 million acres are lots of 10 or more acres.

About 349 million acres in the U.S. are planted for crops. This is about 4 times the size

of Montana. Four crops, namely, feeder corn (80 million acres), soybeans (75 millionacres), alfalfa hay (61 million acres), and wheat (62 million acres), make up 80% of totalcrop acreage. All but wheat are primarily used to feed livestock. The amount of landused to produce all vegetables in the U.S. is less than 3 million acres.

Some 788 million acres, or 41.4% of the U. S. excluding Alaska, are grazed by livestock.This is an area the size of 8.3 times the size of Montana. Grazed lands include rangeland,pasture, and cropland pasture. More than 309 million acres of federal, state, and otherpublic lands are grazed by domestic livestock. Another 140 million acres are forestedlands that are grazed. Forest lands comprise 747 million acres. Of these lands, some 501million acres are primarily forest (minus lands used for grazed forest and other specialcategories).

The USDA report concludes that urbanization and rural residences (subdivisions) "donot threaten the U.S. cropland base or the level of agricultural production." This doesnot mean sprawl doesn't have impacts where it occurs. But the notion that sprawl is thegreatest threat to biodiversity is false. Figure I.14 shows the urban sprawl in Las Vegas,Nevada.



25

Figure I.14. Urban Sprawl (Las Vegas, NV).



26

II. PRINCIPLES OF MASS AND ENERGY BALANCES

Mass (material) and energy balances are key tools in understanding the fate of acontaminant in the environmental compartments under a wide variety of conditions.The law of conservation of mass states that mass can neither be produced nor

destroyed . This law means that if the amount of a chemical somewhere (e.g., in a lake)increases, then that increase cannot be the result of some magical formation. Thechemical must have been either carried into the lakefrom elsewhere or produced viachemical and/or biological reactions from other compounds that were already in thelake. Similarly, if reactions produced the mass increase of this chemical, they must alsohave caused a corresponding decrease in the mass of some other compound(s). Inaddition to the utilization of materials for various purposes, modern society also dependson the extensive uses of energy which require transformations in the form of energy andcontrol of energy flows. Energy takes many forms and the movement of energy andchanges in its forms can be tracked using energy balances, which are analogous to mass

balances. The 1st law of thermodynamics states that energy can neither be producednor destroyed . However, all energy balances are treated as conservative; as long as allpossible forms of energy are considered (in the absence of nuclear reactions), there is noterm in energy balances that is analogous to the chemical/biological reaction term inmass balances.

CONTROL VOLUME (c.v.). Mass and energy balances are only meaningful in termsof a specific region in the space, which has boundaries across which the terms of input(s)output(s), heat flow(s), and work flow(s) (i .e., flow work and shaft work) can be defined,determined, and quantified. Theoretically, any volume of any shape and location can beused as a control volume (c.v.). Realistically, however, certain control volumes are moreuseful than others. The system boundary defined can be real, imaginary, stationary, ormoving. Figure II.1 illustrates the schematic of a control volume along with the terms to

be included.

Figure II.1. Schematic of a control volume.

MASS BALANCES. The law of conservation of mass for a chemical species i can beapplied to the control volumes illustrated in Figure II.1 and written in the followinggeneral way. However, the characteristics of the c.v. need to be clearly defined prior to

CONTROL VOLUME◙ Accumulation◙ Generation◙ Consumption

Input(s) Output(s)

System Boundary



27

the formulation of the mass balance equations that will be useful for practicalapplications.

Accumulation rate of species i in the c.v. (mass/time, or M/T) = Input(s) of species i tothe c.v. (M/T) – Output(s) of species i from the c.v. (M/T) + Generation rate of species i in the c.v. (M/T) - Consumption rate of species i in the c.v. (M/T).

A Completely Mixed Control Volume. In many cases large quantities of energy in various forms can be transported, via artificial or natural means, across the system boundary from the surroundings into the control volume. Such energy input(s) renderthe contents in the control volume to be completely mixed and uniform. As a result, thecharacteristics of the output(s) will be identical to those observed in the control volume.

A number of assumptions can be made to facilitate the mathematical derivation of mass balance equations for a completely mixed control volume:

1. Species i is being carried into and out of the c.v. by a single slow of a specific

environmental medium (e.g., water or air) at a constant volumetric flow rate ( ,

V

volume/time, or L3

/T).2. Species i is being consumed in the c.v. at a rate that is proportional to its mass m according to the first-order rate expression –km = –kVC , where k is the first-order degradation rate coefficient, T-1; V is the volume of the c.v., volume, or L3;and C is the concentration of species i in the c.v., M/L3.

3. No generation of species i in the c.v.4. The volume of the c.v. (V ) is constant.

Therefore,

V

m

V

mC

kC C V

V C

V

V

dt

dC

kVC C V C V dt

dC V

dt

VC d

kmmmdt

dm

i

i

vcou t invc

)(

....

(II.1)

where

m is the mass rate (or mass flux) of species i , M/T; and iC is the input

concentration of species i , M/L3. The first-order rate expression for the consumptionterm is commonly used to describe the rate at which a contaminant is removed in aspecific environmental compartment. A similar expression can also be used to define thegeneration term. (II.1) is the mass balance equation for a control volume which is



30

Example II.2 ( A Smoking Bar). A bar (V = 500 m3) has 50 smokers in it, eachsmoking 2 cigarettes per hour. An individual cigarette emits about 1.4 mg formaldehyde(HCHO). HCHO converts to CO2 at a first-order rate of 0.4 hr-1. Fresh air enters the barat 1,000 m3/hr, and stale air leaves at the same rate. Assume complete mixing, estimatethe steady-state HCHO concentration in the air inside the bar. At 25oC and 1 atm, howdoes the result compare with the threshold for eye irritation of about 0.5 ppmv?

1. At steady state, input = output + consumption (or degradation).2. Input = (50)(2)(1.4) = 140 mg/hr (note that this input is the internal emission

and that the incoming fresh air is free of HCHO).3. Output = (1,000 m3/hr)(C mg/m3) = 1,000C mg/hr.4. Consumption = kCV = (0.4 hr-1)(500 m3)(C mg/m3) = 200C mg/hr.5. Therefore, 140 mg/hr = (1,000 + 200 m3/hr)(C mg/m3) → C = 0.117 mg/m3.

6. According to (I.2),)30)(15.273(

)414.22)(2515.273)(117.0( C = 0.095 ppmv > 0.05

ppmv → the steady-state HCHO concentration in the air inside the bar will causeeye irritation.

Example II.3. The air in the bar of Example II.2 is clean when it opens at 5PM. Ifformaldehyde is emitted from cigarette smoking at a constant rate of 140 mg HCHO/hrstarting at 5 PM, what would the concentration be at 6PM?

1. This is a non-steady-state problem. Since formaldehyde is introduced directly viacigarette smoking at a rate of 140 mg/hr, (II.1) should be modified to:

kVC C V dt

dC V

140 (a)

2. (a) can be integrated with initial condition (C = 0 at t = 0), then

t V

t kV V

ee

kV V

t C 4.21117.01140

)(

( b)

3. If 5 PM is taken as t = 0, then 6 PM is t = 1 hr. Therefore, C = 0.106 mg/m3.

Note that state stead yC C

kV V

)(140

.

Example II.4 (CFSTR vs PFR ). Calculate the respective volumes of a CFSTR system and aPFR system that are required to remove 95% of a pollutant. The first-order decay ratecoefficient is 0.2 day -1.

1. For a CFSTR system,

05.01

1

k C

C

i

95 days.



31

2. For a PFR system, *05.0)( *

t eC

xC kt

i

= 15 days. Therefore, a first-order

PFR system is always more efficient than a first-order CFSTR system, i .e.,

.33.615

95

PFR

CFSTR

V

V

Example II.5 (Burning of Biogas). Biogas, which is about 70% CH4 and 30% CO2, isproduced from the anaerobic fermentation of organic sludge. The biogas can be used as a fuelif the corrosive sulfur compounds (e.g., H2S) are in low concentrations or are removed.

Assuming that this is done, make the mass balance for combustion in the presence of 10%excess O2 to insure complete combustion. The source of O2 is air, which is 79% by weight N2 and 21% by weight O2. Use 1,000 kg input of dry biogas as the basis for calculations. Thecombustion reaction is: O H COOCH 2224 22 .

1. 700 kg CH4 will consume

16

)64)(700( = 2,800 kg O2 → Feed O2 = (2,800)(1.1) =

3,080 kg O2 → Feed N2 =21

)79)(080,3( = 11,586 kg N2. CO2 produced =

16

)44)(700( = 1,925 kg CO2. H2O produced =

16

)36)(700( = 1,575 kg H2O.

Therefore,

Inputs Outputs

700 kg CH4 0 kg CH4

300 kg CO2 2,225 (1,925 + 300) kg CO2

3,080 kg O2 280 (3,080 − 2,800) kg O2

11,586 kg N2 11,586 kg N2

0 kg H2O 1,575 kg H2O

Total: 15,666 kg gas Total: 15,666 kg gas

4. Inputs = Outputs.

ENERGY BALANCES. When dealing with the energy problems, two fundamentalthermodynamics laws must prevail:

1st Law of Thermodynamics: Although energy assumes many forms, the totalquantity of energy is constant, and when energy disappears in one form it appearssimultaneously in other forms.

2nd Law of Thermodynamics: The 2nd law of thermodynamics can be stated indifferent manners. Two most cited statements are:1. No equipment can be designed and operated for the sole purpose of

converting all energy inputs it received into useful work.



32

2. It is impossible to devise a machine that can convert the energy from oneform to another with 100% efficiency.

Energy Forms: Various forms of energy are summarized in Table II.1. u is the velocityof the working fluid, z is the elevation of the working fluid measured against a referencepoint, and U is the internal energy of the working fluid (or the energy content of the

working fluid). The absolute value of U is unknown and as a result, a reference statemust be defined at which U is arbitrarily set at zero. For instance, the reference state forcalculating U values for water at different conditions is liquid water [H2O(l )] @ 0oC and 1

atm.

X is the time rate of X , e.g.,

W is referred to as power which has the unit ofenergy/time. The terms of heat and work refer only to energy that is being transferred (i .e., the amount of heat energy added to or extract from the c.v.). It is meaningless tospeak of the heat and work possessed by or contained within the c.v. Heat is the energythat flows as a result of temperature difference between a c.v. and its surroundings.

Work is the energy that flows in response to any driving force other than a temperaturedifference (e.g., a force, a torque, or a voltage). Heat is defined as positive when it istransferred to the c.v. from the surroundings. Work is defined as positive when it is done

by the c.v. on the surroundings.Table II.1. Energy Forms.

Kinetic Energy2

2mu E k

2

2um E k

Potential Energy mgz E p gz m E p

Internal Energy U

U Heat Q

Q

Work W W

Energy and Power Units:

1. 1 Joule (J) = 1 N-m = 107 ergs = 107 dyne-cm = 0.23901 cal (cal: calories) =9.486×10-4 Btu (Btu: British thermal unit) (1 kJ = 103 J).

2. 1 Watt (W) = 1J/s (1 kW = 103 W = 103 J/s).3. 1 kW-h (kWh) = (103 J/s)(3,600 s) = 3,600 kJ.4. 1 horsepower (hp) = 746 W = 746 J/s.

Table II.2 summaries power consumptions of common household appliances.

Energy Balances on Steady-State, Continuous-Flow Systems. It is convenientto evaluate the rate of change of a specific energy form as the fluid streams ( i .e., air or

water) flow across the c.v. from the inlet locations to the outlet locations. If Δ = outlet −inlet, then

Table II.2. Power Consumptions of Household Appliances. Appliance Power (W) Hours Used/Year (year 2000)



33

Clock 2 17Clothes Dryer 4,600 1,049

Hair Dryer 1,000 60Light Bulb 100 108

Compact Fluorescent Light Bulb 18 19T.V. 350 504

Water Heater (150 L) 4,500 4,698Energy –Efficient Water Heater (150 L) 2,800 2,900Toaster 1,150 552

Washing Machine 700 1,008Refrigerator 360 2,160

Energy –Efficient Refrigerator 180 1,100

Q

W V P V P W W W

U U U

gz m gz m E

umum E

s

in j

j j

out j

j j s fl

in j

j

out j

j

in j

j j

out j

j j p

in

j

j j

out

j

j j

k

22

22

(II.5)

where the subscript j represents the j th stream, fl W

is the rate of flow work which is the

power required to move the fluid streams through the c.v., sW

is the shaft work thataccounts for the work transferred across the boundary of the c.v. by devices with moving

parts located at the boundary of the c.v. (e.g., a pump or a turbine); and

Q is the rate at

which heat is transferred across the system boundary. The definition of fl W

is based on

the observation that , Fut

x F

t

x A P

t

V P V P

where F is the force in the

direction of flow, A is the cross-sectional area which is perpendicular to F and across which the fluid flows, Δ x is the displacement in the direction of flow, and u is the fluid

velocity in the direction of flow. in j j V P )(

, which is the work done by the surroundings

on the j th fluid stream entering into the c.v., is negative (i .e., the fluid is being pushed

into the c.v. from the surroundings. ou t j j V P )(

is the work done by the c.v. on the j th

fluid stream exiting the c.v. and therefore, it is positive (i .e., the fluid is being pushed out

of the c.v.). By the same token, sW

is positive if the devices with moving parts extract

the energy from the c.v., and sW

is negative if the devices with moving parts deliver theenergy from the surroundings into the c.v. At steady state, the sum of the rates of change



34

in kinetic, potential, and internal energies between input and output locations is equal tothe sum of work and heat flows across the system boundary, or

Accumulation rate of energy in the c.v. = Input(s) of energy to the c.v. − Output(s) of energy out ofthe c.v.→ Input(s) = Output(s) (steady state).

Therefore,

input pjkj joutput pjkj j

input pjkj j j j

output pjkj j j j s

E E H E E H

E E V P U E E V P U W Q

)()(

])[(])[( (II.6a)

s pk W Q E E H

(II.6)

where H = U + PV is termed as enthalpy that can be taken as the total energy value of the working fluid at given P and T . Also,

V P U H (II.7)

where

m

H

m

H H is the specific enthalpy which is an intensive property that is

independent of the quantity of the substance present. The absolute values of

H and

U

m

U

m

U are not known, only

H and

U can be determined by referring to a

reference state that can be arbitrarily selected for convenience.

V is the reciprocal ofdensity (i .e.,

V =m

V or

n

V ) (n is the number of moles of the species concerned). For

liquid water [H2O(l )] at low pressures (0−100oC), T C m H p

, where C p is the

specific heat capacity of water at constant pressure (4.184 J/g-oK, or 75.4 J/mol-oK) and∆T is the temperature difference between input and output locations.

Mechanical Energy Balances. In some flow processes the internal energy values ofthe fluid streams remain constant as they enter into and exit from a c.v. As a result,energy exchanges across a c.v. occur among kinetic, potential, and pressure energy terms(e.g., water flow in a pipe, hydropower generations from reservoirs, etc.). In such cases,

(II.6) is modified to either (II.8) or (II.9) when a single fluid stream ( j = 1) with aconstant mass density ( ) flows across a c.v. at a constant mass rate (i .e.,

mmm input output ):



35

m

W

m

QU z g

u P s

2

2

(II.8)

mg

W

g m

Q

g

U z g

u

g

P s

2

2

(II.8a)

02

2

z g u P

(Bernoulli’s Equation) (II.9)

Note that

1

V and

m

QU is the friction loss.

U is the amount of energy

exchanges that is caused by the viscous action between fluid particles.

m

Q is the amount

of energy dissipated from the system that is caused by the frictional resistancesencountered by the flowing fluid stream (e.g., water that flows in a conduit willencountered the frictional resistances at the inner wall of the conduit). Each term in

(II.8a) has the unit of length and as a result, g

P

is the pressure head,

g

u

2

2

is the

velocity head, z is the potential head, and

g m

Q

g

U is the head loss. The sum of

pressure head, velocity head, and potential head is total head. The left side of (II.8a)must be < 0 in order for the energy stored in the incoming fluid stream to be recovered

using the device such as a turbine. If the left side of (II.8a) is > 0, then the fluid streamcannot flow from the inlet to outlet without using the device such as a pump to obtainadditional energy input. If (II.8a) = 0, then the fluid stream can flow from the inlet tooutlet by it self but no energy can be extracted from the fluid stream. (II.9) applies toideal flow conditions (i .e., no friction loss) and without moving parts located at the

boundary of a c.v.

Thermal Processes. Enthalpy changes for processes that involve phase changes (e.g.,from liquid phase to vapor phase), chemical reactions (e.g., combustion of fuels), or largetemperature changes are usually very large as compared to kinetic and potential energychanges. As a result, (II.6) can be modified to:

in j

j j

out j

j j

s

H m H m H

W Q H

(II.10)

If the c.v. is well insulated from the surroundings, then it is adiabatic and

Q = 0. sW

=0 if there are no moving parts located at the boundary of the c.v. Note that the numberof incoming fluid streams may be different from that of outgoing fluid streams. TableII.1 shows the specific enthalpy values of water and vapor at selected temperatures and



36

pressures. The P -T pair in Table II.3 represents the conditions at which liquid water(saturated water) and water vapor (saturated vapor) can coexist in differentproportions according to:

satur ated vO H satur ated l O H

satu rated vO H

satu rated vO H satu rated l O H

satu rated vO H

mm

m x

mm

m x

)()(

)(

)()(

)(

22

2

22

2

(II.11)

where x is the quality of the water-vapor mixture, satu rated vO H

m)(2

[ saturated vO H m )(2

] is the

mass (mass rate) of saturated water vapor in the mixture, and satur ated l O H

m)(2

[ saturated l O H m )(2

] is the mass (mass rate) of saturated liquid water in the mixture. x = 1

indicates that saturated liquid water is absent in the mixture whereas x = 0 indicates the

absence of saturated water vapor in the mixture. The reference state for Table II.1 is

liquid water at 0oC and 1 atm [i .e.,

C H ol water 0@)( 0 kJ/kg ( P = 1 atm)]. For instance,

at P = 100 kPa and T = 99.6oC water can exist as liquid water (saturated water) with aspecific enthalpy value of 417.5 kJ/kg, or as vapor (saturated vapor) with a specificenthalpy value of 2,675.4 kJ/kg, or a mixture of water and vapor with a specific enthalpy

value calculated by (II.11) and (II.12). The specific enthalpy value of the mixture

[ ),( P T H mixture

] is:

),()1(),(),( )()( P T H x P T H x P T H l water vwater mixture

(II.12)

Sensible and Latent Heats. The specific enthalpy changes (

H ) associated withtemperature changes for a single-phase system is referred to as the sensible heat ofthe temperature change. The specific enthalpy changes associated with thetransition of a substance from one phase to another at constant temperature andpressure is referred to as the latent heat of the phase change. Figure II.2 illustratesthe heat needed to convert 1 kg ice to steam at 1 atm. To change the temperature of 1 kgof ice, 2.1 kJ/oC are needed (sensible heat requirement that change the temperature ofthe ice which remains as a solid). To completely melt that ice requires another 333 kJ(latent heat of melting, since the temperature remains constant at 0oC while the icechanges from the solid state to liquid state). Raising the temperature of that liquid waterrequires 4.184 kJ/oC (sensible heat), and converting it to steam requires another 2,257kJ (latent heat of vaporization). To raise the temperature of 1 kg steam (at 1 atm)requires another 2.0 kJ/oC. Table II.4 shows the latent heats of evaporation of water

[ v H

] at selected temperatures which are independent of P (low pressures). The

reference state for Table II.2 is liquid water at 0oC [i .e.,

C H ol water 0@)( 0 kJ/kg ( P = 1

atm)]. The latent heat of evaporation of water is calculated as:



37

Table II.3. Specific Enthalpy Values of Saturated Water [water(l )] and Vapor [water(v)]. P (kPa) T (oC)

)(l water H

(kJ/kg) )(vwater H

(kJ/kg)

1 7.0 29.3 2,514.45 32.9 137.8 2,561.6

10 45.8 191.8 2,584.8

50 81.3 340.6 2,646.0100 99.6 417.5 2,675.4200 120.2 504.7 2,706.3400 143.6 604.7 2,737.6600 158.8 670.4 2,755.5800 170.4 720.9 2,767.5

1,000 179.9 762.6 2,776.22,000 212.4 908.6 2,797.25,000 263.9 1,154.4 2,794.2

10,000 311.0 1,408.0 2,727.715,000 342.1 1,611.0 2,615.020,000 365.7 1,826.5 2,418.4

◙ 1 atm = 100kPa. 1 Pa = 1 N/m2. 1 atm = 1.01325×105 Pa = 101.325 kPa = 1.01325 bar.

◙ ).,()0(),(),( )()()()( P T H C H P T H P T H vwater o

l water vwater vwater

A: mixture of ice/water; B: liquid water (subcooled liquid ); C : mixture of water/vapor (saturatedwater/vapor); D: steam (superheated steam), 1: 100% saturated water (liquid), and 2: 100%saturated steam (vapor).

Figure II.2. Heat needed to convert 1 kg ice to 1 kg steam.

),(),(),( )()( P T H P T H P T H l water vwater v

(II.13)

For instance, at P = 100 kPa and T = 99.6oC, )(l water H

= 417.5 kJ/kg and )(vwater H

=

1 1 2 2

A A

B B

C C

D D

T T

H H eeaat t A Ad d d d eed d



38

2,675.4 kJ/kg (Table II.1). Therefore, v H

= 2,675.4 − 417.5 = 2,258 kJ/kg, which isthe value in Table II.2 at T = 100oC.

Example II.6 (Power for the Hydrologic Cycle). Global rainfall has beenestimated to average about 1 m per year across the entire 5.10×1014 m2 of the earth’s

surface. Find the energy required to cause that much water to evaporate each year(average global temperature is about 15oC). Compare this to the estimated 1987 worldenergy consumption of 3.3×1017 kJ and compare it to the average rate at which sunlightis absorbed at the surface of the earth, 168 W/m2.

1. The energy required to evaporate 1 kg of water at 15oC is 2,466.2 kJ (Table II.2,the average of values @ 14 and 16oC).

2. The total energy requires to vaporize all of that water is: (1 m/yr)(5.10×1014 m2)(1,000 kg/m3) (2,466 kJ/kg) = 1.25×1021 kJ/yr, which is roughly 4,000times of the energy consumed worldwide in 1987. Averaged over the globe, theenergy required to power the hydrologic cycle is: (1.25×1021 kJ/yr)(1 W/J/s)(365 days/yr)-1(86,400 s/day)-1 = 78 W/m2 (~ 46% of the average rate at which

sunlight is absorbed at the earth’s surface). Table II.4. Latent Heats of Evaporation of Water at Low Pressures.

T (oC) Latent Heat of Evaporation (kJ/kg) T (oC) Latent Heat of Evaporation (kJ/kg) 0 2,501.6 52 2,3772 2,496.8 54 2,3734 2,492.1 56 2,3686 2,487.4 58 2,3638 2,482.6 60 2,35810 2,477.9 62 2,35312 2,473.2 64 2,34814 2,468.5 66 2,34316 2,463.8 68 2,338

18 2,459.0 70 2,33320 2,454.3 72 2,32922 2,449.6 74 2,32324 2,444.9 76 2,31825 2,442.5 78 2,31326 2,440.2 80 2,30828 2,435.4 82 2,30330 2,430.7 84 2,29832 2,425.9 86 2,29334 2,421.2 88 2,28836 2,416.4 90 2,28238 2,411.7 92 2,27740 2,406.9 94 2,272

42 2,402.1 96 2,26744 2,397.3 98 2,26246 2,392.5 100 2,25848 2,387.750 2,382.9



39

III. POWER GENERATION AND ENERGY EFFICIENCY

U.S. ENERGY CONSUMPTIONS (sources).

Source %

Oil 40Coal 23Natural Gas (NG) 23

Nuclear 8Hydropower 2.7

Others 3.3

THE COST OF A KILOWATT-HOUR (kW-h).

Source ¢/kW-hCoal 5.0NG 5.0

Wind 6.0

Nuclear 7.0Solar 22.0

NEW YORK CITY WITHOUT FOSSIL FUELS (60% of NYC’s electricityis produced from coal and natural gas).

Solar 74 mi2 (145,225,714 64”×32” 175W -panels) Wind 10.6 mi2 (6,800 1.5MW-turbines)

Nuclear 2 mi2 (4 1,000-MW reactors)

ENERGY CONTENTS OF MSW (Table III.1) The energy content of MSW depends on the materials that it contains as well as its moisture content. The standardtest used to determine the heating value of a material involves completely burning asample in a bomb calorimeter and then measure the rise in temperature of a surrounding

water bath.

Table III.1. MSW Energy Contents (without removing moisture). Material kg kJ/kg kJ

Paper & Paperboard 31.7 15,800 500,860 Yard Trimmings 16.2 6,300 102,060

Plastics 11.5 32,800 377,200Food Wastes 8.5 5,500 46,750

Wood 7.6 16,000 121,600Metals 7.4 -- --

Glass 6.6 -- --Rubber & Leather 3.6 22,300 80,280Textiles 3.3 18,700 61,710

Miscellaneous 3.6 -- --TOTAL 100 -- 1,290,460



40

The following equations are useful in estimating the heating values of MSW :

HHV (kJ/kg) = 53.5( F + 3.6CP ) + 372 PLR (III.1)

Q L = 2,440(W + 9 H ) (III.2)

LHV = HHV − Q L (III.3)

where HHV is the higher heating value, kJ/kg; F is mass % of food, %; CP is the mass %of cardboard and paper, %; PLR is the mass % of plastics, rubber, and leather, %; QL is

the latent heat of vaporization of water, kJ; W is the amount of moisture in the waste, kg;

H is the amount of hydrogen in the dry waste, kg; and LHV is the lower heating value,kJ/kg. Only in unusual circumstances can the HHV be captured. Most likely, Q L is lost

to the atmosphere and therefore, unrecoverable. (III.2) indicates there are two sources of

water vapor losses: moisture in the waste and hydrogen in the waste that reacts withoxygen to form water. 2,440 kJ/kg (or 44 kJ/mol) is the latent heat of vaporization of

water at 25oC.

Example III.1. Typical MSW has a moisture content of around 20%, and roughly 6% of thedry mass of MSW is hydrogen. Estimate the lower heat value.

1. The total heating value of 100 kg MSW is 1,290,460 kJ and therefore, the HHV is1,290,460 kJ/100 kg or 12,900 kJ/kg.

2. In 1 kg waste, there will be 0.2 kg moisture and 0.8 kg of dry waste. In that dry waste, there will be 0.8×0.06 = 0.048 kg of hydrogen.

3. From (III.2), Q L = 2,440(0.2 + (9)(0.048)] = 1,540 kJ/kg.4. Therefore, LHV = 12,900 − 1,540 = 11,360 kJ/kg.

HYDROPWER GENERATION. Hydraulic turbines (i .e., water-driven turbines)are the modern incarnation of one of the oldest of energy conversion technologies, the

water wheel. Modern hydraulic turbines are designed to more efficiently capture theenergy of flowing water at hydroelectric dams and natural waterfalls like Niagara Falls.Figure III.1 illustrates a simple sketch of a hydroelectric facility. Water collected behinda dam is carried through large ducts to the inlet of one or more hydraulic turbines at the

base of the dam. Since the reservoir behind a dam covers a very large surface area, the water level in the reservoir remains unchanged as a result of the release of water forhydropower generation. Therefore, the energy available for the conversion to electricityis the potential energy of the impounded water. Moreover, since the internal energy of

water changes very little between the air-water interface behind a dam (point 1) and tail water (point 2), and the fact that the surface area of tail water is also large because ofthe requirements of reducing the velocity of water after it exits the turbine house tominimize the potential damages to the downstream infrastructure, then

z z z z

uuu

P P P P P atmatm

12

2

1

2

2

2

12

0

0

(III.4)



41

Figure III.1. Schematic of a hydropower plant.

Therefore, the mechanical energy balance equation can be expressed as:

g m

Q

g

U gz mW s (III.5)

The actual power delivered by the hydroelectric plant is sturbine W , whereturbine is the

turbine efficiency.

Example III.2. A hydroelectric plant produces 150 MW of electricity. The operatinghead is 35 m, and the water flow rate into the turbines is 500 m3/s. Determine theturbine efficiency. Assume that the friction loss between reservoir and tail water isnegligible

1. 000,500)000,1)(500(

m kg/s.

2.

turbineturbine

s

turbine

W 610150)35)(81.9)(000,500( 0.87.

A major attraction of hydropower generation is that water is a renewable energy source which is replenished by the natural cycle of evaporation and precipitation. At the sametime, however, there are only limited number of natural waterfalls and river flow capableof generating significant amounts of electricity. In the U.S., the most attractive sites forhydropower generation have already been exploited to provide the generating capacity of76,700 MW (as of 1997). Locations still remain where smaller, low-head hydropowergeneration facilities could be constructed, but such sites are typically in remote locations



42

where the difficulty and expense of transmitting the electricity diminish the benefits ofsuch projects, even in the absence of environmental concerns (e.g., long-term impacts onland use and ecosystem in the vicinity of hydroelectric dams and displacement ofpopulation and communities due to creation of reservoirs).

STEAM ELECTRIC (THERMAL POWER ) PLANTS. Steam turbines are the

most prevalent method used worldwide for spinning the shaft of an electromechanicalgenerator. High-temperature steam (superheated steam) can be generated to produce ahundredfold increase in the electrical output of a turbine generator compared to liquid

water alone. Steam turbines provide ~3

2 of all the electric power generated in the U.S.

Figure III.2 illustrates a basic steam electric plant schematic (i .e., the Rankine cycle).The closed-cycle arrangement permits the turbine exit pressure to fall below 1 atm

because a vacuum forms in the condenser as vapor turns to liquid. At the lowerpressure, condensation begins at temperatures below 100oC (see Table II.1). More usefulenergy can therefore be extracted in the turbine without encountering the problems of

water condensation.

Figure III.2. The schematic of a thermal power plant (the Rankine cycle).

There are four basic pieces of equipment in a typical steam electric plant: boiler, turbine,condenser, and pump (see the diagram attached). The boiler is used to generate thesuperheated steam at high P and T . Fuel and air (in excess of the stoichiometric amountrequired) are introduced through burners protruding from the walls or corners of the

boiler. The boiler itself resembles a large box which can be up to several stories high.The energy released from fuel combustion inside the combustion chamber heats the

boiler feed water, which flows through pipes embedded in the boiler walls. As the waterturns to steam, it is collected in a manifold and routed to the steam turbine. Tomaximize the boiler efficiency, the hot (~ 425oC) .combustion gases exiting thecombustion chamber [flue gas which contains O2, NOX, SO2, N2, H2O(v), andflyash] are passed through a heat exchanger, i .e., air preheater) to heat the incoming airstream. This reduces the amount of fuel needed to heat the air inside the combustionchamber. The boiler efficiency ( boiler ~ 88%) is:

H Q

C Q

sW

C W



44

(III.8), where 3

H is the specific enthalpy value of the saturated water exiting thecondenser. Note that T 2 = T 3 and P 2 = P 3.

32 H H mQC

(III.8)

The pump is used to pressurized the saturated water to that of the superheated steamexiting the boiler. The work required to pressurize the saturated water is given by:

g

P P mW C

34 (III.9)

where P 4 is the absolute pressure at the exit of the pump, N/m2; P 3 is the absolutepressure at the entrance of the pump, N/m2; and ρ is the mass density of water, kg/m3.(III.9) is derived by assuming that changes in internal energy, kinetic energy, andpotential energy across the pump are negligible. The pump is also assumed to befrictionless.

The overall plant efficiency ( overall ) is:

fuel fuel

s

fuel fuel

C s

Rankin eoverall

LHV m

W

LHV m

W W

)()(

(III.10)

Note that overall is always <<boiler

t environmen

T

T 1 = max (Carnot theorem) (T is in oK).

Table III.1 lists the specific enthalpy values of superheated steam at different P and T .

Example III.3. The gross electrical output of a steam electric plant is 500 MW. Thesteam flow rate through the boiler and turbine is 642 kg/s. Steam leaves the boiler assuperheated steam at 15 MPa and 560oC, after entering the boiler as saturated water at apressure of 15 MPa. The boiler fuel is bituminous coal which is burned at a rate of 47.5kg/s. Calculate the boiler efficiency and the overall plant efficiency.

1. From (III.6), 886.0)400,28)(5.47(

)0.611,11.472,3)(642(

boiler , or 88.6%. The specific

enthalpy values of superheated steam at 15 MPa and 500o

C and 600o

C are3,310.6 and 3,579.8 kJ/kg, respectively (Table III.1). Therefore, the specificenthalpy value of superheated steam at 15 MPa and 560oC is

1.472,360100

)6.310,38.579,3(6.310,3

kJ/kg (intrapolation).

2. From (III.10),

100)10)(400,28)(5.47(

105003

6

overall 37.6%.



45

3. If the ambient temperature is 20oC (293.15oK), then

56015.273

15.2931max

65%.

Table III.1.

