# Methods for Estimating Air Emissions from Chemical Manufacturing Facilities

August 2007 Final

Prepared for Emission Inventory Improvement Program Prepared by Mitchell Scientific, Inc. Westfield, NJ RTI International Research Triangle Park, NC 

## Volume II: Chapter 16EIIP Volume II ii 

## Disclaimer As the Environmental Protection Agency has indicated in Emission Inventory Improvement Program (EIIP) documents, the choice of methods to be used to estimate emissions depends on how the estimates will be used and the degree of accuracy required. Met hods using site-specific data are preferred over other methods. These documents are non-binding guidance and not rules. EPA, the States, and others retain the discretion to employ or to require other approaches th at meet the requirements of the applicable statutory or regulatory requirements in individual circumstances.Chapter 16 – Chemical Manufacturing Table of Contents

EIIP Volume II iii 

## Table of Contents 1.0 Introduction................................................................................................................... ..............16.1- 1 2.0 Source Category Description .................................................................................................... ..16.2- 1 2.1 Process Description ...........................................................................................................1 6.2- 1 2.2 Emission Sources ..............................................................................................................1 6.2- 1 2.2.1 Process Operations ...............................................................................................16.2- 2 2.2.1.1 Material Charging Emissions ..................................................................16.2- 2 2.2.1.2 Emissions from Process Heating.............................................................16.2- 2 2.2.1.3 Emissions from Process Depressurization ..............................................16.2- 2 2.2.1.4 Emissions from Gas Evolution Processes ...............................................16.2- 2 2.2.1.5 Emissions from Gas Sweep and Purge Operations .................................16.2- 2 2.2.1.6 Surface Evaporation ................................................................................16.2- 3 2.2.2 Miscellaneous Operations ....................................................................................16.2- 3 2.2.2.1 Solvent Reclamation ...............................................................................16.2- 3 2.2.2.2 Cleaning ..................................................................................................16.2- 3 2.2.3 Wastewater Treatment..........................................................................................16.2- 3 2.2.4 Storage Tanks.......................................................................................................16.2- 3 2.2.5 Equipment Leaks..................................................................................................16.2- 3 2.2.6 Spills.....................................................................................................................16.2

- 4 2.3 Process Design and Operating Factors Influencing Emissions .........................................16.2- 4 2.3.1 VOC Control Systems..........................................................................................16.2- 4 2.3.2 PM/PM 10 Control Systems ...................................................................................16.2- 6 3.0 Basic Air Emi ssion Models...................................................................................................... ...16.3- 1 3.1 Vessel Filling................................................................................................................. ....16.3- 1 3.1.1 Charging to an Empty Vessel...............................................................................16.3- 1 3.1.2 Charging to a Partially Filled Vessel with Miscible Contents .............................16.3- 5 3.1.2.1 Subsurface Addition................................................................................16.3- 6 3.1.2.2 Above Surface Addition..........................................................................16.3- 8 3.1.2.3 Addition of Immiscible Liquids ............................................................16.3- 11 3.2 Purge/Gas Sweep Models................................................................................................16.3- 12 3.2.1 Purge or Gas Sweep – Empty Vessel Purge.......................................................16.3- 12 3.2.2 Purge or Gas Sweep – Par tially Filled Vessel ....................................................16.3- 14 3.3 Vacuum Operations.........................................................................................................16.3- 17 3.4 Gas Evolution.................................................................................................................. 16.3- 20 3.5 Depressurization..............................................................................................................1 6.3- 23 3.6 Heating ........................................................................................................................ ....16.3- 26 3.7 Evaporation Models ........................................................................................................16.3- 31 3.7.1 Evaporation from an Open Top Vessel or a Spill...............................................16.3- 31 3.8 Emission Model for Liquid Material Storage..................................................................16.3- 33 3.9 Emission Model for Wastewater Treatment....................................................................16.3- 33 3.10 Using Sampling and Test Data to Validate Emission Studies.........................................16.3- 33 3.10.1 Correlating Standard Emission Models to Test Results .....................................16.3- 35Chapter 16 – Chemical Manufacturing Table of Contents

EIIP Volume II iv 3.11 Emission Calculations Usi ng Material Balance ..............................................................16.3- 38 3.12 Emission Calculations Using Emission Factors ..............................................................16.3- 38 4.0 Single Stage Vent Devices: Condensers, Vacuum Pumps, and Vacuum Steam Jets..................16.4- 1 4.1 General Calculation Considerations..................................................................................16.4- 1 4.1.1 Material Balance Considerati ons for the Vent Device .........................................16.4- 1 4.1.2 Equilibrium Relationships Between th e Exit Liquid and Vapor Streams ............16.4- 2 4.1.3 Checking to See If Condensation Takes Place in the Vent Device ......................16.4- 2 4.1.4 General Calculation Approach for Solving Single Stage Vent Device Problems...............................................................................................................16.4- 3 4.1.5 Partial Condensers................................................................................................16.4- 3 4.1.6 Total Condensers..................................................................................................16.4- 4 4.1.7 Vent Condensers ..................................................................................................16.4- 4 4.1.8 Liquid Ring Vacuum Pumps ................................................................................16.4- 4 4.1.9 Vacuum Steam Jets ..............................................................................................16.4- 5 4.2 Illustrations.................................................................................................................. ......16.4- 5 4.2.1 Condensers with One Condensable Component ..................................................16.4- 5 4.2.2 Partial Condensers..............................................................................................16.4- 11 4.2.3 Vent Condensers with Single-Phase Condensate ...............................................16.4- 15 4.2.3.1 Option 1 – Modified Rachford-Rice Method ........................................16.4- 15 4.2.3.2 Option 2 – Partial Pressure Correction Method ....................................16.4- 21 4.2.3.3 Option 3 – Material Balance Correction Method..................................16.4- 28 4.2.3.4 Non-ideal Liquid ...................................................................................16.4- 34 4.2.4 Vent Condenser with Two Phase Condensate....................................................16.4- 40 4.2.5 Liquid-Ring Vacuum Pump ...............................................................................16.4- 46 4.2.6 Vacuum Steam Jets ............................................................................................16.4- 50 5.0 Basic Process Operations ....................................................................................................... .....16.5- 1 5.1 Emission Calculations from Solv ent Reclamation Systems..............................................16.5- 1 5.1.1 Emissions from Charging the Distillati on Vessel with Cold Solvent for Recovery...............................................................................................................16.5- 1 5.1.2 Heating the Solvent for Distillation......................................................................16.5- 2 5.1.3 Receiving the Solvent from Distillation...............................................................16.5- 2 5.1.4 Drumming the Purified Toluene...........................................................................16.5- 3 5.1.5 Solvent Emissions from the Overall Toluene Recovery Process .........................16.5- 3 5.2 Filtration Operations.......................................................................................................... 16.5- 3 5.3 Centrifuge Operations .......................................................................................................16. 5- 4 5.4 Vacuum Dryer Model........................................................................................................16.5- 6 5.4.1 Information Needed to Model a Vacuum Tray Drying Operation .......................16.5- 6 5.4.1.1 Placing the material to be dried into the vacuum oven ...........................16.5- 6 5.4.1.2 Depressurization Step..............................................................................16.5- 7 5.4.1.3 Heating Step ............................................................................................16.5- 7 5.4.1.4 Distillation Step.......................................................................................16.5- 7 6.0 Physical Property Relationships................................................................................................ ..16.6- 1 6.1 Basic Physical Properties Relationships............................................................................16.6- 1 6.1.1 Unit Conversations...............................................................................................16.6- 1 6.2 Basic Physical Property Relationships ..............................................................................16.6- 1Chapter 16 – Chemical Manufacturing Table of Contents

EIIP Volume II v 6.2.1 Ideal Gas Law.......................................................................................................16.6- 1 6.2.2 Dalton’s Law........................................................................................................16.6- 2 6.2.3 Mole Fraction in a Liquid.....................................................................................16.6- 3 6.3 Pure Component Vapor Pressure.......................................................................................16.6- 4 6.3.1 Clapeyron Vapor Pressu re Equation ....................................................................16.6- 4 6.3.2 Antoine Equation..................................................................................................16.6- 6 6.3.3 Other Vapor-Pressure Equation Forms ................................................................16.6- 6 6.4 Component Vapor Pressure over Solutions.......................................................................16.6- 7 6.4.1 Equilibrium Conditions (Ideal) and Raoult’s Law ...............................................16.6- 7 6.4.1.1 Using Raoult’s Law to determine molar concentration...........................16.6- 9 6.4.2 Non-Ideal Equilibrium Conditions and Activity Coefficients............................16.6- 10 7.0 Refere nces..................................................................................................................... ..............16.7- 1Chapter 16 – Chemical Manufacturing Table of Contents

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## Table of Illustrations 3-1 Charging a pure solvent to an empty vessel. .............................................................................16. 3- 2 3-2 Charging a solvent mixture to an empty vessel.........................................................................16.3- 3 3-3 Charging a mixture to a partially filled vessel (subsurface addition)........................................16.3- 6 3-4 Charging a mixture to a partially filled vessel (above surface addition). ..................................16.3- 8 3-5 Charging a mixture to a partially fi lled vessel (immiscable liquids).......................................16.3- 11 3-6 Purging an empty vessel or reactor containing solvent vapors................................................16.3- 13 3-7 Gas sweep with a vessel containing with a single volatile solvent..........................................16.3- 16 3-8 Gas sweep with a vessel containing with a volatile solvent mixture.......................................16.3- 16 3-9 Vacuum operation with vessel filling......................................................................................1 6.3- 18 3-10 Vacuum operation without vessel filling.................................................................................16 .3- 19 3-11 Reaction involving a gas ev olution of one component............................................................16.3- 21 3-12 Reaction involving gas evol ution of two components.............................................................16.3- 21 3-13 Reaction involving multicomponent g as evolution and nitrogen purge..................................16.3- 22 3-14 Vessel depressurization invo lving one volatile component.....................................................16.3- 24 3-16 Heatup losses from a vessel containing a single volatile component......................................16.3- 27 3-17 Heatup losses from a vessel containing a volatile mixture......................................................16.3- 28 3-18 Heatup losses from a vessel with a vol atile mixture and nitrogen sweep................................16.3- 29 3-19 Evaporation from a vessel with an open top............................................................................16.3

- 32 3-20 Evaporation losses from a spill........................................................................................... .....16.3- 32 3-21 Using emission measurements to represent productio n operations.........................................16.3- 34 3-22 Using material balance to est imate emissions from operations...............................................16.3- 38 4-1 Condenser with solvent vapor heptane and nitrogen.................................................................16.4- 6 4-2 Condenser with toluene solvent vapor and nitrogen at vacuum conditions ..............................16.4- 8 4-3 Condenser with toluene vapor and nitrogen at atmospheric pressure .......................................16.4- 9 4-4 Rachford-Rice Method ....................................................................................................... .....16.4- 12 4-5 Modified Rachford-Rice Method for Vent Condenser............................................................16.4- 17 4-6 Partial Pressure Correction Method for Vent Condenser ........................................................16.4- 23 4-7 Material Balance Correction Method for Vent Condenser......................................................16.4- 31 4-8 Vent condenser with azeotrope, mate rial balance correction method. ....................................16.4- 34 4-9 Vent condenser with two phase condensate ............................................................................16.4- 40 4-10 Liquid-Ring Vacuum Pump with multi-component process stream. ......................................16.4- 47 4-11 Vacuum steam jet – material balance correction method........................................................16.4- 50 5-1 Estimating emissions from a batch distillation operation..........................................................16.5- 1 6-1 Calculating the moles of gas from volume, temperature, and pressure.....................................16.6- 2 6-2 Calculating molar quan tities for gas mixtures...........................................................................16. 6- 3 6-3 Calculating mole fracti ons for liqiud mixtures..........................................................................16. 6- 4 6-4 Estimating Clapeyron vapor pr essure model coefficients. ........................................................16.6- 5 6-5 Calculating pure component vapor pressures from the Antoine model.....................................16.6- 6 6-6 Calculating vapor pressures using the AIChE DIPPR database model. ....................................16.6- 7 6-7 Estimating component vapor pressures using Raloult’s Law....................................................16.6- 8 6-8 Estimating liquid composition based on vapor pressure measurements....................................16.6- 9 6-9 Determining the molar composition of a liquid from vapor pressure data................................16.6- 9 6-10 Estimating activity coefficients from solution measurements.................................................16.6- 11 6-11 Estimating activity coefficients from azeotropic mixtures......................................................16.6- 11 6-12 Calculating vapor compositions using activity coefficients. ...................................................16.6- 12Chapter 16 – Chemical Manufacturing Table of Contents

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## List of Figures Figure Page 3-1 Saturation Level (S) Plotted as a Functio n of Exit Vent Gas Flow Rate.................................16.3- 37 3-2 Toluene Emission Rate Plotted as a Functi on of Exit Gas Sweep Rate and Saturation Level.......................................................................................................................... ..............16.3- 37 4-1 Diagrams of singl e stage vent devices...................................................................................... .16.4- 1 4-2 Generalized Rachford-Rice equation.......................................................................................16 .4- 11 4-3 Rachford-Rice Plot showing solution at V/F = 0.236044........................................................16.4- 14 4-4 Rachford-Rice Plot showing solution at V/F = 0.90076..........................................................16.4- 19 4-5 Activity coefficients calculated for ethyl acetate and ethanol from the van Laar equation for the complete composition range. .......................................................................................16.4- 36 4-6 Component vapor pressures calculated for ethyl acetate and ethanol based on liquid composition, activity coefficients, and pure component vapor pressures. ..............................16.4- 36 4-7 Ethyl acetate vapor composition plotted as a function of ethyl acetate composition in the liquid......................................................................................................................... ...............16.4- 37 5-1 Example filtration equipment arrangement. ..............................................................................16.5

- 4 5-2 Typical centrifuge equipment arrangement...............................................................................16.5

- 5 5-3 Diagram of a typical vacu um tray dryer system........................................................................16.5- 6Chapter 16 – Chemical Manufacturing Table of Contents

EIIP Volume II viii 

## List of Tables Table Page 3-1 Calculated Values for Use in Eq. 3-14 ....................................................................................16 .3- 17 3-2 Iterative Trial and Error Results for Si using Eq. 3-14............................................................16.3- 17 3-3 Calculated Emission Rates .................................................................................................. ....16.3- 17 3-4 Partial Pressure Calculates for 20°C and 70°C........................................................................16.3- 28 3-5 Final Calculated Results for This Problem..............................................................................16.3- 29 4-1 Inlet Feed Stream Analysis................................................................................................. .....16.4- 13 4-2 Iteration Results for Illustration 4-4 ..................................................................................... ...16.4- 14 4-3 Product Outlet Stream Analysis ............................................................................................. .16.4- 15 4-4 Analysis of the Conden ser Inlet Feed Stream..........................................................................16.4- 17 4-5 Check to See If Condensati on Will Take Place at 2°C............................................................16.4- 18 4-6 Preparation for Applying the Modified Rachford-Rice Method .............................................16.4- 18 4-7 Iteration results for Illustration 4-5..................................................................................... .....16.4- 20 4-8 Product Outlet Stream Analysis (mole fractions and rates in lb-moles)..................................16.4- 21 4-9 Produce Str eams Analysis ................................................................................................... ....16.4- 21 4-10 Analysis of the Conden ser Inlet Feed Stream..........................................................................16.4- 23 4-11 Initial Estimate of Condensate............................................................................................ .....16.4- 24 4-12 Convergence Determination................................................................................................. ...16.4- 27 4-13 Material Balance for Condenser ............................................................................................ ..16.4- 28 4-14 Analysis of the Conden ser Inlet Feed Stream..........................................................................16.4- 31 4-15 Initial Estimate of Condensate............................................................................................ .....16.4- 32 4-16 Convergence Determination................................................................................................. ...16.4- 33 4-17 Material Balance for Condenser ............................................................................................ ..16.4- 33 4-18 Condenser Inlet Stream Analys is for 1.0 hour of Operation ...................................................16.4- 35 4-19 Calculated Mole Fractions for Only the Condensable Portion of the Inlet Process Vapor Stream......................................................................................................................... .............16.4- 37 4-20 Condensate Feasibility Analysis........................................................................................... ...16.4- 38 4-21 Iterative Solution to Determ ine Condensate Composition ......................................................16.4- 39 4-22 Condenser Material Balance at Completion of Iterations........................................................16.4- 39 4-23 Exit Vent Stream Analysis Table ........................................................................................... .16.4- 40 4-24 Inlet Stream Analysis over 1.0 Hour ....................................................................................... 16.4- 40 4-25 Liquid Phase Solubility Data for 20°C....................................................................................1 6.4- 41 4-26 Equilibrium Mole Fraction Concentrations.............................................................................16.4- 42 4-27 Estimated Activity Coefficients........................................................................................... ....16.4- 42 4-28 Aqueous Phase Condensat e Feasibility Analysis ....................................................................16.4- 43 4-29 Non-Aqueous Phase Conden sate Feasibility Analysis............................................................16.4- 44 4-30 Composition and Vapor Pressure Anal ysis of the Two Phase Condensate.............................16.4- 44 4-31 Exit Liquid Stream Analysis............................................................................................... .....16.4- 46 4-32 Condenser Material Balance at Completion of Iterations........................................................16.4- 46 4-33 Exit Vent Stream Analysis Table ........................................................................................... .16.4- 46 4-34 Analysis of the Inlet Stream to the Liquid-Ring Vacuum Pump.............................................16.4- 47 4-35 Analysis of the Inlet Stream and In itial Guess of the Exit Liquid Composition .....................16.4- 48 4-36 Material Balance Accountability........................................................................................... ..16.4- 49 4-37 Partial Pressure Equ ilibrium Accountability ...........................................................................16.4

- 49 4-38 Steam Jet Process Stream Analysis for 1.0 Hour of Operation ...............................................16.4- 51 4-39 Equilibrium Mole Fraction Concentrations.............................................................................16.4- 51 4-40 Estimated Activity Coefficients........................................................................................... ....16.4- 51Chapter 16 – Chemical Manufacturing Table of Contents

EIIP Volume II ix List of Tables (continued) Table Page 4-41 Aqueous Phase Condensat e Feasibility Analysis ....................................................................16.4- 52 4-42 Non-Aqueous Phase Conden sate Feasibility Analysis............................................................16.4- 52 4-43 Iterative Solution to Vacuum Steam Jet Problem with Aqueous Phase Condensate...............16.4- 53 4-44 Condenser Material Balance at Completion of Iterations........................................................16.4- 54 4-45 Exit Vent Stream Analysis Table ........................................................................................... .16.4- 54 6-1 Standard Conditions for Ideal Gases ........................................................................................ .16.6- 2Chapter 16 – Chemical Manufacturing Table of Contents

EIIP Volume II x [This page intentionally left blank.]Chapter 16 – Chemical Manufacturing 1.0 – Introduction

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## 1.0 Introduction The purpose of this guideline docum ent is to describe emission estima tion techniques for point sources in an organized manner and to provide concise example calculations to aid in the preparation of emission inventories. While emissions estimates are not provide d, the information presented in this document may be used to select an emission estimation technique best suited to a particular application. This chapter describes the procedures and recommended approaches for estimating emissions from batch chemical manufacturing operations and is intended to assist in dustry as well as regulatory agency personnel. As EPA has indicated in this and other EIIP documen ts, the choice of methods to be used to estimate emissions depends on how the estimate will be used and the degree of accuracy required, and methods using site-specific data are preferred over other methods. Because this document provides non-binding guidance and is not a rule, EPA, the States, and others retain the discretion to employ or require other approaches that meet the specific requirements of the a pplicable regulations in individual circumstances. Section 2 of this chapter identifies some of the types of emission events in chemical manufacturing operations. Section 3 provides an overview of availa ble emission estimation methods for these types of emission events, and section 4 provides an overview of emission estimation methods for emission streams that are routed to condensers or to vacuum devices in which water mixes with the emission stream (i.e., liquid ring vacuum pumps and vacuum steam jets). It should be noted that the use of site-specific emissions data is always preferred over the use of default values developed through use of industry emission averages. Section 5 provides an overview of considerations that should be used when assessing process vent emissions for basic process unit operations. Section 6 describes many of the underlying physical pr operty relationships that are used in support of the basic models that are presented in earlier sections of this document. The techniques presented in this document repr esent significant advances over previous emissionestimating methods for chemical manufacturing processes. New methodologies are, however, continuously being developed a nd published by EPA for subseque nt incorporation into national environmental policies and programs. The reader is thus encouraged to periodically check the Clearinghouse for Inventories and Emission Factors (CHIEF) website (http://www.epa.gov/ttn/chief) to learn about such new information and developments.Chapter 16 – Chemical Manufacturing 1.0 – Introduction

EIIP Volume II 16.1-2 [This page intentionally left blank.]Chapter 16 – Chemical Manufacturing 2.0 – Source Category Description

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## 2.0 Source Category Description 2.1 Process Description This section provides a brief overview of batch chemical manufacturing operations. Chemical processes vary widely from one industry to another with respect to the types of chemicals that are used, batch and production sizes, and unit operations th at are involved. However, most chemical processes include at least one or more combinations of four basic operations: preassembly, reaction, isol ation, and purification. Solvent recovery operations are al so important because they enable th e chemical operator to reuse basic raw materials and reduce the manufacturing cost a nd environmental impact. Additionally, cleanout operations are important since they enable productio n equipment to be reused for other manufacturing operations.



###  Preassembly. A preassembly (or premixing) is normally the initial step of the process and typically involves charging, mixing, or dissolving various liquids, solids, and/or gases. Essential equipment for this step might include agitated v essels or tanks with charge chutes, liquid inlet lines, and utility connections for temperature and pressure control. For certain continuous chemical processes, feed preparation might involve inline mixers with heat exchangers.



###  Reaction. The purpose of the reaction step is to f acilitate the actual chemical synthesis. A reaction may be carried out by applying heat or by addi ng specific reactants to the batch. The batch composition changes as the reaction takes place although many of the compounds such as process solvents and other materials remain unchanged. Eq uipment that is used to carry out reactions includes a batch, semi-continuous stirred tank or tubular reactor. The actual reactor used must meet the specific chemical, physical, and productivity needs of the process design.



###  Isolation. Once chemical products have been form ed from reaction, they must be recovered or isolated from basic process impurities that also formed or from unreacted materials and/or process solvents. In many cases the product is solid portion of a batch slurry. Isolation can be achieved through the use of spray driers coupled with various dust collectors. Extraction, crystallization filtration, or distillation might be applied in cases when the batch product is a homogeneous solution. Distillation is often used for collecting liquid products when the vapor pressure/temperature relationships can be exploited.



###  Purification. Once isolated, chemical products must be further processed through purification equipment to obtain the desired high purity level. Products from this purification step are to be used either as the final product or as a key ingred ient in the next step of a multi-step synthesis. For example, in a pharmaceutical process a lo w quality product might be purified by carbon treatment, additional extractions, ion exchange, chromatography, or crystallization. The overall purification process involves other preassembly, purification, and final isolation steps. 

### 2.2 Emission Sources The majority of emissions that occur from batch chemical manufacturing operations are from volatile organic solvents that evaporate during manufacturing. Particulate matter emissions may also occur from the handling of solid powders that are used in manufacturing. Several air emission sources have been identified fo r chemical manufacturing operations; they are as follows:



###  Process operationsChapter 16 – Chemical Manufacturing 2.0 – Source Category Description

EIIP Volume II 16.2-2 

###  Storage tanks



###  Equipment leaks



###  Wastewater collection and treatment



###  Cleaning



###  Solvent recovery



###  Spills 2.2.1 Process Operations 2.2.1.1 Material Charging Emissions Volatile Organic Compounds (VOC) emissions may occu r during material loading of equipment due to the displacement of organic vapors. Particulate matter (PM) and PM equal to or less than 10 micrometers in diameter (PM10) emissions may also occur during the material loading process from handling of solids in powder form. VOC and PM emissions during material loading emissions may occur as point source or fugitive, depending on whether a PM emissions collection system is in place. 2.2.1.2 Emissions from Process Heating Many processes involve batch heating in conjunction with a reaction or in preparing for distillation. As the batch temperature is increased to a new level, the molar capacity of the vessel headspace is reduced due to the ideal gas law (PV=nRT). Additionally, th e vapor pressures of volatile materials in the batch also increase. Vapors from vessel headspace are emitted through the process vent until the final temperature is reached. 2.2.1.3 Emissions from Process Depressurization Reducing the system pressure is one way that solv ents can be recovered from the batch at a lower temperature than would normally be possible. In some cases it is desired to replace the primary process solvent with a different one at reduced temperature. In other cases it may be desired to concentrate the batch through solvent stripping at reduced temperat ure to avoid thermal decomposition of compounds in the batch. When the pressure of the batch is redu ced then solvent vapors are drawn from the vessel (and connected equipment) by the vacuum system. 2.2.1.4 Emissions from Gas Evolution Processes Some reactions produce off gases such as hydrogen chlori de, sulfur dioxide, and others that evolve from the batch and exit the process through the vessel vent . These off gases will also carry solvent vapors from the batch with them. 2.2.1.5 Emissions from Gas Sweep and Purge Operations Nitrogen is frequently applied to the process vessel as a means of establishing inert conditions for safety purposes or to prevent moisture from entering the syst em and avoiding undesirable chemical reactions to take place. As nitrogen enters the vessel it must exit the vessel through the process vent along with solvent vapors from the vessel.Chapter 16 – Chemical Manufacturing 2.0 – Source Category Description

EIIP Volume II 16.2-3 2.2.1.6 Surface Evaporation Surface evaporation may occur during mixing and blendi ng operations if the vessel contents are exposed to the atmosphere. 2.2.2 Miscellaneous Operations 2.2.2.1 Solvent Reclamation Solvent reclamation refers to the purification of c ontaminated or spent solvent through distillation. VOC emissions occur from the solvent charging and the normal distillation equipment operation. 2.2.2.2 Cleaning Cleaning is an important ancillary part of the ch emical manufacturing process. Process equipment may be cleaned with solvent as often as after each batch. VOC emissions will result from any of the normal process operations such as charging, heating, gas sw eep, and others. Additionall y, emissions will result from the wiping of equipment with solvent wet clothes. In addition to this type of cleaning, small items used in the chemical manufacturing process may be clean ed by washing with solvents in a cold cleaner or open-top vapor degreaser. 2.2.3 Wastewater Treatment A chemical manufacturing facility may use a wastewater treatment system to treat contaminated water generated during the chemical manufacturing process (e.g ., water that has been used to clean equipment, extractions, crystallizations, and other operations). W astewater treatment systems generally consist of a series of surface impoundments that are used for equaliza tion, neutralization, aeration, and clarification of the waste stream. Fugitive VOC emissions may occu r from each type of basin. Procedures used to estimate emissions from wastewater treatment facilities are described in detail in Volume II, Chapter 5, Preferred and Alternative Methods for Estimating Air Emissions from Wastewater Collection and Treatment

. 2.2.4 Storage Tanks Various types and sizes of storage tanks are used to store solvents and resins used in the chemical manufacturing process. Most of these tanks have a fixed-roof design. The two significant types of emissions from fixed-roof tanks are breathing an d working losses. Breathing loss is the expulsion of vapor from a tank through vapor expansion and cont raction that result from changes in ambient temperature and barometric pressure. This loss occurs without any liqui d level change in the tank. The combined loss from filling and emptying tanks is called working loss. Evaporation during filling operations results from an increase in the liquid leve l in the tank. As the liquid level increases, the pressure inside the tank exceeds the relief pressure and vapors are expelled from the tank. Evaporative emissions during emptying occur when air drawn into the tank during liquid removal becomes saturated with organic vapor and expands, expelling vapor through the vapor re lief valve (EPA, 1995a). Emissions from tanks are characterized as a point source because VOCs are released through a vent. 2.2.5 Equipment Leaks In order to transport stored materials (e.g., organic so lvents and resins) from storage tanks to the chemical manufacturing operation, a network of pipes, pumps, va lves, and flanges is employed. As liquid material is pumped from the storage tanks to the particular process area, the pipes and supporting hardwareChapter 16 – Chemical Manufacturing 2.0 – Source Category Description

EIIP Volume II 16.2-4 (process line components) may develop leaks over time. When leaks occur, volatile components in the transported material are released to the atmosphere . This generally occurs from the following process line components:



###  Pump seals



###  Valves



###  Compressor seals



###  Safety relief valves



###  Flanges



###  Open-ended lines



###  Sampling connections. Emissions from equipment leaks can be characterized as fugitive and are described in detail in Volume II, Chapter 4, Preferred and Alternative Methods for Estimating Fugitive Emissions from Equipment Leaks

. 2.2.6 Spills Solvents, resins, or product may be accidentally spilled during manufacturing or cleaning activities. Materials that are spilled onto the ground may spread over an area, vaporize, and thus result in an air emission (EPA, 1987). Such an emission w ould be characterized as fugitive. 

### 2.3 Process Design and Operating Factors Influencing Emissions VOC and PM emissions from chemical manufacturing may be reduced through the use of add-on control systems or through equipment and process modifications. 2.3.1 VOC Control Systems A VOC control system typically consists of a captu re device and a removal device. The capture device (such as a hood or enclosure) captures the VOC-laden ai r from the emission area and ducts the exhaust air stream to removal equipment such as a recovery device or a destructive control device. In either case, the purpose of the control device is to remove VOCs from the exhaust air stream. The overall efficiency of a control system is a function of the specific removal efficiency for each device in the system. Example recovery devices:



###  Condensers are one of the most frequently used control devices in industry. They work by reducing the temperature of the emission exhaust gas to a cold enough temperature so that VOC vapors are recovered through condensation.



###  Adsorption Devices that incorporate activated carbon are capable of removing VOC vapors from exhaust emission streams to very low levels in the final gas stream. La rge scale adsorption based recovery systems normally have two or more activated carbon adsorption chambers. One carbon chamber is being used to remove VOCs from em ission stream while the spent carbon chamber is being regenerated. VOCs are recovered from the system during the regeneration phase. Steam is routed into the saturated carbon bed to cause the VOCs to desorb from the carbon and condense at the condenser. Once VOCs liquids have been coll ected then they may be recycled or further purified prior to reuse in the manufacturing operation.Chapter 16 – Chemical Manufacturing 2.0 – Source Category Description

EIIP Volume II 16.2-5 

###  Dust collectors are used to collect particulate ma tter from the emission stream. Dust collectors are constructed in many different designs. A bag hou se consists of a large rectangular housing with many internal banks of vertically mounted filter bags. The emission stream enters the bag house through the side inlet, passes through the bag filte r media, and exits the unit through the discharge port at the top. Particulate matter builds up on th e filter media until it is shaken off by pulses of compressed air from within each bag. The dust th at falls from the bags during the pulsing process is collected at the lower section of the bag ho use and finally discharged through the solids outlet to a drum or other container. When designing a bag house for an installation it is important to select the appropriate filter media and surface area for the particulate matter to be collected. The pore size of the filter cloth will determine the removal efficiency of the overall unit.



###  A floating roof on a storage tank helps to reduce solvent emissions by eliminating the headspace that is present in conventional storage tanks. For th e conventional storage tank air that is saturated with solvent vapors exits the vessel as the su rrounding temperature increases during the day. Outside air then reenters the vessel during the evening hours as the surrounding temperature decreases and the daily cycle prepares to be rep eated. Additionally, when a conventional storage tank is filled periodically then emission occur by way of displacement. A floating roof moves up and down the vessel vertical walls as the level of the storage tank changes. Since the vessel contains no headspace all breathing and filling losses are avoided. Example destructive control devices:



###  Catalytic Incinerators are used to eliminate VOCs from process exhaust gases from a broad variety of process operations. Catalytic incineration is a technology used in selective applications to greatly reduce emissions due to VOCs, hydro carbons, odors, and opacity in process exhaust. The catalyst section operates at between 315°C to 400°C to convert VOC to CO2 and H2O. A properly designed and installed syst em can achieve a VOC destruction efficiency of greater than 95%.



###  Thermal Incinerators control VOC levels in a gas stream by passing the stream through a combustion chamber where the VOCs are burned in air at temperatures between 700°C to 1,300°C. Fuel is burned in the unit to supply the necessary heat for decomposition of the VOC’s. Heat exchangers may also be installed as part of the unit to conserve energy by warming the inlet air stream with the hot exhaust gases.



###  Venturi Scrubbers are used to remove particulat e material from vent exhaust streams. These units normally incorporate a spray nozzle section where liquid is discharged at a high velocity, a mixing section where liquid droplets contact the incoming emission gas stream, and a settling/separation section where scrubber fluid is recycled to the inlet spray nozzle and the exit gas is discharged to the atmosphere or to a secondary control device.



###  Enclosed Oxidizing Flares convert VOCs into CO2 and H2O by way of direct combustion. Normally an enclosed oxidizing flare is used when the waste gas is rich enough in organic content to be its own fuel source. If the process gas st ream does not contain an adequate level of combustible VOCs then additional fuel must be supplied for effective operation. The removal efficiency for each control device is a function of the specific design of the unit and how well its capability matches the intended application. Before selecting pollution equipment one should consult different manufacturers and/or engineering firm s to determine the most appropriate control device solution for a given application.Chapter 16 – Chemical Manufacturing 2.0 – Source Category Description

EIIP Volume II 16.2-6 2.3.2 PM/PM 10 Control Systems PM/PM 10 control systems for the chemical industry cons ist of a capture device paired with a control device that is typically a fabric filter (bag house) . These systems are typically employed to reduce PM emissions from charging pigments and other solids in to mixing and grinding devices. The captured dust may be recycled or sent for off-site disposal or treatment. Bag Houses remove particulate material from an emission gas stream by passing the emission stream through engineered fabric filter tubes, envelopes, or cartridges. Particulate material is retained on the filter media as the clean air is discharged to the atmosphere . Vibrators or timed air blast are used for removing and discharging the dust that has been collected in the unit. When identif ying a bag house for an application it is important to consid er the particle size in the emissi on stream, the particle size control requirements, the air flow rate of the emission stream, and the bag filter surface area requirements. Additionally, it is important to identify the appropria te chemical resistance requirements for the materials of construction in the unit. Fabric filters are least efficient with particles 0.1 to 0.3 μm in diameter and w ith emission streams of high moisture content. When operated under optimum conditions, they can generally achieve control efficiencies of up to 99+ percent (EIIP, 2000). Howeve r, typical control efficiencies range from 95 to 99 percent.Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-1 

## 3.0 Basic Air Emission Models Processes for chemical manufacturing consist of di fferent unit operations including filling, mixing, heating, depressurization, gas sweep, gas evolution, dispersing, milling, and others. A mathematical approach to estimating air emissions from these types of processes is to model them as collection of separate unit operations. This section contains several models that can be applied to many of the operations within these processes. For example, the f illing model can be used to estimate the emissions from charging the primary raw materials or transferri ng the batch from one vessel to a second vessel. The heating and filling models can be u sed to model distillation operations. 

