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 16 
EIIP 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. Methods 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 that 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 ...........................................................................................................16.2-1

2.2 Emission Sources ..............................................................................................................16.2-1

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/PM10 Control Systems ...................................................................................16.2-6

3.0 Basic Air Emission 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 – Partially Filled Vessel ....................................................16.3-14

3.3 Vacuum Operations.........................................................................................................16.3-17

3.4 Gas Evolution..................................................................................................................16.3-20

3.5 Depressurization..............................................................................................................16.3-23

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 .....................................16.3-35
Chapter 16 – Chemical Manufacturing Table of Contents 

EIIP Volume II iv 

3.11 Emission Calculations Using 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 Considerations for the Vent Device .........................................16.4-1

4.1.2 Equilibrium Relationships Between the 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 Solvent Reclamation Systems..............................................16.5-1

5.1.1 Emissions from Charging the Distillation 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 ..............................................................................16.6-1
Chapter 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 Pressure 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 References...................................................................................................................................16.7-1
Chapter 16 – Chemical Manufacturing Table of Contents 

EIIP Volume II vi 

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 filled 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......................................................................................16.3-18

3-10 Vacuum operation without vessel filling.................................................................................16.3-19

3-11 Reaction involving a gas evolution of one component............................................................16.3-21

3-12 Reaction involving gas evolution of two components.............................................................16.3-21

3-13 Reaction involving multicomponent gas evolution and nitrogen purge..................................16.3-22

3-14 Vessel depressurization involving 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 volatile 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 production operations.........................................16.3-34

3-22 Using material balance to estimate 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, material 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 quantities for gas mixtures...........................................................................16.6-3

6-3 Calculating mole fractions for liqiud mixtures..........................................................................16.6-4

6-4 Estimating Clapeyron vapor pressure 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. ...................................................16.6-12
Chapter 16 – Chemical Manufacturing Table of Contents 

EIIP Volume II vii 

List of Figures 

Figure Page 

3-1 Saturation Level (S) Plotted as a Function of Exit Vent Gas Flow Rate.................................16.3-37

3-2 Toluene Emission Rate Plotted as a Function of Exit Gas Sweep Rate and Saturation 

Level........................................................................................................................................16.3-37

4-1 Diagrams of single 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 vacuum tray dryer system........................................................................16.5-6
Chapter 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 Condenser Inlet Feed Stream..........................................................................16.4-17

4-5 Check to See If Condensation 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 Streams Analysis .......................................................................................................16.4-21

4-10 Analysis of the Condenser 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 Condenser 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 Analysis 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 Determine 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....................................................................................16.4-41

4-26 Equilibrium Mole Fraction Concentrations.............................................................................16.4-42

4-27 Estimated Activity Coefficients...............................................................................................16.4-42

4-28 Aqueous Phase Condensate Feasibility Analysis ....................................................................16.4-43

4-29 Non-Aqueous Phase Condensate Feasibility Analysis............................................................16.4-44

4-30 Composition and Vapor Pressure Analysis 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 Initial Guess of the Exit Liquid Composition .....................16.4-48

4-36 Material Balance Accountability.............................................................................................16.4-49

4-37 Partial Pressure Equilibrium 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...............................................................................................16.4-51
Chapter 16 – Chemical Manufacturing Table of Contents 

EIIP Volume II ix 

List of Tables (continued) 

Table Page

4-41 Aqueous Phase Condensate Feasibility Analysis ....................................................................16.4-52

4-42 Non-Aqueous Phase Condensate 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 .........................................................................................16.6-2
Chapter 16 – Chemical Manufacturing Table of Contents 

EIIP Volume II x 

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Chapter 16 – Chemical Manufacturing 1.0 – Introduction 

EIIP Volume II 16.1-1 

1.0 Introduction 

The purpose of this guideline document is to describe emission estimation 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 provided, 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 industry as well as regulatory agency personnel. 

As EPA has indicated in this and other EIIP documents, 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 applicable 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 available 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 property 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 represent significant advances over previous emissionestimating methods for chemical manufacturing processes. New methodologies are, however, 

continuously being developed and published by EPA for subsequent 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 

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Chapter 16 – Chemical Manufacturing 2.0 – Source Category Description 

EIIP Volume II 16.2-1 

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 that are involved. However, most chemical processes include at least 

one or more combinations of four basic operations: preassembly, reaction, isolation, and purification. 

