ge text solutions manual

Solutions Manualfor

Green Engineering:Environmentally Conscious Design of Chemical Processes

David R. ShonnardMichigan Technological University

David T. AllenUniversity of Texas at Austin

Other AuthorsFred Arnold

Scott ProtheroU.S. Environmental Protection Agency

Kirsten RosselotProcess Profiles

Prentice Hall, 2001

Contents

I. An Introduction to Environmental Issues, page 3

II. Risk Concepts, page 6

III. Environmental Law & Regulations: from End-of-Pipe to Pollution, page 10

IV. Roles & Responsibilities of Chemical Engineers, page 13

V. Evaluating Environmental Fate: Approaches Based on Chemical Structure, page 15

VI. Evaluating Exposures, page 27

VII. Green Chemistry, page 36

VIII. Evaluating Environmental Performance During Process Synthesis, page 41

IX. Unit Operations & Pollution Prevention, page 46

X. Flowsheet Analysis for Pollution Prevention, page 54

XI. Evaluating the Environmental Performance of a Flowsheet, page 69

XII. Environmental Cost Accounting, page 75

XIII. Life-Cycle Concepts, Product Stewardship & Green Engineering, page 78

XIV. Industrial Ecology, page 83

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Chapter 1. An Introduction to Environmental Issues

1. Electric Vehicles: Effects on Industrial Production of FuelsThere are two main points to be addressed by this question of electric vehicles versus conventional gasoline-powered vehicles: 1) what are the changes likely to occur in industrial fuels production, and 2) what are the likely changes in environmental impact as a result of this change due to combustion of these fuels. To address the first question, information is needed on the average mix of energy sources in the United States for electricity generation. According to the Department of Energy (DOE) in a report “GREET 1.5 – Transportation Fuel-Cycle Model” (http://www.transportation.anl.gov/ttrdc/greet/), the average mix is 53.8% coal, 1.0% oil, 14.9% natural gas, 18% nuclear, and 12.3% others (hydroelectric, wind, solar, etc.). Thus, if electric vehicles replace conventional gasoline-powered vehicles for personal transportation, fuels production and import would switch from petroleum and petroleum products to more coal, natural gas, nuclear, and other. There would be more mining activities for the extraction of coal and uranium and less reliance on foreign oil. The second question, regarding the environmental impacts of the combustion processes to supply the electricity, is more complicated. A study using the GREET model indicate that on a per mile traveled basis comparing electric vehicles compared to conventional gasoline-powered vehicles, CO2 emissions would decrease by about 25%, volatile organic compounds (VOCs) and CO decrease by about 80%, NOx would increase by about 60%, and SO2 would increase by about 240%.

2. Global Energy Balance: No Atmosphere

a) Energy Balance: “Rate of Solar Energy Absorbed” = “Rate of Infrared Energy Emitted”

(1-A) S RE2 = E 4 RE

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b) Global Average Surface Temperature:

Compared to 280 K, the actual average surface temperature. The calculated value is low because of the greenhouse effect was omitted.

3. Global Energy Balance: with a Greenhouse Gas AtmosphereAt the earth’s surface, a radiation balance requires that(irradiance in = irradiance out)

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0.9E + y = x

while for the atmosphere layer, the radiation balance is

E + x = 0.9E + 2y + .2x

3. Solving these equations simultaneously for y and x results in

4. Surface temperature:

Atmosphere temperature:

5. In order for the global average surface temperature of the earth to rise by 1 ˚C above the value calculated in part b (285.52K), the infrared absorbtivity would need to increase to 0.8166 from 0.80.

4. Global Carbon Dioxide Mass Balancea) A mass balance for CO2 at the earth's surface is

Accumulation of CO2 in atmosphere = rate of CO2 release from surface - rate of CO2 removal by surface

Accumulation of CO2 in atmosphere (metric tons C/yr) = (60+6+1.6) - (60+2+.5+1.8) = 3.3 metric tons C/yr

=

6. +10% change in emissions from fuel combustion is +.6% for a total of 6.6%

Accumulation of CO2 in atmosphere (metric tons C/yr) = (60+6.6+1.6) - (60+2+.5+1.8)

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= 3.9 metric tons C/yr

=

+1% change in emissions from release by microorganisms is +.6% for a total of 60.6%

Accumulation of CO2 in atmosphere (metric tons C/yr) = (60.6+6+1.6) - (60+2+.5+1.8) = 3.9 metric tons C/yr

=

7. Rate of change for CO2 concentration in atmosphere (ppm/yr)

Change in number of moles CO2 from part a =

Change in mole fraction (ppm) of CO2 =

This rate of change compares well with the observed rate of change of 0.5%/yr, which at the current concentration of CO2 is (.005)(360 ppm) = 1.80 ppm/yr.

8. The rate of CO2 accumulation would decrease if the processes of CO2 fertilization and forest growth were enhanced by a future global temperature rise. On the other hand, the rate of CO2 accumulation would increase if the processes of CO2 release were accelerated, for example, by microbial metabolism in soil.

5. Ozone Depletion Potential of Substitute RefrigerantsA discussed in chapter 1, bromine is a much more potent ozone depletion substance compared to chlorine on a per-atom basis. Fluorine is thought to have no adverse ozone depletion effects. Therefore, substituting fluorine for chlorine on alternative refrigerants would help solve the ozone depletion problem. On the other hand, fluorine on hydrocarbon refrigerants is a potent greenhouse gas and fluorinated compounds have been found to accumulate in the body fat of animals as far away as the arctic, well away from any known sources of these compounds. Apparently the persistence of these fluorinated compounds combined with atmospheric transport over long distances creates exposure to these remote creatures.

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Chapter 2. Risk Concepts

1. Risk of Death in Automobile AccidentsWe can express the risk of death in an automobile accident in the context of chemical risk as presented in Chapter 2.

Risk of Death by Car Accident = (Exposure) (Hazard)

In this equation, risk is the number of excess deaths by car accidents, exposure is the number of people involved over a specified period of time, and hazard is the number of deaths per unit of population per unit of time. This is very similar formulation as for carcinogenic risk due to exposure to harmful chemicals. Restating this risk equation with the data in the problem statement gives,

Solving for Hazard we calculate,

The number of deaths expected in Minneapolis-St. Paul over a three-day weekend is

2. Fault Tree AnalysisThe fault tree for the collection sump is

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SumpOverflows

No Warn-ing Alarm

Valve failsto OpenAnd

Sensor #1Fails

AlarmFails

Sensor #2Fails

SolenoidFailsOr Or

3. Toxicity Testing: Establishing a Reference Dose for Human HealthIf an animal study yields a no observable adverse effects level (NOAEL), the formula for determining a human health acceptable dose is

where FA is a safety factor of 10 for using animal studies and FH is another safety factor of 10 to account for variability in toxic response in a human population. If the animal studies were based on subchronic exposure, another safety factor of 10 would be used in the denominator to extrapolate the results to chronic exposure in humans.

4. Reference Dose CalculationFor tris(2-chloroethyl) phosphate, NOAEL = 22 mg/kg-day and LOAEL = 44 mg/kg-day for increased weights of the liver and kidneys in rats.

Using the NOAEL:

Using the LOAEL:

The lesser of the two values would be chosen for the RfD.

5. Choice of a Safe Solvent for a Photoresist

a) Rank solvents based on toxicity from higher to lower hazard

PEL (ppm)diethylemine 25monomethyl ether 25furfuryl alcohol 50ethylene glycol 50n-butyl acetate 150

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methyl ethyl ketone 200ethyl acetate 400

b) Rank from higher to lower exposure potential using vapor pressure

Vapor Press. (kPa @ 25˚C)diethylemine 30.1ethyl acetate 12.6methyl ethyl ketone 12.1monomethyl ether 1.3n-butyl acetate 1.3furfuryl alcohol 0.1ethylene glycol 0.008

c) Considering hazard and exposure, rank from higher to lower riskIf we take the inverse of the PEL as a measure of potential hazard and multiply by the vapor pressure (a measure of exposure), we could estimate the overall risk potential by inhalation.

1/PEL • Vapor Press.diethylemine 1/25 • 30.1 = 1.20methyl ethyl ketone 1/200 • 12.1 = 0.06monomethyl ether 1/25 • 1.3 = 0.05ethyl acetate 1/400 • 12.6 = 0.03n-butyl acetate 1/150 • 1.3 = 0.009furfuryl alcohol 1/50 • 0.1 = 0.002ethylene glycol 1/50 • 0.008 = 0.00016

Ethylene glycol is the solvent with the lowest risk.

d) To lower the risk even more, process modifications could be placed on the integrated circuits fabrication process to reduce the amount of vapor generated and to collect and recover any vapors prior to exposure to workers.

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6. Carcinogenic Risk Assessment Near a Refinerya) Inhalation dos, IINH:

b) Inhalation Carcinogenic Risk

c) Risk of cancer due to inhalation is greater than the recommended range of < 10-4 to 10-6.

d) This calculation might actually over predict the risk because RR and ABS were assumed to be 1 when actually they would be less than one but greater than zero. Also, we used the outside concentration of benzene in the air, not the inside (the home) concentration which will be lower. In addition, we assumed a 70 year exposure duration (ED) when often people move and actually spend much less time in any given home.

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Chapter 3. Environmental Law and Regulations: From End-of-Pipe to Pollution Prevention

1. Provide definitions of the following termsa) pollution prevention: Any act of source reduction, in-process recycle, on-site recycle, and off-site recycle that reduces the amounts of releases and the hazardous characteristics of those releases which ultimately reach the environment. b) source reduction: Any modification of a manufacturing process or of production procedures which reduces the amount of components entering a waste stream or the hazardous characteristics of those components entering waste streams prior to recycle, treatment, or disposal.c) in-process versus on-site versus off-site recycle: In-process recycle is the recovery and return of components that would otherwise become waste to the process unit where these components were generated, usually immediately after they are generated. Examples would be unconverted reactants leaving a reactor that are separated and returned to the reactor inlet. On-site recycle is the recovery of valuable stream components using process units within the same facility where those components were generated. Off-site recycle is the recovery of valuable components at a remote location from waste streams generated at a facility and the return of the valuable components to the facility. d) waste treatment: Any process that renders a waste stream less hazardous prior to disposal or direct release through physical, biological, or chemical means. Examples are primary, secondary, and tertiary wastewater treatment, adsorption of volatile organic compounds from air, and landtreatment of petroleum hydrocarbon sludges from tank bottoms. e) disposal: Long-term isolation of raw or treated waste components in a secured landfill. Examples include landfills for domestic and industrial hazardous and non-hazardous waste. f) direct release: The direct release of components from processes to the air, land, or water. An example of this includes the release of volatile organic compounds from fugitive emission sources in chemical or petroleum refinery processes (from valves, fittings, pumps, flanges, connectors, etc.).

2. Solvent Recovery and Recycle for Automobile Assembly Paint OperationsThis recycle operation is an example of off-site recycle. The location of the solvent recovery facility is remote from the locations of automobile manufacture. The feed for the solvent recovery facility is the waste streams from these cleaning processes in automobile assembly painting. This activity would be considered pollution prevention using the expanded definition in this text but would not be considered pollution prevention by the federal definition, which only includes source reduction and in-process recycle.

3. Capital and Operating Costs for Federal Regulations

a) The Toxic Substances Control Act (TSCA) of 1976Chemical manufacturers must submit information on existing chemicals. This information includes chemical identity, name and molecular structure, categories of use, amounts manufactured or processed,

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byproducts from manufacture, processing, use, or disposal, environmental/health effects of chemical and byproducts, and exposure information. Companies must also keep records of any incidents involving the chemical that resulted in adverse health effects or environmental damage. Chemical manufacturers, importers, and processors are required to notify EPA within 90 days of introducing a new chemical into commerce by submitting a Premanufacturing Notice (PMN).

b) The Federal Insecticide, Fungicide, and Rodenticide Act (FIFRA) of 1972The decision by EPA to register a pesticide is based on the data submitted by the pesticide manufacturer in the registration application. The data in the registration application is difficult and expensive to develop. A registration is valid for five years, upon which time it automatically expires unless a re-registration petition is received. FIFRA requires older pesticides that were never subject to the current registration requirements to be registered if their use is to continue.

c) The Occupational Safety and Health Act (OSH Act) of 1970The OSH Act’s Hazard Communication Standard requires that several standards be met by manufacturers or importers of chemicals and also for the subsequent users of them. These requirements include the development of hazard assessment data, the labeling of chemical substances, and the informing and training of employees in the safe use of chemicals. This information must be assembled in a material safety data sheet (MSDS) in accordance with OSH Act standards and accompany any sale or transfer of the chemical. Employers must also develop a written hazards communication plan which outlines the implementation plan for informing and training employees on the safe handling of chemicals in the workplace. Employers must keep records of all steps taken to comply with OSH Act requirements and employers must keep records of all work-related injuries and deaths and report them periodically to OSHA.

d) Clean Air Act (CAA) of 1970Prior to the completion of new facilities or additions to existing facilities, a permit must be obtained from the state air quality authority. Manufacturers must comply with the state-mandated source-specific emission limits on mobile and stationary sources at a sufficient level to assure compliance with federal quality standards. Meeting these requirements is achieved by installing pollution control equipment on waste streams leaving the process or by modifying the process to eliminate the waste streams.

e) The Clean Water Act (CWA) of 1972The National Pollutant Discharge Elimination System (NPDES) permit program requires any point source of pollution to obtain a permit. NPDES permits contain effluent limits, either requiring the installation of specific pollutant treatment technologies or adherence to specified numerical discharge limits. NPDES permit holders must monitor discharges, collect data, and keep records of the pollutant levels of their effluents. These records must be submitted to the agency that granted the NPDES permit in order to assure that the point source is not exceeding the effluent discharge limits. A permit must be obtained from the United States Army Corp of Engineers before any discharge of dredge or fill materials occurs into navigable waterways, including wetlands.

f) Resource Conservation and Recovery Act (RCRA) of 1976A generator of hazardous waste must obtain an EPA identification number within 90 days of the initial generation of the waste. RCRA requires generators to properly package hazardous waste for shipment off-site and to use approved labeling and shipping containers. Generators must maintain records of the

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quantity of hazardous waste generated, where the waste was sent to for treatment, storage, or disposal, and file this data in biennial reports to the EPA. Generators must prepare a Uniform Hazardous Waste Manifest, which is a shipping document that must accompany the waste at all times. A copy of the manifest will be sent back to the generator by the treatment facility to assure that the waste reached its proper destination.

g) The Comprehensive Environmental Response, Compensation, and Liability Act (CERCLA) of 1980After a site is listed in the National Priority List, EPA identifies potentially responsible parties (PRPs) and notifies them of their potential CERCLA liability. If a clean up is conducted by the EPA, the PRPs are responsible for paying their share of the clean up costs. If the clean up has not begun, PRPs can be ordered to complete the clean up of the site.

h) The Emergency Planning and Community Right to Know Act (EPCRAEPCRA requires facilities with more than 10 employees who either use more than 10,000 pounds or manufacture or process more than 25,000 pounds of one of the listed chemicals or categories of chemicals to report annually to EPA. The report must contain data on the maximum amount of the toxic substance on-site in the previous year, the treatment and disposal methods used, and the amounts released to the environment or transferred off-site for treatment and/or disposal.

i) Pollution Prevention Act of 1990The only mandatory provisions of the PPA requires owners and operators of facilities that are required to file a form R under the SARA Title III (the TRI) to report to the EPA information regarding the source reduction and recycling efforts that the facility has undertaken during the previous year.

4. Source Reduction Categoriesa) Specialty Aromatic Compound (SAC): On-line instrumentation for pH control on the raw materials process is an example of equipment or technology modification. Lowering reactor temperature in the SAC process is an example of process or procedure modifications. b) Specialty Alcohol (SA) process: Improved reaction step that reduced impurity levels and eliminated the wash step waste is an example of process or procedure modifications.

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Chapter 4. Roles and Responsibilities of Chemical Engineers

1. Compare Inherently Safer Design with Pollution Prevention

Inherently Safer Design Concepts Pollution Prevention ConceptsIntent: reduce catastrophic releases of hazardous chemicals and the hazards of those releases by modifying the process to eliminate the source of those hazards within the process.

Intent: reduce chronic releases of hazardous chemicals and the hazards of those releases by modifying the process to make them more efficient at using energy and mass.

Minimize: reduce the inventory of hazardous chemicals within the process.

Source Reduction: reduce the generation and release of hazardous wastes from a process.

Substitute: replace hazardous materials with safer chemicals to reduce impacts of catastrophic releases.