H Values (kJ/kg) of Superheated Steam1. P (kPa) 100oC 200oC 300oC 400oC 500oC 600oC 700oC 800oC

1 2,688.6 2,880.1 3,076.8 3,279.7 3,489.2 3,705.6 3,928.9 4,158.75 2,688.1 2,879.9 3,076.7 3,279.7 3,489.2 3,705.6 3,928.8 4,158.7

10 2,687.5 2,879.6 3,076.6 3,279.6 3,489.1 3,705.5 3,928.8 4,158.750 2,682.6 2,877.7 3,075.7 3,279.0 3,488.7 3,705.2 3,928.6 4,158.5

100 2,676.2 2,875.4 3,074.5 3,278.2 3,488.1 3,704.8 3,928.2 4,158.3200 2,870.5 3,072.1 3,276.7 3,487.0 3,704.0 3,927.6 4,157.8400 2,860.4 3,067.2 3,273.6 3,484.9 3,702.3 3,926.4 4,156.9600 2,849.7 3,062.3 3,270.6 3,482.7 3,700.7 3,925.1 4,155.9800 2,838.6 3,057.3 3,267.5 3,480.5 3,699.1 3,923.9 4,155.0

1,000 2,826.8 3,052.1 3,264.4 3,478.3 3,697.4 3,922.7 4,154.12,000 3,025.0 3,248.7 3,467.3 3,689.2 3,916.5 4,149.45,000 2,925.5 3,198.3 3,433.7 3,664.5 3,897.9 4,135.3

10,000 3,099.9 3,374.6 3,622.7 3,866.8 4,112.015,000 2,979.1 3,310.6 3,579.8 3,835.4 4,088.620,000 2,820.5 3,241.1 3,535.5 3,803.8 4,065.330,000 2,161.8 3,085.0 3,443.0 3,739.7 4,018.5

1 Reference State: H2O(l ) at 0oC, i .e.,

)(2 l O H H 0 kJ/kg (@ 0oC).

Environmental Control (Thermal Pollution). The heat released from thecondenser must eventually be transferred back to the environment. There are twoapproaches that can be used to accomplish the required task. The once-through cooling approach uses the cooling water that is circulated through the condenser to absorb theheat released from the water/vapor mixture. Usually, the cooling water is drawn from a

lake or river, heated in the condenser, and returned to that body of water. As a result, nocooling water is lost in the condenser. On the other hand, the water temperature of thereceiving water body downstream from the point of discharge will be higher than that atthe intake of cooling water. In many cases, the rise in water temperature will alter thelocal ecological balance and create thermal pollution problems. A more expensiveapproach involves use of cooling towers (or cooling ponds) that transfer the heat directlyinto atmosphere by converting the cooling water into vapor rather than into a receiving

water body. The cooling water used will be lost as vapor that is released into theatmosphere, but the temperature of the water body from where cooling water is drawn

will not be altered. The cooling water requirements can be calculated via (III.11):

CW p

C throughonceCW

T l O H C

Qm

)(2 (III.11a)

)( CW v

C lost CW

T H

Qm

(III.11b)

where throughonceCW m

is the mass rate of cooling water required in a once-through cooling

system, kg/s; CW T is the allowable temperature increase in the cooling water stream,



46

oC; lost CW m

is mass rate of water lost in a cooling tower or a cooling pond, kg/s; and

)( CW v T H

is the latent heat of evaporation at the cooling water temperature T CW (oC),

kJ/kg.C

Q

is in kJ/s. Figure III.3 shows typical cooling towers as well as a steam

turbine used in thermal power plants.

Figure III.3. Cooling towers and a steam turbine.

Example III.4. A coal-fired power plant converts3

1 of the coal’s energy into electrical

energy. The electrical power output of the plant is 1,000 MW. The other 2/3 of the energycontent of the coal is rejected to the environment as waste heat. About 15% of the wasteheat goes up the smokestack and the other 85% is taken away by cooling water that isdrawn from a nearby river. The river has an upstream flow of 100.0 m3/s and atemperature of 15oC. If the cooling water is only allowed to rise in temperature by 10oC,

what flow rate from the stream would be required? The heat capacity of water is 4.184kJ/kg-oC.

1. Input Power = (3)(Output Power) = 3,000 MW.2. Total losses to cooling water and stack = 2,000 MW, with stack losses = 300 MW

and cooling water losses = 1,700 MW.

3. The rate of change in enthalpy of the cooling water is T C m H p

= 1,700 MW.

4. Now, C p = 4,184 J/kg-oC, ΔT = 10oC, and 1 W = 1 J/s, then

3 MW Steam Turbine



47

)10)(184.4(

10700,1 6

throughonceCW m = 40.6×103 kg/s = 40.6 m3/s (cooling water flow

rate).5. Since 1,700 MW must be returned back to the river with a flow rate of 100 m3/s,

then the

temperature rise in river at the point of discharge of cooling water is

)100)(184.4(

10700,1 6T = 4.1oC.

6. Since the cooling water flow rate is about 41% of the river flow rate (too high), itcreates an unacceptable increase in river temperature (i .e., 4.1oC) downstreamfrom the power plant. Therefore, alternatives for removing waste heat may benecessary.

7. Now, assume that the waste heat is removed in a cooling tower (or cooling pond) by evaporating the water at 15oC. Since the latent heat of vaporization at 15oC is

2,465 kJ/kg, then456,2

10700,1 6

lost CW m = 689.7 kg/s = 0.69 m3/s (makeup

water flow rate).8. Since the makeup water flow rate is < 1% of the river flow rate, this option

appears to be attractive. However, the capital costs of building the cooling facilityand the annual operating and maintenance (O&M) costs of operating the facilityneed to be considered before making the final recommendations.

9. If the cooling tower (or cooling pond) is being operated at a higher temperature,say, 50oC, will this help to reduce the makeup water flow rate? Since the heat of

vaporization at 50oC is 2,382.9 kJ/kg (Table II.2), the amount of heat required to bring the water temperature from 15oC (makeup water) to 50oC (water to beevaporated) (sensible heat) is (4,184)(50 − 15) = 146.44 kJ/kg.

Therefore, the amount of makeup water needed is 44.1649.382,2

10700,1 6

lost CW m =

0.67 kg/s or 0.67 m3/s (no much reduction!).

Environmental Control (Flyash Emissions). Modern steam electric plants areequipped with a variety of technologies to reduce or eliminate pollutant emissions. Thefollowing pollutant emissions are typical for coal-fired plants: flyash (fine ash particles),CO2, SO2, NOX (NO + NO2) heat, water pollutants, and solid wastes (i .e., bottom ashescollected from the boiler). Most modern coal-fired plants employ electrostatic

precipitators (ESPs) to capture nearly all of flyash emissions (Figure III.4). A DCelectric field is imposed between a wire and a pair of plates suspended in the flue gasflow path. As the flue gas flows between two plates, ash particles in the flue gas are

bombarded with negative ions, taking on negative electrical charges. Electrostaticattraction pulls the charged particles toward the positively charged plates, where they arecollected. At intervals the electric field is momentarily relaxed, allowing the ash particlesto fall into a collection hopper. Commercial ESPs contain many plates to ensure that allthe flue gas into a collection hopper is treated. The total ESP plate area ( A, m2) neededto achieve a given overall removal efficiency ( ESP ) is:



48

1: Smoke particles pick up a negative charge. 2: Smoke particles are attracted to the collectingplates. 3: Collecting plates are knocked to remove the smoke particles.

Figure III.4. The schematic of a typical electrostatic precipitator (ESP).

667.1

1

1ln

ESP

ga s

w

V A

(Deutsch-Anderson Equation) (III.12)

where ga sV

is the volumetric gas flow rate (m3/s) and w is the effective particle drift velocity perpendicular to the direction of gas flow (m/s). w can be expressed as afunction of sulfur content, with all other parameters held fixed at typical values (TableIII.2). ESPs use a large number of plates spaced about 30 cm apart, with both sides ofeach plate serving as collectors. As a result, a large surface area can be attained. Fabricfilters (baghouse) also are effective in capturing PM .

Application w (cm/s) Application w (cm/s)Utility Flyash 4.0-20.4 Open-Hearth Furnace 4.9-5.8

Pulverized Coal Flyash 10.1-13.4 Blast furnace 6.1-14.0Pulp & Paper Mills 6.4-9.5 Hot Phosphorous 2.7Sulfuric Acid Mist 5.8-7.6 Flash Roaster 7.6

Cement (wet process) 10.1-11.3 Multiple-Hearth Roaster 7.9Cement (dry process) 6.4-7.0 Catalyst Dust 7.6

Gypsum 15.8-19.5 Cupola 3.0-3.7Smelter 1.8

Table III.2. w Values.



49

Environmental Control (SO2 Emissions). Coal combustion is the major source ofSO2 emissions at coal-fired plants. Flue gas desulfurization (FGD) systems have beenrequired on all new coal-fired plant in the U.S. since 1978 to achieve the highest levels ofSO2 removal. The most widely used FGD technology employs a slurry of pulverizedlimestone mixed with water to remove SO2 via chemical reactions taking place in a vesselcommonly known as a scrubber:

gypsum sO H CaSO

sO H CaSO g CO g Ol O H sCaCO g SO

)(2

)(2)()(2

1)(2)()(

24

2422232 (III.13)

Example III.5 . An existing 500 MW coal-fired power plant (bituminous coal) operatesat an overall efficiency of 36% and an annual capacity factor of 65%. To comply withnational acid rain control requirements, the plant decides to install a wet limestonescrubber that is capable of reducing SO2 emissions by 95%. The scrubber requires areagent stoichiometry of 1.03, i .e., 3% more limestone than theoretically needed toremove a mole of SO2. Calculate for this plant (a) its new annual SO2 emissions, and (b)

the annual quantity of limestone required.

a.1. The maximum energy output of the plant is: (500 MW)(1,000 kW/MW)(24hrs/day)(365 days/yr) = 4.380×109 kW-hr/yr.

a.2. Since the annual capacity factor is 65%, the actual annual power output is(0.65)(4.380×109) = 2.847×109 kW-hr/yr.

a.3. Therefore, the rate of fuel energy input =

139

10847.236.0

)600,3)(10847.2(

kJ/yr.

a.4. The amount of bituminous coal burned each year = 913

10002.1400,28

10847.2

kg/yr.a.5. Since the sulfur content of bituminous coal is 1.5% (weight), then the amount of

sulfur emitted each year is: (0.015)(1.002×109) = 1.50×107 kg/yr, or the amount

of SO2 emitted each year is: 77

1091.232

)97.0)(1050.1)(64(

kg/yr (about 2 to

5% of sulfur in coal are trapped in ash; 64 is the molecular weight of SO2; and 32is the atomic weight of sulfur).

a.6. Therefore, the new SO2 emissions = (0.05)(2.91×107) = 1.46×107 kg/yr.

b. (III.13) shows that the theoretical reagent requirement is 1.00 mole CaCO3/molSO2 removed, versus 1.03 moles CaCO3/mol SO2 removed, or

6094.164

)100)(03.1( kg CaCO3/kg SO2 removed. Therefore, the annual amount

of limestone required = (1.6094)(0.95)(2.91×107) =4.45×107 kg/yr.

Environmental Control (NOX Emissions). To reduce NOX emissions, two generalmethods are used. The 1st method, combustion modification, alters the design of thecombustion chamber to affect the temperature, time, and other parameters that controlNOX formation. Typically, combustion modification methods reduce NOX up to ~ 50%.The 2nd method of NOX control employs a chemical treatment system to remove NOX



50

from the flue gas. This type of system is analogous to the FGD system for SO2 removal.The most prevalent postcombustion NOX control technology is selective catalyticreduction (SCR). A chemical catalyst, which is similar to the ones used in automobilecatalytic converters, is used to selectively reduce NOX in the flue gas to N2. SCRoperation requires temperatures of roughly 400oC, so the SCR reactor is usually locatedat the boiler exit. In power plants, however, ammonia (NH3) is injected into the flue gas

stream to achieve good NOX reduction:

O H N O NH NO

O H N O NH NO

22232

2223

32

3

2

12

2

3

4

1

(III.14)

In practice, some excess ammonia is required to obtain desired NOX reduction (70 to90% reduction).

Environmental Control ( Water Pollution). See the Section on Environmental

Control (Thermal Pollution).

Environmental Control (Solid Waste Management). The large quantities of ashparticles collected at coal-fired power plants represent a major solid waste managementproblem. For plants with large ash ponds, the final ash disposal site is often the ponditself. Otherwise, the pond is periodically dredged, and the ash is trucked to anengineered landfill site. In recent years many utilities have begun to utilize dry flyashcollection systems to avoid the use of water altogether, and the collected ash is disposedof directly in an engineered landfill. The ash and FGD waste can be recycled and used asaggregates for road construction, additives for cement manufacturing, or other usefulpurposes (~ 30% of the roughly 100 million tons of ash collected in the U.S. power plantsis sold). The gypsum produced in modern FGD systems, after the impurities are

removed, can be sold as a commercial by-product to produce wallboard or other building materials.

Example III.6. Calculate the mass of solid waste produced each year by the FGD systemdescribed in Example III.5 . Assume that the FGD waste consists of gypsum plusunreacted limestone. Further assume that the final waste product contains 15% moisture(occluded water) by weight, typical of the dewatering technology used to separate FGDsolids from the wet scrubber slurry.

1. Because one mole of SO2 removed produces one mole of gypsum, then

64

172

2

SO

gyp sum

m

m = 2.6875 kg gypsum/kg SO2 removed.

2. Unreacted limestone = 0469.064

1006094.1 kg limestone/kg SO2 removed.

3. Therefore, total dry waste = 2.6875 + 0.0469 = 2.7344 kg dry waste/kg SO2

removed or 2168.385.0

7344.2 kg wet waste/kg SO2 removed →

(3.2168)(2.765×107)= 8.89×107 kg FGD waste/yr.



51

WIND POWER (Figure III.5). Wind turbines are used to convert the kineticenergy of air into rotation of the rotor blade to drive electric generator. Wind turbine

blades are designed in the shape of an airfoil, similar to the wings of an airplane, so as to be able to take maximum advantage of the aerodynamic forces. The following equationscan be used to rate a wind turbine:

2

593.0

2

3

max

3

air air

electurbine

wind

air air wind

u A

P

P P

u A P

(III.15)

where P wind is the wind power (or the equivalent electrical power of the wind), W;

4

2 D

A

, where D is the tip-to-tip diameter of the circle swept out by the rotating blades, m2; ρair is the density of air under the conditions quoted, kg/m3 (e.g., 1.205 kg/m3 for dry air @ 25oC); uair is the wind velocity, m/s; P max is the maximum useful poweravailable from the wind, W; 0.593 is the Betz limit; ηturbine is the overall efficiency of the

wind turbine; and P elec is the electrical power output from the wind turbine generatorsystem, W. Commercial wind turbines have P elec’s ranging from 250 to 750 kW.Typically, a number of wind turbines are connected together to produce a wind farmcapable of generating large amounts of power. A typical 25 MW wind farm mightinclude 50 wind turbines, each rated at 500 kW. The total land area required for such afacility would be roughly 20 – 40 hectares/MW, depending on terrain and turbineconfiguration. The turbine structures themselves, however, occupy only about 5 to 10%of the total land area.

Example III.7 . A wind turbine with a rotor diameter of 50 m operates on a hilltop where the wind is blowing at 9 m/s. The air temperature is 10oC, and the correspondingair density is 1.25 kg/m3. Determine (1) the equivalent electrical power of the windunder these conditions, (2) the maximum power for the wind turbine, and (3) the actualelectrical power generated if the overall turbine generator efficiency is 34%.

1.)4)(2(

)9)(25.1()50)((

2

323 air air

wind

u A P = 895 kW .

2. )895)(593.0(593.0max wind P P = 531 kW .

3. )895)(34.0(2

3

air air

turbineelecu A P = 304 kW .



52

Figure III.5. Wind turbines and configurations.

GEOTHERMAL ENERGY. Geothermal power is energy generated by heat stored beneath the Earth's surface. Geothermal power supplies 0.416% of the world's energy.Geothermal comes from the Greek words geo, meaning earth, and therme, meaningheat. Prince Piero Ginori Conti tested the first geothermal power plant on 4 July 1904,at the Larderello dry steam field in Italy. The largest group of geothermal power plantsin the world is located in The Geysers, a geothermal field in California. Threegeothermal power plant technologies are currently used to convert hydrothermal fluidsto electricity, depending on the state of the fluid (whether steam or water) and its temp:dry steam, flash, and binary cycle (Figure III.6).

Dry steam power plants systems were the oldest type of geothermal power generationplants built. They use the steam from the geothermal reservoir as it comes from wells,and route it directly through turbine/generator units to produce electricity. These plantsemit only excess steam and very minor amounts of gases. Flash steam plants are themost common type of geothermal power generation plants in operation today. They use

water at temperatures greater than 360°F (182°C) that is pumped under high pressure tothe generation equipment at the surface. Fluid is sprayed into a tank held at a muchlower pressure than the fluid, causing some of the fluid to rapidly vaporize, or "flash."

Windmill for Low Wind Areas



53

Figure III.6. Geothermal power technologies.

The vapor then drives a turbine, which drives a generator. If any liquid remains in thetank, it can be flashed again in a second tank to extract even more energy. Binary cyclegeothermal power generation plants differ from dry steam and flash steam systems inthat the water or steam from the geothermal reservoir never comes in contact with theturbine/generator units.

Most geothermal areas contain moderate-temperature water (below 400°F). Energy isextracted from these fluids in binary-cycle power plants. Hot geothermal fluid and a

secondary (hence, "binary") fluid with a much lower boiling point than water passthrough a heat exchanger. Heat from the geothermal fluid causes the secondary fluid toflash to vapor, which then drives the turbines. Because this is a closed-loop system,

virtually nothing is emitted to the atmosphere. Moderate-temperature water is by far themore common geothermal resource, and most geothermal power plants in the future will

be binary-cycle plants.

Steam and hot water reservoirs are just a small part of the geothermal resource. Theearth's magma and hot dry rock will provide cheap, clean, and almost unlimited energy



54

as soon as the technologies are developed to use them. In the meantime, because they'reso abundant, moderate-temperature sites running binary-cycle power plants will be themost common electricity producers. Before geothermal electricity can be considered akey element of the U.S. energy infrastructure, it must become cost-competitive withtraditional forms of energy. The U.S. DOE is working with the geothermal industry toachieve $0.03 to $0.05 per kilowatt-hour. There will be about 15,000 megawatts of new

capacity within the next decade.

SOLAR ENERGY. The earth receives 174 petawatts (174×1015 J/s) of incoming solarradiation (insolation) at the upper atmosphere (Figure III.7). While traveling throughthe atmosphere, 6% of the insolation is reflected and 16% is absorbed. Averageatmospheric conditions (e.g., cloud, dust, and pollutants) further reduce insolation by20% through reflection and 3% through absorption. These atmospheric conditions notonly reduce the quantity of light reach the earth’s surface but also diffuse roughly 20% ofthe incoming light and filter portions of its spectrum. After passing through theatmosphere roughly half the insolation is in visible electromagnetic spectrum with theother half mostly in the infrared ( IR) spectrum, and a small part of ultraviolet (UV )radiation.

The absorption of solar energy by atmospheric convection (sensible heat transport) and by the evaporation and condensation of water vapor (latent heat transport) drive the winds and the water cycle. Upon reaching the surface, sunlight is absorbed by theoceans, earth, and plants. The energy captured in the oceans is ultimately responsiblefor temperature-driven ocean currents such as the thermohaline cycle and wind drivencurrents such as the Gulf Stream. The energy absorbed by the earth in conjunctionthat recycled by the greenhouse effects warms the surface to an average temperature ofapproximately 14oC. The small portion of solar energy captured by plants and otherphototrophs is converted to chemical energy via photosynthesis. In North America, theaverage insolation at the ground level over an entire year (including nights and periodsof cloudy weather) lies between 125 and 375 W/m2.

Water Heating (Figure III.8). Solar hot water systems use sunlight to heat water.Commercial solar water heaters began appearing in the United States in the 1890s.These systems saw increasing use until the 1920s but were thereafter gradually replaced

by relatively cheap and more reliable conventional heating fuels. The economicadvantage of conventional heating fuels has varied over time resulting in periodicinterest in solar hot water; however, solar hot water technologies have yet to show thesustained momentum they lost in the 1920s. That being said, the recent price spikes anderratic availability of conventional fuels is renewing interest in solar heatingtechnologies. As of 2005, the total installed capacity of solar hot water systems is 88GW-hr and growth is 14% per year. China is the world leader in the deployment ofsolar hot water systems with 80% of the market. Israel is the per capita leader in the use

of solar hot water with 90% of homes using this technology. In the United States heatingswimming pools is the most successful application of solar hot water. Solar waterheating is highly efficient (up to 86%) and is particularly appropriate for lowtemperature (25-65°C) applications such as domestic hot water, heating swimming poolsand space heating.



55

Figure III.7. Earth’s energy fluxes.

The basic components of a solar water heating systems are solar thermal collectors, astorage tank and a circulation loop. There are three basic classifications of solar waterheaters. Batch systems which consist of a tank that is directly heated by sunlight. These The basic components of a solar water heating systems are solar thermal collectors, astorage tank and a circulation loop. There are three basic classifications of solar waterheaters. Batch systems which consist of a tank that is directly heated by sunlight. These are the oldest and simplest solar water heater designs, however; the exposed tank can besusceptible to cooldown. Active systems which use pumps to circulate water or a heattransfer fluid. Passive systems which circulate water or a heat transfer fluid by naturalcirculation. These are also called thermosiphon systems.

Solar Ponds (Figure III.9). A solar pond is a pool of salt water that collects andstores solar energy. A solar pond consists of layers of water that successively increasedfrom a weak salt solution at the top to a high salt solution at the bottom. This in turn

yields a density gradient that prevents convective currents. As a result, the temperatureat the bottom layer could be as high as 100oC.

Generally, there are three main layers. The top layer is cold and has relatively little saltcontent. The bottom layer is hot, up to 100°C and is very salty. Separating these two

http://upload.wikimedia.org/wikipedia/en/c/c7/Breakdown_of_the_incoming_solar_energy.jpg



56

Figure III.8. Solar water heaters.

layers is the important gradient zone (i .e., halocline) where the salt content increases

with depth. Water in the gradient cannot rise because the water above it has less saltcontent and is therefore lighter. The water below it has a higher salt content and isheavier. Therefore, the stable gradient zone suppresses convection and acts as atransparent insulator, permitting sunlight to be trapped in the hot bottom layer from

which useful heat may be withdrawn or stored for later use. If the water is relativelytransluent, and the pond's bottom has high optical absorption, then nearly all of theincident solar radiation (sunlight) will go into heating the bottom layer. The heattrapped in the salty bottom layer can be used for many different purposes, such as theheating of buildings or industrial hot water, or to drive a Rankine cycle turbine, or aStirling engine for generating electricity. The solar ponds are particularly attractive forrural areas in developing countries. Ponds with vary surface areas can be set up for justthe cost of the clay or plastic pond liner. However, the evaporated surface water needs to

be constantly replenished and the accumulating salt crystals have to be removed and can be both a valuable by-product and a maintenance expense.

Figure III.9. Solar pond.



57

Photovoltaics. A solar cell or photovoltaic cell is a device that converts light intoElectricity using the photovoltaic effect. The 1st working solar cells were constructed byCharles Fritts in 1883. These prototype cells were made of selenium and achievedefficiencies of around 1%. Following the fundamental work of Russel Ohl in the 1940s,researchers Gerald Pearson, Clavin Fuller, and Daryl Chapin created the silicon solar cellin 1954. Until recently, their uses have been limited because of high manufacturing

costs. One cost effective use has been in very low-power devices such as calculators withLCDs. Another use has been in remote applications such as roadside emergencytelephones, remote sensing, cathodic protection of pipe lines, and limited "off grid" homepower applications. A third use has been in powering orbiting satellites and spacecraft.

To take advantage of the incoming electromagnetic radiation from the sun, solar panelscan be attached to each house or building. The panels should be mounted perpendicularto the arc of the sun to maximize usefulness. The easiest way to use this electricity is byconnecting the solar panels to a grid tie inverter. However, these solar panels may also

be used to charge batteries or other energy storage devices. Solar panels produce morepower during summer months because they receive more sunlight. Total peak power ofinstalled PV cells is around 6,000 MW as of the end of 2006. Installed PV cells are

projected to increase to over 9,000 MW in 2007. This is only one part of solar-generatedelectric power. Declining manufacturing costs (dropping at 3 to 5% a year in recent years) are expanding the range of cost-effective uses. The average lowest retail cost of alarge photovoltaic array declined from $7.50 to $4 per watt between 1990 and 2005.

With many jurisdictions now giving tax and rebate incentives, solar electric power cannow pay for itself in five to ten years in many places. "Grid-connected" systems, thosesystems that use an inverter to connect to the utility grid instead of relying on batteries,now make up the largest part of the market. In 2003, worldwide production of solarcells increased by 32%. Between 2000 and 2004, the increase in worldwide solar energycapacity was an annualized 60%. 2005 was expected to see large growth again, butshortages of refined silicon have been hampering production worldwide since late 2004.

Analysts have predicted similar supply problems for 2006 and 2007.

Solar Power Plants. Solar power plants use a variety of methods to collect sunlight.Concentrating solar thermal (CST ) systems (Figure III.10) use lenses or mirrors andtracking systems to focus a large area of sunlight into a small beam. CST technologiesrequire direct insolation to perform properly. This requirement makes theminappropriate for significantly overcast locations. The three basic CST technologies arethe solar trough, solar power tower, and parabolic dish. Each technology is capable ofproducing high temperatures and correspondingly high thermodynamic efficiencies butthey vary in the way they track the sun and focus light. A solar trough consists of a linearparabolic reflector which concentrates light on a receiver positioned along the reflector'sfocal line. These systems use single-axis tracking to follow the sun. A working fluid (oil,

water) flows through the receiver and is heated up to 400°C before transferring its heat

to a distillation or power generation system. Trough systems are the most developedCST technology. The Solar Electric Generating System ( SEGS ) plants in California andPlataforma Solar de Almería's SSPSDCS plant in Spain are representatives of thistechnology. A power tower consists of an array of flat reflectors (heliostats) whichconcentrate light on a central receiver located on a tower. These systems use dual-axistracking to follow the sun. A working fluid (air, water, molten salt) flows through thereceiver where it is heated up to 1,000°C before transferring its heat to a powergeneration or energy storage system. Power towers are less advanced than troughsystems but they offer higher efficiency and energy storage capability. The Solar Two in



58

Figure III.10. Solar power plants.

Daggett, California and the Planta Solar 10 (PS10) in Sanlucar la Mayor, Spain arerepresentatives of this technology. A parabolic dish or dish/engine system consists of astand-alone parabolic reflector which concentrates light on a receiver positioned at thereflector's focal point (see the attached photo). These systems use dual-axis tracking tofollow the sun. A working fluid (hydrogen, helium, air, water) flows through the receiver

where it is heated up to 1,500°C before transferring its heat to a sterling engine for powerconversion.



59

BIOGAS PRODUCTION FROM ORGANIC WASTE MATEWRIALS. Theproduction of biogas (primarily CH4 and CO2) production from organic matter isachieved via anaerobic fermentation and oxidation processes. Figure III.11 illustrates 4key steps are involved in the overall anaerobic fermentation and oxidation of wastematerials: (1) hydrolysis, (2) acidogenesis, (3) acetogenesis, and (4) methanogenesis.Since complex organic materials generally contain large molecules such as arbohydrates,

fats, and proteins, these molecules must be broken down into their smaller constituentparts in order for the microbes to tap into the energy potential in the subsequentdegradation steps. The process of breaking these large molecules and dissolving thesmaller constituents into solution by extracellular enzymes (e.g., cellulases,hemicellulases, amylases, lipases, and proteases) is referred to as hydrolysis. In thehydrolysis step, the complex organic molecules are broken down into simple sugars, fattyacids, and amino acids, as well as small amounts of acetic acid, H2, and CO2.

Figure III.11. Reaction pathways in anaerobic ferementation and oxidation.

Acetic acid and H2 produced in the hydrolysis step can be used directly by methanogens.Other molecules such as amino acids, sugars, and some fatty acids with a chain lengththat is greater than acetic acid must first be catabolized into compounds that can be

directly utilized by methanogens. The biological process of acidogenesis, which is similarto the way that milk sours, is where these reactions are carried out the acidogenic (orfermentative) microbes. Here volatile fatty acids are created along with NH3, CO2, andH2S as well as other by-products. The third step is acetogenesis where simple moleculescreated through the acidogenesis step phase are further digested by acetogens to producelargely acetic acid as well as H2 and CO2. For instance, β-oxidation of butyrate producesacetate and H2 whereas decarboxylation of propionate produces acetate, H2, and produceCO2, which are the precursors of methanogenesis [see (III.16) and (III.17),respectively]. The free energy changes associated with the conversion of propionateand butyrate to acetate and H2 requires that low H2 concentrations in the system (i .e.,< 10-4 atm) or the reaction will not proceed.

H H COOCH O H COOCH CH CH 232223 222 (III.16) H H HCOCOOCH O H COOCH CH 233223 33 (III.17)

Non-methanogenic microbes that carry out the first three steps in anaerobic fementationand oxidation includes Clostridium ssp., Peptococcus anaerobus, Bifidobacterium spp.,

Lactobacillus, Actinomyces, Staphylococcus, and Escherichia coli .



60

An important immediate product of the anaerobic fermentation and oxidation is H2. The biologically-mediated reaction pathways of certain compounds may be altered if H2 ispresent at a high concentration. For instance, CH3COOH may react with H2 and CO2 toform CH3CH2COOH. As a result, the reaction as depicted in (III.18) would hinder theprogress of methanogenesis which utilizes CH3COOH as a major substrate.

O H COOH CH CH CO H COOH CH 223223 23 (III.18)

Methanogenesis is carried out by a group of organisms known collectively asmethanogens. Methanogens are classified as archaea, a group quite distinct from

bacteria. They are common in wetlands and in the guts of animals such as ruminhantsand humans. Moreover, in marine sediments methanogens are generally proliferate

below the top layers where sulfates are absent. Methanogens also exist in extremeenvironments such as hot springs and hydrothermal vents at the deep ocean floor. Twogroups of methanogens are involved in CH4 production. One group, termed aceticlasticmethanogens, split acetate into CH4 and CO2. The second group, terms hydrogen-utilizing methanogens, use H2 as the electron donor and CO2 as the electron acceptor toproduce CH4. Acetogens are also able to use CO2 to oxidize H2 and form acetic acid(CH3COOH). An example of methanogens is Methanosarcina barkeri (Figure III.13)

which has been isolated from mud samples in lakes and bogs and municipal wastewatersamples. M. barkeri also lives in the rumen of cows where it helps digest the organicmatter for the cow. It is highly adaptable to its environment because of its ability toutilize a variety of substrates, as shown in (III.19). As a result, it is potentially useful asan agent to carry out the desired biochemical reactions. Although methanogens do notfunction under aerobic conditions, however, they can survive under the oxygen stressconditions for a prolonged period of time. M . barkeri , which possesses a superoxidedismutase (SOD) enzyme, is able to survive longer than the others when oxygen ispresent.

3423

3

2432

2

434233

4342223

434233

3322333

2423

3433

2343

3

34

433994

2332

4334

//

244

34

HCOCH O H COOCH

COOCH

O H CH HCO H H

CO

NH H HCOCH O H NH CH



NH CH NH CH NH CH

O H CH H OH CH

H HCOCH COOCH OH CH

O H H HCOCH OH CH

OH CH

(III.19)

HCOO-: Formic Acid, CH3OH: Methanol, (CH3)3N: Methylamine, and CH3COOH: Acetic Acid.



61

Factors Affecting Anaerobic Fermentation and Oxidation. A variety of factorsaffect the rates of anaerobic fermentation and oxidation (Table III.3). Anaerobic

bacteria consortia can endure temperatures ranging from < 0oC to > 57oC. Optimaltemperatures are at about 37oC (mesophilic) and 54oC (thermophilic). Microbialactivities fall off significantly between about 39o and 52oC and gradually from 35oC to0oC. Studies conducted in the north-central U.S. suggest that maximum net biogas

production can occur in anaerobic sludge at temperatures as low as 22oC. The bioreactorcontents should be void of dissolved oxygen and free from inhibitory concentrations ofheavy metals and sulfides. Also, the aqueous pH in the bioreactor should range from 6.6to 7.6. Sufficient alkalinity should be present to ensure that the aqueous pH will notdrop below 6.2, because methanogens cannot function properly below this point.

Alkalinity ranging from 1, 000 to 5,000 mg/L as CaCO3 and the total VOA (volatileorganic acid) concentration less than 250 mg/L as CH3COOH should be maintained inthe anaerobic bioreactor to ensure satisfactory results. Other factors affecting the rateand amount of biogas output include pH, water/solid (W/S) ratio, carbon/nitrogen(C/N) ratio, degree of mixing, particle matter in the materials being treated (i .e., amountand size distribution), and bioreactor/digester retention time. Pre-sizing and mixing ofthe feed materials with high solid contents for consistency permit the bacteria to work

more quickly. Bicarbonate (HCO3-

) in various forms can be added to maintain a desiredand consistent pH. The feed stream with a high nitrogen content may need to be diluted with the anaerobic bioreactor effluent stream to ensure an adequate feed C/N ratio at20/1 to 30/1. Moreover, the addition of essential nutrients may be required to ensurehealthy methanogenic growth (Table III.4).

Table III.3. Factors Affecting Anaerobic Fermentation and Oxidation.

Environmental Factor Range

pH 6.8 to 7.2

Temperature 30 to 40oC (optimal: 35oC) (mesophilic)50 to 60oC (optimal: 55oC) (thermophilic)

Total Alkalinity 3,000 to 5,000 mg/L as CaCO3

Total VOA < 250 mg/L as CH3COOH

Cu+2 < 150 mg/L

Ni+2 < 200 mg/L

Zn+2 < 350 mg/L

Cr+6 < 200 mg/L

ORP -510 to -540 mv

Biogas. Biogas, which primarily consists of CH4 and CO2, is a major end-product of

anaerobic fermentation and oxidation. Table III.5 lists typical constituents found in biogas. Large quantities of H2S can be produced from the anaerobic treatment of liquid waste streams from industries such as molasses fermentation, yeast production,monosodium glutamate production, and petroleum refining. In addition to its toxicity,the combustion of H2S produces H2SO4 which is highly corrosive. Therefore, theremoval of H2S from the biogas prior to its uses in CHP (combined heat and power)generation is required. The amount of CH4 produced can be accounted for by theamount of COD removed during anaerobic fermentation and oxidation. Thestoichiometric COD of CH4 is the amount of oxygen needed to oxidize CH4 to CO2 and



62

H2O:

O H COOCH 2224 22 (III.20)

According to (III.20), the stoichiometric COD of CH4 is 2(32) = 64 g O2/mol CH4. CODis the preferred concentration parameter for the waste streams with unspecifiedcharacteristics (see Example III.7).

