homeworl. desicrn problems - p2infohouse.org · new engineers with insights into industrial...

Homeworl. Desicrn problems

les

C E N T E R , ; F O R -.-.;.-,2c"y-& J&*++.:; ;IAA4 i ' r - ,

1 $11 f f '

2& :.--II_ z-.ib;ILd - h - 1 W%rn+Sbl,'i q <. -?--,.?.,*-%Y-- ,- - - T ' E 2 C 1 4 ~ o L o ~ r E S

ACKNOWLEDGMENTS

One of the primary driving forces behind the creation of this collection of honlework and design problems was the American Institute for Pollution Prevention (AIPP). The AIPP was founded in 1989 to assist the U.S. Environmental Protection Agency in developing and implementing pollution prevention. The AIPP is composed of a group of experts in pollution prevention, nominated by professional societies and tsade organizations. Members of the Institute strongly suggested that the education of all engineers, not just a small group of environmental engineers, needs to incorporate pollution prevention concepts. This book represents the AIPP's initial efforts to address this problem in engineering education. Members of the Institute contributed case studies that were crafted into problems and provided valuable initial reviews of manuscript drafts.

Once the problem development was underway, several other organizations contributed substantially to the publication and distribution of the volume. The American Institute of Chemical Engineers' Center for Waste Reduction Technologies (CWRT), assisted by rep- resentatives from AIChE's Environn~ental Division, provided final review of the publication. CWRT also provided fi~nding for publishing and took responsibility for widely distributing the manual to chemical engineering departments of colleges and universities throughout the U.S.

The Ralph M. Parsons Foundation and the University of California Toxic Substances Research and Teaching Progran~ also provided funding.

While we are grateful for the advice and assistance of these organizations, the final responsibility for errors or on~issions in this volume rests with the authors. This collection of pollution prevention problems is far fsom comprehensive, and much work remains to be done in producing curricular materials focusing on pollution prevention.

We close by expressing our gratitude to a few of the Inany individuals who contributed to this manuscript. Phyllis Gilbert translated scribbled notes and garbled text files into a usable format. Dave Stephan provided access to key resources in the Environmental Protection Agency. Jerry Kotas, Priscilla Flattery and Terry Foecke publicized and promoted the work, and the entire American Institute for Pollution Prevention provided intellectual and moral support.

David T.' Allen N. Bakshani

Kirsten Sinclair Rosselot Los Angeles, California 1992

FOREWORD

One of the greatest challenges facing business today is maintaining industrial com- petitiveness while ensuring a healthy environment. To remain competitive in environmentally- driven world markets, American industry must develop new and improved process technologies designed to prevent pollutants at the source. We can no longer rely on end-of-the-pipe pollution control solutions.

Given the call for increasingly denlanding standards of environlnental improvement, a new generation of engineers must learn to consider environmental costs in their decisions. A key to this effort is ensuring that the environmental education of this new generation of "pollution preve~ltion" engineers begins in the classroom, incorporating cost-effective pollution prevention technologies and methodologies into standard engineering curricula. Although the problems to be solved are multidisciplinary, at their core, many embody chemical and materials engineering principles.

To meet this need, the Center for Waste Reduction Technologies and the American Institute for Pollution Prevention are introducing Pollcttio~ Preve/~tio~z: Honzevvork arid Desigrl Probleins for Engineering Curriccrln. This manual was designed to be integrated into existing courses leading to chemical and other undergraduate engineering degrees, ranging from sophomore-level material balances to senior-level process design. The problems are directed at increasing students' awareness of, interest in, and knowledge of waste reduction concepts that must become a permanent feature of industrial practice.

Pollutioll prevention, figured in as an integral part of engineering solutions, can provide new engineers with insights into industrial situations they are likely to encounter. Pollcrtioiz Prevention: Honzevvork nrzd Design Prol?lenzs for Eilgineering C~trriculu may also be used as a reference for graduates beginning industrial careers, and as industrial inhouse courses for new engineering einployees to stimulate creative problem solving abilities.

Finally, through use of this material, we hope to instill increased recognition and acceptance of the professional and ethical responsibilities which engineers must have as they develop the next generation of environmentally-compatible, cost-effective process technologies.

Lawrence L. Ross American Institute of Chemical Engineers Center for Waste Reduction Technologies

CONTENTS

INTRODUCTION ......................................................................................................... 1

PROBLEMS

Life Cycle Analyses

1 . Paper or plastic? A life cycle inventory comparing unbleached paper grocery sacks and polyethylene grocery sacks ........................................ 12

2 . A life cycle inventory for soft drink containers ........................................ 24

3 . A life cycle inventory for polystyrene and paper containers .................. 31

4 . A life cycle inventory for three diapering systems .................................. 36

Identifying and Prioritizing Pollutants from Industrial Sites

5 . Estimating and reducing fugitive emissions ............................................. 41

6 . Estimating and reducing secondary emissions ........................................ 50

................... 7 . Estimating and reducing emissions from an API separator 54

8 . Prioritizing pollution prevention options ................................................... 68

Selecting Environmentally Compatible Materials

9 . The use of chlorinated solvents: implications for global warming. stratospheric ozone depletion and smog formation ................................ 82

10 . Choosing a degreasing solvent: an environmental dilemma ............... 92

11 . Thermodynamic constraints in the reprocessing of commingled plastics 96

........... 12 . Additives for enhancing the miscibility of commingled plastics 105

Design of Unit Operations for Minimizing Waste

13 . Minimizingsolventemissionsfromvapordegreasers ............................ 107

14 . The design of distillation column reboilers for pollution prevention .... 112

15 . Reducing wastes during batch processing changeovers ........................ 118

16 . Reaction pathway optimization for waste reduction ............................. 121

Economics of Pollution Prevention

.................. 17 . The effect of future liability costs on return on investment 125

..... 18 . Economic analysis of a pollution prevention process modification 128

...................................................... 19 . The economics of newsprint recycling 3 2

Process Flowsheeting for Minimization of Waste

20 . Mass exchange networks: equilibrium. operating and load lines ....... 137

21 . Mass exchange networks: composition interval diagrams and composite load lines .................................................................................... 143

22 . Mass exchange networks: pairing the rich and lean streams and determining the pinch ................................................................................. 149

Pollution Prevention: Homework and Design Problems for Engineering Curricula

INTRODUCTION

The management of hazardous materials is one of the most serious problems confronting our society. At risk are public health, environmental quality and our standard of living. Engineers will play a crucial role in meeting the challenges of hazardous materials management, but currently, most engineers receive very little background on environmental issues in their formal education. One mechanism for introducing environmental issues into engineering curricula is to incorporate hoinework and design proble~ns on environmental topics into existing courses. This book is a collection of such ho~nework and design problems.

The environmental issues addressed in this collection of problems focus on what has come to be called pollution prevention (also referred to by the related terms of waste reduction, source reduction and waste minimization). Briefly stated, pollution prevention means avoiding the generation of pollutants, rather than controlling them as they escape from the end of the pipe, the top of a stack or into the trash can. While the concept of pollution prevention is generally understood, a precise definition of what is and what is not pollutio~~ prevention senlains a matter of serious debate. As this book is written in late 1991 and early 1992, the clearest description of pollution prevention comes from the waste management hierarchy outlined in the federal Pollution Prevention Act of 1990. The act declared that:

pollution should be prevented or reduced at the source wherever feasible; pollution that cannot be prevented should be recycled in an environ~nentally safe manner whenever feasible; pollution that cannot be prevented or recycled should be treated in an environn~entally safe manner whenever feasible; and disposal or other release into the environment should be enlployed only as a last resort and should be conducted in an environmentally safe manner.

The strictest definitions of pollution prevention encolnpass only the first element of this hierarchy--preventing the generation of pollutants. Morc open definitions of pollution prevention include the first two elements of the hierarchy--preventing generation and recycling. Regardless of the definition, however, the engineering tools required for pollution prevention remain the same. The wastes must be identified and their flow rates must be quantified. Since i t is thermodynamically infeasible to completely eliminate all wastes, it is essential that the polluta~lts be prioritized and prevention targets established. Once the targets are in place, the tools required for pollution prevention are the tools of chemical engineering: process flowsheeting, reaction engineering, separation processes, equipment design, and product design. The problenls included in this book illustrate both waste stream targeting issues and the elements of design for minimizing pollutants. Before proceeding to the problems, however, it is instructive to step back and examine the nature of our waste problem. After all, if we are to recycle, reduce or eliminate wastes, we must know what the wastes are.

The Waste Problem Hundreds of millions of tons of wastes are generated annually in the United States. The

wastes are generally grouped into broad categories, including municipal solid waste and industrial hazardous waste. In this section, we will give a brief overview of some of these waste categories. A Inore comprehensive inventory is available elsewhere (Allen and Jain, 1992).


Throughout this description, we will use terms that have a precise legal meaning (e.g., "hazardous waste"). We will not dwell on the detailed and, at times, arcane definitions of these terms. Rather, we will attempt to provide the reader with a picture of the nature and extent of the waste problem; readers interested in the evolving legal issues associated with hazardous materials should consult other references, soine of which are cited at the end of this brief introduction (Environmental Law Handbook, 1989).

Municipal Solid Waste In 1989, Anlericans generated mire than 4 pounds of garbage per person, per day (EPA,

1990). This translates to almost 200 million tons of waste per year, enough garbage to fill a line of trash trucks that encircles the earth three times. As shown in Figure 1 , the amount of trash generated i n the United States has doubled in the last 30 years, a rate of increase that is slightly greater than the rate of population increase. Although the volume of municipal solid waste has increased, the composition of the waste has remained relatively constant. Paper, yard wastes, food wastes, plastics, iiietal and glass are the major components sf the waste stream. Mechanisms for handling these wastes have also remained fairly constant for the past several decades. As shown in Figure 2, about 70% of municipal wastes are sent to landfills; about 15% of the material is incinerated and about 15% is recycled. The heavy dependence on landfills, coupled with increasing public resistance to incineration and new landfill construction, has created a municipal solid waste crisis in many parts of the United States.

Hazardous Waste Wastes are defined as liazardous if they exhibit high levels of ignitability, corrosivity,

reactivity or toxicity. The Environinental Protection Agency may also list a waste as hazardous based on the following factors:

a General toxicity of the waste • Degree of toxicity of the constituents of the waste a Concentration of the hazardous constituents or the potential for these constituents to

migrate fro111 the waste into the environment + a Persistence and potential for bioaccumulation of the waste once it is in the environment

• Susceptibility of the waste to improper management a Quantity of the waste generated • Whether the waste has a past history of dainage to the environment or to human health.

The quantity of hazardous waste generated a~i~iually in the United States is between 250 and 750 inillion tons, depending on the definition of hazardous waste (EPA, 1988). Figure 3 indicates which industrial sectors generate these wastes. The majority of hazardous waste is generated by the chemical manufacturing industries and by petroleunl and coal processing. The generation is not broadly distributed througliout these industries, but rather, as Figure 4 shows, a few dozen facilities account for most of the waste generation. While i t is striking that a Sew dozen inanufacturing facilities generate [nost of the country's hazardous waste, these waste generation rates nlust be viewed in context. The 250-750 lnillion tons of hazardous waste that are generated annually are over 90% wastewater, as shown in Figure 5. The hazardous constituents in that wastewater make up only a few percent of the total wastewater mass. Thus, the rate of generation of hazardous constituents in the waste is probably on the order of 10-100 million tons


YEAR

Figure 1. Municipal Solid Waste Generation in the United States (EPA, 1990)

recycle

combustion

landfill/other

1960 1965 1970 1975 1980 1985 1988

year

Figure 2. Municipal Solid Waste Management in the United States (EPA, 1990)


Chemical Products

Petroleum/Coal

Electrical/Gas/Sanitary

Primary Metals

Machinery

Other

Figure 3. Hazardous Waste Generation in 1986, Classified by Industry Sector (EPA, 1988)

top 50 units

El top 40 units

E l top 30 units

E l top 20 unlts

El top 10 units

Figure 4. Percentages of Hazardous Waste Managed in the 50 Largest Facilities in 1986 (EPA, 1988)


per year. Compare this to the 300+ million tons of co~n~nodity che~nicals produced annually and the 1000 million tons of petroleum refined annually (Chemical and Engineering News, 199 1 ) and it becomes clear that the chemical and petroleu~n industries are, by some measures, quite efficient. The inass of hazardous constituents in waste is probably less than 570 of cheislical production. The engineering challenges associated with substantially reducing waste generation will be substantial.

The Engineering Challenge Substantially reducing the generation of municipal solid waste, hazardous waste and other

pollutants will require the redesign of Illany of our products and chemical processes. Designing environmentally compatible processes and products will be a challenging new frontier for chemical engineers. As stated in the National Research Council report on Frontiers in Chemical Engineering (NRC, 1988)

"Chemical plants have been designed in the past principally to maximize reliability, product quality and profitability. Such issues as chronic emissions, waste disposal and process safety have often been treated as secondary factors. It has becotne clear, however, that these considerations are as important as the others and nus st be addressed during the earliest design stages of the plant."

Incorporating environmental constraints into all plant design stages will require a new generation of design procedures, which allow the systematic rninitnization of wastes. The primary elements of these new design procedures will be

o identificatiolz crr~d l~rioritizntioli of waste st~+enms jkoni yroce,s,ses and prorlucts. We must consider the life cycle analysis of products, examining all of the energy requirements and waste generation from raw material acquisition to postconsumer disposal. Tools for identifying and prioritizing all of the waste streams from industrial tnanufacturing sites will also be required.

0 procedures ,for selecting er?viror?nzel~tall~ cor71pcrtiOle ~nrrterials. We nus st have tools for predicting not only the properties relevant to product or process function but also the properties influencing environ~nental impact.

0 clesigri of ~i17it ol?eratiorzs tlznt mi1?ir7zize waste. This is a variation on traditional chenlical engineering design procedures.

o econonzics of y o l l ~ ~ t i o l ~ prevention. Quantification of avoided costs is essential in the justification of pollution prevention projects; yet avoided costs are rarely considered in engineering econo~nic analyses.

0 171-ocess f lo~vslieeti~~g for 17zinimizing wastes. Just as process flowsheets can be for~nulated to maximize energy efficiency using well- established design methods, process flowsheets can be formulated to maximize mass efficiency.


The collection of proble~ns presented in this book illustrates these elelnents of design for preventing pollution. As shown below, several problerns are available in each of these topical areas.

IDENTIFICATION AND PRIORITIZATION OF WASTE STREAMS FROM PRODUCTS AND PROCESSES

Life Cycle Analyses Problem 1: Paper or Plastic? A Life Cycle Inventory Comparing Unbleached

Paper Grocery Sacks and Polyethylene Grocery Sacks Problem 2: A Life Cycle Inventory for Soft Drink Containers Probleln 3: A Life Cycle Inventory for Polystyrene and Paper Containers Problem 4: A Life Cycle Inventory for Three Diapering Systems

Identifying Waste Streams from Processes Probleln 5 : Estimating and Reducing Fugitive E~nissions Proble~n 6: Estimating and Reducing Secondary Emissions Problem 7: Estimating and Reducing Emissions from an API Separator

Prioritizing Pollution Prevention Options Probleln 8: Prioritizing Pollution Prevention Options

SELECTING ENVIRONMENTALLY COMPATIBLE MATERIALS Problem 9: The Use of Chlorinated Solvents: Implications for Global

Warming, Stratospheric Ozone Depletion and Smog Formation Problem 10: Choosing a Degreasing Solvent: An Environmental Dilemma Problem 11: Thermodynamic Constraints in the Reprocessing of Commingled

Plastics Problem 12: Additives for Enhancing the Miscibility of Colnmingled Plastics

DESIGN OF UNIT OPERATIONS FOR POLLUTION PREVENTION Problem 13: Minimizing Solvent Emissions from Vapor Degreasers Problem 14: The Design of Distillation Column Reboilers for Pollution

Prevention Proble~n 15: Reducing Wastes During Batch Processing Changeovers Problem 16: Reaction Pathway Optimization for Waste Reduction

ECONOMICS OF POLLUTION PREVENTION Problem 17: The Effect of Future Liability Costs on Return on Investment Problem 18: Economic Analysis of a Pollution Prevention Process Modification Probleln 19: The Economics of Newsprint Recycling

PROCESS FLOWSHEETING FOR POLLUTION PREVENTION Probleln 20: Mass Exchange Networks: Equilibrium, Operating and Load Lines Problenl 21: Mass Exchange Networks: Composition Interval Diagrams and

Composite Load Lines


Problem 22: Mass Exchange Networks: Pairing the Rich and Lean Streams and Determining the Pinch

The problems can also be organized by chemical engineering subject. Listed below are some of the core courses normally found in chemical engineering curricula, followed by a list of problems suitable for use in each course.

MASS AND ENERGY BALANCES Problem 1 : Con~parison of environmental effects of paper and plastic sacks; involves

mass and energy balances with recycle streams. Problem 2: Comparison of environmental effects of soft drink containers; involves

mass balances with recycle streams, focuses on choosing a basis for mass balance comparisons; completion of Proble~n 1 is a prerequisite.

Problem 3: Comparison of environ~nental effects of drinking cups; involves mass and energy balances, focuses on data quality; completion of I'roblen~ 1 is a prerequisite.

Problem 4: Comparison of environmental effects of diapering systems; involves mass and energy balances with recycle streams, focuses on tradeoffs in environmental impacts; con~pletion of Problem 1 is a prerequisite.

Problenl 15: Analysis of plan to reduce resin pan waste; requires the student to perform mass balances and simple economic calculations; appropriate for students who have not taken a course on process economics.

THERMODYNAMICS Problem 1 1: Prediction of plastic miscibility in binary mixtures; appropriate for a course

on chemical thermodynamics or polymer processing, requires an understanding of enthalpies and entropies of mixing. Solution should be performed with the aid of a colnputerized spreadsheet.

Problem 12: Prediction of plastic miscibility in ternary mixtures; appropriate for a course on chemical thermodynamics or polymer processing, requires an understanding of enthalpies and entropies of mixing; completion of Problem 1 1 is a prerequisite.

TRANSPORT PHENOMENA Problem 6: Emission estimate for a wastewater junction box; requires calculation of

overall mass transfer coefficient from individual mass transfer resistances, also requires a mass balance on the wastewater junction box.

Problem 7: Emission estimates for an American Petroleum Institute separator; involves the use of correlations and requires the student to develop an expression for the mass flux of an evaporating liquid.

Problem 13: Emission estimate and methods for reduction of emissions fro111 a vapor degreaser; requires the student to model the mass flux of an evaporating liquid.

Problem 14: Choice of appropriate heat exchanger design; requires students to calculate heat transfer surface areas, reboiler costs, shutdown costs, and payback periods; appropriate for students who have not taken a course on process economics.


REACTION ENGINEERING Problen~ 9: Comparison of the impacts of chlorinated solvents on the environment;

involves an understanding of residence times, energy balances, and chemical reaction rates.

Problem 10: Choice of degreasing solvent; involves an understanding of residence times, energy balances, and chemical reaction rates; completion of Problem 9 is a prerequisite.

Problem 16: Optimal reaction rates for waste reduction; requires students to find economically optimum reaction rates for a three-step pathway where a hazardous waste is generated; appropriate for students who have not taken a course on process economics.

PROCESS ECONOMICS Problem 17: Economic analysis of the effects of future liability; requires the student to

calculate the present value of a pollution prevention project. Problem 18: Economic analysis of a process modification; requires the student to create

a cash flow diagram, calculate present value, and iterate to find the internal rate of return for a process modification.

Problem 19: Analysis of newspaper recycling; requires students to perform a mass balance and economic analysis; understanding of return on investment necessary.

PROCESS DESIGN Problem 5 : Emission estimates from typical process units, such as valves, flanges,

compressors, and pumps; requires the student to use correlations. Problem 8: Prioritization of waste streams from a petroleum refinery; presents

qualitative and semiquantitative prioritization procedures; could be presented in a design course along with Hazard and Operability Analyses.

Problems 20-22: Mass exchange network design; presents mass exchange network synthesis as a method for ininilnizing wastes; should be presented after students have been exposed to heat exchange network synthesis; Problems 20-22 should be assigned as a unit.

A summary of the chemical engineering and pollution prevention content of the problems is given in the matrix of Table 4. The columns of the matrix are the principles of pollution prevention. The rows of the matrix are chemical engineering courses in which the problems could be used. Problem numbers are listed as the elements of the matrix. Thus, the first column and first row of the matrix tells us that problems 1, 2, 3 and 4 illustrate life cycle analyses and could be used in a mass and energy balances course.

Each problem contains background material, a problein statement, questions for discussion, suggestions for further reading, and a solution. The problen~s have been tested in classes at UCLA and other universities. It is our hope that these problelns will catalyze student and faculty interest in preventing pollution.


References

Allen, D.T., and Jain, R., eds., "Special Issue on National Hazardous Waste Databases," Hclzardous Waste and Hc~zarclo~ls Materials, Vol. 9, No. 1, pp. 1 - 1 1 1, 1992.

"Facts and Figures for the Chemical Industry," Clzer~zical and Engineering News, 69(25) 30-37, June 24, 199 1.

Enviror211zerztal Law Handbook, Government Institutes, Inc., Rockville, MD, 1989.

U.S. Environmental Protection Agency, " 1986 National Survey of Hazardous Waste Treatment, Storage, Disposal and Recycling Facilities," EPA/530/sw-881035, 1988.

U.S. Environmental Protection Agency, "Characterization of Municipal Solid Waste in the United States: 1990 Update," EPA 530-sw-90-042, June, 1990.

National Research Council, Frontiers in Clze~~zicnl Engineering, National Academy Press, Washington, D.C., 1988.

Resource Conservation and Recovery Act, U.S.C., 6901, et seq., 1982, and Supplelnent 111, 1985.


PROBLEM 1

Paper or Plastic? A Life Cycle Inventory Comparing Unbleached Paper Grocery Sacks and Polyethylene Grocery Sacks

Chemical Engineering Topics: mass and energy balances, recycle streams Pollution Prevention Concepts: life cycle inventories, product substitution

BACKGROUND At the supermarket checkstand, consumers are asked to choose whether their purchases

should be placed in unbleached paper grocery sacks or in polyethylene grocery sacks. Many consumers make their choice based on their perception of the relative environn~ental impacts of these two products. This mass balance problem will quantitatively examine the relative environmental impacts of paper and plastic grocery sacks by comparing the energy used and wastes generated in the production, use, and disposal of these products.

When making comparisons between different products that have the same use, it is important to have a consistent framework for keeping track of the flows of material, energy and waste. One such framework is shown in Figure 1- 1. This framework for collecting data on material, energy and waste flows from raw material acquisition to product disposal is commonly referred to as a life cycle inventory. Figures 1-2 and 1-3 give graphic representation to the life cycle stages of grocery sacks. For paper sacks, there are steps involving timber harvesting, pulping, paper making, product use and waste disposal. For polyethylene sacks there are steps involving petroleum extraction, ethylene manufacture, ethylene polymerization, polymer processing, product use and waste disposal. In most of these steps, energy is required and wastes are generated. The only exception may be the product use portion of the life cycle which results in little energy use or emissions. For some products, however, this part of the life cycle is significant. Paint, for example, has air emissions while drying, and energy use may be associated with its application. Conlparing Figure 1 - 1, the general framework for life cycle inventories, and Figures 1-2 and 1-3 reveals some small differences. Most notably the product reuse and product remanufacture loops present in Figure 1- 1 are missing in Figures 1-2 and 1-3. They are ignored because almost all recycled grocery sacks are returned to the material manufacture stage. Also note that the only wastes considered in this problem are air emissions, so solid waste and wastewater do not appear in Figures 1-2 and 1-3.

Having introduced the general concept of a life cycle inventory, consider the use of the inventory in comparing paper and plastic grocery sacks. Life cycle waste and emission inventories of paper and polyethylene grocery sacks have resulted in the data given in Table 1- 1 , and these data will serve as a basis for co~nparing these two products. Before a quantitative comparison between the two products can be made, differences in product use must be considered. Although both types of sacks are designed to have a capacity of 116 barrel, fewer groceries are generally placed in polyethylene sacks than in paper sacks, even if the practice of double bagging paper sacks (one sack inside the other), used in some stores, is taken into account. There is no general agreement on the number of polyethylene grocery sacks needed to hold the volume of groceries usually held by a paper sack. Reported values range from 1.2 to


natural resources

Energy + Raw Material

Air Emissions, Solid Acquisition Waste, Wastewater

I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Energy

Material Manufacture Waste, Wastewater

. . . . . . . . . . . . . . . . . . . . . . . . . : a. -.

0 ' : A: - 8:

Product Manufacture 2 : CI). 2: 2: .2f C(

rr. a. 1. U

E= ' 2; . . . . . . . . . . . . . . . . . . . . 2:

2: a: - C1' : 0.

0 . 3' 2: a: 2 U. 0. a:

3 . -a. 2: P,

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Air Emissions, Solid Product Disposal

Waste, Wastewater

Figure 1-1. An Analysis Template for the Life Cycle of Manufactured Goods


Figure 1-2. Polyethylene Grocery Sack Life Cycle


Manufacture

I I=..,.--.. I Product Dimsal I

Figure 1-3. Paper Grocery Sack Life Cycle


3. In this problem a value of 2.0 polyethylene grocery sacks required to replace a paper grocery sack will be used.

Table 1 - 1 . Air Emissions and Energy Require~llents for Polyethylene and Paper Grocery Sacks (Franklin Associates, Ltd., 1990)

Note: These data are based on past practices and may not be current.

PROBLEM STATEMENT a) Using the data in Table 1 - 1 and Figure 1-4, determine the amount of energy required and

the quantity of air pollutants released per 1000 lb of production of polyethylene sacks. Also determine the amount of energy required and the quantity of air pollutants released for the quantity of paper sacks capable of carrying the same volume of groceries as the 1000 lb of polyethylene sacks. Both the air emissions and the energy requirements are functions of the recycle rate, so perform your calculations at three recycle rates: 0%, 50% and 100% recycled. Note that a 50% recycle rate indicates that half of the sacks are disposed of and the other half recycled after the product use stage of their life cycle. Polyethylene sacks have a mass of 0.2632 oz each; paper sacks have a mass of 2.144 oz each.

b) Plot the energy requirements calculated in part a) as a function of recycle rate on the same axes for both types of sacks. Do the same for the air emissions found in part a). Con~pare the energy requirements and air emissions of the paper sacks with those of the polyethyleile sacks at different recycle rates.

c) Discuss the relative environmelltal impacts of the two products. Do the results of part a) allow for a co~nprehensive cornparisoil? Note that the raw material for paper (i.e., wood) is a renewable resource. In atte~npting to answer this question, note that ill part b) the total quantities of air emissions were compared. As shown in Table 1-2, the qualitative characteristics of the air emissions due to paper sacks are different than those due to plastic sacks. In your discussion you should consider whether or not i t is valid to directly compare the mass of at~nospheric e~llissions due to the two products. You are not expected to comment on the environ~nental effects of specific chemicals.