Material Substitution: replace hazardous materials or materials that use excessive energy with more benign materials.

Moderate: Use less hazardous conditions to lessen the impacts of any catastrophic releases.

Process/Procedure Modifications: change process conditions to reduce waste generation and energy consumption.

Simplify: design facilities to be less complex and less error prone, and forgiving of any errors that are made.

Process Modification: change the process control strategy to more precisely achieve the desired operating conditions.

2. Effects of Chemical Properties

Safety Concerns: Chemical properties that would indicate the potential for catastrophic release would be boiling point, reactivity with air or water, vapor pressure, corrosivity to process vessels, endothermic or exothermic heat of reaction, ..

Pollution Prevention Concerns: Chemical properties that would reduce the potential for chronic release from process operations include boiling point, vapor pressure, solubility in water, emulsion forming with water, Henry's constant, compatibility with process materials such as gaskets and valve stems, corrosivity to process vessels, …

3. New "Natural" Product from Botanicals: Ethanol vs. n-Hexane

Physical/Toxicological Properties (reference needed)Properties Ethanol n-HexaneToxicity (LD50, oral rat, mg/kg) 7060 >5000Toxicity (OSHA PEL, ppm) 1000 50

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Carcinogenic no noBoiling Point (˚C @ 1 atm) 78.3 68.7Melting Point (˚C) -114 -95Vapor Pressure (mm Hg @ 20˚C) 44.6 129Vapor Density (air = 1.0) 1.59 3Liquid Density (specific gravity) 0.789 0.659Solubility in Water Infinite NegligibleFlash Point (˚C) 8.9 -23Lower Explosive Limit (%) 3.3 1.2Upper Explosive Limit (%) 24.5 7.7Reactivity Stable Stable

Federal Environmental Permits/Reporting (reference needed)Environmental Regulation Ethanol n-HexaneTSCA PMN application yes yesOSH Requirements yes yesFIFRA no noCAA permit yes yesCWA permit yes yesRCRA Reporting no noEPCRA Reporting no yesCERCLA Reporting no yesPPA Reporting no yes

a) If the decision is based on a comparison between the solvents, then ethanol would be chosen for the process. Ethanol has a lower toxicity judging from the MSDS information and has fewer reporting requirements compared to n-hexane. Of course, more information would be needed to make a reasonable decision. We must know what effects the solvent choice will have on operating costs, on energy consumption, and quality of product. In addition, we must know what effects of solvent choice will have on overall environmental impact of the process, including energy-related emissions and impacts, which can be significant relative to the impacts of the process materials. Methods for evaluating environmental impacts during process design are the topics of discussion for Part II of the book.

b) A Chemical Engineer’s responsibilities are to him or her self, to the employer, the community, and the customer. Each of these responsibilities must be incorporated into any decision regarding product and process design. A potential plant closure resulting in loss of jobs should not influence the decision as to whether a product will be produced or how it will be produced. The primary decision variables are whether the process and product are profitable, which means that society values it, and whether the process and product causes unacceptable damage to the environment and human health.

c) Same as b).

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Chapter 5. Evaluating Environmental Fate: Approaches Based on Chemical Structure

1. Estimate properties for nitrobenzene.Nitrobenzene has the molecular formula

Boiling Pt., from eqn. 5-1, Tb (K) = 198.2 + ni gi

group gi contributionNO2 113.991 aaC 30.76 (a substituted carbon bound to 2 aromatic carbons)5 aaCH 5(28.53)

Tb(K) = 198.2 + 113.99 + 30.76 + 5(28.53) = 485.60 K estimated= 484.12 K actual

Tb (corrected) = Tb – 94.84 + 0.5577Tb – 0.0007705(Tb)2 [Tb 700K] (Eqn. 5-2)= 485.6 – 94.84 + 0.5577(485.6) – 0.0007705(485.6)2 = 479.9 K

Tb (corrected) is in error by 0.87% compared to the actual value.

Vapor pressure (Pvp): ln Pvp = [A(Tb – C) 2 ] * [1/(Tb – C) - 1/(T – C)] (Eqn. 5-7) [0.97 R Tb ]

A = KF (8.75 + R ln Tb ) = 1.05(8.75 + 1.987 ln(485.6)) = 22.09C = -18 + 0.19 Tb = -18 + 0.19(485.6) = 74.26

ln Pvp = [22.09(485.6 – 74.26) 2 ] * [1/(485.6 – 74.26) - 1/(298 – 74.26)] = -8.139 [0.97 (1.987) (485.6) ]Pvp = exp(-8.13) = 2.92x10-4 atm = 0.22 mm Hg.

Henry’s Law constant (H) -log H = log (air-water partition coefficient) = ni hi + nj cj (H unitless)1 Caromatic-NO2 ni hi = 2.4966 Caromatic - Caromatic ni hi = 6(0.2638)5 Caromatic-H ni hi = 5(-0.1543)No corrections factors-log H = 2.2496 + 6(0. 2638) + 5(-0.1543) = 3.06H = 10-3.06 = 8.69x10-4 unitless = 2.13x10-4 atm-m3/moleH (experimental) = 2.40x10-4 atm-m3/mole

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NO2

Octanol-water partition coefficient (Kow):log Kow = 0.229 + ni fi + nj cj

6 Caromatic ni hi = 6(0.2940)-NO2 (aromatic attach.) ni hi = -0.1823No corrections factorslog Kow = 0.229 + 6(0.2940) + -0.1823 = 1.81Kow = 101.81 = 64.67log Kow (experimental) = 1.85

Water solubility (S): (S in mole/L)log S = 0.796 – 0.854 log Kow - 0.00728(MW) + hj (used when melting pt. not available)Corrections factors hj = -.390 (aromatic nitro group)log S = 0.796 – 0.854(1.85) - 0.00728(123.11) + -.390 = -2.07 S = 10-2.07 =8.51x10-3 moles/L = 1.05 g/L = 1,047.5 mg/L

Soil sorption coefficient (Koc): log Koc = 0.531 + 0.62 + njPj (Koc is ratio of g/g carbon to g/ml solution)1 = (i* j)-0.5 , the first order molecular connectivity indexi ,the connectedness of carbon or other heteroatom i)Bond connectedness (i* j)(N-O) (N=O) (N-C) 2(C-C) 4(C-C)(3,1) (3,1) (3,3) 2(3,2) 4(2,2)1 = (i* j)-0.5 1 = = 4.305

log Koc = 0.53(4.305) + 0.62 - 0.632 = 2.27Koc = 102.27 = 186

Atmospheric half life:The only significant reaction is addition to aromatic ring = 0.2437x10-12 cm3/molecule-sec. t1/2½ = ln (2) / (k [OH.]) = 0.693/((0.2437x10-12 cm3/molecule-sec)(1.5x106 molecules/ cm3))t1/2½ = 1.896x106 sec = 22 days.

Biodegradability:I = 3.199 + a1f1 + a2f2 +…… + anfn + amMWUnsubstituted phenyl group a = 0.022Aromatic (NO2) a = -.134 (used aromatic amine as substitute)Molecular weight a = -0.00221I = 3.199 + 0.022 - .134 + 123.11(-0.00221) = 2.81 (weeks to a month)

2. Estimate properties for 2-ChloroanilineFrom the EPIWIN software, enter the CAS Number of 95-51-2. The following summary is obtained with the software output listed directly below.

16

NO O

1 13

3

2

2

2

2

2

Boiling point (Tb) 216.05 ˚CVapor pressure (Pvp) at 300 K 0.16 mm Hg @ 25˚CHenry’s Law constant (H) 1.41x10-6 atm-m3/mole using bond methodOctanol-water partition coefficient (Kow) Log Kow = 1.72Bioconcentration Factor (BCF) Log BCF = 0.763 using BCFwinWater solubility (S) 2241 mg/L - no melting pt. formula used.Soil sorption coefficient (Koc) Log Koc = 1.869Atmospheric half life 0.340 Days (12-hr day; 1.5E6 OH/cm3)Biodegradability 2.5757 (weeks-months)

SMILES : Nc(c(ccc1)CL)c1CHEM : Benzenamine, 2-chloro-CAS NUM: 000095-51-2MOL FOR: C6 H6 CL1 N1 MOL WT : 127.57------------------------------ EPI SUMMARY (v2.40) -------------------------- Physical Property Inputs: Water Solubility (mg/L): ------ Vapor Pressure (mm Hg) : ------ Henry LC (atm-m3/mole) : ------ Log Kow (octanol-water): ------ Boiling Point (deg C) : ------ Melting Point (deg C) : ------

Log Octanol-Water Partition Coef (SRC): Log Kow (KOWWIN v1.60 estimate) = 1.72 Log Kow (Exper. database match) = 1.90

Boiling Pt, Melting Pt, Vapor Pressure Estimations (MPBPWIN v1.28): Boiling Pt (deg C): 216.05 (Adapted Stein & Brown method) Melting Pt (deg C): 24.41 (Mean or Weighted MP) VP(mm Hg,25 deg C): 0.16 (Mean VP of Antoine & Grain methods)

Water Solubility Estimate from Log Kow (WSKOW v1.30): Water Solubility at 25 deg C (mg/L): 2241 log Kow used: 1.90 (expkow database) no-melting pt equation used

ECOSAR Class Program (ECOSAR v0.99b): Class(es) found: Aromatic Amines

Henrys Law Constant (25 deg C) [HENRYWIN v3.00]: Bond Method : 1.41E-006 atm-m3/mole Group Method: 1.86E-006 atm-m3/mole

Probability of Rapid Biodegradation (BIOWIN v3.63): Linear Model : 0.2706 Non-Linear Model : 0.0615 Expert Survey Biodegradation Results: Ultimate Survey Model: 2.5757 (weeks-months)

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Primary Survey Model : 3.3900 (days-weeks )

Atmospheric Oxidation (25 deg C) [AopWin v1.87]: Hydroxyl Radicals Reaction: OVERALL OH Rate Constant = 31.4297 E-12 cm3/molecule-sec Half-Life = 0.340 Days (12-hr day; 1.5E6 OH/cm3) Half-Life = 4.084 Hrs Ozone Reaction: No Ozone Reaction Estimation

Soil Adsorption Coefficient (PCKOCWIN v1.63): Koc : 74.04 Log Koc: 1.869

Aqueous Base/Acid-Catalyzed Hydrolysis (25 deg C) [HYDROWIN v1.64]: Rate constants can NOT be estimated for this structure!

BCF Estimate from Log Kow (BCFWIN v2.10): Log BCF = 0.763 (BCF = 5.794) log Kow used: 1.90 (expkow database)

Volatilization from Water: Henry LC: 1.86E-006 atm-m3/mole (estimated by Group SAR Method) Half-Life from Model River: 536.6 hours (22.36 days) Half-Life from Model Lake : 3986 hours (166.1 days)

Removal In Wastewater Treatment: Total removal: 2.27 percent Total biodegradation: 0.09 percent Total sludge adsorption: 2.07 percent Total to Air: 0.11 percent

------------------------------------------------------------------------------

3. Estimate properties. EPIwin results are shown below.

Property Ethanol 1-propanol 1-hexanol n-propane n-hexane

Boiling point (Tb)

(˚C)

65.11 89.96 159.09 -7.76 71.53

Vapor pressure (Pvp)

(mm Hg @ 25˚C)

116 33.6 0.809 2.29x103 135

Henry’s Law constant 5.67x10-5 7.52x10-6 1.76x10-5 7.3x10-1 1.71

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(H)

(atm-m3/mole)

Octanol-water partition

coefficient, Log (Kow)

-0.14 0.35 1.82 1.81 3.29

Water solubility (S)

(mg/L)

7.92x105 2.72x105 6,885 369 17.24

Soil sorption

coefficient Log (Koc)

0.00 0.122 0.920 1.367 2.17

Atmospheric half life

(days)

2.99 1.95 1.10 8.43 1.96

Biodegradability

(Primary Survey

Model)

3.91

(days)

3.89

(days)

4.10

(days)

3.78

(days)

3.99

(days)

For each of the properties, comment on whether molecular weight or the presence of a hydrogen bonding group has a more pronounced effect on chemical properties.

Boiling point (Tb) hydrogen bondingVapor pressure (Pvp) hydrogen bondingHenry’s Law constant hydrogen bondingOctanol-water partition coefficient bothWater solubility (S) hydrogen bondingSoil sorption coefficient hydrogen bondingAtmospheric half life molecular weightBiodegradability neither

4. Benzene discharge to a river. Total discharge rate to POTW = 2,000 kg benzene / dayRemoval efficiency in POTW = 85%Net benzene discharge to river = (1-.85) 2,000 kg benzene/day = 300 kg benzene/day

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a) Estimate fraction of benzene in water, sediment, and biota at discharge point.From EPIwin, CAS no. 71-43-2Log Koc = 2.219, Koc = 165.5 L water/kg or kg water/kg sediment carbonLog BCF = 0.940, BCF = 8.712 L water/kg or kg water/kg biota

Based on the river flow rate the total flow rates of water, sediment and biota are:Water: (1,250 x106 liter/day * 1 kg/liter) = 1,250 x106 kg/daySediment: 1,250 x106 kg/day * 15 kg sediment carbon/ 106 kg water = 18,750 kg sediment C/dayBiota: 1,250 x106 kg/day * 0.1 kg biota/ 0.1 x106 kg water = 1,250 kg biota/day

Mass balance: “rate of benzene discharge to river” = “rate of benzene flow downriver in water, sediment, and biota”.

300 benzene kg/day = 1,250x106 kg water/day (Cwater) + 18,750 kg sediment carbon/day (165.5 kg water/kg sediment carbon) Cwater + 1,250 kg biota/day (8.712 kg water/kg biota) Cwater

Cwater = 2.39x10-7 kg benzene/kg water = .239 ppm benzene in water. Fraction benzene in water = 1,250x106 kg water/day (Cwater) / 300 kg benzene/day = 99.75%Fraction benzene in sediment = 18,750 kg sediment carbon/day (165.5 kg water/kg sediment carbon) Cwater /300 = .247%Fraction benzene in biota = 1,250 kg biota/day (8.712 kg water/kg biota) Cwater / 300 = .0009%

b) Is volatilization of benzene from the river important?Assume air above river to 10 m height is in equilibrium with benzene in water at discharge point concentration for the full 100 km length of the river. Use Henry’s constant to estimate the air concentrationFrom EPIwin, H = 5.39x10-3 atm-m3/moleAir Concentration = H Cwater when expressed in units of moles benzene / m3

= (5.39x10-3 atm-m3/mole)(2.39x10-7 kg benzene/kg water)(1 mole/.07811 kg benzene)(1000 kg water / m3) = 1.65x10-5 atm

= 1.65x10-5 atm (1 mole benzene/mole air•atm)(.07811 kg benzene/mole bz) (1 mole air/.0224 m3 air) = 5.75x10-5 kg benzene / m3 air

Total Benzene in air = (5.75x10-5 kg /m3 air)(10 m)(30 m)(43.2 km)(1000 m/km) = 745 kg benzNote – 43.2 km is the distance that the river flows in 1 day. Because this estimate is much greater than the input of benzene into the river, volatilization is important.

c) Estimate the fraction that biodegrades prior to reaching the water inlet 100 km downriver. Time to reach 100 km downriver is (100 km)/(43.2 km/day) = 2.3 daysFrom EPIwin, biodegradation takes weeks – months, so fraction biodegraded is insignificant.

d) Estimate the potential toxicity of the releases to aquatic lifeFrom EPIwin, Log Kow = 1.99, MW = 78.11A correlation for toxicity to guppies for chlorobenzenes is given bylog (1/LC50) = 0.871 log Kow - 4.87, where LC50is expressed in units of mol/L. log (1/LC50) = 0.871 (1.99) - 4.87 = -3.14

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1/LC50 = 10-3.14 = 7.30x10-4

LC50 = 1,370 mol/LCwater = 2.39x10-7 kg benzene/kg water from part a) = 2.39x10-4 g benzene/L water = 3.06 mol/L, thus this concentration is not likely to pose a risk to fish since it is much less than the LC50.