Table III.4. Nutritional Requirements of Methanogens. Substance Concentration (mg/L)

Na+ 125-500K + 200-400

Ca+2 100-200Mg+2 75-125

N 60-130Fe+3 1-10Co+2 1-10

P 1-5

Thiamine (C12H17N4OS) 1-5

Table III.5. Typical Biogas Composition.Component % (volume)

Methane (CH4) 50-80Carbon Dioxide (CO2) 20-40

Nitrogen (N2) 0-5Hydrogen (H2) 0-1.0

Hydrogen Sulfide (H2S) 0.05-1.0 Ammonia (NH3) 0.02-0.5

Oxygen (O2) 0-0.5

Example III.7 . Determine the amount of methane produced per kg COD stabilized.Use glucose, methanol, and acetate as the starting compounds. COD: chemical oxygendemand.

◙ Glucose (C6H12O6):

2226126

246126

666

33

COO H OO H C

COCH O H C

1. 1 g C6H12O6 consumes 1.067 g O2 (i .e.,180

192) when it is completely oxidized.

Therefore, the COD of C6H12O6 is 1.067 g O2/g C6H12O6. 1 g C6H12O6 produces

0.267 g CH4 (i .e.,180

48). Therefore, 0.25 g CH4 is produced per g COD stabilized

(i .e., 25.0067.1

267.0 ).

2. Since 1 mole (16 g) CH4 occupies 22.4 L of volume at STP, 0.25 g CH4 will have a



63

volume of 0.35 L at STP (i .e., 35.016

)25.0)(4.22( ). In other words, 0.35 m3 CH4

is produced per kg COD stabilized at STP.

◙ Methanol (CH3OH):

2223

2243

2432

234

COO H OOH CH

COO H CH OH CH

1. 1 g CH3OH consumes 1.5 g O2 (i .e.,64

96) when it is completely oxidized.

Therefore, the COD of CH3OH is 1.5 g O2/g CH3OH.

2. 1 g CH3OH produces 0.375 (128

48) g CH4 (i .e.,

128

48). Therefore, 0.25 g CH4 is

produced per g COD stabilized (i .e.,5.1

375.0), or 0.35 m3 CH4 is produced per kg

COD stabilized at STP.

◙ Acetate (CH3COOH):

2223

243

222 COO H OCOOH CH

COCH COOH CH

1. 1 g CH3COOH consumes 1.067 g O2 when it is completely oxidized, i .e., the CODof CH3COOH is 1.067 g O2/g CH3COOH.

2. 1 g CH3COOH produces 0.267 g CH4, or 0.25 g CH4 is produced per g CODstabilized, or 0.35 m3 CH4 is produced per kg COD stabilized at STP.

Therefore, 0.35 m3 CH4 is produced per kg COD stabilized at STP that isindependent of the starting compound used.

Heat Value Calculations (Mesophilic Case). Since biogas is a major end-product ofthe anaerobic fermentation and oxidation, the cost-effectiveness of the anaerobic

biotechnology for a specific application is very often determined by the net energy valuethat can be recovered. In addition to satisfying the heating and power requirements ofrunning the anaerobic treatment facility, the extra amount of biogas produced canconceivably be utilized at other locations to defray the energy costs involved. Example

A.1. in the Appendix presents detailed energy balance calculations involved in thecombustion of biogas. The enthalpy tables are prepared and used as the means toaccount for the energy values of all species participate in the combustion reaction. Theheat energy recovered is then used to produce a saturated steam stream in a boiler.Example A.2. in the Appendix illustrates the calculations of heating requirements of amesophilic anaerobic bioreactor.



64

ENERGY CONVERSION EFFICIENCIES.

Electric Power Plants.

Hydroelectric 90%Fuel Cell (Hydrogen) 80%

Coal-Fired Generator 38%Oil-Burning Generator 38%

Nuclear Generator 30%Photovoltaic Cell 10%

Space Heating.

Electric Resistance 99%*

High-Efficiency Gas Furnace 90%Typical Gas Furnace 70%Efficient Wood Stove 65%Typical Wood Stove 40%

* Note that 60-70% of the energy in the original fuel is lost in electric power generation.

Lighting.

Sodium Vapor Lamp 66%* Fluorescent Bulb 25%*

Incandescent Bulb 5%* Gas Flame 1%

* Note that 60-70% of the energy in the original fuel is lost in electric power generation.

Inefficiencies in Electricity Generation and Consumption (Figure III.12). Athermal power plant can only cature ~35% of energy value in fuel as electricity. The restis released back to the environment as waste heat. The electricity generated is

transmitted to individual consumers via high-voltage transmission lines which are highlyefficient with ~ 10% losses. If incandescent light bulbs are used, then only ~ 5% of theelectricity delivered a household is used. The overall efficiency in energy generation andconsumption ( overall ) in this example is therefore:

)05.0)(90.0)(35.0( nutilizatioontransmissi generatio noverall = 0.016.

Three Alternatives for Meeting Household Energy Needs (example).Electricity supplies are very efficient when electrical power is needed. When only heat isneeded, the electrical supply is a very inefficient form of delivery. The followingalternatives are illustrated for a house which requires 10 kW of heating power and 1 kWof electrical power. A natural gas ( NG ) thermal power plant is used to generateelectricity with 33% efficiency. If this house is powered only by electricity, then 33 kW ofgas is used to meet 11 kW of electrical needed of the house (alternative 1). If NG is usedto heat the house with a 95% efficient gas heater, then 3 kW of gas is used to meet 1 kW

of electrical needed of the house. The total NG requirement is therefore: 95.0

103 13.5

kW (alternative 2). If a 60% efficient, NG -fed fuel cell is used to generate the electricityfor the house, then 1.67 kW NG is needed to run the fuel cell. The fuel cell will produce0.67 kW waste heat which can be captured for heating the house. Therefore, the



65

total NG requirement for the house is: 1.67 + (10 − 0.67) = 11 kW (alternative 3).Evidently, alternative 2 is best because a fuel cell is rather expensive as compared to agas heater.

Figure III.12. Inefficiencies in electricity generation and consumption.

Improving Energy Efficiency. A key means of reducing environmental impacts inthe operation of steam electric plants is to improve the efficiency of the energyconversion process. If less primary energy (i .e., fuel energy) is needed to produceelectricity, the magnitude of environmental impacts will be reduced accordingly. Theopportunities of improving the overall energy efficiency of the steam electric plants areexamined herein.

Increasing Operating Temperatures. The laws of thermodynamics (e.g., Carnottheorem) pose a substantial barrier for power plants using conventional Rankine cycles

to achieve high efficiencies. As previously discussed, overall is always <<boiler

t environmen

T

T 1 =

max (T is in oK). However, overall can be increased by increasing the operating

temperature of the boiler ( bolier T ). For example, gas turbines today operate at inlet

temperatures of approximately 1,300oC, compared to about 1,100oC a decade ago. Each

35.01

90.02

05.03



66

55oC increment adds about 1 percentage point to the overall efficiency. For coal-firedpower plants, the use of supercritical (the supercritical fluid cannot be classified aseither a steam or a liquid) boilers can achieve an

overall ≈ 43% (including full

environmental controls) by increasing the steam pressure and temperature to 31 MPaand 590oC, compared to typical conditions of 17 MPa and 540oC for conventional boilers

with an overall ≈ 37%. Advances in materials science and engineering are thereforecritical to achieving significant advances in boiler design.

Cogeneration. It is sometimes unnecessary to dump all the condenser heat into a nearby water body, and the heat recovered can provide hot water to nearby residences orindustries thereby yielding significant savings on their energy bills. If temperatures areneeded, steam can be extracted from the turbine at higher temperatures and pressures.This type of combined heat and power (CHP ) system is known in the U.S. ascogeneration. An economic price must be paid, however, for the additional piping andother equipment needed to supply the steam or hot water to customers outside thepower plant. In most situations, this cost far exceeds the value of the heat supplied

because most potential customers are remote from the power plant. Therefore,

cogeneration systems have found limited applications in the U.S., typically at largeindustrial complexes that require both heat and electricity continuously. Cogenerationtypically increases the efficiency of fuel utilization up to 80%.

Advanced Cycles. The use of advanced cycles is another important way to increase theefficiency of electric power generation. One promising method to make coal-fired powerplants more efficient is a combined-cycle system integrated with coal gasification, or anintegrated gasification combined cycle ( IGCC ) system (Figure III.13). In this system,coal reacts chemically with steam and oxygen at high temperature and pressure to form afuel gas mixture consisting mainly of H2, CO, and CH4 (synthetic gas or syngas). O2 levels in the gasifier are kept low to create a chemically reducing environment to avoidthe formation of H2O(v) and CO2. Impurities such as sulfur and ash are removed byefficient environmental control systems, leaving a clean fuel gas whose energy content isabout 15 to 20% that of natural gas. This fuel gas is then used to generate electricity. An

IGCC system can achieve an overall ≈ 40%. The fuel gas produced can also be used to

synthesize liquid fuels and chemicals as well as produce electricity. IGCC systemsalso reduce solid wastes by converting sulfur impurities into H2SO4 or elemental sulfur

by-products. At the present time, the principle drawback of the IGCC system is its highcapital costs.

Another promising technology is the integrated gas turbine/fuel cell technology. Thehighest-efficiency plants under development combine advanced gas turbines withadvanced fuel cells. A high-temperature fuel cell would first generate electricityelectrochemically using natural gas as the feed gas. The exhaust gas from the fuel cell

would then drive a gas turbine to generate additional power. An alternative design usingcoal would first employ a coal gasifier to generate a clean fuel gas for the fuel cell. Another advantage of gasification is that it offers an efficient route for separating andcapturing CO2 as a potential carbon management strategy to address the issue of global

warming. Such systems are at least a decade or two away for commercialization.



67

Figure III.13. The schematic of an IGCC system.

Resource Recovery from Wastes.

Refused-Derived Fuel ( RDF ). RDF is a fuel produced by shredding and dehydratingsolid MSW with a waste converter technology (Figure III.14). RDF consists largely oforganic components of MSW municipal waste such as plastics and biodegradable wastes.

RDF processing facilities are normally located near a source of MSW and, while anoptional combustion facility is normally close to the processing facility, it may also belocated at a remote location. Non-combustible materials such as glass and metals areremoved during the post-treatment processing cycle with cycle with an air knife or othermechanical separation processing. The residual materials can be sold in their processedforms (depending on the process treatment) or they may be compressed into pellets,

bricks or logs and used for other purposes either stand-alone or in a recursive recyclingprocess.

Figure III.14. RDF logs.



69

Although both CO2 and CH4, the latter is 20 times more potent (though shorter-lived inthe atmosphere) than the former. Capturing CH4 from landfills, wastewater treatment,and manure lagoons prevents CH4 from being vented to the atmosphere and allows theenergy to be used to generate electricity or power motor vehicles. All crops, including

biomass energy crops, sequester carbon in the plant and roots while they grow, providinga carbon sink. In other words, CO2 released while burning biomass is absorbed by the

next crop growing. This is called a closed carbon cycle. In fact, the amount of carbonsequestered may be greater than that released by combustion because most energy cropsare perennials, they are harvested by cutting rather than uprooting. Thus the rootsremain to stabilize the soil, sequester carbon and to regenerate the following year.

Acid rain is caused primarily by the release of sulphur and nitrogen oxides from thecombustion of fuels like coal. Acid rain has been implicated in the killing of lakes, as wellas impacting humans and wildlife in other ways. Since biomass has practically nosulphur content, and easily mixes with coal, “co-firing” is a very simple way of reducingsulphur emissions and thus, reduce acid rain. “Co-firing” refers to burning biomass

jointly with coal in a traditionally coal-fired power plant or heating plant.

Biomass crops can reduce water pollution in a number of ways. Energy crops can begrown on more marginal lands, in floodplains, and in between annual crops areas. In allthese cases, the crops stabilize the soil, thus reducing soil erosion. They also reducenutrient run-off, which protects aquatic ecosystems. Their shade can even enhance thehabitat for numerous aquatic organisms like fish. Moreover, because energy crops tendto be perennials, they do not have to be planted every year. Since farm machineryspends less time going over the field, less soil compaction and soil disruption takesplace. Another way biomass energy can reduce water pollution is by capturing themethane, through anaerobic digestion, from manure lagoons on cattle, hog and poultryfarms. These enormous lagoons have been responsible for polluting rivers and streamsacross the country. By utilizing anaerobic digesters, the farmers can reduce odor,capture the CH4 for energy, and create either liquid or semi-solid soil fertilizers which

can replace petroleum-based fertilizers. Table III.7 summarizes a variety of biomasssources that are being studied for biodiesel productions, and Table III.8 summarizes theprocessing technologies that can be used to produce biomass-derived products.

Table III.7. Biomass Sources for Biodiesel Productions. Feedstock Yield (Liters/Hectare1)Soya 60-75

Rapeseed 165-220Mustard 220

Palm 975 Algae 15,000

1 1 hectare = 104 m2 = 2.5 acres.



70

Table III.8. Technologies to Produce Biomass-Derived Products.



72

that caused Asiatic cholera, but his discovery was largely unnoticed.1856 Thomas Hawksley, a civil engineer, advocated continuously pressurized water systems as a strategy

to prevent external contamination.1874 Slow sand filters were installed in Poughkeepsie and Hudson, New York.1880 Karl Eberth isolated the organism ( Salmonella typhosa) that caused typhoid fever.1881 Robert Koch demonstrated in the laboratory that chlorine would inactivate bacteria.1883 Carl Zeiss marketed the first commercial research microscope.

1884 Professor Escherich isolated organisms from the stools of a cholera patient that he initially thought werethe cause of cholera. Later it was found that similar organisms were also present in the intestinal tracts oevery healthy individual as well. Organism was eventually named for him ( Escherichia coli ).

1892 Robert Koch proved that Asiatic cholera was due to a bacterium, Vibrio cholera, which he called thecommon bacillus because of its comma-like shape.

1893 First sand filter built in America for the express purpose of reducing the death rate of the populationsupplied was constructed in Lawrence, Massachusetts. To this end, the filter proved to be a great success

1897 G.W. Fuller studied rapid sand filtration (2 gallons/ft2-day) and found that bacterial removals were much better when filtration was proceeded by good coagulation and sedimentation.

1902 The first drinking water supply was chlorinated in Middlekerke, Belgium. Process was actually a“Ferrochlor” process wherein calcium hypochlorite and ferric chloride were mixed, resulting in bothcoagulation and disinfection

1903 The iron and lime process of treating water (softening) was applied to Mississippi River water supplied toSt. Louis, Missouri.

1906 First use of ozone as disinfectant in Nice, France. First use of ozone in the U.S. occurred four decadeslater.

1908 George Johnson, a member of Fuller’s consulting firm, helped install continuous chlorination in JerseyCity, New Jersey.

1911Johnson published “Hypochlorite Treatment of Public Water Supplies,” in which he demonstrated thatfor contaminated supplies filtration alone was not enough. Adding chlorination to the process of watertreatment greatly reduced the risk of bacterial contamination.

1914U.S. Public Health Service (U.S. PHS) used Smith’s fermentation test for coliform to set standards for the

bacteriological quality of drinking water. The standards applied only to water systems that provideddrinking water to interstate carriers like ships and trains

1941 85% of the water supplies in the U.S. were chlorinated, based on a survey conducted by U.S. PHS.1942 U.S. PHS adopted the first comprehensive set of drinking water standards1974 Dutch and American studies demonstrated that chlorination of water formed trihalomethanes.

Passage of the Safe Drinking Water Act (SDWA).

WATER SUPPLY AND DISTRIBUTION SYSTEMS (Figure IV.1). Watersupply and distribution system (or network) is a system of engineered hydrologic andhydraulic components which supply and distribute the water. Such a system typicallyincludes:

1. The watershed (or catchment) ( 1) which is an area of land that catches waterfrom precipitation and snowmelt. The water then drains to a common waterway,such as, a stream, lake, aquifer, or wetland;

2. Raw (untreated) water collection point (above or below ground) where the wateraccumulates, such as a lake, a river, or groundwater from an undergroundaquifer. Raw water (usually water being transferred to the water purification

facilities) may be transferred using uncovered ground-level aqueducts, coveredtunnels, or underground water pipes ( 3a).



73

Figure IV.1. The schematic of a water supply and distribution system.

3. Water purification facilities ( 2) where raw water is treated using moderntreatment technologies/practices to ensure that the quality of the treated watermeets stringent standards.

4. A pipe network for distribution of treated water (usually underground) ( 3, 3a- 3d )to the consumers such as private houses; industrial, commercial, or

institution establishments; and other usage points (e.g., such as fire hydrants)( 4).

5. Water storage facilities such as reservoirs, water tanks, and water towers ( 5 ) areused to store water to meet demands that are highly time dependent. Smaller

water systems may store the water in cisterns or pressure vessels. Tall buildingsmay also need to store water locally in pressure vessels in order for the water toreach the upper floors.



74

6. Additional water pressurizing components such as pumping stations may need to be situated at the outlets of underground or above ground reservoirs or cisterns ifgravity flow is unfeasible.

7. Connections to the sewers (underground pipes, or aboveground ditches in somedeveloping countries) are generally found downstream of the water consumers,

but the sewer system is considered to be a separate system, rather than part ofthe water supply and distribution system.

WATER TREATMENT METHODS. To better understand how the treatmentmethods developed, it is appropriate to consider their evolution. A summary of themethods used for the treatment of water at the beginning of the 20th century (except for themembrane separation) is presented in Table IV.1.

Table IV.1. Water Treatment Methods. Method Agent/Objectives

Mechanical Separation Gravitational Sedimentation, Screening (screens, scrubbers, and filters),and Adhesion (scrubbers and filters).

CoagulationChemical treatment resulting in drawing matters together into groups,thereby making them more susceptible to removal by mechanicalseparation but without any significant chemical changes in the water.

Chemical Purification Lime Softening, Iron Removal, and Neutralization of Objectionable Acids.

DisinfectionThe processes that utilize ozone (O3), chlorine (Cl2), or cupric sulfate(CuSO4) to poison or kill objectionable organisms without at the sametime adding substances objectionable or poisonous to the users of the

water.

Biological Processes

◙ Oxidation of organic matter as food by organisms thereby resulting in itsdestruction.◙ Death of objectionable organisms, resulting from the production ofunfavorable conditions, such as absence of food (removed by the

purification processes) and killing by antagonistic organisms.

Aeration◙ Evaporation of (1) dissolved gases that produce objectionable tastes andodors, and (2) carbonic acid that is the carbon source supporting thechemoautotrophic bacterial growths.◙ Addition of oxygen for certain chemical purifications and growth of

water-purifying organisms.Boiling Best household method of protection from disease-carrying waters.

Membrane Separation◙ Complete rejection of pathogens of zoonotic origin that can survive a

water supply that is completely free of wastewater contamination.◙ Desalination.

It is interesting to note that these methods are still in use today. The most importantmodern technological development in the field of water treatment is the use of membranetechnology. Membrane filtration was widely used for beverages in the mide-1960s toreplace heat pasteurization as a means of purification and microbial stabilization. In

virtually all of these applications the membranes were treated as disposal items. The idea oftreatment large volumes of drinking water in this manner was untenable. The developmentof membrane filtration fibers in the mid-1980s that could be backwashed after each usepermitted uses on a continuous basis for a long period of time. In the last decades of the20th century there were numerous manufacturers of commercial membrane filtrationsystems and municipal water treatment plants as large as 3×105 m3/day were under



75

construction. Membranes are arguably the most important development in the treatmentof drinking water since the year 1900 because they offer the potential for complete andcontinuous rejection of microbiological contaminants on the basis of size exclusion.Desalination membranes offer the opportunities of converting salty and brackish water intodrinking water for semi-arid and arid areas where fresh water resources are scarce. Twoexamples of water treatment systems are shown in Figures IV.2 and IV.3.

Figure IV.2. A conventional water treatment system.

EFFECTS OF DEVELOPMENT ON STREAM FLOW/STORMAWATERMANAGEMENT. As precipitation hits the ground there is a fork in the pathway ofthe water cycle. Water from rainfall or snowmelt may infiltrate into the ground, or itmay runoff the surface. Several factors such as surface characteristics (porosity andslope), nature of soil, amount of precipitation, and rate of precipitation influence whichdirection the water goes. River flow curves in Figure IV.4 are recorded for three similarstorms (i .e., similar intensity and duration) on Brays Bayou in Houston, Texas, before,during, and after development. The increasing height of the surge (i .e., crest in riverflow) occurring with the progress of development clearly indicates that surface runoff is

becoming predominant, because the paved surfaces have much less capacities inintercepting and retaining rainfalls as compared to surfaces covered with trees andshrubs.

Stormwater management deals with problems such as excessive runoffs, river bankerosion, and flooding. In a heavy rainfall a hectare (2.5 acres) parking lot may generaterunoff at the rate of 6 m3/min. Therefore, traditional stormwater management practicesfocus on rapid drainage. Recent stormwater management practices promote reversephilosophy – hold stormwater at or near where it falls and let it drain away slowly orinfiltrate. As a result, the techniques of building parking lots with porous surfaces, anddesigning rooftops, parking lots, and other flat surfaces so that they “pond” the water andlet it trick away slowly are becoming popular. Stormwater may be a cheap source of water



76

(i .e., gray water) for many uses; e.g., runoff from the paved park lot of a car dealership can be collected and stored in an underground storage tank and then used to wash cars.

Figure IV.3. Desalination scheme using reverse osmosis membranes.

POINT/NON-POINT POLLUTION SOURCES (Figure IV.5). Non-point (ordistributed) source ( NPS ) pollution, unlike pollution from industrial and municipal

wastewater treatment plants, comes from many diffuse sources. NPS pollution is caused by rainfall or snowmelt moving over and through the ground. As the runoff moves, itpicks up and carries away natural and human-made pollutants, finally depositing theminto lakes, rivers, wetlands, coastal waters, and even our underground sources ofdrinking water. These pollutants include: (1) excess fertilizers, herbicides, andinsecticides from agricultural lands and residential areas; (2) oil, grease, and toxicchemicals from urban runoff and energy production; (3) sediment from improperly

Figure IV.4. The effects of land development on peak river flows.

Time (hours)

0 20 40 60 80 100

RiverFlow(m

3/s)

0

20

40

60

80

100

120

140

September 23-25, 1941

May 12-19, 1953

June 23-27, 1960

Brays Bayou, Houston, Texas



78

Inorganic Suspended Solids Solids/Silt/Clay.

Organic Colloidal MaterialsFecal Matter (undigested materials, cellulose, and bacteria).Garbage (from disposals).Cloth Fibers (from laundry).Paper Fibers (from tissues)

Dissolved MaterialsNutrients: Detergents (phosphates) and urine (phosphate,ammonium, nitrate, potassium, sodium, calcium, sulfate, etc.).Toxic Wastes (mostly from industries): heavy metals (copper,mercury, cadmium, chromium, tin, zinc, lead, arsenic, etc.).Non-Biodegradable Organic Compounds.

physical characteristics include solids (total, dissolved, suspended, and settleable),turbidity, color, and temperature. Important chemical characteristics include pH,metals, organic matter, nitrogenous matter, phosphorous, sulfur, and dissolved gases.Important biological characteristics include pathogenic organisms, bacteria, protozoa,helminthes, and viruses.

Organic compounds are normally composed of a combination of carbon, hydrogen, and

oxygen, together with nitrogen, phosphorous, and sulfur in some cases. Theorganicmatter in municipal wastewaters typically consists of protein (40-60%), carbohydrates(25-50%), and oils and fats (8-12%). Urea, the major constituent of urine, is anotherimportant organic compound contributing to fresh municipal wastewaters. Because ureadecomposes rapidly, it is seldom found in other than very fresh municipal wastewaters.In addition, municipal wastewaters contain small quantities of a very large number ofsynthetic organic molecules, with structures ranging from simple to extremely complex.Over the years, a number of different analyses have been developed to determine theorganic content of municipal wastewaters. In general, the analyses may be classified intothose used to measure aggregate organic matter comprising a number of organicconstituents with similar characteristics that cannot be distinguished separately, andthose analyses used to quantify individual organic compounds. The concentration of

organic matter in municipal wastewaters is often measured in terms of it oxygendemand, since dissolved oxygen (DO) is essential to sustaining the aquatic life and manyorganics consume oxygen as they undergo degradation. The depletion of oxygeninventories in surface water bodies will decimate fish and other forms of aquatic life andcreate septic or anaerobic conditions. The oxygen demand of organic matter can bemeasured as either biochemical oxygen demand (BOD) or chemical oxygen demand(COD). The measurements are good with DO concentrations > 1 mg/L.

Biochemical Oxygen Demand. BOD is a gross parameter that is widely used as ameans to measure the concentration of a biodegradable organic chemical. BOD is theamount of oxygen required by the microorganisms to fully oxidize a given organicchemical to CO2 and H2O. Therefore, BOD is always proportional to the organic

chemical concentration. BOD is a convenient parameter to use, especially when a groupof unknown but biodegradable organic chemicals is present. So, always think BOD asthe concentration parameter (i .e., C = BOD).

There are two types of BOD: CBOD and NBOD. See the following reaction expressions.

3222 )2

3

2()

4

3

24( cNH O H

canCOO

cban N O H C Bacteria

cban (IV.1)



79

O H H NOO NH Bacteria

2323 2 (IV.2)

In (IV.1), the amount of oxygen required by heterotrophic bacteria to remove theorganic material present (i .e., carbon oxidation) is referred to as CBOD. In (IV.2), theamount of oxygen required by chemoautrotrophic bacteria to oxidize ammonia to nitrate

(nitrification) is referred to as NBOD. 4.57 g O2 are needed to oxidize 1 g NH3-N to NO3--N. As a result, any discharge containing significant amounts of CBOD and/or NBOD willdeplete the dissolved oxygen (DO) inventories in the surface water body downstreamfrom the point of discharge, primarily caused by the indigenous bacterial activities in

water.

Chemical Oxygen Demand. The equivalent amount of oxygen required to oxidizeany organic matter in a municipal wastewater sample by means of a strong chemicaloxidizing agent (e.g., potassium dichromate, K 2Cr2O7) is called COD. The sample isoxidized with excess amounts of K 2Cr2O7 and concentrated H2SO4. A catalyst, silversulfate (AgSO4), is used to assist the oxidation reaction and mercuric sulfate (HgSO4) isused to prevent the oxidation of chloride (Cl-). The sample mixed with the chemicals (in

a capped vial) are digested at 130oC for 2 hours to convert practically all of the organicmatter to CO2 and H2O. The difference between the initial amount and the final amountof K 2Cr2O7 is the COD of the sample tested.

2363

2

22

38)8( 3

422

2

72

cband

dCr cNH O H cd a

nCO H cd OdCr N O H C cban

(IV.3)

Wastewater treatment and management terminologies that are commonly used aresummarized in the Appendix.

Municipal Wastewater Treatment Technologies (Figures IV.6 and IV.7).municipal wastewater treatment as it is commonly practiced in the U.S. consists of threestages: pretreatment , primary treatment , and secondary treatment . These stagesremove most of the organic matter including microorganisms, but they do not removedissolved nutrients (i .e., nitrogen and phosphorous). Additional treatment, referred toas advanced treatment or tertiary treatment , is now being practiced to removedissolved nutrients and other constituents (e.g., dissolved solids) to render the treatedmunicipal wastewater reusable.

Pretreatment involves the steps required before removal of actual contaminants can begin. The municipal wastewater stream may carry large pieces of debris such as rags,plastic bags, and coarse sand (or grit). If these materials remain, they will clog or hinder

later treatment stages. They are removed by letting the municipal wastewater streamflow through a bar screen, a row of bars mounted about 2.5 cm apart. Debris ismechanically removed from the screens and taken to be incinerated or landfilled. Coarsesand or grit are then removed by passing the municipal wastewater stream through gritchambers where its velocity is slowed just enough the permit such materials to settle.The settled materials are mechanically removed from these tanks and then landfilled.

Primary treatment involves the gravitational settling of heavier particles or organicmatter that are removed in grit chambers in large primary clarifiers where the



80

Figure IV.6. Flow scheme of a typical municipal wastewater treatment plant.

municipal wastewater stream flows very slowly to render it nearly motionless for severalhours. About 30 to 50% of the total settleable solid matter is removed. At the same

time, fatty or oily materials float to the top and are skimmed off. The materials removedare known as the primary sludge which requires separate treatment and disposal.

Secondary treatment is also called biological treatment because it employs organismsthat literally consume the organic matter and convert it through cellular respiration andreproduction to CO2, H2O, and new cells. Either attached-growth processes (e.g.,trickling filters) or suspended-growth processes (e.g., activated sludge processes) areused for secondary treatment. Through primary and secondary treatments, 85 to 90% ofthe total organic matter is removed. The excess cells produced are removed in final



81

Figure IV.7. A municipal wastewater treatment plant. DAF: dissolved air flotation.

clarifiers (physically similar to primary clarifiers) as the secondary sludge or biological sludge. The municipal wastewater stream after secondary treatment is disinfected witheither chlorine gas (Cl2) or chlorine dioxide (ClO2) to reduce disease hazards andenhance public health safety prior to its final discharge into the environment. Secondarytreatment is unable to remove nutrients such as nitrogen and phosphorous that maycause eutrophication in surface water bodies. When eutrophication is a problem, thensecondary treatment is needed to be upgraded to advanced treatment or tertiarytreatment . Novel treatment technologies are available to render the municipal

wastewaters reusable after advanced treatment.

Recall that 30 to 50% of the organic matter in a municipal wastewater stream is removedin the primary treatment as the primary sludge. Primary sludge is a black, foul-smelling,syrupy liquid of about 98% water and 2% organic matter that includes many pathogenicorganisms. Both primary and secondary sludge streams can be treated in anaerobic

sludge digesters (at either 35oC or 55oC) to produce methane (CH4) as a gaseous fuel.The quantity of CH4 that can be produced is roughly 0.03 m3 per day per person served(or 0.35 m3/kg COD removed at 25oC and 1 atm). The stabilized sludge after anaerobicdigestion is usually odorless with little hazards. It is essentially the same as humusresulting from decay in natural eco-systems. It can be used as the soil conditioner toimprove the soil quality.

RIVERS. Rivers serve as the major routes by which continental rain and thecontaminants reach the oceans. Rivers do not contain a very large percentage of the total



82

water on the earth (i .e., 0.0001%), but their importance cannot be overlooked since theyplay a significant role in the transport of water as well as dissolved and suspended matter(including contaminating materials). Rivers have played an important role in humandevelopment as well. As result of human proximity, rivers have been considerablyaffected by activities ranging from agriculture and flood control to the input of humanand industrial wastes. In general, the rivers can be classified on the basis of their water

quality that is safe for the intended uses. For instance, the water quality of a class Ariver should be sufficiently good to satisfy the requirements for water contactrecreational activities. Contaminants commonly found in surface water bodies include:(1) oxygen-demanding organic and inorganic materials, (2) nutrients such as nitrogen(organic and inorganic) and phosphorous (organic and inorganic), (3) pathogens, (4)suspended solids (stormwater and soil erosion), (5) salts (deicing applications andseawater intrusion), toxic chemicals and metals (organic and inorganic), and heat(cooling water discharges). These contaminants may come from a variety of sources thatinclude: (1) municipal discharges (wastewaters and stormwaters), (2) industrialdischarges, (3) agricultural runoffs (fertilizers, pesticides, and animal wastes), andisolated areas (e.g., office parks, apartment complexes, etc.).

Class Remarks

A Water contact recreation (including swimming)

B Able to support fish and wildlife

C Public water supply

D Agricultural and industrial uses

Applications of CFSTR/PFR Models to Surface Water Bodies. The following 2examples illustrate that both CFSTR and PFR models can be applied to assess the waterquality in surface water bodies (i .e., rivers, streams, lakes, and reservoirs).

Example IV.1. An industry is applying for permission to discharge its wastewater into anadjacent pond that is linked to a river at point 1 (see the attached diagram). The pond water isthen mixed with the river water at point 2. A fish hatchery located 2 km downstream frompoint 2 obtains its makeup water directly from the river at point 3 (i .e., the water quality atpoint 3 is a major concern in this application). Recommendations are sought to satisfy bothparties (i .e., industry and fish hatchery). It was previously determined that the wastewater to

be discharged contains an undesirable compound that needs to be strictly regulated. (1) Ponddimension: L:W : D = (50 m):(20 m):(3 m), (2) wastewater discharge rate: 2,000 m3/day, (3)average river flow rate: 100,000 m3/day, (4) average river dimension: W : 10 m, D: 3 m, (5)compound concentration in wastewater: 100 mg/L, and (6) desired compound concentrationat point 3: < 0.01 mg/L.

Moreover, (1) the compound undergoes 1st-order degradation in water [in situ 1st-order rateconstants are 1.2 day -1 (pond) and 0.7 day -1 (river)], (2) the pond is completely mixed undermost circumstances, (3) the river is free of the compound in question upstream from point 2,and (4) the estimated annual treatment cost to remove the compound prior the discharge into

the pond is: )000,12())(($ 356.0

V R year

, where R is fractional removal and

V is in m 3/day.



83

1. V (pond) = (50)(20)(3) = 3,000 m3. The HRT of wastewater in the pond

000,2

000,3 = 1.5 days.