### 3.1 Vessel Filling When a solvent or volatile process mixture is charged into a process vessel then material losses will occur though the process vent in the form of solvent vapors . The amount of solvent that is emitted during this displacement operation is a function of the (1) volum e of liquid entering the tank, (2) the equilibrium vapor pressure of each component that is contained in the inlet stream and/or present in the vessel before the filling operation begins, and (3) degree of saturation of the associated vapors. The resulting emission rate is simply a function of how quickly the filling operation takes place. The equilibrium vapor pressure for each volatile com ponent in the system is calculated by applying Raoult’s Law to the pure vapor pressure and the mo le fraction of each compound in the inlet stream and/or initial vessel contents. If non-idealities exist between molecules in the system then activity coefficient may be used to adju st the vapor pressures accordingly. The calculated equilibrium vapor pressure represents the gas space composition assuming that the degree of vapor saturation is 100%. It is highly possible that the actual saturation level of th e solvent vapors may be less than the assumed 100% level. For example, if representative samples have been taken of the gas space and analyzed then this data may be substituted for the calculated vapor pressure values. However, the conservative approach is normally applied and a 100% saturation level is assumed in most cases. Displacement emissions that are caused by charging operations may be calculated using the ideal gas law on the volume of gas that is emitted. This equa tion assumes that the partial pressure of component i in the vent gas is at saturated levels.
Ideal Gas Law: 

### RT V p E i i

- n 

### = Eq. 3-1
Where E n-i are the moles of component i that are emitted due to vapor displacement p i is the saturated vapor pressure of component i. V

is the displacement volume that was caused by the filling operation. R is the ideal gas constant in consistent units, T

is the temperature of the liquid being charged 3.1.1 Charging to an Empty Vessel When a solvent mixture is being charged to an empty vessel then the vapor composition for the displacement calculation may be made based entirely upon the inlet stream composition.

i i i 

### P p γ x = i Eq. 3-2Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-2 Where p i = effective vapor pressure for component i x i = mole fraction of component i

i 

### γ = component activity coefficient (Beco mes 1.0 when Raoult’s Law applies) P i = pure component pressure i Illustration 3-1: Charging a pure solvent to an empty vessel. A 5,000 gallon reactor is filled at ambient conditions ( 25°C and 1 atm) with 3,60 0 gallons of hexane in one hour. The empty vessel was previously made iner t with nitrogen, and the vessel is vented to atmosphere. Calculate the vapor emissions from this process. Step 1. The displaced gas is defined by the following conditions. T = 25°C = 298°K (System temperature) P System = 1.0 atm = 760 mm Hg (Total system pressure) V displacement = 3600 gal = 481.28 ft 3 (Displacement volume) Time = 1 hr (Time for event) Constants and Relationships: Universal Gas Constant: K mole lb ft

mmHg R ° ⋅ − ⋅ = 3 9

. 998 Antoine Equation: ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = c T b a P

i exp

Gas Law: 

### RT PV n = also 

### RT V P n i

i 

### = for component i in the gas space. Sum of the partial pressures in the gas space: 

# ∑

= 

### = N i i T 

### p P 1 Sum of component moles in the gas space: 

# ∑

= 

### = N i i T 

### n N 1 Step 2. Calculate the amount of each component in the displaced gas. Hexane is the only component in the liquid, so the vapor pressure for hexane is only a function of the system temperature, 25°C. The partial pressure of n itrogen is determined by the difference between the total system pressure, 760 mm Hg, and the partial pressu re of hexane. The vapor pressure of hexane may be calculated from the Antoine equation as follows: 

## ( ) mmHg p hexane 

### 28

. 151 019

. 5 exp 78

. 48 15

. 298 55

. 2697 8366

. 15 exp = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − =
therefore 

### mmHg mmHg mmHg p P p hexane T N 

### 719

. 608 28

. 151 760 2 

### = − = − =
Ideal Gas Law: moles lb C ft mmHg RT V p E hexane hexane n − = + ° ⋅ = =

− 244

. 0 ) 15

. 273 25 )( 9

. 998 ( 28

. 481 281

. 151 3Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-3 moles lb ft mmHg RT V p E N N n − = ⋅ = =

− 984

. 0 ) 15

. 298 )( 9

. 998 ( 28

. 481 719

. 608 3 2 2
Emission level: 

## ( ) lb mole lb lb moles lb E hexane

Wt 03

. 21 17

. 86 244

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − =

− 

## ( ) lb mole lb lb moles lb E N

Wt 562

. 27 01

. 28 984

. 0 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − =

− Step 3. Calculate the emission rate based on the 1 hour addition. Emission Rates: hr lb hr

lb E hexane

R 03

. 21 0

. 1 03

. 21 = =

−

hr lb hr

lb E N

R 56

. 27 0

. 1 56

. 27
2 =

=

− Illustration 3-2: Charging a solvent mixture to an empty vessel. A 50-50 volume percent solvent mixture of heptane and toluene is charged to a surge tank at a rate of 50 gal/min. A total of 1,500 gal of mixed solvent is charged at 20°C. Step 1. Define conditions of the displaced gas: T = 20°C = 298.15°K (System temperature) P T = 1.0 atm = 760 mm Hg (Total system pressure) V disp = 1,500 gal = 200.53 ft3 (Displacement volume) Time = 5 min (Time for event) Constants and Relationships:
Universal Gas Constant: K mole lb ft

mmHg R ° ⋅ − ⋅ = 3 9

. 998 Antoine Vapor Pressure Equation: P i = exp a − b T + c ⎛ ⎝ ⎞ ⎠
Gas Law: RT PV n = , also RT V p n i

i = for a single component i in the gas space. Sum of the partial pressures in the gas space: P p T i i N = = 

## ∑ 1 Sum of component moles in the gas space: N n T i i N = =



## ∑ 1Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-4 Step 2. Calculate the vapor-phase mole fractions:
VOC Mol. Wt. lb/lbmole Density lb/gal Volume Charged Weight Charged lb-moles Xi Heptane 100.205 5.6977 750 4,273.3 42.65 0.42 Toluene 92.13 7.2138 750 5,410.3 58.72 0.58 Total 101.37 1.00 In this problem, heptane and toluene coexist in a miscible liquid. The vapor space partial pressure for each compound may be estimated from the pure component pressure and liquid composition using Raoult’s Law. It is assumed that the vessel cont ains nitrogen as the remaining gas component. Pure component vapor pressures for the liquid components may be estimated using the Antoine equation.
P

heptane = exp 15.8737

− 2911.32 293.15

− 56.51 ⎛ ⎝ ⎞ ⎠ = exp(3.571) = 35.55 mmHg

P

toluene = exp 16.0137

− 3096.52 293.15

− 53.67 ⎛ ⎝ ⎞ ⎠ = exp(3.0835) = 21.84 mmHg
VOC Xi Pi (mm Hg) pi (mm Hg) Heptane 0.42 35.55 14.93 Toluene 0.58 21.84 12.67 Nitrogen 0.0 732.40
Ideal Gas Law: moles lb ft mmHg RT V p E sys heptane heptane

- n − = ⋅ = = 0102

. 0 ) 15

. 293 )( 9

. 998 ( 53

. 200 93

. 14 3

moles lb ft mmHg RT V p n sys toluene toluene n − = ⋅ = =

− 0087

. 0 ) 15

. 293 )( 9

. 998 ( 53

. 200 67

. 12 3

n p V RT mmHg ft lb moles nitrogen air sys = = ⋅ = − 732 40 20053 998 9 29315 05015 3

. .

( . )( . ) . Emission level: 

## ( ) lbs mole lb lbs moles lb E heptane

- Wt 02

. 1 205

. 100 0102

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − = 

## ( ) lbs mole lb lbs moles lb E toluene

Wt 80

. 0 13

. 92 0087

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − =

− 

## ( ) lbs mole lb lbs moles lb E nitrogen

Wt 05

. 14 0134

. 28 5015

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − =

− Step 3. Calculate the emission rate in lb/hr units. Since 1,500 of solvent mixture is charged at 50 gpm the complete charge is completed in 30 minutes.Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-5 hr lbs min hr min/ lbs E heptane

R / 04

. 2 30 60 02

. 1 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

−
hr lbs min hr min/ lbs E toluene

R / 6

. 1 30 60 80

. 0 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

−
hr lbs min hr min/ lbs E nitrogen

R / 1

. 28 30 60 05

. 14 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

− 3.1.2 Charging to a Partially Filled Vessel with Miscible Contents When a liquid mixture is charged to a vessel that already contains process material from a prior process operation then the vessel contents composition will dynamically change as the charging operation takes place. The equilibrium vapor composition above the ba tch will also change in accordance to the batch composition. At any point in the filling operation one may cal culate the batch composition from the initial vessel contents and the amount of material that has been added. Let n A represent the moles of inlet mixture that are to be added to the vessel and let N B represent the total number of moles of mixture that are initially contained in the vessel regardless of composition. Fo r example, if the inlet stream contained ethanol, water, and methanol then n A would represent the total moles of etha nol, water, and methanol that have been charged at any point in the operation.

B A A A 

### N n n + = φ Eq. 3-3
Where A 

### φ is the degree of dilution of the inlet stream mixture at any point during the addition, n A are the moles of inlet steam mixture charged to the vessel, and N B are the moles of mixture that were initia lly contained in the vessel prior to the addition. The average dilution A 

### φ of the inlet stream A from being mixed with the contents of the vessel may be calculated by integrating a differential expression for A 

### d

φ with respect to moles of inlet mixture A and then dividing the results by the total number of moles of mixture A that were charged. [Hatfield, 2003d]. 

## ( ) A N

B A A A A dn N n n N A 

## ∫ + = 0 1 φ Eq. 3-4
⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + = B A B A B A N N N N N ln 1 φ Eq. 3-5 A similar calculation may be made for the average dilu tion factor of mixture B (the initial vessel contents) that exists during the filling process. 

## ( ) A N

B A A B B dn N n N N A 

## ∫ + = 0 1 φ Eq. 3-6Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-6 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − = B A B A B B N N N N N ln φ Eq. 3-7
Once A 

### φ and B 

### φ are determined then the average batch composition that exists during the filling operation may be calculated by multiplying the composition of each mixture by its corresponding integrated average dilution factor. When the filling operation involves subsurface addition then the inlet stream is exposed to the vessel headspace only as it mixes with the vessel contents. In this case only the average batch composition that exists during the filling operation is used to calcu late the average vessel he adspace vapor composition.

A 

### φ and B 

### φ are calculated However, if the operation involves above surface addition then the inlet stream is exposed directly to the headspace in the vessel. The equilibrium vapor pressure of the inlet stream must then be considered as an independent source of vapors in additio n to the average batch composition. 3.1.2.1 Subsurface Addition Illustration 3-3: Charging a mixture to a partially filled vessel (subsurface addition). Three hundred gallons of acetone at 20°C are to be added to the vessel featured in Illustration 2 by way of subsurface addition. For this problem, the initial conten ts of the vessel are 1,500 gallons of a mixture of heptane (42% mole fraction) and toluene (58% mole fraction) at 20°C. The system pressure is 760 mm Hg and the addition process is complete in 0.5 hour. Step 1. Define conditions of the displaced gas. T = 20°C = 298.15°K (System temperature) PT = 1.0 atm = 760 mm Hg (Total system pressure) Vdisp = 300 gal = 40.1 ft3 (Displacement volume) Time = 30 min (Time for event) Acetone = 6.5632 lb/gal @ 20°C Constants and Relationships: Universal Gas Constant: 

### K mole lb ft

mmHg R ° ⋅ − ⋅ = 3 

### 9

. 998
Antoine Vapor Pressure Equation: 

### P i 

### = 

### exp a − 

### b T + 

### c ⎛ ⎝ ⎞ ⎠

Gas Law: RT PV n = , also RT V p n i

i = for a single component i in the gas space. Sum of the partial pressures in the gas space: 

### P p T i i N 

### = =



# ∑ 1 Sum of component moles in the gas space: 

### N n T i i N 

### = = 

# ∑ 1Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-7 Step 2. Calculate the dilution factors for the inlet stream and the initial vessel contents. Inlet Stream Analysis:

VOC Mol. Wt. lb/lbmole Density lb/gal Volume Charged Weight Charged lb-moles Xi Acetone 58.08 6.5632 300 1968.96 33.90 1.00 Total 33.90 1.00 Initial Vessel Contents Analysis:

VOC Mol. Wt. lb/lbmole Density lb/gal Volume Charged Weight Charged lb-moles X i Heptane 100.205 5.6977 750 4,273.3 42.65 0.42 Toluene 92.13 7.2138 750 5,410.3 58.72 0.58 Total 101.37 1.00
Calculate the inlet stream dilution factor: From Eq 3.5: ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + = B A B A B A N N N N N ln 1 φ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + = 37

. 101 90

. 33 37

. 101 ln 90

. 33 37

. 101 1 A φ 

## ( ) 14

. 0 7494

. 0 ln 99

. 2 1 =

+ = A φ Calculate the dilution factor for the initial vessel contents: From Eq 3.7 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − = B A B A B B N N N N N ln φ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = 37

. 101 90

. 33 37

. 101 ln 90

. 33 37

. 101 B φ 

## ( ) 86

. 0 7494

. 0 ln 99

. 2 = − = B φ Step 3. Calculate the average batch and vapor-phase compositions.

VOC Xi φ A, φ B i 

### X P i (mm Hg) p i (mm Hg) Acetone 1.00 0.14 0.14 184.80 25.87 Heptane 0.42 0.86 0.36 35.55 12.80 Toluene 0.58 0.86 0.50 21.84 10.92 Nitrogen 710.41 Total 1.00 1.00 760.00Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-8 Ideal Gas Law: moles lb ft mmHg RT V p E sys acetone acetone
- n − = ⋅ = = 00354

. 0 ) 15

. 293 )( 9

. 998 ( 1

. 40 87

. 25 3

moles lb ft mmHg RT V p E sys heptane heptane

- n − = ⋅ = = 00176

. 0 ) 15

. 293 )( 9

. 998 ( 1

. 40 83

. 12 3

moles lb ft mmHg RT V p E sys toluene toluene n − = ⋅ = =

− 0015

. 0 ) 15

. 293 )( 9

. 998 ( 1

. 40 92

. 10 3

moles lb ft mmHg RT V p E sys nitrogen nitrogen n − = ⋅ = =

− 0973

. 0 ) 15

. 293 )( 9

. 998 ( 1

. 40 41

. 710 3

Emission level: 

## ( ) lbs mole lb lbs moles lb E acetone
- wt 21

. 0 08

. 58 00354

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − = 

## ( ) lbs mole lb lbs moles lb E heptane

- wt 18

. 0 205

. 100 00176

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − = 

## ( ) lbs mole lb lbs moles lb E toluene

wt 14

. 0 13

. 92 0015

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − =

− 

## ( ) lbs mole lb lbs moles lb E nitrogen

wt 73

. 2 0134

. 28 0973

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − =

− Step 4. Calculate the emission rate in lb/hr units.
hr lbs hr lbs E acetone

R / 42

. 0 5

. 0

21

. 0 = =

− hr lbs hr lbs E heptane

R / 36

. 0 5

. 0 18

. 0 = =

− hr lbs hr lbs E toluene

R / 56

. 0 5

. 0

28

. 0 = =

− hr lbs hr lbs E nitrogen

R / 46

. 5 5 , 0

73

. 2 = =

− 3.1.2.2 Above Surface Addition Illustration 3-4: Charging a mixture to a partially filled vessel (above surface addition). Three hundred gallons of acetone at 20°C are to be adde d to the vessel featured in Illustration 3-2 by way of above surface addition. For this problem, the in itial contents of the vessel are 1,500 gallons of a mixture of heptane (42% mole fraction) and toluene ( 58% mole fraction) at 20°C. The system pressure is 760 mm Hg and the addition process is complete in 0.5 hours. For this problem, the inlet stream is in direct contact with the vessel head space as it enters the vessel. Therefore, the equilibrium vapor composition for the inlet stream will be based on the exact composition of the inlet stream. A dilution factor A 

### φ for the inlet stream will be assumed to be 1.0. Step 1. Define the conditions of the displaced vent gas: T = 20°C = 298.15°K (System temperature) PT = 1.0 atm = 760 mm Hg (Total system pressure) V disp = 300 gallons = 40.1 ft3 (Displacement volume)Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-9 Time = 30 min (Time for event) Acetone = 6.5632 lb/gal @ 20°C Constants and Relationships: Universal Gas Constant: K mole lb ft

mmHg R ° ⋅ − ⋅ = 3 9

. 998 Antoine Vapor Pressure Equation: P i = exp a − b T + c ⎛ ⎝ ⎞ ⎠
Gas Law: RT PV n = also RT V p n i

i = for a single component i in the gas space. Sum of the partial pressures in the gas space: P p T i i N = =



## ∑ 1 Sum of component moles in the gas space: N n T i i N = =



## ∑ 1 Step 2. Calculate the dilution factor for the initial vessel contents. Inlet Stream Analysis:

VOC Mol. Wt. lb/lbmole Density lb/gal Volume Charged Weight Charged lb-moles X i Acetone 58.08 6.5632 300 1968.96 33.90 1.00 Total 33.90 1.00 Initial Vessel Contents Analysis:

VOC Mol. Wt. lb/lbmole Density lb/gal Volume Charged Weight Charged lb-moles X i Heptane 100.205 5.6977 750 4,273.3 42.65 0.42 Toluene 92.13 7.2138 750 5,410.3 58.72 0.58 Total 101.37 1.00
Calculate the dilution factor for the initial vessel contents: From Eq 3-7 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − = B A B A B B N N N N N ln φ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = 37

. 101 90

. 33 37

. 101 ln 90

. 33 37

. 101 B φ 

## ( ) 86

. 0 7494

. 0 ln 99

. 2 = − = B φChapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-10 Step 3. Calculate the average batch and vapor-phase compositions.
VOC Xi φ A, φ B i 

### X P i (mm Hg) p i (mm Hg) Acetone 1.00 1.00 1.00 184.80 184.80 Heptane 0.42 0.86 0.36 35.55 12.80 Toluene 0.58 0.86 0.50 21.84 10.92 Nitrogen 551.48 Total 1.00 1.00 760.00

Ideal Gas Law: moles lb ft mmHg RT V p E sys acetone acetone
- n − = ⋅ = = 0253

. 0

. 0 ) 15

. 293 )( 9

. 998 ( 1

. 40 80

. 184 3

moles lb ft mmHg RT V p E sys heptane heptane

- n − = ⋅ = = 00176

. 0 ) 15

. 293 )( 9

. 998 ( 1

. 40 83

. 12 3

moles lb ft mmHg RT V p E sys toluene toluene n − = ⋅ = =

− 0015

. 0 ) 15

. 293 )( 9

. 998 ( 1

. 40 92

. 10 3

moles lb ft mmHg RT V p E sys nitrogen nitrogen n − = ⋅ = =

− 0755

. 0 ) 15

. 293 )( 9

. 998 ( 1

. 40 48

. 551 3

Emission level: 

## ( ) lbs mole lb lbs moles lb E acetone
- wt 47

. 1 08

. 58 0253

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − = 

## ( ) lbs mole lb lbs moles lb E heptane

- wt 18

. 0 205

. 100 00176

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − = 

## ( ) lbs mole lb lbs moles lb E toluene

wt 14

. 0 13

. 92 0015

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − =

− 

## ( ) lbs mole lb lbs moles lb E nitrogen

wt 12

. 2 0134

. 28 0755

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − =

−
Step 4. Calculate the emission rate in lb/hr units. hr lbs hr lbs E acetone

R / 94

. 2 5

. 0 47

. 1 = =

− hr lbs hr lbs E heptane

R / 36

. 0 5

. 0 18

. 0 = =

− hr lbs hr lbs E toluene

R / 56

. 0 5

. 0

28

. 0 = =

− hr lbs hr lbs E nitrogen

R / 24

. 4 5 , 0 12

. 2 = =

−Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-11 3.1.2.3 Addition of Immiscible Liquids Illustration 3-5: Charging a mixture to a partially filled vessel (immiscable liquids). Seven hundred fifty gallons of toluene at 20°C are to be added to the vessel that contains 500 gallons of water at 20°C. The system pressure is 760 mm Hg and the addition process is complete within 0.5 hours. For this problem, the two mixtures are insoluble and form two distinct liquid phases as the addition process takes place. Therefore, dilution of one stream by the other does not occur and the equilibrium vapor composition is determined based on the initial composition of each mixture. Step 1. Define conditions of the displaced gas: T = 20°C = 298.15°K (System temperature) P T = 1.0 atm = 760 mm Hg (Total system pressure) V disp = 750 gal = 100.2 ft3 (Displacement volume) Time = 30 min (Time for event) Constants and Relationships: Universal Gas Constant: K mole lb ft

mmHg R ° ⋅ − ⋅ = 3 9

. 998 Antoine Vapor Pressure Equation: P i = exp a − b T + c ⎛ ⎝ ⎞ ⎠ Gas Law: RT PV n = , also R T V p n i

i = for a single component i in the gas space. Sum of the partial pressures in the gas space: P p T i i N = =



## ∑ 1 Sum of component moles in the gas space: N n T i i N = =



## ∑ 1 Step 2. Calculate the dilution factor for the initial vessel contents. Inlet Stream Analysis:
VOC Mol. Wt. lb/lb-mole Density lb/gal Volume Charged Weight Charged lb-moles X i Toluene 92.13 7.2138 750 5,410.4 58.72 1.00 Total 58.72 1.00 Initial Vessel Contents Analysis:
VOC Mol. Wt. lb/lb-mole Density lb/gal Volume Weight lb-moles X i Water 18.02 8.33 500 4,165 231.13 1.00 Total 231.13 1.00Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-12 Step 3. Calculate the average batch and vapor-phase compositions.

VOC X i P i (mm Hg) Toluene 1.00 21.84 Water 1.00 17.35 Nitrogen 720.81 Total 1.00 760.00
Ideal Gas Law: moles lb ft mmHg RT V p E sys toluene toluene

- n − = ⋅ = = 00747

. 0

. 0 ) 15

. 293 )( 9

. 998 ( 2

. 100 84

. 21 3

moles lb ft mmHg RT V p E sys water water
- n − = ⋅ = = 00594

. 0 ) 15

. 293 )( 9

. 998 ( 2

. 100 35

. 17 3

moles lb ft mmHg RT V p E sys nitrogen nitrogen n − = ⋅ = =

− 24665

. 0 ) 15

. 293 )( 9

. 998 ( 2

. 100 81

. 720 3

Emission level: 

## ( ) lbs mole lb lbs moles lb E toluene

wt 69

. 0 13

. 92 00747

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − =

− 

## ( ) lbs mole lb lbs moles lb E water

wt 11

. 0 02

. 18 00594

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − =

− 

## ( ) lbs mole lb lbs moles lb E nitrogen

wt 91

. 6 0134

. 28 24665

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − =

− Step 4. Calculate the emission rate in lb/hr units.
hr lbs hr lbs E toluene

R / 38

. 1 5

. 0

69

. 0 = =

− hr lbs hr lbs E water

R / 22

. 0 5

. 0 11

. 0 = =

− hr lbs hr lbs E nitrogen

R / 82

. 13 5 , 0

91

. 6 = =

− 

### 3.2 Purge/Gas Sweep Models 3.2.1 Purge or Gas Sweep – Empty Vessel Purge When a gas purge is applied to an empty vessel that still contains residual vapors from a previous process operation then compound emissions may be de termined by the following expression. ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− = −

− V Ft i i n e RT V p E 1 1 , Where E n-i are the moles of component i that are emitted due to vapor displacement, p i,1 is the saturated vapor pressure of component i at initial conditions,Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-13 V

is the gas space volume of vessel when empty, R is the ideal gas constant in consistent units, T

is the temperature of the liquid being charged, F

is the purge gas flowrate, t is the elapsed time for the purge operation. Illustration 3-6: Purging an empty vessel or reactor containing solvent vapors. A 2,000 gallon reactor vessel was cooled to 20°C and the contents, an acetone solvent, were pumped out leaving only vapors. If this vessel is then purged with 1,000 standard cubic feet of nitrogen at 20°C, how much acetone is in the vented nitrogen? Step 1. Determine the initial partial pressure of acetone in the vessel gas space. 

### ( ) p mmHg

acetone C , exp . .

. . exp . . 20 16 6513 2940 46 29315 3593 52196 184 86 ° = −

− ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = = 

## ( ) F t scf acf

⋅ = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1000 27315 20 27315 1073

.

. V g a l gal ft ft = = 2000 7 48 267 3 3 . /

n F t V acf ft = ⋅ = = 1073 267 4 02 3 . Step 2. Calculate the acetone loss from the vessel by conducting a material balance around the vessel. Where N acetone is the amount of acetone displaced from the vessel: ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− = −

− V Ft i i n e RT V p E 1 1 , 

# ( ) 02

. 4 1 ) 15

. 293 )( 9

. 998 ( ) 267 )( 86

. 184 (

−

− −

= e E i

n 

# ( ) moles lb e E acetone

n − = = − = −

− 166

. 0 ) 9821

. 0 )( 1686

. 0 ( 1 ) 1686

. 0 ( 02

. 4 lb mole lb lb moles lb E acetone

wt 58

. 9 ) / 08

. 58 )( 166

. 0 ( = − − =

− Step 3. Calculate the amount of nitrogen emitted from the vessel. Before the empty vessel purge began, the gas space w as saturated with acetone vapors at a partial pressure of 184 mm Hg. At the completion of the purge operati on, the acetone partial pressure has been reduced to 3.32 mm Hg. The amount of nitrogen that leaves the vessel through the vent is equal to the inlet amount minus the accumulation amount. .) ( ) ( ) ( accum N in N out E nitrogen nitrogen nitrogen n − =

−
RT V p p RT PV out E vessel i i purge nitrogen n ) ( ) ( 2 , 1 , −

− =

− moles lb out E nitrogen

n − = −

− =

− 62

. 2 ) 15

. 293 )( 9

. 998 ( ) 267 )( 32

. 3 86

. 184 ( ) 15

. 293 )( 9

. 998 ( ) 1073 )( 760 ( ) (Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-14 lb mole lb lb moles lb E nitrogen

wt 4

. 73 ) / 28 )( 62

. 2 ( = − − =

− 3.2.2 Purge or Gas Sweep – Partially Filled Vessel Air or another non-condensable gas is directed into the vessel at a controlled rate. The discharge vapors from the vessel during this operation are normally assumed to be at equilibrium or saturated with the vessel’s liquid contents within certain flow rate criteri a. Further, it is assumed that the inlet purge rate is known. Eq. 3-8 is used to calculate amounts of condensable components. The mixing factor S i represents the degree of VOC saturation for the vent gas, and it is normally between 0 and 1.0. A value of 0.25 for S i implies that the exit vent gas is at 25% saturation le vel with respect to the solvent vapors in the tank, while a S i value of 1.0 implies that the exit vent gas is at equilibrium with the volatile contents of the vessel. [Hatfield, 2003a]

sat nc

sat i i nc R i R p p S E E

− − = Eq. 3-8
Where E R
- = moles of volatile component i emitted per unit time, S i = saturation level of the exit vent gas stream, E nc = moles of non-condensable gas emitted pe r unit time (= inlet purge rate),

sat i 

### p = partial pressure of component i at saturated conditions,

sat nc 

### p = partial pressure of the non-condensable gas (i.e., air, nitrogen) at saturated solvent pressure conditions. The saturation factor for a solvent vapor is a func tion of the evaporation mass transfer coefficient K i , the liquid surface area in the vessel, and the solvent part ial pressure in the vessel headspace. When the exit emission rate for a vessel is set equal to the evapor ation rate in the vessel then the following expression results for the saturation level S i. Saturation level S I: sat i i nc i i i i Sat i i i F S F A K A K F A K A

K p p S + + = + = = Eq. 3-9

Where 3 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= i o o i M M K K , Eq. 3-10

and 

# ( ) sat i sys sat i nc sat nc sat i nc sat i p P p F p p F F

− = = . Eq. 3-11

i 

### S = saturated vapor pressure for compound i, K i = mass transfer coefficient i, K o = mass transfer coefficient of a reference compound, M i = molecular weight of compound i ,Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-15 M o = molecular weight of the reference compound o , A = surface area of the liquid, F nc = volumetric flow rate of the non-condensable gas (i.e. air, nitrogen),
sat i 

### F = volumetric flow rate of component i (i.e. VOC) at saturated vapor pressure,

sat i 

### p = saturated vapor pressure of component i,

sat nc 

### p = partial pressure of the noncondensable gas (i.e. air, nitrogen) at saturated solvent pressure conditions,

sys 

### P = system pressure. The saturated partial volumetric flow rate for each component is estimated from the component saturated vapor pressure, the inlet gas purge rate, and the pa rtial pressure of the non-condensable at saturated conditions. S

i may be solved using the standard quadratic soluti on. Although the standard qua dratic equation contains two roots, only the one solution shown in Eq. 3-12 produces a realistic value since i 

### S must be a positive number between 0 and 1.0. Quadratic solution: sat i i sat i nc i nc i i F A K F F A K F A K S 2 4 ) ( ) ( 2 + + + +
− = Eq. 3-12 Finally, the emission rate for the volatile component 

### i may be calculated using Eq. 3-13 which allows for the use of i 

### S , where sat i i i 

### p S p = .

RT P F S M E sys sat i i i i = Eq. 3-13 For multi-component liquid mixtures, Eq. 3-9 may be e xpanded to provide partial volumetric flow levels for each volatile component in the liquid.

sat n n sat j j sat i i nc i i i F S F S F S F A K A K S + + + + + = +

... 1 Eq. 3-14 where i is the compound for which the saturation level is being calculated, and terms j through n represent the other components in the liquid. Eq. 3-14 is solved in an iterative trial and error manner with the initial value of S for each component assigned to 1.0. The value of S that is calculated for each component is then used as the starting point for the next iterati on. Finally, when the saturation level S of each compound remains the same for subsequent iteratio ns then the calculation process is stopped. Emission rates for each solvent com pound are then calculated based on the partial volumetric flow rate as calculated from the saturation level and the saturated volumetric flow rate sat i i i 

### F S F 

### = . Solvent emission rate RT P F S M E sys sat i i i i = (wt per unit of time)Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-16 Illustration 3-7: Gas sweep with a vessel containing with a single volatile solvent. A vertical process vessel with a 6-ft cross sectional diameter is at 1 atm pressure and contains a volume of heptane at 25°C. The vessel is being purged with 10 scfm (standard cubic feet per minute) of nitrogen gas. Calculate the emission rate of heptane during the purge operation. The molecular weight of heptane is 100.2. The mass transfer coefficient is estimated using Eq. 3-10 with the known mass transfer coefficient for water of 0.83 cm/s. Other variables are also calculated from established relationships.

min 92

. 0 min 48

. 30 60 4685

. 0 2

. 100 02

. 18 83

. 0 3 1 3 1 ft cm ft s s cm s cm M M K K i o o i =

⋅ ⋅ ⋅ ⋅ × = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

. 86

. 45 tan mmHg P sat e

hep = . 14

. 714 86

. 45 760 mmHg P sat

nitrogen = − = min / 92

. 10 15

. 273 15

. 298 10 3 ft K K scfm F

nitrogen = ° ° =
min / 70

. 0 14

. 714 86

. 45 92

. 10 3 tan ft mmHg mmHg F sat e

hep = =

2 2 2 26

. 28 4 36 14

. 3 4 ft ft d A = ∗ = = π

70

. 0 2 26

. 28 92

. 0 70

. 0 4 ) 92

. 10 26

. 28 92

. 0 ( ) 92

. 10 26

. 28 92

. 0 ( 2
tan ∗ ∗ ∗ ∗ + + ∗ + + ∗
− = e hep S
69

. 0 40

. 1 97

. 0 40

. 1 79

. 72 0

. 1363 92

. 36 tan = = + +
− = e hep S 

## ( ) ( ) 

# ( ) ( ) 

# ( ) ( ) R R mol lb atm ft mmHg ft mole lb lb RT P F S M E L sys sat i i i m ° °

− − = = 537 / 98

. 554 760 min / 70

. 0 69

. 0 / 2

. 100 3 3
hr lb lb E m

R / 2

. 7 min / 12

. 0 = =

− As an illustration on how the saturation factor for he ptane could have been calculated using the iteration technique Eq 3-14 can be applied directly.

i i sat i i nc i i i S S F S F A K A K S ∗ + = + + ∗ ∗ = + + = 7

. 0 92

. 36 99

. 25 70

. 0 92

. 10 26

. 28 92

. 0 26

. 28 92

. 0
Starting with an initial guess for S i to equal 1.0, the convergence occurs rapidly with the final result 0.69 being equal to earlier results us ing the quadratic equation. Iteration 0 S i = 1.00 Iteration 1 S i = f (1.00) = 0.69 Iteration 2 S i = f (0.69) = 0.69 Illustration 3-8: Gas sweep with a vessel containing with a volatile solvent mixture. Suppose the vessel featured in Illustration 7 containe d a solvent mixture consisting of 20% heptane, 70% toluene, and 10% methanol. Assuming that the compos ition is specified in mole percents, calculate theChapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-17 saturation factor for each component using Eq. 3-14. Be fore applying Eq. 3-14 to this solution several values must be calculated for each component as shown in the following table: Table 3-1. Calculated Values for Use in Eq. 3-14 Compound MWt VP@25C Mole fraction Vp Fi(sat) Ki KiA Heptane 100.21 45.86 0.2 9.17 0.14 0.47 13.24 Toluene 92.13 28.44 0.7 19.91 0.30 0.48 13.62 Methanol 32.04 126.88 0.1 12.69 0.19 0.69 19.36 The calculated values in Table 3-1 are then substitute d into Eq. 3-14 for an iterative trial and error procedure.

sat n n sat j j sat i i nc i i i F S F S F S F A K A K S + + + + + = +

... 1 Eq. 3-14
Table 3-2. Iterative Trial and Error Results for Si using Eq. 3-14 Compound KiA FiSat S i (iteration 0) S i (iteration 1) S i (iteration 2) S i (iteration 3) Heptane 13.24 0.14 1.00 0.53 0.54 0.54 Toluene 13.62 0.30 1.00 0.54 0.55 0.55 Methanol 19.36 0.19 1.00 0.63 0.63 0.63 Finally, the solvent emission rate for each com ponent may be calculated from the relationship.