5. Estimate the amount of hexachlorobenzene (Hx) that would be ingested if a person were to eat a 0.5 kg fish from the pond.1 kg hexachlorobenzene added to a pondPond water content = 105 m3 or 108 kg waterOrganic carbon sediment content = 10 kg organic carbon sediment / 106 kg water = 10-5 kg/kgFish content = 100 g fish / 100 m3 or 0.1 kg fish / 105 kg waterFrom EPIwin, with CAS no. of 118-74-1Koc = 3,388 L water/kg org. CBCF = 5,152 L water / kg fishMass balance: 1 kg Hx = Mw + Ms + Mf where Mw is the mass in water = (108 kg water)Cwater, Ms is the mass in sediment carbon = (10-5 kg org. C/kg water)( 3,388 kg water/kg org. C) Cwater,Mf is the mass in fish = (0.1 kg fish/105 kg water)( 5,152 kg water/kg fish) Cwater. Solving for Cwater = 1x10-8 kg Hx/kg water. Concentration in fish = BCF x Cwater

= (5,152 kg water / kg fish)( 1x10-8 kg Hx/kg water) = 5.15x10-5 kg Hx/kg fish.

Ingestion dose to humans = (0.5 kg fish consumed)(5.15x10-5 kg Hx/kg fish) = 2.58x10-5 kg Hx

6. The Great Lakes Basin partitioning of tetrachloroethylene (PERC)Total emissions to the basin are to air (1.8x107 pounds per year), land (2.6x106 pounds per year) and water (8.4x102 pounds per year) = 2.06x107 pounds / yr = 9.34x106 kg/yr

Landscape propertiesProperty air water soil sedimentVolume (m3) 7.6*1014 2.3*1013 2.6*1010 4.8*108

Area (m2) 7.6*1011 2.4*1011 5.2*1011 2.4*1011

Organic fraction 0.02 0.04Density (kg/m3) 1.2 1000 1500 1280Residence time (hr) 130 272,000 550 1700

a) Calculate the equilibrium partitioning of tetrachloroethylene in the air, water, soil and sediment of the Great Lakes Basin .EPIwin provides H and Koc, CAS no. 127-18-4H (atm•m3/mole) = 1.65x10-2 Koc (L (or kg) water/kg org. C) = 106.8

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Mass balance: Assume that PERC is only significant in organic carbon in soil and sedimentMPERC = 9.34x106 kg = Mair + Mwater + Msoil + Msediments

Mair (kg) = mass of PERC in air = Vair Cair = Vair Pair (1 mole PERC/mole air•atm)(.16583 kg PERC/mole PERC)(1 mole air/.0224 m3 air)

= Vair H (Cwater)(1 mole PERC/mole air•atm) (.16583 kg PERC/mole PERC) (1 mole air/.0224 m3 air) = (7.6x1014 m3 air)(1.65x10-2 atm•m3 water/mole)(1 mole PERC/mole air•atm) (.16583 kg PERC/mole PERC)(1 mole air/.0224 m3 air) Cwater = 9.38x1013 m3 water Cwater

Mwater (kg) = mass of PERC in water = Vwater Cwater where Cwater has units of kg PERC/m3 waterMsoil (kg) = mass of PERC in soil = Vsoil Densitysoil Csoil where Csoil is kg PERC/kg soil

= Vsoil Densitysoil (Fract. Org. C) soil Koc (10-3 m3 water / kg water) Cwater

= (2.6x1010 m3)(1500 kg soil/ m3)(.02 kg org.C/kg soil)(106.8 kg water/kg org. C) (10-3 m3 water / kg water) Cwater = 8.33x1010 m3 water Cwater

Msediment (kg) = mass of PERC in sediment = Vsed Densitysed Csed where Csed is kg PERC/kg sed = Vsed Densitysed (Fract. Org. C) sed Koc (103 kg water / m3 water) Cwater

= (4.8x108 m3)(1280 kg sed./m3)(.04 kg org.C/kg sed.)(106.8 kg water/kg org. C)(10-3 m3 water / kg water) Cwater = 2.62x109 m3 water Cwater

Cwater = 9.34x106 kg/(9.38x1013 +2.3x1013+8.33x1010 +2.62x109) m3 = 7.99x10-8 kg/m3waterFraction in air = Mair/9.34x106 kg = (9.38x1013 m3 water) Cwater/9.34x106 kg = 0.802Fraction in water = Mwater/9.34x106 kg = Vwater Cwater /9.34x106 kg = 0.197Fraction in Soil = Msoil/9.34x106 kg = 8.33x1010 m3 water Cwater = 7.13x10-4

Fraction in Sediment = Msediment/9.34x106 kg = 2.62x109 m3 water Cwater = 2.24x10-5

b) The system modeled in part a) will never reach steady-state. PERC will continue to accumulate because the model environment is a closed system and there are no reaction or other loss mechanism included.

c) Steady-state concentration in each environmental compartment with reactionFrom EPIwinAtmosphere: half-life (t1/2) is 50 days (or .137 yr) due to hydroxyl radical oxidation. The 1st order reaction rate constant is, kair = ln2/t1/2 = ln2/(.137 yr) ~ 5 yr-1. Biodegradation: half-life () is about 2-4 weeks (0.1 yr) due to biodegradation. kwater = ln2/t1/2 = ln2/(.1 yr) ~ 7 yr-1.Assume that biodegradation only takes place in the water phase of the soil and sediment and in the water compartment and the volume of water in the soil and the sediment is approximately the same as the soil and sediment volumes respectively, in the absence of other information. Mass balance: at steady-state

Accumulation rate = 0 = input rate - (reaction loss rates)0 = 9.34x106 kg - Vair kair Cair - Vwater kbio Cwater - Vsoil kbio Cwater - Vsed kbio Cwater

0 = 9.34x106 kg/yr - 9.38x1013 m3 water kair Cwater - Vwater kbio Cwater - Vsoil kbio Cwater - Vsed kbio Cwater

Cwater = 9.34x106 kg/yr / (9.38x1013 kair + kbio(2.3x1013 +2.6x1010+4.8*108) m3 water) = 1.48x10-8 kg PERC/ m3 water

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Concentrations in each environmental compartmentAir: Cair = H (Cwater)(1 mole PERC/mole air•atm) (.16583 kg PERC/mole PERC)

(1 mole air/.0224 m3 air) = (1.65x10-2 atm•m3 water/mole)( 1.48x10-8 kg PERC/ m3 water) (1 mole PERC/mole air•atm) (.16583 kg PERC/mole PERC)(1 mole air/.0224 m3 air) = 1.81x10-9 kg PERC/ m3 air

Water: Cwater = 1.48x10-8 kg PERC/ m3 waterSoil: Csoil = (Fract. Org. C)soil Koc (10-3 m3 water / kg water) Cwater

= (.02 kg org.C/kg soil)(106.8 kg water/kg org. C)(10-3 m3 water / kg water) Cwater = 3.16x10-11 kg PERC/kg soil

Sediment: Csed = (Fract. Org. C)sed Koc (10-3 m3 water / kg water) Cwater

= (.04 kg org.C/kg soil)(106.8 kg water/kg org. C)(10-3 m3 water / kg water) Cwater

= 6.32x10-11 kg PERC/kg sediment

d) Effects of advection by wind and water flow on concentrationsAdvection is an additional mass loss mechanism from our environmental model region. Advection loss is represented by the term GiCi where G is the volumetric flow rate of air or water (m3/yr), C is the mass concentration of PERC in the air or water, and the subscript i is either water or air. These terms would be included in the reaction loss term in the steady-state mass balance equation above. The value of G for air and water can be obtained from the residence time values in the data table by dividing the compartment volume by the residence time. Thus for air, Gair = 7.6x1014 m3 air / 130 hr•1yr/8760hr = 5.12x1016 m3 air/yr. Similarly for water, Gwater = 2.3x1013 m3 water / 272,000 hr•1yr/8760hr = 7.4x1011 m3 water/yr.

7. Design a solvent molecule that has a vapor pressure greater than 1 mm Hg, a molecular weight between 75 and 150, and will biodegrade in less than one month.

Vapor pressure: depends on boiling point, assume 300K as the chosen temperatureln Pvp (atm) = [A(Tb – C) 2 ] * [1/(Tb – C) - 1/(T – C)]

[0.97 R Tb ]C= -18 + 0.19 Tb

A = KF (8.75 + R ln Tb )

From Figure P5.7-1below, we see that the hydrocarbon must have a Tb less than about 460 K and greater than 300 K, our lower limit constraint.

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Figure P5.7-1

Boiling Point, Tb (K) = 198.2 + ni gi uncorrected value is adequate for designFor aliphatic hydrocarbons, gi for a -CH3 group is 21.98 and for a >CH2 group it is 24.22. Figure P5.7-2 below shows Tb versus number of carbons over the range allowed (75<MW<150). Thus, the aliphatic hydrocarbon must be between 6 and 10 carbons to meet MW and Pvp constraints.

Figure P5.7-2

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Biodegradability must be greater than 3 according to the equationI = 3.199 + a1f1 + a2f2 +…… + anfn + amMW For aliphatic hydrocarbons, the only parameter that is relevant is the molecular wt. parameter, am = -0.00221. Figure P5.7-3 below shows the biodegradability index as a function of carbon number for aliphatic hydrocarbons

Figure P5.7-3

The only solvent candidate to satisfy all of the constraints is the 6-carbon hydrocarbon, n-hexane. Other classes of organic compounds could have been chosen for this design.

8. Design a solvent molecule that has a vapor pressure less than 1 mm Hg (at 300 K), a molecular weight between 75 and 150, and will biodegrade in less than one month.

Rather than taking a systematic approach similar to problem 7, we will instead apply our knowledge of the effects of functional groups on boiling point, vapor pressure, molecular weight, and biodegradability. The boiling point must be greater than 460 in order to meet the vapor pressure constraint, as shown in Figure P5.7-1. One of the functional groups in Table 5.2-2 that has a large effect on boiling point is the hydroxyl group. Adding this functional group to an aliphatic hydrocarbon will dramatically increase boiling point and decrease vapor pressure. This same functional group also enhances biodegradability, as shown by the positive contribution in Table 5.3-4. Adding the hydroxyl functional group achieves both goals simultaneously. Therefore, we estimate the properties for 1,4-butanediol, HO-CH2-CH2-CH2-CH2-OH.

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Using EPIwinCAS No. 110-63-4The molecular weight is 90.12, within the required rangeBoiling Pt. = 195.6 C = 468.75 KVapor Pressure will be less than 1 mm Hg according to Figure P5.7-1Biodegradability Index = 3.32

This choice satisfies all of the constraints on the solvent design.

9. Magnitude of effect on boiling point of group contributions for the following groups: -OH, -CH3, -Cl (aliphatic).-OH (106.27)>-Cl (62.63)>- CH3 (21.98)Reasoning: hydrogen bonding has a larger influence on boiling point than molecular mass.

10. Magnitude of effect on Kow of group contributions for the following groups: -OH, -CH3, -Cl (aliphatic).-CH3 (0.5473) > -Cl (0.3201) > -OH (-1.4068)Reasoning: -CH3 and -Cl groups are hydrophobic and is hydrophylic.

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Chapter 6. Evaluating Exposures

1. Steady Evaporation Rate of Ethylbenzene From a Pool of Liquid into a Room.

where G is the evaporation rate (g/sec-cm2)MW is the molecular weight of the evaporating species (g/mole) = 106 g/mole for ethylbenzeneVP is the vapor pressure of the evaporating contaminant (atm)v is the air velocity parallel to the surface of the evaporating liquid (cm/sec) = 50 cm/sT is the surface temperature of the evaporating liquid (˚K) = 299.8 ˚Kx is the length of the evaporating pool in the direction of airflow (cm) = 113 cm for a 1 m2 circular pool of liquid.P is the ambient pressure (atm) = 1 atm.

Boiling Pt., from eqn. 5-1, Tb (K) = 198.2 + ni gi

group gi contribution1 aaC 30.76 (a substituted carbon bound to 2 aromatic carbons)5 aaCH 5(28.53)-CH3 21.98>CH2 24.22

Tb(K) = 198.2 + 21.98 + 24.22 + 5(28.53) + 30.76 = 417.81 K estimated

Tb (corrected) = Tb – 94.84 + 0.5577Tb – 0.0007705(Tb)2 [Tb 700K] (Eqn. 5-2)= 417.81 – 94.84 + 0.5577(417.81) – 0.0007705(417.81)2 = 421.48 K

Tb (actual) = 409.15 K

Vapor pressure (Pvp): ln Pvp = [A(Tb – C) 2 ] * [1/(Tb – C) - 1/(T – C)] (Eqn. 5-7) [0.97 R Tb ]

A = KF (8.75 + R ln Tb ) = 1.06(8.75 + 1.987 ln(409.15)) = 20.7C = -18 + 0.19 Tb = -18 + 0.19(409.15) = 59.74

ln Pvp = [20.7 (409.15– 59.74) 2 ] * [1/(409.15 – 59.74) - 1/(299.82 – 59.74)] = -4.18 [0.97 (1.987) (409.15)]Pvp = exp(-4.18) = 1.53x10-2 atm

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G = (8.79x10-5)(106).833(1.53x10-2)[(1/106+1/29).25](50).25]/((299.8 K).05(113).5(1) .5) = 5.63x10-6 g/sec/cm2

The total volatilization rate = GxA, where A is (1 m2)(104 cm2/ m2) = 104 cm2.= (5.63x10-6 g/sec/cm2)( 104 cm2) = 0.0563 g/sec

2. Estimate the Steady-State Concentration for the Emission Source in Problem 1.C = Co + G/(k Q), where C is the concentration in the room, k is the mixing factor (.5 for average ventilation and 0.1 for poor ventilation), and Q is the volumetric flow rate (0.5 m/s x 4 m x 2.45 m = 4.9 m3/s).

Average ventilation: C = 0 + (0.0563 g/s)/((0.5)(4.9 m3/s)) = 0.0230 g/ m3.

Poor ventilation:C = 0 + (0.0563 g/s)/((0.1)(4.9 m3/s)) = 0.115 g/ m3.

PEL = 100 ppm Converting to mass concentration:

Both of the calculated concentrations in the room are less than the PEL, so the exposure is safe.

3. Confirm that equation 6.2.3 is the solution to equation 6.2.1 for a transient emission source.Show that

equation 6.2.3

is the solution to the mass balance equation for a room

equation 6.2.1

where C is the concentration of airborne contaminant in the work area (mass/length3)V is the volume of the work area (length3)t is the time during which the contaminant has been emitted,G is the emission rate of the contaminant to the air (mass/time),Q is the ventilation rate in the work area (length3/time)

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k is a mixing factor to account for incomplete mixing in the work area (unitless)Co is the concentration of the airborne contaminant entering the work area (mass/ length3).

Solution: Rearranging equation 6.2.1,

An integrating factor multiplies both sides of the equation above yielding,

Multiplying both sides by dt and integrating results in,

for constant G and Co. The solution to this is,

Both sides of the equation above is divided by the integrating factor, yielding

and evaluated at the initial condition of C = Co when t = 0.

The solution is then equation 6.2.3

4. Derive equation 6.2.7, which describes ventilation of a two-compartment room. Describe how you would extend the analysis to a three compartment room.The conceptual model of the two compartment room is shown in the schematic diagram below in cross section. The concentration in compartment 1 is C1 and in the larger compartment 2 is C2. Mass is input

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into compartment 1 at a rate G (mass/time) and mass exchange occurs between the compartments at a rate of B (length3/time). The ventilation rate of compartment 2 is Q (length3/time).

At steady-state, Compartment 1: 0 = G – B(C1– C2)Compartment 2: 0 = B(C1– C2) – Q C2

0 = G– Q C2 or C2 = G/QNow compartment 1 again: 0 = G – B(C1– C2) = G – B(C1– G/Q)Solving for C1: B C1 = G + BG/Q

C1 = G/B + G/Q = G(1/B + 1/Q) = G(B+Q)/(BQ)

For a three compartment, a similar analysis would occur using the following diagram as a conceptual model.

5. Assume that the room described in problems 1 and 2 is characterized as a cube, 1.5 meter on each side centered over the pool of ethylbenzene, that exchanges air at a rate of 1 m3/s with the rest of the room. Develop expressions for the transient and steady state concentrations of ethylbenzene in the air immediately above the pool, and in the rest of the room.

At steady-state, Compartment 1: 0 = G – B(C1– C2)Compartment 2: 0 = B(C1– C2) – Q C2

0 = G– Q C2 or C2 = G/Q

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G

C1

B Q

C2

Co=0

G

C1

B1

Q

C2

Co=0

C3

B2

Now compartment 1 again: 0 = G – B(C1– C2) = G – B(C1– G/Q)Solving for C1: B C1 = G + BG/Q

C1 = G/B + G/Q = G(1/B + 1/Q) = G(B+Q)/(BQ)

C1 = G(B+Q)/(BQ) = (.0563 g/s)(1 m3/s + 4.9 m3/s)/((1 m3/s)(4.9 m3/s)) = 0.0678 g/m3. C2 = G/Q = (0.0563 g/s)/(4.9 m3/s) = 0.0115 g/m3.