2. The distance between 2 and 3 x 23 = 2,000 m.3. Average cross-section area of the river A = (10)(3) = 30 m2.

4. Travel time between 2 and 3)000,2000,100(

)30)(000,2(*

t = 0.59 days.

5. Since the river flow rate is >> the wastewater flow rate and the average cross-

sectional area of the river is small, we can assume that both flows will becompletely mixed at point 2. Therefore, 1

12 02.0

000,2000,100

)000,2(C

C C

, where

C 1 is the compound concentration at point 1 and C 2 is the compoundconcentration at point 2.

6. Assuming that the river can be modeled as a PF system (why?), then

7. 76.001.0013.0)02.0( 11

)59.0)(7.0(

1

*

23 C C eC eC C kt mg/L.

8. In other words, the compound concentration leaving the pond must be < 0.76mg/L, say, let C 1 = 0.5 mg/L ( a better margin of safety).

Option 1: The wastewater, after treatment, is discharged into a completely mixed pond.

1. In this case, the compound concentration after treatment is:)5.12.11)(5.0()1(1 k C C in = 1.4 mg/L. The required fraction removal =

0.986.2. Annual treatment cost is: (0.986)(2,000)0.356(12,000) = $177,103/yr.

Option 2: Modify the hydraulic characteristics of the pond through partition, i .e., thepond is divided into 10 narrow channels (each channel has a width of roughly 5 m and a

River ( PFR System)

1

2 3

Pond (CFSTR System)

Fish Hatchery

Wastewater



84

length of 20 m). (see the attached diagram). As a result, the pond behaves like a PFRsystem. The cost of partition should be modest since the pond is shallow and is relativelysmall.

1. In this case, the compound concentration after treatment is:)5.1)(2.1(

1 )5.0( eeC C k

in = 3.02 mg/L. The required fraction removal is: 0.970.

2. Annual treatment cost is: (0.970)(2,000)0.356

(12,000) = $174,229/yr.3. Annual savings in treatment costs as compared to Option 1 are: $177,103 -$174,229 = $2,874/yr (too small!).

Option 3: Dig a new, partitioned pond with a suitable size. Make the wastewater to flowthrough these two ponds in series. Find the size of the new pond such that no treatmentis required. In this case, 0.3)5.0(100 )5.1)(2.1)(( xeC x

in . In other words, the new

pond should have a size twice the existing pond and it can be partitioned into 20 narrowchannels. Again, the costs of digging and partitioning this new pond with 20 narrowchannels should not be excessive. Therefore, it is evident that Option 3 is the best one torecommend.

Example IV.2. The river and tributary shown below have flow rates of 4.0 and 0.5

m3/s, respectively. The river flow splits at point A that yields ABV

= 2.8 m3/s and AC V

=1.2 m3/s. The physical and flow characteristics of the system are summarized as follows:

Section Volume (m3) Flow Characteristics

AB 15,000 Plug Flow

BC 10,000 Plug Flow

AC 30,000 Plug Flow

B 135,000 Completely Mixed

(1) Determine the steady-state concentration of a pollutant at point C (C C ) for thefollowing

conditions: r (degradation rate) = −kC , k = 0.5 day -1, C 1 = 50 g/m3 (concentration atpoint 1), and C A = 100 g/m3 (concentration at point A). (2) If the steady-stateconcentration at point C (C C ) is to be reduced to 50% of the level calculated in (1), whichof the following options will produce the best result: (a) reduce C 1 to 25 g/m3 and C A 50g/m3, (b) reduce C 1 to zero, and (c) reduce C A to 25 g/m3?



85

1.1. Section A-B: )400,86)(9)(7.0(

)000,15)(5.0(*

100

eeC C B Akt

A B A B = 97 g/m3 (PFR system).

1.2. Section B:

B B

B B A B C C

C C C 0400,86

))(000,135)(5.0(]5.0)4)(7.0[()5.0)(())(4)(7.0( 1 =

72.7 g/m3 (CFSTR system).

1.3. Section B-C: )400,86](5.0)4)(7.0[(

)000,10)(5.0(*

7.72

eeC C C Bkt

BC BC = 71.4 g/m3 (PFR

system).

1.4. Section A-C: )400,86)(4)(3.0(

)000,30)(5.0(*

100

eeC C C Akt

AC AC = 86.5 g/m3 (PFR system).

1.5. Therefore,5.00.4

))(4)(3.0())(4)(7.0(

C AC C BC

C

C C C = 75.4 g/m3 (CFSTR

system).

2. Now, the required C C is 37.7 g/m3. The calculations follow the same steps as 1.1

to 1.5.a. C 1 = 25 g/m3 and C A = 50 g/m3 → C C = 37.7 g/m3.

b. C 1 = 0 g/m3 and C A = 100 g/m3 → C C = 63.7 g/m3.c. C 1 = 50 g/m3 and C A = 25 g/m3 → C C = 19.7 g/m3.

3. Therefore, c > a > b (option b will not meet the requirement).

Superposition Technique. Recall that a river can be modeled as PFR system because

ISLAND

Tributary

River A

B

C

1



86

its longitudinal dimension is significant as compared to its lateral dimensions. If aspecies i is consumed in the river at a first-order rate, then at steady state,

u

kx

iu

kx

eC eC xC kC dx

dC u

)0()(0 (IV.4)

where C ( x ) is the concentration of species i at the downstream location x , C i = C (0) is theconcentration of species i at the location x = 0, k is the first-order decay rate coefficient,and u is the average river flow velocity. Any river that can be modeled by (IV.4) isreferred to as a homogenous linear system. A mathematical technique applicable to ahomogeneous linear differential equation such as (IV.4) is the superposition technique.Let C 1( x ) and C 2( x ) be the solutions of (IV.4). By substituting C ( x ) = C 1( x ) + C 2( x ) and itsderivative into (IV.4), then

0

)()(

22

11

2121

kC dx

dC ukC dx

dC u

C C k dx

C C d ukC

dx

dC u

(IV.5)

Therefore, (IV.5) suggests that C ( x ) is also a solution of (IV.4). See the followingexample.

Example IV.3. A river receives two pollution loads as shown in the attached diagram. Thefollowing information is available:

1. Wastewater discharge rate: 2,000 m3/day (location 1) and 1,000 m3/day (location 2).2 Wastewater compound concentration: 200 mg/L (200 g/m3).3. Average river flow rate upstream from location 1: 50,000 m3/day.4. Average river dimension: W : 10 m, D: 3 m.5. In situ first-order rate constant k = 0.7 day -1.6. The river is free of the compound in question upstream from point 1.7. The wastewater stream mixes immediately and completely with the river flow at the

point of discharge.

Find the concentration of the compound in river 4,000 m downstream from location 1.

River

1 ( x = 0 m) 2 ( x = 1,000 m)



87

I. Solution Approach 1.

I.1. The compound concentration in river @ location 1 is:000,50000,2

000,2200)0(

C =

7.69 mg/L.

I.2.310

000,50000,2)21(

u = 1,733.3 m/day.

I.3. 3.733,1

)000,1)(7.0(

69.7)000,1(

eC = 5.13 mg/L (immediate upstream @ location 2).

I.4. The compound concentration in river @ location 2 is:000,1000,52

000,1200000,5213.5

= 8.81 mg/L.

I.5.310

000,1000,52)32(

u = 1,766.7 m/day.

I.6. 7.766,1

)00 0,3)(7.0(

' 81.8)000,3()000,4(

e xC xC = 2.68 mg/L.

II. Solution Approach 2 (Superposition Technique).

II.1. 3.733,1

)000,4)(7.0(

1 69.7)000,4(

e xC = 1.53 mg/L.

II.2. 310

000,1000,50

)000,3)(7.0(

2000,1000,50

200000,1)000,3(

e xC = 1.14 mg/L.

II.3. Therefore, C (4,000) = 1.53 + 1.14 = 2.67 mg/L.

BIOACCUMULATION AND BIOMAGNIFICATION. Some of the chemicals,e.g., heavy metals and halogenated hydrocarbons, may be ingested with food or water areneither broken down nor excreted readily. Instead, they accumulate in the body,concentrating in the liver or other specific organs until they reach damaging levels. Thisphenomenon is called bioconcentration or bioaccumulation. When the phenomenon of

bioaccumulation is put into the context of a food chain, each successive consumer receives amore contaminated food supply and, in turn, accumulates the contaminant to yet a higherlevel. For instance, scientists have observed that the concentration of the pesticide DDT

was magnified some 10 million-fold as it passed through the food chain (biomagnification)(Figure IV.8).

One of the most distressing aspects of bioaccumulation One of the most distressing aspectsof bioaccumulation One of the most distressing aspects of bioaccumulation and

biomagnification is that the diseases which they cause, e.g., cancer, liver and kidneydysfunction, and birth defects, may not appear many years after the initial exposure. Bythat time is extremely difficult, if not impossible, to determine the real cause of the diseaseand too late for the individual concerned to do much, if anything, about it. The

bioconcentration process can be seen as a balance between two kinetic processes, uptake(e.g., inhalation, ingestion, and dermal contact) and depuration (e.g., excretion, urination,and perspiration) as quantified by first-order rate constants k1 and k2, respectively. The rateof change of contaminant concentration in an organism is given by:



88

Figure IV.8. Bioconcentration and biomagnification.

M M B B M B

B M B C BCF C

k

k C C k C k

dt

dC C k C k

dt

dC )(0

2

12121 (IV.6)

where C B is the biotic phase concentration (i .e., the concentration in the organism), C M isthe concentration in the environmental medium, and BCF is the bioconcentration factor.BCF expresses the bioconcentration (or bioaccumulation) of hydrophobic compounds thattend to accumulate in the fat of animals. Although BCF is defined under the equilibriumconditions when the organism has been exposed to the contaminant for a prolong period oftime, however, BCF is a good indicator of the bioaconcentration potential of a contaminant.The bioaccumulation of some anthropogenic, hydrophobic compounds in variousorganisms is summarized as follows where environmental medium is water (i .e., C M is C W ).

Compound Organism BCF C W (μg/L) Aldrin Catfish//Buffalo Fish 1,590//30,000 0.044//0.007

Chlordane Algae 302 6.6×10-3 DDT Algae//Oyster//Trout 500//70,000//200 0.016×10-3//0.1//20

Dieldrin Trout 3,300 2.3PCB Yellow Perch 17,000 1.0

)( 5914 Cl H C DDT

Ambient Environment (e.g., water)

Organism Uptake Depuration



89

PCBs IN THE AQUATIC ENVIRONMENT. PCBs (polychlorinated biphenyls)are stable, non-flammable oils (also a poor conductor of electricity) once usedextensively as transformer and hydraulic fluids, especially for transformer applications,in which power surges can cause arcs and ignite insulation oils. Aroclor is the tradename used by Monsanto for PCBs. The basic unit of PCBs is the aromatic hydrocarbon

biphenyl (Figure IV.9). The biphenyl molecule has 10 sites where chlorine can be

substituted for hydrogen atoms; 5 position in one ring (left) are numbered 2 through 6(counterclockwise), and 5 positions in another ring (right) are numbered 2' through 6'(clockwise). In assigning location numbers to the chlorine atoms, lower number (primedor unprimed) take precedence over higher numbers, and unprimed number are listedover equivalent primed numbers. Chlorination yields up to 207 products (calledcongeners because they are all formed together by the same basic process ofchlorination) with the substitution of one to ten chlorine atoms on the biphenylmolecule.

◙ Biphenyl (C12H10)

Figure IV.9. Polychlorinated biphenyl (PCB) molecule.



90

PCBs are often considered as groups of molecules containing the same number ofchlorine atoms, although the locations of chlorine substitution may be different (i .e.,PCBs with the same number of chlorine atoms are known as isomers). In the U.S., PCBshave been marketed by Monsanto under the trademark Aroclor. Internationalnames include Fenchlor (Italy), Kaneclor (Japan), Pyralene (France), and Clophen(Germany). The difference in chlorination are noted by four-digit numbers after the

name of Aroclor, such as Aroclor 1221 or Aroclor 1232. The first two digits refer to 12carbons in biphenyl. For all Aroclor but one, the last two digits of the four-digit termrepresent the % of chlorination by mass of the PCB mixtures (e.g., Aroclor 1248 is amixture containing 48% by weight of chlorine). The one exception is Aroclor 1016, whichis 41% chlorinated. Since PCBs have a very low water solubility and a very high affinitytowards particles (especially Those with a high organic carbon content) (Table , theenvironmental fate of PCBs is strongly influenced by these important properties. WhenPCBs are released into a surface water body (e.g., a river), they are either (1) present asoil droplets in water (due to a low water solubility) or (2) adsorbed on the suspendedparticles in water (due to a low water solubility and a small Henry’s law constant). As aresult, the settling velocities of PCB droplets and/or PCB-coated suspended particles andthe flow velocity of the water in the river will determine whether or not the PCBs

released will stay in the control volume (i .e., reach the bottom sediments and stay there)or be carried away from the control volume (i .e., move with the river water todownstream locations outside the control volume). These phenomena should beanalyzed before the environmental fate of PCBs can be assessed.

Aroclor 1016 Aroclor 1232 Aroclor 1248 Aroclor 1260

Vapor Pressure1 (mm Hg) @ 20oC

4×10-4 4×10-4 4×10-4 4×10-4

Henry’s Law

Constant2 (atm-m3/mol) @ 25oC

3.3×10-4 8.64×10-4 3.5×10-4 7.1×10-4

LD503 (oral rat) (mg/kg)

2,300 4,470 11,000 1,315

Water Solubility(μg/L)

~ 100 ~ 100 ~ 100 ~ 100

1 The vapor pressure of a species is a measurement of its volatility (i .e., the tendency toaccumulate in thegas phase). 2 Henry’s law constant is a measurement of the concentration distribution of a species

between the liquid phase and the gas phase under the equilibrium conditions. The concentration

in the gas phase isexpressed in terms of the partial pressure of the species. A species with a high Henry’s lawconstantshows that it tends to accumulate in the gas phase. 3 LD50 is a parameter used for defining thedegree of acute toxicity. It is a calculated dose of a material that is expected to cause the death of50% of an entire defined experimental animal population. It is determined from the exposure tothe material by any route other than inhalation of a significant number from that population. Bycomparison, 2,3,7,8-TCDD (2,3,7,8-tetrachlorodibenzo- p-dioxine) has an LD50 = 0.02 mg/kg (i .e.,a taste of 2,3,7,8-TCDD will probably kill a 70-kg person).



91

Adsorption on the Water-Soil Interface. Adsorption is the physical and/orchemical

process in which a substance accumulates at the solid-liquid interface or gas-liquidinterface.

Examples include the adsorption of a gas (adsorbate) on a solid (adsorbent ) or a liquidon a

solid. Sorption is the combined process of adsorption of a solute (sorbate) at a solidsurface

(sorbent ) and partitioning of the solute into the organic carbon (OC ) that has coated thesolid

surface. Electrostatic forces, van der Waals force, and hydrogen bonding are the basicphysical/

chemical forces that cause adsorption to occur. Hydrophobic bonding can also accountfor the

interaction with a surface in soil and sediment environments. (IV.7) and (IV.8) are twocommonly used expressions that describe the adsorption phenomena occurring on

water-solidinterfaces:

ne KC q

1

(Freundlich Isotherm Equation) (IV.7)

eC K q ' (Linear Isotherm Equation) (IV.8)

where q is the equilibrium solid-phase concentration, mg sorbate/g sorbent; K is the

Freundlich capacity parameter, (mg sorbate/g sorbent)(L water/mg sorbate) n

1

; C e is theequilibrium liquid-phase concentration, mg/L; n is the Freundlich intensity parameter;and ' K is the equilibrium constant between an aqueous (or gaseous) and a solid phase,

mL/g or L/kg. (IV.4) is very useful in describing the fate of many anthropogenic organiccompounds in soil (or sediment)-water environmental compartments. However, ' K is

chemical-specific which will be tedious and expensive to measure. As a result, thefollowing procedure has been developed to estimate ' K . For an organic chemical A,

))((

094.0)log(903.0)log(

'

tan

OC OC

OW OC

water

ol ocOW

f K K

K K

A

A K

(IV.9)

where OW K is the octanol-water partition coefficient; OC K is the soil-water partition

coefficient normalized to organic carbon, mL/g organic carbon or L/kg organic carbon;and OC f is the fraction od organic carbon for a specific soil or sediment. OW K is easily

measurable in the laboratory. If an organic chemical, A, is mixed in a sealed containerthat contains equal volumes of octanol (C8H17OH) and water, after equilibration andseparation of the water and octanol phases, the concentrations of A in these phases aremeasured (i .e., [ A]octanol and [ A]water) and OW K is then calculated by (IV.9). Octanol is



93

settles through a fluid (e.g., water or air) by gravity, its settling velocity is dependent onthree

forces acting on the particle: gravitational force, buoyancy force, and drag force. Whenthese

forces are in equilibrium, the settling velocity of the spherical particle will becomeconstant

which is defined as the terminal settling velocity (vs, m/s):

18

)(2 f p p

s

gd v

(Stokes’ law ) (IV.10)

where d p is the diameter of the particle (m), ρ p is the density of the particle (kg/m3), ρ f isthedensity of the fluid (kg/m3), and μ is the dynamic viscosity of the fluid (N-s/m2).

A fluid carrying particles enters a control volume (e.g., a settling chamber or the airpocket above

a city) at a volumetric flow rate of

V . The volume of the c.v. is V and the top area of thec.v. is

Atop. Then, the depth of the c.v. is h A

V

top

. The HRT of the fluid in the c.v. is

V

V .

Thefarthest distance any particle has to fall is h, and the particle must fall this distance

within θ in

order to be removed. Therefore, the fraction of the particles withV

V hhv s

= sov can

becompletely removed, regardless of the location upon entering the c.v. A fraction of theparticles

with

hv s can also be removed, but depending on their location upon entering the c.v.

It is

seen from Figure VI.9 that, A represents the trajectory of particles withV

V hhv s

which can

be removed in the c.v. B represents the trajectory of particles with

hv s which enters

the c.v. atthe upper left corner and as a result, cannot be removed. C represents the trajectories of

particles

with

hv s , but enter the c.v. at locations other than the upper left corner. The particles

following the top C trajectory cannot be removed in the c.v., whereas the particlesfollowing the

bottom C trajectory can be removed in the c.v.



94

Therefore, the percentage that particles with so s vv is removed due to gravity (

sv R )

depends on the location where the particles enter the c.v. The entry location is expressedin terms of a vertical height '

h , which is measured from the bottom of the c.v. Particles

with a small sv should enter the c.v. at a small 'h in order to be removed, i .e.,

%100%100'

so

s

svv

v

h

h R (IV.11)

Therefore, the total fraction of particles removed in the c.v. due to gravity ( RT ) is:

Figure IV.9. Particle trajectory in a horizontally flowing environmentalmedium.

top

so

C X

so

sC T

A

V

V

V hhv

dX v

v X R

0

)1(

(IV.12)

where (1 – X C ) is the fraction of particles with so s vv , and the integral term in (IV.12) is

the fraction of particles with so s vv that is removed. X is referred to as the weight

fraction of remaining which is the weight fraction of particles with sv < the stated value.

Example IV.6. A suspension of PCB droplets entering into a specific section of a river

A: Complete Removal. B: No Removal.

C : Partial Removal.

Fluid Flow Path

Particle Settling

Path Actual Particle Trajectory

h

h′

Atop



95

has the following size distribution and settling characteristics:

sv

(m/min)

3.05 1.52 0.61 0.31 0.23 0.16

X 0.55 0.46 0.35 0.21 0.11 0.03

It was estimated that sov for the suspension of PCB droplets is 2.83 m/min. How much

PCB droplets can be removed in the c.v. due to gravity?1. Prepare an X versus sv plot as shown below. From the plot it is found that

roughly 53% of PCB droplets in the suspension have a sv < 2.83 m/min.

Therefore, 1 – X C = 1 – 0.53 = 0.47.2. The integral term in (IV.8) is the area, which is bound by the Y -axis, the

horizontal line at X = 0.53, and the curve linking the data points, divided by sov .

The area is graphically calculated as 0.396. Therefore,83.2

396.047.0 T R = 0.61.

vs (m/min)

0.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2

WeighFractionofRemaining(X)

0.0

0.1

0.2

0.3

0.4

0.5

0.6

2.83 m/min

0.53

It is often assumed that, when a waste stream is released into a river, an immediate andcomplete mixing occurs at the point of discharge. In reality, this seldom happens.Instead, dilution and mixing occur slowly as the contaminants flow downstream. Thedistance required for complete mixing laterally ( Lm) depends on river geometry and flowproperties. Lm can be estimated as:

h

uw Lm

2

6.2 (side bank discharge)h

uw Lm

2

3.1 (midstream discharge)

(IV.13)

where Lm is the distance from the source to the zone where the discharge has been wellmixed laterally, ft; u is the average river flow velocity, ft/s; w is the average river width,ft; and h is the average river depth, ft. (IV.13) should always be used to the cases thatinvolve PCBs because of concerns about the toxicity of PCBs. A large Lm value impliesthat the PCB concentration in the vicinity of waste discharge may be very high to requireimmediate attention. The units used in (IV.13) are specified and cannot be changed.



97

step that the part undergoes. Circuit boards and other electronic equipment must also be scrupulously clean, both before and after application of the printed circuit. Acommonly used cleaning method consists of placing the part to be cleaned in a rinse tankor consecutively in several rinse tanks containing either stagnant or flowing water. Theobjective is to transfer the contaminants on the surface of the manufactured part to the

water by dissolution and dilution. This is the typical process used to remove excess

plating solution dragged out of a plating bath. However, it is useful only for removingmaterials that are water soluble.

A single stagnant rinse tank can serve as a cleaning device, but it is not very effective andcleaning ability decreases dramatically with time as the level of contaminants in the rinsetank builds. Large quantities of water are needed to effect cleaning. A series of stagnantrinse tanks improves the situation since each subsequent rinse tank in the progressioncontains fewer contaminants than the previous one, but this is still not very efficient. Byusing a flowing tank or series of tanks, cleaning efficiency can be significantly increasedand water use decreased. Running the rinse water through the rinse tankscountercurrent to the flow of the pieces being cleaned provides the best and mostefficient cleaning possible. Each additional cleaning step in the series contains cleaner

water and improves the cleaning operation, but there is a limit to this improvement.Generally, little benefit is achieved by operating more than 3 countercurrent tanks inseries.

Single-Stage Rinsing Operation. Rinse water requirements needed to achieve agiven

allowable contaminant concentration on the rinsed workpiece (or dragout) can bedetermined by

use of mass balances. For a single-stage rinsing operation as shown below, the mass balance

equation can be written as:



98

11

1

11

)(

)(

C

C V

C C

C C V V

C V C V C V C V

P D

R

P D R

R D R R P D

(IV.15)

where C P is the constituent concentration in the plating bath, C 1 is the constituentconcentration in the completely-mixed rinse tank (and waste effluent stream), C R is the

constituent concentration in the clean rinse water (C R ≈ 0), DV

is the equivalent

volumetric flow rate of dragout from the plating bath and from the rinse tank, and RV

isthe volumetric flow rate of the rinse water to the rinse tank. DV

can be estimated as:

))(( C dragout D V N V

(IV.16)

where dragout N

is the number of dragouts being processed per unit of time, and C V is the


P

C 1

C

C V

0

R

R

C

V

1C

V R

Desired Concentration



99

constituent volume on the individual dragout which is assumed to be constant.Remember that C 1 is specified at a low value.

Multi-Stage Rinsing Operation (3 rinsing tanks in series with separateinflows of

rinse water). The mass balance equation for the first rinse tank (RT1) is the same as

(IV.15).

The mass balance equations for RT2 and RT3 are (C R ≈ 0 and the volumetric flow of

rinse water to each rinse tank is constant at RV ):

3

32

2

21

332

221

C

C C

C

C C

V

V

C V C V C V C V

C V C V C V C V

D

R

R D R R D

R D R R D

(IV.17)

Combining the 3rd equation in (IV.17) with the 2nd equation in (IV.15), then

02 2

313

2

13

3

1 P P C C C C C C C C (IV.18)

Since both C P and C 3 are specified, then (IV.18) can be solved for C 1. By substituting C 1into the

2nd equation in (IV.15), RV

can be determined for a given DV

.

P C 1C

))((C dragout D V N V

1C

V R

2C

V R 0

R

R

C

V

0

R

R

C

V

2C




100

Multi-Stage Rinsing Operation (3 countercurrent rinsing tanks in series with a

single inflow of rinse water).

For a countercurrent rinse tank series shown below, a single rinse water stream isintroduced to

the last rinse tank in the series and then flows in a countercurrent fashion to the first

rinse tank before being discarded. In this case, RV

and DV

are the same for each rinse tank. Themass

balance equations for the 1st rinse tank and the remaining rinse tanks (note that C R ≈ 0)are:

3

2

1

332

2231

112

RT C V C V C V

RT C V C V C V C V

RT C V C V C V C V

R D D

R D R D

R D R P D

(IV.19)

2. For a specified C 3 values, the equations in (IV.19) can be solved simultaneously

for C 1, C 2, and RV

.

P C 1C


2C

V R 0

R

R

C

V

1C

V R

2C




101

Example IV.7 . A metal plating firm operates a chromium plating bath that has 85,000 mg/Lchromium and a dragout of 0.060 L/min. What rinse water flow rate is needed to keep thechromium concentration on the rinsed workpiece at 25 mg/L? Assume that the clean rinse

water contains no chromium. Compare (1) a single running rinse tank; (2) series rinsing withthree rinse tanks; and (3) a countercurrent rinsing process with three rinse tanks.

1.1. From (IV.15),)250(

)000,8525)(060.0(

RV = 204 L/min.

2.1. From (IV.18), C 1 is estimated to be 5,900 mg/L (by iteration).

2.2. From the 2nd equation in (IV.15),900,5

)000,85)(060.0(

RV = 0.86 L/min.

2.3. Therefore, total rinse water requirement is: (0.86)(3) = 2.58 L/min.

2.4. From the 3rd equation in (IV.17),060.0860.0

)000,85)(060.0(2

C = 385 mg/L.

3.1. From (IV.19), RV

is estimated to be 0.88 L/min (by iteration).

3.2. From the 3rd equation in (IV.19),060.0

)060.0880.0)(25(2

C = 392 mg/L.

3.3. From the 1st equation in (IV.19),060.0

)880.0)(25()060.0880.0)(392(1

C =

5,775 mg/L (waste effluent stream concentration).

V. EMERGING AIR POLLUTION ISSUES

INDOOR AIR POLLUTION.

Sources and Exposure Guidelines of Indoor Air Pollutants.

Pollutants & Indoor Sources Guidelines Friable Asbestos: Fireproofing, thermal and acoustic insulation, anddecoration.

Hard Asbestos: Vinyl floor and cement products.

0.2 fibers/mL for fibers longer than 5

Carbon Monoxide(CO): Kerosine and gas space heaters, gas stoves,

wood stoves, fireplaces, and smoking.

10 mg/m3 (8 hours).

40 mg/m3 (1 hour). Formaldehyde (HCHO): Particleboards, paneling, plywood, carpets,ceiling tiles, urea-formaldehyde foam insulation, and otherconstruction materials.

120 μg/m3.

Inhalable Particle Matter: Smoking, vacuuming, wood stoves, andfireplaces

50 to 110 μg/m3 (annually).150 to 350 μg/m3 (24 hours).

Nitrogen Dioxide (NO2): Kerosine and gas space heaters and gasstoves.

100 μg/m3 (annually).

Ozone (O3): Photocopy machines and electrostatic air cleaners. 235 μg/m3 (once a year).



102

Radon (Rn-222) & Radon Progeny: Diffusion from soil, groundwater,and building materials.

0.01 working level (annually).

Sulfur Dioxide (SO2): Kerosine space heaters. 80 μg/m3 (annually).365 μg/m3 (24 hours).

Volatile Organics (VOCs): Cooking, smoking, room deodorizers,cleaning sprays, paints, varnishes, solvents, carpets, furniture, anddraperies.

None available.

Some Interesting Statistics:

U.S. LDC1 (urban) LDC (rural ) Indoors: 89%.Outdoors: 5%.

In Vehicles: 6%.

Indoors: 79%.Outdoors: 21%.

Indoors: 65%.Outdoors: 35%.

1 LDC: less developed countries.

PM Emissions (U.S.) PM Exposure (U.S.)Coal-Fired Power Plants: 97%.

ETS 1: 3%.Coal-Fired Power Plants: 2%.

ETS : 98%.1

ETS: environmental tobacco smoking. Per capita emissions of PM (particulate matter): 1.6kg/person(coal-fired power plants), 0.05 kg/person (ETS).

Benzene Emissions (U.S.)

Benzene Exposure (U.S.)

Cars: 82%. Industry: 14%. Personal and Homes: 3%.Cigarettes: 0.1%.

Care Exhaust : 18%. Personal Activities (e.g., driving, consumer products): 18%. Home Sources (e.g., paints, petrol): 16%.Cigarettes: 40%.

ETS : 5%. Industry: 3%.

Important Indoor Air Pollutants.

Radon. Radon is the invisible, radioactive atomic gas that results from radioactive decayof some

forms of uranium that may be found in rock formations beneath buildings or in certain building

materials themselves. Radon is probably the most pervasive serious hazard for indoorair in the

U.S. and Europe, probably responsible for tens of thousands of lung cancer deaths perannum.

There are relatively simple tests for radon gas, but these tests are not commonly done,even in

areas of known systematic hazards. Radon is a very heavy gas and thus will tend toaccumulate atthe floor level. Building materials can actually be a significant source of radon, but very

littletesting is done for stone, rock or tile products brought into building sites. The half life

for radonis 3.8 days indicating that once the source is removed, the hazard will be greatly reduced

within afew weeks.



103

Molds and Other Allergens. These biological agents can arise from a host of means, butthere

are two common classes: (a) moisture induced growth of mold colonies and (b) naturalsubstances released into the air such as animal dander and plant pollen. Moisture

buildup inside buildings may arise from water penetrating compromised areas of the

building envelope or skin, from plumbing leaks, from condensation due to improper ventilation, or from ground moisture penetrating a building slab. In areas wherecellulosic materials (paper and wood, including drywall) become moist and fail to dry

within 48 hours, mold colonies can propagate and release allergenic spores into the air.In many cases, if materials have failed to dry out several days after the suspected waterevent, mold growth is suspected within wall cavities even if it is not immediately visible.Through a mold investigation, which may include destructive inspection, one should beable to determine the presence or absence of mold. In a situation where there is visiblemold and the indoor air quality may have been compromised, a mold remediation may

be needed. Indoors, mold growth can be inhibited by keeping humidity levels below fiftypercent and by eliminating leaks or moisture condensation and accumulation. There aresome varieties of mold that contain toxic compounds (mycotoxins). However, exposure

to hazardous levels of mycotoxin via inhalation is not possible in most cases, as toxinsare produced by the fungal body and are not at significant levels in the released spores.The primary hazard of mold growth, as it relates to indoor air quality, comes from theallergenic properties of the spore cell wall. More serious than most allergenic propertiesis the ability of mold to trigger episodes in persons that already have asthma, a seriousrespiratory disease.

Carbon Monoxide. One of the most acutely toxic indoor air pollutants is carbonmonoxide (CO), a colorless, odorless gas that is a byproduct of incomplete combustion offossil fuels. Common sources of carbon monoxide are tobacco smoke, space heatersusing fossil fules, defective central heating furnaces and automobile exhaust.Improvements in indoor levels of CO are systematically improving from increasing

numbers of smoke-free resutrants and other legislated non-smoking buildings. Bydepriving the brain of oxygen, high levels of carbon monoxide can lead to nausea,unconsciousness and death.

Legionella. Legionellosis or Legionnaire's Disease is caused by a waterborne bacteriumthat grows best in slow moving or still warm water. The primary route of exposure isaerosolization, most commonly from evaporative cooling towers or showerheads. Acommon source of Legionella in commercial buildings is from poorly placed ormaintained evaporative cooling towers, which often release aerosolized water that mayenter nearby ventilation intakes. Outbreaks in medical facilities and nursing homes,

where patients are immuno-suppressed and immuno-weak, are the most commonlyreported cases of Legionellosis. More than one case has involved outdoor fountains in

public attractions. The presence of Legionella in commercial building water supplies ishighly under-reported, as healthy people require heavy exposure to acquire infection.

Legionella testing typically involves collecting water samples and surface swabs fromevaporative cooling basins, shower heads, faucets, and other locations where warm watercollects. The samples are then cultured and colony forming units (cfu) of Legionella arequantified as cfu/L. Legionella is a parasite of protazoans such as amoeba, and thusrequires conditions suitable for both organisms. The bacterium forms a biofilm which isresistant to chemical and antimicrobial treatments, including chlorine. Remediation for



104

Legionella outbreaks in commercial buildings vary, but often include very hot waterflushes (70°C), sterilization of standing water in evaporative cooling basins, replacementof shower heads, and in some cases flushes of heavy metal salts. Preventative measuresinclude adjusting normal hot water levels to allow for 50°C at the tap, evaluating facilitydesign layout, removing faucet aerators, and periodic testing in suspect areas.

Asbestos Fibers. Many common building materials used before 1975 contain asbestos,such as some floor tiles, ceiling tiles, taping muds, pipe wrap, mastics and otherinsulation materials. Normally significant releases of asbestos fiber do not occur unlessthe building materials are disturbed, such as by cutting, sanding, drilling or buildingremodelling. There are particularly stringent regulations applicable to schools.Inhalation of asbestos fibers over long exposure times is associated with increasedincidence of lung cancer. While smokers have a greater risk of lung cancer than asbestos

workers that do not smoke, smokers that are exposed to high levels of asbestos overmany years have a much greater risk of developing lung cancer than either smokers thathave not been exposed to asbestos, or persons that have been exposed to high levels ofasbestos that do not smoke.