RT P F S M Q sys sat i i i i = (weight per unit of time) to produce the following emission rate results. Table 3-3. Calculated Emission Rates Compound MWt S i F i Sat Q i (lb/min) Q i (lb/hr) Heptane 100.21 0.54 0.14 0.0192 1.15 Toluene 92.13 0.55 0.30 0.0389 2.33 Methanol 32.04 0.63 0.19 0.0100 0.60 

### 3.3 Vacuum Operations The application of vacuum is used in many dis tillation or drying operations as a means of reducing the boiling point temperature of a given process mixture. In the case of vacuum distillation, solvent is vaporized in the still vessel, condensed at a low temper ature, and collected in a receiving vessel. In the case of a solids drying operation, wet product cake is placed in a rotary or tray dryer and vacuum is applied to the entire drying system. Heat is then app lied to the dryer and solvent vapors are condensed at a low temperature and collected in the receiving vessel.Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-18 Vent emissions that occur from vacuum operations are the result of air being removed from the system by the vacuum pump or ejector. For solids drying, a nitr ogen sweep may be applied at the dryer as a means of accelerating the drying process. The vacuum pump or ejector must remove this additional nitrogen along with any air that has leaked into the system due to the lower pressure. When the initial distillation phase is underway then th e vent emissions are calculated for the system as being from the distillate receiver based on the co ndensate volume, composition, and temperature. Additionally, the non-condensable gas flow rate (air leak rate and/or nitrogen) and operating vacuum level must be taken into account. It is assumed th at the exiting vent gas from the vacuum receiver is saturated with vapors from the liquid condensate. The moles of each volatile component in the exit vent gas are calculated using the following relationship.

nc i nc n i n p p E E

− − = Eq. 3-15
Where E n-i = moles of volatile component i emitted from the process, E n-nc = total moles of noncondensable gas emitted from the process, p i = partial pressure of volatile component i , p nc = partial pressure of the noncondensable gas (i.e. air, nitrogen) at saturated solvent pressure conditions. E n-nc represents the total moles of non-condensable gas component that are removed from the system by the vacuum pump.

sweep gas nc n nt displaceme nc n leakage nc n nc n E E E E

− − − − − − − + + = Eq. 3-16 Where E n-nc-leakage are the moles of air due to leakage into the system, E n-nc-displacement are the moles of air that are displaced by the condensate, E n-nc-gas sweep are the moles of air or nitrogen admitted as a gas sweep. If the distillation phase has been completed and th e receiver continues to hold condensate while remaining under vacuum, then the non- condensable gas flow rate is based on the air leak rate and/or any nitrogen sweep flow that might exist. In many cases the air leak rate might be expressed in acfm (ft 3 /min at actual temperature and pressure conditions) because it relates to either a known vacuum pump capacity at the 25 mm Hg condition or may have been measured through prior vacuum leak test on the equipment. The nitrogen gas sweep is normally specified in scfm (ft 3 /min at standard temperature and pressure conditions) because it relates to nitrogen gas flow meter that might be used for control purposes. Illustration 3-9: Vacuum operation with vessel filling. 400 gallons of toluene are distilled from a process mi xture under vacuum conditions in 2.5 hours. The equipment consists of a 1000 gallo n still, condenser, and 1000 gallon r eceiver. A liquid ring vacuum is used to reduce the operating pressure of the equipment system to 100 mm Hg. The air leak rate is known to be 10 acfm at these conditions. The condenser is cooled with 5°C chilled glycol and the toluene condensate is measured to be 10°C. Calculate th e vent emissions leaving the equipment system.Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-19 For this problem, toluene is being co llected at 10°C in the receiver and an air leak rate has been specified. Air leak rate: 10 acfm. Receiver displacement volume: 400 gallons Operating pressure: 100 mm Hg. Vapor pressure of toluene at 10°C: 12.43 mm Hg. Antoine equation Process time: 2.5 hours Calculations: Partial pressure of non-condensable: P nc = 100 mm Hg – 12.43 mm Hg. = 87.57 mm Hg. Displacement volume in ft 3 : 3 3 44

. 53 / 485

. 7 400 ft ft gal

gal V =

=
moles lb K ft mmHg RT V p E sys nc nt displaceme nc n − = ° ⋅ = =

− − 25

. 0 ) 15

. 293 )( 9

. 998 ( 44

. 53 57

. 87 3
Leakage volume in ft 3 : 3 1560 ) min/ 60 )( 5

. 2 )( 10 ( ft hr hr acfm V

leakage = = moles lb K ft mmHg RT

VT p E K m nc leakage nc n − = ° ⋅ = =

° − − 53

. 0 ) 15

. 293 )( 9

. 998 ( 1560 100 3

Therefore
sweep gas nc n nt displaceme nc n leakage nc n nc n E E E E

− − − − − − − + + = moles lb E sweep gas nc n nt displaceme nc n leakage nc n nc n − = + + =

− − − − − − − 78

. 0 0 25

. 0 53

. 0 Finally moles lb p p E E

nc i nc n toluene n − = = =

− − 11

. 0 57

. 87 47

. 12 78

. 0 lb 10.23 mole)

- lb/lb 92.13 )( 11

. 0 ( = − = =

− − moles lb Mw E E toluene toluene n toluene wt Illustration 3-10: Vacuum operation without vessel filling. 100 gallons of toluene have been collected from a product solids drying operation. The equipment consists of a 200 ft 3 tray drier, condenser, and 250 gallon r eceiver. A liquid ring vacuum is used to maintain an operating pressure of 25 mm Hg. The air leak rate is known to be 1 acfm under these conditions. For this final phase of the drying operation a 1 scfm nitrogen gas sweep is applied to the tray dryer to help accelerate the final drying phase. Alt hough the distillation phase of the operation has ended, the recovered toluene remains in the receiver and is maintained at 7°C. Calculate the toluene emissions from the equipment system to the vacuum pump over a 1.0 hour period. Since the distillation phase has ended the only the air l eak rate and nitrogen gas sweep rate need to be considered. Air leak rate: 1 acfm.Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-20 Nitrogen gas sweep rate: 1 scfm Operating pressure: 25 mm Hg. Vapor pressure of toluene at 7°C: 10.40 mm Hg. Antoine equation Calculations: Leakage (air): 3 60 min) 60 )( 1 ( ft

acfm V

leakage = = moles lb K ft mmHg RT

VT p E K m nc leakage nc n − = ° ⋅ = =

° − − 0054

. 0 ) 15

. 280 )( 9

. 998 ( 60

25 3
Gas sweep (nitrogen): ) ( 60 min) 60 )( 1 ( 3 stp ft scfm V sweep

gas = = moles lb K ft mmHg RT V P E stp stp sweep gas nc n − = ° ⋅ = =

− − 17

. 0 ) 15

. 273 )( 9

. 998 ( 60 760 3
Therefore

sweep gas nc n nt displaceme nc n leakage nc n nc n E E E E

− − − − − − − + + = moles lb E sweep gas nc n nt displaceme nc n leakage nc n nc n − = + + =

− − − − − − − 18

. 0 17

. 0 0 0054

. 0 Finally mmHg p P p toluene sys nc 6

. 14 4

. 10 0

. 25 = − = − = moles lb p p E E

nc i nc n toluene n − = = =

− − 13

. 0 60

. 14 40

. 10 18

. 0 lb 11.98 mole)

- lb/lb 92.13 )( 13

. 0 ( = − = =

− − moles lb Mw E E toluene toluene n toluene wt 

### 3.4 Gas Evolution Certain processes generate off gases as function of th e reaction chemistry. Vent emissions from these types of operations may be estimated by assuming that the exit vent off gas containing the reaction off gas is fully saturated with vapors from the volatile co mponents in the batch. The partial pressure of each component is calculated based on the pure componentvapor pressure, mixture composition, and any nonidealities that might exist (activity coefficients). The moles of each volatile component in the exit vent gas are calculated using the following relationship.

rxn i rxn n i n p p E E

− − = Eq. 3-17 where: E n-i = moles of volatile component i emitted from the process, E n-rxn = total moles of reaction off gas emitted from the process, p i = partial pressure of volatile component i , p rxn = partial pressure of the noncondensable gas (i.e. air, nitrogen) at saturated solvent pressure conditions.Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-21 The stoichiometric amount of off gas is usually de termined by the process chemistry. However, other considerations may need to be taken into account when estimating the actual amount of off gas that leaves the system. For example, if the off gas is partially so luble in the process solvent then the portion that does not dissolve in the batch will exit the vessel through the vent. If the solubility of the off gas is not known the one could conservatively assume that 100% of the reaction off gas exits the process through the vessel vent. Illustration 3-11: Reaction involving a gas evolution of one component. A reaction takes place in a vessel at 50°C with toluene as the primary solvent. Ei ght pounds of hydrogen chloride is generated based on the chemistry and vented out of the vessel over a one-hour period. Calculate the compound emissions that occur from the reaction, if the system pressure is 760 mm Hg. Also, assume that the batch consist of 95% toluene (mole/mole) and 5% compounds that are nonvolatile. Step 1. Determine the hydrogen chloride that is discharged from the vessel. mole lb mole lb lb lbs MWt

Wt E HCl HCl HCl n − =

− = =

− 219

. 0 / 461

. 36 8
Step 2. Calculate the vapor pressure of toluene at 50°C using the Antoine equation. 

## ( ) mmHg P C

toluene 11

. 92 52298

. 4 exp 67

. 53 15

. 323 52

. 3096 0137

. 16 exp 50 , = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − = ° mmHg mmHg p C

toluene 50

. 87 11

. 92 95

. 0 50 , = × = ° Step 3. Calculate the toluene emission rate by the ratio of vapor pressures. ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

− HCl toluene HCl toluene n p p E E

moles lb mmHg mmHg mmHg moles lb E toluene

n − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

− − =

− 0285

. 0 50

. 87 760 50

. 87 219

. 0 mole lb lb moles lb E toluene

wt − × − =

− / 13

. 92 0285

. 0 lbs E toluene

wt 63

. 2 =

− Illustration 3-12: Reaction involving gas evolution of two components. A reaction takes place in a vessel at 50°C with toluene as the primary solvent. Ei ght pounds of hydrogen chloride along with an equal molar quantity of sulfur dioxide is gene rated and vented out of the vessel over a one-hour period. Calculate the compound emissions that occur from the reaction, if the system pressure is 760 mm Hg. Also, assume that the batc h consist of 95% toluene (mole/mole) and 5% compounds that are nonvolatile. Step 1. Determine the hydrogen chloride and sulfur dioxide that is discharged from the vessel. The molar amount of HCl is calculated from the mol ecular weight and the quantity of HCL emitted. The molar amount of SO 2 is set equal to the calculated molar amount of HCl. mole lb mole lb lb lbs MWt

Wt E HCl HCl HCl n − =

− = =

− 219

. 0 / 461

. 36 8Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-22 moles lb n E HCl SO n − = =

− 219

. 0 2 lbs 14.03 mole

- lb/lb 64.063 219

. 0 2 = × − =

− moles lb E SO

wt Step 2. Calculate the vapor pressure of toluene at 50°C using the Antoine equation. 

## ( ) mmHg P C

toluene 

### 11

. 92 52298

. 4 exp 67

. 53 15

. 323 52

. 3096 0137

. 16 exp 50 , 

### = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − = ° 

### mmHg mmHg p C

toluene 

### 50

. 87 11

. 92 95

. 0 50 , 

### = × = ° Step 3. Calculate the toluene emission rate by the ratio of vapor pressures. 

# ( ) ( )
2 2 2 SO HCl toluene sys toluene SO HCl SO HCl toluene toluene n 

### n n p P p n n p p p E + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

− = + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + =

−


### moles lb mole lb mmHg mmHg mmHg E toluene

n 

### − = − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

− =

− 

### 057

. 0 438

. 0 50

. 87 760 50

. 87


### mole lb lb moles lb E toluene

wt 

### − × − =

− 

### / 13

. 92 057

. 0 

### lbs E toluene

wt 

### 26

. 5 

### =

− 

### lbs E HCl

wt 

### 0

. 8 

### =

− 

### lbs E SO

wt 

### 03

. 14 2 

### =

− Illustration 3-13: Reaction involving multicomponent gas evolution and nitrogen purge. A reaction takes place over a 1.0 hour period in a v essel at 50°C with toluene as the primary solvent. Eight pounds of hydrogen chloride along with an equa l molar quantity of sulfur dioxide is generated and vented out of the vessel over a one-hour period. Ca lculate the compound emissions that occur from the reaction, if the system pressure is 760 mm Hg and a nitrogen purge is being applied at 30 SCFH. Also, assume that the batch consist of 95% toluene (m ole/mole) and 5% compounds that are nonvolatile. Step 1. Determine the hydrogen chloride, sulfur dioxide, and nitrogen that are discharged from the vessel over the one hour period. The molar amount of HCl is calculated from the molecular weight and the quantity of HCL emitted. The molar amount of SO 2 is set equal to the calculated molar amount of HCl. mole lb mole lb lb lbs MWt

Wt E HCl HCl HCl n − =

− = =

− 219

. 0 / 461

. 36 8
moles lb n E HCl SO n − = =

− 219

. 0 2 lbs moles lb lb moles lb E SO

wt 03

. 14 / 64.063 219

. 0 2 = − × − =

− The molar amount of N 2 is calculated based on the 30 SCFH (sta ndard cubic feet per hour) flow rate or nitrogen for the 1.0 hour reaction period.Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-23 mole lb mole lb SCF HR SCFH E N

n − =

− × =

− 084

. 0 / 359.046 0

. 1 0

. 30
2 

### lbs moles lb E N

wt 

### 35

. 2 moles

- lb/lb 28.013 084

. 0 2 

### = × − =

− The total moles of non-condensable compounds are calculated by summing the molar amounts of HCl, SO2, and N2. 

### 084

. 0 219

. 0 219

. 0 2 2 

### + + = + + =

− − − − N n SO n NCl n nc n 

### E E E E moles lb E nc

n 

### − =

− 

### 522

. 0 Step 2. Calculate the vapor pressure of toluene at 50°C using the Antoine equation. 

## ( ) mmHg P C

toluene 

### 11

. 92 52298

. 4 exp 67

. 53 15

. 323 52

. 3096 0137

. 16 exp 50 , 

### = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − = ° 

### mmHg mmHg p C

toluene 

### 50

. 87 11

. 92 95

. 0 50 , 

### = × = ° Step 3, Calculate the toluene emission rate by the ratio of vapor pressures.

nc n toluene sys toluene nc n nc toluene toluene n 

### E p P p E p p E

− − − 

### ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

− = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= 

# ∑
moles lb mole lb mmHg mmHg mmHg E toluene

n 

### − = − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

− =

− 

### 13

. 0 522

. 0 50

. 87 760 50

. 87


### mole lb lb moles lb E toluene

wt 

### − × − =

− 

### / 13

. 92 13

. 0 lbs E toluene

wt 

### 98

. 11 =

− 

### lbs E HCl

wt 

### 0

. 8 =

− 

### lbs E SO

wt 

### 03

. 14 2 

### =

− 

### lbs E N

wt 

### 35

. 2 2 

### =

− 

### 3.5 Depressurization Estimating solvent emissions from the depressurization of a batch pressure filter for solids discharge or for the evacuation of a vessel that contains a volatile liquid mixture and a noncondensable gas-phase component, such as air or nitrogen, requires cer tain assumptions and approximations be made:



###  The system pressure is decreased linearly over time.



###  Air leakage into the vessel durin g the operation is negligible.



###  The liquid and gas space temperature remains constant during the operation.



###  The vapor space of the vessel rema ins in equilibrium with the vol atile liquid contents during the depressurization process.Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-24 Since the system temperature is assumed to remain constant during the depr essurization operation then the equilibrium vapor pressure of the vessel liquid contents remains constant as well. The moles of solvent vapor that exist within th e vessel headspace during the depressuri zation remain constant for this reason. However, the volatile solvent vapor occupi es a greater fraction of the vessel headspace and exit vent gas as the depressurization takes place since the syst em pressure is being reduced toward the solvent vapor pressure level. As the depressurization unfolds th en more and more solvent must evaporate in order to maintain the equilibrium vapor pressure. Ther efore, the solvent emissions that occur during depressurization are equal to the net solvent evapor ation within the vessel, b ased on the assumptions for this model. Eq. 3-18 is based on a material balance for th e non-condensable in the vessel headspace and the assumption that the total system pressure is equal to the partial pressure of the volatile plus the partial pressure of the non-condensable. Eq. 3-18 may be integr ated as shown in Eq. 3-19 to result in Eq. 3-20. . [Hatfield, 1998b]

nc nc i i dp p p RT V out dn ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− = ) ( Eq. 3-18 

## ∫ ∫ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

− = 2

1 2

1 , ) ( p

p nc nc V T i p

p i p dp RT Vp out dn Eq. 3-19
⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= 2 , 1 ,
, ln

nc nc i out i p p RT Vp n Eq. 3-20
Where: n i,out = moles of volatile component i leaving the vessel V = vessel headspace volume p i = partial pressure of the volatile component R = Universal gas constant T = system temperature p nc,1 = partial pressure of the non-conden sable component at initial conditions p nc,2 = partial pressure of the non-conden sable component at final conditions Illustration 3-14: Vessel depressurization involving one volatile component. A 1,000 gallon nutsche filter is used to compress a slurry containing acetone and nonvolatile solids at 80°F (26.7°C). A pressure of 35 psig is imparted ont o the slurry until the desired filtration is achieved (approximately 40 minutes). The nutsche filter is then depressurized prior to the discharging of its contents. Residual solids in the filter are estimated to occupy 500 gallons of the filter volume. Calculate the acetone emissions from the depressurization operation. Given: T = 26.7°C = 299.85°K P1 = 35 psig = 2570 mm Hg (Initial pressure) P2 = 0.0 psig = 760 mm Hg (Final pressure) V = 500 gal = 66.843 ft3 (Gas space volume)Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-25 Step 1. Determine the saturated vapor pressure for acetone at 26.7°C and the non-condensable partial pressure at initial (P nc,1 ) and final conditions (P nc,2 ). 

## ( ) p mmHg

acetone C , . exp . .

. . exp . . 26 7 16 6513 2940 46 299 85 35 93 55098 247 11 ° = −

− ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = = Therefore: Pnc,1 = (2570 – 247.11) = 2,322.89 mm Hg Pnc,2 = (760 – 247.11) = 512.89 mm Hg Step 2. Calculate the amount of acetone vapor that is emitted from the depressurization operation. moles lb p p RT Vp E

nc nc i acetone n − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

− 0827

. 0 89

. 512 89

. 2322 ln ) 300 )( 9

. 998 ( ) 11

. 247 )( 843

. 66 ( ln 2 , 1 ,

E

wt-acetone = (0.0827 lb-moles) ( 58.08 lb/lb-mole) = 4.8 lbs Illustration 3-15: Vessel depressuriza tion involving a solvent mixture. A 1,200 gallon process vessel contains 700 gallons of so lvent mixture that is being prepared for vacuum distillation. The solvent mixture is at 20°C and has a molar composition of 20% acetone, 50% toluene, and 30% methanol. Calculate the emissions from the de pressurization operation if the pressure is reduced from 760 mm Hg to 100 mm Hg . Given: T = 20.0°C = 293.15°K P1 = 760 mm Hg (Initial pressure) V = 500 gal = 66.843 ft3 (Gas space volume) Universal Gas Constant: 

### K mole lb ft

mmHg R ° ⋅ − ⋅ = 3 

### 9

. 998
Gas Law: 

### n PV RT = , also 

### n p V RT i i



### = for a single component, i , in the gas space.
Step 1. Determine the saturated vapor pressure composition for the process material at 25°C. For this illustration the equilibrium vapor pressure composition for the process mixture may be calculated based on the pure vapor pressure and mole fraction of each component in the mixture. The total saturated vapor pressure for the mixture is calculated to be 77.08 mm Hg. as shown in the following table.

Compound Pure Vapor Pressure (mm Hg) Mole Fraction X i Pi=X i *P i (mm Hg) Acetone 184.86 0.2 36.97 Methanol 97.30 0.3 29.19 Toluene 21.84 0.5 10.92 Total 77.08Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-26 moles lb p p RT Vp E

nc nc i acetone n − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

− 0286

. 0 92

. 22 92

. 682 ln ) 15

. 293 )( 9

. 998 ( ) 97

. 36 )( 843

. 66 ( ln 2 , 1 ,

moles lb p p RT Vp E

nc nc i methanol n − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

− 226 0

. 0 92

. 22 92

. 682 ln ) 15

. 293 )( 9

. 998 ( ) 19

. 29 )( 843

. 66 ( ln 2 , 1 ,

moles lb p p RT Vp E

nc nc i toluene n − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

− 85 00

. 0 92

. 22 92

. 682 ln ) 15

. 293 )( 9

. 998 ( ) 92

. 10 )( 843

. 66 ( ln 2 , 1 , 

### 3.6 Heating Heat-up losses that occur during the operation of react ors, distillation equipment, and similar types of processing equipment may be estimated by application of the Ideal Gas Law and vapor-liquid equilibrium principles. Emissions are calculated using the following assumptions: The vessel is closed during the operation but vapors are vented through a process vent. Additional material is not added to the vessel during heat-up. Th e displaced gas is assumed to be saturated with VOC vapor in equilibrium with the process material. When a vessel containing a volatile liquid and a noncondensable gas (e .g. air) is heated at constant pressure, the vapor space gas undergoes expansion and a portion of the gas phase leaves the vessel through the vent. Additionally, the saturated vapor pressures for the volatile liquid components increase. The calculation is based on the premise that the amo unt of the non-condensable component (air, nitrogen, etc.) that is displaced from the vessel is determin ed by the initial and final gas space composition. If anitrogen purge or sweep is placed on the vessel during the heating step, then the amount of noncondensable component that is displaced from the ve ssel is increased by the total amount of purge gas that passes through the vessel during the heating. In the heating model, rising vapors from the vessel liquid contents displace the non-condensable gas components from the headspace through the process vent . As the liquid mixture reaches the boiling point, all of the non-condensable component is purged fr om the vapor space. This model assumes that the average molar headspace volu me remains constant relative to changes in the molar composition of the vessel headspace. Eq. 3-21 is derived from performing material balances around the vessel headspace for
the non-condensable component and for component i during the heating. [Hatfield, 1998c] 

# ( ) vessel i i nc nc avg out i n 

### n n p p N E 1 , 2 , 2 , 1 ,
, 

### ln

− − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

− Eq. 3-21Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-27 Where: 

## ( ) 2 1 

### 2 1 n n N avg 

### + = Eq. 3-22

n i,out = moles of volatile component i leaving the vessel process vent N

avg = average gas space molar volume during the heating process p nc1 = partial pressure of non-condensable in the vessel headspace at temperature T 1 p nc2 = partial pressure of non-condensable in the vessel headspace at temperature T 2 n i,1 = moles of volatile component i in the vessel headspace at temperature T 1 n i,2 = moles of volatile component i in the vessel headspace at temperature T 2 n 1 = total moles of gas in the vessel headspace at temperature T 1 n 2 = total moles of gas in the vessel headspace at temperature T 2 Illustration 3-16: Heatup losses from a vessel containing a single volatile component. A 1250 gallon reactor containing 750 gallons of a toluene solution is heated from 20°C to 70°C. The reactor is vented to the atmosphere during th e heat up. How much toluene will be emitted? Step 1. Calculate the average molar volume of the vessel headspace. T i = 20°C = 293.15°K (Initial temperature) T f

= 70°C = 343.15°K (Final temperature) P T = 1.0 atm = 760 mm Hg (Total system pressure) V gas = 500 gallons = 66.843 ft 3 (Gas space volume) R = 998.9 (mmHg-ft 3 )/(lb-mole- o K) (Universal gas constant) Gas Law: 

### n PV RT = , also 

### n p V RT i i



### = for a single component, i , in the gas space. 

## ( ) 2 1 2 1 n

n N

avg + =
moles lb RT PV RT PV N avg − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + × × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= 1608

. 0 15

. 343 1 15

. 293 1 9

. 998 2 843

. 66 760 2 1
2 1 Step 2. Calculate the initial and final partial pressures of nitrogen. Use the Antoine equation to calculate the partial pressure of toluene: P

toluene, 20 o C = exp 16.0137

− 3096.52 293.15

− 53.67 ⎛ ⎝ ⎞ ⎠ = exp(3.0835) = 21.84 mmHg
P

toluene,70 o C = exp 16.0137

− 3096.52 343.15

− 53.67 ⎛ ⎝ ⎞ ⎠ = exp(5.317) = 203.74 mmHg The partial pressure of nitrogen is the difference between the total system pressure and the partial pressure of toluene: 

### mmHg p p toluene

nc 

### 16

. 738 84

. 21 760 760 1 , 1 , 

### = − = − = mmHg p p toluene

nc 

### 26

. 556 74

. 203 760 760 2 , 2 , 

### = − = − =Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-28 Step 3. Calculate the initial and final number of moles of toluene in the vessel headspace. Use the Gas Law to calculate the moles of toluene: moles RT V p n toluene

toluene 00499

. 0 15

. 293 9

. 998 843

. 66 84

. 21
1 1 , 1 , = × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=


### moles RT V p n toluene

toluene 

### 0397

. 0 15

. 343 9

. 998 843

. 66 74

. 203
2 2 , 2 , 

### = × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= Step 4. Calculate the toluene emission using Eq. 3-21. The moles of toluene that are disp laced from the vessel are calculated by substituting values that have been calculated prior to this point into Eq. 3-21. 

## ( ) vessel i i nc nc avg i n n n p p N E 1 , 2 , 2 , 1 , ln

− − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

− 

## ( ) ( ) moles lb E toluene

n − = − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
=

− 0108

. 0 00499

. 0 0397

. 0 26

. 556 16

. 738 ln 1608

. 0 

## ( )( ) lbs mole lb lb moles lb E toluene

wt 995

. 0 / 13

. 92 0108

. 0 = − − =

− Illustration 3-17: Heatup losses from a vessel containing a volatile mixture. A 2000 gallon reactor contains 1,500 gallons of a so lvent mixture. The solvent mixture has a molar composition of 60% toluene, 30% methyl ethyl ketone, and 10% methylene chloride. The solvent mixture is heated from 20°C to 70°C and the reactor is vented to atmosphere (760 mm Hg) during the heat up operation. How much of each component will be emitted from the process? Step 1. Calculate the gas space partial pressure for each compound in the liquid using Raoult’s Law and residual partial pressure of nitrogen. Table 3-4. Partial Pressure Calculates for 20°C and 70°C Compound Xi P i @ 20C (mm Hg) X i P i (20C) (mm Hg) P i @ 70C (mm Hg) X i P i (70C) (mm Hg) Toluene 0.60 21.835 13.101 203.74 122.25 Methyl Ethyl Ketone 0.30 74.908 22.472 555.52 166.66 Methylene Chloride 0.10 355.540 35.554 2005.2 200.52 Totals 1.00 71.127 489.43 (mm Hg) (mm Hg) Nitrogen (Residual) 688.873 270.57 Step 2. Calculate the average gas space molar volume. 

## ( ) 2 1 2 1 n

n N

avg + =Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-29 moles lb RT PV RT PV N avg − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
× × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= 1608

. 0 15

. 343 1 15

. 293 1 9

. 998 2 843

. 66 760 2 1
2 1 Step 3. Calculate the initial and final number of moles of toluene in the vessel headspace. Use the Gas Law to calculate the moles of toluene: moles RT V p n i

i 01623

. 0 15

. 293 9

. 998 843

. 66 12

. 71
1 1 , 1 , = × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=
moles RT V p n i

i 09544

. 0 15

. 343 9

. 998 843

. 66 43

. 489
2 2 , 2 , = × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= 

## ( ) vessel i i nc nc avg i n n n p p N E 1 , 2 , 2 , 1 , ln

− − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

− 

## ( ) ( ) moles lb E i

n − = − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
=

− 07106

. 0 01623

. 0 09544

. 0 57

. 270 873

. 688 ln 1608

. 0
Table 3-5. Final Calculated Results for This Problem Compound avg p i (mm Hg) Fraction To total n i (lb-moles) MWt i wt i (lbs) Toluene 67.67 0.242 0.0172 92.13 1.58 Methyl Ethyl Ketone 94.566 0.337 0.0240 72.1 1.73 Methylene Chloride 118.037 0.421 0.02992 84.94 2.54 Totals 280.273 1.000 0.0687 Illustration 3-18: Heatup losses from a vessel with a volatile mixture and nitrogen sweep. A 1250 gallon reactor containing 750 gallons of a soluti on of a raw material in toluene is heated from 20°C to 70°C over a one hour period. The vessel has a known gas sweep of 3 scfm of air. The reactor is vented to the atmosphere during the heat up. Assumi ng a 25% vapor saturation of the gas sweep vapors, how much toluene will be emitted? This problem differs from the prior heating illustra tion because we wish to take into account air or nitrogen that is entering the head space of the vessel dur ing the operation. The basic approach used is to first calculate the vent losses of toluene and air (as in the prior example) and then calculate the total toluene losses based on the relative net change in the exit air flow rate due to the gas sweep while at the same time taking into account the saturation level. Step 1. Calculate the average molar volume of the vessel headspace. T i = 20°C = 293.15°K (Initial temperature) T f

= 70°C = 343.15°K (Final temperature) P T = 1.0 atm = 760 mm Hg (Total system pressure) V gas = 500 gal = 66.843 ft 3 (Gas space volume) R = 998.9 (mmHg-ft 3 )/(lb-mole- o K) (Universal gas constant)Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-30 Gas Law: 

### n PV RT = , also 

### n p V RT i i



### = for a single component, i , in the gas space. 

## ( ) 2 1 2 1 n

n N

avg + =
moles lb RT PV RT PV N avg − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + × × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= 1608

. 0 15

. 343 1 15

. 293 1 9

. 998 2 843

. 66 760 2 1
2 1 Step 2. Calculate the initial and final partial pressures of nitrogen. Use the Antoine equation to calculate the partial pressure of toluene: P

toluene, 20 o C = exp 16.0137

− 3096.52 293.15

− 53.67 ⎛ ⎝ ⎞ ⎠ = exp(3.0835) = 21.84 mmHg
P

toluene,70 o C = exp 16.0137

− 3096.52 343.15

− 53.67 ⎛ ⎝ ⎞ ⎠ = exp(5.317) = 203.74 mmHg The partial pressure of nitrogen is the difference between the total system pressure and the partial pressure of toluene: mmHg p p toluene

nc 16

. 738 84

. 21 760 760 1 , 1 , = − = − = mmHg p p toluene

nc 26

. 556 74

. 203 760 760 2 , 2 , = − = − = Step 3. Calculate the initial and final number of moles of toluene and air in the vessel headspace. Use the Gas Law to calculate th e moles of toluene and air: moles RT V p n toluene

toluene 00499

. 0 15

. 293 9

. 998 843

. 66 84

. 21
1 1 , 1 , = × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=
moles

RT V p n nc

nc 1685

. 0 15

. 293 9

. 998 843

. 66 16

. 738
1 1 , 1 , = × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=
moles RT V p n toluene

toluene 0397

. 0 15

. 343 9

. 998 843

. 66 74

. 203
2 2 , 2 , = × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=
moles

RT V p n nc

nc 1085

. 0 15

. 343 9

. 998 843

. 66 26

. 556
2 2 , 2 , = × × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= Step 4. Calculate the toluene emission using Eq. 3-21. The number of moles of toluene displaced from the v essel is equal to the moles of nitrogen that are displaced from the vessel during the heating ope ration multiplied by the average molar ratio. 

## ( ) vessel i i nc nc avg i n n n p p N E 1 , 2 , 2 , 1 , ln

− − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

− 

## ( ) ( ) moles lb E toluene

n − = − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
=

− 0108

. 0 00499

. 0 0397

. 0 26

. 556 16

. 738 ln 1608

. 0Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-31 

### ( )( ) lbs mole lb lb moles lb E toluene

wt 995

. 0 / 13

. 92 0108

. 0 = − − =

− Step 5. Calculate the emission of air without the gas sweep present. moles n n E nc nc nc n 06

. 0 1085

. 0 1685

. 0 2 , 1 , = − = − =

− Step 6. Calculate the moles of gas sweep. 

## ( ) air moles scfm E sweep gas n 501

. 0 15

. 273 9

. 998 min 60 3 760 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ × × × =

− Step 7. Calculate the toluene emissions while taking the gas sweep into account with a 25% saturation level. 

## ( ) moles lb E toluene

n 0333

. 0 0108

. 0 08

. 3 0108

. 06

. 0 501

. 0 25

. 0 06

. 0 = × = × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × + =

− 

## ( )( ) lbs mole lb lb moles lb E toluene

wt 068

. 3 / 13

. 92 0333

. 0 = − − =

− moles lb E nc

n − = + =

− 561

. 0 501

. 0 06

. 0 

## ( )( ) lbs mole lb lb moles lb E nc

wt 25

. 16 / 97

. 28 561

. 0 = − − =

− 

### 3.7 Evaporation Models 3.7.1 Evaporation from an Open Top Vessel or a Spill The rate of vaporization of a liquid can be modeled as a function of several characteristic factors of the compound being considered. [Crowl & Louvar, 2002] 

# ( )

L i sat i i i i n RT P P A K M E − =

− Eq. 3-23
Where E n is the evaporation rate (mass/time). M i is the molecular weight of the volatile substance, K i is a mass transfer coefficient (length/time), A is the evaporation surface area, sat i 

### P is the saturated solvent vapor pressure,

i 

### P is the actual vapor pressure near the liquid surface, R is the ideal gas constant, and T

L is the absolute temperature of the liquid. For many cases, Psat >> p, and Eq. 3-23 may be simplified to

L sat

i i i n RT AP K M E =

− 1 Eq. 3-24 Eq. 3-24 may be used to estimate the vaporization rate of a volatile liquid from an open vessel or a liquid spill.Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-32 The ratio of the mass transfer coefficients betw een the compound of interest K and reference compound K o is expressed as follows:

3 2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=
o i

o i D D K K Eq. 3-25 The gas-phase diffusion coefficient D for a compound is estimated from the ratio of molecular weight of the compound of interest and a known compound (normally water) as follows:

2 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= i o

o i M M D D Eq. 3-26 Combining Eq. 3-25 and Eq. 3-26 results in a relationship that can be used to estimate the mass transfer coefficient of a given volatile compound.

3 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= i o o i M M K K Eq. 3-27 Water is commonly used as a base reference for est imating the mass transfer coefficient for many compounds of interest. The mass transfer coefficient of water at 77 F and 760 mm Hg. is 0.83 cm/s. Illustration 3-19: Evaporation from a vessel with an open top. A large open top vertical tank with a 6-ft diameter co ntains heptane. Estimate the evaporation rate from the tank at 25 C and 1 atm pressure. The molecular weight of heptane is 100.2. The mass transfer coefficient is estimated using Eq. 3-24 with the known mass transfer coefficient for water of 0.83 cm/s.

hr ft cm hr ft s s cm s cm M M K K i o o i 33

. 55 48

. 30

3600 4685

. 0 2

. 100 02

. 18 83

. 0 3 1 3 1

=

⋅ ⋅ ⋅ ⋅ × = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

. 86

. 45 tan mmHg P sat e

hep =

2 2 2 26

. 28 4 36 14

. 3 4 ft ft d A = ∗ = = π 

## ( ) ( ) 

# ( ) ( ) 

# ( ) ( ) K K mol lb atm ft mmHg ft hr ft mole lb lb RT

AP K M E L sat i i i i n ° °

− − = =

− 15

. 298 / 9

. 998 86

. 45 26

. 26 / 33

. 55 / 2

. 100 3 2
hr lb E i

wt / 42

. 24 =

− Illustration 3-20: Evaporation losses from a spill. Toluene is spilled onto the ground outside of a buildin g. Determine the toluene evaporation rate based on the following data: The ambient temperature (T) is 25°C or 298.15°K. ( °K = °C + 273.15)Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-33 The surface area (A) of the spill is 100 ft 2

. The molecular weight of toluene is 92.13 lb/lb-mole. The vapor pressure of toluene is 28.445 mm Hg.

hr ft cm hr ft s s cm s cm M M K K i o o i 50

. 68 48

. 30

3600 58

. 0 13

. 92 02

. 18 83

. 0 3 1 3 1

=

⋅ ⋅ ⋅ ⋅ × = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∗ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= 

## ( ) ( ) 

# ( ) ( ) 

# ( ) ( ) R K mol lb mmHg ft mmHg ft hr ft mole lb lb RT p

A K M E L sat i i i n ° °

− − = =

− 15

. 298 / 9

. 998 44

. 28 0

. 100 / 50

. 68 / 13

. 92 3 2
hr lb E i

n / 60 =

− 

### 3.8 Emission Model for Li quid Material Storage The preferred method for calculating emissions from stor age tanks is the use of equations presented in AP-42 . EPA has developed a software package (TANKS ) for calculating these types of emissions. The reader is referred to Chapter 1 of this volume, Introduction to Stationary Point Source Emissions Inventory Development , for more information on using the T ANKS program. Additionally, the reader should consult their state agency and/or the EPA’s Clearinghouse for Inventories and Emission Factors (CHIEF) website for the most recent version of TANKS. 

### 3.9 Emission Model for Wastewater Treatment VOC emissions from a wastewater treatment system may be estimated using equations presented in Air Emissions Models for Waste and Wastewater (EPA, 1994a), and Chapter 5, Preferred and Alternative Methods for Estimating Air Emissions from Wastewat er Collection and Treatment Facilities, of this volume. These documents, as well as models su ch as WATER9 are available on the EPA’s CHIEF website. 