6. Estimate the dermal and inhalation exposures that might be associated with collecting a sample of ethylbenzene from the pool described in problems 1, 2 and 5. Make reasonable assumptions about the time required to collect the sample and inhalation rates. Is dermal or inhalation the dominant exposure route?

Dermal exposure:DA = (S)(Q)(N)(WF)(ABS) Equation 6.2.10where DA is the dermal absorbed dose rate of the chemical (mass/time),S is the surface area of the skin contacted by the chemical (420 cm2) – assume one handQ is the quantity deposited on the skin per event (2.1 mg/cm2/event),N is the number of exposure events per day (1 event/day),WF is the weight fraction of the chemical of concern in the mixture (1.0), andABS is the fraction of the applied dose absorbed during the event (1.0).

DA = (420 cm2)(2.1 mg/cm /event)(1 event)(1.0)(1.0) = 882 mg/day

Inhalation exposure:InhA = (C1)(CR)(CD)InhA is the inhalation absorbed dose rate of the chemical (mg/day)C1 is the concentration of ethylbenzene in the air above the pool (mg/m3)CR is the contact rate (10 m3 breathed/hr)CD is the contact duration (8 hr/day)

InhA = (0.0678 g/m3)( 10 m3 breathed/hr) (8 hr/day) = 5.4 g/day = 5,400 mg/day.

Inhalation is the dominant exposure route.

7. Exposure to Acrolein and Methyl Ethyl Ketone from Leaking EquipmentQ = ventilation rate = 200 m3/hr\

Acrolein: (2-Propenal) C3H4O (MW=56.06)Solution data 4 %(wt) acrolein in waterLeak rate 1 milliliter solution/minGacrolein, Emission rate to air (1 ml solution/min)(1 g soln/ml)(0.04 g acrolein/g soln)(60 min/hr) = 2.4 g acrolein/hr

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Cacrolein = Gacrolein/(kQ) = (2.4 g/hr)/((0.5)(200 m3/hr)) = 0.024 g acrolein/m3

PELacrolein = 0.1 ppmConverting to mass concentration:

Because the predicted concentration of acrolein in air (0.024 g/m3) is greater than the allowable workplace concentration derived from the PEL (2.5x10-4 g/m3), this release poses a danger to people in the workplace.

Methyl Ethyl Ketone (MEK): (2-butanone) C4H8O (MW=72.11)Solution data 5 %(wt) methyl ethyl ketone in waterLeak rate 30 milliliter solution/minGMEK, Emission rate to air (30 ml solution/min)(1 g soln/ml)(0.05 g MEK/g soln)(60 min/hr) = 90.0 g MEK/hrCMEK = GMEK/(kQ) = (90 g/hr)/((0.5)(200 m3/hr)) = 0.90 g MEK/m3

PELacrolein = 200 ppmConverting to mass concentration:

Because the predicted concentration of MEK in air (0.90 g/m3) is greater than the allowable workplace concentration derived from the PEL (0.644 g/m3), this release poses a danger to people in the workplace.

8. Use the Green Chemistry Expert System (www.epa.gov/greenchemistry/gces.htm) to identify structural modifications that can be made to nitriles to reduce their toxicity.The following comments were extracted from the Green Chemistry Expert System.

Nitriles are organic substances that contain the cyano (CN) group. Certain nitriles are quite potent in causing acute lethality in humans and animals. As a class nitriles vary broadly in their ability to cause acute lethality, and subtle differences in structure can dramatically affect toxic potency. It has been observed that exposure of humans and experimental animals to the more acutely toxic nitriles results in signs and symptoms similar to that of cyanide poisoning, implicating free cyanide as the cause of lethality. Cyanide release from nitriles results from their enzymatic metabolism in the liver, following exposure and subsequent absorption. In the liver, the metabolism of nitriles involves hydroxylation of one of the aliphatic carbon atoms by cytochrome P450 enzymes. Cytochrome P450-mediated hydroxylation of the carbon atom alpha (i.e., adjacent) to the cyano group yields a cyanohydrin intermediate, which rapidly decomposes to liberate cyanide and the corresponding carbonyl compound. For example acetonitrile is not very acutely lethal (mouse oral LD50 = 6.55 mmol/kg) whereas its homolog propionitrile is ten-fold greater in toxicity (LD50 = 0.65 mmol/kg). Large differences in toxicity are known for other nitriles.

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1. When designing a nitrile, add structural features that will prevent or minimize cytochrome P450-mediated hydroxylation at the alpha-carbon. One strategy is to place a hydroxyl group on a carbon remote from the alpha carbon. This will divert Cytochrome P450-mediated hydroxylation away from the alpha carbon, thus reducing cyanide release. 2. Unless absolutely necessary for intended use, do not add hydroxy or amino groups (substituted or unsubstituted) to the alpha-carbon, because these nitriles will be highly acutely toxic.3. When considering structural modifications that are intended to reduce acute toxicity (lethality), be careful not to choose and incorporate structural modifications that will reduce acute toxicity, but will bestow osteolathyrism.

9. In designing chemicals that will minimize human uptake, you may wish to consider properties such as volatility, octanol-water partition coefficient and water solubility. For high, medium and low values of each of these parameters, characterize whether exposure due to inhalation, ingestion, and dermal contact are likely to be important.

Water SolubilityExposure Route High Water Solubility Moderate Water

SolubilityLow Water Solubility

Inhalation Potentially high uptake

Lower uptake rate Lower uptake rate

Ingestion Lower uptake due to poor mass transfer within g.i. tract

Potentially high uptake

Low uptake due to poor mass transfer within g.i. tract

Dermal Contact Lower uptake due to poor mass transfer through the skin


Low uptake due to poor mass transfer through the skin

VolatilityExposure Route High Volatility Moderate Volatility Low VolatilityInhalation Potentially high

uptakePotentially lower uptake rate

Low uptake rate

Ingestion Low uptake rate Low uptake rate Potentially high uptake rate if moderately water soluble

Dermal Contact Potentially low uptake Potentially low uptake Potentially low uptake

Octanol-Water Partition CoefficientExposure Route High Kow Moderate Kow Low KowInhalation Potentially low uptake Potentially low uptake

ratePotentially high uptake rate

Ingestion Lower uptake due to poor mass transfer


Low uptake due to poor mass transfer

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within g.i. tract within g.i. tractDermal Contact Lower uptake due to

poor mass transfer through the skin


Low uptake due to poor mass transfer through the skin

10. Exposure to Hazardous Chemicals from a Primer Production Facilitya) Sources of Occupational ExposureReactor 1: Inhalation exposure during cleanout

Dermal exposure during sampling and cleanoutFiltration: Dermal exposure during disposal of spend filtersReactor 2: Inhalation exposure during cleanout

Dermal exposure during sampling and cleanoutDistillation: Dermal exposure during samplingMixer: Inhalation exposure during cleanout

Dermal exposure during sampling and cleanout

b) Sources of Releases to the EnvironmentReactor 1: Releases to air, water, and/or land from cleanout

Releases to water or land during samplingFiltration: Releases to land during spent filter disposalReactor 2: Releases to air, water, and/or land from cleanout

Releases to water or land during samplingDistillation: Releases to air due to excess solvent treatment or disposal

Releases to water or land during samplingMixer: Releases to air, water, and/or land from cleanout

Releases to water or land during sampling

11. Exposure to Hazardous Chemicals from the Coloring of Leather and Dyesa) Sources of Occupational ExposureWeighing: Inhalation exposure during weighing of dye dustDissolving: Inhalation exposure during cleanout

Dermal exposure during sampling and cleanoutPump: Dermal exposure during cleanoutDyeing: Inhalation exposure during cleanout

Dermal exposure during sampling and cleanoutDryer: Inhalation exposure during venting of heated air

b) Sources of Releases to the EnvironmentWeighing: Releases to air after purification filtersDissolving: Releases to air, water, and/or land from cleanout

Releases to water or land during samplingPump: Releases to air during cleanout

Releases to water or land during samplingDyeing: Releases to air due to excess spent solution treatment or disposal

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Releases to air, water, and/or land from cleanoutReleases to water or land during sampling

Transfer: Release to air during transferDryer: Releases to air during venting

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Chapter 7. Green Chemistry

1. Atom and Mass Efficiency Calculations: Calculate mass and atom efficiencies

a) (Addition reaction) Isobutylene + methanol methyl,tert-butyl etherC4H8 + CH3OH (C4H9)-O-CH3

Mass efficiency: Feedstocks – 5(12) + 12(1) + 1(16) = 88 Product – 5(12) + 12(1) + 1(16) = 88 …….. efficiency = 100%

Carbon efficiency: Feedstocks – 5(12) = 60 Product – 5(12) = 60 …….. efficiency = 100%

Hydrogen efficiency: Feedstocks – 12(1) = 12 Product – 12(1) = 12 …….. efficiency = 100%

Oxygen efficiency: Feedstocks – 1(16) = 16 Product – 1(16) = 16 …….. efficiency = 100%

b) (Substitution reaction) Phenol + ammonia aniline + waterC6H5-OH + NH3 C6H5-NH2 + H2OMass efficiency: Feedstocks – 6(12) + 9(1) + 1(16) + 1(14) = 111

Product – 6(12) + 7(1) + 0(16) + 1(14) = 93 …….. efficiency = 83.8%Carbon efficiency: Feedstocks – 6(12) = 72

Product – 6(12) = 72 …….. efficiency = 100%Hydrogen efficiency: Feedstocks – 9(1) = 9

Product – 7(1) = 7 …….. efficiency = 77.8%Oxygen efficiency: Feedstocks – 1(16) = 16

Product – 0(16) = 0 …….. efficiency = 0%Nitrogen efficiency: Feedstocks – 1(14) = 14

Product – 1(14) = 14 …….. efficiency = 100%

c) (Elimination reaction) Ethylbenzene styrene + hydrogenC6H5-C2H5 C6H5-C2H3 + H2

Mass efficiency: Feedstocks – 8(12) + 10(1) = 106 Product – 8(12) + 8(1) = 104 …….. efficiency = 98.1%



d) Other industrially important examples of additions, substitutions, and elimination reactions (see Wittkcoff, H.A. and Reuben, B.G. “Industrial Organic Chemicals”, John Wiley & Sons, New York, 1996; and Weissermel, K. and Arpe, H.-J. “Industrial Organic Chemistry”, VCH Verlagsgesellschaft mbH a Wiley company, Weinheim Germany, 1997).

Addition Reactions:Ethanol from ethene

36

CH2=CH2 + H2O C2H5OH (hydration of ethylene using a phosphoric acid catalyst, low per pass conversions on the order of 10%)Mass efficiency: Feedstocks – 2(12) + 6(1) + 1(16) = 46

Product – 2(12) + 6(1) + 1(16) = 46 …….. efficiency = 100%Carbon efficiency: Feedstocks – 2(12) = 24


Product – 6(1) = 6 …….. efficiency = 100%Oxygen efficiency: Feedstocks – 1(16) = 16


Ethylene oxide from etheneCH2=CH2 + 0.5O2 CH2—CH2 (silver catalyst, 15 bar and 250˚C, yield > 80%)

\ / O

Mass efficiency: Feedstocks – 2(12) + 4(1) + 1(16) = 44 Product – 2(12) + 4(1) + 1(16) = 44 …….. efficiency = 100%




Vinyl chloride from acetyleneCHCH + HCl CH2=CHCl Mass efficiency: Feedstocks – 2(12) + 3(1) + 1(35.45) = 62.45

Product – 2(12) + 3(1) + 1(35.45) = 62.45 …….. efficiency = 100%Carbon efficiency: Feedstocks – 2(12) = 24


Product – 5(1) = 5 …….. efficiency = 100%Chlorine efficiency: Feedstocks – 1(35.45) = 35.45

Product – 1(35.45) = 35.45…….. efficiency = 100%

Substitution Reactions:Vinyl chloride from etheneCH2=CH2 + Cl2 CH2ClCH2Cl CH2=CHCl + HClMass efficiency: Feedstocks – 2(12) + 4(1) + 2(35.45) = 98.45

Product – 2(12) + 3(1) + 1(35.45) = 62.45 …….. efficiency = 63.4%Carbon efficiency: Feedstocks – 2(12) = 24


Product – 3(1) = 3 …….. efficiency = 75%Chlorine efficiency: Feedstocks – 2(35.45) = 70.9

37

Product – 1(35.45) = 35.45…….. efficiency = 50%

Chlorine from oxidation of HCl4HCl + O2 2H2O + 2Cl2

Mass efficiency: Feedstocks – 4(1) + 4(35.45) + 1(32) = 217.8 Product – 2(70.9) = 141.8 …….. efficiency = 65.1%


Chlorine efficiency: Feedstocks – 4(35.45) = 141.8 Product – 2(70.9) = 141.8 …….. efficiency = 100%


Vinyl chloride from ethylene using oxychlorination (By integrating the above two reactions, we can increase the mass efficiency of the overall production of vinyl chloride)

Mass efficiency: Feedstocks – 6(12) + 15(1) + 6(35.45) + 1.5(16) = 333.7 Product – 6(12) + 15(1) + 6(35.45) + 0(16) = 309.7…….. efficiency = 92.8%



Chlorine efficiency: Feedstocks – 6(35.45) = 222.7 Product – 6(35.45) = 222.7 …….. efficiency = 100%

Oxygen efficiency: Feedstocks – 1.5(16) = 24 Product – 0(16) = 0 …….. efficiency = 0%

Elimination Reactions:Phenol from dehydrogenation of a cyclohexanone/cyclohexanol mixture

Mass efficiency: Feedstocks – 12(12) + 11(1) + 2(16) = 187

38

1.5CH2=CH2 + 1.5Cl2 1.5CH2ClCH2Cl

1.5CH2=CH2 + 3HCl + 0.75O2 1.5CH2ClCH2Cl + 1.5H2O

3CH2ClCH2Cl 3CH2=CHCl + 3 HCl

OH=O + OH + 9/2 H22

Product – 12(12) + 2(1) + 2(16) = 178 …….. efficiency = 95.2%Carbon efficiency: Feedstocks – 12(12) = 144


Product – 2(1) = 2 …….. efficiency = 18.2%Oxygen efficiency: Feedstocks – 2(16) = 32


2. Confirm mass efficiencies in Table 7.1

MnO2: Mass efficiency: Feedstocks – 1(54.94) + 2(16) = 86.94

Product – 1(16) = 16 …….. efficiency = 18.4%

PhIO: Mass efficiency: Feedstocks – 1(30.97) + 1(1) + 1(126.9) + 1(16) = 174.9


H2O2: Mass efficiency: Feedstocks – 2(1) + 2(16) = 34


t-BuOOH: (CH3-CH2-CH2-COOH)Mass efficiency: Feedstocks – 4(12) + 8(1) + 2(16) = 88


NaOCl: Mass efficiency: Feedstocks – 1(22.99) + 1(16) + 1(35.45) = 74.44


39

K2Cr2O7: Mass efficiency: Feedstocks – 2(39.1) + 2(52) + 7(16) = 294.2


KMnO4: Mass efficiency: Feedstocks – 1(39.1) + 1(54.94) + 4(16) = 158.0


40

Chapter 8. Evaluating Environmental Performance During Process Synthesis

1. Compare the carbonylation of dinitrotoluene and the amine-phosgene routes for the production of toluene diisocyanate (TDI) using a Tier 1 economic and environmental performance evaluation.

Amine - phosgene route:C6H3(CH3)(NH2) 2 + 2 COCl2 C6H3 (CH3)(-N=C=O) 2 + 4 HClCarbonylation of nitrobenzene:C6H3 (CH3)(NO2) 2 + 6 CO C6H3 (CH3)(-N=C=O) 2 + 4 C O2

Table of stoichiometric and toxicity data for each reaction pathway

Environmental and Economic Evaluation

Environmental evaluation

(PEL can be used instead of TLV) (Equation 8-2)

(Equation 8-3)

Amine-phosgene routeTLV Index = (.76)(1/.1)+(.01)(1/350)+(.4)(1/7)+(1.26)(1/.4)+(1)(1/.14) = 17.95EPA Index = (.01)(100)+(.4)(100)+(1)(100,000) = 100,041

41

Cost of Raw Materials = (.76)(.576)+(.01)(.550)+(1.26)(.610) = $1.21/lb TDI

Carbonylation RouteTLV Index = (1.04)(1/1.5)+(1)(1/55)+(1)(1/.14)+(1)(1/9000) = 7.85EPA Index = (1.04)(1,000)+(1)(100,000) = 101,040Cost of Raw Materials = (1.04)(.365)+(1)(.04) = $0.42/lb TDI

Discussion: The TLV Index indicates that the carbonylation route is superior to the amine-phosgene route. The EPA Index indicates that both routes are about the same, but data is lacking for this index, so we should rely less on this index. The cost analysis indicates that the carbonylation route is superior.