Life Cycle Stages

Materials Manufacture plus Product Manufacture plus Prodi~ct Use

Raw Materials Acquisition plus Product Disposal

Energy Required, B tulsack

Paper

905

724

Air Emissions, ozlsack

Poly- ethylene

464

185

Paper

0.05 16

0.05 10

Poly- ethylene

0.0 146

0.0045


Table 1-2. Profile of Atmospheric Emissions for Paper and Plastic Grocery Sacks (Franklin Associates, Ltd., 1990)

Note: These data are based on past practices and ]nay not be current.

Pollutant Category

Particulates Nitrogen Oxicles Hydroc a1 .b ons Sulfilr Oxides Carbon Monoxide Aldehydes Other Organics Odorous Sulfilr Am~lionia Hydrogen Fluoride Lead Mercury Chlorine

d) The material and energy requirements of the polyethylene sacks are primarily satisfied using petroleum, a non-renewable resource. In contrast, the paper sacks rely on petroleum only to a limited extent and only for generating a small fraction of the manufacturing energy requiren~ents (Hocking, 1991). Most of the energy requirements of pulp and paper manufacturing are met by burning wood waste. Compare the amount of petroleum required for the manufacture of two polyethylene sacks to the amount of petroleum necessary to provide 10% of the energy required in the manufacture of one paper sack. Assu~ne 0% recycle, and that 1.2 Ib of petroleum is required to manufacture 1 Ib of polyethylene. The higher heating value of petroleuln is 20,000 BTUJlb.

QUESTIONS FOR DISCUSSION 1) Com~nent on the concept of "100% recycle." Is 100% recycle really feasible for the

products being analyzed or for any consumer products? Consider at least two points in your analysis: i) contaminants on or within the sacks and ii) mechanical wear and tear of the grocery sacks.

2) In this proble~n you have considered only two choices for delivery of groceries: paper and plastic sacks. Can yo11 suggest other alternatives?

Atmospheric Emissions Per 10,000 Sacks (Ib)

REFERENCES AND SUGGESTIONS FOR FURTHER READING Federal Office of the Environment, "Comparison of the Effects on the Environment fro111 Polyethylene and Paper Carrier Bags," Bis~llarckplatz 1, 1000 Berlin 33, RFG, August 1988, English version.

Polyethyler~e Sacks

0% Recycling 100% Recycling

0.8 0.8 2. I 1.7 5.8 3.2 2.6 2.7 0.7 0.G 0.0 0.0 0.0 0.0

0.0 0.0

0.0 0.0

Franklin Associates, Ltd., Reso~rrce n l d El~vironl~zental Profile Analysis of Polyethyle~ie and Unblencl~erl Plrper Grocely Sacks (report prepared for the Council for Solid Waste Solutions), Prairie Village, Kansas, June 1990.

Paper Sacks

0% Recycling 100% Recycling

24.6 2.8 9.2 8.0 4.9 3.9 13.6 10.6 7.0 6.5 0.1 0.1 0.3 0.2 4.5 0.0 0.0 0.0

0.0 0.0


Hocking, M. B., "Paper versus Polystyrene," Science, Vol. 25 1, pp. 504-505, February 199 1 (see also letters commenting on this paper, Science, Vol. 25, pp. 1361-1363, June 1991).

Riggle, D., "Recycling Plastic Grocery Bags," BioCycle, pp. 40-41, June 1990.

Society of Environmental Toxicology and Chemistry, "A Technical Framework for Life-Cycle Assessments," SETAC Foundation for Environmental Education, Inc., 1101 14th Street, NW, Suite 1 100, Washington, D.C. 20005, 1991.


SOLUTION TO PROBLEM I a) The energy requirements and total mass of atmospheric pollutants for both paper and

polyethylene (PE) grocery sacks calculated from Table 1-1 are listed in the table below. All values pertaining to PE sacks are based on 1000 lb of product or 60,800 PE sacks. Values for the paper sacks are based on 60,80012 = 30,400 sacks, the number required to hold an equivalent volume of groceries.

When a sack is recycled, the raw material acquisition and product disposal stages of the life cycle do not occur.

Sample calculations: Air emissions for paper sacks at 0% recycle:

30,400 sacks(0.0516+0.0510)

Air emissions for paper sacks

oz 0.5 (30,400 sacks)(0.05 16t0.05 10) - ! + 0.5(30,400 sacks)0.05 16 - sack sack

Air emissions for paper sacks at 0% recycle:

30,400 sacks 0.0516 - - = 98.0 lb [ s ~ r k ) ,QbOz

Table S 1 - 1. Energy Requirements and Atmospheric Emissions for an Equivalent Capacity of Paper and PE Sacks

b) The data from part a) are plotted in Figures S 1- 1 and S 1-2. These figures show the effect of recycle rate on energy requirements and atmospheric pollutants. At 0% recycle, PE sacks (on an equal use basis, 2 PE sacks per paper sack) require approxilnately 20% less energy than paper sacks. However, as the recycle rate increases, this difference in energy requirement decreases linearly. At recycle rates above 80% there appears to be no significant difference in energy requirements for PE and paper sacks. Therefore, on the basis of energy alone, paper sacks would be considered competitive with PE sacks at high (>80%) recycle rates.

Energy Required, MM Btu

Atmospheric Pollutants, Ibs

The plot for total atmospheric emissions shows a similar declining difference between the

0% Recycle

Polyethylene

60,800 sacks

39.5

72.6

Paper

30,400 sacks

49.5

195

50% Recycle

Polyethylene

60,800 sacks

33.8

64.0

100% Recycle

Paper

30,400 sacks

38.5

146

Polyethyle~ie

60,800 sacks

28.2

55.5

Paper

30,400 sacks

27.5

98.0


(basis: 60,790 plastic sacks, 30,395 paper sacks)

25 I I I I

0 2 0 4 0 6 0 8 0 100

recycle rate, %

Figure S1-1. Energy Requirements for Grocery Sacks

recycle rate, %

Figure S1-2. Atmospheric Emissions for Grocery Sacks


products with increasing recycle rates. At 0% recycle, total atinospheric emissions are 60-70% lower for PE sacks; this difference gradually declines to 40% at 100% recycle.

In making these comparisons, it is important to recognize the uncertainties inherent in the analysis. One glaring uncertainty apparent in the calculations of this problem is the issue of how many plastic sacks must be used to match the function of one paper sack. In this problem, two plastic sacks were assumed to be equivalent to one paper sack. If a value of 1 or 3 were used, or if either sack had a subsequent use (e.g., using the paper sack as a trash liner), then the comparison would be quite different.

c) PE sacks generate lower amounts of atmospheric emissions at all recycle rates--a fact that may be significant if there are no qualitative differences between the emissions. However, the emission composition data of Table 1-2 show significant differences in the types of emissions assigned to PE and paper. 111 the case of paper sacks, the amount of particulates, nitrogen oxides and sulfur oxides is higher than for PE. As might be expected, higher levels of hydrocarbon emission are assigned to PE sacks. These hydrocarbons are also very likely to be qualitatively different from the hydrocarbon emissions generated by paper sack production. It would be difficult to assess the respective environmental impacts of the hydrocarbon emissions without a I I I U C ~ more detailed description of the emissions. Also, lack of emission data from other sources within the life cycle (i.e., incineration and enlissions from landfills) makes the comparison of PE and paper sacks incomplete and any comprehensive coinparison difficult.

d) Petroleum requirements of polyethylene sacks:

3 9 . 5 ~ 1 0 "TU 1 lb petroleum 1 lb petroleu~n Fuel :

60,790 sacks 2x 10 9 T U sack

Material: 0.2632 oz [ sack lg) 1.2) = 0.020 lb petroleuin

sack

Total = 0.052 Ib petroleun~/sack

Total for two PE sacks = 0.104 Ib petroleum. Petroleum requirements of paper sacks:

( ""'""' 10. I)[ 1 lb petroleii~n i lb petroleunl Fuel: = 0.008

30,395 sacks 2x104 BTU sack

Two polyethylene sacks require more than an order of magnitude nlore petroleu~n than a paper sack.


QUESTIONS FOR DISCUSSION 1) The term "100% recycle" ilnplies that all of the ~naterial in a grocery sack can be

recovered, but complete material recovery is generally impossible to achieve. In the case of polyethylene and paper sacks, manufacturers invariably print identification labels or advertisements on the sack. The printing is usually done with an ink or dye that is not desirable in the remanufacturing process and is not easily removed. In addition, a variety of consumer items, such as foods and beverages, can contaminate the sacks in a similar manner. In both cases, the contaminants could lower the quality of remanufactured sacks to a point where the sacks are unusable. Therefore, in order to meet quality specifications, some of the recycled material containing the contaminants at concentrated levels is removed and additional raw material and energy are required. Normal wear and tear of the sacks in use makes 100% recycle difficult to attain.

2) Many nations have adopted the reusable grocery sack concept with significant success, where success is measured by the number of people actively practicing the concept. Shoppers may reuse their durable sacks made out of nylon, jute or thick cotton string netting hundreds of times. The effect of grocery sack reuse as opposed to sack recycle is illustrated in Figure 1 - 1 . Sack reuse is represented by the product recycle loop and generally there is less energy consumed, atmospheric pollutants released and waste generated in the product recycle loop than in the materials recycle loop. All material and manufacturing steps are bypassed for the life of the sack. However, because the manufacture of typical durable grocery sacks involves an order of magnitude more energy use and emissions than the manufacture of a paper or plastic sack, the consumer must use the sack at least 10-20 times before any environmental benefit is achieved.


PROBLEM 2

A Life Cycle Inventory for Soft Drink Containers

Chemical Engineering Topics: mass balances, recycle streams Pollution Prevention Concepts: life cycle inventories, choosing a basis for product

comparison

BACKGROUND Soft drinks are commonly delivered to grocery stores and other distribution outlets in

glass, aluminum, and polyethylene terephthalate (PET) containers. This problem will examine the relative environmental impacts of glass, aluminum and PET beverage containers using the life cycle analysis framework shown in Figure 2-1. As this figure shows, several types of container recycling are possible. Some beverage containers, such as refillable glass beverage bottles, can be reused, which is represented by the product recycle loop of Figure 2-1. Other containers, such as aluminum cans, PET bottles and non-refillable glass bottles must be recycled to a remanufacturing process. In this problem, the focus will be on the latter type of container. The life cycle analysis therefore simplifies to the form shown in Figure 2-2. Air emissions, solid waste generation, and energy requirements for aluminum, glass, and PET containers as a function of recycle rate are presented in Table 2-1. Wastewater generation will be neglected.

Table 2- 1. Energy and Environmental Impacts as a Function of Recycling Rate for Three Soft Drink Container Groups (Franklin Associates, Ltd., 1989)

Container Group Recycling Rates

Energy Required11000 gal (MM Btu)

100% 0% 50%

PET (64 f l . oz. bottles) Aluminum (12 fl. oz. cans) Glass (16 fl. oz. bottles)

17.9 32.9 35.0

21.2 50.0 49.0

14.6 15.9 20.9

Atmospheric Emissions/1000 gal (lb)

44.8 48.3 73.5

PET (64 fl. oz. bottles) Aluminum (12 fl. oz. cans) Glass (16 fl. oz. bottles)

62.0 137.0 217.4

Solid Waste/1000 gal (lb)

53.4 91.7 145.4

189.5 198.2 762.3

PET (64 fl. oz. bottles) Aluminum (12 fl. oz. cans) Glass (16 fl. oz. bottles)

513.1 1938 7000

35 1.3 1068 388 1


natural resources

. . . . . . . . . . . . . . . . .

Figure 2-1. An Analysis Template for the Life Cycle of Manufactured Goods

Raw Material

Material Manufacture Waste, Wastewater

. . . . . . . . . . . . .

Product Manufacture

I

. . . . . . . . . . . . .

Energy t

Product Use Air Emissions, Solid Waste, Wastewater

. . . . . . . . . . . . . . .

. . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . 1

Energy C

Air Emissions, Solid Waste, Wastewater

Product Disposal


natural I resources Energy Raw Material Acquisition 1

I

Emissions i ......................... Solid Waste

Materials Manufacture, Product Manufacture,

and Product Use I

Energy

Atmospheric I

Emissions t Solid Waste

Product Disposal Atmospheric I

Emissions i

I

. . . . . . . . . . . . . . . . . . . . . . . . . . .

Solid Waste

Figure 2-2. A Simplified Analysis Template for the Life Cycle of Soft Drink Containers


The data are based on the delivery of 1000 gallons of soft drinks, using 1987 as the baseline year. As might be expected, energy consumption, atmospheric emissions, and solid wastes decline in value as the rate of recycling increases.

PROBLEM STATEMENT a) Calculate the energy requirements and the atmospheric emissions and solid wastes

generated per container for aluminum, glass and PET beverage containers. Perform your calculations at O%, 50% and 100% recycle. Report the energy requirements in Btu and the atmospheric emissions and solid waste generated in ounces.

b) Based on the calculations of part a), what are the energy requirements, atmospheric emissions and solid wastes generated per container in the process of recycling? Compare these results to the energy requirements and wastes generated when virgin materials (no recycled materials) are used.

c) Using the container weights listed in Table 2-2, estimate the mass of raw materials required to nlanufacture each type of container. Note that since mass is conserved, the amount of raw material required is the mass of waste generated during the life cycle, including the disposal mass of the container. Compute the ratio of the mass of the product to the mass required to manufacture the product, which is effectively the mass efficiency of the manufacturing process. Compare the mass efficiencies of the manufacturing processes for the three types of beverage containers. Assume 0% recycling in your calculations and note that the amount of solid waste generated per container at 0% recycle includes the disposal of the container.

Table 2-2. Container Weight and Volume

QUESTION FOR DISCUSSION The data in Table 2-1 are reported on the basis of 1000 gallons of beverage delivered.

In the problem you calculated energy requirements and wastes generated per container. Which of these bases is more appropriate for product comparisons? Does reporting on a per unit volume basis favor any of the container types?

REFERENCE Franklin Associates, Ltd., Cor~zl~nrcrtive Energy crrzd Erzvirorzr~lentcrl I17zpacts for Soft Drir~k Delivery Systen~s (prepared for the National Association for Plastic Container Recovery), Prairie Village, Kansas, March 1989.

Volume (fluid oz)

64 12 16

Container Type

PET Aluminum Glass

Weight (oz)

2.320 0.584 10.664


SOLUTION TO PROBLEM 2 a) To calculate the energy requirements, atmospheric emissions and solid waste generated

per container, it is necessary to first determine the number of containers required to deliver one gallon of soft drink. This number is calculated from

128 oz 1 container N = 1 gal

9

v (oz)

where N is the number of containers per gallon and V is the volume of the container in ounces. N = 2.0 for 64 oz. PET bottles N = 10.7 for 12 oz. aluminum cans N = 8.0 for 16 oz. glass bottles

Sample calculation of energy requirements and wastes generated per container: Energy required (E) per PET container at 0% recycle, using the data in Table 2-1:

E = 21.2x10"TU

x = 10.600 BTU/container 1000 gal 2 containers

Atmospheric en~issions (AE) and solid waste (S) are obtained similarly:

62 lb 16 oz I gal AE = x x 1000 gal lb 2 containers

AE = 0.50 oz/ container at 0% recycle S = 4.1 oz/container at 0% recycle

Table S2-1 shows the results for E, AE and S for all three container groups at the three recycling rates.


Table S2-I. Energy and Environmental Impacts per Container as a Function of Recycling Rate for Three Soft Drink Container Groups

b) The amount of energy required and the masses of wastes generated in recycling a container are simply the values given in Table S2-I under 100% recycle. The requirements for virgin materials are listed under 0% recycle.

Container Group

Table S2-2. The Effect of 100% Recycling on Container Manufacture

Recycling Rates

c) The Inass of virgin material that goes into the manufacture of an unrecycled product is the s u ~ n of the product weight plus the weight of all wastes created in the manufacture of the product (solid waste and atmospheric e~llissions in this problem). Table S2-3 shows the total weight of virgin material (product weight + (total solid waste - disposal weight) + at~nospheric emissions) required for product manufacture and the percentage

0%

Energy RequiredIContainer (BTU)

Container Group

PET

Aluminum

Glass

PET (64 oz. bottles) ~lurrlinurn (12 oz. cans) Glass (16 oz. bottles)

50%

Percentage Change in Impact from 0% Recycle

100%

10,600 4,687 6,125

Energy Required

-31%

-68%

-57%

Atmospheric EmissionsIContai~~er (oz)

8,950 3,084 4,375

PET (64 oz. bottles) Aluminum ( 12 oz. cans) Glass (16 oz. bottles)

Solid Waste Generated

-63%

-90%

-89%

7,300 1,49 1 2,6 12

Air E~nissions Generated

-28%

-67%

-65%

0.50 0.21 0.43

Solid WasteIContainer (oz)

0.43 0.14 0.29

PET (64 oz. bottles) Alu~ninurn (12 oz. cans) Glass (16 oz. bottles)

0.36 0.07 0.15

4.1 2.9 14.0

2.8 1.6 7.8

1.5 0.3 1.5


of virgin tnaterial in the product (container) itself. (Note that the disposal weight per container at 0% recycle is the product weight.)

Table S2-3. Virgin Material Required per Container @ 0% Recycle

The column titled "% of Virgin Material that Appears in the Product" gives the mass efficiencies for the production processes of the three types of containers. Glass container processing has the highest Inass efficiency at 74%. Alu~ninum container processing has the lowest mass efficiency of the three at 19%.

QUESTION FOR DISCUSSION Since containers vary in size, it is desirable to compare them on an equal basis. However,

it is difficult to establish what that basis should be. Comparison of the different container groups on the basis of delivery of equal volcr~lze is quick and simple, but this basis tends to favor larger volume containers which have a lower surface to volutne ratio (PET bottles in this problem).

% of Virgin Material that Appears in the Product

50%

19%

74%

% of Virgin Material that

Is Lost as Waste in Production

50%

81%

26%

Container Group

PET

Aluminum

Glass

Virgin Material Weight (oz)

4.6

3.1

14.4


PROBLEM 3

A Life Cycle Inventory for Polystyrene and Paper Containers

Chemical Engineering Topic: mass and energy balances Pollution Prevention Concepts: life cycle inventories, variation in data from different

sources

BACKGROUND Containers made of polystyrene and paper are used extensively in the food service

industry to deliver and serve food and beverages. Although the container materials, notably polystyrene, can be reprocessed, they are not curre~ltly recycled to any significant degree and are either incinerated or sent to sanitary landfills. Thus, the product life cycle consists of raw material acquisition, product manufacture, and disposal. (A more detailed discussion of life cycle analyses is presented in Problem 1 .) This problem will exanline the energy requirements, wastes and emissions associated with the product life cycle of two types of hot drink containers. To highlight the difficulties, controversies and ambiguities of such product life cycle comparisons, data from two sources will be examined. The data are given in Tables 3-1 and 3-2. Not all the information presented in these tables is necessary for co~npletion of the problem statement, but it is included in order to impart a sense of the differences between the reports.

PROBLEM STATEMENT a) For both sets of data (Tables 3-1 and 3-2), estimate the mass of solid waste generated and

the atmospheric emissions generated in grams per 16 oz polystyrene cup. Note that in Table 3-1, the only waste that is clearly defined as solid is the product disposal in the landfill. This represents a lower bound for solid waste generation. An upper bound can be obtained by assuming that all raw materials used in manufacturing a cup eventually wind up as solid waste. Also, Table 3-1 gives no information about cup capacity. Therefore, in order to compare data from Table 3-1 with data From Table 3-2, nortnalize to a 16 oz cup Inass of 4.4 g (Franklin Associates, Ltd., 1990) by multiplying the per cup data in Table 3-1 by 4.411.5. No adjustment of the values in Table 3-2 is required.

b) Repeat the calculations of part a) for 16 oz LDPE-coated paper cups with a mass of 10.4 g each (Franklin Associates, Ltd., 1990). The norinalization factor for per cup data in Table 3-1 is 10.4110.1.

c) Compare the atmospl~eric elnissions and solid wastes created per cup by the two types of hot drink containers. How does the data of Table 3-1 differ from that of Table 3-2?

REFERENCES AND SUGGESTIONS FOR FURTHER READING Hocking, M. B., "Paper versus Polystyrene," Science, Vol. 251, pp. 504-505, Feb. 1991 (see also letters comnlenting or. this paper, S c i e ~ c e , Vol. 252, 136 1 - 1363, June 199 1).

Franklin Associates, Ltd., Resoclrce a11d E11viror1171enml P ~ v f i l e A ~ ~ n l y s i s of Foal71 Polystyrer?e a11d Bleached Paperboar-d Conta i~~ers (prepared for the Council for Solid Waste Solutions), Prairie Village, Kansas, June 1990.


Table 3-1. Raw Material, Utility, and E~lvironmental Sumlnary for Hot Drink Containers (Hocking, 199 1 )

After use

Iten1

Raw materials: Wood and bark (g) Petroleum fractions (g) Other chemicals (g)

Finished weight (g) Wholesale cost

Utilities Steam (kg) Power (kwh) Cooling water (m')

Water effluent Voluine (m3) Suspended solids (kg) BOD (kg) Organochlorines (kg) Metal salts (kg)

Air elmissions Chlorine (kg) Chlorine dioxide (kg) Reduced sulfides (kg) Particulates (kg) Pentane (kg) Sulfur dioxide (kg)

To primary user

Proper incineration Heat secovery (MJIkg) Mass to landfill (glcup)

Biodegradable

Paper Cup':' Polystyrene Cup?

Per. ccip

33 (28 to 37) 0 4.1 (2.8 to 5.5) 3.2

1.8 0.05 10.1 1.5 2 . 5 ~ x

Pet- nletric to17 of'fi~zisherl rnnterinl

9,000 to 12,000 -5000 980 120 to 180 5 0 154

50 to 190 0.5 to 2 35 to 60 Trace 30 to 50 0.07 5 to 7 0 1 to 20 20

0.5 0 0.2 0 2.0 0

5 to 15 0. I 0 35 to 50

-10 -10

Recycle potential

Possible, though washing Easy, negligible water uptake can destroy

Low, hot lnelt adhesive or High, resin reuse in other coating difficulties applications

Clean Clean 20 40

10.1 1.5 Yes, BOD to leachate, No, essentially inert

methane to air

'"Made from fi~lly bleached kraft pulp; information from British Colulnbia pulp mills. Note that these data have been severely criticized; see Letters to Scierlce, June 7, 1991. -i-Made from molded polystyrene foarnable beads.

Atmospheric Waterborne Industrial Postconsumer Emissions Wastes Solid Waste Solid Waste

Type of Cup (lb) (lb) (lb) (CU ft) (lb) (cu ft)

PROCESS POLLUTANTS Foam Polystyrene 5 .0 1.7 5.2 0.1 120.3 13.7 LDPE-Coated Paperboard 7.4 2.0 28.1 0.6 2 18.3 7.3 Wax-Coated Paperboard 7.0 3.1 32.5 0.7 266.2 8.8

FUEL-RELATED POLLUTANTS Foam Polystyrene 6.8 0.5 13.4 0.3 -

LDPE-Coated Paperboard 10.7 1 .0 26.2 0.5 Wax-Coated Paperboard 14.8 1.4 38.5 0.8

TOTAL POLLUTANTS Foam Polystyrene 11.8 2.1 18.6 0.4 120.3 13.7 LDPE-Coated Paperboard 18.1 2.9 54.3 1.1 2 18.3 7.3 Wax-Coated Paperboard 21.8 4.5 71.0 1.4 266.2 8.8


Jegr, K., AII Oven~ievv of tlze Nortl1 A~~zericarz P~i lp a17d Paper. I ~ z d ~ i s t ~ y , The Forest Economics and Policy Analysis Project, University of British Columbia, Report 85-1 1, December 1985.

Ewing, A., E I I ~ I * ~ ~ EfSicier~cy ir? tlze Pull? n~zcl Paper. Ir~c1~istr.y vvitl? E~nphasis on Developi~~g Co~ i~~ t r i e s , World Bank Technical Paper No. 34.

U.N. Food and Agriculture Organization, Pulp, Paper arzcl Pelperboar-el Ccq2ncitj~ Survej) 1988- 1993, 1988.

Office of Technology Assessment, "Facing American Trash: What Next for Municipal Solid Waste?," 142, OTA-0-424, 1988.


SOLUTION TO PROBLEM 3 a) and b). The solid waste (sum of postconsumer and industrial wastes) and atmospheric

emissions are obtained from Tables 3-1 and 3-2 and are calculated on a per cup basis. Table 3- 1 data have been normalized to the cup weights reported by Franklin Associates, Ltd.

Sample calculations: Table 3-1 air emissions for polystyrene cup (0.1 kg + (35 to 50) kg + 10 kg)/1000 kg (4.4 glcup) = 0.20 to 0.26 glcup

Table 3- 1 solid wastes for polystyrene cup 1.5 g to 3.25 g (4.411.5) = 4.4 to 9.5 glcup

Table 3-2 solid wastes for polystyrene cup [(18.6 + 120.3 lb)/10,000 cup](1000 gl2.2 lb) = 6.3 glcup

Table S3-1. Polystyrene and Paper Cup Manufacture: Solid Waste and Atmospheric Emissions

c) Paper cups create more solid waste than plastic (2 to 4 times as much) and maybe more atmospheric emissions (1 to 1.5 times as much). Also, the air emissions for both polystyrene and paper cups are relatively low in the Hocking data--:! to 5 times less than that indicated by the Franklin Associates. Solid wastes from the two sets of data are in agreement.