### 3.10 Using Sampling and Test Data to Validate Emission Studies Use of stack and/or industrial hygiene test data is likely to be the most accurate method of quantifying air emissions from chemical manufacturing operations. Howe ver, collection and analysis of air samples from manufacturing facilities can be very expensive and esp ecially complicated for chemical manufacturing facilities where a variety of VOCs are emitted and where mo st of the emissions may be fugitive in nature. Test data from one specific process may not be repr esentative of the entire manufacturing operation and may provide only one example (a snapshot) of the facility’s emissions. To be representative, test data would need to be collected over a period of time that covers production of multiple chemical formulations. It may be necessary to sample multiple production areas. In addition, these methods do not address fugitive emissions that occu r outside of a building. If testing is performed, care should be taken to ensure that a representative operational cycle has been selected. If possible, full cycles should be monitored as opposed to portions of cycles. For example, in some facilities the typical process vessel used in mixing or dispersion operations may have hinged lids or other covers that are loosely f itting. Additionally, these vessels may have top mounted agitators that can be raised or lowered depending up on mixing elevation requirements. A gas tight agitator seal may not be practical for this type of pro cess vessel. In other cases, the mixing vessel my be completely open top. Quantifying the gas sweep ra te and volatile vapor satur ation level for the overallChapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-34 emission process would not be possible in these cases. Developing a reliable emission model using classic modeling techniques presented in this docume nt may not be reliable in this case. US EPA Method 204D (Fugitive VOCs from Temporary To tal Enclosure) may be used as a means of quantifying air emissions from specific equipmen t systems for one or more processes. Results from temporary total enclosure testing is considered to be more accurate than from using standardized emission models since the data is from accurate measure ment sources. The results from validated emission measurement studies can be used in developing emission standards that in turn can be used to estimate emissions from partial or comp lete production operations. Illustration 3-21: Using emission measurements to represent production operations. A large mixing tank is used as part of a solvent blendi ng operation. The vessel has a flat lid that contains several smaller hinged covers for hand holes and char ge shoots. Additionally, the vessel has an agitator but no agitator seal. The tank vent is loosely connected to a large duct system that is powered by a remote exhaust fan near the roof. The exhaust manifold is m onitored for air flow rate and solvent concentration using standardized instrumental analysis. During the ti me of testing the atmospheric pressure is measured to be 760 mm Hg (1.0 atm). The flow rate and te mperature of the exhaust gas are measured to be 10 ft 3 /min (or 600 ft 3 /hr) and 25°C, respectively. Analytical measurements made during the process operation show the toluene and heptane content to be 2.8 mm Hg and 3.7 mm Hg in the exhaust gas, respectively. Calculate the average toluene and he ptane emission rate during this process operation. Using the partial pressure for each compound in conjun ction with the ideal gas law and exhaust gas flow rate the following expression can be used to calcu late the emission rate for a given pollutant.



### RT F

p M Q i i i 

### = where M i = molecular weight of the pollutant, p i = partial pressure of the pollutant in mm Hg, F = exhaust gas flow rate in ft 3 /hr, T = exhaust gas temperature in °K, R = universal gas constant for mm Hg, ft 3 , and °K. For Toluene: h r lbs R T F

p M Q toluene toluene toluene 2

. 5 15

. 298 9

. 998 600 8

. 2 13

. 92 = ∗ ∗ ∗ = = For Heptane: h r lbs RT F

p M Q e hep e hep e hep 5

. 7 15

. 298 9

. 998 600 7

. 3 21

. 100 tan tan tan = ∗ ∗ ∗ = = Note that the actual concentration and temperature of the process mixture that is contained in the process vessel are not required in this calculation since the emissions from the process are being entirely characterized from the analysis of the exhaust gas from the system. Also , it is important that only the process being studied be in operation during the me asurement study and that contaminates from other parts of the facility not be present.Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-35 3.10.1 Correlating Standard Emission Models to Test Results Results from total enclosure testing can then be u sed to correlate with conventional batch modeling formulas. Process variables such as v essel gas sweep rate and/or saturation levels may be estimated based on the test data. For example, Eq. 3-28 relates the evaporation rate for a volatile compound i within a vessel to its molecular weight, mass transfer coefficient, surf ace area, and other known variables. [Crowl & Louvar, 2002] 

# ( ) i Sat i m 

### p P RT MKA Q

− = Eq. 3-28
where Q m = evaporation rate (lb/min) M = molecular weight of compound i K = mass transfer coefficient (ft/min) A = surface area (ft 2 ) R = ideal gas constant T = temperature of liquid P Sat = saturated vapor pressure of compound i p i = actual vapor pressure of compound i next to the liquid surface. Eq. 3-29 is the basic equation for calculating the emission rate for compound i from a gas sweep or purge operation based on the exit gas flow rate, partial pressure of compound i , molecular weight, and other known variables.



### RT MFp M P p RT FP Q i T i T v 

### =

= Eq. 3-29 where Q v = emissions from vessel vent (lb/min) F = exit gas flow rate (ft 3 /min) P T = overall system pressure R = ideal gas constant T = temperature of liquid p i = actual vapor pressure of compound i. M = molecular weight of compound i. For a vessel at steady state conditions, the emission rate from the gas sweep activity is equal to the evaporation rate or compound i from the liquid within the vessel. [Hatfield, 2003a] These two equations can be set equal and solved for the saturation level S i as follows:

v m 

### Q Q = Substitute basic expressions 

# ( ) RT MFp p P RT MKA i i Sat i 

### = −
Cancel out common terms 

# ( ) i i Sat i 

### Fp p P KA = − Expand expression i i Sat i 

### Fp KAp KAP = −Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-36 Rearrange terms i i i Sat i 

### p KA F KAp Fp KAP ) ( + = + = Divide by (F+FA) 

### KA F KAP p Sat i i 

### +

=

Rearrange to obtain saturation. 

### KA F KA P

p S Sat i i i 

### + = = Eq. 3-30 Eq 3-30 is consistent with respect to units. If K (ft/min), A (ft 2 ), and F (ft 3 /min) then S i is dimensionless and represents the level of saturation as a decimal fraction between 0 and 1.0. When the purge rate becomes zero, the saturation level inside the tank b ecomes 1.0. The resulting emissions through the vent becomes zero because a purge does not exist through the vessel. When the purge rate increases to the point that F = KA then the saturation level becomes 0.5 (or 50%). Eq 3-30 relates the saturation level of a volatile comp onent the exit gas sweep rate (F) and the evaporation rate (KA). However, Eq 3.30 represents a wors t case condition because it assume s that the gas space in the vessel is perfectly mixed. When F is equal to KA then the partial pressure of the volatile component is only 50% of the saturated vapor pressure. When F is gr eater than KA then the saturation level is greater than 50% and when F is smaller than KA th en the saturation level is less than 50%. Suppose a vessel with a diameter of 5 ft is partially filled with toluene at 25°C. The vapor pressure and molecular weight of toluene are 28.2 mm Hg and 92. 13 lb/lb-mole, respectively. The cross sectional area of the tank or liquid surface area is calculated to be 19.6 ft 2 . Assuming that the gas space in the vessel is mixed then the following calculations can be made.

min 949

. 0 min 48

. 30 60 0.482 13

. 92 02

. 18 83

. 0 3 1 3 1 ft cm ft s s cm s cm M M K K i o o i =

⋅ ⋅ ⋅ ⋅ × = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

=

59

. 18 59

. 18 ) 6

. 19 )( 949

. 0 ( ) 6

. 19 )( 949

. 0 ( + = + = + = F F A K F A

K S i i i Saturation S i is plotted for this vessel as a function of F in Figure 3-1 and shows the relationship between exit gas sweep rate and saturation level of the exit vent gas with respect to toluene vapor pressure.Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-37 Vapor Pressure Saturation Level in Exit Purge Gas as a Function of F, K, A

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

1

0 20 40 60 80 100 120 Purge Rate (ft3/min)Saturation Level

Figure 3-1. Saturation Level (S) Plotted as a Function of Exit Vent Gas Flow Rate. Once the saturation factor S is know for any exit vent gas flow rate then the es timated emission rate for compound i can be plotted as a function of F as by substituting Sat i i i 

### p S p = .



### RT p MFS RT MFp Q Sat i i i v 

### =

= Eq. 3-31 The emission rate for this specific toluene example is plotted in Figure 3-2 in lb/hr as a function of exit gas sweep rate in SCFM. Emission Rate vs Gas Sweep Rate

0 1 2 3 4 5 6 7 8 9 10

0 20 40 60 80 100 120 Gas Sweep or Purge Rate (scfm)Emission Rate (lb/hr)

Figure 3-2. Toluene Emission Rate Plotted as a Function of Exit Gas Sweep Rate and Saturation Level. The vessel had a loose fitting hinged co ver and an over sized hole in the top where the agitator entered the vessel. The exit vent gas rate for the vessel could not be determined due to this equipment configuration. A Temporary Total Enclosure test was conducted on this vessel as it was holding toluene at 25°C whileChapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models

EIIP Volume II 16.3-38 mixing was underway. A steady state toluene emission rate was measured to be 5.14 lb/hr. From Figure 3-2, a 5.14 lb/hr toluene emission rate corresponds to an exit vent gas rate of 20 scfm. From Figure 3-1, a 20 scfm exit gas rate corresponds to a saturate level of 0.48 (or 48%). If the process material is changed with respect to solvent composition then th e established gas sweep and saturation level can be used in support of a re-evalu ation of the revised operation. In this case the vapor pressure and other physical properties of the new pro cess material would be incorporated into the basic purge model for emission estimates. 

### 3.11 Emission Calculations Using Material Balance If the equipment or process operation is such that stan dard modeling techniques can not be applied then a material balance approach may be used to estimate air emissions. Processes that fall into this category might include parts cleaning or degreasing systems wher e the equipment is open to the atmosphere and/or does not fit typical process vessel designs. In such cases, the quantity of initia l cleaning solvent would be weighed prior to being charged to the equipment system. Once the process operation has been completed then all remaining spent solvent would be collected and weighed. If non-volatile compounds su ch as oil or other materials are contained in the residual spent solvent then the material would n eed to be assayed for volatile solvent content. If the test results are to be used for developing emission standards for a specific process operation then additional tests should be implemented to arrive at a statistically relevant emission estimation. Illustration 3-22: Using material balance to estimate emissions from operations. Fresh toluene solvent is charged to an equipment parts degreaser/cleaning unit. The initial amount of toluene charged is 350 lb. At the conclusion of the operation 347.5 lb. of spent toluene (contaminated with waste oil) is collected. A sample of the spen t toluene is assayed using loss on drying (LOD) analysis to be 98.8% toluene by weight. Calculate th e toluene evaporation losses from the operation.

Stream Weight (lbs.) Purity Weight (lbs.) Initial Toluene Charge 350.00 100.0% 350.00 Residual Toluene 347.50 98.8% 343.33 Toluene Emitted 6.67


### 3.12 Emission Calculations Using Emission Factors Emission factors are commonly used to calculate em issions from chemical manufacturing facilities. EPA maintains a compilation of approved emission factors in AP-42 for criteria pollutants and hazardous air pollutants (HAPs). Emission factors for equipment leaks may be found in Protocol for Equipment Leak Emission Estimates (EPA, 1995g). Chapter 4 of this volume discusses emission estimates from equipment leaks. The most comprehensive source for toxic air pollutant emission factors is the Factor Information and Retrieval (FIRE) data system, which also contai ns criteria pollutant emission factors (EPA, 1995h).Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-1 

## 4.0 Single Stage Vent Devices: Co ndensers, Vacuum Pumps, and Vacuum Steam Jets There are many different equipment devices that can be installed and operated in the vent path depending upon what is needed by the process. A condenser would be specified if it was required to collect process vapors from the vent stream in the form of liquid co ndensate while a vacuum pump or steam jet would be used if vacuum was needed at th e process vessel. In some cases the process vent path can contain one or more condensers followed by a vacuum pump followed by another condenser. All of these devices including condensers, vacuum pumps, and vacuum steam jets have several things in common. As process vapors pass through the device the temperature and/or pressure changes. In most cases liquid condensate forms in the device and is di scharged along with the remaining vapor. In the case of a vacuum steam jet or a liquid ring vacuum pump, wo rking fluid (steam, water, or other fluid) enters the device and is allowed to mix with the process st ream as it passes through the unit. A liquid phase is discharged with any remaining vapors similarly to th e condenser. Simple diagrams of each type of device are presented in Figure 4-1. For calculation purposes, one tends to consider that th e liquid and vapor streams are in equilibrium with each other as they exit the device. Additionally, it is assumed that the material balance around the device is satisfied. The rate that material that enters the devi ce is equal to the rate that material exits the device. These two assumptions are what enable one to solve the single stage device problem and to determine the exit stream flow rates and compositions.

V

L F p F p
F b

F f V

L F f F b F p V

L Condenser Liquid Ring Vacuum Pump Vacuum Steam Jet Figure 4-1. Diagrams of single stage vent devices. 

### 4.1 General Calculation Considerations 4.1.1 Material Balance Considerations for the Vent Device
f b p 

### F F F F + + = Eq. 4-1 Where F p is the inlet process rate, F b is any air or nitrogen bleed, and F f is any working fluid such as steam, water, or other external fluid. V is the exit vapor stream and L is the exit liquid stream. 

### V L F + = Eq. 4-2Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-2 

### V y L x F z i i i 

### + = Eq. 4-3 Where F is the combined inlet molar flow rate, L is the molar flow rate of the exit liquid stream and V is the molar flow rate of the vapor st ream. z, x, and y are the mole fractions of component i for the respective stream. If air, nitrogen, or another noncondensable gas with a low solubility is present then one can assume that all of the non condensable gas exits the device by way of the exit gas stream V and that none is present in the exit liquid stream L. 

### V y F z nc

nc 

### = where 

### 0 = nc 

### x Eq. 4-4 For most vent devices, Eq. 4-4 is a very important relationship because it provides a straightforward way to calculate the amount of the non condensable co mponent contained in the exit vapor stream V. Additionally, Eq. 4-4 provides a way to calculate th e moles of component i in the exit vapor stream V from the moles of the non condensable component nc and the partial pressure ratio of component i and the non condensable component.

nc i nc i 

### p p n n = Eq. 4-5
4.1.2 Equilibrium Relationships Between the Exit Liquid and Vapor Streams Equilibrium relationships that exist between the ex it liquid stream L and the exit vapor stream V are needed in order to relate the relative amount of compone nt i that exists between L and V. If equilibrium is achieved between L and V then Eq. 4- 6 and Eq. 4-7 must be satisfied.

i i i i 

### p P x = γ Eq. 4-6

sys i i i sys

i i 

### P P x P p y γ 

### = = Eq. 4-7 Where x and y are mole fractions of component i in streams L and V. P i and p i are the pure vapor pressure and partial pressure of component i. P sys is the system pressure and ( i is the activity coefficient of component i. 4.1.3 Checking to See If Condensation Takes Place in the Vent Device If the process stream contains a noncondensable co mponent (air, nitrogen, or other gases) then it’s important to check to see if condensation will likel y occur before proceeding to solve the condenser problem. This is not normally an issue with calcula tions involving liquid ring vacuum pumps and vacuum steam jets because of the significant presence of the working fluid (water, steam, or another fluid). However, if the process stream from the process cont ains a large amount of no ncondensable component then it is possible that condensation may not take place in a standard condenser if the dew point temperature of the process stream is lower than the condenser operating temperature. It is relatively easy to test for condensation by fi rst assuming that condensate does form containing all of the condensable components from the process stream . Then calculating the mole fraction that each component would need to have at the condenser temp erature in order for the partial pressure to be theChapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-3 same as the partial pressure at the inlet conditions. The sum of the resulting liquid mole fractions is calculated. If the sum of the x i ’s is equal to or greater than 1.0 then one can assume that condensation takes place in the condenser. However, if the sum of the x i ’s is less than 1.0 then one can assume that condensation does not take place and that all of th e process exits the condenser as a vapor stream.

i i i i 

### p P x =

γ Eq. 4-6

i i i i 

### P p x γ = Eq. 4-8
Where P i is the pure vapor pressure at condenser temperature, p i is the partial pressure at the condenser inlet. 

# ∑ ≥ 0

. 1 i 

### x 0

. 1 < 

# ∑ i 

### x Condensation takes place at T condenser . Condensation does not take place at T condenser 4.1.4 General Calculation Approach for Solving Single Stage Vent Device Problems If the inlet process stream to the single stage devi ce contains only one condensable component and one noncondensable component then the composition of the exit vent stream V is fairly straight forward calculation. The concentration of condensable component in the liquid stream L is 100%, therefore, p i is equal to P i for the device temperature. Eq. 4-5 can then be used to directly calculate the moles of the condensable component in the exit vapor stream V. However, if the process contains two or more c ondensable components then the calculation becomes much more involved because there are not enough independent equations to solve the problem algebraically. Instead, equations are constructed to express the component material balance while other equations are written for the equilib rium relationships that exist between exit streams L and V. The general solution to the problem is found by iterati ng back and forth between the material balance equations and the equilibrium equations. The estimated quantity of each component that is contained in the liquid and vapor streams is improved with each ite ration. Iterations contin ue until all equations are satisfied within an acceptable tolera nce. When an iterative approach is needed then the Newton-Raphson or a material balance correction approach can be applied as a means to achieving convergence. Both convergence procedures will be presented in the illustrations that are presented. 4.1.5 Partial Condensers Partial condensers are frequently used to separate compounds from a process that have a much lower boiling point than remaining compounds in the process. For example, suppose a process that is carried out in a toluene solvent (T b = 110°C) chemically produces methyl chloride (T b = -30°C) along with several other compounds having normal boiling points that are greater than 50°C. Now suppose that it is desired to remove methyl chloride from the batch. If the pr ocess stream is in the vapor state then a partial condenser could be used at a temperature that is a bove the boiling point of the methyl chloride and below the boiling point of the toluene and other compounds in the process so that the methyl chloride would remain as a gas and the remaining compounds would te nd to condense. A majority of the methyl chloride would exit the partial condenser as a vapor and the rest of the process stream would exit as a liquid condensate. Only minor amounts of methyl chloride would be left with the batch and only a smallChapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-4 quantity of toluene and other batch compounds would be left in the vapor discharge with the methyl chloride. The general approach to solving the partial conden ser problem is to use the Rachford-Rice equation which combines the material balance constraints with the equilibrium constraints. The Newton-Raphson procedure is used until convergence is achieved. This procedure will be explained further in a worked illustration later in this section. 4.1.6 Total Condensers Generally, when one refers to condensers used in indu stry he or she might be thinking about condensers where everything in the process condenses to a liquid. For example, in a batch distillation a solvent is placed in a still vessel and heated to its boiling point. Solvent vapors leave the st ill and enter an overhead condenser where all of the solvent condenses back into a liquid. Condensate formed in the condenser is then collected in a receiving vessel. Heat continues to be applied to the still until all of the solvent has been distilled. In this example, 100% of the solven t vapor that entered the condenser was condensed and collected in the receiving vessel and none of the solv ent left as a vapor. A condenser operated in this manner would be referred to as a total condenser sin ce the entire inlet process vapor was converted into a liquid condensate. Emissions would not occur from th e condenser directly since the process vapor does not contain a non condensable gas component. Howe ver, emissions would occur from the distillate receiver where the condensate is being collected and non condensable gases which are saturated with solvent vapors are exiting the receiver vent. 4.1.7 Vent Condensers Many process operations (heating, charging, purging, etc) generate solvent emissions even though the batch is well below the boiling point. This is the resu lt of non condensable gases (air, nitrogen, etc) that are leaving the process vessel which carry solv ent vapors in the process. When vapors from these operations flow through a condenser that is colder than the vapor stream then the dew point is reached and condensation takes place. Non condensable gases exit the condenser with a reduced level of solvent vapors due to effect that the colder temperatur e has had on the vapor pressure of the condensable components. Vent condensers represent a large percentage of how most condensers are operated in the field. When a total condenser is not being used to condense so lvent vapors from a distillation operation then they typically operate like a vent condenser at other tim es during the process. Three procedures are presented for solving the vent condenser problem including the Modified Rachford-Rich equations, the vapor pressure correction method, and the ma terial balance correction method. 4.1.8 Liquid Ring Vacuum Pumps Liquid ring vacuum pumps are used to supply vacu um for certain process activities such as vacuum distillations, solids drying, and other operations. Liqui d enters the pump and opera tes in conjunction with the rotating asymmetric internals to produce a low pressure side and a high pressure side by way of several vanes and chambers. In most cases, a large percentage of the liquid that passes through the pump is recycled to the inlet while the remainder is disch arged to waste treatment or for subsequent processing. Fresh liquid makeup is added as required by the vacuum pump. In many cases, the operating liquid for the vacuum pump is water although other process fluids including vacuum pump oil are also used.Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-5 Liquid ring vacuum pump problems can be solved using the same approach that is used to solve vent condenser problems. All inlet streams to the liquid ring pump (process vapor stream, external air bleed, and working fluid throughput) are combined as a si ngle process stream. The final discharge temperature and pressure are used for the equilibrium calculatio ns. A worked illustration is presented which applies the vapor pressure correction method to solve the emission calculation. 4.1.9 Vacuum Steam Jets Vacuum steam jets are also used to supply v acuum to certain process activities such as vacuum distillations, solids drying, and other operations. Steam jets will normally enable lower pressures to be achieved than liquid ring pumps (assuming that water is being used as the working fluid). High pressure steam enters the jet through a speci al nozzle where it is released and allowed to flow through a chamber with an increasing cross sectional area. As the cham ber cross sectional area increases, the steam must undergo a reduction in velocity which causes vacuum to be created at the process inlet. Process vapors enter the steam jet through the inlet vacuum port a nd mix with the working steam. The combined steam and the process vapor stream leave the vacuum cham ber and pass through an internal condenser where the steam and other vapors are condensed and the re maining vapor discharged to the atmosphere. Vacuum steam jets problems can be solved using the sa me approach that is used to solve vent condenser problems. All inlet streams to the vacuum steam jet (process vapor stream, external air bleed, and steam) are combined as a single process stream. The final di scharge temperature and pressure are used for the equilibrium calculations. A worked illustration is presented which appli es the material balance correction method to solve the emission calculation. 

### 4.2 Illustrations Several worked illustrations are pr esented as a general guide for perfo rming condenser calculations in a variety of configurations. As mentioned earlier, a major difference between the calculation of partial condensers and vent condensers is in how the material balance is applied to the inlet and outlet streams. For the partial condenser, the material balance is applied to inlet and outlet streams as shown in Eq. 4-2. For the vent condenser, the material balance is nor mally applied directly to the noncondensable gas as shown in Eq. 4-4. The composition of the exit vapor stream and the exit liquid stream is then established through vapor pressure calculations. If the process invol ves only a single condensable component then the calculation of the exit vapor stream is straight forwar d. However, if two or more condensable components are contained in the process then the calculation of the exit vapor stream normally requires iterations between the vapor pressure equilibrium calculations and the material balance calculations until both relationships are satisfied. 4.2.1 Condensers with One Condensable Component Three examples are presented for condensers with inle t gas streams that consist of a noncondensable and one condensable component. In illustration 4-1 the entire system is at atmospheric conditions. In illustration 4-2 the process vessel and condenser are under vacuum. In illustration 4-3, the process vessel is under vacuum, but the condenser is located on th e outlet side of the vacuum pump. Each of these examples uses Eq. 4-5 to calculate the amount of th e condensable component leaving in the condenser outlet vapor stream. The fraction of the inlet condensab le that leaves in the condenser exit vapor is also calculated in each example, although note that two approaches are used.Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-6 Illustration 4-1: Condenser with solvent vapor heptane and nitrogen Nitrogen gas at 25°C and partially saturated with he ptane vapor at a 69% level is vented from a process vessel (see Illustration 3-7 in section 3.2.2 of th is document) into a condenser. The nitrogen gas constituent has a flow rate of 600.0 SCFH. The gas st ream flows through a condenser and is discharged at 5°C and 760 mm Hg. Calculate the heptane emissions from the condenser over a one hour period. Calculate the material balance for the condenser and the percent removal of heptane. Step 1. The condenser has the following operating conditions. Heptane molecular weight 100.205 Nitrogen molecular weight 28.013 P heptane 45.69 mmHg @ 25°C P heptane 15.43 mmHg @ 5°C T inlet 25°C, 298.15°K Condenser gas inlet T exit 5°C, 278.15°K Condenser gas exit P sys 760 mm Hg System pressure Time 1 hr Time for event Universal Gas Constant: K mole lb ft

mmHg R ° ⋅ − ⋅ = 3 9

. 998 Step 2. Calculate the amount of nitrogen that passes through the condenser. We can assume that 100% of the nitrogen gas exits the condenser in the gas phase since little if any would condense with the heptane at 5°C. The amount of nitrogen flowing through the system over a 1.0 hour period is 600.0 ft 3 (at 273.15°C & 760 mm Hg).
moles lb K ft mmHg RT V P n N

N − = ° ⋅ = = 1.671 ) 15

. 273 )( 9

. 998 ( 600 760 3 2 2 

### lb E N 

### 46.817 MWt 28.013 * 671

. 1 2 

### =

= Step 3. Calculate the amount of heptane that enters the condenser in one hour. The vent stream enters the condenser at 25°C with a he ptane saturation level of 69%. In order to calculate the heptane content in the inlet gas stream we mu st first know the pure component vapor pressure for heptane at the 25°C stream temperature. Then the part ial pressure of the heptane can be calculated for the 69% saturation level. The moles of heptane in the inle t gas stream can be calculated from the ratio of the partial pressure of heptane to the partial pressure of nitrogen and the mola r rate of nitrogen. 

### mmHg mmHg 31.53 45.69 69

. 0 = ×Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-7 The heptane is 69% saturated and so the vapor pressure of heptane is 

### 31.53 mm Hg. The partial pressure of nitrogen is determined by the difference between the total system pressure, 760 mm Hg, and the partial pressure of heptane. 

### mmHg mmHg mmHg p P p H sys N 

### 728.47 31.53 760 2 

### = − = − = The moles of heptane entering the condenser over a one hour period may now be calculated from the moles of nitrogen in the inlet stream and the ratio of the partial pressure of heptane and the partial pressure of nitrogen.

2 2 N H N H p p n n = therefore 2 2 N N H H n p p n =

moles lb moles lb mmHg mmHg n H 0.07233 671

. 1 47

. 728 53

. 31 = =
lb MWt moles lb F in

H 7.247 100.205 7233 0

. 0

, = × = Step 4. Calculate the partial pressure of the heptane at condenser exit temperature. The exit gas condenser temperature is given to be 5° C. The vapor pressure of heptane in the liquid state can be calculated using the Antoine equation at 5°C (or 278.15°K). The heptane vapor pressure at 5°C is 15.43 mm Hg and less than the 31.53 mm Hg vapor pressure at the condenser inlet. Therefore, condensation must have taken place. The partial press ure of the nitrogen component at the condenser exit can be calculated by subtracting the heptane vapor pressure from the total system pressure. 

### mmHg mmHg mmHg p P p H sys C N 

### 744.57 15.43 760 5 , 2 

### = − = − = ° Step 5. Calculate the heptane emissions from the condenser over the one hour period. The final heptane emission level E H,exit may be calculated by multiplying the moles of nitrogen leaving the condenser by the ratio of the heptane and nitrogen partial pressures at the condenser exit conditions. Once E H,exit is known then L H,exit may be calculated by subtracting E H,exit from F H,inlet .

2 2 N H N H p p n n = therefore 2 2 N N H H n p p n =

moles lb moles lb mmHg mmHg n C

H 0.03463 671

. 1 744.57 43

. 15 5 , = = ° 

## ( ) ( ) lb moles lb E exit

H 3.47 MWt 100.205 0.03463

, = = lb lb lb L exit

H 3.78 47

. 3 25

. 7

, = − =
47.86% 7.25

47

. 3 * % 100 % tan = = lb lb E e hepChapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-8 52.14% % 86

. 47 % 100 % tan = − = e hep Contdensed Summary Results Condenser Inlet Condenser Condensate Condenser Exit Vapor

heptane 7.25 lb 3.78 lb, 52.14% 3.47 lb, 47.86% nitrogen 46.82 lb 0.0 lb, 0.0% 46.82 lb, 100.0% Illustration 4-2: Condenser with toluene solvent vapor and nitrogen at vacuum conditions A vacuum receiver in a distillation operation is vent ed through a condenser to a vacuum pump operating at 175 mm Hg. The process vapor entering the condenser consists of nitrogen (85 mm Hg) and toluene (90 mm Hg). 1. At what temperature must the condenser ope rate before condensation will take place? 2. If the condenser operates at 10°C, what percent of the toluene vapor to the condenser would exit as vapor? 3. If the nitrogen component of the vapor stream is 10 lb/hr, what would the toluene emission rate be from the condenser?
Universal Gas Constant: 

### K mole lb ft

mmHg R ° ⋅ − ⋅ = 3 

### 9

. 998

Gas Law: also for component i in the gas space.
Step 1. In order for condensation to take place, the pure vapor pressure of toluene would need to be below 90 mm Hg. The Antoine equation may be solved for the temperature that would result in a 90 mm Hg vapor pressure for toluene. Antoine Equation:

Step 2. If the condenser operates at 10°C, what percent of the toluene inlet condenser process stream would exit as vapor? The molar ratio between the toluene and nitrogen in th e inlet stream can be calculated from the respective partial pressures. The same calculation can be made at the condenser exit. The percent toluene emittedChapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-9 from the condenser may then be calculated by dividing th e molar ratio of toluene to nitrogen at the exit by the molar ratio at the inlet. 

## ( ) nitrogen mole toluene mole mmHg mmHg p p

N T 1.059 0

. 90 0

. 175 0

. 90
2 =

− = 

## ( ) nitrogen mole toluene mole mmHg mmHg p p

N T 76 0

. 0 12.43

- 175.0 12.43
2 =

=


### 7.18% % 100 * 059

. 1 076

. 0 % = = T 

### E Step 3. If the nitrogen component of the vapor stream is 10 lb/hr, what would the toluene emission rate be from the condenser? We can assume that 100% of the nitrogen exits th e condenser in the gas phase since little if any would condense with the toluene. The rate that nitr ogen flows through the system over 10.0 lb/hr.
hr moles

- lb 0.357 28.013 0

. 10 2 =

= MWt hr lb n N


### hr mmHg mmHg p p n n N T N T 

### moles

- lb 0.027 ) 43

. 12 175 ( 43

. 12 * 357

. 0 * 2 2 

### =

− = =



### h r lb 2.49 92.13 * 027

. 0 =

− − = moles lb lb hr moles lb E T Illustration 4-3: Condenser with toluene vapor and nitrogen at atmospheric pressure If the condenser described in Illustration 4-2 were lo cated at the atmospheric discharge of the vacuum pump instead of on the vacuum inlet side what w ould the condenser performance been? The vacuum pump is a dry pump and the discharge vapor temper ature is sufficiently warm due to adiabatic compression that none of the toluene condenses at the 760.0 mm Hg discharge pressure. The partial pressure for the toluene component is 391 mm Hg and the nitrogen flow rate is 10 lb/hr. 1. At what temperature must the condenser ope rate before condensation will take place? 2. If the condenser operates at 10°C, what percent of the toluene vapor to the condenser would exit as vapor? 3. If the nitrogen component of the vapor stream is 10 lb/hr, what would the toluene emission rate be from the condenser?
Universal Gas Constant: 

### K mole lb ft

mmHg R ° ⋅ − ⋅ = 3 

### 9

. 998

Gas Law: also for component i in the gas space.Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-10 Step 1. In order for condensation to take place, the pure vapor pressure of toluene would need to be below 391 mm Hg. The Antoine equation may be solved for the temperatur e that would result in a 391 mm Hg vapor pressure for toluene. Antoine Equation:



### ) 78

. 48 ( ) 391 ln( 0127

. 16 52

. 3096 ) ln( − −

− = −

− = c p a b T i T = 361.9°K = 88.8°C Step 2. If the condenser operates at 10°C, what percent of the toluene vapor to the condenser would exit as vapor? The molar ratio between the toluene and nitrogen in th e inlet stream can be calculated from the respective partial pressures. The same calculation can be made at the condenser exit. The percent toluene emitted from the condenser may then be calculated by dividing th e molar ratio of toluene to nitrogen at the exit by the molar ratio at the inlet. 

## ( ) nitrogen mole toluene mole mmHg mmHg p p

N T 1.060 0

. 391 0

. 760 0

. 391
2 =

− = 

## ( ) nitrogen mole toluene mole mmHg mmHg p p

N T 017

. 0 12.43

- 760.0 12.43
2 =

=


### .60% 1 % 100 * 060

. 1 017

. 0 % = = T 

### E Step 3. If the nitrogen component of the vapor stream is 10 lb/hr, what would the toluene emission rate be from the condenser? We can assume that 100% of the nitrogen exits th e condenser in the gas phase since little if any would condense with the toluene. The rate that nitr ogen flows through the system is 10.0 lb/hr.
hr moles

- lb 0.357 28.013 0

. 10 2 =

= MWt hr lb n N


### hr mmHg mmHg p p n n N T N T 

### moles

- lb 594 00

. 0 ) 43

. 12 760 ( 43

. 12 * 357

. 0 * 2 2 

### =

− = =



### h r lb 0.55 92.13 * 00594

. 0 =

− − = moles lb lb hr moles lb E TChapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-11 4.2.2 Partial Condensers The traditional approach to solving the partial condenser problem is known as the Rachford-Rice Equation [Wankat], shown in equation Eq. 4-9. Many references exist for this procedure and the reader is encouraged to review the literature for additional insights and understanding. [Rachford & Rice][Henley & Seader] 

# ( ) ( ) ( ) 0 1 1 1
1 

### =

− + − = 

# ∑

= C

i i i i 

### F V K z K F V f Eq. 4-9
Where

sys i i i i i 

### P P x y K γ 

### = ≡ If the system is ideal then the activity coefficient γ i can be set equal to 1.0 and the value for K i becomes a constant for all compositions at the fixed system temperature and pressure.

sys i i 

### P P K = , were 

### 0

. 1 = i 

### γ Eq. 4-10 Assuming that the partial condenser is being operated at a temperature that is between the dew point and the liquid boiling point of the inlet process stream then V/F must have a value that is between 0.0 and 1.0. For example, if V/F = 0.4 then this would mean that 40% of the exit process stream will be vapor and 60% will be liquid. By solving the Rachford-Rice function we are simply looking for the value of V/F (0.0 < V/F < 1.0) that enables f(V/F) to become zero. When V/F is too low then f(V/F) will be less than zero and when V/F is too high then f(V/F) will be greater than zero. Figure 4-2 presents a typical example of a plot of f(V/F) versus V/F. In this case, the solution to Eq. 4-9 would occur when V/F is approximately 0.43.

Figure 4-2. Generalized Rachford-Rice equation. Rachford-Rice Function plotted vs V/F

-3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

V/Ff(V/F)Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-12 Solving the Rachford-Rice condenser model requires that a numerical search routine be applied since there are not enough independent equations to solve for V/F algebraically. Numerical search routines such as False Position and Newton’s method can be applied until V/F converges to the desired degree of accuracy or tolerance. If the slope of f(V/F) is sufficiently large near the solution then the False Position and Newton’s method will tend to converge rapidly [Henley & Seader][Rachfo rd & Rice]. Standard spreadsheet software programs generally offer built in goal seeking routines that can be used to solve the Rachford-Rice condenser model. Once V/F has been determined where f(V/F) = 0 then the composition of the exit liquid L and vapor V streams can be directly calculated using Eq. 4-11 through Eq. 4-14. 