2. Tier 2 environmental assessment for the production of maleic anhydride (MA) from n-butane.

a) Emission sources1. Venting from the crude MA and refined MA product storage tanks2. Reactor emissions3. Absorber column off-gas4. Vacuum Stripper5. Solvent purification (distillation column)6. Batch distillation7. Boiler emissions8. Fugitive emissions

b) Estimate wastes and emissions for the processesMA storage tanks, one tank for crude MA and one for refined MA. Since crude MA is 99.8% pure, we assume pure MA for this emission estimation. To proceed, assume that the process produces 50,000 tons MA/yr, which is the throughput for the tanks. To use the TANKS software, the volumetric flow rate is needed in gallons/yr. The density of MA is 12.34 lb MA/gal MA, so the volumetric flow rate is 8.1x106 gal MA/yr. A turnover rate of 2-3 days for the tank is assumed, so the tank working volume should be about 67,500 gal. A tank with a diameter of 28 ft and height of 15 ft will work for this volume of tank. We assume a vertical fixed-roof tank with no pollution control of white color in good condition. Also, the average liquid height is 13 ft and the maximum is 14.65 ft. Because the TANKS software does not have data for MA, we choose a surrogate compound with a similar vapor pressure (0.1 psia at ambient conditions), 1,1,1,2-tetrachloroethane. The annual emissions are 1,356 lb/yr for each tank yielding an emission per unit of MA production of 1,356 lb MA/yr/((50,000 tons MA/yr)(103 lb MA/0.5 ton MA) = 0.0136 lb MA/103 lb MA. For both tanks, the total tank emissions of MA is 2(0.0136 lb MA/103 lb MA) = 0.0271 lb MA/103 lb MA. Reactor Emissions: The reactor is a packed bed of small diameter tubes filled with catalyst. We assume there are no emissions from the reactor directly, but will include the reactor in the estimation of fugitive emissions from the process. For lack of information, we assume no catalyst-related waste generation. Absorber column off-gas: The absorber oil is highly selective for MA, but is not selective for the reactants n-butane and air and the byproducts CO and CO2. We assume therefore that reaction stoichiometry dictates emissions from the absorber. According to literature sources (Kirk - Othmer Encyclopedia of Chemical Technology, Vol. 15, pg 910), typical conversions of n-butane are 85% and yields of MA are

42

50-60% on a molar basis. For 50,000 tons MA/yr, molar flow rate of product is 1.02x106 lbmole MA/yr. n-butane feed rate is 1.02x106/0.85 = 1.2x106 lbmole n-butane/yr or 57.6x106 lb n-butane/yr. n-butane emission rate from the absorber is 0.15(1.2x106 lbmole n-butane/yr) = 0.18x106 lbmole n-butane/yr or 8.64x106 lb n-butane/yr )/((50,000 tons MA/yr)(103 lb MA/0.5 ton MA) = 86.4 lb n-butane/103 lb MA. Assuming that yield of MA is 60% on a molar basis and that byproduct formation is equally split between CO and CO2, reaction stoichiometry will dictate emissions of these byproducts from the absorber column. Reactions for CO and CO2 areC4H10 + 6.5 O2 4 CO2 + 5 H2OC4H10 + 4.5 O2 4 CO + 5 H2OMolar emission rates for each byproduct are (4)(20%) of the molar MA product flow rate, (4)(0.2)(1.02x106 lbmole MA/yr) = 8.16x105 lbmole CO or CO2/yr. Converting to mass emission rates for each results in, for CO2, 44(8.16x105 lbmole/yr)/((50,000 tons MA/yr)(103 lb MA/0.5 ton MA) = 359.2 lb CO2/103 lb MA and for CO, 28(8.16x105 lbmole/yr)/((50,000 tons MA/yr)(103 lb MA/0.5 ton MA) = 228.4 lb CO2/103 lb MA.

Vacuum Stripper: According to the average emission factors for chemical process units shown in Table 8.3.2, the emission rate of 0.2 lb MA/103 lb MA.

Solvent Purification Column: Little information is given regarding solvent purification, but we may assume that distillation of residual MA is separated from the heavier absorber solvent. According to Table 8.3.2, distillation emissions are 0.1 lb/103 lb MA and we assume that these emissions are MA. (Note that there is an error in the entry in Table 8.3.2 for distillation emission factor. The correct factor should be 0.1 not 0.7). Batch Distillation: The emission rate for MA for batch distillation are again 0.1 lb/103 lb MA.

Bioler Emissions: To estimate these emissions, an estimate of energy consumption per mass of product is required.Rudd, et al., (1981 in chapter 8) provide estimates of energy consumption for a number of processes and suggest a value of 0.41 metric tons of fuel oil equivalent per metric ton (103

kg) of product. Assuming that #6 fuel oil with 1% sulfur is used and that no emission controls are in place leads to estimates (based on Table 8.3-5) of:SO2 emissions:(19 kg/103

L fuel oil)/(0.8 kg/L) (410 kg fuel oil/ 103 kg MA product) = 9.3 kg SO2 /103 kg MA.

SO3 emissions:(0.69 kg/103

L fuel oil)/(0.8 kg/L) (410 kg fuel oil/103 kg MA product) = 0.33 kg SO3 /103 kg MA

NOx emissions:(8 kg/103

L fuel oil)/(0.8 kg/L) (410 kg fuel oil/103 kg MA product) = 4.1 kg NOx / 103

kg MAParticulate Matter emissions:(1.5 kg/103

L fuel oil)/(0.8 kg/L) (410 kg fuel oil/103 kg MA product) = 0.8 kg PM / 103

kg MACarbon Monoxide emissions: (0.6 kg/103

L fuel oil)/(0.8 kg/L) (410 kg fuel oil/ 103 kg MA product) = 0.3 kg CO /103 kg MA

Carbon Dioxide emissions: (3,025 kg/103

L fuel oil)/(0.8 kg/L) (410 kg fuel oil/ 103 kg MA product) = 1,550.3 kg CO2 /103 kg MA

43

Fugitive emissions:To accurately estimate fugitive emissions requires a count of valves, flanges, fittings, pumpsand other devices that are used in the process. Such counts are not generally available forpreliminary process designs, however, rough estimates can be made based on experience.Typically, fugitive emissions for chemical processes total 0.5-1.5 kg per 103 kg product. In this case, we have probably already accounted for some of the fugitive emissions through the emission factors for the reactors and distillation column, therefore, an estimate of 0.5 kg per 103 kg product is appropriate, with the emissions being primarily MA.

Loading and Unloading Losses: The loading and off-loading emissions can be estimated using Equation 8.3-4. Assuming a saturation factor of 0.6 (Table 8.3.9), a vapor pressure of 0.25 mm Hg for MA (from ChemFate, the database on the Syracuse Research Corporation web site - see Appendix F), a molecular weight of 98 and a temperature of 530 R gives a loading loss of : LL = 12.46 (0.6*0.25*(14.7/760)*98/530) = 0.0067 lb/103 gal = 5.4x10-4 lb/103 lb product (specific gravity of MA is 1.48).

A summary of emissions is provided in the table below (kg emitted / 103 kg MA)

Release Source Maleic Anhydride(MA)

n-butane Criteria Pollutants

CO2

Venting from storage tanks 0.027

Absorber column86.4 228.4 (CO) 359.2

Vacuum Stripper0.2

Distillation Columns0.2

Boiler emissions9.3 (SO2)0.3 (SO3)4.1 (NOx)0.8 (PM)0.3 (CO)

1,550.3

Fugitive Emissions 0.5

Loading/unloading operations 0.0005

Total 0.93 86.4 243.2 1,909.5

44

Chapter 9. Unit Operations and Pollution Prevention

1. Solvent Choice for Extraction of Caffeine from Coffee BeansData for carbon dioxide, ethyl acetate, and dichloromethane can be found in MSDSs. A good web site for MSDSs is http://msds.pdc.cornell.edu/msdssrch.asp

Compound Name CAS Number PEL (ppm) LD50 (mg/kg) Cancer EffectsMethylene Chloride 75-09-2 500 2136 1Carbon Dioxide 124-38-9 5000 not known 2Ethyl Acetate 141-78-6 400 5620 3Trichloroethylene 79-01-6 100 4900 4

1. Methyulene Chloride Cancer EffectsCarcinogenicity - NTP: YES Carcinogenicity - IARC: YES Carcinogenicity - OSHA: NO Explanation Carcinogenicity: METHYLENE CHLORIDE: GROUP 2B(IARC), GROUP2(NTP).

2. Carcinogenicity - NTP: NO Carcinogenicity - IARC: NO Carcinogenicity - OSHA: NO Explanation Carcinogenicity: THIS COMPOUND CONTAINS NO INGREDIENTS ATCONCENTRATIONS OF 0.1% OR GREATER THAT ARE CARCINOGENS OR SUSPECTCARCINOGENS.

3. Carcinogenicity - NTP: NO Carcinogenicity - IARC: NO Carcinogenicity - OSHA: NO Explanation Carcinogenicity: THERE ARE NO INGREDIENTS ABOVE 0.1% WHICH AREIDENTIFIED AS CARCINOGENS BY NTP, IARC OR OSHA.

4. Carcinogenicity - NTP: NO Carcinogenicity - IARC: NO Carcinogenicity - OSHA: NO Explanation Carcinogenicity: PER MSDS:+CARC RESPONSE OCCURRED ONLY IN MICEGIVEN LG DOSES.DATA SUGGEST SHOULD POSE LITTLE/NO CARC HAZ FOR MAN.

The least preferred solvent is methylene chloride because the LD50 is the smallest and the cancer risk is the greatest of the three choices. The most preferable is carbon dioxide because the PEL is highest (least toxic) and no cancer risk. Note that it is best to check several MSDSs as variations and errors appear in different company reports.

45

2. Optimum Plug Flow Reactor Volume for a Series ReactionThe three equations describing the dynamics of the reacting system are

Reactant concentration, CA

In this equation, we can separate variables directly and integrate.

. The solution to this equation subject to the initial condition is

Product concentration, CB

In this equation, we must substitute in for CA from the solution above.

. If this equation is rearranged, the result is

. To solve this equation, we multiply both sides by the integrating factor,

. After multiplying and then simplifying, we arrive at an equation to integrate,

. After integrating and evaluating the integration constant using the initial condition above, the concentration of product is

.

Waste concentration, CC

The equation for CC can be solved using the result for product concentration above,

, by first separating variables and then integrating

. After evaluating the constant of integration, we get

.

a) To maximize the yield of B, the volume of the reactor must be such that the residence time in the

reactor is optimum. To determine this optimum reactor residence time, we must determine and

solve for the residence time.

. Multiplying through by we get

46

, and simplifying further

and further still until .

Solving for the residence time, we arrive at

. The reactor volume required is the volumetric

flow rate divided by the residence time, (100 gal/min)/6.93 min = 14.43 gal or 2.24 ft3. The concentrations of A, B, and C for this residence time are

b) If S were not present in large excess and actually was a limiting reagent, then not as much of the waste byproduct C would be formed because S would not be present to react with product B.

3. Energy Efficient Extraction Coupled with DistillationAssumption: The flow rates of raffinate and extract phases in the extractor are constant. The ratio of raffinate to extract flow rate must be 0.8 times the minimum ratio; (R/E) = 0.8 (R/E)min. Assume that the feed to the distillation column is a saturated liquid. Use a reflux ratio (L/D) of 1.2 times the minimum (L/D)min..

a) Determine the change in reboiler duty, , for Kd values of 0.3, 0.5, and 0.7 in the extractor for a constant relative volatility in the distillation column of P,S = 3.

Kd of 0.3, P,S = 3.First we must find the slope of the operating line that makes a “pinch” with the equilibrium line in the extraction column. The slope of that line is R/Emin.

47

The ratio of raffinate to extract flow rate is

and the extract flow rate is

The mass fraction of our desired component in the extract leaving the column is the sum of the component entering the column in the extract stream plus that recovered from the raffinate stream, assuming constant flow of E (an assumption).

The feed to the distillation column is equal to the extract exiting the extraction column,

and the mass fraction of the distillation column feed is

The flow rate of the distillate can be obtained as

The bottoms product flow rate can be easily obtained.

We must find the “pinch” point where the feed line for the distillation column touches the equilibrium curve for the distillation column. This mass fraction is

The minimum reflux ratio is

The reflux ration is 1.2 times the minimum.

From the definition of the reflux ratio, we calculate the liquid flow rate at the top of the distillation column. We assume constant flows in each column section for this problem.

The vapor flow rate at the top of the column is

The vapor flow rate at the bottom of the column is equal to that at the top of the column since the feed is a saturated liquid.

The reboiler duty is the product of the latent heat of vaporization and the vapor flow rate at the bottom of the column.

These calculations may be repeated for each value of Kd of 0.5 and 0.7 and of P,S = 3 in a similar fashion.

48

b) Determine the change in reboiler duty for P,S = 3, 5, and 7 in the distillation column for a constant value of Kd = 0.4 in the extractor.Similar to part a) values of the reboiler duty are obtained and are listed in the following table as a summary.

c) Which parameter, Kd or P,S, has the greatest influence on QR? Both parameters strongly influence the value of QR. Increasing Kd and increasing P,S cause a significant decrease in the reflux ratio in the distillation column, thereby reducing flow rates in the column and the reboiler duty.

d) Would higher or lower values of Kd or P,S result in a more energy-efficient process?Choosing a solvent with high valued for both of these parameters would decrease the reboiler duty for this separation.

49

Kd 0.3 0.5 0.7 0.4 0.4 0.4 3 3 3 3 5 7

0.322 0.544 0.767 0.433 0.433 0.433

3.879 2.296 1.630 2.885 2.885 2.885

0.024 0.040 0.056 0.032 0.032 0.032

3.879 2.296 1.630 2.885 2.885 2.885

0.024 0.040 0.056 0.032 0.032 0.032

0.091 0.091 0.091 0.091 0.091 0.091

3.788 2.205 1.539 2.794 2.794 2.794

0.069 0.112 0.152 0.091 0.143 0.189

20.439 12.298 8.792 15.357 7.673 5.112

24.527 14.757 10.550 18.429 9.208 6.135

2.232 1.343 0.960 1.677 0.838 0.558

2.323 1.434 1.051 1.768 0.929 0.649

2.323 1.434 1.051 1.768 0.929 0.649

kcal/hr 232.30 143.39 105.11 176.80 92.89 64.93

REmin

E R

0.8 REmin

Instructor’s note: We may relax the assumption of constant flows in the extraction column and ask the students to perform parts a) and b) while analyzing this problem with the more realistic situation of varying flows of R and E in the extraction column. After calculating the first three steps in the analysis of part a), the exiting raffinate concentration can be calculated by assuming that R does not transfer to the E stream. Thus,

and solving for RN, we obtain the following equation.

Similarly, from an equation for conservation of solvent in the extract stream and a component balance for the column we can calculate the exit extract flow rate, E1, and concentration, y1. The remaining calculations are the same. The resulting values are very similar as shown in the table below.