Material

Polystyrene

Paper (LPDE)

Table 3-2 Data

Cup Weight

(9)

4.4

10.4

Solid Waste (glcup)

6.3

12

Atmos. Emissions

CUP)

0.54

0.82

Table 3-1 Data

Solid Waste ( ~ I c ~ P )

4.4 to 9.5

10 to 40

Atmos. Emissions

(~ IcuP)

0.20 to 0.26

0.18 to 0.29


PROBLEM 4

A Life Cycle Inventory for Three Diapering Systems

Chemical Engineering Topics: mass and energy balances, recycle streams Pollution Prevention Concepts: life cycle inventories, tradeoffs in environmental impacts

BACKGROUND Disposable diapers, manufactured from paper and petroleu~n products, are one of the no st

convenient diapering systems available, while cloth diapers are often believed to be the most environmentally sound. Indeed, many of the cloth diaper service industry's advertising campaigns attempt to convince consumers that making the "green" choice is worth a sacrifice in convenience. The truth is not so clear cut, however. This problem will quantitatively examine the relative energy requirements and the rates of waste generation associated with the diapering systems, illustrating the complexity of the environmental issue.

There are three types of diapering systems considered in this problem: home laundered cloth diapers, commercially laundered cloth diapers, and disposable diapers containing a superabsorbent gel. The results of life cycle inventories for the three systems are given in Table 4-1. Energy requirements and waste inventories for all phases of the life cycle, from raw material acquisition to final disposal, are included. (A more detailed discussion of life cycle analyses is presented in Problem 1.)

Table 4-1. Energy Requirements and Waste Inventory Data per 1000 Diapers for Three Diapering Systems (Franklin Associates, Ltd., 1990)

PROBLEM STATEMENT a) The authors of the report from which the data of Table 4-1 were taken found that an

average of 68 cloth diapers were used per week per baby. Disposable diaper usage is

Impact

Energy Requireinents, 1 O6 Btu

Solid Waste, ft"

Atn~ospheric Emissions, lb

Waterborne Wastes, lb

Water Volume Requirements, gal

Disposable Diapers

3.4

17

8.3

1.5

1300

Cloth Diapers

Commercially Laundered

2.1

2.3

4.5

5.8

3400

Home Laundered

3.8

2.3

9.6

6.1

2700


expected to be less because disposable diapers are changed less frequently and never require double or triple diapering. In order to compare the different diapering systems, determine the number of disposable diapers required to match the performance of 68 cloth diapers per week. For disposables, the number of diapers required per week per baby can be determined fro111 the following: I. 15.8 billion disposable diapers are sold annually (3.04x108/week), 11. 3,787,000 babies are born each year, . . . 111. children wear diapers for the first 30 months, and iv. disposable diapers are used on 85% of children.

b) Complete Table 4-2 using the data in Table 4-1. Remember to apply the factor for equivalent product use found in part a). Compare the three diapering systems quantitatively. For example, the data already in Table 4-2 indicate that home laundered cloth diapers require roughly twice the energy of either disposable or commercially laundered cloth diapers. If the values in Table 4-1 have an accuracy of &lo%, which of the ratios in Table 4-2 indicate a real difference between systems? An accuracy of + lo% indicates that the true value lies within 10% of the reported value.

Table 4-2. Ratio of Impact to Home Laundered Impact

c) Using the data in Table 4-3, determine the percentage of disposable diapers that would have to be recycled in order to make the solid landfill requirements equal for cloth and disposable systems.

Iiilpact

Energy Requirements Solid Waste Atnlospheric Emissions Waterborne Wastes Water Volume Requirements

Table 4-3. Recycle Ratios and Their lnipact on Solid Waste for Disposable Diapers (Franklin Associates, Ltd., 1990)

Disposable

0.50

1

Percentage of Diapers Recycled

0 25 50 7 5 100

Cloth

Solid Waste per 1000 Diapers (ft3)

17 13

9.0 4.9

0.80


0.55

Home Laundered

1 .O 1 .O 1 .O 1 .O 1 .O


QUESTIONS FOR DISCUSSION 1. Do you think that the differences between the two cloth diaper systems are more

meaningful than the differences between cloth and disposable diapers? Why? Is it possible to determine which of the three systems causes the least environmelltal impact? What might be a more meaningful comparison?

2. Water volulne used is not equal to net water consumption. Often, treated water is returned to the process stream from which it came. Does this negate the importance of the water volunle comparison?

3. Reuse of cloth diapers as rags when they are no longer suitable for use as diapers was not considered in the Franklin Associates report. How might this reuse affect the environmental impacts? Assunle that as rags the cloth diapers replace cloth towels.

REFERENCE Franklin Associates, Ltd., El~ergy uncl Elzvirol~merztal Profile Allalysis of Chilclren's Disl2osnDle nncl Clotlz Dicrl2er-.Y (report prepared for the American Paper Institute's Diaper Manufacturers Group), Prairie Village, Kansas, July 1990.


SOLUTION TO PROBLEM 4

15.8x109 disposables [ yr 1 Yr diapers

a) Yr 1 1 1 = 38 52 wks (0.85)3.787x 10 "babies 2.5 yr wk baby

68 cloth = 1.8

38 disposable b) Sample calculation: ratio of disposable net energy requirements to home laundered net

energy requirements

3 . 4 ~ 1 0 "ti1 y 1 disposable 1000 disposable diapers 1 1.8 cloth diapers

= 0.50 3 . 8 ~ 1 0 "tu

1000 home laundered

Table S4- 1 . Ratio of Impact to Honle Laundered Impact

Home laundered cloth diapers require twice the energy of the other two systems. Disposable diapers create four t i~nes as much solid waste as cloth diapers. Home laundered diapers produce roughly twice the at~nospheric emissions of the other two systems. Cloth diapers p rod~~ce about seven times as much waterborne waste as disposables. Disposable diapers have about 114 of the water volunle requirements of hoine laundered, while cotnlnercially laundered diapers require about 30% more water than hoine laundered. Assu~ning an accuracy of 10% in the data, the significance of the differences reported in the table can be assessed. For net energy required, disposable and co~i~mercially laundered diapers are significantly different than home laundered. Disposable diapers have a real solid waste difference from cloth diapers. For atmospheric emissions, disposable and cominercially laundered are different fro111 home laundered. For waterborne wastes, disposables are different than cloth. All systems are different in water volurne I-equirements.

c) First find the aiiiount of solid waste from disposable diapers that corresponds to the solid waste from cloth diapers given in Table 4-1.

Iinpact

Energy Requirements Solid Waste Atlnospheric E~nissions Waterborne Wastes Water Voluine Requirements

Interpolating between 75% and 100% recycle in Table 4-3 gives

Disposable

0.50 4.1 0.48 0.14 0.27

Cloth


0.55 1 .O

0.47 0.95 1.3

Home Laundered

1.0 1 .O 1 .0 1 .O 1 .O


2.3 f t 1 L8cloth 1 1000 cloth diapers 1 disposable

= 4.1 ft waste equivalent of 1000

disposable diaper uses

where x is the % recycle at which solid landfill requirements for disposable diapers are the same as for cloth. Solving gives

x = 80% .

QUESTIONS FOR DISCUSSION 1 ) Comparisons of the differences between the two cloth diaper systems might be more

meaningful because their waste streams are more likely to contain similar compounds, while the waste streams from disposable diaper manufacture would be completely different. It is not possible to determine from the given information which of the three systems causes the least environn~ental impact. To make a more meaningful comparison one would have to study the risk due to the waste streams and the effects of water use on the ecological well-being of the surroundings.

2) In some geographical areas, water is plentifiil and the water volume required may not be important, while in other areas withdrawing large amounts of water could adversely affect the environment.

3) The cloth towels would have to be manufactured using the same resources that produce cloth diapers, so environmental impacts due to the production of cloth diapers would be reduced if this reuse was taken into account.


PROBLEM 5

Estimating and Reducing Fugitive Emissions

Chemical Engineering Topics: process design, process flowsheeting Pollution Prevention Concepts: identifying emission sources, methods for estimating

emissions

BACKGROUND Fugitive emissions are unintentional releases from pumps, valves, flanges and other con-

nectors, pressure relief valves, compressors, open-ended lines, and sampling connections. The quantity of these emissions is hard to measure, but in some cases they may account for 70% to 90% of the air emissions from a chemical manufacturing operation. The U.S. Environmental Protection Agency (EPA) allows five methods for estimating fugitive emissions from synthetic organic chemical manufacturing industry (SOCMI) facilities. Given in increasing order of accuracy and cost, they are 1) average emission factors, 2) leaklno leak emission factors, 3) stratified emission factors, 4) EPA correlation curves, and 5) process-specific correlation curves. Fugitive emission factors were empirically generated from data gathered at a number of petroleum refinery and SOCMI facilities.

In the first method, a count of all the equipment is multiplied by the factors shown in Table 5-1 to determine total hydrocarbon en~issions. In each of the remaining methods, every piece of equipment, except for possibly some flanges, is screened to determine the concentration of organics outside the equipment. For the leaklno leak method, the equipment is considered to be leaking if the organic concentration in the air surrounding the equipment is >10,000 ppm. Table 5-2 gives the emission factors for this method of estimation. The stratified factor method is similar to the leaklno leak method, the difference being that there are three ranges of screening concentration used to estimate emissions rather than two. Emission factors for this method are given in Table 5-3. Methods 4 and 5 make use of curves of emission rate as a continuous function of screening concentration. Facilities can use either EPA-provided correlation curves (shown in Figure 5-1) or sample their own equipment to create process-specific emission curves. When estimating emissions of a single compound, it is assumed that the mass fraction of the chemical is the same in the fugitive emissions as it is in the volatile organic compound fraction of the fluid serviced by the equipment. As shown in Tables 5-1 to 5-3, emission factors for pumps and valves increase as the volatility of the fluid they service increases.

Fugitive emissions can be reduced either by making equipment changes or by instituting a leak detection and repair (LDAR) program. Equipment control techniques include leakless technology for valves and pumps, plugs for open-ended lines, rupture disks for safety relief valves, dual mechanical seals with nonvolatile barrier fluidldegassing vent systems for rotary equipment, closed loop sampling systems, and incineration of emissions from a process vent. Use of well-designed and maintained equipment can result in a significant reduction of emissions. The effectiveness of LDAR programs depends greatly on the frequency of monitoring sources for leak detection (U.S. EPA, 1986).

This problem will examine fugitive e~nissions from an acrolein manufacturing facility. Acrolein is a relatively high-value chemical that is used in the formation of acrylic acid.


Methods 1, 2, and 5 for estimating fugitive emissions will be applied to the acrolein manufacturing plant whose process diagram is given in Figure 5-2.

Table 5-1. Average Emission Factors for SOCMI Fugitive Emissions (US EPA, 1986)

Table 5-2. LeakINo Leak Emission Factors for SOCMI Fugitive Emissions (US EPA, 1986)

Equipment

Valves

Pumps

Compressors

Pressure Relief Valves

Flanges and Other Connectors

Open-Ended Lines

Sampling Connections

Service

Gas Light Liquid Heavy Liquid

Light Liquid Heavy Liquid

Gas

Gas

All

All

All

Equipment

Valves

Pumps

Compressors



Open-Ended Lines

Emission Factor, kg/hr/source

0.0056 0.007 1

0.00023

0.0494 0.02 14

0.228

0.104

0.00083

0.00 17

0.0 15

Service



Gas

Gas

All

All


Leak (> 10,000 ppm)

0.045 1 0.0852

0.00023

0.437 0.3885

1.608

1.69 1

0.0375

0.01 195

No Leak (< 10,000 ppm)

0.00048 0.00 17 1 0.00023

0.0 120 0.0 135

0.0894

0.0447

0.00006

0.00 150


Screening Concentration, ppmv

Valves in gas service, Emissions (kghr) = 1.7 12(10~535)(ppmv)0~3

Valves in light liquid service, Emissions (kghr) = 3.735(1@Y2)(ppmv)0"7

Pumps in light liquid service, Emissions (kg/hr) = 1.333(l~'")(ppmv)os~

Flanges, Emissions (kg/hr) = 0.9 1 8(10"m)(ppmv)0d'8

Figure 5-1. Published Curves Correlating Screening Concentration and Fugitive Emissions (Schaich, 1991)


OXYGEN

PROPYLENE

B = ABSORPTION SYSTEM C = INCINERATOR D = STRIPPING COLUMN E = REFINING COLUMN

UNREACTED WASTE + GASES

Figure 5-2. Acrolein Production Process (Berglund et al, 1989)

C + REFINED ACROLEIN

E

STEAM

T +

A HOT

PRODUCT

REFINED. PRODUCT

PRODUCT

1;


Table 5-3. Stratified Emission Factors for SOCMI Fugitive Emissions (Schaich, 1991)

PROBLEM STATEMENT a) A component count of plant equipment containing acrolein revealed that there were 1400

valves, 3048 flanges and other connectors, 27 pumps, 20 pressure relief valves, 21 open- ended lines, and 20 sampling connections (Berglund et al., 1989). Determine the fugitive acrolein emissions in lblyr using the average SOCMI emission factors from Table 5-1. Assume that on average the volatile organic compound fraction of the process fluid contains 87% by mass acrolein. For this case, estimated fugitive acrolein emissions for each type of equipment are equal to

Equipment

Compressors

Pumps

Valves



Open-Ended Lines

where fi is the average emission factor for equipment type i and Ni is the number of pieces of equipment type i. Of the valves, 168 are in gas service; all other valves and pumps are in light liquid service. Tabulate your results by equipment type and give the percentage of total emissions due to each equipment type.

b) When each component was screened with an organic vapor analyzer (OVA), it was discovered that the equipment in acrolein use had a far lower leak rate than the equipment used in developing the average SOCMI factors. Only one pump and one valve in liquid service gave readings corresponding to a hydrocarbon concentration greater than 10,000 ppm (Berglund et al., 1989). Use the leaMno leak emission factors given in Table 5-2 to determine acrolein emissions in lbslyr making the same assumptions as before and using the average emission factor for sampling connections. Report the results as in part a). How do the estimates differ? How does the distribution of emissions differ?

Service

Gas



All

Gas

All


0- 1,000 ppm

0.0 1 132

0.00198 0.00380

0.000 14 0.00028 0.00023

0.00002

0.01 14

0.000 13

1 ,00 1 - 10,000 PPm

0.264

0.0335 0.0926

0.00 165 0.00963 0.00023

0.00875

0.279

0.00876

> 10,000 PPm

1.608

0.437 0.3885

0.045 1 0.0852 0.00023

0.0375

1.69 1

0.01 195


c) For a more accurate estimate of acrolein emissions, a statistically significant number of valves and flanges were "bagged" to develop correlation curves relating screening values in ppm to actual hydrocarbon emission rates. During "bagging," a component is completely enclosed in mylar through which a nitrogen purge stream is blown. After the enclosure reaches steady state, a portion of the purge stream is captured and analyzed. For valves and flanges with screening values below the OVA detection limit of 0.1 ppm, the hydrocarbon emission rate was assumed to be 3x10.' Iblhr for valves and 8 ~ 1 0 - ~ Iblhr for flanges. For screening values above the detection limit, the correlation curves were given by

and E = 0.00033C0.86 (for valves)

E = 0.00006C0.44 (for flanges)

where E is the hydrocarbon emission rate in glhr and C is the screening value in ppm (Berglund et al., 1989). There was no statistical difference between valves in light liquid service and valves in gas service. Screening values (simplified) are as follows:

Table 5-4. Data from Screening Valves and Flanges

Find the acrolein emissions in Iblyr for valves and flanges using these correlations. Compare to the emissions for valves and flanges from parts a) and b).

Screening Value, ppm

20,000 5000 1000 25 0 7 5 5 1

not detected

QUESTION FOR DISCUSSION If you were to make equipment changes to lower acrolein emissions in this production

unit, what equipment would you add or alter first? Use the percentages of emissions due to each component type that you developed in part b) and assume that there are no differences in costs between different leakless equipment types.

REFERENCES AND SUGGESTIONS FOR FURTHER READING Berglund, R. L., D. A. Wood, and T. J. Covin, "Fugitive Emissions from the Acrolein Production Industry," paper presented at the Air and Waste Management Association International Specialty Conference: SARA Title 111, Section 3 13, Industry Experience in Estimating Chemical Releases, April 6, 1989.

Number of Component at Screening Value

Valves

1 10 10 10 10 10 2

remainder

Flanges

0 10 10 10 10 10 2

remainder


Berglund, R. L., W. J. Mollere, C. R. Ritter, L. W. Salathe, "A Quality Program to Reduce Episodic and Fugitive Emissions from an Ethylene Oxide Unit," paper for presentation at the AIChE Annual Sum~ner Meeting, Aug. 1990.

Schaich, J. R., "Estimate Fugitive Emissions from Process Equipment," Chein. Eng. Prog., Vol. 87,No. 8 ,pp . 31-35,Aug. 1991.

U.S. Environmental Protection Agency, "Emission Factors for Equipment Leaks of VOC and HAP," EPA-45013-86-002, Research Triangle Park, NC, 1986.


SOLUTION TO PROBLEM 5 a) Estimate total fugitive emissions using the emission factors in Table 5-1.

For valves: (0.87) 168~0.0056 (kg/hr)(2.2 lb/kg)(24 hl-/day)(365 daylyr) = 15,774 lblyr (0.87) 123 1 x0.007 1 (kg/hr)(2.2 lb/kg)(24 hr/day)(365 daylyr) = 14,666 lblyr total emissions from valves = 162,000 lblyr

Table S5- 1. Results from Equipment Count

b) Using the 1eakJno leak factors of Table 5-2 for valves: no leak gas service:

0.87(168x0.00048 kg/hr)(2.2 lb/kg)(24 hr/day)(365 daylyr) = 1352 lblyr no leak light liquid service:

0.87(123 1~0.00171 kg/hr)(2.2 lb/kg)(24 hr/day)(365 daylyr) = 35,294 Iblyr leak liquid service:

0.87( 1 x0.0852 kg/hr)(2.2 lb/kg)(24 hr/day)(365 daylyr) = 1429 lblyr for pumps: leak: lx0.87x0.437 kglhr(2.2 lb/kg)(24 hr/day)(365 daylyr) = 7327 lblyr no leak:

26x0.87x0.012 kglyr(2.2 lb/kg)(24 hr/day)(365 daylyr) = 523 1 lblyr

Table S5-2. Results for LeakINo Leak Emission Factors

% of Emissions by Equipment Type

60.6 15.9 8.4

13.1 0.2 1.9

100

Equipment

Valves Flanges Purnps PRVs Open lines Samplers TOTAL

Emissions, lblyr

162,000 42,400 22,400 34,900

599 5,030

267,000

% of Emissions by Equipment Type

51.3 4.1

17.0 20.2 0.7 6.8

100

Equipment

Valves Flanges Pumps PRVs Open lines Samplers TOTAL

Emissions, lblyr

38,100 3,070

12,600 15,000

528 5,030

74,300


Emission estimates are close to four times lower using this procedure than the procedure of part a). The distribution of sources is also different. In part a) it was valves > flanges > PRVs > pumps > samplers > open lines. In part b) it was valves > PRVs > punlps > samplers > flanges > open lines.

c) Using the equation for valves given in the problem statement and the one valve screened at 20,000 ppm as an example, E = (0.870)0.00033(20,000 ppm)0.86 g/hr(kg/1000 g)(2.2 lb/kg)(24 hr/day)(365 daylyr) = 27.7 lblyr. Number of valves below the detection limit = 1347; number of flanges below the detection limit = 2996.

Table S5-3. Results from Correlation Curves

This emission estimate is allnost 300 times less than leaWno leak results for valves and over 1000 times less than leaWno leak results for flanges. The bagging correlation curves give an emission estimate for valves that is 1000 times less than the average emission factor and an estimate for flanges that is 16,000 times less.

Screening Value

20,000 5,000 1,000

250 7 5

5 1

not detected TOTAL

QUESTION FOR DISCUSSION Based on the factors in Table 5-2, the first piece of equipment to alter or to lower

emissions would be the leaking pump, then the leaking valve, then the pressure relief valves, and so on. It is important to note that even conventional (as opposed to no leak design) coinponents rarely leak significantly. A prudent equipment replacement strategy would be to substitute ,

chronically leaking equipment with replacements of the same general design.

Emissions, lblyr

Valves

27.7 84.0 21 .o

6.39 2.27 0.221 0.01 11 3.08

145

Flanges

0 0.427 0.210 0.114 0.0672 0.0204 0.00201 1.83 2.67


PROBLEM 6

Estimating and Reducing Secondary Emissions

Chemical Engineering Topic: mass transfer Pollution Prevention Concepts: identifying emission sources, emission mechanisms for

waste water collection components

BACKGROUND Emissions from a manufacturing plant can be classified as primary or secondary. Primary

emissions are released from process units such as reactors and refining towers; emissions that occur as the result of the construction or operation of a major stationary source but that do not come from the source itself are called secondary emissions. One major source of secondary emissions is wastewater treatment systems. The components of wastewater collection and treatment systems are often partially or fully uncovered, which allows volatile organic compounds (VOCs) to escape. Some common wastewater collection and treatment components are junction boxes, sumps, lift stations, equalization basins, clarifiers, treatment tanks, and aeration basins. The emission mechanism from each of these components is similar: mass transfer driven by a concentration gradient with resistance occurring in both the air and water side of the airlwater interface.

In this problem you will estimate the emissions of toluene from the junction box pictured in Figure 6-1. Junction boxes are used to combine waste streams. Although they are not mechanically aerated, the wastewater flowing through them is assumed to be turbulent, and a correlation for estimating the liquid phase mass transfer coefficient in mechanically aerated impoundments can be used. Equilibrium is assumed to exist at the airlwater interface, so the equilibrium partition coefficient of the compound between air and water must also be determined. It is found from

K,,, = HIRT,

where H is the Henry's constant for the compound, R is the ideal gas law constant, and T is the temperature in K. In this context, Henry's constant is defined by the equation

where pi is the partial pressure of colnpound i in equilibrium with the concentration of compound i in water, Ci. Once these coefficients have been determined, the overall inass transfer coefficient, K, can be calculated using the equatiofi

where k, is the liquid phase mass transfer coefficient and k, is the gas phase mass transfer coefficient. Flux is then found from

J = KC,,


where C,, is the bulk concentration of the compound in the waste stream.

PROBLEM STATEMENT a) What is the value of the overall mass transfer coefficient, K, for toluene in this typical

junction box? The values for the liquid and vapor phase mass transfer coefficients are k, = 4.6~10." m/s and k, = 0.01 1 m/s (U.S. EPA, 1990). The Henry's constant for toluene is 0.01 1 atm-m3/mol at 3S°C.

b) Because the contents of the junction box are assumed to be well mixed, the bulk concentration in the box is equal to the effluent concentration. Use mass balance techniques to develop an equation relating K, the overall mass transfer coefficient, Q, the flow rate, A, the air/wates interfacial area, C,,, the inlet concentration, and C,, the bulk concentration in the junction box.

c) If the inlet concentration of toluene is 5x10." g/cm" the airlwater interfacial area is 0.656 m2, and flow rate through the junction box is 2.52~10.' mvs, find the bulk concentration of toluene in the junction box.

d) Estimate the emissions of toluene from the junction box described in part c).

QUESTIONS FOR DISCUSSION 1) What resistance dominates this mass' transfer scenario? Why? What resistances can be

neglected? 2) What would lower the mass transfer coefficients, thus lowering emissions? Consider

temperature, turbulence, the dimensions of the junction box, and air movement. How might emissions be reduced without affecting the mass transfer coefficients?

REFERENCE U.S. Environmental Protection Agency, "Industrial Wastewater Volatile Organic Compound Emissions: Background Information for BACTJLAER Determinations," Control Technology Center, Research Triangle Park, NC, January 1990.

ground level *

- \ inlets @ wastewater - outlet

I

Figure 6-1. A Junction Box


SOLUTION TO PROBLEM 6 1 1 1

4 m K = 4 . 2 ~ 1 0 - - S

b) E = emissions = JA = KC,A Perform a mass balance around the dashed box in Figure S6-I. QC, = KC,A + QC,, QC, = C,(KA+Q)

m 2 . 5 2 ~ 10 -' - i S

m 3 f4.2110 :)0.656 1n ')+2.52x10 -' -

sec

d) E = 4 . 2 ~ 1 0 -4 - 4 . 5 ~ 1 0 -4 - 0.656 ~n ') 3600 s 24 hr 365 day ( [ h r 1 ~ ~ y r )

lb x 1 0 1 c n l g ) = 8600 - Yr

QUESTIONS FOR DISCUSSION 1) The liquid film resistance dominates so the assumption that the liquid is well mixed will

significantly influence the results. The liquid phase mass transfer coefficient is based on a mechanically aerated basin. It is not clear whether this really represents the behavior of a junction box. Bulk liquid and air resistances can be neglected.

2) Lower temperature and less turbulence would reduce the mass transfer coefficients. Strategies for reducing emissions should concentrate on increasing liquid film resistance. A smaller aidwater interfacial area would reduce emissions without affecting the mass transfer coefficients. The snlaller airlwater inferfacial area could be achieved by altering the geometry of the box or by placing a floating cover on the water phase.


emissions

Figure S6-1. Mass Balance around a Junction Box


PROBLEM 7

Estimating and Reducing Emissions from an API Separator

Chemical Engineering Topic: mass transfer Pollution Prevention Concepts: identifying emissions sources, emission measurement

techniques

BACKGROUND Oil refining processes such as coking, sulfur recovery, steam cracking, hydrocracking, and

crude desalting create large volumes of wastewater. The wastewater stream must be treated before release to make it less toxic and to recover any useful material it contains. A common unit in a refinery's wastewater treatment system is an American Petroleum Institute (API) separator, which is a large rectangular gravity separator where sludge settles to the bottom while a skimmer crosses the surface at intervals to push accumulated oil into a slotted pipe or drum. A simple API separator is shown in Figure 7-1. If the separator is uncovered, as is frequently the case, volatile organics in the oil layer can escape to the atmosphere and account for a significant fraction of a refinery's air emissions.

In this problem, five methods of estimating the emissions of benzene, toluene, ethylbenzene, and xylene (BTEX) from an API separator will be compared. The methods to be explored make use of 1) flux chamber measurements, 2) mass balances, 3) diffusion model techniques, 4) mass transfer equations, and 5) EPA emission factors. The separator has a surface area of 84 m2 and an inlet/outlet flow rate of 7600 m3 of wastewater per day. The mass fractions of the compounds in the separator's oil layer are given in Table 7-1.