## ( ) F V F V 

### = Eq. 4-11 L = F – V Eq. 4-12 

## ( )( ) 1 1

−

+ = i i i 

### K F V z x Eq. 4-13

i i i 

### K x y = Eq. 4-14 Newton’s Method Using Newton’s method, a predicted value of V/F for iteration k+1 is computed from the recursive relation shown in Eq. 4-15 where f(V/F) is expressed in Eq. 4-9 and f N (V/F) is expressed in Eq. 4-16. 

## ( ) ( ) ( ) ( ) k k k k 

### F V f F V f F V F V 

### ′ − = + 1 Eq. 4-15 

## ( ) ( ) ( ) 

## [ ] 

# ∑

= 

### − + − = ′ C

i k i i

i k 

### F V K z

K F V f 1 2 2 

### ) ( 1 1 1 Eq. 4-16 Eq. 4-15 should be iterated until the desired accuracy of V/F is achieved. Stop iterating when two successive values of V/F satisfy the conditions in Eq. 4-17. 

## ( ) ( ) ( ) level tolerance F V F V F V
k k k 

### < − + 

### /

1 Eq. 4-17 The acceptable tolerance level may vary depending on th e application, but generally it should be small. For example, 0.0001 might be a reasonable value. Illustration 4-4: Rachford-Rice Method A partial condenser is used to separate excess methyl chloride from a process stream for incineration. The process vapor stream enters the condenser containing methyl chloride (1,250 lb/hr), n-hexane (2,585 lb/hr), and toluene (4,125 lb/hr). The partial condenser operates at atmospheric pressure (760 mm Hg) and at 45°C. The process stream may be calculated as an id eal mixture so that Raoult’s Law may be applied.Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-13 Calculate the flow rate and composition of the di scharge vapor and condensate liquid streams from the partial condenser assuming that the exit vapor and liquid stream are at equilibrium. Step 1. Establish characteristics of the inlet feed stream. Feed composition F i (lb-moles/hr) and mole fraction z i are calculated for each component in the inlet stream from the mass flow rates and molecular weights that are given. The pure vapor pressure (Pure VP) for each component in the feed stream at 45°C has been calculated based on the Antoine Equation and is supplied for this problem. K i has been calculated using Eq. 4-10 because the problem stated that the process mixture could be treated as ideal. The results are presented in Table 4-1. Table 4-1. Inlet Feed Stream Analysis
Compound F i lb/hr MWt F i lb-moles/hr z i mole frac. Pure VP i (mm Hg)

45°C K i =P i /P sys methyl chloride 1250.0 50.49 24.757 0.249 7369 9.696 n-hexane 2585.0 86.17 29.999 0.301 338 0.445 toluene 4125.0 92.13 44.774 0.450 74 0.097 Totals 99.6 1.00 Step 2. Add known values to the Rachford-Rice equation. The Rachford-Rice equation (Eq. 4-9) is populated using K i and z i from Table 4-1. Note that a separate term is required for each compound in the inlet gas stream. 

# ( ) ( ) ( ) 0 1 1 1
1 

### =

− + − = 

# ∑

= C

i i i i 

### F V K z K F V f 

## ( ) ( ) ( ) ( ) ( ) ( ) F V K z K F V K z K F V K z K

t t t h h h

mc mc mc 

### / 1 1 1 / 1 1 1 / 1 1 1 0

− + − +

− + − +

− + − = 

## ( )

( ) ( ) ( ) ( ) ( ) F V F V F V / 1 097

. 0 1 45

. 0 1 097

. 0 / 1 445

. 0 1 301

. 0 1 445

. 0 / 1 696

. 9 1 249

. 0 1 696

. 9 0

− + − +

− + − +

− + − = Eq. 4-18
Step 3. Add known values to the derivative of the Rachford-Rice equation. The derivative of the Rachford-Rice equation (Eq. 4-16 ) is populated using the same data from Table 4-1. 

## ( ) ( ) ( ) 

## [ ] 

# ∑

= 

### − + − = ′ C

i k i i

i k 

### F V K z

K F V f 1 2 2 

### ) ( 1 1 1 

## ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 

### / 1 097

. 0 1 45

. 0 1 097

. 0 / 1 445

. 0 1 301

. 0 1 445

. 0 / 1 696

. 9 1 249

. 0 1 696

. 9 F V F V F V F V f k



### − + − +

− + − +

− + − = ′ Eq. 4-19Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-14 Step 4. Solve for V/F using Newton’s method. Once Eq. 4-18 and Eq. 4-19 have been configured then guess an initial value for V/F and use Newton’s method as shown in Eq. 4-15 until the accuracy as per Eq . 4-17 is less than 0.0001. The results of this iterative process are shown in Table 4-2. The initial gu ess for V/F was assumed to be 0.5. Inserting this value in Eq. 4-18 and Eq. 4-19 resulted in initial estimates for f(V/F) and f N (V/F) of 0.567262 and 2.053191, respectively. The next estimate of V/F w as calculated using these values in Eq. 4-15, and then the initial and second estimates were compared using Eq. 4-17. This process was repeated for four iterations. At that point the value of V/F had converged within the specified tolerance level.
Table 4-2. Iteration R esults for Illustration 4-4

Iteration 

# ( ) k 

### F V 

# ( ) F V f 

# ( ) F V f ′ 

# ( ) 1 + k 

### F V 

## ( ) ( ) ( ) k k k F V F V F V

− + 1 1 0.5 0.567261 2.053191 0.223717 0.552565 2 0.223717 -3.45E-02 2.86477 0.235756 5.38E-02 3 0.235756 -7.89E-04 2.736357 0.236044 1.22E-03 4 0.236044 -4.16E-07 2.733474 0.236045 6.45E-07 The Rachford-Rice equation for this problem has been plotted in Figure 4-3. The solution to the problem where f(V/F) crosses the axis at V/F = 0.236 and f(V/F) = 0 has been marked with the vertical line.

Figure 4-3. Rachford-Rice Plot sh owing solution at V/F = 0.236044. Step 5. Determine the outlet vapor and liquid stream flows and compositions. After V/F has been determined by trial and error, as s hown in Table 4-2, then the molar flow rates of the exit liquid and vapor streams are calculated using Eq. 4-11 and Eq. 4-12. For example, the total moles of vapor is calculated as F x V/F (99.6 x 0.236045 = 23.4 94 lbmoles/hr). The mole fractions of components in each exit stream are calculated using Eq. 4-13 and Eq. 4-14. Finally, the molar flow rates of each component are calculated by multiplying the total moles by the applicable mole fractions. The results of these calculations are shown in Table 4-3. Rachford-Rice Function, f(V/F) vs V/F

-2.00 -1.00 0.00 1.00 2.00 3.00 4.00 5.00

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 

# ( ) ( ) ( ) 

# ∑

= 

### − + − = N

i i i 

### F V K z K F V f 1 1 

### 1 1 1Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-15 Table 4-3. Product Outlet Stream Analysis Compound x i Li (lb-moles/hr) y i Vi (lb-moles/hr) methyl chloride 0.081 6.196 0.79008 18.562 n-hexane 0.347 26.375 0.15427 3.624 toluene 0.572 43.466 0.05566 1.308 Totals 1.000 76.037 1.000 23.494 4.2.3 Vent Condensers with Single-Phase Condensate When the process stream contains two or more c ondensable components and also a noncondensable gas component (air, nitrogen, etc.) then the Rachford-Ric e equation that was discussed in section 4.2.2 must be modified in order to solve the condenser problem. As in illustration 4-4, the value of V/F that solves the equation must be determined ite ratively. Alternatively, two othe r iterative approaches may be used. Both of these alternatives involve making an in itial guess of the condensate composition. This information is used to solve material balances or e quilibrium relationships. If the relationships are not met within a specified level of tolerance, then the condensate composition is adjusted and the process is repeated until the specified accuracy is achieved. These two altern atives to the modified Rachford-Rice equation are not needed to solve the single liquid phase problems in this section. However, they are introduced here to show that they ach ieve the same results. They will also be used to solve problems with multiple liquid phases in section 4.2.4, which are not readily solved using the modified Rachford-Rice method. All three options are illustrated in this section fo r a condenser with the same inlet stream. These illustrations use an inlet stream with three condensab le components, but the procedures could be applied to any stream with two or more condensable components. A fourth illustration for a system with liquid that is not ideal is also presented in this section. 4.2.3.1 Option 1 – Modified Rachford-Rice Method Including noncondensable in the inlet gas stream to a condenser complicates the determination of emissions from the condenser. Typically, the solubility of a noncondensable gas in the condensate that would form in the condenser is either very small or not known. One can simplify the problem by assuming that all of the noncondensable gas exits the vent device through the exit vapor stream and that none is contained in the exit condensate stream. If the solubility of the noncondensable gas in the condensate is set to 0.0, then x i = 0, and K i becomes infinity. When K i equals infinity then the standard Rachford-Rice model becomes indeterminate since K i appears in the numerator and the denominator. This problem may be solved by dividing both numerator and denominator by (K i -1) as shown. 

## ( ) ( ) 

# ∑

= 

### − + − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ C

i i i i 

### F V K z K F V f 1 

### 1 1 1 Eq. 4-9 

## ( ) ( ) ( ) ( ) 

# ∑

= 

### ⎥ ⎦ ⎤ ⎢ ⎣ ⎡

− − ×

− + − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ C

i i i i i i 

### K K F V K z K F V f 1 

### 1 1 1 1 1 1 1 Eq. 4-20 

## ( ) 

# ∑

= 

### +
− = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ C

i i i 

### F V K z F V f 1 

### 1 1 Eq. 4-21Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-16 For the case when component i is a nonconden sable gas (nitrogen, air, etc) where x i = 0 then:

i i i 

### x y K = Eq. 4-10 

## [ ] ∞ = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡

= → → i i
x i x 

### x y K
i i 0

0 

### lim lim Eq. 4-22 The modified Rachford-Rice expression for the no ncondensable component may now be evaluated as shown. 

## ( ) nc i

nc i

nc i i K 

### F V

z F V z F V K z

i 

### ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ∞ = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +
− ∞ → 

### 1 1 1 lim Eq. 4-23 Combining Eq. 4-21 (for the condensable com ponents) with Eq. 4-23 (for the noncondensable components) leads to a modified Rachford-Rice equa tion that can be used to solve condenser problems when non-condensable components such as air or nitrogen are present in the process stream. 

## ( ) 

# ∑ ∑

=

= 

### + +
− = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ nc c C
j j C

i i i 

### F V

z F V K z F V f 1 1 

### 1

1 Eq. 4-24 The derivative of the modified Rachford-Rice equation, with respect to V/F, is shown in Eq. 4-25 and can be applied within the framework of Newton’s method using Eq. 4-15. 

# ( ) ( ) ( ) ( ) 

# [ ] [ ] 

# ∑ ∑

=

= 

### − + +
−

− = = ′ nc c C
j j C

i i i 

### F V z F V K z F V d F V f F V f 1 2 1 2 

### 1 1 Eq. 4-25 

## ( ) ( ) ( ) ( ) F V f F V f F V F V k k 

### / ' / 1 

### −

= + Eq. 4-15 After V/F is determined, the composition of each ex it stream may be calculate d using Eq. 4-11 through Eq. 4-14. 

## ( ) F V F V = Eq. 4-11 L = F - V Eq. 4-12 

## ( )( ) 1 1

−

+ = i i i 

### K F V z x Eq. 4-13

i i i 

### K x y = Eq. 4-14Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-17 Illustration 4-5: Modified Rachford-Rice Method for Vent Condenser 500 gallons of condensate from Illustration 4-4 was cool ed to 30°C and is being held in a 1,000 gallon receiver. Methyl chloride, n-hexane, and toluene are assumed to be an ideal mixture. The vessel is under a nitrogen gas sweep of 1.0 scfm which is vented to a total condenser (standard conditions based on temperature of 0°C). The nitrogen off gas exits the co ndenser at 2°C and is assumed to be saturated for the purposes of this illustration (the actual saturation level could be determined using the procedure in illustration 3-8 in section 3.2.2). What would the emissi ons and the material balance for the condenser be over a one hour period under these conditions? Step 1. Establish characteristics of the inlet gas stream. An analysis of the process gas stream entering the co ndenser from the receiver is shown in Table 4-4. Table 4-4. Analysis of the Condenser Inlet Feed Stream
Compound F i lb/hr MWt F i lb-moles/hr z i Mole frac. Partial pressure p i mm Hg Noncondensable nitrogen 4.68 28.01 0.1671 0.354 269 Condensable

methyl chloride 12.67 50.49 0.2509 0.532 405 n-hexane 3.47 86.17 0.040252 0.0854 65 toluene 1.20 92.13 0.013006 0.0276 21 Totals 0.47126 1.0 760 Step 2. Check to confirm wh ether condensation will occur Before proceeding to calculate the material balan ce for the vent condenser it’s a good idea to perform a quick test to see if condensation is likely to o ccur in the condenser. In some cases the quantity of noncondensable gas component (air, nitrogen, etc) could be too high for condensation to occur. For this test we can calculate x i for each condensable component in hypothetical condensate at the condenser temperature (2°C in this ex ample). If the sum of the calculated x i values for all of the condensable components is greater than 1.0 then we can conclude that condensation will take place. However, if the sum of the calculated x i values is less than 1.0 then condensation is not possible at the current condenser operating conditions.

C i C i C i i C

i i 

### P p P p x ° °

° °


### = = 2 30 2 30 

### γ Eq. 4-26
where P i is the pure component vapor pressure at 2°C, p i is the component partial pressure at th e condenser inlet conditions (30°C), and

i 

### γ is 1.0 for the ideal mixture. The partial pressure p i for each component in the inlet stream to the condenser was established in step 1. The pure component vapor pressure (P i ) for each component in the feed st ream at 2°C has been calculated based on the Antoine Equation, and the resulting valu es are shown in Table 4-5. The hypotheticalChapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-18 condensate mole fraction for each component x i has been calculated using Eq. 4-26 because the problem stated that the process mixture could be treated as idea l. All of the component mole fractions and the sum are shown in Table 4-5. The sum is greater than 1.0. Therefore, condensation will occur in this condenser. Table 4-5. Check to See If Condensation Will Take Place at 2°C Inlet Partial Pres. (mm Hg) P i @ 2°C x i Estimated Condensable

methyl chloride 405 2,089 0.19 n-hexane 65 50 1.3 toluene 21 8 2.63 

# ∑ i 

### x = 4.1 Step 3. Calculate Values to use in the Modified Rachford-Rice Equation Values for z i are needed for all four components; these va lues were established in step 1. Values for K i are needed for methyl chloride, n-hexane, and toluene; a K value for nitrogen is ignored since it is now assumed that the solubility of nitrogen is 0. The K i values are calculated using Eq. 4-10 with the pure component vapor pressures at 2°C, which were calculated in step 2. Table 4-6 shows all of the input data and the calculated K i values. Table 4-6. Preparation for Applying the Modified Rachford-Rice Method Compound F i lb-moles/hr z i Mole frac. P i (mm Hg) 2°C K i =P i /P sys nitrogen 0.1671 0.354 --- --- methyl chloride 0.2509 0.532 2,089 2.749 n-hexane 0.040252 0.0854 50 0.066 toluene 0.013006 0.0276 8 0.011 Totals 1 Step 4. Add known values to the modified Rachford-Rice equation. The modified Rachford-Rice equation (Eq. 4-24) is populated using K i and z i from Table 4-6. Note that a separate term is needed for each component in the inlet stream. 

## ( ) 

# ∑ ∑

=

= 

### + +
− = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ nc c C
j j C

i i i 

### F V

z F V K z F V f 1 1 

### 1

1 

## ( ) ( ) ( ) F V z F V K z F V K z F V K z n t t h h

mc mc 2 

### / 1 1 / 1 1 / 1 1 0 + +
− +

− + +
− = 

## ( ) ( ) ( ) F V F V F V F V 

### 354

. 0 / 1 011

. 0 / 1 0276

. 0 / 1 066

. 0 / 1 0854

. 0 / 1 749

. 2 / 1 532

. 0 0 + +
− + +
− + +
− =Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-19 Step 5. Add known values to the derivative of the modified Rachford-Rice equation. The derivative of the modified Rachford-Rice equation (Eq. 4-25) is populated using the same data from Table 4-6. 

# ( ) ( ) ( ) ( ) 

# [ ] [ ] 

# ∑ ∑

=

= 

### − + +
−

− = = ′ nc c C
j j C

i i i 

### F V z F V K z F V d F V f F V f 1 2 1 2 

### 1 1 

# ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 

### / 354

. 0 / 1 011

. 0 / 1 45

. 0 / 1 066

. 0 / 1 301

. 0 / 1 749

. 2 / 1 249

. 0 F V F V F V F V F V f − + +
− − + +
− − + +
− − = ′
Step 6. Solve for V/F using Newton’s method. An initial guess of V/F is needed to start the itera tive process of solving the modified Rachford-Rice equation. Rather than picking a value arbitrarily, so metimes it’s helpful to graphically plot f(V/F) as a function of V/F as shown in Figure 4-4 for this proble m. The objective is to identify the value of V/F where f(V/F) is crossing the horizontal line where the y axis = 0. Using an initial value of V/F that is close to this point will help ensu re that the method will converge on a valid solution. Newton’s method relies on the slope of f(V/F) to help locate the solu tion. At points where the slope of the function is sufficiently low the method has been known to extrap olate into areas that are out of bounds for the solution of V/F (i.e., less than 0 or greater than 1) . This problem can be avoided by making an initial guess of V/F close to the value where the function crosses the x-axis. By inspection of Figure 4-4 one would know that the value of V/F that solves the R achford-Rice equation for this condenser problem is close to 0.9.

Figure 4-4. Rachford-Rice Plot sh owing solution at V/F = 0.90076. Based on Figure 4-4, an initial guess for V/F was assu med to be 0.85. Newton’s method as shown in Eq. 4-15 is applied until the V/F remains the same value within a 0.0001 tolerance, as determined using Eq. 4- 17. The results of this iterative process are shown in Table 4-7. Inserting the initial guess for V/F in Eq. 4- 24 and Eq. 4-25 resulted in initial estimates for f(V/F) and f N (V/F) of 0.23186 and -3.32408, respectively. 

### f(V/F) Plotted vs V/F

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-20 The next estimate of V/F was calculated using these values in Eq. 4-15, and the initial and second estimates were compared using Eq. 4-17. This process w as repeated for four iterations. At that point the value of V/F had converged within the specified tolerance level.
Table 4-7. Iteration re sults for Illustration 4-5

Iteration 

# ( ) k 

### F V 

# ( ) F V f 

# ( ) F V f ′ 

# ( ) 1 + k 

### F V 

## ( ) ( ) ( ) k k k F V F V F V

− + 1
1 0.85 0.23186 -3.32408 0.91975 0.08206 2 0.91975 -0.12876 -7.54609 0.90269 0.01855 3 0.90269 -0.01159 -5.85834 0.90071 0.00219 4 0.90071 0.00030 -5.70487 0.90076 0.00006 After V/F has been determined by trial and error, as s hown in Table 4-7, then the molar flow rates of the exit liquid and vapor streams are calculated using Eq. 4-11 and Eq. 4-12. 

## ( ) F V F V = 

## ( ) ( ) 90076

. 0 47126

. 0 = 

### hr moles lb / 42454

. 0 = V F L

− = 

### 42454

. 0 47126

. 0

− = 

### hr moles lb / 0467

. 0 = The mole fractions of components in ea ch exit stream are calculated using Eq. 4-13 and Eq. 4-14. For example, the mole fractions for meth yl chloride are calculated as follows: 

## ( )( ) 1 1

−

+ =
mc mc mc 

### K F V z x 

## ( )( ) 1 749

. 2 90076

. 0 1 532

.

− + = 20676

. 0 =

i i i 

### K x y = 

## ( ) ( ) 749

. 2 20676

. = 

### 56831

. 0 =Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-21 The molar flow rates of each component are calculate d by multiplying the total moles by the applicable mole fractions. For example, the molar flow ra tes of methyl chloride are calculated as follows: 

## ( ) ( ) hr moles lb n L

mc 

### / 00967

. 0 0467

. 20676

. , 

### = = 

## ( ) ( ) hr moles lb n V

mc 

### / 24127

. 0 42454

. 56831

. , 

### = = Finally, the mass flow rates for each component are calculated by multiplying the molar flow rates by the molecular weight of the component. The results of all of the calculations in step 7 are shown in Tables 4-8 and 4-9. Table 4-8. Product Outlet Stream Analysis (mole fractions and rates in lb-moles) F i lb-moles/hr Z i K i L i lb-moles/hr X i V i lb-moles/hr Y i nitrogen 0.1671 0.354 --- 0.00000 0 0.16708 0.39356 methyl chloride 0.2509 0.532 2.749 0.00967 0.20676 0.24127 0.56831 hexane 0.04025 0.0854 0.066 0.02521 0.53903 0.01506 0.03546 toluene 0.01301 0.0276 0.0276 0.01189 0.25417 0.00114 0.00268 Totals 0.47126 0.0467 0.99996 0.42454 1.00000 Table 4-9. Product Streams Analysis

Component MW t F i Inlet Stream lb/hr L i Condensate lb/hr V i Vapor Emissions lb/hr nitrogen 28.01 4.6800 0.0000 4.6800 methyl chloride 50.49 12.6700 0.4883 12.1817 hexane 86.17 3.4700 2.1727 1.2973 toluene 92.13 1.2000 1.0954 0.1046 Totals 22.0200 3.7563 18.2637 Additional Considerations Regardi ng Numerical Convergence Techniques It was mentioned earlier that Newton’s method had the potential to extrapolate the search into places that are out of bounds with respect to a realistic soluti on. In some cases a damping factor may be used to improve the tendency for Newton’s method to cont inually move toward the desired solution. Additionally, other numerical methods exist includi ng the False Position, Bisection, and Fibonacci Methods that can be applied to either the standard Rachford-Rice or the modifi ed Rachford-Rice solution model. If it is desired to use standard spreadsheet soft ware then the Goal Seek function can be applied to solve the condenser problem. 4.2.3.2 Option 2 – Partial Pressure Correction Method A second approach to solving the vent condenser prob lem when two or more condensable components are present in the system involves making an initial gu esstimate of the condensate composition and then placing the balance of the process in the vapor phase. Component vapor pressures are then calculated forChapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-22 the condensate and compared to the component par tial pressures of the vapor phase. Since equilibrium between the condensate liquid phase and the vapor phase is assumed, both partial pressures that have been calculated by different means should be the same for each component. The amount of each component in the condensate is adjusted up or down until the partia l pressures are the same for each component (within an acceptable tolerance). Material balance accounti ng is assured since 100% of each component in the feed is allocated to be split between the condensate phase and the vapor phase. Condensable component material balance:

V i L i F i 

### n n n
, , , 

### + = Eq. 4-27 The partial pressure of each condensable component in the exit liquid stream may be calculated using Raoult’s Law which may be extende d with activity coefficients.



### L P n P x p i i L i i i i L i 

### γ γ ,
, 

### = = Eq. 4-28 The partial pressure of each condensable compone nt in the exit gas phase may be calculated using Dalton’s Law.

sys i sys V i V i 

### P y P V n p = = ,
, Eq. 4-29 At equilibrium both partial pressures should be the same for each condensable component.

V i L i 

### p p
, , 

### = Eq. 4-30 However, since we are making only an initial guesstimat e, the two partial pressures will likely not be the same. An error of estimate can be calcula ted by subtracting one from the other.

V i L i i 

### p p n f

, , 

### ) (

− = Eq. 4-31 Newton’s method is applied to all condensable com ponents simultaneously until the two partial pressures (from the condensate and the gas phase) agree with in acceptable tolerance for each component. 

# ( ) ( ) k i k i k L i k L i 

### n f n f n n

, ,
, , 1 , , 

### ' − = + Eq. 4-15 

### ) ( ' i



### n f may be approximated using the calculus of fin ite differences where only condensable component i is varied and the other components are held constant. [Constantinides] 

# ( ) ( ) ( )

j P T i i j P T k L i k L i k L i 

### n p n f n f n f

, , , , , , , ,
, , 

### 2 ' ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ

Δ = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡

− − + ≅ δ δ δ Eq. 4-32 If the process contains only two condensable com ponents then Eq. 4-15 would be required for both components. If the process contains 5 condensable compounds then five separate equations would beChapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-23 required and so on. For each equation f ( n i,L,k ) and f N ( n i,L.k ) are calculated with only n i being varied while all other compound quantities held constant. Once f ( n i,L,k ) and f N (n i,L,k ) have been calculated n i,L,k+1 is calculated for each condensable compound in the system as shown. Component 1 

# ( ) ( ) 1 , , , , 1 , , 1

, , 1 1 , , 1 

### ' ≠ + 

### − = j P T k L k L k L k L 

### n f n f n n

Component 2 

# ( ) ( ) 2 , , , , 2 , , 2

, , 2 1 , , 2 

### ' ≠ + 

### − = j P T k L k L k L k L 

### n f n f n n

· · · 

### • • • Component n 

# ( ) ( )
n j P T k L n k L n k L n k L n 

### n f n f n n
≠ + 

### − =

, , , , , ,
, , 1 , , 

### ' Illustration 4-6: Partial Pressure Correction Method for Vent Condenser 500 gallons of the condensate from Illustration 4-4 was c ooled to 30°C and is being held in a 1,000 gallon receiver. A 1.0 scfm nitrogen gas sweep enters that v essel headspace and exits fully saturated with vapors from the condensate. The condenser is operated so process gas stream exits at 2°C. What would the emissions and the material balance for the conden ser be over a one hour period under these conditions? Step 1. Establish characteristics of the inlet gas stream. An analysis of the process gas stream entering the co ndenser from the receiver is shown in Table 4-10. This information is the same as for illustration 4-5. Table 4-10. Analysis of the Condenser Inlet Feed Stream
Compound F i lb/hr MWt F i lb-moles/hr z i Mole frac. Partial Pressure mm Hg Noncondensable nitrogen 4.68 28.01 0.1671 0.354 269 Condensable

methyl chloride 12.67 50.49 0.2509 0.532 405 n-hexane 3.47 86.17 0.04025 0.0854 65 toluene 1.20 92.13 0.01301 0.0276 21 Totals 0.47126 1.0 760 Step 2. Check to confirm wh ether condensation will occur. This procedure is presented in illustra tion 4-5 and is not repeated here.Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-24 Step 3. Make initial estimate of composition of condensate. In setting up to solve this problem using Newton’s me thod, it improves the efficiency of convergence if one first approximates of the molar composition of the condensate to as near to the final solution as possible before beginning the iteration process. Sin ce the pure vapor pressure of toluene (8 mm Hg @ 2°C) is lower than the partial pressure of toluene in the feed stream (21 mm Hg) then all of the toluene moles can be placed in the condensate. Similarly, since the pure vapor pressure of n-hexane (50 mm Hg @ 2°C) is lower than the partial pressure of n-hexane in the feed (65 mm Hg) then all of the n-hexane moles in the condensate. However, the pure vapor pre ssure of methyl chloride (2089 mm Hg) at 2°C is greater than the inlet partial pressure of methyl chlo ride (405 mm Hg) and so only a fraction of the inlet moles should be allocated for the condensate. If one assumes that the condensate will be an ideal mi xture where the activity coefficient equals 1.0 for each component in the condensate, th en an initial guess for the concentra tion of methyl chloride can be estimated using Eq. 4-8, where x i is the mole fraction of methyl chloride that would be needed in the liquid to result in the partial pressure of 405 mmHg.


### 0.19 2089 * 1 405 2 

### = = = ° C i i feed

i i 

### P p x γ Eq. 4-8
Once x i is known, then the moles of methyl chloride th at should be allocated to the liquid condensate can be calculated by the ratio of the methyl chloride mo le fraction and the non methyl chloride portion of the condensate as shown in Eq. 4-33, wher e j represents n-hexane and toluene.

L j i i L i 

### n x x n ,

, 

### 1 ∗

− ≈ Eq. 4-33



### 0533

. 0 19

. 0 1 19

. 0 ∗

− = hr moles lb / 01249

. 0 = The resulting initial estimate of the condensate is presented in Table 4-11. Table 4-11. Initial Estimate of Condensate Compound F i lb-moles/hr p i at Feed mm Hg P i at 2°C mm Hg L i , estimated lb-moles/hr Noncondensable nitrogen 0.1671 269 Condensable

methyl chloride 0.2509 405 2,089 0.01249 n-hexane 0.04025 65 50 0.04025 toluene 0.01301 21 8 0.01301 totals 0.06575Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-25 Step 4. Solve for the partial pressures using Newton-Raphson method to iterate on the number of moles of each component in the condensate. Once an initial guess of the condensate has been made , then the iterative processing can begin as shown below. For the first iteration, the mole fraction of each compound in the liquid is calculated based on the initial composition from Table 4-11. For example, the mole fraction for toluene is estimated as 0.01301/0.06575 = 0.2. The partial pressure based on this liqui d composition and the pure component vapor pressure at 2°C is calculated using Eq. 4-28. Fo r example, the calculation is as follows for toluene: 

## ( )( )( ) Hg mm

P x P i i i L i 

### 6

. 1 8 1 2

. 0 ,



### =

= = γ

For the first iteration, the vapor out of the condenser is assumed to consist of only nitrogen and methyl chloride. The amount of methyl chloride in the ou tlet vapor is estimated as the difference between the methyl chloride in the feed and the methyl chloride in the condensate. The mole fractions are estimated based on the total number of moles of nitrogen and meth yl chloride in the outlet vapor stream. The partial pressures are then estimated using Eq. 4-29. For exampl e, the calculation is as follows for methyl chloride in the first iteration: 

## ( )( ) mmHg P y p sys i V i 

### 447 760 59

. 0 ,



### =

=

=

The difference in the partial pressures (i.e., the f unction) is calculated using Eq. 4-31. For methyl chloride: 

## ( ) mmHg p p n f V i L i i 

### 50 447 397 , ,



### − = − = −

=

The derivative of the function is estim ated using finite differences as d escribed in Eq. 4-32. This means the calculations of the partial pressures described above need to be repeated using slightly different estimates of the number of moles in th e condensate. In this illustration the δ used in Eq. 4-32 was equal to 0.05 percent of the original estimate of the number of moles in the condensate. For example, for methyl chloride in the first iteration the calculations were repeated using 0.012484 lbmoles and 0.012496 lbmoles (i.e., δ = 0.0005 x 0.01249 = 0.000006245). Using these values for n i results in f(n i ) equal to -50.188 and - 49.857, respectively. The difference between these values (0.331) is the numerator in Eq. 4-32, and 0.00001249 is the denominator. Dividing 0.331 by 0.00001249 results in 26,507 for the derivative of the function. This process is repeated for all of the other components in the condensate. The numerical values are not tabulated in this document due to space considerations. The number of moles of each component in the cond ensate for the second iteration is estimated using Eq. 4-15. For example, the next second estimate fo r methyl chloride is estimated as follows:Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-26 

# ( ) ( )

hr moles lb n f n f n n k i k i k L i k L i



### / 0144

. 0 507 , 26 50 01249

. 0 , ,
, , 1 , ,



### = −

− = ′ − = +

This process is continued until the difference in th e partial pressures estimated based on the liquid and vapor moles fractions for all components agree with in an acceptable level of tolerance. For this illustration the analysis stopped after 7 iterations , when the differences in vapor pressures for all compounds appeared to be in good agreement fo r all compounds. One can calculate the percent agreement for each component by dividing Δ pi by the average of both partial pressures. For Iteration 7 the % agreement is shown in Table 4-12.
Iteration 1 Compound n i.L,k x i p i,L V i y i p i,V ∆ p i ∆ p i / ∆ n i n i,L,k+1 nitrogen 0 0.167 0.41 313 N/A N/A N/A methyl chloride 0.01249 0.19 396.91 0.238 0.59 446.82 -50.2 26507 0.0144 n-hexane 0.04025 0.61 30.6 0 0 0 30.6 2169.6 0.0261 toluene 0.01301 0.2 1.6 0 0 0 1.6 1972.3 0.0122
Iteration 2 Compound n i.L,k x i p i,L V i y i p i,V ∆ p i ∆ p i / ∆ n i n i,L,k+1 nitrogen 0 0.167 0.41 313 N/A N/A N/A methyl chloride 0.0144 0.27 570 0.237 0.57 430 140 29609 0.0096 n-hexane 0.0261 0.5 24.8 0.014 0.03 25.6 -0.8 2233.1 0.0265 toluene 0.0122 0.23 1.9 8E-04 0 1.5 0.4 1929.4 0.012
Iteration 3 Compound n i.L,k x i p i,L V i y i p i,V ∆ p i ∆ p i / ∆ n i n i,L,k+1 nitrogen 0 0.1671 0.395 300.2 N/A N/A N/A methyl chloride 0.00964 0.2 418.3 0.2413 0.57 433.4 -15 35464.5 0.01007 n-hexane 0.02651 0.551 27.5 0.0137 0.032 24.7 2.9 2205 0.02522 toluene 0.012 0.249 2 0.001 0.002 1.8 0.2 1917.1 0.0119Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-27 Iteration 4 Compound n i.L,k x i p i,L V i y i p i,V ∆ p i ∆ p i / ∆ n i n i,L,k+1 nitrogen 0 0.1671 0.394 299.5 N/A N/A N/A methyl chloride 0.01007 0.213 445.6 0.2408 0.568 299.5 14 35598.2 0.00967 n-hexane 0.02522 0.534 26.7 0.015 0.035 26.9 -0.22 2222.3 0.02532 toluene 0.0119 0.252 2 0.0011 0.003 2 0.04 1914.7 0.01188
Iteration 5 Compound n i.L,k x i p i,L V i y i p i,V ∆ p i ∆ p i / ∆ n i n i,L,k+1 nitrogen 0 0.1671 0.394 299.2 N/A N/A N/A methyl chloride 0.00967 0.206 431.1 0.2412 0.568 432 -0.9 36143.2 0.0097 n-hexane 0.02532 0.54 27 0.0149 0.035 26.7 0.26 2218.7 0.0252 toluene 0.01188 0.253 2 0.0011 0.003 2 0.01 1913.9 0.01187
Iteration 6 Compound n i.L,k x i p i,L V i y i p i,V ∆ p i ∆ p i / ∆ n i n i,L,k+1 nitrogen 0 0.1671 0.394 299.2 N/A N/A N/A methyl chloride 0.0097 0.207 433.1 0.2412 0.568 431.8 1.29 36176.4 0.00966 n-hexane 0.0252 0.539 26.9 0.0151 0.035 27 -0.01 2220.4 0.0252 toluene 0.01187 0.254 2 0.0011 0.003 2 0 1913.7 0.01187
Iteration 7 Compound n i.L,k x i p i,L V i y i p i,V ∆ p i ∆ p i / ∆ n i n i,L,k+1 nitrogen 0 0.1671 0.394 299.2 N/A N/A N/A methyl chloride 0.00966 0.207 431.9 0.2412 0.568 431.9 -0.03 36228.9 0.00966 n-hexane 0.0252 0.539 26.96 0.0151 0.035 26.94 0.024 2220.1 0.02519 toluene 0.01187 0.254 2.03 0.0011 0.003 2.03 0.001 1913.6 0.01187 Table 4-12. Convergence Determination Component p i avg. mm Hg ∆ p i mm Hg ∆ p i /p i % nitrogen N/A N/A methyl chloride 431.86 -0.03 -0.007% n-hexane 26.95 0.02 0.089% toluene 2.031 0.001 0.049% Step 5. Perform material balance check. The final results after Iteration 7 are presented in Tabl e 4-13 with the material balance displayed in pound moles and in pounds for the exit va por and liquid streams. The composition of the exit vapor streams is also listed by the average partial pressu re for each component being emitted.Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-28 Table 4-13. Material Balance for Condenser