50

4. Benzene Emissions from Various Storage Tank Types

51

Kd 0.3 0.5 0.7 0.4 0.4 0.4 3 3 3 3 5 7

0.322 0.544 0.767 0.433 0.433 0.433

3.879 2.296 1.630 2.885 2.885 2.885

0.909 0.909 0.909 0.909 0.909 0.909

0.095 0.093 0.093 0.094 0.094 0.094

3.970 2.387 1.721 2.976 2.976 2.976

0.024 0.039 0.054 0.032 0.032 0.032

3.970 2.387 1.721 2.976 2.976 2.976

0.024 0.039 0.054 0.032 0.032 0.032

0.092 0.092 0.092 0.092 0.092 0.092

3.878 2.295 1.630 2.884 2.884 2.884

0.068 0.109 0.146 0.089 0.140 0.186

20.718 12.661 9.192 15.688 7.839 5.222

24.861 15.193 11.030 18.826 9.407 6.267

2.283 1.395 1.013 1.729 0.864 0.575

2.375 1.487 1.105 1.821 0.956 0.667

2.375 1.487 1.105 1.821 0.956 0.667

kcal/hr 237.48 148.69 110.47 182.05 95.56 66.73

REmin

E R

0.8 REmin

Using the TANKS version 4.0, the following results were obtained using the input data provided in the problem statement.

a)VFRT – total = 1,046.43 lb/yr, Working Losses = 848.41 lb/yr, Breathing Losses = 198.02 lb/yrIFRT – total = 208.79 lb/yr, Rim Seal Losses = 39.64 lb/yr, Withdrawal Losses = 6.40 lb/yr, Deck Fitting Losses = 162.75 lb/yr, Deck Seam Losses = 0.0 lb/yr. DEFRT – total = 124.83 lb/yr, Rim Seal Losses = 39.64 lb/yr, Withdrawal Losses = 6.40 lb/yr, Deck Fitting Losses = 78.80 lb/yr, Deck Seam Losses = 0.0 lb/yr. Roof Type (pontoon), Tank Construction (welded)

b) Percent reduction in emissions compared to VFRTFor IFRT, % = (VFRT-IFRT)/VFRTx100 = (1,046.43-208.79)/1,046.43x100 = 80.1%For DEFRT, % = (VFRT- DEFRT)/VFRTx100 = (1,046.43-124.83)/1,046.43x100 = 88.1%

c) Comparison to Example 9.6-1Toluene (lb/yr) Benzene (lb/yr)

Vertical Fixed-Roof Tank 337.6 1,046.4Internal Floating-Roof Tank 66.2 208.8Domed External Floating-Roof Tank 42.8 124.8

Benzene, being more volatile than toluene, emits at roughly a factor of three higher for each tank type compared to toluene. Percentage reductions are similar for toluene as for benzene.

5. Storage Tank Pollution Prevention by Applying New PaintUsing the TANKS 4.0 program, the following results were obtained for the vertical fixed-roof storage tank with gray/medium paint in poor condition for toluene emissions.

a) Gray/medium paint in poor conditionVFRT – total = 509.29 lb/yr, Working Losses = 345.54 lb/yr, Breathing Losses = 163.75 lb/yrThe percentage of the total emissions caused by standing (breathing) losses is 32.2%.

b) Emissions reductionAfter the application of white paint, the annual emissions of toluene are VFRT – total = 337.23 lb/yr, Working Losses = 273.36 lb/yr, Breathing Losses = 63.87 lb/yr. This emission rate is much lower (33.8%) than the original gray paint in poor condition. However, there was an emission of toluene during the painting and paint curing process. This amount of toluene emission is determined by estimating the surface area of the 20 ft-tall (H) and 12 ft diameter (d) tank ( d H = shell area (754 ft2) + d2/4 = dome area-approximate (113.1 ft2)) and assuming that the density of the paint is 6 lb/gal of paint. Emissions during painting are (754+113.1 ft2)(1 gal paint/100 ft2)(6 lb paint/gal paint)(0.5 lb toluene emitted/lb paint) = 26.01 lb toluene.

52

The net emissions reduction in the year of painting the tank are 509.29 – (337.23+26.01) = 146.05 lb/yr. In the second and subsequent years, assuming that the paint remains in good conditions, the emissions reductions are greater, 509.29 – 337.23 = 172.06 lb/yr.

53

Chapter 10. Flowsheet Analysis for Pollution Prevention

1. Heat Exchange Network for Conserving Hot Water in the Homea) Identify warm water streams

i. shower/bath tub (2) T = 35˚C (95˚F), Fshower/tub = 568 kg/day (150 gal/d)ii. dishwasher (1) T = 50˚C (122˚F), Fdishwasher = 100 kg/day (26 gal/d)iii. washing machine (1) T = 30˚C (86˚F), Fdishwasher = 200 kg/day (53 gal/d)

b) Annual energy savings if warm drain water contact inlet cold water entering hot water heater.Data: heat capacity of water CP,W = 4.2 kJ/(kg•˚C), temperature of entering cold water = 15˚C, minimum temperature driving force at heat exchangers = 5 ˚C. Final temperature of hot water heater water = 60˚C.

Solution: A pinch diagram such as shown in Figure 10.4 is not needed to solve this problem. We simply need to calculate the amount of energy transferred from each drain stream per day and compare this to the energy needed for each use, and then convert to an annual basis. With a 5˚C minimum temperature difference in each heat exchanger, the maximum temperature to which the cold water can be heated is 5˚C less than the temperature of the warm drain water streams.

Shower/tub exchanger: Heat exchangedQshower/tub = CP,W Fshower/tub T = (568 kg/day)( 4.2 kJ/(kg•˚C))((35-5)-15)˚C = 35,784 kJ/dNormal Heat Requirement

= CP,W Fshower/tub T = (568 kg/day)( 4.2 kJ/(kg•˚C))((60-15)˚C = 107,352 kJ/dPercent savings on energy: = 35,784 kJ/d / 107,352 kJ/d (100) = 33.33%Annual Energy Savings: = (35,784 kJ/d)(365 d/yr) = 1.31x107 kJ/yr

Dishwasher: Heat exchangedQdishwasher = CP,W Fdishwasher T = (100 kg/day)( 4.2 kJ/(kg•˚C))(( 50-5)-15)˚C = 12,600 kJ/dNormal Heat Requirement

= CP,W Fdishwasher T = (100 kg/day)( 4.2 kJ/(kg•˚C))((60-15)˚C = 18,900 kJ/dPercent savings on energy: = 12,600 kJ/d / 18,900 kJ/d (100) = 66.66%Annual Energy Savings: = (12,600 kJ/d)(365 d/yr) = 4.60x106 kJ/yr

Washing Machine: Heat exchangedQwash = CP,W Fwash T = (200 kg/day)( 4.2 kJ/(kg•˚C))(( 30-5)-15)˚C = 8,400 kJ/dNormal Heat Requirement

= CP,W Fwash T = (200 kg/day)( 4.2 kJ/(kg•˚C))((60-15)˚C = 37,800 kJ/dPercent savings on energy: = 8,400 kJ/d / 37,800 kJ/d (100) = 66.66%Annual Energy Savings: = (8,400 kJ/d)(365 d/yr) = 3.07x106 kJ/yr

Annual Savings on EnergyCost of electricity = $.06/kW•h

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Efficiency of electric hot water heater = 50% (including heat losses from the tank)Conversion factor; 1 kW•h = 3,600.9 kJTotal energy savings = (1.31x107 + 4.60x106 + 3.07x106)kJ/yr / 0.5 = 4.15 x107 kJ/yrAnnual savings on energy = (4.15 x107 kJ/yr)( 1 kW•h/3,600.9 kJ)( $.06/kW•h) = $692

c) Calculate time required to recover installation costs from savingsCost per exchanger = $500Number of exchangers needed = 4Total cost of installing exchangers = (4)($500) = $2,000Time needed to recover installation costs for exchangers = $2,000/($692/yr) = 2.9 yrs

2. Pinch Point Determination for a Mass Exchanger NetworkStream InformationRich Stream Flow Rate (kg/s) yin yout Equilibrium

R1 3 0.45 0.1 y = 0.5 x0.8

R2 3 0.3 0.1 y = 1.5 xLean Stream Flow Rate (kg/s) xin xout

L1 5 0.1 0.45

Mass Fraction Conversion TableRich Stream Flow Rate (kg/s) xin* xout*

R1 3 0.8766 0.13375R2 3 0.2 0.06667Lean Stream Flow Rate (kg/s) xin xout

L1 5 0.1 0.45

Mass Exchange Summary TableMass Cummulative

Rich Streams Exchanged Massx* Interval (kg/s) Comments (kg/s)

.877 to .2 -0.936 y = .45 to .138 for Stream R1 -0.936

.2 to .13375 -0.414 y = .138 to .1 for Stream R1 and y = .3 to .2 for R2 -1.35

.13375 to .06667 -0.3 y = .2 to .1 for Stream R2 -1.65

L1 Stream 1.75 for x, from .1 to .45

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A plot of the composite load line diagram is shown in the next figure, where the mass exchanged for the rich streams is negative, indicating mass loss from the rich stream, and is positive for the lean stream, indicating mass gain by this stream.

In order to create the pinch on the combined load line diagram, we must "move" the rich composite curve vertically until it touches the lean curve. The pinch will occur at a mass fraction of 0.2, where the value of the mass exchanged on the lean curve is 0.5 kg/s. Therefore, we need to add the difference between the two curves at x = 0.2 to each point on the rich composite load line curve (0.5 -(-.936) = 1.436. The combined load line diagram with the pinch is shown on the next figure.

Extra Credit: From the diagram below, we can calculate the percentage of mass transfer from the rich and lean streams that can be accomplished by the internal mass exchange network versus the percentage of mass transfer that must be accomplished using external mass separating agents. The internal network of mass exchangers is able to transfer at a rate of 1.436 kg/s. This represents 1.436/1.75*(100) = 82.1% of the load of the lean stream and 1.436/1.65*(100) = 87.0% of the load of the rich streams. An external mass separating agent will be required to increase the mass fraction of the lean stream from 0.393 to 0.45 with a mass loading of 1.75-1.436=.324 kg/s. Similarly, for rich stream 2, an external mass separating

56

agent will be required to remove 0.214 kg/s from this stream in order to reduce its concentration from x* = 0.114 to x* = .06667.

3. Styrene Process: Mass Exchange Network Pinch Analysis

Styrene

Formula: C8H8 Molecular Weight: 104.15 CAS Registry Number: 100-42-5

Stream InformationRich Stream Flow Rate (kg/hr) yin yout Equilibrium

R1 to stripper 70,607 1.77x10-3 1.5x10-4 y = 0.001 x

57

R2 to adsorber 70,607 1.5x10-4 5.7x10-6 y = 0.00071 xLean Stream Flow Rate (kg/hr) xin xout

L1 to stripper 70 0.0 1.62 y = 0.001 xL2 to adsorber 53 3x10-5 0.2 y = 0.00071 x

a) Composition Interval Diagram for Rich Streams

b) Composition Interval Diagram for Lean Streams

58

1.77x10-3

1.5x10-4

5x10-6

y scale

Interval #1

Interval #2

Rich Streams

#1 #2

1.62

.200

3x10-5

x scale

Interval #1

Interval #2

Lean Streams

#1 #2

Interval #3

0

c) Pinch DiagramIn this problem, because the rich streams are both aqueous based and therefore similar, we will convert the lean stream mass fractions to a rich mass fraction basis using the equilibrium relationships.

Mass Fraction Conversion TableRich Stream Flow Rate (kg/h) yin yout

R1 70607 0.00177 0.00015R2 70607 0.00015 0.0000057Lean Streams Flow Rate (kg/h) y*in y*out

L1 to stripper 70 0 0.00162L2 to adsorber 53 2.13x10-8 0.000142

Mass Exchange Summary Table

A plot of the composite load line diagram is shown in the next figure, where the mass exchanged for the rich streams is negative, indicating mass loss from the rich stream, and is positive for the lean stream, indicating mass gain by this stream.

59

In order to create the pinch on the combined load line diagram, we must "move" the rich composite curve vertically until it touches the lean curve. The pinch will occur at a mass fraction of 0.0000057, where the value of the mass exchanged on the lean curve is 0.82 kg/s. Therefore, we need to add the difference between the two curves at x = 0. 0000057 to each point on the rich composite load line curve (0.82 -(-124.57) = 125.39 kg/h. The combined load line diagram with the pinch is shown on the next figure.

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MaximumBenzeneExchange = 123.18 kg/hr

The rate of benzene mass exchange can be calculated from the pinch diagram by taking the difference between the maximum value of mass exchanged on the lean curve (123.998 kg/h) and subtract the minimum value of mass exchanged on the rich curve at pinch (0.82 kg/h), as shown on the figure above.

The value of the benzene recovered is ($0.2 / kg benzene) (123.18 kg benzene/hr) = $24.64 / hr.

For the case of a minimum mass transfer driving force of 0.001, we must move the rich composite curve down until the horizontal distance on the pinch diagram between the lean and rich curves is 0.001, which is at the pinch point in the composite load line diagram w/pinch, above. The horizontal distance of 0.001 mass fraction units is measured from the terminus of the lean composite curve at coordinates (y=2.13x10-

8, Mass Exchanged=0.001491). To determine the vertical distance, we must multiply the horizontal distance (0.001+2.13x10-8) by the slope of the rich composite curve (70,607 kg/h) which is equal to 70.61 kg/h. The modified diagram is shown below.

The maximum rate of benzene transfer when the minimum driving force is 0.001 can be obtained from the figure above, (54.78-0.001491) kg/h = 54.78 kg/h.

The value of the benzene recovered is ($0.2 / kg benzene) (54.78 kg benzene/hr) = $10.96 / hr.

e) The recovery rate of benzene from the steam stripper calculated using the analysis in part c is 123.18 kg benzene/hr and this compares with 113 kg/hr recovered from the decanter plus 11 kg/hr recovered from the decanter downstream from the regenerator of the adsorption regenerator unit for a total of 124 kg/hr. The agreement is very close.

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4. Copper Etching from Circuit Boards

Mass Fraction Conversion TableRich Stream Flow Rate (kg/s) yin yout Equilibrium

R1 0.25 0.13 0.10R2 0.02 0.06 0.02Lean Streams Flow Rate (kg/s) Max X, Max y*in X, y*out

S1 ?? 0.07 0.053 0.03 0.0233 y = 0.734 x+0.001S2 ?? 0.20 0.40 0.001 0.1015 y = 1.5 x+0.1

a) Composition Interval Diagram for Rich and Lean Streams

Composition Interval Diagram for Lean Streams

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0.13

0.10

0.02

y scale

Interval #1

Interval #2

Rich Streams

#1 #2

0.06

b) Maximum amount of copper removed using S1

Mass Exchange Summary Table

Mass CummulativeRich Streams Exchanged Massy Interval (kg/s) Comments (kg/s)0.13 to 0.1 -0.0075 -0.00750.06 to 0.02 -0.0008 -0.0083

The plot of the load line diagram for the rich streams is shown in the next figure. The lean curve for S1 is missing because the flow rate is not specified, and should be calculated as part of the solution. We can move the rich curve up until it touches the lean curve, which terminates at the point on the diagram (0.0233, 0 kg/s). In order to do this, we add the vertical difference between the two curves at this pinch point (0.0083 kg/s) to each point on the rich curve. The other end of the lean curve for stream S1 must terminate at the point on the diagram (0.053, .00823 kg/s) so that the maximum copper concentration for stream S1 is not exceeded while using the minimum possible flow rate of S1.

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0.40

0.1015

0.053

y* scale

Interval #1

Lean Streams

#1 #2

Interval #2

0.0233

The combined load line diagram with pinch is shown in the next figure. The pinch occurs at the bottom of the lean curve.

64

The required flow rate of stream S1 is obtained from the maximum rate of copper mass exchange (0.00823 kg/s) and the actual mass fraction values of the inlet (x=.03) and outlet stream (x=.07), equal to (0.00823 kg/s)/(.07-.03) = 0.206 kg/s.

c) Maximum amount of copper removed using S2The plot of the load line diagram for the rich streams is again the figure first figure in part b. The lean curve for S2 is missing because the flow rate is not specified, and should be calculated as part of the solution. We can move the rich curve up until it touches the lean curve, which terminates at the point on the diagram (0.1015, 0 kg/s). In order to do this, we add the difference between the two curves at this pinch point (0.00720 kg/s) to each point on the rich curve. The other end of the lean curve for stream S2 must terminate at the point on the diagram (0.13, .00720 kg/s) so that the maximum copper concentration for stream S2 does not violate equilibrium constraints while using the minimum possible flow rate of S2.

The combined load line diagram with pinch for stream S2 is shown in the next figure. Both composite curves lay on top of each other. The required flow rate of stream S2 is obtained from the maximum rate of copper mass exchange (0.00720 kg/s) and the actual mass fraction values of the inlet (x=.001) and outlet stream (x=.02), equal to (0.00720 kg/s)/(.02-.001) = 0.0362 kg/s.

d) The maximum amount of copper that can be recovered using both S1 and S2 is the greater of the amounts calculated in parts b and c, 0.00823 kg Copper/s.

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e) If the cost of S2 is 1/2 the cost of S1, determine the optimum ratio of the streams that still recovers the maximum amount of copper.