Table 7-1. Mass Fractions of Chemicals

PROBLEM STATEMENT a) A flux chamber, pictured in Figure 7-2, can be used to directly measure emissions as it

rests on the surface of the separator. When the contents of the chamber reach steady state, they are sampled and analyzed to determine the concentration in ppmW of carbon for benzene, toluene, ethylbenzene, and xylene (BTEX). Parts per million by weight of carbon, or ppmW of carbon, is the Inass of carbon belonging to the species per million mass units of material sampled. Develop an equation relating the measured concentration, C (in ppmW of carbon), to the mass flux from the separator, J. Assume the gases behave ideally and that Q, the sweep air flow rate of the flux chamber, and A, the area the flux chamber shares with the surface of the separator, are known.

Compound

Benzene Toluene Ethylbenzene Xylene

Mass Fraction in Oil Layer

0.00 1 0.018 0.014 0.075


inlet outlet ____e

ra

water layer

slotted pipe

skimmer

sludge hopper

Figure 7-1. An API Separator (not to scale)

to analyzer

sweep thermocouple air inlet

Q

t + Q , c

____)

sweep air - outlet

API separator J 1 ,c '

surface

Figure 7-2. A Flux Chamber


b) Using the equations developed in part a) and the data in the following table, determine the emissions of the separator in kglhr. Use a sweep air rate of 5 Llrnin for a flux chamberlsurface area of 0.13 m2 and an ambient temperature of 25°C.

Table 7.2. Flux Chamber Data

Measurements taken using flux chambers are affected by the positioning of the chamber, by surfacing gas bubbles, and by the position of the skimmer. Also, flux chambers perturb the conditions at the surface of the separator by their presence. However, they represent a direct measurement of e~nissions from a wastewater treatment pond surface.

c) It is not possible to determine emissions by performing a Inass balance around the API separator because wastewater flow rates are close to 10 orders of magnitude larger than emission rates. To prove this, define n~ , , , ,~~ ,~ as the mass flow rate of organic co~npound i in the emissions. A mass balance around the separator gives

Compo~lnd


where m,,ll,t,i and n~,,,,,,,,~ are the inlet and outlet mass flow rates of component i in the oil layer. Dividing both sides by the oil layer mass flow rates yields

Concentration, ppmW of Carbon Measured in the Flux Chamber Effluent

2 1 120 3 1 142

where x~,,,,~,~ and x,,,,,,~~ are the Inass fractions of compound i in the oil layer, and m,,, is the oil layer mass flow rate. Using the results from part b), find m,,l,it,i/moi, for the four compounds, given that the oil layer is 10% by volume of the total wastewater flow and that the amount of BTEX in the water layer is negligible. Use an approximate density of 1 kg/L for both oil and water. If mass fraction can be measured to the nearest thousandth, is it possible to estimate emissions by determining the difference between the inlet and outlet Inass fractions?

d) Find a theoretical lower bound for emissions in kglhr by performing a mass transfer analysis neglecting convective effects. Assume that a stagnant air layer one meter deep exists above the oil layer and that above that stagnant air layer the concentration of the four chemicals is effectively zero. The equation for mass flux through a stagnant gas film is


-cD,, dx, NAz = - - '

1-x, dz

where NAz is the molar flux of species A in the z direction, c is the total molar concentration of the stagnant layer, D,, is the diff~~sivity of compound A in the stagnant layer, and x, is the mole fraction of species A (Bird, Stewart, and Lightfoot, 1960). Also, a mass balance around the incremental volume height Az in Figure 7-3 gives

as Az goes to zero. Combine the diffusion equation with the mass balance equation to arrive at a second order differential equation relating x, and z. Assume that the concentration of BTEX at the surface of the oil layer is equal to the mixed oil concentration. Also assume that the oil forms an ideal mixture and that the gases behave ideally so that at the oillair interface

where x, is the vapor mole fraction of component A, y, is the liquid mole fraction of component A, and P is pressure. (For this order of magnitude estimate, assume that mass fraction and mole fraction are equal.) Vapor pressures and diffusivities are given in Table 7-3.

Table 7-3. Diffusivities in Air and Vapor Pressure

"Property data for 111-xylene.

Cornpound

Benzene Toluene Ethylbenzene Xylene5' Water Vapor

e) If measurements of compound concentration, wind speed, and temperature are taken using a device like the one pictured in Figure 7-4 at intervals within two meters of the separator surface, e~nissions can be estimated using a diffusion model (Thibodeaux et al., 1982). The measurements must be made in the re-established boundary layer pictured in Figure 7-5 so results are not affected by upwind disturbances. Emissions per unit area are given

by

Diffusivity in Air, cm2/s

0.077 0.079

0.0658 0.059 0.22

Vapor Pressure at 25'C, mm Hg

95.19 28.4 9.53 8.3


Figure 7-3. Control Volume for Diffusion Mass Balance

SAMPLING DEVICE

ROTAMETERS

+ WIND DIRECTION

Figure 7-4. Concentration Mast for Diffusion Model Measurements (Chemical Manufacturers Association, 1990)


~4 ,, UNDISTURBED - MODIFIED PROFILE PROFILE

/ /

Undisturbed I /

I I

Boundary I I I

Layer Region of inf luence of levee on pressure f i e l d

//---

/ I--

Re-est ablished Boundary Layer ,'

0 0 L P, -

I + t

Figure 7-5. Boundary Layers above a Wastewater Treatment Pond (Thibodeaux, et a1 1982)


The denominator in this equation is a stability correction factor and can be taken as unity for this example. The term S,, characterizes the velocity distribution and is defined as the slope of the line from a plot of air velocity vs. the natural log of the measurement height in cm; S, is the slope of the line from a plot of concentration vs. the natural log of the lneasurernent height. Subscripts A and W denote the emitted chemical and water, respectively, and D represents diffi~sivity in air, which is given in Table 7-3. In practice, at least six observations of each variable would be made, but for this problem data from two heights will be assumed to provide an accurate picture. The wind velocity at 10 cm was 230 cm/s and at 100 cm was 700 cm/s. Concentration data are given below.

Table 7-4. Concentration Data for the Diffi~sion Model

Find the estimated releases of each cornpound in units of kglhr. f) The US EPA provides emission factors to make reporting of emissions simple for

facilities. For uncovered oillwater separators in refineries, the volatile organic compound (VOC) emission factor is 0.6 kg/103 L wastewater. Assume that benzene, toluene, ethylbenzene, and xylene are the only volatile organic compounds (VOCs) emitted and report their emissions in kglhr. It is standard practice to proportion air emissions among chemicals according to their proportions in the volatile fraction of the waste stream.

g) Prepare a table conlparing the results of parts b), d), e), and f). Which method gives results nearest to the flux chamber results? How do the estimation methods differ from the flux chamber results?

Co~llpound


QUESTIONS FOR DISCUSSION 1) If you were the engineer assigned to report the emissions from an API separator, which

method would you prefer? Why? Is there motivation for refineries to make measurements rather than rely on emission factors provided by the government?

2) How might elmissions from an API separator be reduced? Consider variables important in parts d) and e).

Concentration at 10 cm, ng/L

470 890 240 900

Concentration at 100 cm, ng/L

390 5 90 190 650


REFERENCES AND SUGGESTIONS FOR ADDITIONAL READING Barnes, D., C. F. Forster, S. E. Hrudey, S~lrveys in 111elcrstr.ial Wastewater Trecrtr7zer1t: Petr.oleunz ernel Organic Cl~erniccrls Ir~dustries, Vol. 2, Pitman Publishing Limited, Bath, 1984.

Bird, R. B., W. E. Stewart, E. N. Lightfoot, T~nnsl?orpt Plzenornencr, John Wiley & Sons, Inc., New York, 1960.

Chemical Manufacturers Association, Ir~zl~rovirzg Air Q~rcrliiy: A G~licle to Estir~zntirzg Secor~cl~rr-y E I ~ I ~ S S ~ ~ I Z S , Washington, D. C., 1990.

Thibodeaux, L. J., D. G. Parker, H. H. Heck, "Measurement of Volatile Chemical Elmissions from Wastewater Basins," US EPA, PB83- 135632, Nov. 1982.

U.S. Environmental Protection Agency, Conzl?ilatio~z qf Air. Pol l~~ta~z t Ernission Fc~tors , Vol. I , 4th ed., Research Triangle Park, NC, 1985.


SOLUTION TO PROBLEM 7 a) Let Y be the species concentration in the chamber in gll. Perform a mass balance around

dashed line in Figure S7-I to get

To convert C (ppmW of carbon) to Y (gll): let a = number of carbon atoms/molecule of species

C g carbon = [ lo6 g air air)[ ~nolar volume 1 of air 12a g carbon

mol species 1 ! MWspecies

10" air

1 atm 1no1 K b) Using - = 1 = 0.0409 moll1 and MW,,,,. = 29 glmol RT 0.08206 1 atm 298 K r

separator [ oa: 1 JA\epnl.alol- [60i:in 1 egissisis = E

Table S7- 1. Flux Chamber Results

c) Mass flow rate of oil layer:

Compound


a MW E, kglhr

6 7 8 8

78.1 1 92.13

106.16 106.16

0.005 0.03 0.008 0.04


r - -" - - - - - - - - - - - - - - - - - - - - - - - - - - - - I

I I I I I I

Q I I I + A I

s, Q, Y I I

I I I

surface I I - - - - - - - - - - - - - - - - . . - - - - - - - - - - - - - - - - . of

separator J x A, where A is chamber/surface interfacial area

Figure S7-1. Mass Balance around a Flux Chamber


Table S7-2. Mass Balance Results

It is not possible to get measurable differences between inlet and outlet mass fractions since x,,,,,,,, is less than one thousandth for all compounds.

d) Given:

Compound


and

Substitute for N,, to get

Mass Flow Rate of Emissions

(kglhr)

0.005 0.03 0.008 0.04

where cD,, are constant and drop out. Integrate twice to get

lnenlit.i

Inoil

1 . 6 ~ 1 0 . ~ 9 . 5 ~ 2 . 5 ~ 1 0 . ~ 1 . 3 ~ 1 0-6

and -In(l -x,) = C,z+C,.

The constants of integration are determined from the boundary conditions: at z = 1 m, x , = 0 at z = 0 111, xAO = (P,:,,,/760 mm Hg)y,


Table S7-3. The Constants of Integration

Evaluate N,, at z = 0 (or any z)

Co~npound


where c = n/V = P/RT = 1 m01/82.05 cm3 (298 K) = 4.09~10-' mol/cm3

c I cz -1.25~10-' 1.25~10-" - 6 . 7 3 ~ 10." 6 . 7 3 ~ lo" - 1 . 1 6 ~ 10.' 1 .87~10" - 8 . 1 9 ~ 1 0.' 8.19~10-"

Table S7-4. Molar Flux

Benzene Toluene

Converting to kg/hr of emissions = E

Ethylbenzene Xylene

Table S7-5. Diffusion Results

5x 10.' 2~ 1 0-7

Using benzene as an example:

Co~llpound


E, kglhl.

0.0009 0.006 0.002 0.006


Table S7-6. Characteristic of Concentration

Again, using benzene as an example:

Compound


Table S7-7. Diffusion Model Results

S,, ng/L

-34.74 - 130.3 -21.71

- 108.6

f > First determine R, the fraction of VOC in the oil stream belonging to each compound. The total Inass fraction of VOC in the oil layer is:

0.001 + 0.018 + 0.014 + 0.075 = 0.108. R = Inass fraction10. I08

Coinpound


0.6 kg

1 O3 1 wastewater

E, kglhr

0.002 0.006 0.001 0.004


Table S7-8. Mass Fraction VOC and Emission Factor Results

Table S7-9. Comparison of Estimate Methods

Compound


Diffusion modeling and mass transfer analysis give results nearest to flux chamber measurements. Diffusion modeling estimates were 2 to 10 times lower than flux chamber measurements. Estimates from the mass transfer analysis are 4-7 times less than the flux chamber results but are proportional to flux chanlber results. E from emission factors is 400 to 4000 times higher than flux chamber measurements.

g) E in kglhr

R

9 . 2 6 ~ 1 0-3 0.167 0.135 0.694

Compound


QUESTIONS FOR DISCUSSION 1 ) Flux chamber ineasuren~ents are probably the most reliable, despite the difficulty in

making the measurenlents and the inherent problems like vapor bubbles. The best esti~nation method appears to be mass transfer analysis, followed by diffusion modeling. However, diffi~sioil modeling might not be truly applicable to an API separator because there might not be enough distance to re-establish the boundary layer. Refineries should be motivated to make their own measurements since government-provided factors give very high estimates.

2) En~issions might be reduced by increasing freeboard height cooling the liquid inlet covering the separator.

E (kglhr)

2 3 0 3 0

100

b d e f

0.005 0.0009 0.002 2 0.03 0.006 0.006 3 0 0.008 0.002 0.00 1 3 0 0.04 0.006 0.004 100


PROBLEM 8

Prioritizing Pollution Prevention Options

Chemical Engineering Topics: process economics, process design Pollution Prevention Concepts: prioritizing pollutants from industrial sites, ranking

pollution prevention options

BACKGROUND A large ~nanufacturing facility can have hundreds to thousands of waste streams, each

with different environmental impacts, treatment costs, and technical characteristics. Choosing which streams should be the focus of waste reduction projects is not a straightforward task, not only because of the number of streams involved, but also because widely varying criteria are used in establishing priorities. In addition, even after waste streams are targeted for reduction, there may be several feasible technological options for reducing the waste and these options must be prioritized. Thus, a systematic approach to ranking pollution prevention or waste reduction projects will consist of two steps. The first step identifies the waste streams where the highest gains from pollution prevention might be realized. The second step is to propose and prioritize specific pollution prevention options for the high-ranking waste streams. Part I of this problem provides an overview of the waste stream and option prioritization processes at a large midwestern refinery. Part I1 explores the details of ranking pollution prevention options (the second task) at another refinery using different ranking procedures and discusses how the validity of the ranking results might be tested.

Part I. A pollution prevention audit at a refinery. BACKGROUND

The procedure for conducting a waste audit at a large refinery consisted of the following steps (Balik and Koraido, 199 1):

1) Subdivide the facility and designate contacts for each area. 2) Conduct a waste survey. 3) Screen the waste streams and identify the high-priority streams. 4) Develop pollution prevention options for the high-priority streams. 5 ) Screen the pollution prevention options and identify the high-priority options. 6) Evaluate the high-priority pollution prevention options.

The objectives of this problem are to provide a glimpse of the huge amount of data generated by a pollution prevention audit, to indicate the extensive number of pollution prevention options available, and to show that the only practical way to proceed to final decisions is to devise a simple (and therefore imperfect) system for prioritizing the streams and the pollution prevention options associated with them.

As a first step, the refinery was divided into 22 process units and 28 support areas. Facility Operations Supervisors became the designated contacts for each area. When a team of engineers met with the supervisors to conduct the survey, they found that the refinery had 660 waste streams. Table 8-1-1 gives the waste stream inventory for five of the 22 process units.


Table 8-1- 1 . Partial Inventory of Waste Stseams by Phase

Unit Name

Number of Streams

Gas Liquid Solid

To screen the waste streams, a rating of 0 to 5 points was assigned each stream for each of the following criteria:

a) waste quantity and frequency, b) the cost of managing existing waste, C) possible regulatory impact in the future, d) safety and health risks to the employees and the community, e) ease and cost of implementing pollution prevention options, and f) delnonstrated effectiveness of pollution prevention options.

The points for each stream were added and streams with the highest scores were assigned the highest priority for reduction.

The next step was to propose pollution prevention options. Input from plant personnel and ideas from a literature search produced 250 possible techniques for preventing pollution from the 50 highest priority waste streams. These options were screened by assigning a rating of 0 to 5 points based on each of the following criteria:

1 ) pollution prevention hierarchy: source reduction favored over recycling favored over waste treatment,

2) reduction of waste volume or disposalltreatment cost, 3) ease of implementation, 4) proven performance, 5) safety and health risks, and 6) quantifiable results.

After this preli~ninary screening was done, the top twenty options were evaluated more carefully to provide capital cost and cost savings estimates. Table 8-1-2 gives the results of this evaluation for five of the pollution prevention options. Of these options, using a coke silo, paving the loading area, and switching to reusable bins for lubricating oils are what are called "good housekeeping" practices, which reduce waste generation without substituting raw materials or modifying processes. Housekeeping practices presented the nlost cost-effective opportunities for this refinery to reduce waste.

Alkylation Cokes Fluid Catalytic Cracking Reforlner Crude Vacuum Distillation

3 18 7 3 9 12 2 8 1 1 7 9 9 6 16 1 1


Table 8-1-2. Results of Evaluating High-Priority Pollution Prevention Options

Pollution Prevention Method

Store coke in intermediate silo to allow more controlled railroad car loading.

Install inclined plate separator to increase separation of fines from water

Pipe scrubber water to desalter instead of discharging to water treatment unit.

Pave and curb residue loading area to make recovery of spills possible.

Buy high-volume lubricating oils in bulk and use reusable containers instead of purchasing the oils in 55-gallon drums.

Additional Operating Capital Operating Expenses Cost, $ Expenses, $/yr Saved, $/yr

1,000,000 200,000 1,000,000

negligible negligible 250,000

225,000 negligible 450,000

10,000 negligible 12,000

4,000 negligible 25,000

PROBLEM STATEMENT a) Calculate the number of times a rating from 0 to 5 was assigned during this audit. b) Calculate the annual net savings in operating expenses and the capital payback period for

the pollution prevention techniques given in Table 8-1-2. Because of the relatively short time frames involved, neglect interest, so that the capital payback period is capital cost divided by annual net savings in operating expenses.

QUESTIONS FOR DISCUSSION 1) How would you weight criteria a) to f), used for ranking waste streams? Would you

assign equal weights, or do you think some of the criteria are inore important than others? 2) The "prescriptive" approach used to survey the waste streams at this refinery has been

criticized because it fails to recognize the interconnectedness of different waste streams. Some investigators advocate a different method for developing a pollution prevention strategy where the process is the focus and wastes and releases are considered as losses (Pojasek and Cali, 1991). This is called the "descriptive" approach and entails a complete materials accounting and construction of a process flow diagram for each process, which requires technical knowledge of the process. What advantages do you think the descriptive approach has over the prescriptive? What are its disadvantages?


Part II. Prioritization of pollution prevention options at a petroleum refinery. BACKGROUND

A petroleum refinery in the southeastern United States followed a different procedure for evaluating pollution prevention options than that described in Part I. After a group of pollution prevention options was generated (10 options will be considered in this problem), a hierarchical ranking system was employed. The set of 10 pollution prevention options is listed and briefly described in Table 8-11-1, and the criteria chosen for prioritizing these options are shown in Figure 8-11- 1. Figure 8-11- 1 also defines the hierarchical structure of the ranking criteria. For example, it can be seen that the main criteria are risk reduction, technical characteristics, and cost factors. Cost factors are comprised of three criteria: operating and maintenance costs, capital costs, and liability costs. Liability costs are in turn made up of remedial, catastrophic, and product liability. To begin ranking the pollution prevention options, evaluations of each criterion were made for each option. The bases for evaluating the criteria of Figure 8-11-1 are described briefly in Table 8-11-2. These bases are not always obvious. Cost factors, again, are a useful example. As stated in Table 8-11-2, capital costs and operating and maintenance costs were evaluated quantitatively while liability costs, which are particularly difficult, if not impossible, to estimate, were evaluated as either marginally decreasing, not changing, marginally increasing, or significantly increasing. The cost factor evaluations are given in Table 8-11-3, where capital costs are in dollars, operating and maintenance costs are in dollars per year, and codes of -, 0, +, and ++ are assigned to the four evaluation levels for the categories of liability. For technical characteristics, only the release reduction amount was evaluated quantitatively. All other technical characteristics were given a qualitative rating, as can be seen in Table 8-11-4.

Once evaluations were complete, unitless scores were established for each of the criteria. The evaluations of Tables 8-11-3 and 8-11-4 were converted into scores from 0 to 50, with higher scores indicating greater desirability. Assigning unitless scores to the criteria evaluations leads to a systematic prioritization because it allows combinations of characteristics to be compared quantitatively. A detailed explanation of the cost factor scores for upgrading the drainage system will clarify this point. As can be seen in Table 8-11-3, the liability ratings for upgrading the drainage system were a marginal decrease in remedial liability, a marginal increase in catastrophic liability, and no change in product liability. A score of 50 was assigned to a marginal decrease in liability, a score of 33 was assigned to no change in liability, a score of 11 was assigned to a marginal increase in liability, and a score of 0 was assigned to a significant increase in liability. Therefore, the liability scores for upgrading the drainage system are 50, 1 1, and 33 for remedial, catastrophic, and product liability, respectively. The next step was to combine the three liability scores into a single score for total liability. The three types of liability were considered to be equally important in the pollution prevention program, so each liability score was assigned a weight of one third. Therefore, for the drainage system upgrade, the total liability score is equal to

50 x 0.33 + 1 1 x 0.33 + 33 x 0.33 = 31.

As can be seen in Table 8-11-3, the capital costs of upgrading the drainage systenl were by far the highest of any of the options at $25,800,000, and were assigned a score of zero. The operating costs were also the highest of the options, and were assigned a score of five. Unlike the three types of liability, capital costs, operating and lnaintenance costs, and total liability costs were not considered to be equally important in the pollution prevention program. Instead, operating costs were considered to be the most important cost factor and given a weight of 0.42,


Table 8-11-1. Pollution Prevention Options at a Refinery

Reroute desalter water

Option

1 Replace cyclones

Description

Reduce secondary emissions fro111 the refinery wastewater system by installing a line to carry hot desalter effluent water directly to the wastewater treatment units instead of draining i t to the process water sewer, which is ventilated to the atmosphere.

Replace the current particle collection equipment (cyclones) which capture fine catalyst particles generated by a petroleum cracking unit.

Install double seals on tanks

Eliminate cokes blowdown pond

Keep soils out of drain

This pond's ernissions can be elimi~latecl by eliminating the pond.

Upgrade drainage system

Modify sampling systems

Reduce barge loading emissions

Improve sour water system

Install double seals on all storage tanks with external floating roofs and add internal floaters to tanks with fixed roofs to reduce fugitive emissions.

Keep soils out of the sewer by sweeping the roadways and concrete areas and by installing sewer boxes designed to reduce soil movement into the sewer system. Soil in the sewer system not only increases the mass of solids in the wastewater sludge, but also makes hydrocarbons that adsorb onto the soils more difficult to treat.

Reduce secondary emissions by installing above-grade pressurized sewers and segregating storm water and process water.

Current sampling lines are open ended. When flushed prior to sampling, they increase sewer loadings. Replace existing sampling stations with flow-through sampling stations to reduce the oil loading in the sewer and oil drained to the deck.

Barges are used to transport some refinery products. Loading the barges generates fugitive ernissions. Install a marinc vapor loss control system to reduce barge loading emissions.

Refinery sour waters are wastewaters containing NH, and H,S. These gases are typically removed from the water and purified in a sour water stripper. The H2S can cause odor problems, so this improvement would upgrade the sour water stripper to reduce H2S emissions.

Implement quarterly LDAR Reduce fugitive e~nissio~is through early detection of leaking equipment with a Leak Detection and Repair (LDAR) program.


GOAL: Identify the Most Desirable Pollution Prevention Options for the Refinery

Risk Reduction Technical (double weight) Characteristics

Cost Factors

Resource Timeliness Release Transferabilily Operation and ~~i~~~~~~~~ Capilal Liabilil~

Utilizalion Reducdon

A A A Remedial Cataslrophic Producl

Raw Utilities Mode Quanlily Materials

Figure 8-11-1. Criteria Hierarchy


Table 8-11-2. Evaluation of Criteria

Techr~icnl C11crrcrctc.r.i.stic.s Resource utilization

Criteria

Timeliness

Basis for Evaluation

Release reduction

Transferability

Cost Fcrctors Operation and maintenance

Capital

Liability

Determine the relative impacts on human risk each option would have if impleme~~ted .

Find the nu~nber of decreases or increases in raw material and utilities for each option.

Determined as either short (less than three years), ~nediuln (three to six years) or long term (more than seven years). Shorter tern1 options considered Inore desirable.

Quantitative estimate of release reciuction in tonslyr, ancl categorization of thc reduction mode into one of four categories (listed in descending order of preference): source reduction, recycling, treatment, ancl disposal.

Evaluateti in terms of whether or not the technology required to implement the option could be transferred to other facilities near tlie refinery, other refineries, or other industries. Highest priority given to options transferable to other refineries, followed by transferability to tlie commiunity ancl otlicr industries, in that order.

Quantitative estimate in $/yr.

Quantitative esti~ilate in $

Determined qualitatively for thsee types of liability: remedial (for cleanup of current disposal sites in the future), catastrophic, and product. Options determined to cause either a marginal ciecrcase, no change, a marginal increase, or a significant increase in these types of liability.

Table 8-11-3. Cost Factor Evaluations (Klee, 1991)

Reroute desalter water Replace cyclones Eliminate coker blowdown pond Install double seals on tanks Keep soils out of drain Upgrade drainage system Modify sampling systems Reduce barge loading emissions Improve sour water system Implement quarterly LDAR

Option

- = marginal decrease, 0 = no change, + = marginal increase, ++ = significant increase

Table 8-11-4. Technical Characteristic Evaluations (Klee, 199 1 )

Capital, $

Release Reduction Resource Utilization

Operating and Maintenance, $/yr


Liability

Remedial Catastrophic Product

Option

refinery ref~nery

community industry refinery industry industry refinery refinery

Amount, tons/yr Mode'

no change no change

increase H,O, yield increase feed increase feed

no change no change

increase H 2 0 decrease H 2 0

no change S

' S = source reduction, R = recycling, T = treatment, D = disposal 'S = short term (<4 years), M = medium term (4 to 7 years), L = long term (>7 years)

Timing'

industry

no change increase elec

increase elec, H,O, steam no change no change

increase elec, H20, steam, other no change

increase elec, H,O, steam, other increase H 2 0

no change

Transfer- ability Raw Materials Utilities ----


capital costs were considered to be the next most important cost factor and given a weight of 0.32, and liability was considered to be the least important cost factor and given a weight of 0.26. (Note that the sum of the weights must equal unity.) Thus, the total cost factor score for upgrading the drainage system is

5 x 0.42 + 10 x 0.32 + 31 x 0.26 = 10. The unitless combined scores for cost factors and technical characteristics for all the pollution prevention options are given in Table 8-11-5. Risk reduction evaluations also appear in Table 8-11-5.