Carbon Dioxide. Carbon dioxide (CO2) is a surrogate for indoor air pollutants emitted byhumans and correlates with human metabolic activity. Carbon dioxide at levels that areunusually high indoors may cause occupants to grow drowsy, get headaches, or function

at loweractivity levels. Humans are the main indoor source of carbon dioxide. Indoor levels are

anindicator of the adequacy of outdoor air ventilation relative to indoor occupant density

andmetabolic activity. To eliminate most indoor air quality complaints, total indoor carbon

dioxideshould be reduced to below 600 ppmv above outdoor levels. NIOSH (National Institute

of

Occupational Safety and Health) considers that indoor air concentrations of carbondioxide thatexceed 1,000 ppmv are a marker suggesting inadequate ventilation (1,000 ppmv equals

0.1%). ASHRAE (American Society of Heating, Refrigerting and Air-Conditioning Engineers)recommends that carbon dioxide levels not exceed 700 ppmv above outdoor ambient

levels. TheUK standards for schools say that carbon dioxide in all teaching and learning spaces,

whenmeasured at seated head height and averaged over the whole day should not exceed

1,500 ppmv.The whole day refers to normal school hours (i .e., 9.00 A.M. to 3.30 P.M.) and includes

unoccupied periods such as lunch breaks. Canadian standards limit carbon dioxide to3,500 ppmv.

Figure V.1 illustrates sources of indoor air pollutants in homes.



105

Figure V.1. Sources of indoor air pollutants in homes.

Analysis of Indoor Air Pollution. Just with outdoor air, the amount of air availableto dilute

pollutants is an important indicator of indoor air pollution. Indoor air can be exchanged with

outdoor air by any combination of the following three mechanisms: infiltration, naturalventilation, and forced ventilation. Infiltration is the term used to describe the natural

airexchange that occurs between a building and its environment when doors and windows

areclosed. That is, it is leakage that occurs through various cracks and holes that exists in

the building envelope. Natural ventilation is the air exchange that occurs when windows or

doorsare purposely opened to increase air circulation. Forced ventilation occurs when

mechanical airhandling systems induce air exchange using fans or blowers.

Nationwide, roughly 10% of total energy consumption is attributed to the losses caused by infiltration. Since infiltration is quite easily and cheaply controlled, it is notsurprising that tightening buildings has become a popular way of saving energy.However, this process of saving energy also exacerbates indoor air quality problemsunless the sources of indoor air pollution were simultaneously reduced. The infiltrationsites in homes include: chimneys, ducts, vents, dropped ceilings, whole house fans,plumbing, door frames, windows, exhaust vents above stoves, fireplaces, and electricaloutlets. Infiltration is driven by pressure differences between the inside of the buildingand the outdoor air. These pressure differences can be caused by wind, or by inside-to-



106

outside temperature differences. Wind blowing against a building creates higherpressure on one side of the building than the other, inducing infiltration through cracksand other openings in the walls. Temperature induced infiltration (stack effect ) isinfluenced less by holes in the walls than by various openings in the floors and ceilings.In the winter, warm air rises in the building and exits through breaks in the ceilings anddraws in cooler air through floor openings. As a result, infiltration rates are influenced

not only by how fast the wind is blowing and how great the temperature difference is between inside and outside, but also by the locations of the leaks in the buildingenvelope. Greater leakage areas in the floors and ceilings encourage stack-driveninfiltration, while leakage areas in vertical surfaces encourage wind-driven infiltration.Figure V.2 illustrates the infiltration sites in homes.

Figure V.2. Infiltration sites in homes.

While it is usually assumed that increasing the infiltration rates will improve indoor airquality, it may not be the case when radon is emitted from the soil under a building. For

radon, wind-driven infiltration helps reduce indoor radon concentrations by allowingradon-free fresh air to blow into the building. Stack-driven infiltration, which draws airthrough the floor, may actually induce new radon to enter the building, negating thecleaning that infiltration usually causes Figure V.2). Infiltration rates may be expressedin units such as m3/hr or ft3/min (cfm). More often the units are given in air changes

per hour (ach). The air exchange rate in air changes per hour is simply the number oftimes per hour that a volume of air equal to the volume of space in the house isexchanged with outside air (0.5 ach to 1.0 ach in American homes). A basic mass

balance equation of indoor air pollution in the building, assuming that the air mass inthe building is completely mixed, yields:

V

V D

eC ek D

DC V

M

t C

kVC C V M C V dt

dC V

t k Dt k Di

i

)()( )0(1)( ( V.1)



107

where V is the air volume in the building, m3; D is the air exchange rate (or dilution rate),hr-1; M is the source emission rate in the building, mg/h; C is the indoor pollutantconcentration, mg/m3; C i is the ambient pollutant concentration, mg/m3; and k is thefirst-order decay rate, hr-1. Some of the pollutants such as CO and NO can be treated asnon-reactive (i .e., k = 0). Some useful information is listed below.

1. Measured Emission Rates ( M ) for Various Sources (mg/hr).

Source CO NOX-N SO2 HCHO

Gas Range - Oven- One Top Burner

1,9001,840

5283

0.91.5

2316

Kerosene Heater - Convective- Radiant

71590

12215

----

1.14.0

One Cigarette 86 0.05 -- 1.44

2. First-Order Pollutant Decay Rate (k, hr-1).

Pollutant CO NO NOX-N HCHO SO2 PM (< 0.5μm)

Radon

k 0.0 0.0 0.15 0.40 0.23 0.48 7.6×10-3

Example V.1. An unvented, portable, radiant heater, fueled with kerosene, is testedunder controlled laboratory conditions. After running the heater for two hours in a testchamber with a 46 m3 volume and an infiltration rate of 0.25 ach, the concentration ofCO reaches 20 mg/m3. Initial CO in the laboratory is zero, and the ambient CO isnegligible throughout the run. Treating CO as a non-reactive pollutant, find the rate at

which the heater emits CO. If the heater were to be used in a small home to heat 120 m3

of space having 0.4 ach, predict the steady-state indoor CO concentration.

1. With C i = C (0) = k = 0, then the 2nd equation in ( V.1) can be simplified to:

Dt e

t DVC M

1

)(.

2. Therefore,)2)(25.0(1

)20)(46)(25.0(

e

M = 585 mg/hr.

3. From the 2nd equation in ( V.1),)120)(4.0(

585)(

DV

M C = 12.2 mg/m3

(steady-state concentration). This steady-state indoor CO concentration exceeds

the 8-hour ambient standard of 10.0 mg/m3

.URBAN/REGIONAL AIR POLLUTION. Air pollutants in urban/regional areasare from two major sources: stationary (point) and mobile (non-point) sources. As apollutant mixes with a large air mass, its concentration diminishes by dilution. Withsufficient dilution, concentration is reduced to or below a threshold level which isassumed to cause no ill effects. Soil microbes or other abiotic processes may absorb andassimilate pollutants, removing them from the system entirely. Therefore, theassumption that nature will take care of pollutants may be true. Unfortunately, modern



108

civilization produces pollutants in such quantities and kinds that this assumption doesnot hold. A temperature inversion may cause episodes of high concentration of airpollutants. Normally, air temperatures are highest at the ground level and decrease athigher elevations. Since warmer air rises, pollutants are carried upward and diluted inthe air above. A temperature inversion is a situation in which a layer of warmer airoverlies cooler air at the ground level. This in turn blocks the normal updrafts and

causes pollutants to accumulate like cigarette smoke in a closed room.

Urban Smog. When oxides of nitrogen, VOCs (volatile organic carbons), and sunlightcome together, they can initiate a complex set of reactions that produce a number ofsecondary pollutants known as photochemical oxidants. Ozone (O3) is the mostabundant of the photochemical oxidants and it is the one for which an ambient airquality standard has been written. Although ozone is responsible for many of theundesirable properties of photochemical smog, from chest constriction and irritation ofthe mucous membrane in the people to the cracking of rubber products and damage to

vegetation, it is not itself a cause of the eye irritation that is our most common complaintabout smog. Eye irritation is caused by other components of photochemical smog,principally formaldehyde (HCHO), peroxybenzoyl nitrate (PBzN), peroxyacetyl nitrate

(PAN), and acrolein (CH2CHCOH).

7 oxides of nitrogen are known to occur: NO, NO2, NO3, N2O, N2O3, N2O4, and N2O5, theonly two that are important air pollutants are nitric oxide (NO) and nitrogen dioxide(NO2). There are two sources of nitrogen oxides (or NOX) when fossil fuels are burned.Thermal NO X is created when nitrogen and oxygen in the combustion air are heated to ahigh enough temperature (roughly 1,000oK) to oxidize the nitrogen. Fuel NO X resultsfrom the oxidation of nitrogen compounds that are chemically bound in the fuelmolecules. Natural gas has almost no nitrogen compounds and some coal has as muchas 3% (by weight) of nitrogen compounds. Fuel NOX is the dominant source of total NOX emissions. Almost all NOX emissions are in the form of NO, which is a colorless gas thathas no known adverse health effects at concentrations found in the atmosphere.

However, NO can oxidize to NO2 that irritates the lungs, causes bronchitis andpneumonia, and lowers resistance to respiratory infections. NOX can also react with VOCs in the presence of sunlight to form photochemical oxidants that have adversehealth consequences.

VOCs include unburnt hydrocarbons that are emitted from tailpipes and smoke stacks when fossil fuels are not completely combusted, along with gaseous hydrocarbons thatenter the atmosphere when solvents, fuels, and other organics evaporate. In addition,there are natural sources of reactive hydrocarbons such as deciduous trees and shrubsthat emit isoprene (CH2CCH3CHCH2), and conifers that emit pinene and limonene.Transportation sector is responsible for about one-third of anthropogenic VOCemissions. From 1984 to 1993, VOC emissions from motor vehicles decreased by about25%. Reformulated gasoline (which is less volatile) and the decline in the percentage ofdirtier, older vehicles on the road account for such a drop. As emissions from motor

vehicles become less significant, other sources, such as gasoline-powered lawnmowers,outboard motors, barbeque starter fluids, and oil-based paints, begin to look moreimportant and are beginning to be regulated as well. Industrial sources account for two-thirds of VOC emissions, with much of that being caused by vaporization ofhydrocarbons. Less than 2% of VOCs result from fossil-fuel combustion in power plantsand industrial boilers.



109

Formation of Photochemical Smog (or Smog). Important reactions that causethe formation of smog in urban air mass are summarized in ( V.2):

223

32

2

22

22

22

2

NOO NOO

M O M OO

O NOh NO

NOO NO

NOO N

Smog h NOVOCs X

( V.2)

where hν represents a photon (with wavelength λ < 0.39 μm) and M represents amolecule (usually oxygen and nitrogen since they are most abundant in the air). 1st equation: formation of photochemical smog, 2nd equation: combustion of fossil fuel, 3rd equation: oxidation of NO, 4th equation: photolysis, 5th equation: formation of O3, and 6th equation: formation of NO2 (or destruction of O3).

A Box Model for Urban Air Pollution. Valuable insights can be gained bydescribing mathematically the accumulation of pollutants over an urban area. Althoughsophisticated computer models are available, a simple representation of the scenario isoften capable of producing useful information for making critical decisions to rectify theproblems. The box model, which models the air mass over an urban area as a CFSTRsystem (Figure V.3), is a simple but useful first approximation tool for such tasks. The

box has horizontal dimensions Δ x and Δy (the dimensions of the urban area) and height H ( H is also referred to as the atmospheric mixing height and for a typical urban area itis about 1,000 m). H can be much smaller than 1,000 m when thermal inversion occursabove the urban area. Let u be the average wind velocity (L/T) with the wind directionparallel to the x dimension of the urban area, M= mA = mΔ x Δy be the mass emission

rate of the pollutant from the surface (M/T) (m is the emission rate per unit surface area,M/L2-T), R1 be the mass removal rate of the pollutant in the air mass due to theprocesses of dry deposition (M/L3-T), R2 be mass removal rate of the pollutant in the airmass due to other processes (e.g., photolysis) (M/L3-T), and P be the mass generationrate of the pollutant in the air mass (M/L3-T). Then,

P RC H

vC C

H

m

dt

dC

P RC H

v

H

mC C

dt

dC P RC

H

v

y x H

M C C

x

u

t

C

y x PH y x H R y x H R M yC uH yC uH t

C y x H

d

o

d

o

d o

o

2

22

21

0

)(

( V.3)

where C o is the background pollutant concentration (M/L3), C is the pollutantconcentration in



110

1: Pollutant Dry Deposition Velocity (vd ).

2: Total Pollutant Emission Rate from the Surface ( M = mΔ x Δy).

Figure V.3. A box model for the air mass over an urban area.

the urban air mass (M/L3), vd is the pollutant dry deposition velocity (L/T), and

u

x is

the residence time (T). vd is a parameter used to describe how rapidly the pollutant(either solid particles or aerosol particles) is removed from the air mass by deposition

onto the ground (usually by gravity). The mass flux ( F , M/L2-T) of dry deposition (i .e.,the mass rate at which the pollutant is deposited onto a unit surface area) is usuallyexpressed as: F = vd C .

Example V.2. The city of Leadville (dimensions: 120 km × 120 km) is characterized byan average daily emission rate of 5,000 kg/d of lead into the atmosphere from autoexhaust and industrial processes. The background concentration of atmospheric lead is0.1 μg/m3. During a typical hot summer day in Leadville the atmospheric mixing heightis 500 m and the average wind velocity is 2 m/s. A typical lead deposition velocity forsimilar meteorological conditions is 0.05 cm/s. (1) What is the steady-stateconcentration of lead in the urban air? (2) What (if anything) must Leadville citizens doto achieve an atmospheric lead concentration of 0.4 μg/m3? (3)What if the air qualitytarget is 0.15 μg/m3?

1.1.

u

x =

)600,3)(2(

)000,1)(120( 16.7 hrs,

2

6

)000,1)(120()24(

)10)(000,1)(000,5(m = 15 μg/m2-hr, and

vd = 0.05 cm/s = 1.8 m/hr.

1.2. From the 4th equation in ( V.3),

)7.16)(8.1(500

)7.16)(15()500)(1.0(

d

o

v H

m H C C 0.57

μg/m3.2.1. Under the conditions specified, the only variable the citizens of Leadville can

readily control is the lead emission rate, m. From the 3rd equation in ( V.3),

Air Mass

Δ x

Δy

H 1 2

Wind (u, C o) Wind (u, C )

Urban Area



111

7.16

)500)(1.0()7.16)(8.1(500)4.0()(

H C v H C m

o

d 9.7 μg/m2-hr.

2.2. Therefore, the reduction in m is 5.3 μg/m2-hr (≈ 35% reduction). 3. If the target atmospheric lead concentration is 0.15 μg/m3, then m = 1.8 μg/m2-

hr (≈ 90% reduction).

The Photochemical Cycle of NOX and Formation of Ozone. From ( V.2) thefollowing

kinetic expressions for the formation of O3 in the urban areas can be formulated:

][

][][

]][[

][

]][[

3

2232

332

22

213

NOk

NOk O R R

O NOk R

NOk R

OOk R

NO ph oto lysis

NO

ph tolysis

O

( V.4)

The 4th equation in ( V.4) describes the photo-stationary state relationship in the urbanair. First,

[O3] depends on directly on sunlight intensity through the photolytic rate constant, k2.Therefore, [O3] is expected to be highest during the sunny midday and to decline to near

zeroovernight. Second, [O3] depends on the relative amounts of [NO2] and [NO]. Since NOX emissions in an urban area are mainly NO with little NO2, the initial equilibrium O3 concentration as shown by the 4th equation in ( V.4) will be low. As NO is oxidized to

NO2, the

ratio of][

][ 2

NO

NO will change with time. However, the reaction equations in ( V.2) indicate

thateach time a molecule of NO reacts, a molecule of NO2 is produced, and vice versa; the

sum oftheir concentrations will always remain constant (i .e., [NO] + [NO2] = [NOX]). As a

result, theequilibrium ozone concentration can be calculated by:

030

3

2

2

3

2030

3

20303 ][][4][][5.0][][5.0][ O NO

k

k

k

k O NO

k

k O NOO

( V.5) where the subscript 0 indicates the initial concentration at t = 0.

VOCs and Formation of O3 (Figure V.4). The O atoms formed via reactions shownin ( V.2)

can react with water vapor in the atmosphere to form two hydroxyl radicals. Thehydroxyl radical

will react with VOCs ( RH ) in the atmosphere that lead to the conversion of NO to NO2:



113

concentration in theurban air is held constant, the maximum O3 concentration will increase with increasing

NOX concentration until a maximum value is reached (point B in Figure V.5). Beyond that,

themaximum O3 concentration will decrease even the NOX concentration is continuously

increased.This is a typical result when two counter-reaction mechanisms exist in the same system.

If point A is the current situation in the urban air mass, then at a constant VOC level, the

reduction inNOX emissions will actually increase the O3 concentration significantly.

Ozone Isopleth Diagram (Figure V.6). The ozone isopleth diagram is prepared based on the

assumptions that the box model previously presented is applicable and that VOC andNOX

concentrations in the atmosphere are proportional to their respective source emission

rates. It isuseful to devise the strategies for smog control in a given urban area. The diagramseparates the

X : Initial NOX Concentration.Y : Maximum O3 Concentration.◙ The VOC concentration in the urban air mass is constant.

Figure V.5. Predicted maximum O3 concentration as a function of NOX levelfor constant VOC.



114

Figure V.6. The ozone isopleth diagram.

Figure V.6. The ozon isopleths diagram.diagram into two regions: VOC-limited regime (point A) and NOX-limited region (point

B). Labels on curved lines are the peak O3 concentrations in ppbv in an air mass. Theinitial VOC level represents a weighted sum of all VOC mole fractions, where the

weighting factor is equal to the number of carbon atoms per molecule.

Ozone Formation Potentials of VOCs. In the discussions so far, all hydrocarboncompounds have been lumped together and included under VOCs. However, not onlythere are hundreds of unique hydrocarbon compounds in the atmosphere, but differentcompounds are emitted from different sources. The approaches summarized below take

into account of the ozone formation potentials of different hydrocarbon compounds.

Approach Remarks

Carbon Mass

◙ Only consider the mass of carbon in VOC molecules.◙ Reactivity is not considered.

◙ Calculate i

n

i

ic M 1

for both current and new situations and

compare the difference (i is referred to as the i th VOC molecule, M i isthe emission rate of the i th VOC molecule, and ci is the % of carbonin the i th VOC molecule).

Reactive Organic Gas ( ROG )(Carbon Mass)

◙ Use kOH of ethane (C2H6) as the dividing line. Any VOCs with kOH ’s>

kOH (C

2H

6) are considered as reactive and included in the

calculation.

◙ Calculate i

n

i

ic M 1

for both current and new situations and

compare the difference.

kOH (cm3/molecule-s) ◙ Calculate i

n

i

iOH M k 1

)( for both current and new situations and

compare the difference.



116

approach, (2) the ROG carbon mass approach, (3) the kOH reactivity scale, and (4) the MIR reactivity scale. Based on the results, is it a good idea to use the new technology toreduce the O3 formation?

1.1 The current carbon emissions are:

44

245.1

28

242

28

123.1 = 3.2 kg C/day.

1.2. The new carbon emissions are:32

1211

58

363.0

28

1210 = 8.6 kg C/day.

1.3. Therefore, 2.36.8 = 5.4 kg C/day > 0.1.4. By this measure, O3 formation is predicted to increase using the new technology.

2.1. The current ROG carbon emissions are:44

245.1

28

242 = 2.5 kg C/day.

2.2. The new ROG carbon emission is:32

1211 = 4.1 kg C/day.

2.3. Therefore, 5.21.4 = 1.6 kg C/day > 02.4. By this measure, O3 formation is predicted to increase using the new technology.

3.1. The current emissions are: 1210)165.15.8221.01.1( = 41.3 reactivityunits.3.2. The new emissions are: 1210)94.01122.03.021.010( = 12.5 reactivityunits.3.3. Therefore, 8.283.415.12 reactivity units < 0.3.4. By this measure, O3 formation is predicted to decrease using the new technology.4.1. The current MIR is: )3.65.13.82065.02.1( = 26.1 kg O3/day.

4.2. The new MIR is: )65.01149.03.0065.010( = 7.1 kg O3/day.

4.3. Therefore, 0.191.261.7 kg O3/day.4.4. By this measure, O3 formation is predicted to decrease using the new technology.5. The first two approaches indicate that the new technology would increase the O3

formation in the region. However, according to the 3rd and 4th approaches, thenew technology would decrease the O3 formation in the region, especially the

MIR approach predicts a significant decrease of 19 kg O3 per day. So it wouldindeed be a good idea to use the proposed new technology to reduce the O3 formation in the region.

ACID PRECIPITATION. Emissions of sulfur and nitrogen oxides react with water vapor in the atmosphere to form their respective acids (i .e., H2SO4 and HNO3) whichcome back down as dry acid deposition or mixed with water, causing the precipitation to

be abnormally acidic. Coal-burning power plants are primarily responsible for theemissions of sulfur dioxide and nitrogen dioxide. The emissions from the tall stacks ofcoal-burning power plants located in the Ohio River Valley tend to migrate to NewEngland and eastern Canada. Tail-pipe emissions from motor vehicles are the othermajor sources of nitrogen dioxide.

The average acidity of rainfall in Scandinavia, the northeastern U.S., Canada, and partsof Europe has increased steadily over the past 40 years. There seem no question thatthis change is primarily due to the increased emissions of sulfur oxides (SOX) andnitrogen oxides (NOX) that have accompanied the greatly increased economic activities(and hence increased combustion of fuels) in or upwind of these regions. SOX and NOX



117

are slowly oxidized over several hours to several days to sulfuric acid (H2SO4) [( V.8)]and nitric acid (HNO3) [( V.9)]. They then are generally captured by raindrops andreturned to the Earth’s surface as acid precipitation. The common name is acid rain,

but the complete description includes acidic rain, acidic snow or hail, acids adsorbed onfalling dust particles, etc. In addition, hydrochloric acid (HCl) can be directly releasedfrom coal combustion and MSW incineration. Figure V.7 illustrates how acid

precipitation is formed in the atmosphere.

4223

32

322

22

SO H O H SO

SOOSO

OSOOSO

SOhSO

( V.8)

3252

5232

2332

32

23

2 HNOO H O N

O N NO NO

O NOO NO

HNO NOOH

OOhO

( V.9)

Rain falling through a perfectly unpolluted atmosphere will arrive at the earth’s surface with a pH

Figure V.7. The formation of acid precipitation in the atmosphere.

of about 6.5 because of the carbon dioxide in the atmosphere:

33222 HCO H CO H O H CO ( V.10)

Carbonic acid (H2CO3) is a weak acid, and ( V.10) is reversible, with the acidconcentration in the rain depending on the concentration of carbon dioxide in the air.Generally, any rain with a pH < 5.6 (i .e., an H+ concentration > 10-5.6 mol/L) isconsidered acidic, but damages to animals (or fish) and plants are not evident until a pHof about < 4.5 is reached. Although damages to



118

human health have not been shown, acid damages to the ecology of mountain lakes andforests are well documented. The transport distances between emission andprecipitation are generally hundreds of miles, so that local control seems futile in manycases. The effects of acid precipitation on materials are emerging quite clearly. Acidsdegrade building materials, especially limestone, marble, various commonly used metalssuch as galvanized steel, and certain paints. In fact, the increased rate of weathering and

erosion of building surfaces and monuments was one of the first indications of adverseimpacts from acid precipitation. It is important to note however, that adding acid to asolution may have little or no effects on pH, depending on whether or not the solutionhas buffers. Buffers are substances capable of neutralizing added hydrogen ions. Theavailable buffering of an aquatic ecosystem (e.g., a lake) is not only a function of itschemical characteristics, but also nearby soils through which water percolates as ittravels from land to the system concerned. Therefore, information on the pH ofprecipitation alone, without taking into account the chemical characteristics of thereceiving water body and surrounding soils, is a poor indicator of the potential effects ofacid precipitation on an aquatic ecosystem. Most aquatic ecosystems are buffered by

bicarbonate (HCO3-), which is related to carbonic acid by:

332 HCO H CO H ( V.11)

Note that the reaction of hydrogen ions with bicarbonate removes bicarbonate from thesolution so, unless there is a source of new bicarbonate, its concentration will decrease asmore acid is added. At some point, there may be so little bicarbonate left that relativelysmall additional inputs of acid will cause pH to decrease rapidly. This phenomenonleads to one way to classify lakes in terms of their acidification. As shown in Figure V.8,a bicarbonate lake ( I ) shows little decrease in pH as hydrogen ions are added, until pHdrops to about 6.3. As pH drops below this point, the bicarbonate buffering is rapidlydepleted and the lake enters a transitional state (a transition lake, II ). Transition lakes

with pH between 5.0 and 6.0 are very sensitive to small changes in acid. Below pH 5.0,

the lake is unbuffered and chronically acidic (an acid lake, III ). In a Norwegian study,684 lakes were categorized into bicarbonate, transition, and acid lakes. For eachcategory observations of fish populations were made which indicated the strongdependency of fish population on the degree of acidification in lakes.



119

Figure V.8. Lakes and their abilities to resist pH changes.

Fish Population(↓)

Bicarbonate Lakes(129)

Transition Lakes(332)

Acid Lakes(223)

No Fish 10% 25% 75%Sparse 25% 60% 25%Good 55% 10% 0%

Overpopulated 10% 5% 0%

Example V.4. The total annual U.S. emissions of SO2 in 1997 were 20.4 million tons.If it is assumed that 25% of total emissions was in the Midwest-Ohio Valley area, and50% of that came to ground as acid precipitation in a 1.0×106 km2 area in thenortheastern U.S. and southeastern Canada, and that average precipitation over thatarea is 1 m/yr, by how much would this SO2 (if all converted to H2SO4) change the pH ofthe rain water which was 5.6?

1. Annual U.S. SO2 emissions = 20.4×106 ton = 3.2×1011 moles.2. Annual precipitation = (1.0×1012 m2)(1m) = 1015 L.3. Since each mole of SO2 produces two moles of H+ ions, so the increase in [H+] =

(2)(0.25)(0.50)(3.2×1011 moles)(1015 L)-1 = 7.23×10-5 mol/L.4. The original [H+] = 10-5.6 = 2.51×10-6 mol/L.

5. The total new [H+] = 2.51×10-6 + 7.23×10-5 = 7.48×10-5 mol/L or the new pH =4.13.

Control of SOX and NOX (Motor Vehicles). During the exhaust stroke of aninternal combustion engine, combustion gases are pushed through the exhaust manifoldand out the tailpipe, and it is in this exhaustion system that most of the control ofautomobile emissions now occurs. The most commonly used systems for treatment ofexhaust gases are thermal reactors, exhaust gas recirculation ( EGR) systems and

catalytic converters. A thermal reactor is basically



120

an afterburner that encourages the continued oxidation of CO and unburnedhydrocarbons (HC)

after these gases have left the combustion chamber (i .e., engine). The reactor consists ofa multi-

pass, enlarged exhaust manifold with an external air source. Exhaust gases are kept hotin the

thermal reactor and enough oxygen is provided to allow combustion to continue outsidethe

engine itself, thereby reducing CO and HC emissions. Usually, the carburetor isdesigned to

cause the engine to run rich in order to provide sufficient unburned fuel in the thermalreactor to

allow combustion to take place. This has the secondary effects of modestly reducing NOX emissions, but fuel consumption is increased. Some degree of control of NOX can be

achieved byrelatively inert gas that is added to the incoming air/fuel mixture absorbs some of the

heatgenerated during combustion without affecting the air/fuel ratio. The heat absorbed

helpsreduce the combustion temperature, and, therefore, helps decrease the production ofNOX. The

coupling of exhaust gas recirculation with a thermal reactor (Figure V.9), reduces theemissions

of all three pollutants, CO, HC, and NOX, but at the expense of performance and fueleconomy.

The most popular approach to control automobile emissions is the three-way catalyticconverters (three-way just means that it handles CO, HC, and NOX). A three-waycatalytic converter is ableto oxidize CO and HC to CO2 while reducing NOX to N2 all in the same catalyst bed.

These catalytic converters are very effective in controlling emissions, and they have theadvantage of allowing the engine to operate at near optimal conditions. Catalyticconverters can be destroyed very quickly if leaded fuels are burned. Three-way catalyticconverters are very effective once they are warmed up, but when they are cold, as well as

when there are spurts of sudden acceleration or deceleration, they release excessiveamounts of pollution that offset much of their perceived benefits. Those shortcomingscan be addressed by using the new multiple catalytic converters to be on the market verysoon. Another way to help reduce motor vehicle emissions is to provide more oxygen forcombustion by incorporating oxygen-containing additives in gasoline. The mostcommon additives are ethanol or MTBE. MTBE has certain advantages over ethanol: ithas a higher octane number (% volume of C8H18 in the mixture of C8H18 and C7H16) and itis not as volatile. But MTBE may cause nausea, dizziness, and headaches. Benzene,toluene, and xylene (BTX) have been used to substitute tetraethyllead as octaneenhancers. BTX



121

Figure V.9. Exhaust gas recirculation system.

hydrocarbons are more reactive than normal gasoline constituents and they can increase

the formation of photochemical smog.

The utilization of alternative fuels other than gasoline and diesel has been promoted as afeasible means to reduce emissions that cause photochemical smog and acidprecipitation. In addition, a number of bio-based fuels (or biofuels) that can be used toreduce the consumptions of gasoline and diesel in transportation sectors are alsopromoted. The consumption of gasoline in the U.S. would increase to roughly 290

billion gallons in 2050 if the gas mileage requirements remain unchanged. On the otherhand, it is estimated that roughly 67% of the U.S. transportation fuel demands in 2050can be met with biomass ethanol that is produced with current agricultural crop/plantresidue mix if 50 mpg requirements are imposed.

Fuel Remarks

Methanol (CH3OH)

It has a high octane number than gasoline and a lower flametemperature than gasoline and diesel. As a result, it reducesNOX, CO, and HC emissions. However, it is difficult to startengines in cold conditions, because methanol is less volatile. Inaddition to a higher formaldehyde emission (toxic), methanolitself is also is highly toxic.

M85

The fuel has 85% methanol and 15% gasoline. The octanerating is 102 (87-92 for gasoline). The driving distance is about60% of that of gasoline.

Gasoline/Ethanol (Flex Fuel)Gasohol contains 10% ethanol (C2H5OH) and 90% gasoline andE85 contains 85% ethanol and 15% gasoline. The flex fuel has a

lower CO and HC emissions than gasoline. Also, ethanol is arenewable resource.

Compressed Natural Gas (CNG)It is used in the fleets of trucks, delivery vans, and buses. It haslow CO, HC, particulate, and toxic emissions, but higher NOX emissions than gasoline.

R1 r GHG (kg/m3)2,3 Remarks

Corn EthanolThe yield is about 0.46 m3/ha. The U.S. produced1.84×107 m3 @ $288/m3 (2006). Most ethanol is sold as



122

1.3 1,943 a gasoline additive (e.g., to replace MTBE, a suspectedcarcinogen) or, as E85 fuel.

Cane Ethanol 8.0 1,080The yield is about 0.92-1.38 m3/ha. Brazil produced1.84×107 m3 @ $230/m3 (2005). The stalk is 20% sugarand the waste can be burned to power the distillery andthereby, lowering fossil fuel uses.

Cellulosic Ethanol 2.0-36.0 228Can be produce3d from agricultural residues (e.g., stalksand leaves), forestry residues (e.g., wood chips, sawdust,and tree barks), municipal solid wastes (e.g., paperproducts and household garbage), paper pulp, andswitchgrass which can grow on marginal land. Nocurrent commercial productions.

Biodiesel 2.5 912

Biodiesel is made by chemically altering plant oils.Germany is the world’s leading producer of biodieselusing canola oil (2.32×106 m3 in 2005). Biodiesel isprimarily produced from soybeans in the U.S (yield:0.09 m3/ha).

Algal Oil -- --

Algae can grow in the presence of sunlight by absorbingand utilizing CO2 in fossil-fuel power plant gasemissions (theoretical yield: 7.67 m3/ha). No current

commercial productions.

1 R is the energy yield ratio defined asU

P

E

E R , where E P is the energy value of 1 kg biofuel (kJ)

and E U is the fossil-fuel energy consumed per kg biofuel produced (kJ). 2 rGHG is the greenhousegas (GHG) emission rate (production and use). 3 rGHG (gasoline) is 2,448 kg/m3 and rGHG (diesel)is 2,808 kg/m3.

Control of SOX and NOX (Stationary Sources).

Approach Examples

Pre-Combustion Controls

Examples include switching to fuels with less sulfur or

nitrogen contents in power plants (emissions can be reduced by anywhere from 30 to 90%). In some cases fossil fuels can be physically, chemically, or biologically treated to removesome of the sulfur or nitrogen before combustion.

Combustion ControlsExamples include new burners in power plants that reduceNOX emissions and new fluidized bed boilers that reduce bothNOX and SOX.

Post-Combustion Controls

Capture emissions after they have been formed but before theyare released to the air. On power plants, these may becombinations of particulate collection devices (e.g., dust bagsand electrostatic precipitators) and flue-gas desulfurizationtechniques (e.g., dry and wet scrubbers).

OZONE DEPLETION.

Formation and Destruction of O3 (Chapman Mechanism) (Figure V.10). Inthe stratosphere (10 to 45 km), ozone (O3) is created, destroyed, and then recreated byseveral natural chemical reactions [( V.12)]. The 1st reaction in ( V.12) (i .e., photolysis) iscaused by solar irradiation at wavelengths λ < 240 nm (UV-C ). The 3rd reaction (i .e., absorption of UV) is caused by solar irradiation at wavelengths 240 nm < λ < 320 nm



123

(UV-B and UV-C ). The 4th reaction describes the reformulation of ozone.