Compound MWt F i lb/hr F ilbmoles/

hr L ilbmoles/

hr L i lb/hr V ilbmoles/hr V i lb/hr P i mm Hg nitrogen 28.01 4.68 0.167 0 0.167 4.68 299.2 methyl chloride 50.49 12.67 0.251 0.0097 0.49 0.241 12.18 431.86 n-hexane 86.17 3.47 0.040 0.0252 2.17 0.015 1.30 26.95 toluene 92.13 1.20 0.013 0.0118 1.09 0.001 0.11 2.031 Totals 22.02 0.471 0.0467 3.75 0.424 18.27 760.04
The results from the Pressure Balance Method for this problem are virtually identical to the results that were obtained using the Modified Rachford-Rice Method shown in Illustration 4-5. 4.2.3.3 Option 3 – Material Balance Correction Method A third approach to solving the vent condenser pr oblem when two or more condensable components are present involves making an initial guesstimate of th e liquid condensate composition. Component vapor pressures are then calculated for the condensate and used in conj unction with the noncondensable component to estimate the amount of each component that exists in the vapor space. For each condensable component, the moles in the liquid phase are added to the moles that have been calculated to exist in the vapor phase and the ratio between th ese two quantities is calculated. The moles of each component in the liquid phase are then corrected by the ratio value for the next iteration. For example, if the moles of component i in the outgoing vapor and li quid streams is calculated to be 20% greater than the moles of component i in the incoming feed stream th en the amount of i in the condensate is divided by 1.2 for the next iteration. This iterative process for co rrecting the moles of i in the liquid phase continues until the material balance for all components is in agreement within acceptable tolerances. For any condenser, the inlet stream must equal the sum of th e exit vapor and the liquid condensate at steady state operation. Material Balance Considerations Overall material balance: 

### V L F + = Eq. 4-34 A material balance can be written for each condensable component in the process as shown in Eq. 4-35. Balance of condensable components: 

### V y L x F z i i i 

### + = Eq. 4-35

V i L i F i 

### n n n
, , , 

### + = Eq. 4-36 Assuming that all of the noncondensable gases (air, nitr ogen, etc) exit the condenser in the vapor stream then Eq. 4-35 and Eq. 4-36 can be easily simplified. The moles of noncondensable n nc in the exit gas stream are equal to the moles that are contained in the inlet stream. Noncondensable component: 

### V y F z n nc nc V nc 

### = = , Eq. 4-37Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-29 Equilibrium Considerations The vapor pressure of each condensable component in the exit liquid stream can be calculated using Raoult’s Law and extended with activity coefficients.

i i i i 

### P x p γ = Eq. 4-38 At steady state conditions, the moles of each condensable component i in the exit gas stream can be calculated from the vapor pressure of component i in the liquid, the partial pressure of the noncondensable in the exit gas, and the mol es of the noncondensable component as shown.

nc i V nc V i 

### p p n n =

, , Eq. 4-39

nc i i L i nc nc i i i nc nc i nc V i 

### Lp P n n p P x n p p n n γ γ ,
, 

### = = = Eq. 4-40 If the process contains only a single condensable component then Eq. 4-40 can be applied directly to calculate the exit gas emissions from the condenser. This was shown in Illustrations 4-1, 4-2, and 4-3. Combining Material Balance an d Equilibrium Considerations If the process contains two or more condensable co mponents then a trial and error approach will be needed to determine the moles of each condensable component in the liquid and vapor phases. The equilibrium relationship ( Eq. 4-40) can be combined with the material balance ( Eq. 4-36) to result in a single expression that must hold true at single stage steady state conditions where equilibrium exist between each component in the liquid and vapor phase.

nc i nc L i V i L i F i 

### p p n n n n n + = + = , , , , Eq. 4-41 Eq. 4-41 ties the material balance and the equilib rium relationships together for each condensable component in the system and is independent of the number of condensable components present. Description of the Basic Material Balance Correction Procedure In the trial and error approach that follows, an edu cated guess of the condensate liquid is made. Eq. 4-41 is used to calculate the molar inventory of com ponent i from the condensate and calculated vapor phase. Eq. 4-42 is then used to calculate the ratio Ф i between the moles of component i in the feed and the moles calculated to be in the exit vapor and liquid phases. At steady state and equilibrium conditions Ф i must equal 1.0.

i F i V i L i 

### n n n φ 

### = +

, , , Eq. 4-42

F i nc i

i nc L i L i i 

### n L p P n n n , , , 

### 1 ≈ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + γ φ Eq. 4-43Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-30 F i nc i

i nc i L i 

### n L p P n n

,

, 

### 1 ≈ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + γ φ Eq. 4-44

n i,L,k+1 is estimated by dividing n i,L,k by Ф i as shown in Eq. 4-45 and applied to the next iteration until Ф i becomes close to 1.0 within an accep table tolerance. It should be noted that L are the total moles of condensate in the system which includes component i. For this reason Eq. 4-44 and Eq. 4-45 are approximations which narrow as the iterative process converges.

k i k L i k L i 

### n n

, , , 1 , , 

### φ = + Eq. 4-45 For each condensable component in the inlet stream, the following steps are followed until the error of estimate for each component has falle n within an acceptable tolerance. Basic Procedures An initial guesstimate of the condensate liquid must be made before the iteration process is started. One approach involves partitioning the moles of component i in the liquid using a measure of the relative volatility. Components having the greatest volatility w ill tend to concentrate in the vapor phase while compounds with the lowest volatility will tend to concentrate in the liquid phase. 

# ∑

= 

### = N j j i i 

### P P

1 

### ν Eq. 4-46

The moles of component n i,L in the condensate liquid are calculated by multiplying the moles of n i,F in the feed by (1- i 

### ν ) since (1- i 

### ν ) would be a fraction of a compound to remain in the liquid phase. 

# ( ) ( ) F i i k L i 

### n n ,

, 

### * 1 

### ν

− = Eq. 4-47 Also, p nc,k is set equal to the partial pressure of the n oncondensable as contained in the feed stream p nc,F . This is because we know that any condensa tion to occur will lead to value of p nc that is greater than what we started with. So the noncondensable component partia l pressure at the inlet feed is considered to be lowest starting value possible and can only b ecome larger as the convergence takes place. *********************** Step 1. Calculate L, x i , and p i for each component in the liquid phase. 

# ∑ = L i 

### n L

, , 

### L n x L i i ,



### = , i i i i 

### P x p 

### γ = Step 2. Calculate n i,L,k+1 for each condensable component in the vapor phase and calculate V k+1 from the sum of the moles of condensable components in the vapor phase plus the noncondensable moles (which remain constant).

k nc k i k nc k V i 

### p p n n

, ,
, 1 , , 

### = + nc C j k V j k 

### n n V + = 

# ∑

= + + 1 1 , , 1Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-31 Step 3. Calculate p nc,k .

sys k nc k nc 

### P V n p 1 1 , +

+ 

### = Step 4. Calculate Ф i for each condensable component using Eq. 4-39 substituted in Eq. 4-42.

1 ,
, 1 , ,
+ + 

### = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ +
k i F i k nc i nc L i


### n p p n n φ Step 5. Calculate the next iteration value for (n i,L ) k+1 . 

# ( ) ( )

1 , 1 , + + 

### = k k L i k L i 

### n n φ Step 6. Repeat Steps 1 through 5 until Ф is between 0.999 and 1.001 for all condensable components in the system. Illustration 4-7: Material Balance Correction Method for Vent Condenser 500 gallons of the condensate from Illustration 4-4 was c ooled to 30°C and is being held in a 1,000 gallon receiver. A 1.0 scfm nitrogen gas sweep enters that v essel headspace and exits fully saturated with vapors from the condensate. The condenser is operated so process gas stream exits at 2°C. What would the emissions and the material balance from condenser be over a one hour period under these conditions? Table 4-14. Analysis of the Condenser Inlet Feed Stream
Compound F i lb/hr MWt F i lb-moles/hr z i Mole frac. Partial Pressure mm Hg Noncondensables nitrogen 4.68 28.01 0.1671 0.354 269 Condensables methyl chloride 12.67 50.49 0.2509 0.532 405 n-hexane 3.47 86.17 0.040252 0.0854 65 toluene 1.20 92.13 0.013006 0.0276 21 Totals
Solution The efficiency of convergence is always improved if one first approximates of the molar composition of the condensate that might be close to th e final composition as an initial guess.Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-32 Table 4-15. Initial Estimate of Condensate Compound Pi, mm Hg v i 1 - v i n ,i,F n ,i,L x i methyl chloride 2089 0.9730 0.0270 0.2509 0.0068 0.115 n-hexane 50 0.0233 0.9767 0.040252 0.0393 0.666 toluene 8 0.0037 0.9963 0.013006 0.0130 0.219 Totals 2147 1.0000 0.0591
Iteration 1

Components Fi lb-m n i,L lb-m Xi fraction Pi mm Hg Vi lb-m Yi fraction Li+Vi lb-m Ф (Li+Vi)/Fi

lb-m ni, k+1 nitrogen 0.1670 269.00 0.1670 0.4942 0.1670 1.0000 methyl chloride 0.2509 0.0068 0.1151 240.36 0.1492 0.4415 0.1560 0.6218 0.0109 n-hexane 0.0403 0.0393 0.6650 33.25 0.0206 0.0611 0.0599 1.4892 0.0264 toluene 0.0130 0.0130 0.2200 1.76 0.0011 0.0032 0.0141 1.0835 0.0120 Totals 0.0591 1.0000 544.37 0.3380 1.0000 P nc,k+1 = 375.56
Iteration 2

Components Fi lb-m n i,L lb-m Xi fraction Pi mm Hg Vi lb-m Yi fraction Li+Vi lb-m Ф (Li+Vi)/Fi

lb-m ni, k+1 nitrogen 0.1670 375.56 0.1670 0.4330 0.1670 methyl chloride 0.2509 0.0109 0.2217 463.14 0.2059 0.5339 0.2169 0.8644 0.0127 n-hexane 0.0403 0.0264 0.5351 26.75 0.0119 0.0308 0.0383 0.9512 0.0277 toluene 0.0130 0.0120 0.2432 1.95 0.0009 0.0022 0.0129 0.9890 0.0121 0.0493 1.0000 867.39 0.3857 1.0000 P nc,k+1 = 329.06
Iteration 3

Components Fi lb-m n i,L lb-m Xi fraction Pi mm Hg Vi lb-m Yi fraction Li+Vi lb-m Ф (Li+Vi)/Fi

lb-m ni, k+1 nitrogen 0.1670 329.06 0.1670 0.3824 0.1670 methyl chloride 0.2509 0.0127 0.2408 503.11 0.2553 0.5847 0.2680 1.0681 0.0118 n-hexane 0.0403 0.0277 0.5282 26.41 0.0134 0.0307 0.0411 1.0223 0.0271 toluene 0.0130 0.0121 0.2309 1.85 0.0009 0.0021 0.0131 1.0048 0.0121 0.0525 1.0000 860.43 0.4367 1.0000 P nc,k+1 = 290.65Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-33 Iteration 20
Components Fi lb-m n i,L lb-m Xi fraction Pi mm Hg Vi lb-m Yi fraction Li+Vi lb-m Ф (Li+Vi)/Fi

lb-m ni, k+1 nitrogen 0.1670 298.81 0.1670 0.3931 methyl chloride 0.2509 0.0097 0.2070 432.33 0.2416 0.5688 0.2513 1.0016 0.0097 n-hexane 0.0403 0.0252 0.5391 26.96 0.0151 0.0355 0.0403 1.0006 0.0252 toluene 0.0130 0.0119 0.2539 2.03 0.0011 0.0027 0.0130 1.0001 0.0119 0.0468 1.0000 760.13 0.4248 1.0000 P nc,k+1 = 298.76
Iteration 21
Components Fi lb-m n i,L lb-m Xi fraction Pi mm Hg Vi lb-m Yi fraction Li+Vi lb-m Ф (Li+Vi)/Fi

lb-m ni, k+1 nitrogen 0.1670 298.76 0.1670 0.3933 methyl chloride 0.2509 0.0097 0.2068 431.94 0.2414 0.5686 0.2511 1.0008 0.0097 n-hexane 0.0403 0.0252 0.5392 26.96 0.0151 0.0355 0.0403 1.0003 0.0252 toluene 0.0130 0.0119 0.2540 2.03 0.0011 0.0027 0.0130 1.0001 0.0119 0.0467 1.0000 759.69 0.4247 1.0000 P nc,k+1 = 298.88 After Iteration 21, all component values for Ф are between 0.999 and 1.001 and imply that the maximum molar change that each condensable component under goes between iterations is smaller than 0.1%, as shown in Table 4-16. Table 4-16. Convergence Determination Component Ф (Li+Vi)/Fi ( Ф -1)% nitrogen methyl chloride 1.0008 0.08% n-hexane 1.0003 0.03% toluene 1.0001 0.01% The final results after Iteration 21 are presented in Table 4-17 with the material balance displayed in pound moles (lb-m) and in pounds (lb) for the exit va por and liquid streams. The composition of the exit vapor streams is also listed by the average pa rtial pressure for each component being emitted. Table 4-17. Material Balance for Condenser Compound MWt n i,F lb-m/hr F i lb/hr n i,L lb-m/hr Li lb/hr n i,V lb-m/hr V i lb/hr p i mm Hg nitrogen 28.01 0.167 4.68 0 0.1670 4.68 298.76 methyl chloride 50.49 0.251 12.67 0.0097 0.49 0.2414 12.18 431.94 n-hexane 86.17 0.040 3.47 0.0252 2.17 0.0151 1.30 26.96 toluene 92.13 0.013 1.20 0.0119 1.09 0.0011 0.11 2.03Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-34 Totals 0.471 22.02 0.0468 3.75 0.4247 18.27 759.69 The results from the Material Balance Correction Method for this problem are very near the results that were obtained for the same problem using the Modifi ed Rachford-Rice Method (Illustration 4-5) and the Partial Pressure Correction Method (Illustration 4-6). 4.2.3.4 Non-ideal Liquid If a process mixture is known to exhibit non-ideal beha vior then this specific information can be used within the condenser modeling framework. Illustration 4-8: Vent condenser with azeotrope, material balance correction method. A vent condenser, operating at 18.7°C, is used in an atmospheric distillation involving ethyl acetate and ethanol. The process vent stream to the vent condenser is at 30°C and contains ethyl acetate (1.10 lb/hr), ethanol (1.30 lb/hr), and nitrogen (1.56 lb/hr). [Hougen, Watson, & Ragatz]. Ethyl acetate and ethanol are known to exhibit a minimum boiling point azeotrope with a composition of 73.4 % ethyl acetate and 26.6% ethanol (concentrations in mole percent) at the 18.7°C temperature. Estimate the discharge vent emissions from the condenser. Data Provided Pure component vapor pre ssure and other conditions: P ethyl acetate 66.19 mm Hg 18.7°C P ethanol 36.36 mm Hg 18.7°C P system 760 mm Hg discharge pressure Time 1 hr time for event Inlet process stream to the condenser: Compound Rate (lb/hr) Molecular Weight Nitrogen 1.56 lb/hr 28.013 Ethyl Acetate 1.10 lb/hr 88.10 Ethanol 1.30 lb/hr 46.07 Azeotrope Data Temperature 18.7°C Ethyl Acetate 0.734 mole fraction Ethanol 0.266 mole fraction Total Vapor Pressure 77.4 mm Hg ethyl acetate + ethanol Combined Inlet Material Flow to Condenser The combined inlet streams to the condenser are an alyzed using standard gas property relationships:Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-35 i i F i 

### MWt F n =

, 

# ∑ = F F i i 

### n n y , i i i 

### P y p =
Table 4-18. Condenser Inlet Stream Analysis for 1.0 hour of Operation Compound F i lb/hr Molecular Weight n i, F lb-mole y i mole fraction p i mm Hg Nitrogen 1.56 28.013 0.056 0.578 439.070 Ethyl Acetate 1.10 88.100 0.012 0.130 98.445 Ethanol 1.30 46.070 0.028 0.293 222.485 Totals 0.096 1.000 760.00 Equilibrium Considerations and Es timating Activity Coefficients If azeotrope data is available for a two component li quid mixture then either the van Laar or Margules equations may be applied and used to calculate activity coefficients for ethyl acetate and ethanol during the iterative processing of the condenser calculation. Other thermodynamic models may alternatively be used depending upon the availability of data.

2

2 1 1 

### 1 ln ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + =

x x β α α γ 2

1 2 2 

### 1 ln ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + =

x x α β β γ Eq. 4-48

Where α and β are calculated from known experimental data.

1 2

1 1 2 2 

### ln

ln ln 1 

### γ γ γ α ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + = x x 2 2

2 2 1 1 

### ln

ln ln 1 

### γ γ γ β ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + = x x Eq. 4-49
And γ i can be calculated since x i = y i for each component at the azeotrope.

i vp sat i i vp sat i i 

### P P P x P y = = 

### γ Eq. 4-50 From the azeotrope data that has been supplie d, the following calculations can be made:


### 16

. 1 19

. 66 4

. 77 1 

### =

= mmHg mmHg 

### γ 13

. 2 36

. 36 4

. 77 2 

### =

= mmHg mmHg 

### γ ethyl acetate ethanol 

## ( ) ( ) ( ) 202

. 1 16

. 1 ln 16

. 1 ln 734

. 0 13

. 2 ln 266

. 0 1 2



### = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + = 

### αChapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-36 

## ( ) ( ) ( ) 797

. 1 13

. 2 ln 13

. 2 ln 266

. 0 16

. 1 ln 734

. 0 1 2



### = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + = 

### β

2

2 1 1 

### 797

. 1 202

. 1 1 202

. 1 ln ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + =

x x 

### γ 2

1 2 2 

### 202

. 1 797

. 1 1 797

. 1 ln ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + =

x x 

### γ Eq. 4-51

Where 

## ( ) 1 1 

### ln exp γ γ 

### = 

## ( ) 2 2 

### ln exp γ γ 

### = Eq. 4-12 Activity coefficients and vapor pressures for ethyl acetate and ethanol are calculated and plotted in Figure 4-5 and Figure 4-6.

Act ivit y Coef f icient s f rom van Laar Equat ion

0 1 2 3 4 5 6 7

0% 20% 40% 60% 80% 100% Ethyl Acetate M olar Concentration Et Ac Et OH Vapor Pressure vs Condensate Concentration

0 10 20 30 40 50 60 70 80 90

0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100% Ethyl Acetate M olar Concentration EtAc EtOH EtAc + EtOH Figure 4-5. Activity coefficients calculated for

ethyl acetate and ethanol from the van Laar equation for the complete composition range. Figure 4-6. Component vapor pressures calculated for ethyl acetate and ethanol based on liquid composition, activity coefficients, and pure component vapor pressures.
Testing for condensation For this problem, the inlet process stream at the ve nt condenser has been analyzed and is shown in Table 4-18. A van Laar equation has been develope d based on the known azeotrope properties of the ethyl acetate and ethanol mixture (Figure 4-5) and can now be used to estimate component vapor pressures at different condensate compositions (Figure 4-6). One problem that exists for completing a mathematical test for condensation is that the values of the activity coefficients corresponding vapor pressures fo r ethyl acetate and ethanol are not known since the composition of the liquid condensate is not initially know n. However, if condensation is to occur then the first drop of condensate formed would be at equ ilibrium with the process vapor phase at the vent condenser conditions. Therefore, the composition of the first drop of condensate would have the same vapor pressure composition as the vapor stream from the process. Since a van Laar model has been established for the ethyl acetate – ethanol system then it is possible to back calculate the equilibriumChapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-37 condensate composition assuming that the dew point fo r the process vapor stream is reached at the 18.7°C. The vapor composition for the ethyl acetat e-ethanol system was calculated based on the thermodynamic model Eq. 4-51 and is plotted as a function of the liquid composition in Figure 4-7. This plot represents the equilibrium relationship between the vapor and the corresponding liquid and provides a convenient way to gauge the composition of the co ndensate at dew point conditions for the inlet condenser stream. The partial pressures of ethyl acetate and ethanol at th e condenser inlet conditions are used to calculate the relative mole fractions of ethyl acetate and etha nol that would exist at dew point conditions. These calculations are shown in Table 4-19 where the rela tive ethyl acetate concentration is 0.307 or 30.7%. Table 4-19. Calculated Mole Fractions for Only the Condensable Portion of the Inlet Process Vapor Stream Compound p i mm Hg y i Condensables Ethyl Acetate 98.445 0.307 Ethanol 222.485 0.693 Totals 320.930 1.000 We are interested in locating the point on the plot in Figure 4-7 where y = 30.67% ethyl acetate that corresponds to the condensable composition of the inlet process vapor phase from Table 4-19. The x coordinate at the intercept corresponds to the initia l condensate composition at the vapor dew point. By inspection, this value is between 7% and 8% ethyl ace tate. Alternatively, it is possible to use a goal seek routine in a spread sheet to back calculate th e condensate concentration at dew point as 7.66% (mole/mole) ethyl acetate.
van Laar Model for Ethyl Acetate - Ethanol (18.7°C)

0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%

0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100 % Ethyl Acetate Liquid Composition (mole/mole)Ethyl Acetate Vapor Composition

Figure 4-7. Ethyl acetate vapor composition plotted as a function of ethyl acetate composition in the liquid. The condensation test can now be comple ted using the following relationships.Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-38 phase liq n
i i i i n
i i 

### P p x ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ < = 

# ∑ ∑

=

= 

### 0

. 1 1 1 

### γ phase liq n
i i i i n
i i 

### P p x ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ≥ = 

# ∑ ∑

=

= 

### 0

. 1 1 1 

### γ
No Condensation Present Condensation Present Results from this analysis are shown in Table 4-20. Table 4-20. Condensate Feasibility Analysis

Inlet Partial Pressure (mm Hg) Activity Coefficient Pure Component Vapor Pressure mm Hg@ 18.7°C Mole Fraction

i i i i 

### P p x 

### γ = Ethyl Acetate 98.445 2.945374 66.19 0.50 Ethanol 222.485 1.004975 36.36 6.09 

# ∑

= n
i i 

### x 1 = 6.59
The sum of the theoretical liquid phase mole fractions for the condensate is 6.59 and therefore, one can assume that condensation will take place in the vent condenser. Solving for a single phase condensate using the Material Balance Correction Method In the material balance correction method, the tota l amount of each component leaving the single stage device in liquid and vapor form is divided by the amount of that same component that is entering the device. The moles of each component in the condensat e are corrected by the resulting ratio to obtain the corrected value to use for the next iteration, k+1.

i F i V i L i 

### n n n φ = +

, , , Eq. 4-53

k i k L i k L i 

### n n

, , , 1 , , 

### φ = + Eq. 4-54 Eq. 4-53 and Eq. 4-54 are repeated until the error of estim ate for each component has fallen within an acceptable tolerance. The iteration results for the vent condenser problem are presented in Table 4-21. The initial guess for the condensate was set equal to the com position of the inlet gas stream.Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-39 Table 4-21. Iterative Solution to Determine Condensate Composition Iteration 1 n i x i γ i p i n i,V n i,L+V n i,F Φ -1 n i,k+1 Nitrogen 690.8 0.05569 0.05569 0.0557 1.0000 0.05569 EtAC 0.01249 0.3067 2.047 41.6 0.00335 0.01584 0.0125 0.7884 0.00984 EtOH 0.02822 0.6933 1.098 27.7 0.00223 0.03045 0.0282 0.9267 0.02615

Iteration 2 n i x i γ i p i n i,V n i,L+V n i,F Φ -1 n i,k+1 Nitrogen 692.6 0.05569 0.05569 0.0557 1.0000 0.05569 EtAC 0.00984 0.2735 2.155 39.0 0.00314 0.01298 0.0125 0.9619 0.00947 EtOH 0.02615 0.7265 1.075 28.4 0.00228 0.02843 0.0282 0.9924 0.02595

Iteration 3 n i x i γ i p i n i,V n i,L+V n i,F Φ -1 n i,k+1 Nitrogen 693.0 0.05569 0.05569 0.0557 1.0000 0.05569 EtAC 0.00947 0.2673 2.175 38.5 0.00309 0.01256 0.0125 0.9939 0.00941 EtOH 0.02595 0.7327 1.072 28.5 0.00229 0.02825 0.0282 0.9990 0.02593

Iteration 4 n i x i γ i p i n i,V n i,L+V n i,F Φ -1 n i,k+1 Nitrogen 693.0 0.05569 0.05569 0.0557 1.0000 0.05569 EtAC 0.00941 0.2663 2.179 38.4 0.00309 0.01250 0.0125 0.9991 0.00940 EtOH 0.02593 0.7337 1.071 28.6 0.00230 0.02822 0.0282 0.9999 0.02592

Iteration 5 n i x i γ i p i n i,V n i,L+V n i,F Φ -1 n i,k+1 Nitrogen 693.0 0.05569 0.05569 0.0557 1.0000 0.05569 EtAC 0.00940 0.2662 2.179 38.4 0.00309 0.01249 0.0125 0.9999 0.00940 EtOH 0.02592 0.7338 1.071 28.6 0.00230 0.02822 0.0282 1.0000 0.02592

Iteration 6 n i x i γ i p i n i,V n i,L+V n i,F Φ -1 n i,k+1 Nitrogen 693.0 0.05569 0.05569 0.0557 1.0000 0.05569 EtAC 0.00940 0.2661 2.180 38.4 0.00309 0.01249 0.0125 1.0000 0.00940 EtOH 0.02592 0.7339 1.071 28.6 0.00230 0.02822 0.0282 1.0000 0.02592 Table 4-22. Condenser Material Balance at Completion of Iterations Feed Condensate Out Vapor Out Iteration 6 F i, lb n i,L lb-mole L i, lb n i,V lb-mole V i, lb L i + V i Material Balance Nitrogen 1.56 0 0 0.0557 1.56 1.5600 100% Ethyl Acetate 1.10 0.0094 0.8282 0.0031 0.27 1.1000 100% Ethanol 1.30 0.0259 1.1942 0.0023 0.11 1.3000 100%Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-40 Table 4-23. Exit Vent Stream Analysis Table (Based on the final results from Table 4-22 and Iteration #6) Compound p i , mm Hg M. Wt. Lb-Moles V i Nitrogen 693.0 28.013 0.0557 1.56 lb Ethyl Acetate 38.4 88.1 0.0031 0.27 lb Ethanol 28.6 46.07 0.0023 0.11 lb
4.2.4 Vent Condenser with Two Phase Condensate When two liquid phases are generated, the problem ca n be solved algebraically because the two liquid phases must be in equilibrium. This situation is ex amined in illustration 4- 9. The problem is more complicated with additional liquid phases. The modifiedRachford Rice equation would not be applicable, but either the vapor pressure correction method or the material balance correction method could be used. This more complicated situation is beyond the current scope of this document. Illustration 4-9: Vent condenser with two phase condensate A condenser is used at atmospheric pressure (760 mm Hg). The inlet process gas stream contains nitrogen (5.0 lb/hr), water (10.0 lb/hr), and toluene (20.0 lb/h r). The condenser temperature is such that the exit streams are at 20°C. Using the simplified condensate analysis approach, calculate the emissions (V i ) and the material balance for this condenser. Data Provided P t

21.827 mm Hg 20°C P w 17.351 mm Hg 20°C P sys 760 mm Hg System pressure Time 1 hr Time for event Inlet Stream: Compound Rate (lb/hr) Molecular Weight Nitrogen 5.0 lb/hr 28.013 Water 10.0 lb/hr 18.02 Toluene 20.0 lb/hr 92.13 The inlet gas stream at the condenser is analyzed a nd the rate in lb-mole/hr, mole fraction, and partial pressures are calculated using the standard equations.

i i F i 

### MWt F n =

, 

# ∑ = F F i i 

### n n y , i i i 

### P y p =
Table 4-24. Inlet Stream Analysis over 1.0 Hour Compound F i Rate (lb/hr) Molecular Weight n i,F (lb-mole) Mole Fraction Partial Pressure (mm Hg) Nitrogen 5.0 lb/hr 28.013 0.1785 0.188 142.72Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-41 Water 10.0 lb/hr 18.02 0.5549 0.584 443.69 Toluene 20.0 lb/hr 92.13 0.2171 0.228 173.59 Totals 0.9505 1.000 760.00 Equilibrium Considerations For a single liquid phase, the following equation can be used to calculate the equilibrium vapor pressure

i 

### p for component i in the liquid.

i i i i 

### p P x =



### γ Eq. 4-55 Where i 

### x is the liquid mole fraction of component i ,

i 

### γ is the activity coefficient of component i ,

i 

### P is the pure component vapor pressure of component i at the mixture temperature, and For two liquid phases in equilibrium, the vapor pressure i 

### p of component 

### i is the same for each phase.

2 , 1 , i i 

### p p = Eq. 4-56 Where i 

### x is the liquid mole fraction of component i ,

i 

### γ is the activity coefficient of component i ,

i 

### P is the pure component vapor pressure of component i at the mixture temperature. Substituting the

i i i i i i 

### P x P x 2 , 2 , 1 , 1 , 

### γ γ = Eq. 4-57

2 , 2 , 1 , 1 , i i i i 

### x x 

### γ γ = Eq. 4-58 Therefore, if a two phase liquid exists of known composition and the activity coefficient 1 , i 

### γ is known then 2 , i 

### γ may be estimated as shown.

1 , 2

, 2 , 1 , 2

, 2 , 1 , i i i i i i

i i i 

### x x P x P x 

### γ γ γ = = Eq. 4-59 Activity coefficients for water and toluene in a two liquid phase mixture can be estimated using these relationships between the two liquid phases. At 20°C water and toluene form a two liquid phase mixture with the following composition. Table 4-25. Liquid Phase Solubility Data for 20°C NonAqueous Phase Aqueous Phase Toluene 0.9995 (wt/wt) 0.000515 (wt/wt)Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-42 Water 0.0005 (wt/wt) 0.999485 (wt/wt) The equilibrium concentrations listed in Table 4-26 can be converted to equivalent mole fractions using Eq. 4-.

j j i i i i i 

### MWt wt MWt wt MWt wt x + = Eq. 4-13
Table 4-26. Equilibrium Mole Fraction Concentrations Non-Aqueous Phase molar concentration Aqueous Phase molar concentration Toluene 99.745% 0.010% Water 0.255% 99.990% Estimating activity coefficien ts for liquid phase components The activity coefficients for toluene and water can be estimated for each liquid phase using the relationship in Eq. 4-58. For the aqueous phase it is reasonabl e to assume that water exhibits an ideal behavior and that the activity coefficient for water is 1.0 since the concentration of water is close to 100%. Similarly, the activity coefficient for toluene in the non-aqueous phase can be assumed to be 1.0. The activity coefficient for toluene in the aqueous phase may then be calculated.

1 , 2

, 2 , 1 , i i i i 

### x x 

### γ γ = 2 , 1

, 1 , 2 , i i i i 

### x x 

### γ γ = Eq. 4-61 Calculate the activity coefficient for toluene in the aqueous phase (phase #1) by applying Eq. 4-61. 

## ( ) ( ) ( ) 5

. 9974 0001

. 0 0

. 1 99745

. 0 1 , 

### =

= t 

### γ Eq. 4-62 Calculate the activity coefficient for water in the non-aqueous phase (phase #2) using Eq. 4-61. 

## ( ) ( ) ( ) 392.1 00255

. 0 0

. 1 9999

. 0 2 , 

### = = w 

### γ Eq. 4-63
Table 4-27. Estimated Activity Coefficients Non-Aqueous Phase Aqueous Phase Toluene 1.0 9974.5 Water 392.1 1.0Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-43 Evaluating whether condensation will take place in the condenser and if so will one or two liquid phases be formed For this problem, the combined process stream thr ough the condenser (air leakage, toluene vapors, and water vapors) has been analyzed and is shown in Tabl e 4-24. Now that the activity coefficients for toluene and water have been estimated and are shown in Tabl e 4-27 for a two liquid phase mixture, it is important to evaluate whether condensation will occur in the conde nser at all or if so, will one or two liquid phases be formed. At equilibrium conditions, Eq. 4-55 applies for each liquid phase that might be present. Eq. 4-55 can be modified so that x i can be calculated from the conditions at equilibrium. The sum of all components in the liquid phase must equal to 1.0 and so Eq. 4-66 can be written. However, at this stage of the problem, equilibrium compositions have not yet been determined for the problem. Eq. 4-66 can be used in conjunction with the co mponent partial pressures in the inlet stream and the pure component vapor pressures at the condenser te mperature to determine if condensation should be considered for a specific liquid phase. If the calculated value for Eq. 4-66 is greater than 1.0, then one can initially assume that condensation will occur for th e liquid phase being considered. On the other hand, if the calculated value for Eq. 4-66 is less than 1.0 then one can assume that condensate will not occur for the liquid phase being studied.

i i i i 

### p P x =



### γ Eq. 4-64

i i i i 

### P p x γ 

### = Eq. 4-65

phase liquid n
i i i i n
i i 

### P p x ⎥ ⎦ ⎤ ⎢ ⎣ ⎡

= = 

# ∑ ∑

=

= 

### 0

. 1 1 1 

### γ Eq. 4-66 The feasibility that condensate will fo rm in the condenser is determined for an aqueous liquid phase and a non-aqueous liquid phase using Eq. 4-66. Results from these calcula tions are shown in Table 4-28 and Table 4-29. Table 4-28. Aqueous Phase Condensate Feasibility Analysis
Inlet Partial Pressure (mm Hg) Activity Coefficient Pure Component Vapor Pressure mm Hg@ 20°C Mole Fraction
i 

### x Water 443.69 1 17.351 25.57 Toluene 173.59 9974.5 21.827 0.0008 

# ∑

= n
i i 

### x 1 = 25.57
The sum of the theoretical liquid phase mole frac tions for an aqueous phase condensate is 25.57 and greater than 1.0. Therefore, we can assume that a condensate liquid mixture will form and contain at least an aqueous phase.Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-44 Table 4-29. Non-Aqueous Phase Condensate Feasibility Analysis

Inlet Partial Pressure (mm Hg) Activity Coefficient Pure Component Vapor Pressure mm Hg@ 20°C Calculated Mole Fraction
i 

### x Water 443.69 392.1 17.351 0.07 Toluene 173.59 1.0 21.827 7.95 

# ∑

= n
i i 

### x 1 = 8.02
The sum of the theoretical liquid phase mole fracti ons for a non-aqueous phase is 8.02 and also greater then 1.0. Therefore, a condensate mixture from th is condenser will likely have a non-aqueous phase as well as an aqueous liquid phase. During the course of iterating to determine the conden ser solution, it is possible for the sum of the x i values might fall below 1.0 for one of the liquid ph ases, in which case that phase would no longer be considered as feasible. This is because, as itera tions are taking place and the separate liquid phases are being formed (by way of the math), it is possible for the residual amount of material in vapor phase to fall below a level that can continue to suppor t the presence of one of the liquid phases. Solving for a two liquid phase condensate The condenser is being treated as a single stage device and so equilibrium is assumed to exist between the exit vent gas and the liquid condensate. Equilibrium is also assumed to exist between the two separate liquid phases in the condensate as well. Therefore, th e vapor pressure of toluene and water for each liquid phase is identical as shown by Eq. 4-56 and Eq. 4-57. Phase solubility data for toluene and water at 20°C are given in Table 4-25 and Table 4-26 while the activity coefficients are calculated and shown in Table 4-27. As long as there is enough toluene and water in the system to support two separate liquid phases in the condensate then the molar composition of each phase is governed by the phase solubility data. The calculated partial pressure shown in Table 4-30 will be the same regardless how much toluene and water are present in the system as long as there is enough to support two liquid phases in the condensate. Table 4-30. Composition and Vapor Pressure Analysis of the Two Phase Condensate Nonaqueous Phase Aqueous Phase Pure VP i Partial VP i Component X i γ i X i γ i P i p i Toluene 0.9975 1.0 0.0001 9974.5 21.827 21.772 Water 0.0025 392.1 0.9999 1.0 17.351 17.349 One can calculate the partial pressure of nitrogen in the exit vapor stream by subtracting the partial pressure of toluene and water from the system pressure as shown. Exit Vapor Analysis

w t N 

### p p p − − = 0

. 760 2 Eq. 4-67 

### mmHg p N 

### 879

. 720 349

. 17 772

. 21 0

. 760 2 

### = − − =Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-45 

### moles lb p p n n N t N V t 

### − = = = 0054

. 0 879

. 720 772

. 21 1785

. 0 2 2 ,


### moles lb p p n n N w N V w 

### − = = = 0043

. 0 879

. 720 349

. 17 1785

. 0 2 2 , Exit Liquid Analysis The moles of condensate can be determined by subtrac ting the moles of toluene and water that exit in the vapor phase from the moles of toluene and wa ter that exit in the two liquid phases. 