PROBLEM STATEMENT a) Using the data in Table 8-11-5, rank the pollution prevention options from high priority

to low priority. Assign risk reduction twice the importance of technical characteristics and cost factors, so that the sum of 0.25 times the cost factor score, 0.25 times the technical characteristics score, and 0.5 times the risk reduction score gives the ranking value. What do the high-ranking options have in common? What do the low-ranking options have in common? .

b) "The criteria weights and option evaluations used to produce a ranking can be somewhat arbitrary and are in some cases based on order of magnitude estimates. To determine if a ranking is unstable due to the uncertainties inherent in the weight and evaluation assignments, a sensitivity analysis is performed by perturbing the variables and finding the new ranking. If the ranking produced from the perturbed variables is similar to the original ranking, the original ranking is considered to be robust." Test the validity of the ranking produced in part a) by performing a sensitivity analysis of the criteria weights. To do this, assign risk reduction, cost factors, and technical characteristics equal weights and sum the points in Table 8-11-5. Compare this ranking to that of part a). Is the ranking sensitive to criteria weights? Would you recommend making adjustments to the criteria weight assignments?

c) Perform another sensitivity analysis, this time on the criteria evaluations of Table 8-11-5. Assign a value of high (3 points) to evaluations greater than 34, medium (2 points) to evaluations from 17 to 34, and low (1 point) to evaluations less than 17. Determine a ranking using these new assignments and the original criteria weights of part a). Do your results indicate that adjustments need to be made to the original ranking evaluations?

QUESTIONS FOR DISCUSSION I) Suggest additional sensitivity analyses that might be performed. 2) Can you suggest ways to streamline the ranking procedures, based on the results of parts

b) and c)?


Table 8-11-5. Point Assignments for Cost Factors, Technical Characteristics, and Risk Reduction

Option Technical Risk -1

Cost Factors Characteristics Reduction


REFERENCES AND SUGGESTIONS FOR FURTHER READING Balik, J. A. and Koraido, S. M., "Identifying Pollution Prevention Options for a Petroleum Refinery," Pollution Prevention Review, 1 :273-293, Summer 199 1.

Klee, H., Amoco/EPA Pollution Prevention Project, personal communication, 199 1.

Pojasek, R. B., and L. J. Cali, "Contrasting Approaches to Pollution Prevention Auditing," Pollution Prevention Review, 1 :225-235, Summer 199 1.


SOLUTION TO PROBLEM 8 Part I a) # of ratings = # of streams x # of stream criteria

+ # of pollution prevention options x # of option criteria

= 6 6 O x 6 + 5 0 x 6 = 4260 ratings.

Annual net savings in operating expenses = operating expense savings - additional operating expenses.

Payback period = capital costlannual net savings. Values are given in Table S8-I- 1.

Table S8-1-1. Capital Payback Period and Annual Net Savings in Operating Expenses

Method

Silo for coke storage Inclined plate separator Scrubber water to desalter Pave and curb loading area Reusable bins for oil

15 rno. 0

6 mo. 10 mo. 9 wk.

Annual Net Savings, $/yr

QUESTIONS FOR DISCUSSION I ) The relative weights assigned to each criterion vary considerably from individual to

individual. For criteria that are rated as extremely important, it might be useful to develop a pre-screening condition or conditions that must be satisfied befbre a more detailed evaluation occurs. For example, if a stream had a safety and health rank of 0 or 1 it inight be of no interest. The pre-screening score could be determined first. In this way, a more carefill ranking of the important streams could be made ~ I I the sarne amount of time as a quick ranking of all 660 streams.

2) Advantages of the descriptive approach: leads to a better understanding of the fundamental sources of waste, process interrelationships are considered. Disadvantages: detailed process knowledge is necessary.

Payback Period

Part I I a) Example: Reroute desalter water

4 1 ~ 0 . 2 5 + 3 6 ~ 0 . 2 5 + 2x0.5 = 20. Total points for all options are given in Table S8-11-1.


Table S8-11-1. Ranking with Risk Assigned a Weight of Two

All high- and mi'd-ranked options have medium to high cost factor scores and medium to high technical characteristic scores. The barge loading option has a much higher score than any other option because of its risk reduction score. All low-ranking options (coker pond, cyclones, and drainage) had low risk reduction scores and medium to low technical characteristic scores.

b) Example: Reroute desalter water 41~0.33 + 36~0.33 + 2x0.33 = 26.

Total points for all options are given in Table S8-11-2.

1

2 3 4 5 5 7

8 9 10

Table S8-11-2. Ranking with Equal Weights for the Three Groups of Criteria

Total Points Rank

High Rank Reduce barge loading emissions

Mediunz Rank Install double seals on tanks Implement quarterly LDAR Reroute desalter water Keep soils out of drain Improve sour water system Modify sampling systems

Low Rank Eliminate coker blowdown pond Replace cyclones Upgrade drainage system

Rank

Options in Order of Desirability


High Rank Reduce barge loading emissions

Medium Rank Implement quarterly LDAR Reroute desalter water Keep soils out of drain Install double seals on tanks Improve sour water system Modify sampling systems


Total Points


Barge loading emission reduction remains the top choice. The medium-ranked options remain medium ranked but their order changes. The last four options retain the same order. These results suggest that the results of part a) are not highly sensitive to changes in criteria weights.

c) The new assignment of points is given in Table S8-11-3.

Table S8-11-3. New Point Assignments

Option

Reroute desalter water Replace cyclones Eliminate coker blowdown pond Install double seals on tanks Keep soils out of drain Upgrade drainage system Modify sampling systenls Reduce barge loading emissions Improve sour water system Implement quarterly LDAR

Technical Risk Cost Factors Characteristics Reduction

Example: Reroute desalter water 3x0.25 + 3x0.25 + 1x0.5 = 2.

Total points for all options are given in Table S8-11-4.

Table S8-11-4. Ranking for High, Medium, and Low Evaluations

Rank

1

2 3 3 3 3 3

8 8 10


Higlz Rank Reduce barge loading emissions

Medi~lrn Rank Reroute desalter water Keep soils out of drain Modify sampling systems I~nprove sour water systenl Implement quarterly LDAR Install double seals on tanks



As in part b), barge loading emissions remain the top priority. The three lowest ranking options remain the three lowest ranked options, with the drainage system upgrade reinaining at the bottom. Rerouting the desalter water becomes the most important mid- ranked option. This suggests that the ranking of part a) is robust with respect to criteria evaluations.

QUESTIONS FOR DISCUSSION 1) Suggested sensitivity analyses: i) rank with respect to risk alone, or technical

characteristics alone, or cost factors alone. ii) Rank leaving risk out, or technical characteristics out, or cost factors out. iii) Assign evaluations of high and low only, iv) Make changes in criteria weights.

2) The results of parts b) and c) indicate that criteria weights and evaluations need not be determined with a great deal of accuracy, as long as it is known that they fall into a given range.


PROBLEM 9

The Use of Chlorinated Solvents: Implications for Global Warming, Stratospheric Ozone Depletion and Smog Formation

Chemical Engineering Topics: thermodynamics, transport phenomena, reaction engineering

Pollution Prevention Concepts: selection of environmentally compatible materials, development of indices for environmental effects

BACKGROUND Chlorinated solvents have properties that make them suitable for a wide range of industrial

applications. Roughly a million tons are used annually in processes ranging from vapor degreasing to the fabrication of electronic components. Chlorinated solvents are non-flammable and have normal boiling points near or slightly above room temperatures. Data on the production of five of the major chlorinated solvents are given in Table 9- 1 . The table shows that while total production has been relatively stable over the past decade, the production rates of particular solvents have seen major fluctuations. For instance, production of 1,1,2- trichlorotrifluoroethane (CFC-113) doubled between 1979 and 1989, while production of trichloroethylene (TCE) decreased by 40%. Some of the primary driving forces for these fluctuations have been environmental regulations. In particular, regulations related to smog formation, stratospheric ozone depletion and global warming have influenced the patterns of solvent usage. These environmental issues are complex and interrelated. This problem will introduce you to some of the basic concepts behind these issues.

Table 9-1. Current and Historical Production of Chlorinated Solvents (Wolf and Cramm, 1987; Wolf et al., 1991) (thousands of metric tons)

TOTAL

1155 1088 103 1 95 2 939 995 90 1 896 924 943 94 1

Year

1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989

1,1,1- Trichloro-

ethane

325 3 14 279 270 266 306 268 296 315 328 353

CFC- 1 1 3

47 5 0 53 5 6 60 6 8 73 7 3 78 78 7 8

Trichloro- ethylene

145 12 1 117 120 100 8 6 73 8 2 8 2 8 2 82

Tetrachloro- ethylene

35 1 347 3 13 265 248 260 224 188 215 226 215

Methylene chloride

287 256 269 24 1 265 275 263 25 7 234 229 2 13


Consider first the issue of stratospheric ozone depletion. Ozone is formed in the stratosphere (15-50 km in altitude) as the result of the dissociation of molecular oxygen. This reaction is driven by the absorption of radiation of wavelengths shorter than 200 nm. The atomic oxygen subsequently reacts with molecular oxygen to form ozone. The ozone in turn decomposes either by photolysis or by reaction with atomic oxygen. The mechanism for this reaction, as originally proposed by Sydney Chapman in 1930 (Chapman, 1930a,b) is:

net: absorption of radiation h < 350 nin

The net effect of this series of reactions is the formation of an ozone layer in the upper atmosphere which absorbs ultraviolet radiation that could be harmful to life on the surface of the planet. The chlorinated solvents examined in this problem play a role in this chemistry through the catalytic destruction of ozone by the following mechanism (Stolarski and Cicerone, 1974; Wofsy and McElroy, 1974):

net: 0, + 0 = 20,

Chlorinated solvents that reach the stratospheric ozone layer can photodissociate, releasing chlorine atoms that catalyze ozone destruction (Molina and Rowland, 1974). For a compound to cause stratospheric ozone depletion, it must have a lifetime in the atmosphere sufficient to reach the stratosphere and it must contain chlorine or some other halogen, such as bromine. The reactivity in the ozone cycle and the atmospheric lifetime are represented by "ki and ti in the equation

Okiti ODP, =

t (1 mole CFC-11) OkCFC-ll CFC-ll

where "ki is the rate constant for reaction of compound i with atomic oxygen, ti is the atmospheric lifetime of compound i, and ODP, is the approximate ozone depletion potential for compound i, normalized by the ozone depletion potential of CFC-1 I . The rate constants, atmospheric lifetimes, and ODPs for the major chlorinated solvents are given in Table 9-2. Note that the ODP provides a relative measure of the extent to which a coinpound contributes to ozone


destruction. On a molar basis, ODP is highest for CFC-113, which is both highly chlorinated and unreactive in the lower atmosphere.

Table 9-2. Ozone Depletion Potential Indices for the Major Chlorinated Solvents'$

NA = not available.

Compound

CH,CI, (methylene chloride)

Cl3C-CH3 (1,1,1 -trichloroethane)

HClC=CCl, (trichloroethylene)

Cl,C=CCl, (tetrachloroethylene)

C1,FC-CClF, (CFC- 1 13)

CFCl, (CFC- 1 1 )

'"For a more rigorous evaluation of ozone depletion potential, see Fisher et al. (1990a) or Atkinson et al. (1989).

Next consider the greenhouse warming potential of atmospheric gases. So-called "greenhouse" gases, such as carbon dioxide and methane, influence the overall energy balance of our planet. This balance is complex, but it is known to depend strongly on the emission of infrared radiation. When atmospheric gases intercept this re-radiated energy, the earth's energy balance can be upset and the planet warms. While this concept is quite simple, producing accurate and precise estimates of the effect of atmospheric gases on global temperature is very difficult. In this problem an index is defined that describes the global warming potential (GWP) of the five chlorinated solvents listed in Table 9-1. The global warming potential of a chemical depends on both its ability to absorb infrared radiation and the length of time that it remains in the atmosphere. Thus, an approximate global warming potential index for compound i, normalized by the global warming potential of CFC- 1 1, is given by

Rate Constant for Reaction with Atomic

Oxygen at 298 K, cm3/molecule/s

(JPL, 1987)

2 . 1 2 ~ 1 0 - ' ~

3 . 1 8 ~ 1 0 - ' ~

NA

NA

2 . 0 ~ 1 0 - ' ~

2 . 3 ~ lo-''

Atmospheric Lifetime, yrs

(WMO, 1989)

0.6

6.3

0.1

0.6

90

60

Ozone Depletion Potential Index,

l/gmol

0.009

0.15

none

none

1.3

1 .O


Iiti GWP, =

t L C - I I cFc-I I (1 mole CFC-I 1) '

where Ii is the infrared absorption band intensity of co~npound i and ti is its atmospheric lifetime. The intensities, atmospheric lifetimes, and GWPs of the major chlorinated solvents are given in Table 9-3. Note that the index is a measure of the relative rise in global temperature to be expected due to atmospheric emissions. It is highest for CFC-113, which is unreactive in the lower atmosphere and strongly absorbs infrared radiation.

Table 9-3. Global Warming Potential Indices for the Major Chlorinated Solvents#:

'For a more rigorous evaluation of global warming potential, see Fisher et al. (1990b) or Lashof and Aluya (1990).

Compound

CH,Cl, (methylene chloride)

C13C-CH3 ( l , l , 1 -trichloroethane)


Cl,C=CCl, (tetrachloroethylene)


CFCl, (CFC- 1 1)

Finally, consider the smog formation potential of the chlorinated solvents. Anyone who has visited Los Angeles during the summer has experienced the effects of photochemical smog. Smog is, in part, the result of the photochemical oxidation of hydrocarbons such as the solvents considered in this problem. The che~nistry and physics of smog formation are complex and beyond the scope of this problem, just as the chemistry and physics of global warming and stratospheric ozone depletion are complex. Once again, however, an approximate index can be calculated. In the case of smog formation potential, the index is based on the oxidation of the solvent by the hydroxyl radical (.OH). Therefore, an approximate smog formation potential index for compound i, normalized by the smog forlnation potential of 1-octene, is given by

IR Absorption Band Intensity, abs cm-' (Pouchert, 1989)

1604

1300

1978

2226

3907

2389

Atmospheric Lifetime, yrs

(WMO, 1989)

0.6

6.3

0.1

0.6

90

60

Global Warming Potential Index,

I Igmol

0.007

0.057

0.00 1

0.009

2.5

1 .0


K.

SFPi = .OHk 9

I -octene (1 mole 1-octene)

where 'OHki is the rate constant for the reaction of compound i with the hydroxyl radical. Values for this index and the rate constants for the reaction of the major chlorinated solvents with the hydroxyl radical are given in Table 9-4. This index is a measure of the tendency of the chemical to participate in photochemical reactions and is highest for trichloroethylene.

Table 9-4. Smog Formation Potential Indices for the Major Chlorinated Solvents

The three indices for ozone depletion, global warming, and smog formation potential are measures of some of the known environmental impacts of chlorinated solvents. Your assignment is to develop as complete a picture as possible of the environmental impacts of chlorinated solvent use and to determine whether the patterns of solvent usage in 1989 were better for the environment than the patterns of solvent usage in 1979.

PROBLEM STATEMENT a) Assume that the 1979 and 1989 atmospheric emissions of each solvent amounted to 80%

of production. Using the smog formation, ozone depletion and global warming indices, rank the five solvents by their impact on smog formation, ozone depletion and global

Smog Formation Potential Index,

1 /gmol

2 . 9 ~ 1 0 . ~

4 . 4 ~

5 . 1 ~ 1 0 . ~

3 . 8 ~ 1 0 . ~

1 . 1 ~ 1 0 - ~

1 .O

Compound

CH2C12 (methylene chloride)

C13C-CH3 (1,1,1 -trichloroethane)


C12C=CC12 (tetrachloroethylene)

C12FC-CCIF2 (CFC- 1 13)

C8H16 (1 -octene)

Rate Constant for Reaction with Hydroxyl Radical at

298 K, crn3/molecule/s (Atkinson, 1989)

1.3~10-"

2.0x10-'~

2 . 3 ~ 1 0 . ' ~

1 . 7 ~ 1 0 - ~ '

s . o x ~ o - ' ~

4 . 5 ~ 1 0 - ~ ~


warming. Compare the rankings for the three environmental impacts. Compare the overall magnitudes of the indices for the years 1979 and 1989.

b) Suppose that an international body has decided that all CFCs.must be replaced by an equal mass of 1,1,1 -trichloroethane. Repeat the calculations of part a) for this scenario using the 1989 data. Compare the overall magnitudes of the three indices for this case with the overall magnitudes for 1989.

QUESTION FOR DISCUSSION How are ozone depletion potential, smog formation potential and global warming potential

interrelated?

REFERENCES AND SUGGESTIONS FOR FURTHER READING Atkinson, R., Baulch, D. L., Cox, R. A., Hampson, R. F., Ken-, J. A., and Troe, J., "Towards a Quantitative Understanding of Atmospheric Ozone," Planet. Space Sci., 37, 1605- 1620, 1989.

Atkinson, R., "Kinetics and Mechanisms of Gas Phase Reactions of the Hydroxyl Radical with Organic Compounds," Jour~zal of Physical and Clzemical Reference Data, Monograph No. 1, American Chemical Society and the American Institute of Physics, 1989.

Chapman, S., "A Theory of Upper Atmospheric Ozone," Menz. Roy. Met. Soc., 3, 103, 1930a.

Chapman, S., "On Ozone and Atomic Oxygen in the Upper Atmosphere," Philos. Mag., 10, 369, 1930b.

Davidson, J. A., Schiff, H. I., Brown, T. J., and Howard, C. J., "Temperature Dependence of the Rate Constants for Reactions of o('D) Atoms with a Number of Halocarbons," J. Clzenz. Phys., Vol. 69, No. 9, 1 November 1978, pp. 4277-4279.

Fisher, D. A., Hales, C. M., Filkin, D. L., KO, M. K. W., Sze, N. D., Commell, P. S., Wuebbles, D. J., Isaksen, I. S. A., and Stordal, F., "Model Calculations of the Relative Effects of CFCs and Their Replacements on Stratospheric Ozone," Nature, 344, 508-5 12, 1990a.

Fisher, D. A., Hales, C. H., Wang, W-C., KO, M. K. W., and Sze, N. D., "Model Calculations of the Relative Effects of CFCs and Their Replacements on Global Warming," Nature, 344, 5 13- 516, 1990b.

Jet Propulsion Laboratory (JPL), "Chemical Kinetics and Photochemical Data for Use in Stratospheric Modeling, Evaluation Number 8," JPL Publication 87-4 1, 1987.

Lashof, D. A., and Aluya, D. R., "Relative Contributions of Greenhouse Gas Emissions to Global Warming," Nature, 344, 529-53 1, 1990.

Molina, M. J., and Rowland, F. S., "Stratospheric Sink for Chlorofluoro~nethanes: Chlorine Atorn-Catalyzed Destruction of Ozone," Nature, 249, 8 10, 1974.


Pouchert, C. J., The Aldriclz Libmry of FT-IR Spectra, Vapor Plzase, Edition 1, Vol. 3, "Non- , Aromatic Halogenated Hydrocarbons," The Aldrich Chemical Company, Inc., Milwaukee, WI,

1989.

Ramanathan, V., "Greenhouse Effect Due to Chlorofluorocarbons: Climatic Implications," Scierzce, Vol. 90, October 1975, pp. 50-52.

Rogers, J. D., and Stephens, R. D., "Absolute Infrared Intensities for F-113 and F-114 and an Assessment of Their Greenhouse Warming Potential Relative to Other Chlorofluorocarbons," Jo~ir r~al of Geol2lzysicnl Resenrclz, Vol. 93, No. D3, pp. 2423-2428, March 20, 1988.

Stolarski, R. S., and Cicerone, R. J., "Stratospheric Chlorine: A Possible Sink for Ozone," Can. J. Clzer~z., 52, 1610, 1974.

Wofsy, S. .C., and McElroy, M. B., "HO,, NO, and C10,: Their Role in Atmospheric Photochemistry," Can. J. CIze17z., 52, 1582, 1974.

Wolf, K., and Cramm, F., "Policies for Chlorinated Solvent Waste--An Exploratory Application of a Model of Chemical Life Cycles and Interactions," RAND R-3506-JMOIRC, 1987.

Wolf, K., Yazdani, A., and Yates, P., "Chlorinated Solvents: A Case Study of Trade Offs," J. Air Wciste Manage. Assoc., Vol. 41, No. 8, pp. 1055- 1061, Aug. 1991.

World ~ e t e o r o l o ~ i c a l ' Organization (WMO), Global Ozone Research and Monitoring Project-- Report No. 20, "Scientific Assessment of Stratospheric Ozone: 1989 Volume 11," 1989.

Pollution Prevention: Homework and Design Problems for Engineering curricula

SOLUTION TO PROBLEM 9 a) The data presented in Table 9-1 must be converted from a molar basis to a mass basis.

Sample conversion from tons to moles:

1 metric ton CH2C12 = = 1 1,765 gmol.

From Table 9- 1, 80% of production for CH2C12 gives

Atmospheric emissions in 1979 = 287,000x0.8 = 229,600 T

or 2.7x109 moles. Similarly, in 1989 emissions of CH2C12 were 2.0x109 moles. Combining with data in Table 9-4 gives smog formation potential (SFP) for emissions of CH,C12 in 1979 = 2 . 7 ~ 1 0 ~ * 2 . 9 ~ 1 0 - ~ = 7.83x106.

Table S9- 1. Atmospheric Impact of Tonnage of Solvent Emissions; Ranking in Parentheses ( )

Rankings: i 1 Smog Formation Potential

The rankings do not change between 1979 and 1989 despite the wide variation in solvent use. For example, the use of trichloroethylene dropped -40% (from 145,000 tons to 82,000 tons) and the use of trichloroethane increased -9% in the same 10 year period. The critical issue in smog formation is the presence of unsaturated bonds in the chemical structure of the solvent.

Compound

CH,Cl, (methylene chloride)

C1,C-CH, (1,1,1-trichloroethane)

HClC=CCI, (trichloroethy lene)

C1,C=CC1, (tetrachloroethylene)


Total

Smog Formation Potential

1979

(2) 7.8x106

(4) 8.6~105

(1) 4.5x107

(3) 6.4x106

(5) 2.2x103

6.Ox1O7

1989

(2) 5.8~10"

(4) 9.3x105

(1) 2.5x107

(3) 3 . 9 ~ 106

(5) 3.7x103

3.6x107

Global Warming Potential

1979

(3) .9x107

(2) l.lxlOX

(5) 8.8x105

(4) 1 .5x107

(1) 5.0x10X

6.5x108

Ozone Depletion Potential

1989

(3) 1 .4x107

(2) 1 .2x108

(5) 5.0~105

(4) 9.3x106

(1) 8.3x108

9.8x108

1979

(3) 2.4x107

(1) 2.9xIOX

0

0

(2) 2.6x108

5 .8x lP

1989

(3) 1 .8x107

(2) 3.2x10Y

0

0

(1) 4 .3~10"

7 .7~10"


ii) Global Warming Potential Again, the rankings do not change between 1979 and 1989. Solvents with

high atmospheric lifetimes rank high in this category. Unsaturated solvents that react in the troposphere to form smog rank lowest, followed by solvents that are highly chlorinated.

iii) Ozone Depletion Potential TCA and CFC-113 trade places: TCA ranked highest in 1979 and second

in 1989, while CFC- 113 ranked highest in 1989 and second in 1979. Magnitude of indices:

Magnitudes of ozone depletion potential and global warming potential are higher overall in 1989 than in 1979, while smog formation potential decreased. For all three indices, the increase/decrease was in the 30-50% range.

b) All CFCs are replaced by an equal mass of 1,1,1-trichloroethane.

Table S9-2. Impact of CFC Replacement with Trichloroethane for 1989

Adding the sniog formation potentials for 1989 from Table S9-1 (prior to hypothetical CFC-113 ban) gives a total sniog formation potential of

After the ban. the total would be

New Ozone Depletion Potential

3.9x108

0

4.1x108

Therefore, the replacement of CFC- 1 13 with 1,1,1 -trichloroethane would not impact smog formation. All potential totals are listed in Table S9-3.

Cornpound

CI,C-CH, ( 1 , 1 , 1 -trichloroethane)

C12FC-CClF, (CFC- I 13)

New Total

New Smog Formation Potential

1.1x106

0

3 . 6 ~ lo7

New Solvent Mass

(metric tons)

43 1,000

0

New Global Warming Potential

1 .5x108

0

1.7x10X


Table S9-3. Potential Totals Before and After Substitutio~l

Total global warming potential decreases by more than 80% while ozone depletion potential decreases by approximately one half when an equal mass of 1,1,1 -trichloroethane is substituted for CFC- I 13.

Total

Smog Formation Potential

Global Warming Potential

Ozone Depletion Potential

QUESTION FOR DISCUSSION Compounds with high hydroxyl radical reaction rates have short atmospheric lifetimes.

Therefore, there is a tradeoff between low smog levels and high ozone depletion and global warming levels.

Before Substitution

3 . 6 ~ 1 O7

9 . 8 ~ 10"

7 . 7 ~ I OX

After Substitution

3.6x107

1 . 9 ~ 10'

4.1x10X


PROBLEM 10

Choosing a Degreasing Solvent: An Environmental Dilemma

Chemical Engineering Topics: thermodynamics, reaction engineering, transport phenomena

Pollution Prevention Concepts: selecting environmentally compatible materials, choosing between solvents

BACKGROUND Metals are frequently coated with thin films of light hydrocarbon oils in order to prevent

oxidation of the raw metal feedstock during shipping and storage. Heavy oils, esters and particulate matter tend to accumulate in these films during metal-forming operations. A co~nbination of immersion solvent tanks and vapor degreasing tanks is used to remove the oils and dirt and provide clean parts for surface treatment operations. A conceptual diagram of a vapor degreasing operation is shown in Figure 10- 1 . In the degreaser, liquid solvent is vaporized. The parts to be cleaned are placed in the vapor zone and the hot solvent condenses on the cold part. The condensed solvent, contaminated with the dirt from the cleaned part, drips back into the liquid pool. Solvent losses are controlled by the use of condensers and freeboard, as shown in Figure 10-1. However, 90% of the solvent used may eventually escape into the atmosphere.