Figure V.10. Destruction of O3 in the stratosphere.

23

23

32

2

2

2

OOO

OOhO

OOO

OhO

( V.12)34

2

23

23 )230(

CH HCl CH Cl

OCl OClO

OClOOCl

nm F CCl Cl h F CCl

( V.13)

However, the introduction of some anthropogenic chemicals has disrupted ozone’snatural cycle. This has resulted in enhanced destruction rates of ozone, which has led toozone depletion in the stratosphere [( V.13), mid-altitude, 25 – 40 km]. Over a month, achlorine atom in the stratosphere will destroy roughly 100,000 ozone molecules before iteventually reacts with the small amount of natural methane found in the stratosphere[i .e., ( V.13)]. The HCl molecule formed will be washed out by rain and eventuallytransported down to the troposphere (in a matter of months).

The unit introduced by Dr. Gordon M.B. Dobson (1889-1976, a geophysicist at theUniversity of Oxford) is used to measure the amount of trace gases such as O3 in theatmosphere. One DU refers to a layer of O3 that would be 10 μm thick at 0oC and 1 atm.

For instance, 300 DU of O3 brought down to the earth’s surface at 0o

C and 1 atm wouldoccupy a layer only 3 mm thick. One DU is 2.7×1016 O3 molecules per cm2, or2.7×1020/m² (Example. V.5).

Example V.5 . Calculate the number of molecules of O3 per cm2 of the earth surface for1 DU (Dobson Unit) assuming the ideal gas law is applicable.



124

1.3

33

3O

OO

O RT

V pn (ideal gas law).

2.3

3

33

3

33

3

33

3 O

O

aOO

O

aOaO

a

O

OO

aO h

RT

N p

A

V

RT

N p

A

N n N

RT

V p N n , where

3On is the number

of moles of O3,3O p is the pressure of pure O3,

3OV is the volume of pure O3, R is the

universal gas constant,3OT is the temperature of pure O3, A is the area of the earth’s

surface, and3Oh is the height of a layer of pure O3. Note that aO N n

3 is the number of

molecules of O3. Therefore, )10)(10)(01.0()273)(1021.8(

)10023.6)(1( 43

5

23

3

aO N n = 2.69×1016

molecules of O3/cm2.

3. Also,1 DU = 4.4615 x 10-4 mol O3/m2 or 2.1415 x 10-5 kg O3/m2 .

Figure V.11 illustrates the data on Antarctic O3 concentration in October each year from1980 to 2007.

Atmospheric Absorption of Solar Radiation. Atmospheric absorption of solarradiation is governed by the interaction between photons and matter. If radiation ofincipient intensity I o traverses an absorber of unit area and thickness z , then the intensity

will be reduced to I in accordance with Beer-Lambert law:

z

o

z

o

e

I

I T

e I I

( V.14)

where is the absorptivity or extinction coefficient, L-1 (Figure V.12); and T is thetransmittance.

Example V.6. If there is a 1% decrease of O3 in the ozone layer, what is the percentageof increase in UV transmittance? Assume that at 1atm and 0oC the ozone layer thicknessis 0.34 cm.

1. Differentiating ( V.14) yields:



125

TOMS: Total Ozone Mapping Spectrometer.

Figure V.11. Antarctic O3 concentration in October each year from 1980 to2007.

Figure V.12. UV attenuation in the ozone layer.



127

for C2H2F4? What is H-2402? Halo is referred to as chlorine, fluorine, and bromine.The following formulae are useful:

wxyz H

z y x xyz xyz CFC

'''90

' x is the number of C, ' y is the number of H, and ' z is the number of F.w is the number of C, x is the number of F, y is the number of Cl, and z is the number ofBr.

1. CFC-115: adding 90 to 115 gives 205. Therefore, a molecule contains 2 carbons,no hydrogen, and 5 fluorine atoms.

2. Since two carbon have six bonding sites (# of bonding sites = 2n + 2, where n isthe # of carbon), five of which are taken by fluorine, the remaining site is taken

by chlorine. The chemical formula is therefore C2F5Cl.3. C2H2F4 has two carbon atoms, two hydrogen atoms, and four fluorine atoms.

Subtracting 90 from 224 gives 134.4. Since there is no chlorine, this halocarbon is a hydrofluorocarbon, or HFC-134.

5. H-2402 has two carbon atoms, four fluorine atoms, zero chlorine, and two bromine atoms. Therefore, H-2402 is C2F4Br2.

Ozone Depletion in Stratosphere. Hydrochlorofluorocarbons (HCFCs) andhydrofluorocarbons (HFCs) are substitutes for CFCs since their lifetimes are shorter andtheir breakdowns are relatively quick. HCFCs and HFCs have a modest potential toaffect ozone. HFCs contain no chlorine atoms to threaten the ozone layer, but theycontribute to global warming. According to the Montreal Protocol on Substances that

Deplete the Ozone Layer, all CFCs should be banned by 1996 and all HCFCs will be banned by 2030. HFCs will be restricted if other alternatives exist.

CFC Formul

a

Main Uses ppt1 τ

(years)CFC-11 CCl3F Aerosol propellant and foam blowing. 268 50CFC-12 CCl2F2 Aerosol propellant, foam blowing, and refrigerant. 503 102CFC-113 C2Cl3F3 Solvent. 82 85CFC-114 C2Cl2F4 Aerosol propellant and refrigerant. 20 300Carbon

TetrachlorideCCl4 Solvent. 132 42

Methyl Chloroform CH3CCl3 Solvent. 135 4.9H-1211 CBrF2Cl Fire retardant. 7 20H-1301 CBrF3 Fire retardant. 3 65H-2402 C2Br2F4 Fire retardant. 0.7 20

HCFC-22 CHF2Cl Aerosol propellant, foam blowing, refrigerant, fireretardant, and solvent.

100 12.1

HCFC-141b C2H3FCl2 Foam blowing and solvent. 2 9.4HCFC-142b C2H3F2Cl Foam blowing and solvent. 6 18.41 Atmospheric concentration (1992) in ppt = parts per trillion (molecules/molecules).

A Box (Atmosphere) Model on Ozone Destruction by CFCs. As in the case ofmodeling the air pollution phenomena in an urban area, the box model approach canalso be used to describe the accumulation of CFCs in the atmosphere. Since CFCs areemitted across the globe use to their widespread usage, the CFC concentration in the



128

atmosphere is assumed to be uniform. The box model is constructed by the followingsteps.

1. Rate of change of CFC mass in the box over time isdt

dm.

2. CFC flow into the box = 0 (there are no CFCs in the outer space). CFC flow out ofthe box = 0 (there are no CFCs in the outer space).3. Emission within the box = P . Removal of CFCs within the box = R = −km.

Therefore,

t

mi

a

t

mi

a

mi

a

t t

e N M

N P e

N M

N mt C

N M

mN C

k

e P emt m

P kmdt

dm

1)(

1

1)(

0

0

( V.15)

where k is the first-order rate coefficient, m0 is m at t = 0, τ is the atmospheric lifetime, C is the

CFC i concentration as fraction, N a is Avogadro number (6.023×1023 molecules/mole), M i is the

molecular weight of CFC i , and N m is the number of air molecules in the atmosphere(1.1×1044

molecules).

Example V.9. The production of CFC-11 throughout the 1960s averaged about 1.2×1011 g/yr. Assume that t = 0 corresponds to the beginning of 1960 and that m0 = C 0 = 0. Alsoassume that all of the compound produced each year is emitted to the atmosphere in thesame year. If the emission rate of CFC-11 in the 1960s continued indefinitely, calculatethe steady-state mass and concentration of this CFC. How long would it take for theconcentration to reach 98% of the steady-state value?

1. From the 1st

equation in ( V.15), )50)(102.1(0

11

P mdt

dmSS 6×10

12

g.

2. 9

44

2312

10241.0)101.1)(136(

)10023.6)(106()(

mi

aSS

N M

N P C C

= 0.241 ppb (molecule/molecules basis).3. From the 5th equation in ( V.15),



129

.1

98.0)(

11)(

50

t

SS

t

SS

t

mi

a

e

C

t C eC e

N M

N P t C

4. Therefore, t = 196 years.

GREENHOUSE GASES (GHGs) AND GLOBAL WARMING.

Atmospheric Composition (% volume):

Nitrogen 78

Nitrogen 21

Argon 0.9

Carbon Dioxide 0.03

Carbon dioxide (CO2), methane (CH4), and water vapor are transparent to incomingshort-wavelength solar rays, but which retain the long-wavelength infrared rays.Moreover, they also block a large fraction of the earth’s emitted long-wavelengthradiations. Therefore, they are referred to as greenhouses gases (GHGs).

CO2 Emissions Caused by Fuel Burning (1998):

Sources Pg (as C) (1 Pg = 1×1015 g = 1 billion tons)

Petroleum 2.5

Coal 2.2

Gas 1.0Biomass 1.0

∑ 6.7

1. Current atmospheric CO2 concentration is about 360 ppmv (i .e., 360 m3 CO2 per106 m3 air). ppmv is defined as parts per million (volume).

2. 360 ppmv CO2 =)104.22)(10(

)44)(360(36

= 0.707 g CO2/m3 air. Under the standard

conditions, 1 g mole CO2 is equivalent to 44 g CO2 which has volume of 22.4liters.

3. Total atmospheric CO2 inventory (as C) (1998) = 44)29.1)(101.5)(12)(707.0(

18 =

762×1015 g C = 762 Pg C = 762 billion tons C. The atmospheric density isassumed as 1.29 kg air/m3. The total weight of the atmosphere is 5.1×1018 kg air.

4. 1 ppmv CO2 = (762/360) = 2.12 Pg C = 2.12 billion tons C.5. The atmospheric CO2 concentration increases at a rate of 1.5 ppmv/year since

1958. Therefore, the airborne fraction of CO2 =44

245.1 = 0.5 (50%).



130

6. The remaining CO2 emitted is most absorbed by the oceans, which contain largequantities of CO2 in the form of HCO3

-. The Earth’s oceans also support largequantities of phytoplankton cells which perform photosynthesis that consumesCO2. When phytoplankton cells die they eventually settle to the ocean floor

where the CO2 fraction incorporated in phytoplankton cells will be immobilized for a long time without being immediately released back to the atmosphere, i .e.,

the Earth’s oceans is a major CO2 sink. Other possible uptakers of CO2 are theliving green plants. Experimental studies have shown that plant growth increasesin the presence of increased CO2 concentration.

Major CH4 Emission Sources. CH4 accounts for about 15-20% of greenhouse gasesin the

atmosphere and CO2 accounts for about 60%.

Fossil Fuels 26%

Enteric Fermentation 22%

Rice Paddies 16%

Biomass Burning 11%

Landfills 11%

Municipal Wastewaters 7%

Animal Wastes 7%

N2O. Nitrous oxide (N2O) is the result of increasing fossil fuel emissions, biomass burning, and

nitrogen fertilizer applications worldwide. As ammonia-nitrogen is oxidized in the soiland

nitrate is denitrified, N2O represents an intermediate oxidation state, which is volatilized

to theatmosphere. N2O is a difficult gas to control, because fertilizer applications of ammonia,ammonia nitrate, and ammonia sulfate are likely to increase worldwide.

32224 NO NOO N N NH ( V.16)

Chlorofluorocarbons (CFCs). Chlorofluorocarbon concentrations are alsoincreasing, and

they have a long half-life in the atmosphere. They are used as refrigerants, blowingagents, and

cleaning chemicals in the microelectronic industry. Two of the most widely used are

CFC-11(trichlorofluoromethane, CFCl3) and CFC-12 (dichloro-difluoromethane, CF2Cl2). Due to

a slowreaction rate with hydroxyl radical, ●OH, in the lower atmosphere, CFCs are transported

intothe stratosphere (upper atmosphere), where they deplete the ozone layer that shields the

earthfrom harmful UV radiation.



131

Summary of Major Anthropogenic Greenhouse Gases.

Gases Concentration

(ppmv) GreenhouseContributio

n (%)

Rate ofIncrease(%/year)

Half-Life(years)

RelativeGreenhouse Effect

CO2 360 60 0.4-0.5 1 1 (per kg)1 (per mole)

CH4 1.7 15-20 0.7-0.9 70 70 (per kg)25 (per mole)

N2O 0.31 5 0.2 200 200 (per kg)200 (per mole)

O3 0.01-0.05 8 0.1-0.5 1,800 1,800 (per kg)

2,000 (per mole)CFCl3 0.28 (ppbv) 4 0-4 4,000 4,000 (per kg)

12,000 (per mole)

CF2Cl2 0.48 (ppbv) 8 0-4 6,000 6,000 (per kg)15,000 (per mole)

Global Carbon Cycle. Figure V.13 shows the main reservoirs and transfer processes inthe

global carbon cycle. The carbon cycle begins with the primary environmental reservoir of

carbon, the CO2 present in the air and/or dissolved in water. Through photosynthesis,carbon

atoms from CO2 are incorporated into glucose (C6H12O6) and then into other organicmolecules

that make up all plant tissues. They may then be passed to consumers and/ordecomposers

through feeding. As plants themselves or any consumer or decomposer breaks downorganic

molecules through cell respiration to release energy, the carbon atoms are returned tothe air or

water as CO2 molecules. If organic materials are simply burned they also are returnedCO2 to the

air. In either case, it may be reabsorbed by plants and repeat the cycle.

Another interesting and important aspect of the carbon cycle is that in ancient geologicaltimes

(hundreds of millions of years ago) much of the organic matter produced inphotosynthesis was

neither consumed nor decomposed; it accumulated and was buried under sediments. Asa result

of millions of years under heat and pressure in the earth, this detritus has beenconverted to

crude oil, natural gas, and coal − the fossil fuels that are mining or pumping today.



132

33222 HCO H CO H O H CO ( V.17)

Figure V.14 shows upward trends (~ 4%/decade) in mean annual CO2 concentrations atPoint

Barrow, Alaska; Mauna Loa, Hawaii; and the South Pole. The annual rhythm is due tothe season

variation in plant and soil absorption and release, chiefly in the northern hemisphere. Arapid

increase in atmospheric CH4 concentration in the 20th century is also evident by the datashown

in Figure V.15.

Figure V.13. Global carbon cycle.



133

Fundamentals of Greenhouse Effects. The basic processes governing the earth’stemperature and climate are related to radiative heat exchange between the earth and

sun. Any objective with a temperature above absolute zero continuously radiates energy. The

amountof energy radiated depends on the temperature of the material and the nature of its

radiatingsurface. The maximum amount of energy, in the form of heat, can be radiated by a body

at agiven temperature is given by the Stefan-Boltzmann equation:

4

max AT Q

( V.18)

Figure V.14. Upward trends (~ 4%/decade) in mean annual CO2 concentrations at

Point Barrow, Alaska; Mauna Loa, Hawaii; and the South Pole.

wheremax

Q is the maximum rate of energy radiated (W), σ is the Stefan-Boltzmann

constant(5.67×10-8 W/m2-oK 4), A is the surface area of the body (m2), and T is the absolutetemperature of the body (oK). Any body radiating the energy at this maximum rate isknown as a black body.The radiant heat reaching the earth from the sun closely approximates the radiation



134

from a black body at T = 5,800oK. the amount of solar energy incident on the earth’satmosphere averages about 342 W/m2 ( S o) based on the surface area of the earth (4 πR2,

R is the radius of the earth).

A Simple Earth Energy Balance Without Atmosphere. As illustrated in Figure V.16, I is

the incoming solar radiation ( S o). Part of this incoming solar radiation is reflected backout to

space ( II , II = aS o, where a is called albedo which is about 0.31 for the earth’s surface).Therefore, the rate of solar radiation absorbed by the earth’s surface, IV , is (1 − a) S o. If

theearth’s surface is assumed to approximate a black body, then the outgoing rate of

radiation fromthe unit earth’s surface area to space, III (or V ), is 4

eT , where T e is the earth’s surface

temperature. At steady state, therefore,

Figure V.15. Atmospheric CH4 concentrations.

4

1

4 )1()1(

o

eeo

S aT T S a ( V.19)

or T e = 254oK = −19oC, which suggests a cold and inhospitable earth! The actual earth’ssurface

temperature is about 15oC, which is a result of the presence of greenhouse gases (GHGs)in the

earth’s atmosphere.



135

Figure V.16. Energy fluxes of the earth without the atmosphere.

Figure V.17 depicts the absorption spectra for the key gases of concern in theatmosphere, along with their effects on incoming solar radiation and outgoing infrared

radiation emitted by the earth’s surface. Most of the long-wavelength energy radiated bythe earth is absorbed by water vapor, CO2, CH4, O2, N2O, and O3. Water vapor, which is

by far the most important GHG, absorbs thermal radiation with wavelengths < 8 μm or >18 μm. CO2 shows a strong absorption band centered at 15 μm, as well as bands centeredat 2.7 μm and 4.3 μm. Between 7 μm and 15 μm there is a relatively clear sky foroutgoing thermal radiation, referred to as the atmospheric radiative window.Essentially all incoming solar radiation with wavelengths < 0.3 μm (ultraviolet) isabsorbed by O2 and O3. Radiatively active gases that absorb wavelengths >4 μm are called GHGs. This absorption heats the atmosphere, which, in turn, radiatesenergy back to the earth as well as out to space. These GHGs act as a thermal blanketaround the globe, raising the earth’s surface temperature beyond −19oC.



136

Figure V.17. Absorption spectra of atmospheric gases.

A More Realistic Earth Energy Balance. To derive an energy balance, all energyflows are

expressed in terms of heat fluxes. For outgoing radiations, all surfaces are assumed toradiate as

black bodies. According to Figure V.18, I = S o = 342 W/m2 and II = 107 W/m2.Therefore, the

net solar energy input to the atmosphere is S a = 235 W/m2, which is partially absorbed by the

atmosphere and partially by the earth’s surface.



137

Figure V.19. Energy fluxes of the earth with the atmosphere.

Now, III is the radiation from the atmosphere to the earth’s surface which is assumed to be

identical to the radiation from the atmosphere out to space, IV is the part of the netincoming

solar energy that is absorbed by the atmosphere (αS a), V is the part of the net incomingsolar

energy that is absorbed by the earth’s surface [(1 − αS a)], and VI is the radiation from theearth’s

surface to the atmosphere.

Since both the atmospheric temperature (T a) and the earth’s surface temperature (T e) areunknown, two independent equations are needed to solve the problem. At steady state,

then

44

44

)1(

2

eaa

aea

T T S

T T S

( V.20)

For α = 0.29, then T e = 17oC, which is close to the actual earth’s surface temperature atabout

15oC.

Radiative Forcing (Δ F ). Figure V.20 depicts the radiative equilibrium and imbalancesituations in the troposphere where GHGs and aerosols are generally confined. Thelength of each arrow in the diagram represents the magnitude of radiative heat flux(W/m2) at the tropopause, which is the upper boundary of the troposphereapproximately 10 km above the



138

earth’s surface.

Figure V.20. Radiative equilibrium and imbalance situations in the

troposphere.

The top left figure represents the initial radiative equilibrium conditions that result in azero net

energy flux. The top right figure represents the temporary imbalance that results fromthe

addition of a GHG like CO2. More of the outgoing radiation is now absorbed in theatmosphere,

so the incoming solar radiation exceeds the outgoing radiation. Such changes in theaverage net

radiation at the tropopause result in a net decrease of outgoing radiation per unit of area.In this

case, a positive radiative forcing would occur. The bottom left figure shows a change insolar

radiation input that perturbs the earth’s energy balance to produce a radiative forcingeven

nothing is added to the atmosphere. If the solar radiation input were increased by aspecific

amount, a positive radiative forcing would occur, tending to warm the planet.

Radiative forcing also can be induced by changes in earth’s albedo. The addition of



139

aerosols(suspensions of fine particles less than 10 μm in diameters) to the atmosphere tends to

reflectincoming solar radiation back to space thereby increasing the earth’s albedo. As a result,

moreradiation is leaving at the tropopause than coming in, resulting in a net decrease in heat

flux. Inthis case, a negative radiative forcing would occur as shown by the bottom right figure. A

generalexpression for radiative forcing (Δ F ) combines the overall changes in incoming solar

radiation with the overall changes in outgoing radiation at the tropopause:

after ininitial inin

after ou t initial ou t ou t

inou t inou t

qqq

qqq

qq A

Q

A

Q F

( V.21)

where the subscripts in represents the incoming solar radiation, out the outgoingradiation,initial the conditions before perturbations, and after the conditions after perturbations.

Example V.10. (1) An increase in atmospheric CO2 concentration reduces the outgoingIR radiation at the tropopause from 235 to 233 W/m2. The incoming solar radiation andalbedo do not change. Calculate Δ F . (2) An increase in atmospheric sulfate aerosolfrom combustion of coal and oil adds 0.5 W/m2 to the global average flux of radiationreflected by the troposphere (albedo). Calculate Δ F .

1.1. 0

inq (since there are no changes in incoming solar radiation and albedo).

1.2. 2233235

out q W/m2.

1.3. Therefore, Δ F = 2 − 0 = 2 W/m2 > 0 (net warming).2.1. Denote the initial outgoing radiation as x W/m2.

2.2. 0

inq (since there is no change in incoming solar radiation).

2.3. 5.0)5.0(

x xqout W/m2.

2.4. Therefore, Δ F = −0.5 − 0 = −0.5 W/m2 < 0 (net cooling).

Radiative Forcing (Δ F ) versus Concentration (C ).



140

0

3

02

01

ln

)(

C

C k F

C C k F

C C k F

( V.22)

The 1st equation in ( V.22) is for the low concentration conditions, the 2nd equation forthe moderate concentration concentrations, and the 3rd equation for the highconcentration conditions.

GHG Δ F (W/m2) Remarks

CO2

0

ln3.6C

C F

C is in ppmv. C < 1,000 ppmv.

CH4 0036.0 C C F C is in ppbv.C < 5,000 ppbv.

N2O 014.0 C C F C is in ppbv.C < 5,000 ppbv. O3 002.0 C C F C is in ppbv.

CFC-11 (CFCl3) )(22.0 0C C F C is in ppbv.C < 2 ppbv.

CFC-12 (CF2Cl2) )(28.0 0C C F C is in ppbv.C < 2 ppbv.

Global Warming Potential (GWP). The GWP is a weighting factor that enablescomparisons

to be made between the global warming impact of 1 kg of any GHG and 1 kg of CO2. Forinstance,

the 20-year GWP for N2O is 280, which means 1 kg N2O emitted today will exert 280times asmuch global warming over the next 20 years as would 1 kg of CO2 emitted today. The

GWP is aratio of the cumulative radiative forcing for 1 kg of a GHG over some period of time to

thecumulative radiative forcing for 1 kg of CO2, over that same period of time.

Mathematically, theGWP can be expressed as:

T

o CO

T

CO

GHG

T

o CO

T

o GHG

CO

GHG

T

o COCO

T

o GHGGHG

dt t R

e

F

F

dt t R

dt t R

F

F

dt t R F

dt t R F GWP

)(

1

)(

)(

)(

)(

22

22

22

( V.23)



141

whereGHG F is the radiative forcing of the GHG in question per kg (W/m2-kg),

2CO F is

the radiative forcing of CO2 per kg (W/m2-kg), )(t RGHG is the fraction of the 1 kg of GHG

remaining in the atmosphere at time t , )(2

t RCO is the fraction of the 1 kg of CO2

remaining in the atmosphere at time t , T is the time period for cumulative effects (years),

and τ is the atmospheric lifetime (or time constant) of the GHG in question (years). Notethat )0(GHG R 1 kg.

T ( years)

T

CO dt t R0 2

)( ( years)

20 13.2100 43.1500 138.0

Example V.10. 1 kg of HFC-143a (CF3CH2F) in the atmosphere causes a radiativeforcing that is 4,129 times the forcing of 1 kg CO2. The decay of HFC-134a is exponential

with a time constant of 14.6 years. Estimate the GWP over 20-year and 500-yearperiods.

1.

2.13

1)6.14(

1

129,4

6.14

20

20

e

GWP 3,400.

2.0.138

1)6.14(

1

129,4

6.14

50 0

50 0

e

GWP = 435.

3. Notice how the relatively short time constant of HFC-134a results in a muchhigher GWP in the 20-year time period than it does over the much longer 500-

year period.

Reducing GHG Gas Emissions.

Strategy Remarks

Reducing Energy Intensity

◙ A change in the makeup of industries, e.g., a decline inenergy-intensive industries (e.g., steel manufacturing) and arise in light industries (e.g., semiconductor assembly).◙ Efficiency improvements that provide economic incentives,e.g., the replacement of incandescent light bulbs by LEDs.

Reducing Carbon Intensity

◙ Alternative energy sources that emit less carbon than fossilfuels.◙ Carbon sequestration via biomass absorption of CO2 andnew techniques to capture and sequester CO2.

Reducing Non-CO2 Emissions◙ Capture of CH4 leaked from landfills and undergroundcoal seams.◙ Reduction in fertilizer uses to control N2O emissions.◙ Replacements of CFCs by HCFCs and HFCs.

Example V.11. Compare the carbon emissions to heat household water using the



142

following three energy systems: (1) a very good, 37% efficient coal-fired power plantdelivering electricity to a 100% efficient electric water heater, (2) a new, 45% efficientnatural-gas-fired combined cycle power plant delivering electricity to a 100% efficientelectric water heater, and (3) an 85% efficient gas-fired water heater. Each system isprovided with 100MJ of energy. The LHV (low heating value) carbon intensity values forcoal, natural gas, and oil are 25.8, 15.3, and 20.0 g C/MJ, respectively.

1. Since 2,580 g of carbon will be released to generate 37 MJ of electricity,

therefore, the carbon emission is37

580,2 = 69.7 g C/MJ.

2. Since 1,530 g of carbon will be released to generate 45 MJ of electricity, therefore,

the carbon emission is45

530,1 = 34.0 g C/MJ.

3. Since 1,530 g of carbon will be released to capture 85 MJ to heat the water, the

carbon emission is85

530,1 = 18.0 g C/MJ.

Example V.12. Consider two fuel systems each delivering 100 MJ of energy to someend use that burns the fuel releasing CO2. One system delivers the amount of energy inthe form of coal. In the other methane is the fuel, but the pipeline leaks 1 MJ of methanefor every 100 MJ it delivers to the end user. Compare the 20-year, emission-weightedglobal warming potentials (GWPs) of the two systems. HHV and LHV values formethane are 890 and 802 kJ/mol, respectively. Methane has 56 times the warmingpotential of CO2 over the 20-year period.

O H COOCH 2224 22

1. Since the LHV carbon intensity for coal is 25.8 g C/MJ, then CO2 emission from

coal burning is12

)44)(100)(8.25( = 9,460 g CO2.

2. Since burning 1 mole of methane releases 802 kJ of energy (LHV) whileproducing 1 mole of carbon dioxide, then the LHV carbon intensity for methane

is802

)000,1)(12( = 15.0 g C/MJ (1 mole of carbon dioxide has 12 g of C).

3. CO2 emission from burning CH4 =12

)44)(100)(15( = 5,500 g CO2.

4. Methane leaked = (1 MJ)(16 g CH4/mol)(0.802 MJ/mol)-1 = 19.95 g CH4.5. For the coal system all the emissions are in the form of CO2, which by definition

has GWP = 1. Then, for coal, GWP20 = 9,460 g CO2.6. For the methane system, GWP20 = 5,500 + (19.95)(56) = 6,617 g CO2.

Therefore, the ratio of the two system is460,9

617,6 = 0.70 (i .e., methane reduces the

warming impact by 30%).

VI. Environmental Risk Assessment



143

One of the most important changes in environmental policy in the 1980s was theacceptance of the role of risk assessment and risk management in environmentaldecision making. It was assumed in early environmental legislation (e.g., the Clean Airand Clean Water Acts) that contaminants have thresholds, and that exposure toconcentrations below these thresholds would produce no adverse health effects.

However, many toxic substances are carcinogens, and carcinogenesis is assumed toconsist of one or more stages at the cellular level beginning with a single-cell mutation,at which point cancer is initiated, i .e., no threshold exists for cancer-causing chemicals.Therefore, any concentration is a risk.

Hazardous wastes have been generated from essentially all industrial activities. Prior tothe passage and promulgation of federal legislation in the later 1970s, hazardous wastes

were often disposed of in pits, ponds, and lagoons, on surface soils, and in landfills. TheResource Conservation and Recovery Act (RCRA) was passed in 1976 to provide cradle-to-grave management of hazardous wastes; it was amended as Hazardous and Solid

Waste Amendments (HSWA) in 1984. Hazardous waste generators, transporters, andtreatment/storage/disposal facility operators have responsibilities that provide

safeguards against improper hazardous waste disposal. The ComprehensiveEnvironmental Response, Compensation and Liability Act (CERCLA), or Superfund, waspassed in 1980 to provide a mechanism for the mitigation of chronic environmentaldamage, particularly the cleanup of contaminated sites. Superfund was amended in1986 as the Superfund Amendments and Reauthorization Act (SARA). Hazardous

wastes are defined by RCRA, hazardous substances by CERCLA, and hazardousmaterials by Department of Transportation regulations. The current estimate ofhazardous waste generation in roughly 750 million tons (680.4×106 metric tons) per yearin the United States. Most of the waste is classified as corrosive, and can be treated byneutralization.

The assessment of health effects on workers, the general public, and the environment is

often required in hazardous waste management. A multi-step risk assessment process isoften used to assess health and ecological risks at Superfund sites as well as to evaluatethe effectiveness of remedial alternatives in attaining a required level of cleanup. Formost contaminants that are considered hazardous under CERCLA, specific cleanuprequirements have not been established. For instance, not all soils at sites containingPCP (pentachlorophenol, C6Cl5OH) are remediated to a certain level, e.g., 1 mg/kg.Rather, each site is assessed individually and remediated to a predetermined level ofrisk, e.g., one cancer case per million people. The rationale of using risk assessment can

best be illustrated by comparing two very different sites. Figure VI.1 illustrates a high-risk site, although containing only 10 mg/kg PCP, is near a house and a drinking water

well. As a result, remediation to low levels would be necessary to protect public health.On the other hand, at the low-risk site (Site B), a PCP-bearing sludge (10,000 mg/kg)has been buried, but there are no receptors nearby and the low rate of PCP migration inclay subsurface materials (low permeability) would further reduce the risk. Therefore,the best remedial option would be to leave the buried sludge in place, where slow naturalattenuation processes would eventually result in its degradation.

House

Well

Site A



144

Figure VI.1. Risk assessments.

The usual starting point for an explanation of risk is to point out that there is alwayssome risk associated with human activities. Since risk has no units, additionalclarifications on risk may be needed, e.g., whether the risk is a lifetime risk or an annualrisk, whether it is an average risk to the general public or a risk faced by individuals whoengage in some activities, or whether it is being expressed as a percentage or as a decimalfraction. Also, environmental risk assessments deal with incremental probabilities ofsome adverse health effects occurring. For instance, the U.S. EPA attempts to control

the exposure to toxins to levels that will pose incremental lifetime cancer risks to themost exposed members of the public of roughly 10-6 (one additional cancer in millionpeople) to 10-4 (100 additional cancers in million people). Suppose all 300 million

Americans face a 10-6 lifetime risk of cancer from exposure to a particular toxic chemical,then 300 extra cancers would be observed during their lifetimes. For a typical lifetime of70 years, then roughly 4 extra cancers would be observed per year in the U.S. In 1992,roughly 521,000 cancer deaths were reported in the U.S. so 4 extra cancers caused bytoxic exposure would be < 0.001% of the normal rate. Presenting risk as an annualprobability of death to individuals who engage in some activities is a much more

way to express risks than simply looking at the population as a whole. Table VI.1. showsrisk data for some common activities that are a mix of actuarial values and estimates

based on various risk models. It should always be kept in mind that when risks are basedon models, there are generally very large uncertainties in the estimates. Nevertheless,one of the purposes of risk assessment is to provide a starting point in balancing thetradeoffs between an acceptable incremental risk and the cost of controlling risk to thatlevel. For instance, immunizations and phasing out leaded gasoline yield the directsavings in health care that far exceed the costs associated with the implementation of thepractices.

Table VI.1. Risks Associated with Some Common Activities.

Groundwater

10 mg PCP/kg Sandy Soil10 m

Subsurface Clay Material

10,000 mg PCP/kg20 m

100 m

Groundwater

Site B



145

Activity/Exposure Annual Risk (Deaths/100,000 people at risk)Motorcycling 2,000

Smoking (all causes) 300Smoking (cancer) 120

Hang Gliding 80Coal Mining 63

Farming 36Motor Vehicles 24Chlorinated Drinking Water (chloroform) 0.8

4 Teaspoons Peanut Butter Daily (aflatoxin) 0.83 oz Charcoal Broiled Steak Daily (PAH) 0.5

10-6 Lifetime Risk 0.0014

Four steps have been defined by both the National Academy of Sciences and the U.S.EPA for the hazardous waste risk assessment:

Hazard Identification. The process of determining whether or not a particular chemicalis causally linked to particular health effects, such as cancer or birth defects. Sincehuman data are so often difficult to obtain, this step usually focuses on whether a

chemical is toxic in animals or other test organisms.

Dose-Response Assessment. The process of characterizing the relationship between thedose of an agent administered or received and the incidence of an adverse health effect.Many different does-response relationships are possible for any given agent dependingon such conditions as whether the response is carcinogenic or non-carcinogenic and

whether the experiment is a one-time acute test or a long-time chronic test. Since mosttests are performed with high doses, the dose-response assessment must include aconsideration for the proper method of extrapolating data to low exposure rates thathumans are likely to experience. Part of the assessment must also include a method ofextrapolating animal data to humans.