### moles lb n n n V t F t t 

### − = − = − = 2117

. 0 0054

. 0 2171

. 0

, , 2 & 1 , Eq. 4-68 

### moles lb n n n V w F w w 

### − = − = − = 5506

. 0 0043

. 0 5549

. 0

, , 2 & 1 , Eq. 4-69

n T,1&2 and n W,1&2 are the moles of toluene and water that ar e contained in the combined two liquid phases. Equations can be written that relate the compos ition of the aqueous phase and the nonaqueous phase to the total moles of toluene and water as liquid.

2 & 1 , 2 2 , 1 1 , t t t 

### n L x L x = + Eq. 4-70

2 & 1 , 2 2 , 1 1 , w w w 

### n L x L x = + Eq. 4-71 L 1 are the moles of nonaqueous phase and L 2 are the moles of aqueous phase in the condensate. The objective is to now solve Eq. 4-70 and Eq. 4-71 for L 1 and L 2 and then calculate th e moles of toluene and water that are contained in each phase based on th e known solubility data. This can be accomplished through algebraic manipulation of Eq. 4-70 and Eq. 4-71 as shown:

2 & 1 , 2 , 2 2 , 2 , 1 1 , 2 , t w t w t w 

### n x L x x L x x = + Eq. 4-72

2 & 1 , 2 , 2 2 , 2 , 1 1 , 2 , w t w t w t 

### n x L x x L x x = + Eq. 4-73

2 & 1 , 2 , 2 & 1 , 2 , 1 1 , 2 , 1 1 , 2 , w t t w w t t w 

### n x n x L x x L x x − = − Eq. 4-74

1 , 2 , 1 , 2 , 2 & 1 , 2 , 2 & 1 , 2 , 1 w t t w w t t w 

### x x x x n x n x L

− − = Eq. 4-75

L 1 and L 2 can finally be calculated using Eq. 4-76 and Eq. 4-77. Molar quantities of toluene and water can then be calculated once L 1 and L 2 are known. 

### moles lb L

− = = ∗
− ∗ ∗
− ∗ = 0.2117 0.9998 0.2116 0025

. 0 0001

. 0 9975

. 0 9999

. 0 5506

. 0 0001

. 0 2117

. 0 9999

. 0 1 Eq. 4-76


### moles lb L L L

− = = − = 0.5506 0.2067

- 0.7623 1 2 & 1 2 Eq. 4-77Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-46 Nonaqueous Phase 

### moles lb L x n t

t 

### − = ∗ = ∗ = 0.2111 0.2117 9975

. 0 1 1 , 1 , 

### moles lb x L x n w

w 

### − = ∗ = ∗ = -4 1 1 , 1 , 

### 10 5.292 2117

. 0 0025

. 0 Aqueous Phase 

### moles lb L x n t

t 

### − = ∗ = ∗ = -5 2 2 , 2 , 

### 10 x 5.506 5506

. 0 0001

. 0 

### moles lb L x n w

w 

### − = ∗ = ∗ = 0.5506 5506

. 0 9999

. 0 2 2 , 2 , Table 4-31. Exit Liquid Stream Analysis Nonaqueous Aqueous X i n 1 (lb-moles) X i n 2 (lb-moles) Toluene 0.9975 0.2111 0.0001 5.5506 x 10 -5 Water 0.0025 5.292 x 10 -4 0.9999 0.5506 Totals 0.2117 0.5507 Table 4-32. Condenser Material Balance at Completion of Iterations Nonaqueous Aqueous Vapor Total In Total Out Material n i,1 lb n i,2 lb n i,V lb Lb Lb Balance Nitrogen 0.000 0.000 0.1785 5.00 5.000 5.00 100.0% Toluene 0.2111 19.45 5.5506x10 -5 5.11x10 -3 0.0054 0.498 20.000 19.95 99.8% Water 5.292x10 -4 9.54x10 -3 0.5506 9.92 0.0043 0.077 10.000 10.01 100.1% 0.2117 0.5507 Table 4-33. Exit Vent Stream Analysis Table

(Based on the final results from Table 4-32) Compound mm Hg MWt Lb-Moles Lb Nitrogen 720.9 28.013 0.1785 5.000 Toluene 21.8 92.13 0.0054 0.498 Water 17.4 18.02 0.0043 0.077 4.2.5 Liquid-Ring Vacuum Pump In a liquid-ring vacuum pump the gas stream mixes wi th the fluid (typically water) used to operate the pump. As a result some of the soluble compounds in the gas stream will be absorbed into the fluid. The procedures for estimating the amount of each soluble compound that is absorbed and emitted is the same as for the condensers described in previous sections in this document. Just as in the condenser examples, the composition of the exit liquid from the liquid-ring vacuum pump is unknown and must be determined by trial and error. The modified Rachford-Rice equa tion, the partial pressure correction method, or theChapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-47 material balance correction method could be used to solve the problem. Illustration 4-10 uses the partial pressure correction method. Illustration 4-10: Liquid-Ring Vacuum Pump with multi-component process stream. A liquid-ring vacuum pump is being used to main tain a distillation operation at 100 mm Hg absolute pressure. The inlet process vapor stream at the vacuum pump consists of air (0.9 5 lb/hr), methanol (8.56 lb/hr), ethanol (6.35 lb/hr), and isopropyl alcohol (4.25 lb/hr) vapors. The exit liquid and vapor from the vacuum pump discharges to atmospheric conditions at 25°C. External air is being bleed into the vacuum pump at a rate of 2.60 lb/hr for control purposes to help maintain the 100 mm Hg absolute pressure. The liquid -ring vacuum pump is operated using water with a fresh water make-up rate of 480.0 lb/hr. Spent wa ter from the vacuum pump and travels to waste treatment at 25°C. What would the emissions and the material balance for the vacuum pump be over a one hour period under these operating conditions? Assume that liquid phase in the vacuum pump behaves as an ideal mixture. An analysis of the problem requires that the inlet process stream, the air bleed, and the make-up water must all be taken into account. The three streams enter the vacuum pump, become mixed, and then separate out as they exit the unit. The resulting g as and liquid streams that exit the vacuum pump are considered to be in equilibrium. Table 4-34 prov ides a component break down of the combined process stream present in the vacuum pump. Table 4-34. Analysis of the Inlet Stream to the Liquid-Ring Vacuum Pump
Source F i lb/hr MWt n i,F lb-m/hr z i Noncondensable Air Process 0.95 Air External Bleed 2.60 Air (total) 3.55 28.97 0.1225 0.0045 Condensable

Methanol Process 8.56 32.04 0.2672 0.0098 Ethanol Process 6.35 46.07 0.1378 0.0051 Isopropyl Alcohol Process 4.25 60.096 0.0707 0.0026 Water External Make-Up 480.00 18.02 26.6371 0.9780
Solution In preparing to solve this problem using the Partial Pr essure Correction method, we must first arrive at an initial guess for the composition of the liquid that w ill discharge from the vacuum pump. Upon inspection of Table 4-34, it becomes apparent that the bulk of th e inlet stream (97.8%) consists of water (480 lb/hr) and that the remainder of the components (includi ng air) makes up only 2.2%. Additionally, the water exits the vacuum pump at 25°C and at atmospheric pressure which means that approximately 97.8% of the total discharge from the vacuum pump will be in the form of liquid. Additionally, the vacuum pump discharges at 760 mm Hg pressure which is significantly higher than the vapor pressure of water (23.6 mm Hg). Therefore, one would not be very far away from the discharge liquid composition if 100% of the condensable components were chosen as an initial guess.Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-48 Table 4-35. Analysis of the Inlet Stream a nd Initial Guess of the Exit Liquid Composition

Component P i @ 25°C mm Hg Feed from Table 4-34 n i,F lb-moles/hr Initial Guess n i,L lb-moles/hr Noncondensable Air (total) N/ A 0.1225 N/A Condensable

Methanol 122.278 0.2672 0.2672 Ethanol 57.6 0.1378 0.1378 Isopropyl Alcohol 45.152 0.0707 0.0707 Water 23.575 26.6371 26.6371 Newton’s method can readily accommodate this initial guess since it would result in 

# ( ) L i V i L i 

### p p p
, , , 

### = − since V i 

### p , would be initially set to 0. Moles of each component would immediately begin being moved from the liquid phase to the vapor phase upon the first ite ration. Only three iterations were required to calculate a solution to the desired tolerance as shown:
Iteration 1 Components n i,L,k x i p i,L n i,V y i p i,V p i,L – p i,V Δ n i,L n i,L,k+1

Air 735.223 0.1225 1.0000 760.000 Methanol 0.2672 0.00985 1.205 0.0000 0.00E+00 0.000 1.20E+00 1.94E-04 0.2670 Ethanol 0.1378 0.00508 0.293 0.0000 0.00E+00 0.000 2.93E-01 4.72E-05 0.1378 IPA 0.0707 0.00261 0.118 0.0000 0.00E+00 0.000 1.18E-01 1.90E-05 0.0707 Water 26.6371 0.98245 23.161 0.0000 0.00E+00 0.000 2.32E+01 3.69E-03 26.6334
Iteration 2 Components n i,L,k x i p i,L n i,V y i p i,V p i,L – p i,V Δ n i,L n i,L,k+1

Air 735.224 0.1225 0.9688 736.263 methanol 0.2670 0.00985 1.204 0.0002 1.53E-03 1.166 3.78E-02 6.29E-06 0.2670 ethanol 0.1378 0.00508 0.293 0.0000 3.73E-04 0.284 9.18E-03 1.53E-06 0.1378 IPA 0.0707 0.00261 0.118 0.0000 1.50E-04 0.114 3.69E-03 6.15E-07 0.0707 water 26.6334 0.98246 23.162 0.0037 2.92E-02 22.173 9.89E-01 1.68E-04 26.6332
Iteration 3 Components n i,L,k x i p i,L n i,V y i p i,V p i,L – p i,V Δ n i,L n i,L,k+1

air 735.224 0.1225 0.9674 735.240 methanol 0.2670 0.00985 1.204 0.0002 1.58E-03 1.203 1.62E-03 2.71E-07 0.2670 ethanol 0.1378 0.00508 0.293 0.0000 3.85E-04 0.292 4.05E-04 6.76E-08 0.1378 IPA 0.0707 0.00261 0.118 0.0000 1.55E-04 0.118 1.64E-04 2.73E-08 0.0707 water 26.6332 0.98246 23.162 0.0039 3.05E-02 23.148 1.38E-02 2.35E-06 26.6332Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-49 The material balance and composition equilibrium can be evaluated for iteration 3 to ensure that the convergence has been adequately me t. The material balance can be checked for each component by dividing the sum of the moles in the liquid and the vapor by the moles in the inlet stream for each component. A 100% material balance is an integral part of the Partial Pressure Correction method since the moles of each component in the vapor phase are calc ulated by subtracting the moles that have been assigned to the liquid phase from the moles that are in the feed. This is reflected in Table 4-36 under material accountability column. Table 4-36. Material Balance Accountability Component ni,F ni,L ni,V ni,L+ni,V Accountability Air 0.1225 0.0000 0.1225 0.1225 100.00% Methanol 0.2672 0.2670 2.01E-04 0.2672 100.00% Ethanol 0.1378 0.1378 4.88E-05 0.1378 100.00% IPA 0.0707 0.0707 1.96E-05 0.0707 100.00% Water 26.6371 26.6332 3.86E-03 26.6371 100.00% 27.2353 27.1087 0.1267 27.2353 100.00%
The Partial Pressure Correction method works to mini mize the difference between the calculated partial pressure for each component based on the liquid phase composition with the calculated partial pressure of each component based on the vapor phase makeup. This convergence technique was fairly efficient for this vacuum pump problem since the two partial pressures agreed within 0.00213% after only three iterations. Table 4-37. Partial Pressure Equilibrium Accountability Component piL piV %error Air 735.2238 735.2240 -3.43E-05 Methanol 1.2042 1.2042 1.94E-03 Ethanol 0.2928 0.2928 2.10E-03 IPA 0.1178 0.1178 2.13E-03 Water 23.1615 23.1613 9.50E-04 Additional considerations regarding numerical convergence techniques The Partial Pressure Correction method proved to be a very efficient approach for this problem. When this same problem is solved using the Modified Rachford-Rice model then as many as 100 iterations are required to achieve the same degree of convergence. A main reason that the Modified Rachford-Rice method was less efficient was due to the large amount of water that was in the system. Since the process stream consists of 97.8% water then the remaining components (methanol, ethanol, and isopropyl alcohol) became very small numbers to wo rk with. The Modified Rachford-R ice approach adjusts V/F until the complete based equation is solved. However, sin ce V/F includes the methanol, ethanol, and isopropyl alcohol along with the large amount of water then V/ F needs to be evaluated to a much greater precision to accommodate for the loss in accuracy of the convergence.Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-50 The Partial Pressure Correction method focuses on th e convergence of each individual component in a concurrent fashion. So regardless whether the com ponent occupied as much as 97.8% (as in the case of water) of the process or as little as 0.0026% (as in th e case of the isopropyl alcohol) convergence process was carried out to the same degree of precision for each component. 4.2.6 Vacuum Steam Jets In a vacuum steam jet steam is mixed with the pr ocess gas stream and then condensed. As for the condensers described in earlier sections in this chapter, the composition of the condensate may be unknown, which means the composition of the exit gas stre am must be determined through trial and error. The modified Rachford-Rice equation, partial pressure correction method, or material balance correction method could be used to solve this problem. The probl em in illustration 4-11 is solved using the material balance correction method. Illustration 4-11: Vacuum steam jet – material balance correction method. A vacuum steam jet is used to maintain 100 mm Hg for a process. The inlet process gas stream contains air (1.61 lb/hr) and toluene vapors (1.53 lb/hr). Steam is feed to the vacuum jet at a rate of 120 lb/hr and is condensed in the unit as part of the normal operation. An external air bleed (1.0 lb/hr) is used to help maintain the controlled 100 mm Hg ab solute pressure at the process. The liquid condensate and the exit vapors discharge to atmospheric pressure at 55°C. Estimate the discharge vent emissions from the vacuum steam jet. Data Provided: P T 113.8 mm Hg 55°C P W 118.0 mm Hg 55°C P sys 760 mm Hg discharge pressure Time 1 hr Time for event Solubility data at 55°C: X T,ph-1 0.000168 mole fraction aqueous phase X W, ph 2 0.0063 mole fraction non-aqueous phase Inlet process stream to the vacuum steam jet: Compound Rate (lb/hr) Molecular Weight Air 1.61 lb/hr 28.97 Toluene 1.53 lb/hr 92.13 Steam and external air bleed rate to the vacuum steam jet: Compound Rate (lb/hr) Molecular Weight Water 120.0 lb/hr 18.02 Air 1.0 lb/hr 28.97 The inlet gas stream at the condenser is analyzed as follows using standard e quations and relationships:

i i F i 

### MWt F n =

, 

# ∑ = F F i i 

### n n y , i i i 

### P y p =Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-51 Table 4-38. Steam Jet Process Stream Analysis for 1.0 Hour of Operation Compound F i lb/hr Molecular Weight n i,F lb-mole y i mole Fraction p i , mm Hg Air from process 1.61 Air from bleed 1.00 Total Air 2.61 28.970 0.090 0.013 10.12 Water 120.00 18.020 6.659 0.984 748.02 Toluene 1.53 92.13 0.017 0.001 1.86 Totals 6.766 1.000 760.00 Equilibrium Considerations and Es timating Activity Coefficients Table 4-39. Equilibrium Mole Fraction Concentrations Aqueous Phase

mole fraction Non-Aqueous Phase mole fraction Toluene 0.000168 0.993697 Water 0.999832 0.006303 The activity coefficients for toluene and wate r can be estimated for each liquid phase using Eq. 4-78. For the aqueous phase it is reasonable to assume that wate r exhibits ideal behavior and that the activity coefficient is 1.0 since the concentration of water is close to 100%. Similarly, one can assume that the activity coefficient for toluene in a non-aqueous phase would be approximately 1.0. The activity coefficient for toluene in the aqueous phase and the activity coefficient for water in a non-aqueous phase may then be calculated from equations that were previously developed.

1 , 2

, 2 , 1 , i i i i 

### x x 

### γ γ = 2 , 1

, 1 , 2 , i i i i 

### x x 

### γ γ = Eq. 4-78 Where the activity coefficient for toluene in the aqueous phase (ph ase #1) is calculated as follows: 

## ( ) ( ) ( ) 5903 000168

. 0 0

. 1 993697

. 0 1 , 

### =

= t 

### γ Similarly, the activity coefficient for water in the non-aqueous phase (phase #2) is calculated. 

## ( ) ( ) ( ) 159 006303

. 0 0

. 1 999832

. 0 2 , 

### = = w 

### γ Eq. 4-79
Table 4-40. Estimated Activity Coefficients Aqueous Phase Non-Aqueous Phase Toluene 5903.0 1.0 Water 1.0 159.0Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-52 Testing for condensation at steam jet discharge conditions For this problem, the combined process stream through the vacuum jet (air leakage, toluene vapors, and water vapors) has been analyzed and is shown in Tabl e 4-38. Now that the activity coefficients for toluene and water have been estimated and are shown in Tabl e 4-40 for a two liquid phase mixture, it is important to test to know if the process discharge stream will contain one or more liquid phases. Results from this analysis are shown in Table 4-41 and Table 4-42. Table 4-41. Aqueous Phase Condensate Feasibility Analysis
Inlet Partial Pressure (mm Hg) Activity Coefficient Pure Component Vapor Pressure mm Hg@ 20°C Mole Fraction
i 

### x Water 748.02 1 118.0 6.34 Toluene 1.86 5903 113.8 2.8E-06 

# ∑

= n
i i 

### x 1 = 6.34
The sum of the theoretical liquid phase mole frac tions for an aqueous phase condensate is 6.34. Therefore, we can assume that the discharge process st ream from the vacuum jet will at least contain an aqueous phase which will consist mostly of water. Table 4-42. Non-Aqueous Phase Condensate Feasibility Analysis

Inlet Partial Pressure (mm Hg) Activity Coefficient Pure Component Vapor Pressure mm Hg@ 20°C Calculated Mole Fraction
i 

### x Water 748.02 1.0 118.0 0.04 Toluene 1.86 159.0 113.8 0.02 

# ∑

= n
i i 

### x 1 = 0.06
The sum of the theoretical liquid phase mole fractions for a non-aqueous phase is 0.06. Therefore, we can conclude that the discharge process stream will not contain a non-aqueous phase since the sum of the mole fractions is less than 1.0. Solving for a single phase condensate using the Material Balance Correction Method

i F i V i L i 

### n n n φ = +

, , , Eq. 4-80

k i k L i k L i 

### n n

, , , 1 , , 

### φ = + Eq. 4-81Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-53 For each condensable component in the inlet stream, the following steps are followed until the error of estimate for each component has falle n within an acceptable tolerance. Iterating to solve the condenser problem The iteration results are presented in Table 4-43 for the solution. The equilibrium mole fractions in the aqueous solution at 55°C (see Table 4-39) are used to determine the number of moles of condensables (i.e., water and toluene) in the condensate for the firs t iteration. The partial pressures of toluene and water in the outlet vapor, assuming equilibrium with the co ndensate are estimated using Eq. 4-6; the partial pressure of air is determined as the difference between the system pressure and the sum of the partial pressures for toluene and water. Moles of toluene an d water in the outlet vapor are estimated using Eq. 4- 5. The value of Φ , the ratio of the estimated sum of moles of a condensable in condensate and outlet vapor to the known feed moles, is estimated fo r both condensable compounds using Eq. 4-80. New estimates of the number of moles of each condensable in the condensate are calculated using Eq. 4-81. These estimates are used as the starting values for the second iteration. The process repeats until convergence is obtained. The resulting composition of the condensate and the outlet gas stream are shown in Tables 4-44 and 4-45. Table 4-43. Iterative Solution to Vacuum St eam Jet Problem with Aqueous Phase Condensate Iteration 1 n i,L,k x i p i n i,V n i,L+V n i,F Φ i,k n i,L,,k+1 Air 528.9 0.09009 0.09009 0.0901 1.0000 0.09009 Water 6.65927 0.9998 118.0 0.02010 6.67936 6.6593 0.9970 6.63923 Toluene 1.119E-03 1.68E-04 113.1 0.01926 0.02038 0.0166 0.8147 0.00091

Iteration 2 n i,L,k x i p i n i,V n i,L+V n i,F Φ i,k n i,L,,k+1 Air 549.6 0.09009 0.09009 0.0901 1.0000 0.09009 Water 6.63923 0.9999 118.0 0.01934 6.65857 6.6593 1.0001 6.63993 Toluene 9.116E-04 1.37E-04 92.4 0.01515 0.01606 0.0166 1.0339 0.00094

Iteration 3 n i,L,k x i p i n i,V n i,L+V n i,F Φ i,k n i,L,,k+1 Air 546.5 0.09009 0.09009 0.0901 1.0000 0.09009 Water 6.63993 0.9999 118.0 0.01945 6.65938 6.6593 1.0000 6.63982 Toluene 9.425E-04 1.42E-04 95.5 0.01575 0.01669 0.0166 0.9947 0.00094

Iteration 4 n i,L,k x i p i n i,V n i,L+V n i,F Φ i,k n i,L,,k+1 Air 547.0 0.09009 0.09009 0.0901 1.0000 0.09009 Water 6.63982 0.9999 118.0 0.01943 6.65925 6.6593 1.0000 6.63983 Toluene 9.375E-04 1.41E-04 95.0 0.01565 0.01659 0.0166 1.0008 0.00094

Iteration 5 n i,L,k x i p i n i,V n i,L+V n i,F Φ i,k n i,L,,k+1 Air 546.9 0.09009 0.09009 0.0901 1.0000 0.09009 Water 6.63983 0.9999 118.0 0.01944 6.65927 6.6593 1.0000 6.63983 Toluene 9.383E-04 1.41E-04 95.1 0.01567 0.01661 0.0166 0.9999 0.00094Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices

EIIP Volume II 16.4-54 Iteration 6 n i,L,k x i p i n i,V n i,L+V n i,F Φ i,k n i,L,,k+1 Air 546.9 0.09009 0.09009 0.0901 1.0000 0.09009 Water 6.63983 0.9999 118.0 0.01944 6.65927 6.6593 1.0000 6.63983 Toluene 9.382E-04 1.41E-04 95.1 0.01566 0.01660 0.0166 1.0000 0.00094 Table 4-44. Condenser Material Balance at Completion of Iterations
Iteration 6 Total In (lb) n i,1 (lb) n i,V (lb) Total Out (lb) Material Balance Air 2.61 0.00 0.09009 2.61 2.61 100% Water 120 6.63983 119.65 0.01944 0.35 120.00 100% Toluene 1.53 9.382E-04 0.09 0.01566 1.44 1.53 100% Table 4-45. Exit Vent Stream Analysis Table

(Based on the final results from Table 4-44) Compound mm Hg MWt Lb-Moles Lb Air 546.9 28.013 0.09009 2.61 Water 118.0 18.02 0.01944 0.35 Toluene 95.1 92.13 0.01566 1.44Chapter 16 – Chemical Manufacturing 5.0 – Basic Process Operations

EIIP Volume II 16.5-1 

## 5.0 Basic Process Operations 5.1 Emission Calculations from Solvent Reclamation Systems After being collected from manufacturing processes, wast e solvents are frequently purified and reused in the factory. Distillation is one of the most common means of purifying solvents for reuse. Many forms of distillation are used including simple batch, continuous, or steam distillation. For emissions modeling, a typical distillation process involves several separate emission modeling steps: 1. The initial filling step when waste solvent is charged to the empty distillation vessel. 2. A heating step when the waste solvent is rai sed to the solvent boiling point temperature. 3. The recovery phase when purified distillat e is collected in the receiving vessel. 4. At the completion of the distillation process any re maining waste solvent in the still is normally cooled. The emissions from cooling are assumed to be zero unless a nitrogen sweep that would overcome gas contraction is being applied. 5. A final drumming step when the recovered solvent is transferred to a solvent holding area or to drums. Illustration 5-1: Estimating emissions from a batch distillation operation. One hundred gallons of waste toluene are to be ch arged to a batch still for distillation recovery. The toluene to be charged is at 18°C and contains 1.5% (mole/mole) dissolved non-volatile waste solids. A boiling point check shows that the waste toluene h as a normal boiling point of 111.2°C. The still vessel has a gas space volume of 220 gallons when empty. An overhead heat exchanger is used to condense the pure toluene distillate at 20°C. What are the vent emissions from this event? 5.1.1 Emissions from Charging the Distillation Vessel with Cold Solvent for Recovery The standard charging model is used to calculate th e vent emissions that occur from charging the cold waste solvent to the distillation vessel. Basic values that will be used in the calculation are first calculated. Determine T(°K): T(°K) = 18.0°C + 273.15 = 291.25°K Antoine Equation: mmHg P 582

. 19 ) 975

. 2 exp( 67

. 53 15

. 291 52

. 3096 0137

. 16 exp toluene = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

− − = Raoult’s Law: p toluene = 0.985 P toluene = (0.985)(19.582 mm Hg) = 19.288 mm Hg. Displacement Vol. V (ft 3 ) = 100 gallons * 0.13368 (ft 3 /gal) = 13.368 ft 3 Ideal Gas Constant R = 998.9 mmHg• ft 3 / lb-mole•°K Moles of toluene emitted per batch: 000887

. 0 15

. 291 9

. 998 368

. 13 288

. 19 =

⋅ ⋅ = i n lb-moles Weight of toluene emitted per batch: wt i = (0.000887 lb-moles)(92.13 lb/lb-mole) = 0.082 lbsChapter 16 – Chemical Manufacturing 5.0 – Basic Process Operations

EIIP Volume II 16.5-2 5.1.2 Heating the Solvent for Distillation Since the distillation system contains a process conden ser operating at 20°C then we can assume that the exit vent gas caused by heating will leave the equipmen t system saturated with toluene vapors at the 20°C temperature. It should be noted that during the initial heating process between 18°C and 20°C that the toluene content of the condenser exit gas will be less th an the 20°C saturation level. However, this error is considered to be in significant when compared to th e total emissions that occur while heating to the final 111.2°C boiling point temperature prior to distillation.
T i = 18°C = 291.15°K (Initial temperature) P T = 1.0 atm = 760 mm Hg (Total system pressure) V gas = 120 gallons = 16.04 ft 3 (Gas space volume) R = 998.9 (mmHg-ft 3 )/(lb-mole- o K) (Universal gas constant) P toluene, 18°C = 19.906 mm Hg. (Antoine Equation) The partial pressure of nitrogen at 18°C is the differe nce between the total system pressure and the partial pressure of toluene: mmHg p p toluene

nc 4

. 740 6

. 19 760 760 1 , 1 , = − = − =
Gas Law: 

### n PV RT = , also 

### n p V RT i i



### = for a single component, i , in the gas space. Calculate the total amount of nitrogen that is displ aced from the still during the heating activity using the ideal gas law at the initial conditions. When the wast e toluene temperature reaches the boiling point, it is assumed that all of the nitrogen has been expelled and that the head space in the distillation vessel contains only saturated toluene vapors. Therefore the ideal gas law is used to calculate the moles of nitrogen that is displaced. Nitrogen discharged from heating: 

### 041

. 0 15

. 291 9

. 998 04

. 16 4

. 740 2 

### = ∗ ∗ = = 

### RT V p n n

N2 lb-moles The condenser has an outlet vent temperature of 20°C. It is assumed that the outlet vent gas is saturated with toluene vapors. Therefore the quantity of toluen e that is finally emitted from the process system is calculated by multiplying the moles of nitrogen by the ratio of the toluene and nitrogen partial pressures. Partial pressure of toluene at 20°C (Antoine Equation): 21.8 mm Hg Partial pressure of nitrogen at 20°C by difference: (760 – 21.8 ) = 738.2 mm Hg. Final toluene emission: 

### moles lb mmHg mmHg n toluene 

### − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= 

### 00121

. 0 041

. 0 2

. 738 8

. 21

Final toluene emission: 

## ( )( ) lbs mole lb lb moles lb wt toluene 

### 0012

. 0 / 13

. 92 00121

. 0 = − − = 5.1.3 Receiving the Solvent from Distillation If the initial waste toluene to be distilled contains 1.5% non-volatile impurities and the composition of the final still bottoms is estimated to be 50% toluene and 50% non-volatile impurities, then the expected volume of ST-100 CN-100

TA-101Chapter 16 – Chemical Manufacturing 5.0 – Basic Process Operations

EIIP Volume II 16.5-3 toluene to be recovered will be 97 gallons. The emi ssions that are associated with the actual distillation operation are calculated using the filling or charging mode l for TA-101. It is estimated that the distillate toluene enters TA-101 from the condenser at 20°C based on earlier discussions. Vapor pressure of toluene at 20°C: 21.85 mm Hg. Displacement volume: 97 gallons * 0.133680555 (gal/ft 3 ) = 12.967 ft 3 Toluene emitted per batch: 

### 001048

. 0 15

. 291 9

. 998 967

. 12 85

. 21 

### = ∗ ∗ = = 

### RT V P n i

i lb-moles Toluene emitted per batch: 

## ( ) ( ) lbs mole lb lb moles lb wt toluene 

### 097

. 0 / 13

. 92 001048

. 0 = − − = 5.1.4 Drumming the Purified Toluene Once the 97 gallons of toluene have been purified and collected in TA-101, the batch will be drummed in 50 gallon drums. Unless spot ventilation is provided for the drums then the emission losses that occur during drumming would be considered to be fugitive emissions. Em issions for this operation are calculated as a simple filling ope ration. As the 97 gallons of toluene are transferred from the receiver into drums then 97 gallons of saturated solvent vapor are displaced from the drums. The emissions are calculated using the ideal gas law as before. Toluene emitted per batch: 

### 001048

. 0 15

. 291 9

. 998 967

. 12 85

. 21 

### = ∗ ∗ = = 

### RT V P n i

i lb-moles Toluene emitted per batch: 

## ( ) ( ) lbs mole lb lb moles lb wt toluene 

### 097

. 0 / 13

. 92 001048

. 0 = − − = 5.1.5 Solvent Emissions from the Overall Toluene Recovery Process Solvent emissions from the overall toluene recovery process are estimated by adding the emissions from each of the individual steps. Activity Description Amount Initial charging of 100 gallons of waste toluene solvent 0.0820 lbs Heating the batch for distillation 0.0012 lbs Receiving toluene in r eceiver TA-101 0.0970 lbs Drumming 97 gallons of recovered toluene 0.0970 lbs Total Emissions from process 0.2772 lbs 

### 5.2 Filtration Operations Filtration is an operation that is commonly used wh en it is desired to separate the solids portion of a process slurry from the liquid portion. Most filtrations involve a slurry feed vessel, a filter, and a filtrate receiving vessel. In some cases the slurry is passed through the filter by pressurizing the feed tank while in other operations the slurry may be pumped. Diffe rent types of filters may be used depending upon the processing requirements. For example, the filter may be a bag filter, filter press, leaf filter, centrifuge, or another design. ST-100 CN-100

TA-101Chapter 16 – Chemical Manufacturing 5.0 – Basic Process Operations

EIIP Volume II 16.5-4 Examples of batch filtration might include processing wh en crystalline product th at has formed during the process must be isolated from the batch slurry. Other filtration operations are performed when the product is dissolved in the primary processing solvent and the solids are either waste compounds or materials that have added to help purify the dissolved product such as diamatious earth or activated carbon.



### TA-100 TA-105

Filter Press Vent Vent

Receiver Slurry N2

Figure 5-1. Example filtration equipment arrangement. Batch filtration consists of at least three separate modeling activities including sending the process slurry to the filter initially with process mother liquors being directed into the receiver. The charging or filling model is used to calculate the process vent emissi ons that occur during this processing step. Since the filter is connected directly to the filtrate receivi ng vessel then all emissions exit through the receiver process vent. In this case the emissions from the operati on are based on the total volume of slurry being processed forward. A fresh solvent (normally the same solvent that is contained in the process) is passed through the filter cake to the filtrate receiver. The emissions from this operation may also be calculated using a filling model for the filtrate receiver based on the volume of solvent that is used during the wash. Air or nitrogen gas is usually passed through the filte r to displace the residual liquid from the solids filter cake. Displaced liquid from the filter cake is sent to the filtrate receiver. Emissions that occur from this operation may be calculated using the standard filling m odel that also includes a gas sweep for the filtrate receiver. An exit gas saturation level of 100% is used based on the process filtrate composition. 

### 5.3 Centrifuge Operations Centrifugation is an operation that is commonly used when it is desired to separ ate the solids portion of a process slurry from the liquid portion. Most centrif uge operations contain a slurry feed vessel, a centrifuge, and a filtrate receiving vessel. In most cases the slurry is passed to the centrifuge by way of a centrifugal pump. Centrifuge operations normally incl ude a feed step (when slurry is feed to the centrifuge so that the internal bask et can be loaded), a spin step (when filtrate is permitted to exit the centrifuge cake), a washing step (when water or process so lvent is feed to the centrifuge), a final spin step (when the wash liquor is allowed to separate from the centrifuge cake), followed by a plowing step (when the solid cake is removed from the centrifuge).Chapter 16 – Chemical Manufacturing 5.0 – Basic Process Operations

EIIP Volume II 16.5-5 Processing when centrifugation might be used would be for situations when crystallized material that has formed during the process must be isolated from the batch. Several centrifuge loads must usually be processed from a single batch of slurry ma terial depending upon the batch size.