Trichloroethylene (TCE) used to be the most widely used solvent for degreasing and metal cleaning. Since the connection between TCE and photochemical smog was established, 1,1,1- trichloroethane (TCA) has been substituted for TCE. In this problem you are asked to evaluate the decision to substitute TCE with TCA on the basis of not just smog formation potential, but also global warming and stratospheric ozone depletion potentials. See Problern 9 for a discussion of these potentials.

PROBLEM STATEMENT a) Use the information in Tables 9-2 to 9-4 to estimate the smog formation potential, global

warming potential and ozone depletion potential per ton of TCE and TCA emissions. Note that the s~nog formation potential, global warming potential, and ozone depletion potential are reported on a molar basis.

b) Discuss the decision to replace TCE with TCA on the basis of the results in part a). Assume that the degreasing efficiency of TCE and TCA is equivalent on a mass basis.

QUESTIONS FOR DISCUSSION 1 ) The largest expense in vapor degreasing, besides labor, is the solvent. Solvent losses

occur when vapor diffuses into the atmosphere from the degreaser. Dragout, which is the solvent that remains on the degreased part after it has been removed from the vapor, is another cause of solvent loss. Suggest methods for preventing solvent emissions. (HINT: You may find Perry's Cl~emical Engineer's Handbook useful.)

2) Discuss the possibility of replacing the chlorinated degreasing solvents with non-regulated solvents/solutions. What characteristics or properties are required of a degreasing solvent/solution?

Condenser


Freeboard

Heat

Figure 10-1. A Vapor Degreaser


SOLUTION TO PROBLEM 10 a) The values in Tables 9-2 to 9-4 must be converted from a per mole basis to a per ton

basis. The smog formation, global warming and (stratospheric) ozone depletion potentials for trichloroethylene (TCE) and trichloroethane (TCA) are:

Table S 10- 1. Comparison of Potentials for TCE and TCA

b) The increasingly stringent regulation of TCE, a photochemically reactive chlorinated solvent, encouraged industry to convert to TCA. The results in part a) indicate that while TCA plays a relatively minor role in photochemical smog formation, its contribution to global warming and to the depletion of stratospheric ozone is relatively significant. A dilemma arises from the conflicting criteria used in choosing a solvent. For instance, should we replace a photochemically reactive solvent that breaks down easily in the troposphere (lower atmosphere) with a solvent that is persistent and makes its way up into the stratosphere? Once there, the solvent will absorb infrared radiation and contribute to global warming or release a chlorine atom which then participates in the catalytic destruction of the ozone layer.

Clearly, the selection of replace~~ient solvents must be based on multiple criteria that include assessments of all the harmful effects of the solvent.

QUESTIONS FOR DISCUSSION 1 ) The vapor degreasing process can be changed in a variety of ways to reduce solvent loss.

Among the possibilities are: i 1 Lid. The degreasing tank generally stays open as parts are lowered into and raised

from the vapor zone, as well as between uses. An improved operating practice would be to install a sliding lid or cover to prevent evaporation losses between uses and during shutdown.

ii) Cor7derzsation. Installation of a condensation unit above the "freeboard" will cause the vapor to condense and will reduce evaporation loss during shutdown and idle periods. The non-condensed component of the vapor stream will escape into the atmosphere.

iii) Ccrrborz Adsorl7tiorz. The escaping vapor may be captured with an exhaust system that has a carbon filter. The activated carbon is then stripped with steam to recover the solvent. The problem with this method of emission reduction is that it generates a wastewater stream when the carbon is regenerated and a solid waste stream when the carbon is spent and has to be discarded.

Ozone Depletion Potential per gmol

(per ton)

negligible

0.15 (l.0x103)

Global Warming Potential per g~nol

(per ton)

l x 10.' (6.9)

0.057 (390)

Solvent

TCE

TCA

Smog Formation Potential per gmol

(per ton)

5 . 1 ~ 1 0 . ~ (3.5x102)

4 . 4 ~ 1 0 . ~ (3.0)


The substitute degreasing solvent must (i) remove all contaminants and not attack the part to be cleaned, (ii) evaporate rapidly to leave a dry surface (so the solvent will not interfere with subsequent lllanufacturing steps), (iii) have low toxicity, (iv) have low flammability and (v) have negligible effect on smog formation, global warming, and ozone depletion. The requirement that the metal part be safe from attack and be dry generally rules out acidic and basic solutions. The selection criteria create a substitution dilemma. A high evaporation rate leads to high emissions and low evaporation rate will create solvent dragout.


PROBLEM 11

Thermodynamic Constraints in the Reprocessing of Commin- gled Plastics

Chemical Engineering Topics: thermodynamics, polymer processing Pollution Prevention Concepts: selecting environmentally compatible materials, plastics

recycling

BACKGROUND Plastics can be made from a wide range of molecular building blocks called monomers.

Use of different lnonolners gives a variety of plastics with different physical and chemical characteristics. Solne of the commonly used plastics, such as polyvinylchloride, polystyrene, and polyethylene, are produced by the polymerization of one monomer. They are called homopolymers. Other plastics, such as styrene lnaleic anhydride and styrene acrylonitrile, are co~nbinations of more than one monomer within a single polymer. They are called copolymers. Very few plastics are made frorn combinations of different polymers, yet this is the type of material that is frequently generated by recycling operations. A representative homopolynier, copolymer, and commingled plastic are pictured in Figure 1 1 - 1. Often, the different plastics that are commingled during recycling are not miscible when melted and cannot be reprocessed together. Miscibility increases with increasing temperature, but the temperature to which commingled plastics can be raised is limited because at high temperatures the plastics undergo complete degradation. This problem will examine the thermodynan~ic basis for determining whether plastics can be co~n~ningled in reprocessing facilities.

The miscibility of plastics depends on the Gibbs free energy of mixing, which is equal to

where T is temperature in K, AH,,, is the enthalpy of mixing, and AS,,, is the entropy of mixing. A negative AG,,, is one of the necessary conditions for miscibility. Gibbs free energy, entropy, and enthalpy in equation ( 1 ) and the equations that follow are expressed per unit volume. Researchers have demonstrated that high ~nolecular weight plastics, in general, are miscible if there is favorable interaction between the nlonomer building blocks co~nprising the plastics (Brannock and Paul, 1990; Nishimoto et al., 1989; Paul and Barlow, 1984). The case of binary (two component) polymer mixtures has been studied in detail (tenBrinke et al., 1983,1984; Alexandrovich et al., 1977). The enthalpy of mixing for a binary plastic mixture is given by

where B is a measure of the net energy of interaction between the plastics and $;, and $, are the volu~ne fractions of plastics a and b. B is estimated by calculating the energy of interaction of various lnonolner units. For a two-plastic mixture, where plastics a and b contain monomers 1, 2 and 3, the value of B is given by the expression


H H I I . . . . -C-C- . . . .

I I H H

Polyethylene, a homopolymer

I Styrene acrylonimle, a copolymer

H H I I

. . . . -C-C- . 8 . .

I I :N=C H H H H H I 1

. , . . -c -c- . . . I I

. -C-C- . . . .

I I H H

:N=C H H H I I I I

A . . . . -c-c-. . . , -c-c- . . . H H

I I I I 6 . . . . -C-C- . . . . H H I I

H H Commingled Polyethylene and Styrene acrylonitrile

Figure 11-1. A Homopolymer, a Copolymer, and a Commingled Plastic


where $,I is the volume fraction of nlonolner i in polyrner a, $,'I is the volume fraction of monomer i in polynler b, and Bis is a parameter describing the interaction between monomer units i and j. In this problem a two-plastic ~nixture where polymer a contains monomer units 1 and 2 and polymer b contains monomer unit 3 will be considered. In this case equation 3 reduces to

Note that 9," from equation (3) equals one, and that the enthalpy of mixing per unit volurne depends on the average volume fractions of the monoiners in the polymer blend and is assunled to be independent of ~nolecular size. The entropy of mixing per unit volume, on the other hand, is strongly dependent on the molecular size of the plastics and is equal to

where Vi is the molar volurne of plastic i and r is the n ~ ~ ~ n b e r of plastics in the mixture. For a two-component system this becomes

Thus, for a mixture of poly~ner a and poly~ner b, the free energy of mixing per unit volume is represented by the Flory-Huggins expression (Flory, 1953):

A negative value of AG,,, is a necessary but not sufficient condition for nliscibility. For binary mixtures the additional requirement for stability is

where Qi is the volume fraction of either plastic.

PROBLEM STATEMENT a) Calculate the value of the interaction energy, parameter B, for a rnixture of styrene

acrylonitrile (SAN) and polymethyl~nethacrylate (PMMA). SAN and PMMA have the


following monomer interaction values (Nishimoto et al., 1989): monomer 1 = styrene (65 vol% in SAN) monomer 2 = acrylonitrile (35 vol% in SAN) monolner 3 = lnethylmethacrolate B,? = 6.74 cal/cm3 B,, = 0.181 cal/cm3 B2, = 4.1 1 cal/c1n3

b) Develop a free energy diagram (plot AG,,,/RT vs volume fraction @,) for the mixture of the copolymer styrene acrylonitrile (SAN) and the holnopolymer polymethyl~nethacrylate (PMMA). Perform your calculations at 2S°C, 100°C, 200°C and 300°C. Ignore the effects of changes in temperature on density.

V,,, = 1 . 6 ~ 10' ~ m " ~ r n o l V,,,,, = 1 . 5 ~ 10%cm3/gmol

c) From the results of part b), determine the composition range over which SAN and PMMA are miscible. Perform your calculatio~ls at 2S°C, 100°C, 200°C and 300°C.

d) An alternative method for estimating endothermic heats of mixing per unit volume for polymers is

where 6,, and 6, are the solubility parameters for polymer a and polymer b (Billmeyer, 1962). For polystyrene and polymethylmethacrylate, find two expressions for the enthalpy of mixing: one from equation (4) and one from the solubility parameter approach. The solubility parameters for polystyrene and polymethylmethacrylate are 8.6 and 9.1 (cal/cm")"', respectively. Comment on the differences between the two approaches.

REFERENCES AND SUGGESTIONS FOR FURTHER READING Alexandrovich, P., Karasz, F. E., and MacKnight, W. J., "Partial Miscibility in the System Poly(para-chlorostyrene-co-ortho-cl~lorosty1e1e)/poly(2,6-di1nethyl- I ,4-phenylene oxide)," PoIj)171er, 1977, 1 8, pp. 1022- 1026.

Billmeyer, F. W., Textbook of Polynzel+ Science, Wiley Interscience, 1962.

Brannock, G. R., and Paul, D. R., "Phase Behavior of Ternary Polymer Blends," Mcrcr.ol~zoleccrles, 1990, 23, pp. 5240-5250.

Flory, P. J., Prir~ciples qf Polylner Che171istry, Cornell University Press, Ithaca, NY, 1953.

H -: altlson, E. Z., "Plastics in the Waste Stream," Cornell Waste Management Institute, Quarterly, 1990.

Nishimoto, M., Keskkula, H., and Paul, D. R., "Miscibility of Blends of Polymers Based on Styrene, Acrylonitrile and Methyl Methacrylate," Polylne~, 1989, 30, pp. 1279- 1286.


Paul, D. R., and Barlow, J. W., "A Binary Interaction Model for Miscibility of Copoly~ners in Blends," Polynzer, 1984, 25, pp. 487-494.

tenBrinke, G., Bukovic, R., Rubinstein, E., Karasz, F. E., and MacKnight, W. J., "Phase Behavior in Copolymer Blends of Polystyrene and Poly(o-chlorostyrene-co-12-chlorostyrene)," 1. Appl. Plzys., 1984, 56(9), pp. 2440-2443.

tenBrinke, G., Karasz, F. E., and MacKnight, W. J., "Phase Behavior in Copolymer Blends: Poly(2,6-dimethyl- l,4-phenylene oxide) and Halogen-Substituted Styrene Copolymers," Mncronzolecule.~, 1983, 16, pp. 1827- 1832.


SOLUTION TO PROBLEM I I

a> B = ~ , , b : +k3b ; -BI2blb:

Inserting monomer interaction parameters values and @: = 0.65, $5 = 0.35 gives B =

0.0228 cal/cm3. b) Values of AG,,,/RT calculated using Eq. (7) at T = 298 K, 373 K, 473 K and 573 K are

tabulated below and plotted in Figures S 1 1 - 1 to S 1 1-3.

Table S 1 1 - 1. AG,,,/RT for SAN-PMMA Blend, mol/cm% lo6

Volume Fraction SAN

0 0.0 1 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.30 0.40 0.50 0.60 0.70 0.78 0.80 0.82 0.84 0.86 0.88 0.90 0.92 0.94 0.96 0.98 0.99 1 .00

298 K

0 0.027 0.13 0.4 1 0.73 1.1 1.4 1.7 2.1 2.4 2.7 3 .0 4.2 4.9 5.1 4.9 4.1 3.2 2.9 2.6 2.3 2.0 1.7 1.3 1 .0 0.68 0.38 0.1 1 0.012 0

373 K

0 -0.05 -0.018 0.12 0.29 0.49 0.70 0.9 1 1.1 1.3 1.5 1.7 2.5 3 .O 3.2 3.0 2.5 1.8 1.7 1.5 1.3 1.1 0.85 0.64 0.44 0.25 0.078

-0.042 -0.065 0

473 K

0 -0.1 1 -0.15 -0.13 -0.075 0.01 1 0.1 1 0.22 0.34 0.45 0.57 0.68 1.2 1.5 1.6 1.5 1 . 1 0.73 0.62 0.51 0.39 0.28 0.16 0.056

-0.041 -0.12 -0.17 -0.17 -0.13 0

573 K

0 -0.16 -0.23 -0.30 -0.31 -0.30 -0.27 -0.23 -0.17 -0.12 -0.058 0.0022 0.28 0.47 0.53 0.45 0.24 0.0044

-0.058 -0.12 -0.18 -0.23 -0.28 -0.33 -0.35 -0.36 -0.33 -0.25 -0.17 0


0.0 0.2 0.4 0.6 0.8 1 .0 Volume Fraction SAN

Figure S11-1. Free Energy Diagram

0.00 0.05 0.10 0.15 0.20

Volume Fraction SAN

Figure Sl l -2 . Free Energy Diagram for Low Volume Fraction SAN

-0.5 0.75 0.80 0.85 0.90 0.95 1 . O O

Volume Fraction SAN

Figure Sll-3. Free Energy Diagram for High Volume Fraction SAN


C) The two plastics are miscible when the conditions AG,,JRT < 0 a n d d ' ( ~ ~ , ~ / ~ ~ ) / d $ : > 0 are satisfied. The second condition is:

where $, = I-$,, B = 0.0228 cals/cm3, V, = 1 . 6 ~ 10' cm3/gmol and V, = 1 . 5 ~ 10%m"~mol. Values for the second derivative at different temperatures and volume fractions are tabulated below.

Table S 11 -2. Second Partial of AG,,,/RT with Respect to Volume Fraction SAN, mol/cm3

Final results from Tables S 1 1 - 1 and S 1 1-2 are as follows:

Volume Fraction SAN

0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20

0.78 0.80 0.82 0.84 0.86 0.88 0.90 0.92 0.94 0.96 0.98

473 K

270 110 63 3 7 2 1 11 3.9

-1.5 -5.7 -8.9

-10 -7.4 -3.9 0.59 6.4

14 25 42 69

120 290

298 K

240 8 6 3 4

8.4 -7.1

- 17 -25 -30 -34 -37

-39 -36 -32 -28 -22 -14

-3.4 13 4 1 9 6

260

573 K

280 120 7 1 45 30 20 12 6.9 2.8

-0.47

-1.7 1.1 4.6 9.1

15 23 3 4 5 0 78

130 3 00

373 K

260 100 5 0 24

8.4 -1.9 -9.1

-15 - 19 -22

-23 -20 -17 -12

-6.6 1.1

12 29 56

110 280


Table S 1 1-3. Results for the Four Temperatures

d) From the solubility parameter approach:

AH,,, = :I Qb(8 n -8J2 = (8.6-9. 1)2$il$i) = 0.25 cal/cm3$;,~,, from Equation 4:

Temperature

298

373

473

573

= B [ 34 ; iB23$i -B 2$; $; (4) which simplifies to, B = B,, = 0.181 cal/cm\ince $,/ = 0 (polystyrene and polymethylmethacrylate each contain one monomer). Therefore, AH,,, = 0.181 cal/cm3$,$,. Using the solubility parameter approach, the value of interaction parameter B will always be positive. That is why the equation is valid for endothermic systems only. Also, the solubility parameter approach predicts narrower ranges of miscibility for this binary mixture.

AG i 0 met

never

0-0.03, 0.97- 1

0-0.07, 0.9 1 - 1

0-0.19, 0.79- 1

J ~ A G - > O met

aQ2

0-0.09, 0.9 1 - 1

0-0.1 1, 0.87- 1

0-0.15, 0.83- 1

0-0.19, 0.79- 1

~ 0 t h Constraints Met

never

0-0.03, 0.97- 1

0-0.07, 0.9 1 - 1

0-0.19, 0.79- 1


PROBLEM 12

Additives for Enhancing the Miscibility of Commingled Plastics

Chemical Engineering Topic: thermodynamics Pollution Prevention Concepts: selecting environmentally compatible materials, enhancing

plastics miscibility for recycling

BACKGROUND See Problem 11

PROBLEM STATEMENT The plastics polyrnethylmethacrylate (PMMA) and polyacrylonitrile (PAN) are immiscible

at 100°C. It has been suggested that addition of a third plastic could make the ternary system miscible. If the binary mixture of PMMA:PAN has a volume ratio of 10: 1 and the third plastic is styrene-maleic anhydride (SMA), determine the amount of SMA required to make the mixture miscible at 100°C. Use Equations (1) and ( 5 ) and the conditions for miscibility given in Problem 1 1 . For this system, the entropy terms can be neglected, simplifying the equations so that it is possible to solve them analytically. Note that the enthalpy of mixing for a ternary plastic blend is given by

Here, a denotes SMA, b denotes PAN, and c denotes PMMA. The monomer interaction parameter values at 100°C are 4.0 callcm" 5.0 cal/cm" and -0.79 cal1cm"or B,,, B,,, and Bc,, respectively (Nishimoto et al., 1989).

REFERENCE Nishimoto, M., H. Keskkula, and D. R. Paul, "Miscibility of Blends of Polymers Based on Styrene, Acrylonitrile, and Methylmethacrylate," Polyiner, 1989, 30, 1279- 1286.


SOLUTION TO PROBLEM 12 Inserting the values for B,,, B,, and B,, in the ternary equation gives

It is also known that

Combining the above equations gives

AH,,, = 4.0(0. l$,J(l - 1. 10,) 4- 5$,(O. 1 Q c ) - 0.79$,(1 - 1 . Iqc) ,

Taking the derivative of this equation twice gives

The equations above show that AH,,, will be negative at $, < (0.3910.92) = 0.42. If the entropy of mixing is negligible, the plastics will be miscible if $, < 0.42. The second partial derivative of AH,,, is positive for all voluine fractions. Therefore, the limiting value for the additive SMA is $, = 1-1.1$, = 0.54. In the original binary mixture $, = 0.90; in the ternary inixt~~re $, must have a value equal to or less than 0.42 and therefore the volun~e fraction of the additive SMA must exceed 0.54 in the ternary mixture.


PROBLEM 13

Minimizing Solvent Emissions from Vapor Degreasers

Chemical Engineering Topic: transport phenomena Pollution Prevention Concepts: design of unit operations for waste minimization, vapor

losses from degreasers

BACKGROUND Vapor degreasers like the one pictured in Figure 13-1 are widely used for cleaning metal

parts. The parts are suspended in a solvent vapor zone, where solvent condenses on them and drips off, taking contaminants with it. Typically, the piece to be cleaned is held in the vapor until its temperature has equilibrated with the vapor and condensation ceases. Unfortunately, vapor degreasers are often open to the atmosphere for ease of dipping and removing parts and because covering them might generate an explosive mixture. The open tanks can be a significant source of solvent emissions. Solvent losses can be reduced by installing a condenser at the top of the vapor zone or by increasing the length of the freeboard, which is the portion of the degreaser that extends above the condenser.

In this problem you will analyze methods for reducing solvent loss. In order to keep the analysis manageable, it is necessary to make a nurnber of silnplifying assumptions. Consider a cylindrical degreaser with a diameter of eight feet and a freeboard of five feet. The solvent is 1,1,1 -trichloroethane (TCA), which has the following physical properties:

Table 13- 1 . Physical Property Data for I , I , I -Trichloroethane

Assume that the condenser cools the vapor to the ambient temperature of 298 K. Also assume that the gases behave ideally, that the mole fraction of TCA in the air moving above the degreaser is effectively zero, and that diffusivity is independent of concentration.

Property

Normal boiling point, K Vapor pressure at 25OC, mm Hg Molecular weight, g/mol Heat of vaporization, cal/mol Approxilnate diffusivity in air, cm2/s

PROBLEM STATEMENT a) A further simplification can be made to the problem of estimating enlissions by assulning

that temperature is uniform across the vapor surface. Estimate the emissions in pounds per day from this degreaser. Use the following equation for the f lux of TCA through stagnant air:

Value

347 127.3

133.42 80 12.7

0.1


where NTcA, is the molar flux of TCA in the z direction, c is the total molar concentration of the air, D is the diffusivity of TCA in air, and xTcA is the molar fraction of TCA. For a derivation of this equation, see Bird, Stewart, and Lightfoot (1960). Combine the mass flux equation with the results of a mass balance over an incremental column height Az. Using the results of part a), determine how much the mass flux would be reduced if the freeboard were doubled. Using the Clausius-Clapeyron equation, estimate the vapor pressure of TCA at 50°F and calculate the mass flux of TCA from the degreaser if the condenser is operated at 50°F. Compare this result with the emission estimate of part a).

QUESTIONS FOR DISCUSSION 1) If you were the engineer assigned to reducing the emissions from a degreaser, what

design variables would you need to consider? 2) Will actual emissions be higher or lower than your estimates? Why? Do you feel that

the simplified scenarios you solved for revealed the important design variables?

REFERENCE Bird, RB., Stewart, W. E., and Lightfoot, E. N., Transport Phenomena, Wiley, New York, 1960.

Condenser

Free board

z = O

Heat

Figure 13-1. A Vapor Degreaser


SOLUTION TO PROBLEM 13 a) By doing a mass balance over an incremental column height Az we know that:

where N,,,, is the molar flux of TCA in the z direction. From a mass balance and Fick's law (see Bird, Stewart, and Lightfoot, 1960) we know

where c is the total molar concentration, x,,, is the mole fraction of TCA a~ ld D is the diffusivity. Combining these two equations gives

The boundary conditions are: at z = 0 (condenser height)

- Vapor pressure of TCA - - 127.3 'TCA -

= 0.1675 Total pressure 760

and at z = 5 ft

Solving yields

where x,,, is the mole fraction of TCA at a height z (in feet above the condenser). To evaluate the flux, we must determine dx,,,/dz. Evaluating this derivative and using it in

yields


Since

1 atmosphere mol c = = 0.0409 -, RT e

mol NTCAZ = 5x10 -6

ft sec

The total emissions are

5 x 1 0 6 mol emissions = flux (area) = x n(4 ft)2

ft sec

133 g 2.2 lb,,, (3600x24) sec - X X X

lb - 6 -

mol 1000 g day day

b) E~nissions = 1

I [ 1 where z is the freeboard height. It can be zh

1 - 0.01675

seen from this equation that if area, c and D remain constant while freeboard z,, is doubled, the emissions are halved.

c) Estimate the vapor pressure of TCA at 50°F from the Clausius-Clapeyron equation:

Based on the data for 25"C, it can be determined that

Tlie boundary condition at z = 0 becomes


Solving as in part a) yields

Evaluating dxTCA/dz and substituting yields

where

1 atm mol c = - = 0.04306 -

RT e

The flux becomes

mol N,,,, = 2.2~10-"

ft2sec

and the emissions become 3 Iblday which is half as much as in part a).

QUESTIONS FOR DISCUSSION 1) Design variables to consider

freeboard height solvent vapor pressure temperature in the vapor zone condenser temperature degreaser surface area

2) Actual emissions are likely to be higher than these estimates because of convective mass transfer. At least some important design variables are revealed.


PROBLEM 14

The Design of Distillation Column Reboilers for Pollution Prevention

Chemical Engineering Topics: engineering economics, process design, heat transfer Pollution Prevention Concepts: design of unit operations for minimizing waste, avoided

cost

BACKGROUND Fouling of heat transfer tubes in heat exchangers is a maintenance concern in the chemical

and petroleum industries. The fouling, which is due to scale and corrosion products forming on heat exchange surfaces, adds resistance to heat transfer. Often the deposit layer is large enough to reduce the flow of heat transfer fluids and increase the pressure drop through the heat exchanger. The losses in heat transfer efficiency due to fouling result in higher energy requirements. When the energy requirements become too high, or the amount of heat transferred falls below design specifications, the heat exchanger must be shut down and cleaned. If the heat exchanger is a reboiler for a distillation column, shutdown of the heat exchanger can mean shutdown of the column. As the column is shut down, the material in the column (the hold-up) can become off-specification product requiring disposal.

This problem will examine whether product loss and waste disposal costs provide enough economic incentive to modify reboiler design. Two modified designs will be considered. In the first modified design the size of the reboiler is increased, allowing a longer period between reboiler shutdowns. The second modified design employs a spare reboiler, eliminating the need for colun~n shutdown. The design alternatives are shown in Figure 14-1.

For this problem, heat transfer area will be calculated using an overall heat transfer coefficient and the equation

where Q is the rate of heat transfer in Btulhr, U is the overall heat transfer coefficient in Btu/(hr ft2 OF), A is the heat transfer surface area in ft2, and LMTD is the log mean temperature difference across the tube wall ("F). The overall heat transfer coefficient, U, is calculated from the individual heat transfer resistances as shown below:

where h, is the heat transfer coefficient for the outer wall of the heat transfer surface, h, is the inner wall heat transfer coefficient, and R is the resistance due to fouling.