Exposure Assessment. The determination of the size and nature of the population thathas been exposed to the toxicant under consideration, and the length of time andtoxicant concentration to which they have been exposed. Consideration must be given tosuch factors as the age and health of the exposed population, smoking history, thelikelihood that members of the population might be pregnant, and whether or notsynergistic effects might occur due to exposure to multiple toxicants.

Risk Characterization. The determination of a number that expresses risk, e.g., one inone hundred (1×10-2) or one in one million (1×10-6).

Hazard Identification. The 1st step in a risk assessment is to determine whether ornot the chemicals that a population has been exposed to are likely to have ant adverse

health effects. Figurer VI.2 shows the human circulary system that identifies some of theprinciple organs and nomenclature for toxic effects. A toxicant can enter the body usingany of three pathways: by ingestion with food or drink, through inhalation, and bycontact with skin (dermal) or other exterior surfaces, such as the eyes. Once in the body,the toxicant can be absorbed by the blood and distributed to various organs and systems.Chemicals that can cause liver damages are called hepatotoxins (e.g., carbontetrachloride, CCl4; chloroform, CHCl3; trichloroethylene, C2HCl3; DDT; arsenic; iron;acetaminophen; and anabolic steroids). Toxicants that damage kidneys are callednephrototoxins (e.g., cadmium, mercury, lead, and some chlorinated hydrocarbons).



146

Hematotoxicity is the term used to describe the toxic effects of substances on the blood(e.g., CO and NO3

-).

Figure VI.2. The human circulary system and the nomenclature for toxiceffects.

Classification of Toxic Responses. The most important factor that influencestoxicity is the dose, and it has been said that the dose makes the poison. A secondimportant factor is the time period of exposure. Toxicity is often classified by number orduration of exposure. Repeated exposures may be classified as acute (< 5% of theorganism’s life span), subchronic (5-20% of the life span), and chronic (> 20% of the lifespan). For chemicals that are both acutely and chronically toxic, the mechanisms oftoxicity for these two categories are often different. For instance, acute toxicity from alarge dose of chloroform is caused by effects on the central nerve system, which results inand narcosis. However, consumption of drinking water containing trace concentrationsof chloroform over a lifetime causes liver damage and cancer. Many chemicals that areacutely toxic may not be chronically toxic and vice versa. For instance, vitamin D takenin pure form exhibits high acute toxicity. However, low doses of vitamin D such as in thenormal intake of milk are not only nontoxic, but essential for good health. In quantifyingtoxic responses, numerical values may be absolute or normalized for body weight ( BW ).

Dosage is often defined as the total mass of chemical to which an organism is exposed. Dose, on the other hand, is chemical dosage normalized for body weight. A dose of 8mg/kg being administered to a 3-kg animal results in a dosage of (8 mg/kg)(3 kg) = 24mg. There are different ways of defining dose. Administered dose, which is the mostpractical for environmental applications is the concentration of chemical to which theorganism is exposed. Intake or uptake dose is the actual amount of chemical absorbed



147

by the organism. Target or effect dose is the amount of chemical reaching the targetorgan.

Acute Toxicity. Acute toxicity refers to effects that are caused within a short period oftime after a single exposure to the chemical. One way to describe the acute toxicity of achemical is by the amount that is required to kill the organisms. Table VI.2 summarizes

a conventional rating system for the acute toxicity of chemicals in humans.

Table VI.2. Conventional Rating System for Acute Toxicity in Humans.

Toxicity Rating Probable Lethal Oral Dose (mg/kg BW ) Average AdultPractically Nontoxic > 15,000 > 1 quart

Slightly Toxic 5,000 to 15,000 1 pint to 1 quartModerately Toxic 500 to 5,000 1 ounce to 1 pint

Very Toxic 50 to 500 1 teaspoon to 1 ounceExtremely Toxic 5 to 50 7 drops to 1 teaspoon

Supertoxic < 5 < 7 drops

It is noteworthy that not every member of an exposed population will react the same wayto a toxin. As a result, the variations in respo0nses are illustrated with a dose-responsecurve that shows the percentage of a population that is affected as a function of the dosereceived. S -shaped curves will be yielded when the dose is plotted on a logarithmic scale(Figure VI.3). The dose is normalized with BW to permit it to be extrapolated toindividuals with different sizes. The dose-response curves generated using animal testsalso permit the extrapolation of the likely effects on a human.

Figure VI.3. Dose-response mortality curves for acute toxicity.

The dose that will kill 50% of a population is designated as LD50 or LC 50 ( LD: lethal;dose, LC : lethal concentration). It is seen from Figure VI.3 that chemical A has la lower LD50 than chemical B and it is always more toxic. However, just because one chemicalhas a lower LD50 than another does not necessarily means it is always more toxic.Chemical A has a lower LD50, which would normally suggest that it is more toxic thenchemical C, but notice it is not as toxic as Chemical C at low doses. So the dose-responsecurve does provide more information than a simple table of LD50 doses. LC 50 is oftenused in the assessment of aquatic toxicity.

A LD50 B LD50 C LD50



148

Initial Screening and Selection of Surrogates. The identities, nomenclatures,concentrations, and properties for all contaminants should be obtained at a site orfacility. Some of the specific tasks may include sampling, installation of monitoring

wells, chemical analysis, assurance/quality control plans, and data analysis. Since mostsites may contain tens to hundred of chemicals, data collection can become unrealisticand/or overwhelming. Therefore, the hazard identification usually focuses more on

chemicals that pose the greatest risks, the surrogates. The use of surrogates significantlyreduces the data input for pathway studies. The two most important sourcecharacteristics that are used in screening a large number of chemicals at a site or facilityare their concentrations and toxicities. The procedure for the initial screening ofcontaminants to identify surrogates is outlined as follows.

1. Sort the contaminant data by medium (i .e., soil, miscellaneous solids such assludges, surface water, groundwater, and air).

2. Tabulate the mean and range of concentrations at the site or facility.3. List the reference doses ( RfDs) for non-carcinogens and slope factors ( SF s) for

carcinogens.4. Determine the chemical score (i .e., a risk factor) for each contaminant. In

general, the chemical score is:

iij j

iijij

i

ij

ij

ijijij

R R

SF C R

RfD

C R

T C R

( VI.1)

where Rij is the chemical score for chemical i in medium j ; C ij is the concentration

of chemical i in medium j ; T ij is the toxicity value for chemical i in medium j ; Rij isthe chemical score for chemical i in medium j ; RfDi is the reference dose forchemical i (non-carcinogen), mg/kg-day; SF i is the slope factor for chemical i (carcinogen), kg·day/mg; and R j is the total chemical score in medium j .

5. Rank the chemicals by chemical scores for each exposure route (i .e., medium).6. Select chemicals comprising 99% of total score (i .e., surrogates).7. Further discussions on RfD and SF will be presented later.

Example VI.1. The following pesticides and corresponding concentrations (mgchemical/kg soil) are found at an abandoned pesticide formulation site: Aldicab, 140;Captan, 86; Fonofos, 280; Malathion, 65; and Naled, 90. Select the surrogates thataccount for 99% of the risk. The pesticides are considered to be non-carcinogenic andthe corresponding oral RfDs (mg/kg-day) are: Aldicab, 0.001; Captan, 0.13; Fonofos,0.002; Malathion, 0.02; and Naled, 0.002.

1. .i

iiij

RfD

C R R

2.

Pesticide Ri (days) Rank Aldicab 140,000 1



149

Captan 662 4Fonofos 140,000 1

Malathion 3,250 3Naled 45,000 2

3. The two pesticides that should be selected as surrogates are Aldicab and Fonofos.

However, since the total R is 328,912, the two selected surrogates make up only85% of the total R. Therefore, Naled must also be considered as a surrogate forthe site in oder to account for 99% of the risk.

Dose-Response Assessment. The fundamental goal of a dose-response assessmentis to obtain a mathematical relationship between the amount of a toxicant that a humanis exposed to and the risk that there will be an unhealthy response to that dose. Thedose-response relationships (or curves) are the result of chronic toxicity, i .e., theorganism is subjected to a prolonged exposure over a considerable fraction of its life. Asa result, the abscissa is the dose, which is usually expressed as the average milligrams ofsubstance per kg of body weight per day (mg/kg-day) as opposed to the unit of dose (i .e.,mg/kg) used in acute toxicity assessments. The dose is an exposure averaged over an

entire lifetime (70 years for humans). The ordinate is the response, which is the risk (orprobability) that there will be some adverse health effects. Therefore, response has nounits. For instance, if prolonged exposure to a chemical would be expected to produce700 cancers in a population of 1 million, the response could be expressed as 0.0007, or0.07%. The annual risk would be obtained by spreading that risk over an assumed 70-

year lifetime, yielding a risk of 0.00001, or 1×10-5 per year. For carcinogens, it is alwaysassumed that exposure to any amount of the carcinogen will create some likelihood ofcancer, i .e., a plot of response versus dose is required to go through the origin. For non-carcinogens, it is usually assumed that there is some threshold dose, below which there

will be no response. Therefore, the dose-response curves and the methods used to applythem are quite different for carcinogenic and non-carcinogenic effects.

The most controversial aspect of dose-response curves for carcinogens is the methodchosen to extrapolate from the high doses actually administered to test animals to thelow doses to which humans are likely to be exposed. Even with extremely largeof test animals used in a bioassay, the lowest risks that can be measured are usually a fewpercent. Since regulators attempt to control human risks to several orders of magnitudeless than that, there will be no actual animal data anywhere near the range of mostinterest. One model, the one-hit model , is commonly used for the extrapolation to lowdoses. In the one-hit model, the relationship between the dose (d ) and the lifetime risk(probability) of cancer, P (d ), is given as:

0

)10(

1)0(

1)(

q

d qq

e P

ed P

( VI.2)

where P (d ) is the rate of cancer incidence at the d , %; P (0) is the background rate ofcancer incidence, %; and q0 and q1 are parameters. Now, for small x ,

x x x

xe x 1...!3!2

132

( VI.3)

Therefore, for small q0 and q1, P (0) ≈ 1 – (1 − q0) = q0 and



150

d qd A P d P d q P d qqd qqd P 111010 )()0()()0()1(1)( ( VI.4)

where A(d ) is the upper-bound, additional (or incremental) lifetime risk, %. The one-hitmodel predicts that for low doses, the additional lifetime probability of cancer is linearly

related to dose. Another model, the multi-stage model, is based on the observation thattumors are the result of a sequence of biological events:

n

i

ii d q

ed P 01)( ( VI.5)

where the individual parameter qi are positive constants picked to best fit the dose-response data. However, at low doses, the multi-stage model also has the feature ofproducing a linear relationship between additional risk and dose.

Potency Factor ( PF ) or Slope Factor ( SF ). For chronic toxicity studies, a low dose isadministered over a significant portion of the animal’s lifetime. The resulting dose-

response curve has the incremental risk of cancer (above the background rate) on the y-axis and the lifetime average daily dose of toxicant along the x -axis. At low doses, wherethe dose-response curve is assumed to be linear, the slope of the dose-response curve iscalled the potency factor ( PF ), or slope factor ( SF ):

BW

ADI CDI d

d

d ASF

)(

( VI.6)

where ADI is the average daily intake of the chemical, mg/day; and BW is the average

body weight, kg.

Example VI.2. When drinking water is disinfected with chlorine an undesirable byproduct, chloroform (CHCl3), is formed. Suppose a 70-kg person drinks 2 L of waterevery day for 70 years with a chloroform concentration of 0.10 mg/L (the drinking waterstandard). (1) Find the upper-bound cancer risk for this individual, A(d ). (2) For a city

with 500,000 people (70-year lifetime) drinking the same water, how many extra cancerper year would be expected? (3) Compare the extra cancers per year caused bychloroform in the drinking water with the expected number of cancer deaths from allcauses (i .e., 193 per 100,000 per year in the U.S.). The SF (oral route) of chloroform is6.1×10-3 kg∙day/mg.

1.1.70

)1.0)(2( CDI d = 0.00286 mg/kg-day.

1.2. .104.17)00286.0)(101.6())(()( 63 d SF d A So over a 70-year period theupper-bound estimate of the probability that a person will get cancer from thisdrinking water is about 17 in one million.

2. If there are 17.4 cancers per one million people over a 70-year period, then in anygiven year in a population of one-half million, the number of cancers caused by



151

chloroform would be)70)(10(

)4.17)(000,500(6

= 0.12 cancers/year.

3. The total number of cancer deaths that would be expected in a city of 500,000

people would be510

)193)(000,500( = 965 cancer deaths/year >> 0.12 new

cancers/year.

In Example VI.2 it is assumed that everyone drinks 2 L of chloroform-laden waterevery day for 70 years. When a risk assessment is made for exposures that do not last forthe entire lifetime, CDI needs to be redefined as follows.

))((

))()()((

AT BW

C ED EF CRCDI ( VI.7)

where CR is the rate of contact with the contaminated medium, L/day, mg/day, orm3/day, depending on the medium; EF is the frequency of exposure to the contaminant,

days/year; ED is the duration of exposure to the contaminant, years; BW is the average body weight, kg; AT is the period over which the exposure is averaged, days; and C is theaverage contaminant concentration during the exposure period, mg/L for a contaminantin water, mg/mg for a contaminant in soil, and mg/m3 for an airborne contaminant.( VI.7) can be used to calculate contaminant intake rates for representative members ofthe potentially exposed population. Intake rates for each contaminant are calculated foreach potential exposure pathway. EF is given as 350 days/year for a resident, to allowfor vacations, while 250 days/year is used to estimate intake rates for a worker (i .e., 5days/week, 50 weeks/year). The choice of 70 years for AT in cancer calculations is basedon the assumption that cancer effects are cumulative over a lifetime and that high dosesapplied over a short time are equivalent to low doses spread over a longer period. Table

VI.3 summarizes parameter values used in ( VI.7).

Table VI.3. Parameter Values Used in (VI.7).

Parameter Resident Worker

CR

◙ 2 L/day (drinking water)◙ 100 mg/day (soil and dust ingestion)◙ 30 m3/day (air inhalation)

◙ 1 L/day (drinking water)◙ 50 mg/day (soil and dust ingestion)◙ 30 m3/day (air inhalation)

EF 350 days/year 250 days/year

ED Acute event duration or 30 years forchronic effects (cancer calculations)

Acute event duration or 25 years forchronic effects (cancer calculations)

BW 70 kg 70 kg

AT Acute event duration for non-carcinogeniceffects or (365 days/year)(70 years) forcancer calculation

Acute event duration for non-carcinogeniceffects or (365 days/year)(70 years) forcancer calculation

Example VI.3. When chlorine is used for disinfection of drinking water, chloroform can beproduced by the reaction of chlorine with residual organics in the water. Estimate the ingestionintake rate for non-carcinogenic and carcinogenic effects on an adult resident of a homereceiving tap water with an average chloroform concentration of 65 μg/L (0.065 mg/L).



152

1. For non-carcinogenic effects, use 30 years for both ED and AT . Then,

)30)(365)(70(

)30)(350)(2)(065.0( iccarcinogennonCDI = 1.8×10-3 mg/kg-day.

2. For carcinogenic effects, use 30 years for ED and 70 years for AT . Then,

)70)(365)(70()30)(350)(2)(065.0(iccarcinogenCDI = 7.6×10-4 mg/kg-day .

The Reference Dose (RfD) for Non-Carcinogenic Effects. The key assumption for non-carcinogens is that there is an exposure threshold; that is, any exposure below threshold

would be expected to show no increase in adverse health effects above natural background rates. One of the principle goals in toxicity tests is to identify and suchthresholds. Unfortunately, for the usual cases, inadequate data are available to establishsuch thresholds with any degree of certainty and therefore, it is necessary to introduce anumber of assumptions and definitions. Suppose there is a precise threshold for someparticular toxicant for some particular animal species. In a program animals would beexposed to a range of doses. Doses below the threshold would elicit no response; doses

above the threshold would produce responses. The lowest dose administered that resultsin a response is referred to as the lowest-observed-effect level ( LOEL) (Figure VI.4).Conversely, the highest dose administered that does not create a response is referred toas the no-observed-effect level ( NOEL). LOELs and NOELs are further refined by notinga distinction between effects that are adverse to health and effects that are not.Therefore, there are also lowest-observed-adverse-effect-level ( LOAEL) andno-observed-oadverse-effect-level ( NOAEL).

The reference dose, RfD, is introduced to give an indication of a level of human exposurethat is likely to be without appreciable risk. The unit of RfD is mg/kg-day average over aThe reference dose, RfD, is introduced to give an indication of a level of human exposurethat is likely to be without appreciable risk. The unit of RfD is mg/kg-day average over a

lifetime. RfD is obtained by dividing the NOAEL by a safety factor of 10 to account forthe differences in sensitivity between the most sensitive individuals (e.g., pregnant women, babies, and the elderly) and the normal, healthy people in an exposed humanpopulation. Another safety factor of 10 is introduced when the NOAEL is based onanimal data to be extrapolated to humans. Finally, an additional safety factor of 10 isapplied when there are no good human data and the animal data available are limited.

The Hazard Index ( HI ) for Non-Carcinogenic and Carcinogenic Ef fects. Since RfD isestablished at what is intended to be a safe level for non-carcinogenic effects, well belowthe level at which any adverse health effects have been observed, it makes sense tocompare the actual exposure to the RfD to see whether the actual dose is safe. The hazard quotient ( HQ) is based on that concept:

RfD NOAEL Actual T hreshold

Response



153

0.1: Protection of sensitive population subgroups.0.01: Extrapolation uncertainty from animal data to human populations.0.001: Poor data quality.

Figure VI.4. Location of NOAEL and RfD does with respect to the actualthreshold on a typical dose-response curve.

))()((

))()()((

RfD AT BW

C ED EF CR HQ ( VI.8)

A cumulative hazard index ( HI ) at a contaminate site is calculated by adding the HQs for

all chemicals of concern over all potential exposure pathways. An 11

n

i

i HQ HI is

considered unacceptable level of risk for non-carcinogenic health effects. Potentialcarcinogenic health effects are estimated by calculating the individual excess lifetimecancer risk ( IELCR) from the non-carcinogenic intake rate and the slope factor:

))((

))()()()((

AT BW

SF C ED EF CR IELCR ( VI.9)

A cumulative hazard index for a contaminate site is calculated by adding the IELCRs forall chemicals of concern over all potential exposure pathways. An IELCR < 10-6 to 10-4 (proposed by the U.S. EPA) is considered acceptable level of risk for carcinogenic healtheffects.

Example VI.4. Consider the household in Example VI.3, calculate the potential non-carcinogenic health risks associated with ingesting the tap water for an adult resident ofthe house.

1. For non-carcinogenic effects, the HQ for the ingestion pathways is 1.8×10-3 mg/kg-day. Therefore, the chloroform concentration will not pose a significanttoxic risk to the resident of the house.

2. For carcinogenic effects, calculate the IELCR for the ingestion pathways:



155

conversion factor for soil (10-6 kg/mg). The 1st equation in ( V.10) is for ingestion(swimming), 2nd equation is for dermal contact (water), and the 3rd equation is fordermal contact (soil). The following are some actions that increase the risk of death by10-6:

◙ Living 2 days in New York or Boston (air pollution/heart disease).

◙ Living 2 months with a cigarette smoker (cancer/heart disease).◙ Flying 1,000 miles by jet (accident).◙ Living 150 years within 30 kilometers of a nuclear power plant (cancer caused byradiation).

Table VI.5. Parameter Values Used in (VI.10).

Parameter Standard Value Parameter Standard Value

Average Body Weight (a) 70 kg

Skin Surface Available (c)◙ 2 to 6 years (m, f )◙ 6 to 9 years (m, f )◙ 9 to 12 years (m, f )

◙ 12 to 15 years (m, f )◙ 15 to 18 years (m)◙ 15 to 18 years ( f )

0.72 m2 0.925 m2 1.16 m2 1.49 m2 1.75 m2 1.60 m2

Average Body Weight (c)◙ 0 to 1.5 years◙ 1.5 to 5 years◙ 5 to 12 years

10 kg14 kg26 kg

Soil Ingestion Rate (c,1 to 6 years) 200 mg/day

Daily Water Ingested (a) 2 L Soil Ingestion Rate (c, > 6 years) 100 mg/day Daily Water Ingested (c) 1 L Skin Adherence Factor (potting soil to

hands) 1.45 mg/cm2

Daily Air Breathed (a) 20 m3 Skin Adherence Factor (kaoline clay tohands)

2.77 mg/cm2

Daily Air Breathed (c) 5 m3 Exposure Duration◙ Lifetime◙ At one Residence (90th percentile)◙ National Medium

70 years30 years5 years

Daily Fish Consumed (a) 6.5 g Average time 365( ED)

Contact Rate (swimming) 50 m3/hr Exposure Frequency◙ Swimming◙ Eating Fish & Shell Fish

7 days/year48 days/year

Skin Surface Available (a, m) 1.94 m2 Exposure Time◙ Shower (90th percentile) ◙ shower (50th percentile)

12 minutes7 minutes

Skin Surface Available (a, f ) 1.69 m2 a: adult, c: child, m: male, and f : female.

Example VI.5. The average daily consumption of locally caught fish by an individualin the U.S. is estimated be about 54 g. Estimate the lifetime cancer risk from fish takenfrom waters containing 0.1 mg/L (100 ppb) of TCE.

1. The BCF for TCE is 10.6 L/kg.2. Therefore, C fish = (10.6 L/kg)(0.1 mg/L) = 1.06 mg TCE/kg fish. 3. For a 70-kg person consuming 54 g fish perday, 350 days per year for 30 years,

the chronic daily intake (CDI ) can be estimated from ( VI.7), i .e.,



156

))((

))()()((

AT BW

C ED EF CRCDI =

)70)(365)(70(

)06.1)(30)(350)(1054( 3 = 3.36×10-4 mg/kg-day.

4. The oral SF for TCE is 1.1×10-2 (mg/kg-day)-1. Therefore, the upper-bound,incremental lifetime risk is: A(d ) =(3.36×10-4 mg/kg-day)[1.1×10-2 (mg/kg-day)-1] = 3.6×10-6.

Exposure Assessment. Once the exposure pathways have been analyzed, an estimateof the concentrations of toxicants in the air, soil, and food at a particular exposure pointcan be made. Then, human contacts with those contaminants must be estimated.Necessary information includes number of people exposed, duration of exposure, andamounts of contaminated air, water food, and soil that find their ways into each exposedperson’s body. One potential important exposure route is human consumption ofcontaminated fish. It is relatively straightforward to estimate the concentrations ofcontaminants in water, and it also is reasonable to make estimates of consumption offish that individual may consume. However, it is necessary to establish a ling thatpermits the estimation of the concentration of a contaminant in fish based on thecontaminant concentration in water. The bioconcentration factor ( BCF ) is the key link

(Table VI.6).Table VI.6. Bioconcentration Factors.

Chemical BCF (L/kg) Aldrin 28

Arsenic and Compounds 44Benzene 5.2

Cadmium and Compounds 81Carbon Tetrachloride 19

Chlordane 14,000Chloroform 3.75

Chromium III, VI, and Compounds 16

Copper 200DDE 51,000DDT 54,000

1,1-Dichloroethylene 5.6Dieldrin 4,760

Formaldehyde 0Heptachlor 15,700

Hexchloroethane 87Nickel and Compounds 47

PCBs 100,0002,3,7,8-TCDD (Dioxin) 5,000

Tetrachloroethylene 311,1,1-Trichloroethane 5.6

Trichloroethylene (TCE) 10.6 Vinyl Chloride 1.17

C fish can be estimated from the chemical concentration in water (C water) via BCF as:

))(( water fish C BCF C ( VI.11)

Contaminant Degradation. Many toxic chemicals of concern are reactive (i .e., non-



157

conservative) and they degrade with time. Degradation may be result of many processessuch as phase transfer as a chemical volatilizes; chemical transformations if it reacts

with other substances (e.g., with water, hydrolysis, or with light, photolysis); ortransformations mediated by biological activities. The persistence of chemical as itmoves through various environmental compartments may be affected by somecombination of these mechanisms. A convenient way to deal with such complexity is

simply to combine the various degradation processes into a single, overall half-life (t 0.5).Since most degradation reactions occurring in the environment can be modeled as first-order reactions, then

k k t

eC C

eC t C

kt

kt

693.0)2ln(

)0()0(5.0

)0()(

5.0

5.0

( VI.12)

Table VI.7. summarizes t 0.5 values for a variety of chemicals.

Table VI.7. Half-Life (t 0.5) Values for a Variety of Chemicals.

Chemical t 0.5 (days) Benzene 6 (air), 1 to 6 (surface water)

Benzo(a)pyrene 1 to 6 (air), 0.4 (surface water)Carbon Tetrachloride 8,030 (air), 0.3 to 300 (surface water)

Chlordane 40 (air), 420 to 500 (surface water)Chloroform 80 (air), 0.3 to 30 (surface water)

DDT 56 to 110 (surface water)1,1-Dichloroethane 45 (air), 1 to 5 (surface water)

Formaldehyde 0.8 (air), 0.9 to 3.5 (surface water)Heptachlor 40 (air), 0.9 (surface water)

Hexchloroethane 7,900 (air), 1.1 to 9.5 (surface water)PCBs 58 (air), 2 to 12.9 (surface water)2,3,7,8-TCDD (Dioxin) 365 to 730 (surface water)1,1,1-Trichloroethane 803 to 1,752 (air), 0.14 to 7 (surface water)

Trichloroethylene (TCE) 3.7 (air), 1 to 90 (surface water) Vinyl Chloride 1.2 (air), 1 to 5 (surface water)

Example VI.6. An underground storage tank (UST) has been leaking for many years,contaminating the groundwater and yielding a contaminant concentration directly

beneath the site of 0.30 mg/L. The contaminant is flowing at a rate of 0.5 m/day towardto public drinking water well 1 km away. The half-life of the contaminant is 10 years. (1)Estimate the steady-state contaminant concentration at the well. (2) If the slope factorof the contaminant is 0.02 (mg/kg-day)-1, estimate A(d ) if a 70-kg person drinks 2 L ofthis water per day for 10 years.

1.1. The time required to travel from the site to the well is:5.0

000,1 = 2,000 days.

1.2. Assuming a first-order degradation for the contaminant, then



158

)365)(10(

693.0693.0

5.0

t

k = 1.9×10-4 day -1.

1.3. Assuming a plug-flow situation, then )000,2)(4109.1(*

3.0 eeC C kt

sitewell

0.205 mg/L.

2.1. Now,)70)(365)(70(

)365)(10)(2)(205.0(CDI = 8.4×10-4 mg/kg-day.

2.2. Therefore, The upper-bound, incremental lifetime cancer risk, A(d ), is: )02.0)(104.8())(()( 4SF CDI d A 1.7×10-5.

Example VI.7. Chemicals in the environment may undergo natural degradations (i .e., biodegradation, photolysis, etc.). Sometimes degradation products may be moreharmful than parent chemicals. For instance, NO is a constituent in the flue gas streamsproduced from coal-burning power plants. In the presence of UV, NO reacts with oxygenand forms NO2. NO does not have adverse health effects under the normal atmosphericconcentrations, however, NO2 causes lung irritation, bronchitis, pneumonia, andrespiratory infections.

Another example is 1,1,1-trichloroethane which is a common solvent (one of least toxic tohumans) used in metal and electronic industries. When 1,1,1-trichloroethane is releasedinto the an anaerobic environment, it is first degraded via an abiotic reaction step toform C2H2Cl2, which is then biologically degraded to form vinyl chloride (C2H3Cl), acarcinogen.

Risk Characterization. The final step in risk assessment is to bring the variousstudies together into an overall risk characterization. In a simple sense, this step could

be interpreted to mean simply multiplying the exposure (dose, or CDI ) by the potency(i .e., RfD-1 for non-carcinogens and SF for carcinogens) to get individual risk, and thenmultiplying that by the number of people exposed to get an estimate of overall risk tosome specific population. While there are obvious advantages to presenting a simple,single number for extra cancers, or some other risk measures, a proper characterizationof risk should be much more comprehensive. The final expressions of risk derived in thisfinal step will be used by regulatory decision makers in the process of weighing healthrisks against other societal costs and benefits, and the public will use them to decide on

the adequacy of proposed measures to manage the risks. Both groups need to appreciatethe extraordinary leaps of faith that, by necessity, have had to be used to determine thesesimple quantitative estimates. Always remember that these estimates are preliminary,subject to change, and extremely uncertain. A number of questions suggested by theNational Academy of Sciences should be addressed in a final characterization of risk:

1. What are the statistical uncertainties in estimating the extent of health effects?2. What are the biological uncertainties? What are their origins? How will they be

estimated? What effects do they have on quantitative estimates? How will the

22

22

22

2

NOO NO

NOO N

h

Cl Cl H C e H Cl H C

Cl H Cl H C Cl H C

Biotic

Abiotic

32222

222332

2



159

uncertainties be described to regulatory decision makers?3. What dose-response assessments and exposure measurements should be used?4. Which population groups should be the primary targets for protection, and which

provide the most meaningful expression of the health risk?

15 most common groundwater contaminants detected at national

superfund sites.

Rank Contaminant Common Sources

1 Trichloroethylene Dry cleaning, metal degreasing

2 Lead Gasoline (before 1975), mining, pipes, manufacturing

3 Tetrachloroethylene Dry cleaning, metal degreasing

4 Benzene Gasoline, manufacturing

5 Toluene Gasoline, manufacturing

6 Chromium Metal plating

7 Methylene Chloride Degreasing, solvents, paint removal

8 Zinc Manufacturing, mining

9 1,1,1-Trichloroethane Metal and plastic cleaning

10 Arsenic Manufacturing, mining

11 Chloroform Solvents

12 1,1-Dichloroethane Degreasing, solvents

13 Trans-1,2-Dichloroethene Degradation product of trichloroethylene

14 Cadmium Mining, plating

15 Manganese Manufacturing, mining

Hazardous Waste Generator Classifications.

Small QuantityGenerators (SQGs)

Large QuantityGenerators (LQGs)

ConditionallyExempt (SQGs)

Quantity

100-1,000 kg ofhazardous waste permonth and< 1 kg of acutelyhazardous waste per

month.

> 1,000 kg of hazardous waste per month or> 1 kg of acutelyhazardous waste permonth.

< 100 kg of hazardous waste per month and< 1 kg of acutelyhazardous waste permonth.

On-Site StorageTime (days) 180-270 < 90 180-270

Analysis of Hazardous Waste Problems.

Variables Hypothesis to be Tested



160

Sources

ContaminantsConcentrationsTimeLocations

Source existsSource can be containedSource can be removed and disposedSource can be treated

Pathways MediaRate of migration

TimeLoss and gain functions

Pathway existsPathway can be interrupted

Pathway can be eliminated

Receptors

TypesSensitivitiesTimeConcentrationsNumbers

Receptor is not impacted by migration ofcontaminantsReceptor can be relocatedInstitutional control can be appliedReceptor can be protected

VII. ENVIRONMENTAL FATE MODELING

The understanding of the behavior of chemicals (primarily organic chemicals) in the



161

biosphere of air, water, soil, sediments, and the diversity of biota is vital to thepreservation and protection of the multimedia environment. The chemicals behave inaccordance with the laws of nature which dictate chemical partitioning tendencies andrates of transport and transformation. Most fundamentally, the chemicals are subject tothe laws of conservation of mass, or the “mass balance.” Mathematical modelingprovides a framework in which the fate of chemicals in real environments can be

accurately and conveniently accounted for using the mathematical techniques withdifferent levels of difficulty.

The concept of fugacity, which was introduced by G.N. Lewis in 1901 as a criterion ofequilibrium, has proven to be a very convenient and elegant method of calculatingmultimedia equilibrium partitioning. The fugacity approach of environmental fatemodeling requires that the environment be divided into a number of interconnectedabiotic and biotic compartments. The simplest is a simple 4-compartment system whichis easily understood and illustrates the general principles which are applied inmultimedia calculations. A more complex 8-compartment system is more representativeof real environments, and is correspondingly more demanding of data, and lead to morelengthy calculations.

ATMOSPHERE.

Air. The layer of the atmosphere which is in most intimate contact with the surface ofthe earth is the troposphere, which extends to a height of about 10 km. The temperature,density, and pressure of the atmosphere fall steadily with increasing height, which is anuisance in subsequent calculations. If a uniform density is assumed at atmosphericpressure, then the entire troposphere can be viewed as being compressed into a height ofabout 6 km. Exchange of matter from the troposphere through the tropopause to thestratosphere is a relatively slow process and is rarely important in environmentalcalculations. If the environmental model is concerned with a localized situation, forexample, a state, province, or metropolitan region, it is unlikely that most pollutants

would manage to penetrate higher than about 1,000 to 2,000 m during the time the airresides over the region. It may therefore be appropriate to reduce the height of theatmosphere to 1,000 to 2,000 m in such cases. In extreme cases, for example oversmall ponds or fields, the accessible mixed height of the atmosphere may be as low as 10m. Therefore, the evaluative air volume is (1,000 m)(1,000 m)(6,000 m) = 6×109 m3.

Aerosol. The atmosphere contains a considerable amount of particular matter oraerosols which are important in determining the fate of certain chemicals. Theseparticles may range from water in the form of fog or cloud droplets to dust particles fromsoil and smoke from combustion. The concentration of these particles is normallyreported in μg/m3. A rural area may have a concentration of about 5 μg/m3, and a fairlypolluted urban area a concentration of 100 μg/m3. If it is assumed that the particleshave a density of 1.5 g/cm3 and are present at a concentration of 30 μg/m3, then in anevaluative air volume of 6×109 m3, there is thus 0.12 m3 (or 120 L) of solid material.