Stream S1 must recover copper over the rich stream range of 0.02329<y<0.1015 because S2 can not recover copper in that range due to equilibrium constraints. The flow rate of S1 over this range of y is equal to the rate of copper exchanged (.0011 kg copper/s) divided by the mass fraction change (.1015-.02), which gives 0.0135 kg S1/s. Also, over the range of 0.1015<y<0.13, stream S2 should be used because it is the cheapest and its flow rate for each unit of copper recovered is much less than S1. Therefore, the optimum ratio S1/S2 = (0.0135 kg S1/s)/( 0.0362 kg S2 /s) = 0.37.

5. Source-Sink Diagrams for Improved Industrial Water Use

Sink StreamsFlow Rate Maximum

Streams 103 kg/hr Conc. (ppm)1 50 202 100 503 80 1004 70 200

Source StreamsFlow Rate Maximum

Streams 103 kg/hr Conc. (ppm)A 50 50B 100 100C 70 150D 60 250

The source-sink diagram for this network of streams is shown in the next figure. In general, the source concentrations are higher than the sink concentrations, but one source has the target concentration of a sink but a higher flow rate (B and 3) while another source has the target flow rate of a sink but a lower concentration (C and 4). The latter source can be used to satisfy its sink because the concentration targets are maximum only and a lower concentration can be tolerated. A portion of source B (80,000 kg/h) can be used to satisfy sink 3.

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With these matches made, a new source-sink diagram can represent the remaining streams. The remaining 20,000 kg/h of source stream B is indicated with a prime in the diagram.

67

Source streams A and B’ plus 30,000 kg/h of pure water can be combined to satisfy the flow rate requirements of sink 2. The concentration of this combined source is

which is less than the maximum allowable concentration of the sink stream (50 ppm). Now all of the source streams have been fully utilized except for D, which will have to be treated and disposed of. As for sink 1, this could be satisfied with pure water at a flow rate of 50,000 kg/h.

The total fresh water demand of this mass exchange network is 80,000 kg/h and the total flow rate of source wastewater needing treatment is 60,000 kg/h at a concentration of 250 ppm.

This solution is very similar to the network arrived at by Polley and Polley, Chemical Engineering Progress, Feb. 2000, Figures 8 and 9.

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Chapter 11. Evaluating the Environmental Performance of a Flowsheet

1. Maximum Incremental Reactivity of Reformulated Fuel BlendsData: chemical MIR ethanol 1.34 toluene 2.70 xylene 7.10 base fuel 1.5

10% ethanol, 90% base fuel Percent Reduction Mixture MIR = (.1)(1.34)+(.9)(1.5) = 1.48

10% toluene, 90% base fuelMixture MIR = (.1)(2.7)+(.9)(1.5) = 1.62 (1.48-1.62)/1.62(100)= -8.6%

10% xylene, 90% base fuelMixture MIR = (.1)(7.1)+(.9)(1.5) = 2.06 (1.48-2.06)/2.06(100)= -28.2%

2. Hydroxyl Radical Reaction with EtheneData: k = 2nd -order rate constant for reaction of ethene with OH• = 2.7x10-13 (cm3/(molec.•sec))[OH•] = 10-12 moles / L

a) pseudo-1st-order reaction rate constant (k[OH•])[OH•] = (10-12 moles/L)(6.02x1023 molecules/mole)(1 L/1000 cm3) = 6.02x108 molecules/cm3

k[OH•] = (2.7x10-13 (cm3/(molecules•sec)))(6.02x108 molecules/ cm3) = 1.63x10-4 sec-1

= 0.59 hr-1

b) Ethene remaining after 1 hourInitial concentration of ethene is 100 ppm. From Chapter 5, [Ethene] = [Ethene]o exp(-k[OH•]t)

= (100 ppm) exp(-(.59 hr-1)(1 hr)) = 55.7 ppm

3. Global Warming Parameters

a) The chemicals in Table D-1 having the highest values of both and BI can be found by taking the product of these two parameters for each chemical and then ranking the chemicals in descending order. The figure below shows an increasing trend in GWP with the product •BI, indicating the importance of these two parameters on GWP.

69

Chemical Formula (yrs) BI (atm-1 cm-2) •BI GWP

CFC-115 CF3CClF2 400.0 4678 1,871,200 7000CFC-114 CClF2CClF2 200.0 4141 828,200 7000CFC-12 CCl2F2 120.0 3240 388,800 7100CFC-113 CCl2FCClF2 90.0 3401 306,090 4500CFC-11 CCl3F 60.0 2389 143,340 3400HCFC-142b C2F2H3Cl 19.1 2577 59,221 1800Tetrachloromethane CCl4 47.0 1195 56,165 1300HCFC-22 CF2HCl 15.0 2554 38,310 1600HCFC-124 C2F4HCl 6.9 4043 27,897 440HCFC-141b C2FH3Cl2 10.8 1732 18,706 5801,1,1-trichloroethane CH3CCl3 6.1 1209 7,375 100HCFC-123 C2F3HCl2 1.7 2552 4,338 90Dichloromethane CH2Cl2 0.5 1604 802 9

b) We may estimate the magnitude of each partial derivative in the following equation

by inspection of the GWP table and a simple difference calculation.

70

is estimated by using chemicals with nearly the same value of BI but different values of .

The following two chemicals will work.

Chemical Formula (yrs) BI (atm-1 cm-2) •BI GWP

HCFC-142b C2F2H3Cl 19.1 2577 59,221 1800HCFC-22 CF2HCl 15.0 2554 38,310 1600

. We may wish to consider two chemicals more widely separated

in terms of , for example CFC-12 and HCFC-113.

. Finally, using tetrachloromethane and 1,1,1-trichloroethane

. We might use an average of these for a representative number,

55 yr-1.

For the term , our choices from the table above are limited. We choose HCFC-124 and 1,1,1-

trichloroethane for this calculation. .

c)

The residence time in the atmosphere, , appears to have the dominant effect on GWP.

4. Incremental Reactivities for Volatile Organic Compounds

a) Average MIRs for compound classes: If we place the average MIR for compound classes in descending order and include the average MIR for reactive organic gases (ROG), the following list is obtained. Aldehydes, compounds with carbon-carbon double bonds, and aromatics are more potent smog forming chemicals (low level atmospheric ozone formation).

Compound Class MIR aldehydes 7.24other alkenes 6.85secondary alkenes 6.75primary alkenes 5.66aromatics 4.34

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base reactive organic gas mixture 3.10acetylenes 2.30cyclic alkanes 2.06alcohols and ethers 1.32branched alkanes 1.20aromatic oxygenates 0.95ketones 0.87n-alkanes 0.55

b) If one of the main considerations for choosing a compound for a reformulated gasoline is smog formation potential, the compounds to consider would be in the compound classes of alcohols and ethers, aromatic oxygenates, and ketones since they all have MIR values below the base ROG mixture. All of these classes have oxygen atoms integrated into the molecule and the presence of the oxygen will decrease CO emissions.

5. Environmental index calculations for the case study of toluene and ethyl acetate recovery and recycle from Example 11.3-2.

The emissions for the absorber oil flow rates of 0 and 100 kgmole/min are given below. n-C14 is the absorber oil and hexane will be used for TOC, a byproduct of utility boiler combustion.

AbsorberOil Flow Rate (kgmol/hr)

Emission Rate (kg/hr)

TolueneEthyl

Acetate CO2 CO TOC NOx SOx n-C14

0 193.55 193.55 0 0.0 0.0 0.0 0.0 0.0100 0.02 128.07 360 0.129 0.007 0.52 3.39 4.23

Absorber Oil Flow Rate = 0 kgmole/hr

Global warming index:

The global warming potential for each emitted chemical must be calculated using equation 11.3-5,

.

For toluene, GWPi (indirect) = (7 carbons)(44/92) = 3.35 and for ethyl acetate, GWPi (indirect) = (4 carbons)(44/88) = 2.0IGW = (3.35)(193.5 kg/hr) + (2)(193.5 kg/hr) = 1,035 kg/hr

Smog Formation Index:

The maximum incremental reactivity for a chemical is provide in Appendix D, Table D-4.

The smog formation potential for a chemical is

72

For toluene, SFPi = 2.7/3.1=0.87 and for ethyl acetate, SFPi = 1.1/3.1=0.35 (an average of alcohols and ethers and ketones). ISF = (0.87)(193.5 kg/hr) + (0.35)(193.5 kg/hr) = 237 kg/hr

Acid Rain Index: IAR = 0, There are no acid forming chemicals emitted for this case.

Ozone Depletion Index: There are no ozone depleting chemicals emitted for this case.

Non-Carcinogenic Ingestion Toxicity Index:

The ingestion toxicity potential is given by eqn. 11.3-14 or

depending upon the availability of data.

For toluene, data is given in Example 11.3-3; INGTPi = [(4.0x10-7 g/m3)/(0.2 mg/kg/day)] / [(4.0x10-7 g/m3)/(0.2 mg/kg/day)] = 1.0.For ethyl acetate, INGTPi = [(5.0x10-6 g/m3)/(0.9 mg/kg/day)] / [(4.0x10-7 g/m3)/(0.2 mg/kg/day)] = 2.8.IING = (1.0)(193.5 kg/hr) + (2.8)(193.5 kg/hr) = 735 kg/hr.

Non-Carcinogenic Inhalation Toxicity Index:

The ingestion toxicity potential is given by eqn. 11.3-14 or

depending upon the availability of data.

For toluene, data is given in Example 11.3-3; INHTPi = [(1.97x10-7 g/m3)/(4,000 ppm)] / [(1.97x10-7 g/m3)/(4,000 ppm)] = 1.0.For ethyl acetate, INHTPi = [(4.36x10-7 g/m3)/(3200 ppm)] / [(1.97x10-7 g/m3)/(4,000 ppm)] = 2.8.IINH = (1.0)(193.5 kg/hr) + (2.8)(193.5 kg/hr) = 735 kg/hr.

Absorber Oil Flow Rate = 100 kgmole/hr

Global warming index:

IGW = (3.35)(0.02 kg/hr) + (2)(128.07 kg/hr) + (1.0)(360 kg/hr) = 616 kg/hr

Smog Formation Index: For n-C14, SFPi = 0.32/3.1=0.10 and for TOC (hexane), SFPi = 0.98/3.1=0.32. ISF = (0.87)(0.02 kg/hr) + (0.35)(128.07 kg/hr) + (0.10)(4.23) + (0.98)(0.007) = 42 kg/hr

Acid Rain Index:

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The acid rain potentials for a chemical, , are given in Table D-3 in Appendix D.

For SOx, ARPi = 1.0 (assuming SO2), and for NOx, ARPi = 1.07 (assuming NO). IAR = (1.0)(3.39 kg/hr) + (1.07)(0.52 kg/hr) = 4.0 kg/hr

Non-Carcinogenic Ingestion Toxicity Index: For hexane, data is given in Example 11.3-3; INGTPi = [(1.5x10-9 g/m3)/(28,700 mg/kg)] / [(4.0x10-7 g/m3)/(5,000 mg/kg)] = 6.5x10-4.IING = (1.0)(0.02 kg/hr) + (2.8)(128.07 kg/hr) + (6.5x10-4)(0.007) = 359 kg/hr.

Non-Carcinogenic Inhalation Toxicity Index: For hexane, data is given in Example 11.3-3; INHTPi = [(1.97x10-7 g/m3)/(0.2 mg/m3)] / [(1.97x10-7 g/m3)/( 0.4 mg/m3)] = 2.0.IINH = (1.0)(0.02 kg/hr) + (2.8)(128.07 kg/hr) + (2.0)(0.007) = 359 kg/hr.

6. Carbon Dioxide Emission FactorsFuel Oil: assume stoichiometry of C10H22 + 15 1/2 O2 10 CO2 + 11 H2OBasis of 1 mole C10H22 Moles of CO2 = 10 molesMass of CO2 = (10 moles)(44 g/mole) = 440 g CO2 = 0.44 kg CO2

Volume C10H22 = (1 mole C10H22)(144 g/mole)(1 kg/103 g)/(0.73 kg/L) = 0.197 L C10H22

CO2 emission factor = 0.44 kg CO2 / 0.197 L C10H22 = 2.23 kg CO2 / L = 2,300 kg CO2 /103 L.If a specific gravity of fuel oil closer to 1 is chosen, then the emission factor is very close to the value given in Table 8.3-5 of 3,025 kg/103 L.

Natural Gas: assume stoichiometry of CH4 + 2 O2 CO2 + 2 H2OBasis of 1 mole CH4 Moles of CO2 = 1 moleMass of CO2 = (1 moles)(44 g/mole) = 44 g CO2 = 0.044 kg CO2

Volume CH4 at standard conditions = (1 mole CH4)(22.4 L/mole)(10-3 m3/L) = CO2 emission factor = 0.044 kg CO2 / 0.0224 m3 CH4 = 1.96 kg CO2 / m3 CH4

= 1.96x106 kg CO2 / 106 m3 CH4.The value of emission factor for CO2 on Table 8.3-6 is 1.97x106 kg CO2 / 106 m3 CH4.

74

Chapter 12. Environmental Cost Accounting

1. Estimating Tier 1 (Hidden) Environmental Costs for Cyclohexanone/Cyclohexanol Production

Costs: cyclohexanone ($0.73/lb), cyclohexanol ($0.83/lb), from cyclohexane ($0.166/lb)

Raw Materials per mole of cyclohexanone/cyclohexanol Avg. Molecular weight (MW) = 991.1 mole cyclohexane (MW= 84)2 moles oxygen (derived from air - no material acquisition cost)

Wastes generated per pound of product0.060 pounds of organics in the gaseous effluent to be treated0.2 pounds of organic aqueous wastes to sent to water treatment

Raw materials costs per pound of product are:(1.1 lbmole)(84 lb cyclohexane/lbmole)($ 0.166/lb cyclohexane) = $15.34 per 99 lb product

= $0.155 /lb product

Mass balance on gaseous effluent is:Assuming complete utilization of the oxygen in the inlet air stream, the gaseous waste stream will contain N2 in the quantity,(1 mole N2/0.2 moles O2)(2 moles O2 consumed) per mole product = 10 moles N2/mole product.

= (10)(28)/99= 2.83 lb N2/lb product

The total gaseous waste stream mass is 2.83 + 0.06 = 2.89 lb gaseous waste / lb product.

Waste treatment operating costs per pound of product are:(2.89 lb gas/lb product)(1.5x10-4 $/lb gas)+(0.2 lb water/lb product)(7.4x10-5 $/lbwater) = 4.48x10-4 $/lb product.

Waste treatment operating costs are (4.48x10-4/0.155) (100) = 0.3% of raw material costs.

2. Select a process documented in the AP-42 documents at www.epa.gov/chief/ and estimate the costs of waste treatment per pound of product.Chapter 5 Section 1 of the AP-42 documents on the AirChief 8.0 CD contains a description of a generic refinery process. Table 5.1-1 of that chapter lists a number of airborne emission factors (kg pollutant emitted/103 L refinery feed) for several pollutants, including particulate matter, SO2, CO, hydrocarbons, NOx, aldehydes, and ammonia. The total of the emission factor values for all process units and for all pollutants is 66 kg pollutant emitted/103 L refinery feed. This data will be used to estimate the cost of treating these airborne emissions for the refinery facility.

A typical refinery might be expected to process about 50,000 barrels/day of crude oil, or in terms of SI units, (50,000)(42 gal/barrel)(3.785 L/gal) = 8x106 L crude/day, and to generate 1x106 gal wastewater /

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day (Amoco/USEPA Pollution Prevention Project: Summary of Refinery Release Inventory, U.S. Dept. of commerce, National Technical Information Service, PB92228550, March 1991).

Table 12.7 in the text provides order of magnitude estimates of the operating and capital costs per pound of waste to treat the wastewater and airborne emissions. Because the emission factors from AP-42 for airborne emissions are for pollutant loading rather than for the total gaseous waste stream, as an approximation we will adopt the pollutant loading cost factors listed for wastewater in Table 12.7 instead (capital costs, $0.74/lb pollutant; operating costs, $0.25/lb pollutant). In the absence of more information airborne cost factors, this should provide an order-of-magnitude cost estimate. We can also apply the wastewater cost factors assuming total flow (capital costs, $7.4x10-4/lb wastewater flow; operating costs, $7.4x10-5 /lb wastewater flow).

Airborne Emissions:Capital costs = ($0.74/lb pollutant)(66 kg pollutant/103 L crude oil)(8x106 L crude oil/day)

= $390,000 (probably too low)Operating costs = ($0.25/lb pollutant)(66 kg pollutant/103 L crude oil)(8x106 L crude oil/day)

= $132,000/day (probably too high)

Wastewater:Capital costs = ($7.4x10-4/lb wastewater flow)(1x106 gal wastewater/day) = $740 (probably too low)Operating costs = ($7.4x10-5 /lb wastewater flow)( 1x106 gal wastewater/day)

= $74/day (probably too low).