PROBLEM STATEMENT a) Calculate the area of heat transfer surface 'equired if h , = 450 Btu*(hr ft2 OF).', h, = 300

Btu'"hr fthF)-', R = 0.0001 hr ft2 "FIBtu and the log mean temperature difference is 98OF.


ORIGINAL DESIGN:

r

POLLUTION PREVENTION DESIGNS:

ALTERNATIVE 1

Column shutdown eve

ALTERNATIVE 2 Spare Reboiler

I--

Figure 14-1. Reboiler Alternatives


The distillation column reboiler has a duty of 2.0x107 Btulhr. b) The colulnn inust be shut down every 90 days to chemically clean or hydroblast the scale

formed on the heat transfer surfaces. You propose designing a new reboiler that can operate for 360 days between cleanouts. Use the R values in Figure 14-2 to calculate the heat transfer surface area required for the new reboiler. Assume that the log mean temperature difference and the inner and outer wall heat transfer coefficients remain constant.

c) If the original reboiler cost $150,000 and if the cost of the reboiler is linearly proportional to heat transfer area, calculate the cost of the replacement reboiler of part b).

d) The plant produces net revenue of $10,00O/day. Assume that a reboiler cleanout results in I day of downtime and that the product hold-up weight of the column is 10,000 Ib. Calculate the total cost of a shutdown if the off-specification material in the column costs $3.00/lb to produce and disposal costs are $l.OO/lb.

e) Calculate the approximate payback time on the pollution prevention strategy of replacing the old reboiler. Recall that reboiler cleanout will occur every 360 days.

f) Repeat part e) for the installation of a spare reboiler (design alternative 2). Assume that ( i ) the heat transfer surface area for the spare is the same as the original reboiler and (ii) the reboiler changeover occurs every 90 days with no downtime.

g) Compare the spare reboiler strategy with the redesigned reboiler option as potential pollution prevention solutions to the reboiler fouling problem.

QUESTION FOR DISCUSSION What inight be done to hinder heat exchanger fouling and reduce the need for mechanical

cleaning?

SUGGESTIONS FOR FURTHER READING Butterworth, D., and C. F. Mascone, "Heat Transfer Heads into the 21st Century," Clzernicnl Er7gheerir7g Progress, Vol. 87, no. 9, pp. 30-37, September 1991.

Sanatgar, H., and E. F. C. Somerscales, "Account for Fouling in Heat Exchanger Design," Chenzicnl Er~girzeerirzg PI-ogress, Vol. 87, no. 9, pp. 53-59, December 1991.


Time, days

Figure 14-2. Buildup of Fouling Resistance


SOLUTION TO PROBLEM 14 a) To determine the heat transfer surface area A, rearrange equation 1.

where Q and LMTD are known. The overall heat transfer coefficient U is calculated from equation (2):

Inserting values for h,, hi and R gives a value for U of 176.7 ~ t u * ( h r . f t ~ . " ~ ) - ' . Solving equation (1 a) yields

A = 1154 f t 2 .

b) Reboiler Redesign To operate for 360 days between cleanouts, the overall heat transfer coefficient must be recalculated with the new fouling resistance R (h, and hi are constant). The fouling resistance - time curve (Figure 14-2) indicates R = 0.0025 ~ T ~ - ' . h r . f t ~ . " F at 360 days. From equation (2),

or U = 124.1 ~Tu. (hr . f t~ ."F) - ' . Using the new value of the overall heat transfer coefficient (U = 124.1) in equation ( la) (note Q and LMTD are assulned to be constant) gives

c) The cost of the redesigned (or replacement) reboiler is calculated using the ratio of heat transfer surface area of the redesigned reboiler (A,) to the heat transfer surface area of the original reboiler (A,):

Cost = (ARIAo) *$150,000.

Here, $150,000 represents the cost of the original reboiler. Inserting the values of A, and A, yields

Cost of the redesigned reboiler = $214,000


d) Total cost of a shutdown is given by the sum of the following costs: Loss of production for 1 day $10,00O/day 'I: 1 day = $10,000 Loss of product hold-up in colu~nn 10,000 lbs " $3/lb = $30,000 Disposal cost of "product" 10,000 1bs ": $l/lb = $10,000 Total cost per shutdown = $50,000

e) The payback time on the pollution prevention strategy of replacing the original reboiler is obtained by computing the downtime cost data (annual basis) for the two reboilers. Original reboiler:

shutdown cost = # shutdowns/yr 'k cost of 1 shutdown = 4 (shutdown every 90 days) 'k $50,000 = $200,000

Replacement reboiler: shutdown cost = 1 (every 360 days) $50,000

= $50,000 Savings due to pollution prevention strategy

= $200,000 - $50,000 = $150,000

Cost of replacement reboiler = $214,000 The savings in downtime costs for 1.4 years equals the cost of the replacement reboiler, so the payback period is 1.4 years.

f> Spare Reboiler Strategy This strategy requires the addition of a reboiler similar to the original (alternative 2 in Figure 14-l), to be used during cleanout of the original, preventing shutdowns. Payback time on spare reboiler strategy (alternative 2) is calculated as in e). Original reboiler: downtime cost = $200,00O/yr from e) Spare reboiler strategy: downtime cost = none The cost of the spare reboiler is the same as the original - $150,000. Payback period for spare reboiler is

$150,000 3 = - yr or about 9 months

$200,000 4

g) Reboiler fouling causes pollution and often leads to redesigning the reboiler or frequent shutdowns to meet specific performance levels, as shown in parts e) and f). The two important parameters to consider in handling reboiler fouling problems are capital expense and downtime loss. The strategy used in part f), installation of a "spare reboiler," attacks the area of greatest return, which is downtime loss. The ability to switch between reboilers reduces downtime considerably, prevents pollution (from hold-up material disposal) and minilnizes production losses.

QUESTION FOR DISCUSSION Design equipment to have fouling-resistant surfaces and better flows that discourage

deposits. Use chemical treatment programs.


PROBLEM 15

Reducing Wastes During Batch Processing Changeovers

Chemical Engineering Topics: mass balances, engineering economics Pollution Prevention Concepts: design of unit operations for minimizing waste, reducing

unit size to reduce waste

BACKGROUND A facility manufactures sheets of composite material for use in the aerospace and sporting

goods industries. The composites are made by coating fiberglass or kevlar fabrics with a liquid resin. As shown in Figure 15-1, the coating process takes place in a pan containing the resin. The resins are dissolved in solvent, and a heat curing process drives off excess solvent from the composite to make the finished product. At the end of each run, the resin pan must be emptied, rinsed, and cleaned. This results in a hazardous waste (rinsate, leftover solvent, and resin) which is either partially recycled or is incinerated.

Treater pans must be at least ten inches wider than the fabric being coated to provide clearance for machinery, but when the facility's operating records were examined, it was discovered that the pans were excessively wide. This results in unnecessary waste generation. Plant data for the percentage of production at each fabric/treater pan width combination are given in Table 15- 1.

Table 15-1. Fabric and Treater Pan Width

It is proposed that blocks molded to fit into the ends of the pans be used to insure that the effective treater pan width be exactly ten inches wider than the fabric being coated. The blocks would have a one-time cost of $1000 and would not require removal of the pans. An operator would only have to remove the last block and insert the block appropriate for the next run. Your task is to consider the economic viability of this option, given that in a nine-month period, there are 1369 resin treater cleanouts. The pans have a wetted cross-sectional area of 0.22 ft2, and the specific gravity of the resin is 1.1. The cost of the resin is $1.64/lb, while the cost of incinerating resin waste is $0.14/lb.

Fabric Width, in

32 3 8 50 44 50 60

PROBLEM STATEMENT Neglect interest and calculate how long it would take for the molded blocks to pay for

Treater Pan Width, in

60 7 8 84 84 8 6 8 6

Percent of Production

40 20 15 5 9 11


themselves. Note that when the blocks are used, some incineration cost is avoided and less resin is thrown away. Ignore the volume of resin in the recirculation reservoir and assume that the amount of rinsing required is not changed by the blocks. Also assume that all the waste is incinerated.

Tlze material for this problem resulted from a waste nzinirnization acidit sl~orzsored by a gmnt fro111 tlze California Departnzent of Healtlz Services.

scrap fiber

Figure 15-1. Fabric Coating and Heat Curing Process


SOLUTION TO PROBLEM 15 Average treater pan width: (before reduction)

Minimum possible average treater pan width:

Cost of resin and incineration without the use of blocks

Cost of resin and incineration with the use of blocks

Daily savings:

Payback period would be -4 days.


PROBLEM 16

Reaction Pathway Optimization for Waste Reduction

Chemical Engineering Topic: reaction engineering Pollution Prevention Concepts: design of unit operations for minimizing waste, hazardous

byproduct formation

BACKGROUND Selection of reaction conditions in chemical synthesis has traditionally focused on

maximizing the yield of the desired product. It is clear, however, that the formation of hazardous and otherwise undesirable byproducts must also be minimized. In many instances the dual goals of maximizing product yield and minimizing hazardous waste are compatible. This problem will examine a case in which the goals are incompatible.

Using raw materials A, B and C, the product P is formed via the following pathways:

(reaction 2 ) C + H + E

(reaction 3) 2 D + E + P

(net reaction at 2 A + 2 B + C + P + H stoichior7zetric conditiorzs)

Intermediate H, formed in reaction 1 and consumed in reaction 2, is regulated under the Resource Conservation and Recovery Act (RCRA) as hazardous. Stoichiometry would dictate that reaction 1 be run at twice the rate of reactions 2 and 3. However, if we run the reactions at these rates we will produce 1 mole of H per mole of product.

PROBLEM STATEMENT a) Based on the data given in Table 16- 1, develop an expression for net income where

raw material hazardous waste "" ) - [ cost ) - [ disposal cost ' income i In the interest of illustrating the fundamentals of reaction pathway optimization, the assumption is made that the costs of capital, energy, and other factors are negligible and do not impact net income. Given these assumptions, net income is a function of reaction rates. For example, product value would be the rate of production of P, multiplied by its price. Raw material costs are the prices of A, B and C multiplied by their consumption rates. The rates of production and consumption in your expression for net income will be dependent on r, , r, and r,, which are the rates of reactions 1, 2 and 3.


Table 16- 1. Compound Values

b) From stoichiometry we know that two moles of D are required for each mole of P produced. Thus

2r3 I rl .

Compound

A B C D E H P

Similarly, r3 I r2

Valuelmole, $

1 2 3 0 0

-4 (disposal cost) 14

and r2 5 r , .

Plot these constraints on a graph with r,/r, and r,/r, as axes. Label feasible and infeasible regions.

c) The goal is to maximize net income subject to the constraints of part b) by varying the relative rates of ~eeactions 1, 2 and 3. Assume that reaction 1 has a rate of 1 mole A reacted per second. Find the rates of reactions 2 and 3 that result in maximum income and calculate the income generated at these rates. (HINT: The maximum must be on one of the corners of your feasible region.)

d) Repeat parts a) and c) for the case where the disposal cost of H is zero. Compare this result with the answer to part c).


SOLUTION PO PROBLEM 16 net - product raw material hazardous waste

a ) - - -

income value cost disposal cost net

income = r;($ 14) -r,($3 -$4) - r , ( $ l ~ $ 2 -(-$4))

f

b) See Figure S 16- 1 .

C ) Optimum is at r,/r, = 1 , r,/r, = 0.5 Income = (14(0.5) + 1 - 7) = $ 1/sec

d ) I11 this case

1'2

Optimum is at = 0.5 l' I

r - = 0.5

'-1

Note that this is the ratio of rates dictated by stoichiometry. Income - - (14(0.5) - 3(0.5) -3)

- - $2.5/sec This is 2.5 times higher than the income where the disposal cost of compound H is $4/mole.


Figure S16-1. Constraints Shown Graphically


PROBLEM 17

The Effect of Future Liability Costs on Return on Investment

Chemical Engineering Topic: process economics Pollution Prevention Concepts: economics of pollution prevention, future liability

BACKGROUND One of the challenges of calculating the econolllic rate of return for a pollution prevention

project is estimating avoided liability costs. The following case study illustrates the potential importance of avoided costs.

During 1979 and 1980, Conlpany A legally disposed of 104 drums of hazardous waste in a landfill. The landfill subsequently leaked and the landfill operator is now bankrupt. Although its wastes were disposed of legally, under the strict, joint and several liability provisions of the Superfund legislation, Company A is now liable for $100,000 in clean-up costs. These strict, joint and several pr.ovisions mean that a company might be held responsible for all of the clean-up costs associated with a site, even if the company only contributed a small fraction of the wastes disposed of at the site.

When the wastes were generated, at the rate of 1 drum per week in 1979 and 1980, disposal costs were $IO/drum. In 1978 the company evaluated a process modification that would have eliminated this waste stream. Capital costs for the project were estimated to be $2,000. Operating costs were estimated to be $5 per drum of waste avoided.

PROBLEM STATEMENT a) Calculate what the present value for the pollution prevention project would have been at

the start of the project (October 1 , 1978), assuming that the capital equipn~ent was purchased on October I, 1978 and that the unit began operation on January I , 1979. Also assume that the process generating the wastes shut down on December 3 1 , 1980. Ignore liability costs and use a 10% annual interest rate, colnpounded weekly.

b) Repeat the calculation of part a) incorporating the liability costs. Assume that the liability was incurred on October 1 , 1990.

QUESTION FOR DISCUSSION How can one estimate future liabilities?

SUGGESTION FOR FURTHER READING R. T. McHugh, "The Economics of Waste Minimization," in Hnznrclo~is Waste Minimizatiorz, Freeman, H. M., ed., McGraw-Hill, Inc., New York, 1990, pp. 127- 140.


SOLUTION TO PROBLEM 17 a) The present value of the project on October 1 , 1978 would have i~lcluded the following

elements:

Present Value on 10/1/78

Capital costs $-2,000

$5 net savings per drum, 104 drums, 1 drum per week beginning 11 1/79

Net present value on 10/1/78

"'Present value of these costs is determined using the forlnula

r 1

where: PV is present value evaluated on 1/1/79, the date when operations started i = interest (I OWyear) p = n ~ ~ ~ n b e r of periods per year (p = 52) n = number of years (n = 2) OC - = operating cost per period (OC/p = $(lo-5) = $5, a $5 savings per drum).

P

Present value at 1/1/79 is PV = $47 1

Adjust this present value to the value on October 1 , 1978 to get

For 11 = 114, PV = $459.


b) Determine the present value 011 October 1 , 1978, for a $100,000 liability expense that occurred on October 1, 1990 from

For n = 12, PV = $30,154. The net present value of the pollution prevention project, including avoided liability costs, is $28,613.

QUESTION FOR DISCUSSION Predictions of liability costs might be ~nade using data on previous failures of

environmental control systems. The EPA pro-jects that most landfills will leak after approximately 20 years, and information such as tliis may also be used in predicting environmental control failures.


PROBLEM 18

Economic Analysis of a Pollution Prevention Process Modifi- cation

Chemical Engineering Topic: engineering economics Pollution Prevention Concepts: economics of pollution prevention, avoided costs

BACKGROUND The manufacture of polypropylene, a cornlnon polymer used in consumer products,

resulted in a byproduct stream containing high molecular weight waxes with trace metal contaminants. Recycling these waxes back into virgin polypropylene manufacture was not successful because the waxes formed a viscous melt at high temperatures; this melt was difficult to handle and the waxes were therefore treated as waste. The majority of this waste was landfilled and the remainder was pyrolyzed.

Research on preventing the formation of the byproduct waste stream led to a process modification. The modification involves the use of a newly developed catalyst for the manufacture of polypropylene. Using the new catalyst, the amount of byproduct wax formed in the propylene polymerization is negligible. The net effect of the modification is that polypropylene product yield increases and disposal costs for the wax are virtually eliminated. The data sum~narized in Table 18-1 are estimates of the capital and incremental operating costs for the modification.

The income generated from the process modification comes from two sources. First, since the yield in the polymerization unit is now higher, there is no longer a need to pay for the raw materials that created the waste stream. Assume the savings in raw materials is $15,00Olyear. In addition, there is no longer any need to pay for waste disposal costs, which totalled $160,00Olyear. The goal in this problem is to assess the economic viability of the pollution prevention modification. In assessing the cost, new capital (material, installation, and other costs), operating expenses, and avoided costs should be considered.

Definitions Net present value (NPV) = PV - C, , where PV= present value and C,, = initial investment. Present value is found from

where C, is the net income in period t and r is the discount rate. Internal rate of return (IRR) is the rate at which NPV = 0; IRR is a single nu~nber that summarizes the merits of a prqject. - .

Depreciationlyear = CJt where t is in years.

PROBLEM STATEMENT a) Using the cost data in Table 18-1, show the capital outlay and the incremental operating

cost (or savings) for each of the first 5 years on a cash flow diagram. A cash flow


Table 18- 1. New Reactor System for Propylene Polymerization: Cost Data

diagram shows net cash inflow (income) or net cash outflow (expense) per year. (See Sample Cash Flow Diagram.) Note that implernentation of the pollution prevention

\ modification results in a reduction of the use of raw materials and a reduction in the generation of waste; both are regarded as income.

Capital Costs

Equipment/Material $250,000

Installation $53,000

Engineering/Consulting $40,000

Permitting $20,000

Start-up $7,000

Contingency $30,000

Cash Outflow 1

Incremental Operating Costs per Year

No. of Cost per Units Unit

Catalyst one charge $10,000 of catalyst

Maintenance (yrs 3 to 5 ) 175 hours $30

Training (yr 1 only) 200 hours $50

Supervision (yr I only) 350 hours $25

Figure 18-1. Sample Cash Flow Diagram

b) The discount rate is 13%. Calculate the net present value (NPV) for the modification using the data in the cash flow diagram of part a).

c) Use a trial and error procedure or an iterative program to determine the internal rate of return (IRR). Compare this to the discount rate of 13%.

d) The plant has an estimated life of 10 years. Use the straight line depreciation method to determine (i) the yearly capital write-off, (ii) the taxable earnings each year for the first 5 years and (iii) the after-tax cash flow for each of the 5 years. Use a corporate tax rate of 25%.

QUESTION FOR DISCUSSION In calculating after-tax cash flow, should taxes be charged against avoided costs?


SOLUTION TO PROBLEM 18 a) Cash Flow Diagram

Calculate incremental operating costs/year using the data in Table 18- 1. Capital cost C,, = $400,000

raw material avoided waste catalyst -training - -s~i l~i I ! i,i( savings chargcs cost

= ($15,000) + ( 160,000) - 10,000 - 10,000 - 8,750 = $146,250. A similar exercise for years 2 through 5 yields the savings shown in the cash Jl )w diagsanl~.

Cash Outflow

$400,000

Figure S18-1. Cash Flow Diagram

b) Net present value (NPV) or discounted cash flow analysis:

L1 NPV = C : , + -- , where r = 13% 1-1 ( I A s ) '

= $154,042. The 5-yeas analysis indicates the project will provide a net positive cash flow i ~ t the discount rate of 13%.

c) Internal rate of return (IRR) is that rate of return at which the NPV is equal to z(:til. or

LI NPV = C , , - - = O ,=I ( 1 ~ s ) '


Using a trial and error procedure, we find that the NPV is equal to zero at IRR=27.6%. I ) Depreciation = C,/T = $400,000110 = $40,000/year = yearly capital write-off.

Sample Calculations Year 1: earnings from part a ) = $146.250. Taxable earnings are $146,250 - $40,000 = $106,250. For a tax rate of 25%, cash flow = $146,250 - (0.25)$106,250 = $1 19,687.

Table S 18-2. After-Tax Cash Flows for the First Five Years

QUESTION FOR DISCUSSION Yes, taxes should be charged against avoided costs because by definition, taxable income

is gross income minus a set of deductions. The deductions for this project have been lowered by the amount of the avoided costs. If the gross income is unchanged, the company's profit or taxable income is higher. Avoided costs in this case are treated as equivalent to income and increase the tax burden.

Year

1

Taxable Earnings

$106,250

Cash Flow, After Tax

$1 19,687


PROBLEM 19

The Economics of Newsprint Recycling

Chemical Engineering Topics: engineering economics, mass balances, recycle streams Pollution Prevention Concepts: economics of pollution prevention, recycled material

handling

BACKGROUND Americans consume more paper than any other nationality, using twice as much as people

in the European Economic Community. Recycling could have a substantial impact on the 40% of municipal solid waste which is paper and paperboard and on the half of the domestic timber harvest that goes to the pulp and paper industry. Newsprint, which represents 17% of the total domestic use of paper, is an especially good candidate for recycling prograins because the ink is nontoxic and the paper itself is not coated or heavily bleached. Many local govern~nents across the nation have been collecting newsprint and other recyclables, with the expectation that the sale of materials will help to defray the collection costs. However, partly because of such aggressive recycling policies, the price of old newspapers (ONP) collapsed in 1989, falling to a negative $20 to $35 per ton, meaning that there was an effective disposal fee.

This collapse of the ONP market calls for a better understanding of the econoinics of recycling. Basically, there are three tiers to consider: collection, industry remanufacture, and custoiners of recycled goods. In the case of ONP, there has been some resistance on the part of printers to use recycled paper, but it is the shortage of newsprint deinking and reprocessing facilities that is the limiting factor.

Converting a large mill to 100% recycled fiber has been projected to cost $100 million, so ca re f~~ l decisions regarding change are required. Management must know the quantity and type of material available and the size of the market for recycled goods. In this problem, you will examine the ONP generated by the state of Minnesota, which uses curbside collection of recyclables. Table 19-1 contains information about the two largest newspapers in Minnesota, which together comprise 75% by weight of the newspaper circulation in Minnesota. Of course, not all of what the consumer receives is recyclable newsprint. Seventeen percent of the weight is non-newsprint material for glossy advertisements and magazines, and 1 % is ink. Also, 5% of what is printed is overissues and never reaches the consumer, while 3% of the newsprint delivered to the press becomes newsroom scrap and is not printed.

Table 19-1. Newspaper Generation Data for Minnesota (Franklin and Totten, 1990)

Average Weight per Paper, lb

1.08

0.80

Newspaper

Star Tribune

Sf . Paul Piorzeer Press nrzd Disl~ntclz

Papers Circulated per Week

2,973,100

1,229,500


PROBLEM STATEMENT a) Find the liiass of newspaper that reaches consumers in Minnesota each year. How much

of this is recyclable newsprint, including ink but excluding glossy pages? Assume that the average newspaper delivered in Minnesota is similar in mass to a circulation weighted average of the two newspapers described in Table 19- 1.

b) If 65% of consumed newsprint and 100% of overissue newsprint and pressroom scrap is collected for recycling, what is the mass of recycled material sent to pulp mills? Recall that overissue is 5% of circulation and pressroom scrap is 3% of the newsprint required for printing.

c) Newsprint must be deinked before it can be reused. Assume that the deinking processes remove all of the ink, but in doing so 15% of the paper is lost as waste. Calculate the mass of dry recycled pulp produced.

d) If it takes 1.5 cords of wood to create one ton of virgin pulp and there are an average of 21.3 cords per acre of timberland, how many acres of trees are saved annually by this level of recycling?

e) What would the cost of recycled newsprint have to be in order to justify the conversion of a 750 tonlday pulp mill to recycled fiber? Base your calculations on a virgin material cost of $90/ton, a conversion cost of $100 million and a 20% return on investment. Assume that the cost of producing pulp from the two feedstocks is identical, so that the only economic incentive for converting to recycled newsprint is the difference between feedstock costs.

QUESTION FOR DISCUSSION Suggest some ways to increase the market demand for old newspapers. Include both new

uses and possibilities for legislation.

REFERENCES AND SUGGESTIONS FOR FURTHER READING Franklin, W. E., and K. L. Totten, Market Development for Old Newslxq2ers in the Twin Cities Metropolitan Area: Progress and Opportunity, Franklin Associates, Ltd., Prairie Village, Kansas, Feb. 1990.

Haynes, R. W., coordinator, Auz Anal)lsis of the Tinzber Situation irz the Urzited States: 1989- 2040, USDA Forest Service General Technical Report RM-100, Dec. 1990.

Smith, M., "Recycling vs. the Paper Industry: An Environmental Argument for Industrial Restructuring," M.S. Thesis, Graduate School of Architecture and Urban Planning, University of California, Los Angeles, 199 1.


SOLUTION TO PROBLEM 19 a) The mass of newspaper that reaches consumers is

2,973, I00 papers 1, 229,500 papers 52 wk i[ 0.75 wk = 29 1 x 1 O"b/yr.

l o 8 wk

Of this, recyclable newsprint inakes up

b) First deterinine the weight per year of scrap and overissue. Overissue = 0.05 (circulation)

= 0.05 (29 1 x lo6 lblyr) = 14.5x10h Ib/yr

Recyclable amount = (1 -0. 17)(14.5x106 1blyr) Overissue recyclable = 12.1 x 1 0"bIyr To get pressroom scrap, do a mass balance around the dashed box in Figure S19-1. x -I- O.Ol(1 .05y) = 0 . 0 3 ~ + 1 .05y - 0.17(1 .05y) y = 29 1 x 1 0"bIyr (from part a)) => x = 2.58~10"b/~r scrap = 0.03(2.58x 10"b/yr) = 7 . 7 4 ~ lo6 lb/yr (all recyclable) Sum of scrap, overissue and recycled circulation is [7.74x10% 12.1 x 10% 0.65(24 1x 10"] = recycled Inass to pulp mill = 1 7 6 . 7 ~ 10"bsIyr.

c) Dry recycled fiber mass is equal to 85%x(mass to pulp mill-mass of ink) = (0.85)[176.7~10" O.Ol(176.7xl0~7.74~ IO"] lb/yr = 149~10"b/~r

1b [ ton 1 . S o d acre 1 d) Timberland saved is 149x 10 "

y 2000 Ib 21.3 cord = 5240 acres per year.

e) Let X be the price of ONP in $/ton at which there is a 20% return on investment. Then.

2 0.20($100x 1 0 9 => x < $( 17)/ton.

would result in a 20% return on investment.