Deposition Processes. The aerosol material has a very large specific surface area andtherefore absorbs, or adsorbs many pollutants, especially those of very low vaporpressure such as the PCBs or polyaromatic hydrocarbons (PAHs). The chemicalsassociated with aerosol particles are subject to two important deposition processes: drydeposition and wet deposition. In dry deposition the aerosol particles can be regarded asfalling under the influence of gravity to the earth’s surface. The falling velocity (or



162

deposition velocity) is quite slow and depends on the conditions of atmosphere, the sizesand properties of aerosol particles, and the nature of the ground surface, but a typical

velocity is about 0.3 cm/s or 10.8 m/h. The result is deposition of (10.8 m/h)[2×10-11 (volume fraction)](106 m2) or 0.00026 m3/hr or 1.89 m3/yr. The aerosol particles may

be scavenged or swept out of the air by wet deposition with rain drops. As it falls, eachrain drop sweeps through a volume of air about 200,000 times its volume prior to

landing on the land or water surface. It thus has the potential to remove a considerableamount of aerosol particles from the atmosphere. Rain is, therefore, often highlycontaminated with substances such as PCBs and PAHs. Typical rainfall rates lie in therange 0.3 to 1 m/yr. If a value of 0.8 m/yr is adopted, then the wet deposition rate is(0.8 m/yr)(200,000)[2×10-11 (volume fraction)](106 m2) or 0.00044 m3/hr or 3.20m3/yr. In the 4-compartment evaluative environment aerosols are ignored, but they areincluded in the 8-compartment version.

HYDROSPHERE.

Water. Since 70% of the earth’s surface is covered by water, the area of water in theevaluative environments is taken as 700,000 m2. Similarly to the atmosphere, only

near-surface water is accessible to pollutants in the short term. In the oceans, this depthis about 100 m, but since most situations of environmental interest involve fresh orestuarine water, it is more appropriate to use a shallower water depth of perhaps 10 m.This yields a water volume of 7×106 m3. If the aim is to mimic the proportions of waterand soil in a political jurisdiction, such as a state or province, the area of water willnormally be considerably reduced to perhaps 15% of the total, giving a volume of about106 m3.

Water is normally treated as being pure, i .e., containing no dissolved electrolytes, but itdoes contain suspended particles or suspended solids ( SS ). Particular matter in the

water plays a key role in influencing the behavior of chemicals. A very clear water mayhave a concentration of particles as low as 1 g/m3, or 1 mg/L. In most cases, the

concentration is higher, in the range of 5 to 20 mg/L. Very turbid, muddy water maycontain over 100 mg/L. Assuming a concentration of 7.5 mg/L and a density of 1.5g/cm3 gives a volume fraction of particles of about 5×10-6. Therefore, in the 7×106 m3 of

water, there are 35 m3 of particles (or SS ).

The particular matter in the water consists of a wide variety of materials. It containsmineral matter which may be clay or silica in nature. It also contains dead or detritalorganic matter which is often referred to as humin, humic acids, and fulvic acids or,more vaguely, as organic matter (OM ). Typically OM contains 50 to 60% organic carbon(OC ). For convenience, OM = 0.5OC , and the densities for both OM and OC are equal tothat of water. The organic materials in the suspended phases are of great importance

because they have a high sorptive capacity for organic chemicals. It is therefore commonto assign an OC content to these particular phases. In a fairly productive lake the OM content of particles may be as high as 50%, but for illustrative purposes a figure of 35%for OM or 16.6% OC is convenient.

Fish and Aquatic Biota. Fish are of particular interest because they are ofcommercial and recreational importance to users of water, and they tend tobioconcentrate or bioaccumulate metals and organic chemicals from water. Forillustrative purposes, it is assumed that all the materials in the water are fish and thetotal concentration is about 1 part per million by volume, yielding a volume of “fish” of



163

about 7 m3. In summary, the water thus consists of 7×106 m3 of water, containing 35 m3 of particulate matter and 7 m3 of “fish” or biota.

Deposition Processes. The particular matter in water is important because, likeaerosols in the atmosphere, it serves as a vehicle for the transport of chemicals from the

bulk of the water to the bottom sediments. Hydrophobic substances tend to partition

appreciably onto suspended particles, and are thus subject to fairly rapid deposition.This deposition velocity is typically 0.5 to 2.0 m/d or 0.02 to 0.08 m/h. This velocity issufficient to cause removal of most of suspended matter from most lakes during thecourse of a year. Assuming a figure of 5×10-6 m3 particles per m3 of water and adeposition velocity of 200 m/yr, the deposition rate will be 0.001 m3 particles per m2 ofsediment area per year, or for an area of 7×105 m2, a flow of 700 m3/yr.

BOTTOM SEDIMENTS. The bottom sediments typically consist of 95% water and 5%particles, and are often highly organic in nature. The bottom sediments may consist ofdeposited particles and fecal matter from the water column. They are stirred bycurrents, and by the action of the various biota present in this “benthic” region. Thesediments become more consolidated at greater depths, and the water content tends to

drop toward 50%. The top few centimeters of sediment are occupied by burrowingorganisms which feed on the organic matter and on each other, and generally turn over(bioturbate) this entire “active” layer of sediment. Depending on the conditions of the

water column above, this layer may be oxygenated (aerobic or oxic) or depleted ofoxygen (anaerobic or anoxic). This has profound implications on the fate of inorganicsubstances such as metals and arsenic but it is relatively unimportant for organicchemicals, except in that the oxygen status influences the nature of the microbialcommunity, which in turn influences the availability of metabolic pathways for chemicaldegradation. Most of the activity occurs in the top 5 cm of the sediment.

Chemicals present in sediments are primarily removed by degradation, burial, orresuspension and diffusion back to the water column. For illustrative purposes, a

sediment depth of 3 cm is adopted and the sediment consists of 63% water and 37%solids by volume. These solids consist of about 10% OM or 5% OC . Living creatures areincluded in these figures. It is now possible to assemble an approximate mass balancefor the sediment mineral matter ( MM ), organic matter (OM ), and the organic carbon(OC ). On a 1 m2 basis, the deposition rate may be 0.001 m2/year, or 1,000 cm3/year,

which with a particle density of 1.5 g/cm3 corresponds to a deposition rate of 1,500g/year (500 g OM and 1,000 g MM ). If 40% is resuspended (i .e., 200 g OM and 400 g

MM ), then 300 g OM will remain in bottom sediments. Of the remaining 300 g OM , it isassumed that 233 g is degraded to CO2 and 67 g is buried along with the remaining 600 gof MM , then the total burial is 667g, which consists of 600 g or 300 cm3 of MM and 67cm3 of OM (i .e., 10% OM by mass or 18% OM by volume). The total volumetric burialrate of solids is 367 cm3/year. Now associated with these solids is 633 cm3 of pore waterand therefore, the total volumetric burial rate of solids plus water is approximately 1,000cm3/year, corresponding to a rise in the sediment-water interface of 1 mm/year. The

percentage of OC in the depositing and resuspending material is 17% while in the buriedmaterial it is 5%. The bottom sediment bulk density, including pore water, is thus 1,300

kg/m3. On a 7×105 m2 basis the deposition rate is 700 m3/year, resuspension is 280

m3/year, burial is 257 m3/year, and degradation accounts for the remaining 163 m3/year.

SOIL. A typical soil may consist of 50% solid matter, 20% air, and 30% water, by volume. The solid matter may consist of about 2% OC or 4% OM . During and after



164

rainfall, water flows vertically downward through the soil and may carry chemicals withit. During periods of dry weather, water often returns to the surface by capillary action,again moving the chemicals.

For illustrative purposes, the soil covers an area of 1,000 m by 300 m by 15 cm deep, which is about the depth to which agricultural soils are plowed. This yields a volume of

45,000 m3. The soil consists of about 50% solids of 4% OM content, or 2% OC by mass.The porosity of the soil, or void space, is 50%, and consists of 20% air and 30% water.

Assuming a density of the soil solids of 2,400 kg/m3and water of 1,000 kg/m3 givesmasses of 1,200 kg solids and 300 kg water per m3, totaling 1,500 kg, corresponding toa bulk density of 1,500 kg/m3. Rainwater falls on this soil at a rate of 0.8 m/year. Ofthis, perhaps 0.3 m evaporates, 0.3 m runs off, and 0.2 m percolates to depths andcontributes to groundwater flow. This results in water flows of 90,000 m3/year byevaporation, 90,000 m3/year by runoff, and 60,000 m3/year by percolation to depths,totaling 240,000 m3/year.

Terrestrial Biota or Plants. Amounts of biomass per m2 vary continuously from zeroin deserts to massive quantities which greatly exceed soil volumes in tropical rain forests.

They also vary seasonally. If it is desired to include vegetation, a typical depth of plant biomass might be 1 cm. It is convenient to regard the plants as having a volume of3,000 m3, containing the equivalent of 1% lipid-like materials and 50% water. Plantsare ignored in the simple 4-compartment model, treating the soil as only a simple solidphase. Figure VII.1 illustrates both 4- and 8-compartment environmental systems.

Compartment Volume (m3) Density (kg/m3) Composition Air 6×109 1.2 --

Water 7×106 1,000 --Soil (50% solids, 30% water, and 20% air) 4.5×104 1,500 2% OC

Sediment (37% solids) 2.1×104 1,300 5% OC Suspended Solids (S.S.) 35 1,500 16.7% OC

Aerosols 0.12 1,500 30 μg/m3 Aquatic Biota 7 1,000 5% lipid

Terrestrial Biota 3,000 1,000 1 % lipid

Area (m2)◙ Air-Water: 7×105.◙ Water-Sediment: 7×105.◙ Soil-Air: 3×105.

Rain Rate◙ 0.8 m/year or 800,000 m3/year.◙ 560,000 m3/year→ water.◙ 240,000 m3/year→ soil.

Aerosol Deposition Rate (total)◙ 2.16×10-4 m3/hr or 1.89 m3/year (dry).◙ 3.65×10-4 m3/hr or 3.2 m3/year (wet ).

Sediment Deposition Rates ◙ Deposition: 700 m3/year (solids, 17% OC ).◙ Resuspension: 280 m3/year (solids, 17% OC ).◙ Net Deposition or Burial: 257 m3/year (solids, 5% OC ).

Fate of Water in Soil ◙ Evaporation: 90,000 m3/year.◙ Surface Runoff: 90,000 m3/year (→ water).◙ Percolation: 60,000 m3/year (→ groundwater)◙ Solids Runoff: 90 m3/year (erosion).



165

Figure VII.1. 4- and 8-comparment environmental systems.

RELATING CONCENTRATION AND FUGACITY ( Z VALUES). The concept of fugacity is the tendency of a given molecular species to escape from one phase toanother. So the higher the fugacity, the lower the chemical concentration. Its units arepressures. Fugacity is related to the chemical concentration by:

Zf C

Z

C f

( VII.1)

where f is the fugacity, Pa; C is the chemical concentration, mol/m3; and Z is the fugacity

capacity, mol/m3-Pa. Z is analogous to a heat capacity. If two phases are in equilibrium,then f 1 = f 2. A chemical is in equilibrium with several phases, if f AIR = f WATER = f SOIL =

f SEDIMENT = f BIOTA, etc. In other words, under equilibrium conditions there is one f value which is applicable to all phases involved. The equations for calculating Z values are:



166

v AEROSOL

WATEROW OW

OCTANOL

v

pu reSOLUTE

WATER B B B B

BIOTA

WATER pS

pS

SEDIMENT SOIL

v

eWATER

AIR

RTP Z

Z K H

K Z

P V

Z

Z K H

K Z

Z K H K Z

P

C

H Z

RT Z

6

)(

)(

106

1

1

1

( VII.2)

where R is the universal gas constant, Pa-m3/mol-K; T is the absolute temperature, oK; H is the Henry’s law constant, Pa-m3/mol; C e is the aqueous solubility, mol/m3; P v is the

vapor pressure, Pa; K p is the partition coefficient, L/kg; ρ S is the solid density, kg/L; K B is

the bioconcentration factor ( BCF ), L/kg; ρ B is the biota density, kg/L;

V is the solutemolar volume, m3/mol; and K OW is the octanol-water partition coefficient,dimensionless. The following two equations are also useful for estimating relevantcoefficients in ( VII.2):

70.0)ln(85.0)ln()ln(494.7)ln(

OW B

eOW

K K C K ( VII.3)

Level I Calculations. The level I calculation applies to steady-state, no-flow systems(i .e., closed systems) in equilibrium. In these systems the fugacities are equal andunchanging; there is no reaction, inflow, or outflow. The total amount of the materialreleased to the environment is given and equals M (moles). In addition, each phasein equilibrium has a volume of V i (m3) and a fugacity capacity Z i (i denotes thecompartment i ). Moreover, the material is uniformly distributed in each phase. Thefollowing steps can be followed to perform the level I calculation:

n

iiiii

i

n

ii

iii

i

n

ii

in

iii

n

ii

m M V C m

Z V

MZ fZ C

Z V

M f Z V f C V M

1

11

11

( VII.4)



167

where C i is the molar concentration of the material in the phase i , mole/m3; and mi is theamount of the material in phase i , moles. In many cases, the concentration of the

material in the phase i is expressed asi

ii

MW C C

)000,1)()((' , where '

iC is the mass

concentration of the material in the phase i , μg/g; MW is the molecular weight of the

material, g/mol; and ρi is the density of the phase i , kg/m3

.

Example VII.1. Calculate the distribution of 100 moles (20 kg) of a hypotheticalchemical “hypothene” in the simple 6-compartment evaluative environment using theproperties below which applied at 27.5 oC. Assume organic carbon (OC ) contents of 2%and 4% in soil and sediment, respectively.

● Temperature: 27.5oC (300.6oK);● Molecular Mass ( MW ): 200 g/mol;● Water Solubility (C e): 20 g/m3 (0.1 mol/m3);● Vapor Pressure ( P v): 1.0 Pa (7.5×10-3 mm Hg or 9.9×10-6 atm); and● K OW : 100,000.

1. Air-Water Partition Coefficient and Z Values.

1.1. s

s

C

P H = 10 Pa-m3/mol.

1.2. K AW (Air-Water Partition Coefficient) = RT

H = 0.004.

1.3. Z AIR = RT

1 = 4.0×10-4 mol/m3-Pa.

1.4. Z WATER = H

1 = 0.1 mol/m3-Pa.

1.5. ρ A (Air Density) = RT

325,101029.0 = 1.17 kg/m3 (ideal gas law).

2. Compartmental Characteristics.

Name Soil Sediment Suspended Solids Fish

ρi (kg/m3) 1,500 1,500 1,500 1,000

OC or LipidContent f OC (g/g) 0.02 0.04 0.04 0.048

K OC or K OW 41,000 41,000 41,000 100,000

K P = f OC K OC (or f OC K OW )

820 1,640 1,640 4,800

000,1i P

PW

K K

1,230 2,460 2,460 4,800

WATER PW i Z K Z 123 246 246 480



168

3. K OC = 0.41 K OW .

4. The Level I calculation performed are summarized in the table shown below.

4.1. 6

6

1

1024.7

i

ii Z V

M

f , C G (g/m3

) = C i ×(molecular weight, MW), and

i

GU

C C

)000,1)(( .

4.4. Most of the compound is associated with soil and sediment (>77%), 17.4% in theair, and 5.1% in the water. The concentrations on the μg/g basis, on the otherhand, show a different picture. The negligible amount present in fish (i.e.,0.024%) represents a concentration of 0.70 μg/g, which is within a factor of 10 oflevels considered unsafe for human consumption in many cases. The sedimentcould obviously act as a source of biotic contamination for a long time.

Air Water Soil Sediment S.S. Fish TOTALV i (m3) 6.0×109 7.0×106 4.5×104 2.1×104 35 7 --

Z i (moles/m3-Pa) 4.0×10-4 0.1 123 246 246 480 --V i Z i (moles/Pa) 2.4×106 7.0×105 5.54×106 5.17×106 8,610 3,360 1.381×107

C i = fZ i (moles/m3)

2.9×10-9 7.2×10-7 8.9×10-4 1.8×10-3 1.8×10-3 3.5×10-3 --

mi = C i V i (moles) 17.4 5.1 40.0 37.4 0.062 0.024 100% 17.4 5.1 40.0 37.4 0.062 0.024 100

C G (g/m3) 5.8×10-7 1.4×10-4 0.178 0.36 0.36 0.70 -- ρi (kg/m3) 1.17 1,000 1,500 1,500 1,500 1,000 --C U (μg/g) 4.9×10-4 1.4×10-4 0.12 0.24 0.24 0.70 --

Level II. Calculations. The level II calculation applies to steady-state, flow systems(i .e., open systems) in equilibrium. In these systems the fugacities are equal andunchanging; there are significant transformations and persistence caused by reactions,inflow, and outflow. The input of the compound concerned into the evaluativeenvironment is at steady state. Transformations are induced by photolysis, hydrolysis,oxidation, biodegradation, and advection of the compound out of the region of interest.

Advection. Advection means the movement of chemical by virtue of it presence in the

medium which happens to be flowing. The rate of advection of a chemical is simply

n

(mol/hr), the product of the flowrate of the advecting medium,

V (m3/h) and theconcentration of chemical in the medium, C (mol/m3):

C V n

( VII.5)

If pure air and pure water contaminated only by the target chemical are the means thatcarry the chemical into the evaluative environment, then the total influx of the target

chemical, total n

, is:



169

IW W IA Atotal C V C V E n

( VII.6)

where E is the total emission rate of the chemical into the evaluative environment,mol/h; and the subscripts A, I , and W denote air, input, and water, respectively.

Degrading Reactions. A simple first-order kinetic expression for all reactions involved will be used. The basic rate equation is −VkC (mol/hr), where k is the first-order ratecoefficient (1/hr).

Combined Advection and Reactions. In This case, the mass balance equation is:

IW W IA Atotal C V C V E n

= iii

iW W A A C k V C V C V

( VII.7)

Assume a common fugacity f , then in fugacity format,

total n

= iii

iW W A A C k V C V C V

=

iri

iaiii

iiWATERW AIR A D D f k Z V f Z V Z V f

( VII.8)

The subscripts a and r denote advection and reaction, respectively. The following stepscan be followed to perform the level II calculation:

1.WATERW AIR A

iai Z V Z V D

(Advection)

2. iii

ii

ri k Z V D (Reaction)

3. IW W IA Atotal C V C V E n

=

i iriai D D f (Total Influx)

4.

i iriai

total

D D

n f (Fugacity)

5. i

iiiiii mmV C m f Z C

6.

i

ai

iii

WATER A AIR A

iii

i ii

iii

i ii

iii

i ii

a D

Z V

Z V Z V

Z V

Z V f

Z V f

C V

C V

C V

mt

7.

iri

iii

iiii

iii

iiii

iii

iiii

iii

iiii

r D

Z V

k Z V

Z V

k Z V f

Z V f

k C V

C V

k C V

mt

8.r ao t t t

111



170

where t a, t r, and t o denote advection residence time, reaction residence time, and overallresidence time, respectively.

Example VII.2. Calculate the level II calculations of a hypothetical chemical“hypothene” in the simple 4-compartment evaluative environment using the properties

below which applied at 27.5 oC. Assume organic carbon (OC ) contents of 2% and 4% insoil and sediment, respectively.

● First-Order Rate Coefficients (1/hr): 0.001, water; 0.01, soil; 0.0001, sediment;and 0, air.

● AV

= 107 m3/hr, IAC = 10-6 mol/m3, W V

= 1,000 m3/hr, IW C = 0.01 mol/m3, and

E = 100 mol/hr.● Temperature = 27.5oC (300.6oK).● Molecular Weight (MW) = 200 g/mol.● Water Solubility (C s) = 20 g/m3 (0.1 mol/m3).● Vapor Pressure ( P s) = 1.0 Pa (7.5×10-3 mm Hg or 9.9×10-6 atm).

● log( K OW ) = 5 (or K OW = 100,000).

Air Water Soil SedimentV i (m3) 6.0×109 7.0×106 4.5×104 2.1×104

Z i (moles/m3-Pa) 4.0×10-4 0.1 123 246V i Z i (moles/Pa) 2.4×106 7.0×105 5.54×106 5.17×106

ki (1/hr) 0 0.001 0.01 0.0001 Dri = V i Z i ki (moles/Pa-hr) 0 700 5,540 51.7

iiai Z V D

( moles/Pa-

hr)

4,000 1,000 0 0

C i (moles/m3) 4.46×10-6 1.15×10-3 0.14 0.28mi (moles) 2.28×104 8,050 6,400 5,900

% 57.5 16.8 13.3 12.4

1. Reaction half life (ik

t 693.0

5.0 , hrs): 4, air; 693, water; 69.3, soil, and 6,930,

sediment.

2. .0106.07.51540,5700000,1000,4

1010100

i iriai

Total

D D

n f

3. C G (g/m3): 9.2×10-4, air; 0.23, water; 28, soil; and 56, sediment.4. C U (μg/g): 0.78, air; 0.23, water; 19, soil; and 38, sediment.5. m = 48,300 moles.6. t a = 1,018 hrs, t r = 665 hrs, and t o = 402 hrs.



171

VIII. Appendices

Prefixes Used in the SI System.

Prefix Symbol Multiples & Submultiples Yotta Y 1024

Zeta Z 1021 Exa E 1018 Peta P 1015 Tera T 1012 Giga G 109 Mega M 106 Kilo K 103

Hector H 102 Deka Da 101 Deci D 10-1 Centi C 10-2 Milli M 10-3

Micro Μ 10-6

Nano N 10-9 Pico P 10-12

Femto F 10-15 Atto A 10-18 Zepto Z 10-21 Yocto Y 10-24

Common Unit Conversions and Derived SI Units.

1 = 10-10 m.1 m = 100 cm = 3.28 ft = 39.4 in.1 km = 1,000 m = 0.62 mi.

1 mi2 = 640 acres = 259 hectares.1 hectare = 104 m2 = 10-2 km2 = 2.47 acres.1 m3 = 1,000 L = 106 cm3 = 264 gal = 35.3 ft3.1 acre-foot = 1,233.5 m3.1 m/s = 2.24 mph (mi/hr) = 3.28 ft/s = 3.6 km/hr.1 cm/s = 1.97 ft/min.1 m3/hr = 0.59 cfm (ft3/min) = 6.3×10-3 MGD (million gallons per day).1 MGD = 0.0438 m3/s.1 kg = 2.2 lbm.1 tonne (metric tone) = 1,000 kg = 1.1 ton (short tone).1 N = 1 kg·m/s2 = 105 dynes = 105 g·cm/s2 = 0.225 lbf .1 atm = 101.3 kPa = 1.013 bar = 14.7 psi = 760 mm Hg = 407 in H2O.1 Pa = 1 N/m2.

1 J = 1 kg·m2

/s2

= 107

erg = 107

g·cm2

/s2

= 0.239 cal = 9.48×10-4

Btu.1 W = 1 J/s = 1 N·m/s = 107 kg·m2/s3 = 3.41 Btu/hr = 1.34×10-3 hp.1 M = 1 mole/L, 1 N = 1 eq/L, 1% = 10 -2, 1‰ = 10-3, 1 ppm = 10-6, 1 ppb = 10-9, and 1 ppt = 10-12.

The Gas Constant ( R) (oK = 1.8oR, K: Kelvin, R: Rankine).

8.314 m3·Pa/mol·oK 62.36 L·mm Hg/mol·oK 8.314 J/mol·oK0.08314 L·bar/mol·oK 0.7302 ft3·atm/lb-mol·oR 1.987 cal/mol·oK0.08206 L·atm/mol·oK 10.73 ft3·psia/lb-mol·oR 1.987 Btu/lb-mol·oR



172

Properties of the Earth.

Surface Area 5.1×1014 m2 Mass 6.0×1024 kgMass of Hydrosphere 1.4×1021 kg

Mass of Atmosphere 5.3×1018 kgRadius 6,370 kmGravitational Acceleration (g) 9.81 m/s2

Wastewater Treatment and Management Terminologies.

Biosolids

Biosolids are primarily an organic, semi-solid wastewater product thatremain after solids are stabilized biologically or chemically and aresuitable for beneficial uses. Biosolids in which the pathogens (includingenteric viruses, pathogenic bacteria, and viable helminth ova) arereduced below current detectable levels are referred to as class A

biosolids. Biosolids in which the pathogens are reduced to levels that areunlikely to pose a threat to public health and the environment underspecific use conditions are referred to as class B biosolids. Class B

biosolids cannot be sold or given away in bags or other containers orapplied on lawns or home gardens.

Characteristics General classes of wastewater constituents such as physical, chemical, biological, or biochemical.

Composition The makeup of wastewater, including the physical, chemical, and biological constituents.

Constituents Individual components, elements, or biological entities such assuspended solids or ammonia nitrogen.

Contaminants/Pollutants Constituents added to the water supply through use.

Disinfection Reduction of disease-causing microorganisms by physical or chemicalmeans.

Impurities Constituents added to the water supply through use.Non-Point Sources Sources of pollution that originate from multiple sources over a relatively

large area.

Nutrient An element that is essential for the growth of plants and animals.Nutrients in wastewater, usually nitrogen and phosphorus, may causeunwanted algal and plant growths in lakes and streams.

Parameter A measurable factor such as temperature.

Point SourcesPollution loads discharged at a specific location from pipes, outfalls, andconveyance methods from either municipal wastewater treatment plantsor industrial wastewater treatment plants.

Pollutants Constituents added to the water supply through use.Reclamation Treatment of wastewater for subsequent reuse applications or the act of

reusing treated wastewater.Recycling The reuse of treated wastewater and biosolids for beneficial purposes.

Re-Purification Treatment of wastewater to levels suitable for a variety of applicationsincluding indirect or direct potable reuse.

Reuse Beneficial reuse of reclaimed or re-purified wastewater or stabilized

biosolids.Sludge Solids removed from wastewater during treatment.

SolidsMaterials removed from wastewater by gravity separation (by clarifiers,thickeners, and lagoons) and are the solid residues from dewateringoperations.



173

Coupled, First-Order Ordinary Differential Equations (ODEs). The generalproblem for two coupled, first-order ODEs can be written as follows, where

S i, Ai, Bi, and Li are constants.

22

11

1121222

221111

)0(

)0(

B y

B y

y AS y Ldt

dy

y AS y L

dt

dy

( VIII.1)

Rearrange the 2nd equation in ( VIII.1) to solve for y1 and then substitute into the 1st equation in ( VIII.1):

221122

22

2111221212

212

2

2

)0(

)0(

)()(

B L B AS dt

dy

B y

S LS A y A A L L

dt

dy L L

dt

yd

( VIII.2)

The solution to the 1st equation in ( VIII.2) is:

2121

2111,2

,22121

,2

212

,2

2

,2,22

0)()(

)()(

A A L LS LS A y

y A A L Ldt

dy L L

dt

yd

yt yt y

p

h

hh

ph

( VIII.3)

The general solution to the 1st equation in ( VIII.3) is:

21

2

21

,21112212

21

2

21

,22222112

21

2

2121

2

21

2

2121

1

212121

2

21,2

4)(

)()(

4)(

)()(

4)(2

1

2

)(

4)(2

1

2

)(

0)()(

)(

A A L L

yr B AS Lr BG

A A L L

yr Lr B B AS F

A A L L

L L

r

A A L L L L

r

A A L Lr L Lr

Ge Fet y

p

p

t r t r

h

( VIII.4)



174

A similar approach is used to solve for y1 and the result is:

21

2

21

,11221111

21

2

21

,12121221

2121

1222,1

,121

,1,11

4)(

)()(

4)(

)()(

)()(

A A L L

yr B AS Lr B J

A A L L

yr Lr B B AS H

A A L L

S LS A y

y Je He yt yt y

p

p

p

p

t r t r

ph

( VIII.5)

Example A.1. 1,00o moles/s of biogas (70% CH4 and 30% CO2) @ 25oC and 1 atm iscompletely burned in a burner @ 400oC and 1 atm with 10.0% excess dry air preheatedto 100oC. The burner is divided into two chambers: a combustion chamber and a boiler(see the attached diagram). The dry air is preheated in a preheater using the stack gas

(@ 400o

C) released from the combustion chamber. The reaction heat produced in thecombustion chamber is transferred into the boiler where a liquid water stream @ 25oCand 1 atm is heated to produce a saturated steam @ 100oC and 1 atm. Find the heattransferred from the combustion chamber into the boiler (kW) and the mass rate at

which the saturated steam is produced (kg/s). The stoichiometric combustion equationof CH4 is:

)(2)()(2)( 2224 vO H g CO g O g CH

◙ Combustion Chamber:

1. The specific enthalpy of a species i ( i H

) in the combustion chamber can becalculated as:

dT C H H i

r

T

T

i p

i

o

f i

)(

wherei

o

f H

is the standard heat of formation of the species i , kJ/mol; r T is

the reference temperature, oC (or oK); iT is the actual temperature of the species i ,

oC (or oK) and i pC )( is the specific heat capacity of the species i at constantpressure, kJ/mol-oC.

2. The enthalpy table for the stack gas is constructed and shown below. 25oC ischosen as the reference temperature.



175

Reference States: CH4(g), CO2(g), O2(g), N2(g), and H2O(v) @ 25oC and 1 atm.Substanceinn

(moles/s) in H

(kJ/mol) ou t n

(moles/s) ou t H

(kJ/mol)CH4(g)

1

n (700) 1

H (-74.52)-- --

CO2(g)2

n (300) 2

H (-390.61) 5

n (1,000) 5

H (-377.16)O2(g)

3

n (1,540) 3

H (2.24) 6

n (140) 6

H (11.72)N2(g)

4

n (5,793) 4

H (2.19) 7

n (5,793) 7

H (11.15)H2O(v) -- --

8

n (1,400) 8

H (-228.60)

3. )7.0)(000,1(1

n = 700 moles/s.

)3.0)(000,1(2

n = 300 moles/s.

)2)(700)(1.1(3

n = 1,540 moles/s.

21.0

79.0)2)(700)(1.1(4n = 5,973 moles/s = 7

n .

1: 700 moles/s CH4 @ 25oC and 1 atm.2: Dry Air (10% excess) @ 100oC and 1 atm.3: Stack Gas @ 400oC and 1 atm.4: Liquid Water @ 25oC and 1 atm.5: Saturated Steam @ 100oC and 1 atm.



176

)1)(700(3005

n = 1,000 moles/s.

)2)(700(540,16

n = 140 moles/s.

)2)(700(8

n = 1,400 moles/s.

4. 1

H =)(4 g CH

o

f H

= -74.85 kJ/mol.

2

H = )25()100( )()(

)(

22

2

C H C H H o g CO

o g CO

g CO

o

f

= -393.51 + 2.90

= -390.61 kJ/mol

3

H = )25()100( )()(

)(

22

2

C H C H H o g O

o g O

g O

o

f

= 0 + 2.24

= 2.24 kJ/mol

4

H = )25()100( )()(

)(

22

2

C H C H H o g O

o g N

g N

o

f

= 0 + 2.19

= 2.19 kJ/mol.

5

H = )25()400( )()(

)(

22

2

C H C H H o g CO

o g CO

g CO

o

f

= -393.51 + 16.35

= -377.16 kJ/mol.

6

H = )25()400( )()(

)(

22

2

C H C H H o g O

o g O

g O

o

f

= 0 + 11.72

= 11.72 kJ/mol.7

H = )25()400( )()(

)(

22

2

C H C H H o g N

o g N

g N

o

f

= 0 + 11.15

= 11.15 kJ/mol.

8

H = )25()400( )()(

)(

22

2

C H C H H ovO H

ovO H

vO H

o

f

= -241.83 + 13.23

= -228.60 kJ/mol.

5. Therefore,ini

ii

ou t i

ii H n H n H Q

= -475,517 kW (heat is taken

out of the combustion chamber).◙ Boiler:

1. The enthalpy table for water (liquid and steam) is constructed and shown below.25oC is chosen as the reference temperature.

Reference States: H2O(l ) @ 25oC and 1 atm.



177

Substanceinm

(kg/s) in H

(kJ/kg) ou t m

(kg/s) ou t H

(kJ/kg)H2O(l )

m (178) 9

H (0)-- --

H2O(v) -- --

m (178) 10

H (2,676)

2. 9

H = 0 kJ/kg (reference).

10

H = 2,676 kJ/kg.

3. Therefore,676,2

517,475

910

H H

Qm = 178 kg/s.

4. Since the specific volume of the steam @ 100oC and 1 atm is 1.674 m3/kg, the volumetric rate of steam produced is 106.3 m3/s.

Example III.9. A complete-mixed anaerobic bioreactor is designed to treat anindustrial wastewater at a capacity of 1,500 m3/day. The bioreactor contents are to beheated by the circulation of the mixed liquor through an external hot water heatexchanger. Assuming that the following conditions apply, find the heating requirementto maintain the bioreactor temperature at 35oC.

Concrete Bioreactor Dimensions:- Diameter: 20 m.- Side Depth: 6 m.- Center Depth: 9 m.

Heat Transfer Coefficients (U j , W/m2-oC): - Dry Earth Embankment (entire depth): 0.7.- Bioreactor Floor (moist earth): 0.7.- Roof (exposed to air): 0.9.

Temperatures (oC):- Ambient Air: -5.- Earth (next to the wall): 0.- Earth (below the floor): 10.- Feed Wastewater Stream: 10.

Specific Heat Capacities (C p, kJ/kg- oC):- Feed Wastewater Stream/Bioreactor Contents: 4.184.

Feed COD (mg/L): 10,000.COD Removal Efficiency (%): 80.

◙ Total Heating Requirements.1. The rate of enthalpy required to bring the temperature of incoming wastewater

from 10 oC to 35oC ( 1 H ) is:

)25)(184.4)(000,1)(500,1()(1

T C m H w pw = 1.569×108 kJ/day.

where wm

is the mass flow rate of feed waste material, kg/day; w pC )( is the

specific heat capacity of the waste material at constant pressure, kJ/kg-oC (4.184

enviornmental engineering

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