## TA-100

TA-105 Wash

Centrifuge Vent Vent Vent

Receiver Slurry N2

Figure 5-2. Typical centrifuge equipment arrangement. Centrifugation consists of at least three separate modeling activities. Feed Step . The initial feed step is when the centrifuge is being loaded with process slurry from the feed vessel and filtrate is being directed to the filtrate re ceiver from the centrifuge. The charging or filling with gas sweep model is used to calculate the process ve nt emissions that occur from the centrifuge as well as from the filtrate receiver. The volumetric capacity of the centrifuge basket is used as the filling volume for the centrifuge and any additional ventilation rate (as established through plan t data) is used for the gas sweep portion of the centrifuge emission calculation. If the centrifuge is tight ly connected to other ancillary equipment such as an enclosed bottom hopper then the g as sweep rate may be negligible. Similarly, the volume of filtrate that enters the filtr ate receiver from the centrifuge in addition to any gas sweep that is present are the basis for the emission calculation for this vessel. Wash step . Fresh solvent (normally the same solvent as is contained in the process) is passed through the centrifuge cake to the filtrate receiver. During this ph ase of the operation the centrifuge basket is already full of product solids so the emissions that occur from the centrifuge would be from the gas sweep assuming that a gas sweep exists. However, if the cen trifuge is tightly connected to other ancillary equipment items and a gas purge is not being used at the centrifuge then the focus of the emission calculation would be placed on the filtrate receiver. The volume of wash solvent that enters the filtrate re ceiver in conjunction with any gas sweep that also might be present forms the basis for the emi ssion calculation from the filtrate receiver.Chapter 16 – Chemical Manufacturing 5.0 – Basic Process Operations

EIIP Volume II 16.5-6 

### 5.4 Vacuum Dryer Model Examples of vacuum drying include processing when fina l product solids are dried in a vacuum tray dryer or rotary dryer. The vacuum drying process consists of at least four separate activities and includes (1) placing the process material into the dryer, (2) reducing the system pr essure to the design level (3) heating the batch for evaporation to take place, and (4) collecti ng the solvent distillate in the receiver.

CN-100



### TA-200 VP-100 DR-100

Figure 5-3. Diagram of a typical vacuum tray dryer system. 5.4.1 Information Needed to Model a Vacuum Tray Drying Operation 5.4.1.1 Placing the material to be dried into the vacuum oven Product solids that are wet with solvent from the prior filtration or centrifugation step will undergo evaporation losses as they are being prepared for the v acuum dryer. Material to be dried is physically transferred from a hopper, drum, or other container onto trays and then spread evenly so that the drying process will be uniform. It is difficult to predict the evaporation rate fro m wet solids because each process is different and the material being processed varies with respect to partic le size, solvent content, exposure condition, and other variables. A conservative approach to estim ating the evaporation losses from a wet product solids cake would be to apply the basic evaporation model as presented in Section 3.6 of this document. 

# ( )

L i sat i i i m 

### RT P P A K M Q − = Eq. 5-1 where Q m is the evaporation rate (mass/time). M is the molecular weight of the volatile substance, K is a mass transfer coefficient (length/time), A is the evaporation surface area, sat i 

### P is the saturated solvent vapor pressure,

i 

### P is the actual vapor pressure near the liquid surface,Chapter 16 – Chemical Manufacturing 5.0 – Basic Process Operations

EIIP Volume II 16.5-7 R is the ideal gas constant, and T L is the absolute temperature of the liquid. Equation 5-1 would provide a conservative estimate of the evaporation rate because the solvent is distributed throughout the product cake as opposed to ex isting as a continuous liquid with a fixed surface area. As evaporation takes place the amount of solven t that is close to the surface of the cake becomes depleted and any remaining solvent must migrate to the surface for further evaporation. The evaporation process will also remove heat from the product cake and cause sat i 

### P to reduce due to the lower temperature. An alternative approach to quantifying the evaporation rate would be to perform a material balance study by making weight loss measurements using representativ e samples of wet cake. The results from wet loss studies could then be used to make future estimates. 5.4.1.2 Depressurization Step The standard depressurization model is applied to calculate the solvent emissions from vacuum oven during the evacuation operation from one atmosphere down to the planned operating pressure. In many cases the planned operating pressure for the oven is less than the saturation vapor pressure of the solvent. In these cases, the depressurization model would be cal culated between one atmosphere and the solvent saturated vapor pressure. 5.4.1.3 Heating Step If the planned operating pressure of the vacuum oven is less than the solvent saturated vapor pressure at initial temperature conditions then the heating model can be used to estimate the emissions that occur during heat up. The heating model would be calcula ted between the initial temperature conditions and within 2°C of the boiling point temperature for th e solvent at the planned operating pressure of the vacuum oven. 5.4.1.4 Distillation Step Once the drying process is underway, solvent vapors are carried from the vacuum oven and into the heat exchanger where they condense and dr ain into the distillate receiver. Th e distillate receiver continues to fill with solvent as long as solvent is being remove d from the product solids. Air emissions that occur while the vacuum drying process is ongoing originat e from the distillate receiver. The vacuum pump continually removes air from the equipment to maintain the correct pressure of operation. The filling model is used to estimate the solvent emis sions during this part of the drying process. The composition and quantity of the distillate being collect ed is the same as the original amount of solvent that was contained in the wet cake. The fill volume is e qual to the total volume of solvent that is collected. The temperature and pressure of the distillate receiver is used in the calculations as well as any estimated leak rate that the vacuum pump must overcome as part of the process operation.Chapter 16 – Chemical Manufacturing 5.0 – Basic Process Operations

EIIP Volume II 16.5-8 [This page intentionally left blank.]Chapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships

EIIP Volume II 16.6-1 

## 6.0 Physical Property Relationships 6.1 Basic Physical Prope rties Relationships This section describes the equations and mathematical relationships that are used in calculating air emissions from point sources in chemical manufacturing processes. 6.1.1 Unit Conversations The following relationships are helpful when converting from one unit system to another for temperatures, pressures, and volumes. Temperature Conversions T (°R) = T (°F) + 459.69 Eq. 6-1 T (°K) = T (°C) + 273.15 T (°K) = 1.8 * T (°R) T (°K) = [T (°F) + 459.69] / 1.8 T ( C) = T (°K) – 273.15 T ( C) = [T (°F) – 32.0] / 1.8 Pressure Conversions P (mm Hg) = 760.0 * P (atm) Eq. 6-2 P (mm Hg) = 51.7 * P (psia) P (mm Hg) = 25.4 * P (in Hg) P (mm Hg) = P (Pa) / 133.3 Volume Conversions V ft3 = 0.03531467 * V (liter) Eq. 6-3 V ft3 = 0.133680555 * V (gal) 

### 6.2 Basic Physical Property Relationships 6.2.1 Ideal Gas Law The Ideal Gas Law is used to calcu late the total number of moles in a gas space from known variables, such as pressure, volume, and temperature. 

### nRT PV = Eq. 6-4 where: n = moles of gas P = pressure (absolute) V = volume T = temperature (absolute) R = Universal Gas Constant The following properties exist fo r ideal gases at the standard temperature and pressure condition:Chapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships

EIIP Volume II 16.6-2 Table 6-1. Standard Conditions for Ideal Gases
Mass Pressure Volume Temperature Ideal Gas Constant (R) 1.0 lb-mole 14.7 lb/in 2 359.046 ft3 491.69 °R 10.731 1.0 lb-mole 29.92 in Hg 359.046 ft3 491.69 °R 21.8484 1.0 lb-mole 760 mm Hg 359.046 ft3 273.15 °K 998.9 1.0 g-mole 760 mm Hg 22.414 liters 273.15 °K 62.3637 1.0 g-mole 1.0 atm 22.414 liters 273.15 °K 0.8206 Illustration 6-1: Calculating the moles of gas from volume, temperature, and pressure. A process vessel has a void space volume of 1200 gallons when empty of liquid. Calculate the moles of gas that are discharged from the process vent if 550 gallons of liquid are charged into the vessel. Also calculate the moles of gas that are still contained in the vessel headspace at the completion of the filling operation. The batch temperature is 35°C and the system pressure is 760 mm Hg. Assumptions: The vessel temperature and the batch temperature are the same 35°C before and after the addition is made. Additionally, any evaporation that may occur when the liquid enters the initially empty vessel is ignored. Part A (Moles of gas displaced through the process vent). Displacement Volume V d = 550 gal = 0.133680555 ft3/gal * 550 gal = 73.52 ft3 Temperature T = 35°C = 35°C + 273.15°K = 308.15°K
Moles moles

- lb 0.182 15

. 308 99

. 998 52

. 73 760 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
∗ ∗ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= RT PV n Part B (Moles of gas retain ed in the vessel headspace). Gas Space Volume V g = 1200 gal – 550 gal = 650 gal. Volume conversion V g = 0.133680555 ft3/gal * 650 gal = 86.89 ft3 Moles moles

- lb 0.215 15

. 308 99

. 998 89

. 86 760 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

⋅ ⋅ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= RT PV n 6.2.2 Dalton’s Law Equations based on Dalton’s Law ( Eq. 6-5 ) of partial pressure and Amagat’s law ( Eq. 6-6 ) of partial volume are used with the Ideal Gas Law to determin e the number of moles of a specific compound in the vapor space of the vessel. Dalton’s Law 

### RT V p n i

i 

### = Eq. 6-5

Amagat’s Law 

### n i 

### = Pv i 

### RT Eq. 6-6 where: P = pressure V = volumeChapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships

EIIP Volume II 16.6-3 T = temperature R = Universal Gas Constant n i = moles of i p i = partial pressure of i v i = partial volume of I Illustration 6-2: Calculating molar quantities for gas mixtures. The headspace of a vessel contains the vapors of wa ter (20 mm Hg), methanol (45 mm Hg), ethanol (40 mm Hg), and a non-condensable component nitrogen. How many moles of each compound are contained in the headspace if the batch temperature is 35°C, the head space volume is 453 gallons, and the total system pressure is 760 mm Hg. Gas Volume V = 453 gal * 0.133680555 ft3/gal = 60.56 ft3 Temperature (°K) T = 35°C = 35°C + 273.15°K = 308.15°K Dalton’s Law 

### RT V p n i

i 

### =

Moles of water moles

- lb 0.00393 15

. 308 99

. 998 56

. 60 20 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
∗ ∗ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= RT V p n i

water Moles of methanol moles

- lb 0.00885 15

. 308 99

. 998 56

. 60 45 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
∗ ∗ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= RT V p n i

methanol Moles of ethanol moles

- lb 0.00787 15

. 308 99

. 998 56

. 60 40 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
∗ ∗ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= RT V p n i

ethanol Partial pressure of nitrogen mmHg p nitrogen 655 45 40 20 760 = − − − = Moles of nitrogen moles

- lb 0.00885 15

. 308 99

. 998 56

. 60 45 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
∗ ∗ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= RT V p n i

methanol 6.2.3 Mole Fraction in a Liquid The ratio of moles of i to the total moles in a single liquid phase is defined as mole fraction, X i
. The liquid mole fraction of a compound is used later for calcula ting the vapor pressure for the same compound using Raoult’s Law.
Mole fraction total i i 

### n n x = Eq. 6-7 where: x i = mole fraction for component i n i = moles of i in a single liquid phase n total = total moles in a single liquid phaseChapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships

EIIP Volume II 16.6-4 Illustration 6-3: Calculating mole fractions for liqiud mixtures. A process batch consist of methanol (1,435 lb), isopropyl alcohol (546 lb), and acetone (584 lb). Determine the mole fraction (X i ) of each compound in the solution. Compound Weight Molecular Weight Methanol 1,435 lb 32.04 Isopropyl alcohol 546 lb 60.096 Acetone 584 lb 58.08

moles of methanol: moles

- lb 79

. 44 04

. 32 435

, 1 = = mwt lb n

moles of isopropyl alcohol: moles

- lb 09

. 9 096

. 60 546 = = mwt lb n

moles of acetone: moles

- lb 06

. 10 08

. 58

584 = = mwt lb n
Total moles in the batch: 44.79 + 60.096 + 10.03 = 69.93 lb-moles

mole fraction of methanol: moles

- lb 09

. 9 096

. 60 546 = = mwt lb n

mole fraction of isopropyl alcohol: 0.142 93

. 69 09

. 9 =

− − = moles lb moles lb X i

mole fraction of acetone: 0.157 93

. 69 06

. 10 =

− − = moles lb moles lb X i 

### 6.3 Pure Component Vapor Pressure 6.3.1 Clapeyron Vapor Pressure Equation Essentially, all liquids and some solids exhibit a vapor pressure that can be measured. The vapor pressure for any compound is a function of temperature and composition within a given solution. If the compound exists in pure form then the vapor pressure becomes only a function of temperature. Many mathematical relationships ha ve been formulated over the year s that enable the pure component vapor pressure of a compound to be accurately es timated within a given temperature range. The Clapeyron Vapor Pressure Model ( Eq. 6-8 ) is based upon the equality of chemical potential, temperature, and pressure in both liquid and gas phases. [Reid, Prausnitz, & Sherwood, 1977]
Clapeyron model T B A P

i −

= ) ln( Eq. 6-8Chapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships

EIIP Volume II 16.6-5 where P i = pure component vapor pressure T = temperature, degrees Kelvin A = empirical constant B = v v 

### Z R H Δ Δ ( Δ H v - heat of vaporization, Δ Z v - compressibility factor). Although the Clapeyron Vapor Pressure model is based on the heat of vaporization and the compressibility factor for a compound, this thermodynamic data is not needed. If two reliable vapor pressure data points can be obtained then A and B can be determined mathematically. However, one disadvantage to using this model is that the vapor pressure correlation may not be as accurate as other models that contain a greater number of empirical coefficients. Illustration 6-4: Estimating Clapeyron vapor pressure model coefficients. Toluene has a boiling point of 110.6°C at 760 mm Hg. and a boiling point of 31.8°C at 40 mm Hg. Determine A and B for the Clapeyron Vapor Pressure model and then calculate the vapor pressure of toluene at 51.9 C. Clapeyron model 

### T B A P

− = ) ln( (a) Solve for B 

### ) ( ) ln( ) ln( 1 1 1 2 1 2 − −



### − − = − T T B P P T1 = 31.8 C = 31.8 + 273.15 = 304.95 K 1/T1 = 0.0032792 ln(P1) = ln(40) = 3.6888795 ln(P1) = 3.6888795 T2 = 110.6 C = 110.6 + 273.15 = 383.75 K 1/T2 = 0.0026059 ln(P2) = ln(760) = 6.6333184 ln(P2) = 6.6333184
Solve for B: 

### ⎥ ⎦ ⎤ ⎢ ⎣ ⎡

− −

− =

− − 1 1 1 2 1 2 

### ) ln( ) ln( 

### T T P P B



### 4372.7371 0032792

. 0 0026059

. 0 6888795

. 3 6333184

. 6 

### = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡

− −

− = B

Solve for A 

### 18.0281 75

. 383 7371

. 4372 6333184

. 6 ) ln( 2 2 

### = + = + = T B P A Calculate the vapor pressure of toluene at 59.9 C (325.05 K) 

### 4.575567 05

. 325 7371

. 4372 0281

. 18 ) ln( 

### = − = − = T B A P P 51.9 C =97.1 mm Hg. Note that the vapor pressure of toluene from lite rature sources is 100 mm Hg for 51.9 C or 3% higher than the Clapeyron model prediction using this technique.Chapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships

EIIP Volume II 16.6-6 6.3.2 Antoine Equation The Antoine equation is a modification of the Clapey ron model and is one of the most frequently used equations for estimating the vapor pressure of a pure compound. [Reid, Prausnitz, & Sherwood, 1977] The Antoine equation is shown in the following general form: 

# ) ( = ) ln(P i i i i 

### c T b a + − Eq. 6-9
where: P i = pure component pressure of compound i T = absolute temperature a i , b i , c i = Antoine constants
Coefficients a i , b i , and c i are published in several literature sources for many compounds. It is important to note the specific temperature and pressure units that ar e associated with the Antoin e coefficients that are listed in the literature. The reader should keep in mind that the Antoine equa tion is a general mathematical relationship and can be used with temperatures in either Centigrade, Ke lvin, Fahrenheit, or Rankin units. Additionally, the equation can be used for calculating vapor pressures in different units (atm, mm Hg, and psia) and in the natural log (ln) or base 10 log form. It is always a good idea to calculate one or more known vapor pressure data points as a means of verifying that the model is consistent with respect to coefficients, temperature units, and pressure units being used. Illustration 6-5: Calculating pure component vapor pressures from the Antoine model. The Antoine coefficients for tolu ene are a = 16.0137, b = 3096.52, and c = –53.67. These coefficients were obtained from a source where the Antoine equation is in the natural log form, pressure is in mm Hg, and temperature is in Kelvin units . Additionally, these coefficients are valid between 7°C and 137°C. What is the pure component vapor pressure for toluene at 35°C and 70°C? Antoine Equation 

# ) ( = ) ln(P i i i i 

### c T b a + − Temperature (35°C) T = 35°C = 35.0°C + 273.15°K = 308.15°K 

### 3.8457 67

. 53 15

. 308 52

. 3096 0137

. 16 = ) ln(P 

# ) ( i 

# =

− − Hg. mm 46.79 8457

. 3 35 

# = = ° 

### e P C Temperature (70°C) T = 70°C = 70.0°C + 273.15°K = 343.15°K 

### 5.3169 67

. 53 15

. 343 52

. 3096 0137

. 16 = ) ln(P 

# ) ( i 

# =

− − Hg. mm 203.74 3169

. 5 70 

# = = ° 

### e P C 6.3.3 Other Vapor-Pressure Equation Forms Over the years researchers have published many diffe rent equations for calculating the vapor pressure of pure compounds as a function of temperature. [Reid, Prausnitz, & Sherwood, 1977] Among these include:Chapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships

EIIP Volume II 16.6-7 Riedel-Plank-Miller Equation: 3 

### ln 

### DT CT T B A P

vp 

### + + + = Eq. 6-10
Rankine-Kirchhoff Equation: 

### T C T B A P

vp 

### ln ln 

### + + = Eq. 6-11
Riedel Equation: 6 

### ln ln 

### DT T C T B A P

vp 

### + + + = Eq. 6-12 Coefficients that apply to each equation are normally provided through commercially available databases. The DIPPER Database (AIChE) provides modeling coeffi cients for an equation of the general form:

E vp 

### DT T C T B A P + + + = ln ln Eq. 6-13
Where P vp = vapor pressure in Pascal (Pa) units. T = temperature in degrees Kelvin. Illustration 6-6: Calculating vapor pressures using the AIChE DIPPR database model. According to the AIChE DIPPR Database, the vapor pressure coefficients for acetone are as follows: A = 69.006 B = -5599.6 C = -7.0985 D = 0.0000062237 E = 2.0000 What is the vapor pressure of acetone if the temperature of the liquid is 25°C? 

### K C T ° = + ° = 15

. 298 15

. 273 25

2 

### 0000062237

. 0 ln 098

. 7 6

. 5599 006

. 69 ln 

### T T T P

vp 

### +
− − = 334

. 10 553

. 0 444

. 40 781

. 18 006

. 69 ln 

### = +
− − = vp 

### P mmHg Pa mmHg Pa P

vp 

### 697

. 230 35

. 133

1 10 074

. 3 4 

### = × × = 6.4 Component Vapor Pressure over Solutions 6.4.1 Equilibrium Conditions (Ideal) and Raoult’s Law Ideal vapor pressure conditions normally exist for a solution when strong molecular associations between molecules are not present. This is normally the case when the molecular sizes are approximately equal and the attractive forces between like and unlike molecu les are equal as well. Ideal equilibrium conditions are typically present when the solution is made up completely of nonpolar compounds such as organic solvents (toluene, heptane, hexane, etc.). A generalization know as Raoult’s Law states that the equilibrium vapor pressure that is observed for a compound is proportional to the mole fraction of that compound in solution. For example, given an equalChapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships

EIIP Volume II 16.6-8 molar solution (0.5:0.5) of toluene and heptane at 35°C, the equilibrium vapor pressure of the toluene would be 101.87 mm Hg or one half of its pure component vapor pressure of 203.74 mm Hg at 35°C. Raoult’s Law i i 

### P X p =

i where: p i = effective vapor pressure of i P i = pure component pressure i X

i = mole fraction of component i Illustration 6-7: Estimating component vapor pressures using Raloult’s Law. A solution contains 20% toluene (wt/wt), 50% heptan e (wt/wt), 30% acetone. The solution temperature is 35°C. The pure component vapor pressures for tolu ene, heptane, and acetone are 46.79 mm Hg, 74.04 mm Hg, and 347.1 mm Hg respectively for the 35°C condition. The molecular weights for toluene, heptane, and acetone are 92.13, 100. 205, and 58.08, respectively. Calculate the moles and mole fraction of each co mponent in the solution on a 100 lb basis. Moles of toluene moles

- lb 0.22 13

. 92 20 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= lb MWt

wt n toluene toluene toluene Moles of heptane moles

- lb 0.50 205

. 100

50

tan tan tan = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= lb MWt

wt n
e hep e hep e hep
Moles of acetone moles

- lb 0.52 08

. 58 30 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= lb MWt

wt n acetone acetone acetone Sum of moles n total = 0.22 + 0.50 + 0.52 = 1.23 lb-moles Mole fraction of toluene 0.176 23

. 1 22

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= moles total moles X toluene

toluene Mole fraction of heptane 0.405 23

. 1 50

. 0 tan tan = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= moles total moles X e hep e hep Mole fraction of acetone 0.419 23

. 1 52

. 0 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= moles total moles X acetone

acetone Calculate the equilibrium vapor pressure for each component. Vapor pressure for of toluene 

### Hg mm 8.24 46.79 176

. 0 

### = ∗ = toluene 

### p Vapor pressure for of heptane 

### Hg mm 29.97 74.04 405

. 0 tan 

### = ∗ = e hep 

### p Vapor pressure for of acetone 

### Hg mm 145.46 347.1 0.419 

### = ∗ = acetone 

### pChapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships

EIIP Volume II 16.6-9 6.4.1.1 Using Raoult’s Law to determine molar concentration In special cases, Raoult’s Law can be used to dete rmine the molar concentration of a single solvent solution. For example, suppose waste solvent from a process contains dissolved non-volatile or very low volatile compounds (such as heavy oil). The boiling point temperature of waste solvent and the pure solvent is measured. The molar concentration of the primary solvent in the waste solvent solution may be determined by dividing 760 mm Hg by the calculated vapor pressure of the pure solvent at the elevated temperature. Illustration 6-8: Estimating liquid composition based on vapor pressure measurements. The boiling point temperature of a sample of wast e toluene was measured to be 112.3°C. The boiling point temperature of a sample of pure toluene was measured to be 110.8°C. (Note that measuring the boiling point temperature of a pure solvent can result in a different reading than the value that is contained in the literature due to atmospheric elevation differences and measurement accuracy issues.) Boiling point difference between the two samples (112.3°C – 110.8°C): 1.5°C. Normal boiling point temperature of toluene from the literature: 110.6°C. Corrected boiling point of the waste toluene sample (110.6°C + 1.5°C): 112.1°C
675

. 6 67

. 53 25

. 385 52

. 3096 0137

. 16 = ) ln(P 

## ) ( i 

## =

− − 

### Hg. mm 362

. 792 675

. 6 1

. 112 

# = = ° 

### e P C Raoult’s Law i i 

### P p x = i Molar concentration: 

### % 9

. 95 % 100 362

. 792 0

. 760 P p = i i i 

### = × = 

### mmHg mmHg x Illustration 6-9: Determining the molar composition of a liquid from vapor pressure data. A mixture has been prepared for processing at 35°C. Th e contents contain 850 lbs of toluene, 525 lbs of heptane, and 2,500 lbs of other non-volatile material s. Determine the vapor pressure of the mixture at 35°C and the mole fraction of toluene and heptane. The molecular weight of toluene is 92.13 and the molecular weight of heptane is 100.2. The molar quantities of toluene and heptane are calculated Moles of toluene moles

- lb 22

. 9 13

. 92 850 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= lb MWt

wt n toluene toluene toluene Moles of heptane moles

- lb 24

. 5 205

. 100 525

tan tan tan = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛
= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

= lb MWt

wt n
e hep e hep e hep A sample of the process mixture is placed in a vacuum flask equip with a temperature probe, reflux condenser, vacuum pump, vacuum controller, and pressu re gauge. The mixture is heated to 35°C and then placed under increasing vacuum until the mixture begins to boil with reflux present. The boiling point pressure is recorded from this experiment to be 40 mm Hg. The pure vapor pressures for toluene and heptane at 35°C are 45.79 mm Hg and 74.04 mm Hg, respectively. For this mixture the volatile components have been id entified to consist entirely of toluene and heptane. However, the mole fraction of these two components ca nnot be determined because the molecular weightChapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships

EIIP Volume II 16.6-10 data for the remaining components is not available. Therefore, the following general equation can be established that relates the unknown mole fraction to the partial pressure for each volatile compound.
Raoult’s Law i i i i i 

### P n n P x p 

# ∑ = =
Mole fraction is gas space 

# ∑ = p p y i

i
Combining equations 

# ∑ = i i i i 

### nP P n y Eq. 6-14 Finally, the liquid mole fraction for each volatile component may be determined from Raoult’s Law.

i T i i i i 

### P P y P p x = = Eq. 6-15 For this problem only toluene and heptane need to be considered. Eq. 6-14 is customized to accommodate only toluene and heptane. For toluene 521

. 0 ) 04

. 74 24

. 5 ( ) 79

. 45 22

. 9 ( ) 79

. 45 22

. 9 ( = × + × × = + = j j i i i i i P n P n P n y For heptane 479

. 0 ) 04

. 74 24

. 5 ( ) 79

. 45 22

. 9 ( ) 04

. 74 24

. 5 ( = × + × × = j y Finally, For toluene 455

. 0 79

. 45 0

. 40 521

. 0 = × = = i T i i P P y x For heptane 258

. 0 04

. 74 0

. 40 479

. 0 = × = = j T j j P P y x The results of these calculations suggest that 0.286 (or 1.0 – 0.455 – 0.258) mole fraction of the process mixture consist of other nonvolatile components. 6.4.2 Non-Ideal Equilibrium Conditions and Activity Coefficients In many cases, ideal equilibrium relationships do not exist for solutions. This is especially true for solutions containing compounds that are polar in natu re or have significant attraction to other compounds in the same solution. An example of a solution with highly non ideal equilibrium properties is aqueous hydrochloric acid with the vapor pressure of hydroge n chloride being orders of magnitude lower than expected by Raoult’s Law. Other examples of soluti ons with non ideal vapor pressure behavior would include systems with azeotropic properties such as acetone – hexane. Non ideal equilibrium systems should be calculated us ing an activity coefficient (or correction factor) as part of the basic equilibrium calculation. The equ ilibrium vapor pressure equation for ideal solutions has been modified to include the activity coefficient as part of the calculati on. For solutions that exhibit ideal behavior the activity coefficient is defaulted to 1.0. For solutions that exhibit significant non ideal vapor pressure behavior the activity coefficient ma y be estimated from known vapor pressure data. Eq. 6-16Chapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships

EIIP Volume II 16.6-11 shows how the vapor pressure of a component in so lution would be calculated using the Antoine vapor pressure model. p XP X a b c T i ii i i i i i i = γ γ = − + exp 

## ( ) Eq. 6-16 where: p i = effective vapor pressure of i P i = pure component pressure i X

i = mole fraction of component i γ i = activity coefficient for component i T = absolute temperature a i , b i , c i = Antoine constants Illustration 6-10: Estimating activity coefficients from solution measurements. Using Raoult’s Law the pure component vapor pr essure of ammonia over a 19.1% (wt/wt) aqueous solution of ammonia acid at 21.1°C is 1308 mm Hg a nd the vapor pressure of water is 14.9 mm Hg. The measured vapor pressures of ammonia and water are 221.2 mm Hg and 14.5 mm Hg, respectively. What are the activity coefficients for ammonia and water under these conditions? Activity Coefficient 

### 16917

. 0 1308 3

. 221
3 

### =

= mmHg mmHg NH 

### γ Activity Coefficient 

### 972

. 0 9

. 14 47

. 14 

### = = mmHg mmHg Water 

### γ
Assuming that the activity coefficients remain cons tant between 10°C and 30°C, what would the vapor pressure of hydrogen chloride and wate r be for the same solution at 25°C? Based on 100 lbs of solution, Mole fraction of NH 3 : 

### 20

. 0 02

. 18 9

. 79 031

. 17 1

. 19 031

. 17 1

. 19 

### = + Mole fraction of H2O: 

### 80

. 0 031

. 17 1

. 19 02

. 18 9

. 79 02

. 18 9

. 79 

### = + At 25 C the calculated pure vapor pressure for NH 3 and H 2 O are 7,370 mm Hg and 23.58 mm Hg, respectively. Vapor pressure of NH 3: 

### Hg mm P p p NH NH NH NH 

### 3

. 249 7370 * 16917

. 0 * 20

. 0 3 3 3 3 

### = = = 

### γ Vapor pressure of H 2 O: 

### Hg mm P X p O H O H O H O H 

### 35

. 18 58

. 23 972

. 0 80

. 0 = 2 2 2 2 

### = ∗ ∗ = 

### γ Illustration 6-11: Estimating activity coefficients from azeotropic mixtures. The azeotropic composition of a solution containing heptane and methanol is 48.5% (wt/wt) and 51.5% (wt/wt), respectively. The azeotropic boiling point for th e solution is 59.1°C and the system pressure is 760 mm Hg. Based on Raoult’s Law, the calculated vapor pressures for the same solution of heptane andChapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships

EIIP Volume II 16.6-12 methanol are 47.9 mm Hg and 470.1 mm Hg, respectiv ely. Note that the liquid and vapor compositions for an azeotropic system are th e same at the boiling point. For this problem the composition of heptane and meth anol must be converted from weight percent to mole percent. This may be accomplished by basing the calculations for an arbitrary 100 lb of solution and then calculating the moles of each compound using the molecular weight. Finally, the mole fraction of each compound is calculated by dividing the moles of each compound by the total moles in the solution. The 48.5% (wt/wt) for heptane is converted to 23.1% (mole/mole), and the 51.5% (wt/wt) for methanol becomes 76.9% (mole/mole). Calculate the activity coefficients for he ptane and methanol under these conditions. γ heptane 3.672 9

. 47 760 231

. 0 tan = ∗ = mmHg mmHg
e hep γ γ methanol 1.243 470 760 769

. 0 = ∗ = mmHg mmHg methanol γ Illustration 6-12: Calculating vapor compositions using activity coefficients. A solution of heptane and methanol was distilled at its azeotropic composition for 1 atm pressure. If the temperature of the recovered distillate is 30°C, what would be the vapor composition for this solution? The system pressure is 760 mm Hg and the vessel is blanketed with nitrogen. Vapor pressures at 30°C: Heptane = 58.54 mm Hg. Methanol = 163.8 mm Hg. Azetropic composition: Heptane = 48.5% (wt/wt) = 23.1% (mole/mole) Methanol = 51.5% (wt/wt) = 76.9% (mole/mole) Component vapor pressure 

### Hg mm 50 54

. 58 672

. 3 231

. 0 = tan tan tan heptane 

### = ∗ ∗ = e hep e hep e hep 

### P X p 

### γ Hg mm 157 8

. 163 243

. 1 769

. 0 = 

### = ∗ ∗ = methanol methanol methanol methanol 

### P X p 

### γ Hg mm 553 157

- 50

- 760 = 

### = nitrogen 

### p Vapor space molal percentage compositions y i : 

### 6.6% % 100 760 50 tan 

### = ∗ = e hep 

### y 20.5% % 100 760 156 

### = ∗ = methanol 

### y 72.9% % 100 760 553 

### = ∗ = nitrogen 

### yChapter 16 – Chemical Manufacturing 7.0 – References

EIIP Volume II 16.7-1 

## 7.0 References Alkis Constantinides, “Applied Numerical Met hods with Personal Computers”, 1987, ISBN 0-07- 012463-9, McGraw-Hill, Inc., pp 76- 78, pp 132-134, pp 276-277. Crowl, Daniel and Louvar, Joseph. 2002. Chemical Process Safety, Fundamentals with Applications

. Second Edition. Prentice Hall PTR, Upper Saddle River, NJ 07458. www.phptr.com . ISBN: 0-13- 018176-5. EIIP. 2000. How to Incorporate The Effects of Air Pollution Control Device Efficiencies and Malfunctions Into Emission Inventory Estimates. Chapter 12 in EIIP Volume II. Point Sources Preferred and Alternative Methods. U.S. Environm ental Protection Agency, Office of Air Quality Planning and Standards. Research Triangle Pa rk, North Carolina. (EIIP Internet address http://www.epa.gov/ttn/chief/eiip). EPA. 1987. Estimating Releases and Waste Treatment Effic iencies for the Toxic Chemical Release Inventory Form . U.S. Environmental Protection Agency, Office of Pesticides and Toxic Substances, EPA-560/4-88-002. Washington, D.C. EPA. 1994a. Air Emissions Models for Waste and Wastewater. U.S. Environmental Protection Agency, Office of Air Quality Planning and Standards, EPA-453/R-94-080a. Research Triangle Park, North Carolina. EPA. 1994c. Control of Volatile Organic Compound Emissions from Batch Processes -Alternate Control Techniques . U.S. Environmental Protection Agency, Office of Air Quality Planning and Standards, EPA-453/R-93-017. Research Triangle Park, North Carolina. EPA. 1995a. Compilation of Air Pollutant Emission Factors. Volume I: Stationary Point and Area Sources, Fifth Edition, AP-42. Section 7.1, Or ganic Liquid Storage Tanks. U.S. Environmental Protection Agency, Office of Air Quality Planning and Standards. Research Triangle Park, North Carolina. EPA. 1995g. Protocol for Equipment Leak Emission Estimates . U.S. Environmental Protection Agency, Office of Air Quality Planning and Standards, EPA-453/R-95-017. Research Triangle Park, North Carolina. EPA. 1995h. Survey of Control Technologies for Low Concentration Organic Vapor Gas Streams . U.S. Environmental Protection Agency, Office of R esearch and Development, EPA-456/R-95-003. Research Triangle Park, North Carolina. (Internet address, http://www.epa.gov/ttn/catc/dir1/low_vo.pdf ) Ernest J. Henley & J. D. Seader, “Equilibrium-Stage Separation Operations in Chemical Engineering”, 1981, ISBN 0-471-37108-4, John Wiley & Sons. H. H. Rachford, Jr. and J. D. Rice, J. Pet. Tech., 4 (10), Section 1, p. 19, and Section 2, p. 3 (October, 1952). Hatfield, J. A. (a). “A Model for Estimating Pr ocess Emissions from Purge Operations in Batch and Continuous Chemical Operations”, Accepted for Publication in Environmental Progress, August 14, 2003.Chapter 16 – Chemical Manufacturing 7.0 – References

EIIP Volume II 16.7-2 Hatfield, J. A. (b). “Improved Algorithm for Estim ating Process Emissions from Batch Depressurization”, Environmental Progress , Vol. 17 No. 3, pp. 195-198 (Fall 1998). Hatfield, J. A. (c). “Improved Algorithm for Estimating Process Emissions from Batch Heating”, Environmental Progress , Vol. 17 No. 3, pp. 190-194 (Fall 1998). Hatfield, J. A. (d). “Subsurface Addition Emission Model”, Accepted for Public ation in Environmental Progress, November 24, 2003. Hougen, Olaf and Watson, Kenne th and Ragatz, Roland. 1958. Chemical Process Principles, Part I

. Second Edition. John Wiley & Sons, Inc. New York, NY. Hougen, Watson, & Ragatz, “Chemical Process Prin ciples”, 1959, ISBN 0-471-41382-8, John Wiley & Sons. pp 919. Perry. 1984. Perry’s Chemical Engineers’ Handbook, 6 th Edition . 1984, McGraw-Hill Book Company. ISBN 0-07-049479-7. Phillip C. Wankat, “Separations in Chemical Engineering, Equilibrium Stage Separations”, 1988, ISBN 0-13-500968-5, Prentice-Hall P T R. Reid, Robert and Prausnitz, John and Sherwood, Thomas. 1977. The Properties of Gases and Liquids

. Third Edition. McGraw-Hill Book Company. ISDN: 0-07-051790-8.