QUESTION FOR DISCUSSION Legislate that a fixed amount of recycled newsprint be bought by newspaper printers. New uses: insulation

aniinal bedding hydromulch molded pulp containers packing illaterials

Legislate that newspaper delivery materials be linlited to recyclable inaterial (no staples, waxes, toxic inks, etc.).


ink = O.Ol(1.05y) glossy inserts = 1.05y(0.17) I I

Figure S19-1. Mass Balance Diagram

t I

1 y = circulation I

x = incoming ne&sprint I printed newsprint = I * I 1 . 0 5 ~ - 0,17(1,05y) ) I L

I overissue = 0.05~- - - - - - - -

scrap ='0,03x


PROBLEM 20

Mass Exchange Networks: Equilibrium, Operating and Load Lines

Chemical Engineering Topics: process design, separation processes Pollution Prevention Concepts: flowsheeting for waste minimization, thermodynamic

constraints

BACKGROUND The design of chemical ~nanufacturing operations has traditionally focused on lnaxirnizing

reliability, product quality and profitability. Issues such as waste generation have often been treated as secondary factors. It is now clear, however, that minimizing waste generation is a primary design issue and must be addressed in the earliest design stages of chemical manufacturing plants (National Research Council, 1988). This series of problems (20-22) examines quantitative methods for the design of chemical manufacturing operations for ininimurn waste generation.

The objective of the design is to arrive at the process configuration that results in the lowest generation of waste mass. The design problem is somewhat analogous to the design of energy-efficient processes. During the late 1970s and early 1980s, design methods were developed for heat exchange networks (HENS) that minimize the energy lost by processes. A group at UCLA has developed an analogous design method for mass exchange networks (MENs), which allows the waste minimization potential of a process to be evaluated (El-Halwagi and Manousiouthakis, 1990). A well-designed mass exchange network is a system of separators that achieves, in a cost-effective manner, minimal discharge of pollutants. Simply stated, mass exchange network synthesis maximizes the extent to which molecules that are potential pollutants can be transferred to streams in which they have a positive value.

Refinery wastewaters provide an example of how prevention of pollution might be achieved through implementation of an MEN. One of the waste components of interest in refinery wastewaters is phenol. Phenol can be found in the water effluent of catalytic cracking units, desalter wash water and spent sweetening waters. In these streams, phenol is a pollutant, but in other refinery streams phenol can be a valuable additive. A simplified schematic of this inass exchange problem is shown in Figure 20-1. In this case, three phenol-rich streams R, and three phenol-lean streams L, are considered. Proper matching and exchange of phenol between rich streams R, ... R,, and lean streams L, ... L,, within the process network is the key to optimization.

The amount of solute that can be transferred from a rich stream to a lean stream is limited by mass balance constraints and equilibrium constraints, as follows: (i) the total mass transferred by the rich stream must be equal to that received by the lean stream and (ii) mass transfer is possible only if a positive driving force r exists for all rich stream/lean stream matches. The equation describing the first constraint is found by performing a material balance 011 the solute to be transferred from stream i to stream j in Figure 20-2,


Atmospheric Distillation Uriit

Phenol-Rich Streams Phenol-Lean Streanis

R,: Wastewater from cracking unit L,: Light gas, oil R,: Spent sweetcning agent L,: Activated carbon R,: Wash water to desalter L,: Crude oil

Make Up Activated Carbon Refomling

Naptha

Gas, Oil * 4

Cracking 4 L2 L3 7 Gasoline I

Mass Exchange Biotre;~tnient

Sweetening Network 1 1 T k

To Storage

Figure 20-1. A Mass Exchange Network in an Oil Refinery

Figure 20-2. A Mass Exchanger


where R; is the flow rate of rich stream i, L, is the flow rate of lean stream j , j,,, is the mass fraction of solute k in rich strealn i, and x,, is is the mass fraction of the solute k in lean stream j . In this equation and for all of Problellls 20-22, the flow rates of the streams are assumed to be constant. While this is not strictly the case, it is a good approximation if the concentration of solute in the streams is low. Equation I represents the operating line for the Inass transfer of solute from the rich stream to the lean stream. An operating line for contact between a single rich and a si~lgle lean stream is plotted in Figure 20-3. Note that the slope of the operating line is equal to LIR. The second constraint, a positive driving force for mass transfer, is satisfied if the equilibriiul~ line lies to the right of the operating line at every point. Equilibriulu between a rich and a lean stream can be represented by an equation of the for111

where x,, is the Inass fraction of component k in strealn j that is in equilibrium with the mass fraction j;, in stream i. The constants of Equation 2 are thermodynamic properties and lriay be obtained through experimental data. A plot of an ecluilibrium line is shown in Figure 20-3. Equation 2, which establishes the locus of values for which E cquals zero, can be rewritten as

In this problem a Illass transfer unit with a single rich stream and a single lean stream is considered. Table 20- 1 gives the supply compositions and flow rates of the two strean~s. The target exit concentration for the rich stream (the desired concentration of pollutant in the effluent) is also given.

Table 20-1. Stream Compositions and Flow Rates

The equilibrium equation in the region of interest is

S tream

Lean

Rich

PROBLEM STATEMENT a) Calculate the mass of component k that is transfen-ed between the rich and lean streams

if the rich stream exits the exchanger at the target concentration and all mass exchange is accomplished through contact between the two streams.

Flow Rate, kgls

L, = 0.27

R, = 0.20

s U P P ~ Y Concentration, kglkg

x,,"' = 0.006

),,,111 = 0.08

Target Concentration, kglkg

j.,lO1ll = 0.02 i

Figure


Equilibrium and Driving Force Constraints on Mass Exchange Network Design


b) Plot the equilibrium line for component k. In the same figure, plot the operating line. c) Now examine a different approach for plotting the stream conditions. Using the axes

shown in Figure 20-4, plot the amount of inass exchanged as a function of the mass fraction of k in the rich stream. This will be called a load line. (HINT: At y,, = y,,"' the amount of mass that has been exchanged is zero, and at y,, = y,,""' all of the mass initially in the stream has been exchanged. Also, mass is leaving the rich stream, so the values on the y-axis are less than or equal to zero.) Plot the load line for the lean stream on a separate set of axes.

y, rich stream mass fraction

Figure 20-4. Alternative Axes for Plotting Load Lines

REFERENCES AND SUGGESTIONS FOR FURTHER READING Douglas, J. M., Corzcel~t~inl Design of Clzer.~zicnl Processes, McGraw Hill, 1988, pp. 216-260.

El-Halwagi, M. M., and Manousiouthakis, V., "Automatic Synthesis of Mass Exchange Networks with Single Component Targets," Clze~?zicnl Engineering Science, Vol. 45, No. 9, pp. 2813-2832, 1990.

National Research Council, Frontiers of Clzel?zicnl Engineering, National Academy Press, Washington, D.C., 1988.


SOLUTION TO PROBLEM 20 a) The amount of pollutant that is transferred from the rich stream to lean stream is

Rich stream flow rate R, multiplied by y,,"'-y,,""' 0.2 kg/s x (0.08-0.02) = 0.012 kg/s

b) Operating Line: Using the Inass balance equation, solve for x,,""' Ollf

RI(Ykl" - Y,~*"') = LI(xkl - xkli") 0.2(0.08 - 0.02) = 0.27(xk,""' - 0.006)

x,,"'~' = 0.05 Plot and connect points (~,,'",y,,~)"') and (x,,""',y,,'") on x, y axes to generate the operating line. Equilibrium Line: Plot the equilibrium equation

y = 0 . 6 7 ~ ' . See Figure S20- 1.

c) Mass exchanged = flow rate x (mass fraction out - mass fraction in). For the rich stream: at y,,"' no mass is exchanged. Therefore, at kgls = 0, y = 0.08 and at y,,""' = 0.02, -0.012 kg/s of mass is transferred (from part a)). For the lean stream: end points are 0 kg/s, x = 0.006 and 0.012 kg/s, x = 0.05. Plotting these points on the given axes, we get the graphic relationship between mass transferred (kgls) and Inass fraction shown in Figures S20-2 and S20-3. Note that the slope of the load line for the rich stream is (0.012-0)/(0.08-0.02) or 0.20 kg/s, which is the same as its flow rate. The same is true of the lean stream load line.

0.00 0.05 0.10 0.15 lean stream mass fraction

Figure S20-1. Equilibrium and Operating Lines

-

-

-

-

.' ,. ,' .*'

,.*'

,.*' ,'

.'#

4'

,' 4' .'

#'

,,**

- - - . - - - - - - ,

,* equilibrium line .' ,'

,' operating line #*'

I I


I I

y OU' = 0.02 y in = 0.08 y , mass fraction of k in the rich stream

Figure S20-2. Load Line for the Rich Stream

I I x in = 0.006 011[ x = 0.05

x , mass fraction of k in the lean stream

Figure S20-3. Load Line for the Lean Stream


PROBLEM 21

Mass Exchange Networks: Composition Interval Diagrams and Composite Load Lines

Chemical Engineering Topics: process design, separation operations Pollution Prevention Concepts: process flowsheeting for waste minimization, multiple

streams

BACKGROUND Problem 20 demonstrated the construction of equilibrium, operating, and load lines for

mass exchange units. In this problem you will learn how the equilibrium relationship can be used to bring together the rich and lean stream compositions in a combined composition interval diagram (CID). You will also learn how to make composite load lines that depict multiple rich or multiple lean streams.

The first step in constructing a CID is to map the rich and lean streams as shown in Figure 21- 1. The data for this figure are found in Table 21-1. For each rich stream, an arrow is drawn with its tail at the entering mass fraction and its head at the exiting mass fraction. Similar arrows are drawn for lean streams. In Figure 21-1, the con~positions of the rich and lean streams are on separate axes. These axes can be brought together through the equilibrium relationship. As in Problem 20, the equilibrium relationship in the region of interest for the species considered in this problem is given by

y = 0.67~':. This means that a mass fraction of y = 0.1 in the rich stream would be in equilibrium with a mass fraction of x:': = 0.150 in the lean stream. By converting lean stream compositions to the rich stream compositions with which they would be in equilibrium, and vice versa, the axes of the rich and lean stream plots from Figure 21-1 can be brought together in a combined CID as shown in Figure 21-2.

Table 2 1 - 1 . Stream Data

The load lines of part c) in Probleln 20 were for a single rich and a single lean stream. When there is more than one rich stream or more than one lean stream to consider, a composite load line representative of the multiple stream must be constructed. A CID such as the one shown in Figure 21- 1 can be used to develop composite load lines. The composite rich stream load line for the rich streams in Table 21-1 is plotted in Figure 21-3. This load line consists of

Rich Streams Lean Stream

Stream

R,

R2

Stream

L

kgls

5

10

kgls

15

x"'

0.00

x""'

0.05

yil'

0.10

0.07

yoff'

0.03

0.03


Region

Region

Rich Stream # 1 2

y-scale Lean Stream

x-scale

Figure 21-1. Composition Interval Diagrams

Rich Stream # 1 2

y-scale

Region 1

Region 2 - - - 0,033 - -

Region 3

Lean Stream

x-scale ~ - - - - - - - 0,15_0- - - - -

Figure 21-2. Combined Composition Interval Diagram


two segments, corresponding to the regions shown in Figure 21-1. In region A are the rich streams with mole fractions less than 0.1 and greater than 0.07 (0.1 I y I 0.07). Only R, falls into this category, so the total flow rate of the rich streams in this composition range is 5 kgls. The st,arting point for the load line is y = 0.1 and 0.0 kgls transferred. The mass transferred in each CID region is equal to the mass fraction exiting the region minus the mass fraction entering the region, multiplied by the sum of the flow rates in the region. Therefore, at y = 0.07, (5 kg/s)(0.07-0.1) or -0.15 kg/s have been transferred. So, the end point of the load line in this region is y = 0.07 and -0.15 kgls transferred. Note that the slope of the load line equals the mass flow rate of the stream. In region B are the rich streams with mole fractions less than 0.07 and greater than 0.03. Both rich streams fall into this region. When the load line is plotted, it has a slope equal to the sum of the flow rates of all streams in this region. The starting point for this segment of the load line is the termination point of the previous segment. The load line for the lean stream is given in Figure 21-4. In the next problem you will see how combined CIDs, such as the one pictured in Figure 21-2, are used to make combined rich and lean stream composite load line diagrams, which are necessary for designing MENs.

PROBLEM STATEMENT a) Draw a combined composition interval diagram (CID) for the three rich streams and one

lean stream represented in Table 21 -2.

Table 21-2. Stream Data

b) Calculate the mass transferred out of the rich streams in kgls within each region of the CID you created in part a). The mass transferred from the rich streams within each region is equal to ~ ' - y ' " ) x C ~ , , where yo"' and y'" are the exiting and entering rich stream mass fractions, respectively, and CR, is the sum of the rich stream flow rates in the region. Note that mass transferred is negative for the rich streams because they are losing mass.

c) Plot the composite load line for the rich streams. d) Based on the total amount of mass transferred from the rich stream, calculate the

minimum flow rate required for the lean stream. In the next problem, the validity of using this simple overall mass balance technique for determining the minimum lean stream flow rate is examined.

Rich Streams Lean Stream

Stream

R I

R2

R3

xu""

0.14

Stream

L

kgls

5

10

5

yit'

0.10

0.07

0.08

xi"

0.00

yo"'

0.03

0.03

0.01


y, mass fraction of component k in rich streams

Figure 21-3. Composite Rich Stream Load Line

x , mass fraction of k in lean stream

Figure 21-4. Lean Stream Load Line


SOLUTION TO PROBLEM 21 a) The rich and lean streams are mapped (from Table 21-2) to generate the CID shown in

Figure S2 1 - 1.

b) Mass transferred in: region 1 = (ymrrf-yilr) x C R ~ = (0.938-0.10)5 = -0.03 1 kgls

region 2 = (0.08-0.0938)5 = -0.069 kg/s region 3 = (0.07-0.08)(5+5) = -0.10 kg/s region 4 = (0.03-0.07)(5+10+5) = -0.80 kgls region 5 = (0.0 1 -0.03)5 = -0.10 kgls

c) The end points for the segments of the composite load line for the rich streams are: (0.1,0), (0.08,-0.10 kgls), (0.07,-0.20 kgls), (0.03,- 1 .O) and (0.01 ,- I . I). Note that to plot a continuous load line we use cumulative mass transferred; the mass transferred in each interval is added to the mass transferred in all previous intervals. See Figure S21-2.

d) Mass transferred by rich streams = mass gained by lean streams. From part b), total mass transferred by rich streams = 1.10 kgls. From Figure S2 1 - 1 the n~ass gained by the lean stream is equal to L(0.14-0.0), where L is the flow rate of the lean stream. Thus,

L(O. 14-0.0) = 1.10 or L = 7.86 kg/s.

Figure S21-1. Combined Composition Interval Diagram

Rich Stream # 1 2 3

y-scale

Region 2

Region 3

Region 4 0,042 - - - -

Region 5 - - - -

Region 6 - - - o.o_!Oo_ - - - - - - - - - - - - - - - - - ~ 4 0 4 - - - -

- -

Region 1 - -.

Lean Stream

x-scale

- - - - - -

.-, -, -,

- o x - - - - -

-

- O.l_oO! - - - - - - - - - - .

Q.Q931- -. , - - -, - - -


y, mass fraction of component k in rich streams

Figure S21-2. Composite Rich Stream Load Line


PROBLEM 22

Mass Exchange Networks: Pairing the Rich and Lean Streams and Determining the Pinch

Chemical Engineering Topics: process design, separation processes Pollution Prevention Concepts: flo wsheeting for waste minimization, pairing of mass

exchange streams

BACKGROUND The rich and lean streams described in Problem 21 must somehow be matched to develop

a mass exchange network. In this problem you will learn how to do the matching by constructing a combined load line diagram. In the background section of Problem 21, the composite rich and lean stream load lines for the streams described in Table 21-1 were plotted on separate diagrams. (See Figures 21-3 and 21-4.) These diagrams can be combined by making use of the equilibrium relation, which is

y = 0.67x*

in the region of interest. A combined figure can be made following several different conventions, each of which gives the same final results. The convention used here is to construct the combined figure by first plotting the load line of the lean stream as shown in Figure 21-4. For review, the lean stream load line begins at 0 mass fraction of k and 0 kgls mass exchanged and ends at 0.05 mass fraction of k and L,(xU"' - xi") = 0.75 kgls mass exchanged. The rich stream composite load line is added to the figure after converting the rich stream mass fractions into the lean stream mass fractions with which they are in equilibrium. These conversions were made in order to construct the composition interval diagram of Figure 21-2. The rich stream composite load line is free to move vertically; its placement determines the contact between the lean and rich streams. The rich stream load line has this freedom to move vertically because the values for mass exchanged on the y-axis are not absolute: they are useful only in relative terms, i.e., in terms of the differences in mass transferred between points. Therefore, the composite rich stream load line for the streams of Table 21-1 begins at x:' (the x-axis) = lean stream mass fraction with which the rich stream is in equilibrium = 0.1010.67 = 0.15 and continues downward with a slope of 0.67R,. The next point falls at x = 0.0710.67 = 0.10 where the slope changes to. 0.67(R, + R,). The load line ends where x = 0.0310.67 = 0.045. The load lines for the lean and rich streams of Table 2 1 - 1 are plotted together in Figure 22- 1. As stated before, the composite rich stream load line could have been located in any number of different vertical positions.

In Figure 22-1 the lean stream load line falls to the left of the composite rich stream load line at every point. This indicates that the desired mass exchange is thermodynamically feasible, and that the transfer could be accomplished using exchangers of finite size with no need for mass exchange into or out of any other streams. In Figure 22-2, the lean stream load line lies to the left of the rich stream load line, except for a point, called the pinch point, where the lines meet. Mass exchange in a case such as this is thermodynamically feasible, but would require an infinitely large mass exchanger (an infinite number of trays or stages, for example). Therefore, there is a practical requirement that conditions be manipulated so that a positive horizontal E


x, mass fraction of k in the lean stream

Figure 22-1. Combined Lean and Rich Stream Load Lines


Figure 22-2. Load Lines Form a Pinch


exists between the load lines. If at any point the lean stream load line lies to the right of the rich stream load line, as shown in Figure 22-3, mass exchange in the desired direction is not thermodynarnically feasible. In fact, if streams with such characteristics are contacted, mass exchange from the lean stream to the rich stream occurs. This infeasible situation could be made feasible by moving the rich stream load line down, but as you will learn later, there is a utility cost associated with moving the rich stream down in a case such as this.

Diagrams that combine the rich and lean stream load lines also show the amount of excess mass transfer capacity available from the lean stream and the amount of excess mass transfer capacity available from the rich stream. These regions were deliberately, if unrealistically, omitted in Figure 22-1 for the sake of illustrating thermodynamic feasibility. In Figure 22-4, three regions labelled I, 11, and 111 are clearly identified. In region I the lean stream has the capacity to exchange more mass and become richer but there is a "shortage" of rich stream. The lean stream must be brought up to its specified concentration in a manner other than through mass exchange with the rich stream. For instance, solute k may be added to it. In region I1 mass exchange can occur through contact between the rich and lean streams. In region 111, the rich stream is capable of mass exchange, but there is a "shortage" of lean stream. An external lean stream mass separating agent is required to achieve the rich stream's target concentration. For example, an adsorbent such as activated carbon might be used to take up the excess k, which is a pollutant in the rich stream. For the lean and rich streams depicted in Figure 22-4, the least amount of k and external lean stream or mass separating agent is required when the rich stream load line is manipulated to form a pinch point (E = 0). As E increases, operating costs (the cost of k and the cost of the mass separating agent) increase and capital costs (the cost of the network) decrease. As r decreases, operating costs decrease and capital costs increase. It is possible to find the r at which total annualized costs are minimized.

This method of synthesizing MENs is powerful because of its graphic determination of the pinch point and its ability to show whether mass exchange is thermodynamically feasible. The culmination of all the previous steps is the synthesis of a mass exchange network. Combined load line diagrams show where the lean and rich streams contact each other in a mass exchange network. For example, inspection of Figure 22-1 reveals that rich stream #I contacts the lean stream as the lean stream exits the mass exchange network. Then, where the lean stream mass fraction is 0.60/L, = 0.04, the lean stream is split and one third contacts rich stream # I while the remaining two thirds contacts rich stream #2. The lean stream enters this mass exchange network at the same point where both rich streams exit.

In this problem you will continue your exploration of the design of a mass exchange network for the streams described in the problem statement of Problem 21. Data for these streams are restated in Table 22- 1. (Note that a value has been assigned for the flow rate of the lean stream.) Take a moment to review the composition interval diagram and rich stream composite load line that made up parts a) and b) of the solution to Problem 21.

PROBLEM STATEMENT a) Using the CID you generated in part a) of Problem 21, plot the load lines of the rich and

lean streams described in Table 22-1 on the same diagram so that a pinch exists. Can the specified target concentrations be achieved solely through contact between the lean and rich streams? What does this indicate to you about your answer to part c) of Problem 21? Is there a lean stream flow rate at which all the desired mass exchange can occur solely through contact between the streams? If yes, what is that flow rate?


- rich stream


Figure 22-3. Load Lines Depict Thermodynamic Infeasibility


Figure 22-4. The Three Regions of a Mass Exchange Network


Table 22- 1 . Data for the Mass Exchange Network Streams

I Rich Streams I Lean Stream 1 / Flow Rate, kgls Y y""' I Flow Rate, kgls xJfJ x"ul 1

b) Use the graph you created in part a) to find the minimum amount of k (k,.,,) the lean stream will need to have added to it after contact with the rich stream and the minimum amount of k (k,.,,,,) a mass separating agent must remove from the rich stream after contact with the lean stream. At the values you find, the mass exchange network will have an infinitely high capital cost.

c) If the optimum r is determined from an economic analysis to be equal to 0.01, find the optimal amount of k (k,.,,') to add to the lean stream after contact with the rich stream and the optimal amount of k (k,.,,,,') a mass separating agent removes from the rich stream after contact with the lean stream. Add the rich stream corresponding to an r of 0.01 to the combined operating diagram from part a).

d) Describe how the streams contact each other in the optimal mass exchange network of part c). For example, rich stream #I contacts the lean stream when the lean stream exits the mass exchange network. Further down the mass exchange network, the lean stream is split into two parts, with one third contacting rich stream #1 and two thirds contacting rich stream #2. Complete this description for the mass exchange network, and give the lean stream and rich stream mass fractions at which the splits/junctions occur.


SOLUTION TO PROBLEM 22 a) The easiest way to add the composite rich stream load line to the lean stream diagram is

to figure out where the pinch point will occur, plot it, and work outward. From examination of Figure S21-2, it appears that the point where x = 0.0710.67 = 0.104 will be the pinch point. The value for mass exchanged at this point is found using the equation for the lean stream load line, which is

mass exchanged = (8 kg1s)x. Therefore, the first point of the composite rich stream load line to plot is (0.104, 8(0.104)kg/s = 0.832 kgls). See Figure S22-1 for the remaining points on the composite rich stream load line. Target concentrations cannot be achieved solely through contact between the streams because there are two regions where the rich stream does not lay to the right of the lean stream (bottom left and top right). The lean stream flow rate here (8 kgls) is greater than the minimum lean stream flow rate calculated in part c) of Problem 21, but it is insufficient to effect all the necessary mass exchange. There is no lean stream flow rate at which all the desired mass exchange can be achieved solely through contact between the streams.

b) Amount of k required: Top right y-axis coordinate of the lean stream = 1.12 kgls. Top right y-axis coordinate of the rich stream = 1.032 kgls.

Therefore, k ,,, = (1.12-1.032)kgls = 0.088 kgls.

Bottom left y-axis coordinate of the lean stream = 0. Bottom left x-axis coordinate of the rich stream = -0.068 kgls.

Therefore, k,.,,,, = (0-(-0.068))kgls = 0.068 kgls.

c) The composite rich stream load line must be moved downward until the point with a value of x = 0.104 (the former pinch point) is 0.01 to the right of the lean stream load line. The equation of the lean stream load line is

mass exchanged = (8 kgls)x. Therefore, the y-axis value for the composite rich stream load line at x = 0.104 is

mass exchanged = 8 kgIs(0.104-0.01) = 0.752 kgls. This makes the first point to plot on the composite rich stream load line (0.104, 0.752 kgls). The remaining composite rich streain load line points are shown in Figure S22-I. Values for k,,,' and k,,,,,' are calculated as in part b):

k,,,' = (1.12-0.952)kgls = 0.168 kgls and

k,. ,,,,' = 0-(-0.148 kgls) = 0.148 kgls. d) As stated in the problem statement, rich stream #1 contacts the lean stream when the lean

stream exits the mass exchange network. At this point the mass fraction of k in the lean stream is 0.952 kgls 1 8 kgls = 0.1 19 (found from the equation of the lean stream load line) and the rich stream mass fraction is 0.149 x 0.67 = 0.10 (read directly from the composition interval diagram). Where the lean stream mass fraction is 0.85218 = 0.107 and the rich streain mass fraction is 0.1 19 x 0.67 = 0.08, the lean stream is split, with one third contacting rich stream #I and two thirds contacting rich stream #2. The lean stream is split three ways when its mass fraction is 0.75218 = 0.094 and the rich stream mass


fraction is 0.104 x 0.67 = 0.07, with one fourth contacting rich stream # I , one half contacting rich stream #2, and the remaining fourth contacting rich stream #3. To find the mass fraction of k in the lean stream when the rich streams exit the mass exchange network, the equation of the corresponding portion of the composite rich stream load line must first be found. It is known that the slope of this line is

m = 0.67(R, + R, + R,) = 13.4 kgls. The intercept is found by using one of the known points as follows:

0.752 kgls = 13.4 kg/s(.07/.67) + b. Therefore, b = -0.648 kgls. The mass fraction at which this line crosses the x-axis can be found from

0 = (13.4 kg1s)x - 0.648 kgls. This means that the rich streams exit the mass exchange network when the mass fraction of k in them is 0.67 x 0.648113.4 = 0.0324. The lean stream mass fraction at this point is 0.

- rich stream, E = 0

(0.0149, -0.068) / ..3 (0.0448, -0.048) 1

,,....**" (0.0 149, -0.148) r*""

-0.2 1 I I

0.00 0.05 0.10 0.15

x , mass fraction of k in the lean stream

Figure S22-1. Combined Lean and Composite Rich Stream Load Lines

homeworl. desicrn problems - p2infohouse.org · new engineers with insights into industrial...

Documents