3. Net revenue change due to process improvements

A chemical manufacturing facility buys raw material for $0.60 per pound and produces 90 million pounds per year of product, which is sold for $0.75 per pound. The process is typically run at 90% selectivity and the raw material that is not converted into product is disposed of at a cost of $0.80 per pound (by incineration). A process improvement allows the process to be run at 98% selectivity, allowing the facility to produce 98 million pounds per year of product. What is the net revenue of the facility (product sales - raw material costs - waste disposal costs) before and after the change? How much of the increased net revenue is due to increased sales of product and how much is due to decreased waste disposal costs?

Original Process: 90% selectivity.Net Revenue (per lb product) = selling price - raw matls. cost - waste treatment costs

= ($0.75/lb prod.) - (1/.9)($0.60)/lb prod. - (0.11)($0.60)/lb prod.= ($0.75/lb product - $0.667/lb product - $0.067/lb product= $0.0173/lb product

Improved Process: 98% Selectivity.Net Revenue (per lb product) = selling price - raw matls. cost - waste treatment costs

= ($0.75/lb prod.) - (1/.98)($0.60/lb feed) - (0.02)($0.60/lb waste)= ($0.75/lb product - $0.612/lb product - $0.012/lb product= $0.126/lb product

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The net change in revenue fromto this process improvement is due to a reduction in raw materials costs of $0.055/lb product and due to a reduction in waste treatment costs of $0.055/lb product. Both factors contribute equally to the enhanced revenue.

4. Estimated Health Costs for Exposure to Elevated Levels of Fine Particulate MatterReduction: 300 tons/day of fine particulate matterExpected Benefits: 17 less early deaths/yr, and 24 fewer cases of chronic bronchitis per yearCosts: $6,000,000 per early death and $300,000 per case of chronic bronchitis.

Net Cost Reduction = (17 less early deaths/yr)( $6,000,000 per early death) + (24 chronic bronchitis cases per year)($300,000 per case of chronic bronchitis) = $109.2x106/yr.

Social Costs Per Ton Emitted: ($109.2x106/yr)/[(300 tons/day)(365 days/yr)] = $997.26/ton.

In Table 3-27 of the American Institute of Chemical Engineers’ Center for Waste Reduction Technologies Total Cost Assessment Methodology, July 22, 1999, a range of health costs of cancer was shown to span the range of $2/ton to tens of millions of dollars/ton of pollutant emitted to the air. The value calculated in this problem is within that very broad range.

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Chapter 13. Life-Cycle Concepts, Product Stewardship, and Green Engineering

1. Life Cycle Inventory for Paper versus Plastic Grocery Sacks

A simplified diagram of the major life cycle stages for grocery sacks is shown below. the grocery sacks are recycled back to the use stage of the life cycle.

The inventory elements for the sack air emissions and energy consumption are shown in the table below.

a) Effects of recycle rate on the life-cycle air releases and energy for plastic and paper sacks.Functional Unit: 1 paper sack = 2 plastic sacks0% Recycle Rate: Paper sack: Air releases = (0.0516 + 0.051) oz/sack = 0.1026 oz/sack

Energy = (905 + 724) Btu/sack = 1,629 Btu/sackPlastic sack: Air releases = 2(0.0146 + 0.0045) oz/sack = 0.0382 oz/2 sacks

Energy = 2(464 + 185) Btu/sack = 1,298 Btu/2 sacks

50% Recycle Rate: (assume no emissions or energy for recycled sacks)Paper sack: Air releases = .5(0.0516 + 0.051) oz/sack = 0.0513 oz/sack

Energy = .5(905 + 724) Btu/sack = 814.5 Btu/sackPlastic sack: Air releases = (0.0146 + 0.0045) oz/sack = 0.0191 oz/2 sacks

Energy = (464 + 185) Btu/sack = 649 Btu/2 sacks

100% Recycle Rate: (assume no emissions or energy for recycled sacks)Paper sack: Air releases = 0.0(0.0516 + 0.051) oz/sack = 0.0 oz/sack

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Raw Materials Extraction

Materials Manufacture

Grocery Sack

Manufacture

Grocery Sack Use

Grocery Sack

Disposal

Recycle

Energy = 0.0(905 + 724) Btu/sack = 0.0 Btu/sackPlastic sack: Air releases = 0.0(0.0146 + 0.0045) oz/sack = 0.0 oz/2 sacks

Energy = 0.0(464 + 185) Btu/sack = 0.0 Btu/2 sacks

b) Plot air emissions and energy as a function of recycle rate

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The life-cycle air emissions and energy consumption of plastic sacks are about 40% and 80%, respectively, of those for paper sacks on an equal functional unit basis. Only at 100% recycle are the life-cycle inventories equal for these choices of grocery sacks.

c) Paper sacks probably have a higher environmental impact compared to plastic sacks. The air emissions are significantly higher and the energy consumption is slightly more. However, this conclusion is uncertain because with the information given, it is impossible to determine the specific chemical releases and therefore the specific characterization of impacts. Specific impacts of energy consumption will only be known if the fuel type is identified for each sack. The composition of the air releases for each sack is unknown. One type of sack might have significant toxic emissions compared to the other sack, which in many cases would outweigh the differences in mass emission rate.

d) Petroleum consumption comparison: assume 0% recycle ratePaper Sack: 10% of energy needed to manufacture one paper sack is fossil fuel based

= (0.1)(905+724) Btu/sack = 73.4 Btu Petroleum required = 73.4 Btu/20,000 Btu/lb petroleum = 3.67x10-3 lb petroleum

2 Plastic sacks: 0.2 lb petroleum needed for energy to create 1 lb polyethylene for sacksMass of petroleum equivalent for energy needs of 2 sacks = 2(464+185) Btu/20,000

= 6.5x10-2 lb petroleumTotal mass of petroleum for 2 sacks = (6.5x10-2 lb petroleum)(1 lb polyethylene/.2 lb petroleum) = 0.32 lb petroleum

e) Large uncertainty in the functional unit equivalence will change the conclusions of this study. For example, if instead of 2 plastic sacks being equivalent to 1 paper sack, 4 plastic sacks are actually needed, then the air emissions and energy releases of plastic sacks would be the dominant compared to paper.

2. Life-Cycle Assessment of Diapering Systems

a) Diapers per baby per week for disposable diapers – Equivalency of diapers15.8 billion disposable diapers are sold annually3,787,000 babies are born each yearchildren wear diapers for the first 30 monthsdisposable diapers are used on 85% of children.

Number of babies in diapers= (3,787,000 babies born/yr)(30 mo. in diapers/12 mo/yr) = 9,467,500

Number of babies in disposable diapers= 9,467,500 babies(0.85) = 8,047,375

Number of disposable diapers per baby per year= (15.8x109 disposable diapers)/(8,047,375 babies) = 1,963.4 disposable diapers/baby

Number of disposable diapers per baby per week= (1,963.4 disposable diapers/baby)/52 weeks = 39.3

Equivalence = (39.3 disposable diapers/baby/wk)/(68 cloth diapers/baby/wk) = 0.577

b) Complete the Table of Ratio of Impact to Home Laundered Impact

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Impact Disposablediapers

Commerciallylaundered cloth

diapers

Home launderedCloth diapers

Energy requirements (million BTU) 0.52 0.55 1.0Solid waste (cubic feet) 4.26 1.0 1.0Atmospheric emissions (lb) 0.50 0.47 1.0Waterborne wastes (lb) 0.14 0.95 1.0Water requirements (gal) 0.28 1.26 1.0

c) Recycle Percentage for Disposable Diapers

The recycle percentage needed for disposable diapers to equal the solid waste requirements of laundered cloth diapers (2.3 cubic feet) can be obtained from the best-fit equation in the figure above. Rearranging the trendline from able and solving for x, x = (17.04-2.3)/0.163 = 90.43%.

4. Streamlined Fuel Life-Cycle Assessment for Conventional Gasoline and Gasohol in Vehicle TransportationConventional gasoline is a blend of a number of crude oil refinery products with the following average weight percentages (aromatics, 28.6%; olefins, 10.8%; benzene, 1.6%; and the remainder a mixture of alkanes). Gasohol is a blend of 10% vol. ethanol with 90% vol. conventional gasoline. Gasohol is used primarily in winter to reduce CO emissions and smog precursors. The purpose of this problem is to perform a streamlined life-cycle assessment of these two fuel types. The scope of the assessment will be only CO2 emissions over the entire fuel cycle. Other impact categories will be neglected in this analysis. The form of the assessment will be mass of CO2 per gallon of fuel.

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The fuel cycle for any given transportation fuel includes several stages: feedstock production (i.e. crude oil production for conventional gasoline or corn production for ethanol), feedstock transportation and storage (T&S), fuel production, fuel transportation, distribution, and storage (TS&D), and fuel combustion during vehicle use. The inventory of energy utilization during these life-cycle stages will be taken from the document “GREET 1.5 – Transportation Fuel Cycle Model: Volume 1, Methodology, Development, Use, and Results”, Wang, M.Q., Transportation Technology R&D Center, Argonne National Laboratory (www.transportation.anl.gov).

Conventional Gasoline (CG) Inventory: Basis is 1 gallonEnergy efficiencies during various life-cycle stages for the production of CG are as follows; crude oil recovery (98%), crude T&S (99.5%), CG refining (85%), and CG TD&S (98.5%). The overall energy efficiency is (.98*.995*.85*.985)(100) = 81.6% from Table 4.43 of Wang. For each unit of energy contained in CG, an additional quantity of energy of magnitude 1/0.816 – 1 = 0.225 units is required to process and transport the feed materials. For each gallon of CG, the energy content is 120,000 Btu/gal (Table 3.3 of Wang). Per gallon of CG, the energy needed to process this fuel is (0.225)(120,000 Btu) = 27,000 Btu. In the absence of information regarding fuel type used to satisfy this energy load, natural gas will be assumed since most of the energy is consumed during refinery operations and natural gas is likely to be a major component of this energy consumption. From Table 8.3-6 of chapter 8, the emission factor for CO2 is 0.12 lb CO2/ standard ft3 of natural gas and the heating value of natural gas is 1,000 Btu/standard ft3. The total life-cycle emissions of CO2 per gallon of CG combusted in vehicle transportation is= (120,000+27,000)Btu(1 standard ft3/1,000 Btu)( 0.12 lb CO2/ standard ft3) = 17.6 lb CO2.

Gasohol Inventory: Basis is 1 gallonTable 4.21 from Wang shows that the total corn-to-ethanol energy consumption is 40,000 Btu/gallon of ethanol and that 80% is supplied by coal and 20% by natural gas. Because ethanol is 10% by volume ethanol, 4,000 Btu of energy per gallon gasohol must be added to the contribution of CG to the blend. The contribution of CG can be calculated from the result above, (0.9)(17.6 lb CO2) = 15.84 lb CO2. The coal contribution can be determined knowing the heating value of coal (12,000 Btu/lb coal) and the emission factor for coal (5,500 lb CO2/ton coal). Contribution of coal = (0.8)( 4,000 Btu)(1 lb coal/12,000 Btu)(1 ton/2,000 lb)(5,500 lb CO2/ton coal) = 0.73 lb CO2. Contribution of natural gas = (0.2)(4,000 Btu)(1 standard ft3/1,000 Btu) (0.12 lb CO2/ standard ft3) = 0.10 lb CO2. The total CO2 inventory for gasohol is = 15.8 + 0.7 + 0.1 = 16.6 lb CO2. This is 5.7% lower CO2 emissions over the fuel life cycle compared to CG.

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Chapter 14. Industrial Ecology

1. List Processes that Produce and Consume These ChemicalsHydrogen: Information Source – Kirk-Othmer Encyclopedia of Chemical Technology

Processes that Consume Hydrogen Processes that Produce Byproduct HydrogenHeteroatom (S,N,O) Removal from Fuels Catalytic Reforming in RefineriesAmmonia synthesis Steam Pyrolysis of HydrocarbonsMethanol Synthesis Methanol Plant Purge GasHydrogenations of Edible Oils Ethylene Plant Tail GasFloat glass manufacturing CO Plant H2 GasElectronic industry Chloralkali Plant H2

Chemical Synthesis Coke Oven GasNote: Refineries product much byproduct H2, but consume about the same amount of H2 within the facility in other important reactions. Co-location of ammonia plants with methanol plants is one major application of the use of H2 byproducts for industrial sysnthesis of products.

Ammonia: Information Source – Kirk-Othmer Encyclopedia of Chemical Technology

Processes that Consume Ammonia Processes that Produce Byproduct AmmoniaUrea production: CO2+2NH3NH2CO2NH4 Coking of coal: 15-20% of N in coal NH3

Nitric Acid production: oxidation of NH3 Treatment of Refinery Sour Gas: H2S+NH3

Ammonium Nitrate production: HNO3+NH3 Heteroatom (N) Removal from FuelsAmmonium Sulfate production: H2SO4+2NH3

Ammonium Phosphate production: H3PO4+3NH3

Chemical Explosives: 2,4-diamino-1,3,5-trinitrotoluene and othersFibers – Plastics:Note: Less than 1% of ammonia consumption worldwide is satisfied using byproduct ammonia. Therefore, there is little opportunity to construct waste to feed stock transformation of industry for ammonia byproducts.

2. Kalundborg EcoPark Analysis

a) Efficiency of Energy Utilization at a Power Plant4.5x106 tons coal combusted/yrCoal heating value = 10,000 Btu/lb coalEnergy released upon coal combustion

= (4.5x106 tons coal/yr)(2,000 lb/ton)(10,000 Btu/lb coal) = 9.0x1013 Btu/yr

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Power plant rating = 1,500 megawatts = 5.12x109 Btu/hr = 4.48x1013 Btu/yrStatoil refinery steam use

= (140,000 tons steam/year)(2,000 lb/ton)(1,000 Btu/lb) = 2.8x1011 Btu/yrNovo Nordisk steam use

= (215,000 tons steam/year)(2,000 lb/ton)(1,000 Btu/lb) = 4.3x1011 Btu/yrDistrict heating steam use

= (225,000 tons steam/year)(2,000 lb/ton)(1,000 Btu/lb) = 4.5x1011 Btu/yr

Fraction of coal energy:to power plant (4.48x1013 Btu/yr)/( 9.0x1013 Btu/yr) = 0.498to Statoil refinery (2.8x1011 Btu/yr)/( 9.0x1013 Btu/yr) = 0.0031to Novo Nordisk (4.3x1011 Btu/yr)/( 9.0x1013 Btu/yr) = 0.0048to district heating (4.5x1011 Btu/yr)/( 9.0x1013 Btu/yr) = 0.005

Total rate of energy utilization (fraction utilized)= 0.498 + 0.0031 + 0.0048 + 0.005 = 0.5109

b) Variability in Energy DemandDaily variability:

Electricity demand in largest in the morning, during late afternoon for air conditioning, and in the evening. Electricity demand will be highest in summer for air conditioning.

Refinery demand will be greatest in the summer when more automobile travel occurs and in winter months when residential heating needs are greatest.

Novo Nordisk demand is likely higher in the winter months when seasonal illnesses are more common (colds).

District heating demand is more in the winter months.

Response by the power plant to variability in demand:The variability in demand for electricity is the most challenging problem to deal with. Given that the demands for waste heat from the power plant is a very small fraction of total heat available from coal combustion means that meeting the changing demands of these energy sinks will be easy to achieve.

c) Value of Residential Heating SinkDistrict heating energy demand

= 4.5x1011 Btu/yrValue of waste steam for district heating

= (4.5x1011 Btu/yr)/(1.5x105 Btu/gal heating oil)($2/gal heating oil) = $6x106/yr

3. Co-location Opportunities for Industry from the TRI Data

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In Michigan, there are a number of industrial facilities that are located in Rouge River. The following list shows several of these facilities that could, with further consideration, be connected to share materials and energy.

1. AMOCO OIL RIVER ROUGE TERMINAL, 205 MARION ST., RIVER ROUGE, WAYNE COUNTY, MI, 48218

2. DETROIT EDISON - RIVER ROUGE POWER PLANT, 1 BELANGER PARK DR., WAYNE COUNTY, MI, 48218

3. EQUILON LUBRICANTS CO. RIVER ROUGE PLANT, 245 MARION AVE., RIVER ROUGE, WAYNE COUNTY, MI, 48218

4. FRITZ PRODS., 255 MARION, RIVER ROUGE, WAYNE COUNTY, MI, 48218

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