Solvent recovery operations are also important because they enable the chemical operator to reuse basic 

raw materials and reduce the manufacturing cost and environmental impact. Additionally, cleanout 

operations are important since they enable production 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 vessels 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 facilitate the actual chemical synthesis. A reaction 

may be carried out by applying heat or by adding 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. Equipment 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 formed 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 ingredient in the next step of a multi-step synthesis. 

For example, in a pharmaceutical process a low 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 for chemical manufacturing operations; they are as 

follows: 

 Process operations 
Chapter 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 occur 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, the 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 solvents 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 temperature to avoid thermal decomposition of compounds in 

the batch. When the pressure of the batch is reduced 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 chloride, 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 system 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 blending 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 contaminated 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 chemical 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 sweep, and others. Additionally, 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 cleaned 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). Wastewater treatment systems generally consist of a 

series of surface impoundments that are used for equalization, neutralization, aeration, and clarification of 

the waste stream. Fugitive VOC emissions may occur 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 and working losses. Breathing loss is the expulsion of 

vapor from a tank through vapor expansion and contraction that result from changes in ambient 

temperature and barometric pressure. This loss occurs without any liquid 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 level 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 relief 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 solvents and resins) from storage tanks to the chemical 

manufacturing operation, a network of pipes, pumps, valves, and flanges is employed. As liquid material 

is pumped from the storage tanks to the particular process area, the pipes and supporting hardware 
Chapter 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 would 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 capture device and a removal device. The capture device 

(such as a hood or enclosure) captures the VOC-laden air 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. Large scale adsorption based 

recovery systems normally have two or more activated carbon adsorption chambers. One carbon 

chamber is being used to remove VOCs from emission 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 collected 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 matter from the emission stream. Dust collectors are 

constructed in many different designs. A bag house 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 filter media, and exits the unit through the discharge 

port at the top. Particulate matter builds up on the filter media until it is shaken off by pulses of 

compressed air from within each bag. The dust that falls from the bags during the pulsing process 

is collected at the lower section of the bag house 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 the conventional storage tank air that is saturated 

with solvent vapors exits the vessel as the surrounding 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 repeated. 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, hydrocarbons, 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 system 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 particulate 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 stream 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 firms 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/PM10 Control Systems 

PM/PM10 control systems for the chemical industry consist 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 into 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 identifying a bag house for an 

application it is important to consider the particle size in the emission 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 appropriate 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 with emission streams of high 

moisture content. When operated under optimum conditions, they can generally achieve control 

efficiencies of up to 99+ percent (EIIP, 2000). However, 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 different 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 filling model can be used to estimate the emissions 

from charging the primary raw materials or transferring the batch from one vessel to a second vessel. The 

heating and filling models can be used 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) volume 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 component in the system is calculated by applying 

Raoult’s Law to the pure vapor pressure and the mole 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 adjust 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 the 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 equation assumes that the partial pressure of component i in the 

vent gas is at saturated levels. 

Ideal Gas Law: 

RT

p V E i

n-i = Eq. 3-1 

Where En-i are the moles of component i that are emitted due to vapor displacement 

pi 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. 

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

EIIP Volume II 16.3-2 

Where pi = effective vapor pressure for component i 

xi = mole fraction of component i 

i γ = component activity coefficient (Becomes 1.0 when Raoult’s Law applies) 

Pi = 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 3,600 gallons of hexane in 

one hour. The empty vessel was previously made inert 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) 

PSystem = 1.0 atm = 760 mm Hg (Total system pressure) 

Vdisplacement = 3600 gal = 481.28 ft3

 (Displacement volume) 

Time = 1 hr (Time for event) 

Constants and Relationships: 

Universal Gas Constant: 

lb mole K

mmHg ft R − ⋅°

⋅ =

3

998.9

Antoine Equation: ⎟

⎠

⎞ ⎜

⎝

⎛

+ = −T c

b Pi exp a

Gas Law: 

RT

PV

n = also 

RT

PV

n i

i = for component i in the gas space. 

Sum of the partial pressures in the gas space: ∑= = N

i T i P p1

Sum of component moles in the gas space: ∑= = N

i T i N n1

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 nitrogen is determined by the difference between the 

total system pressure, 760 mm Hg, and the partial pressure of hexane. The vapor pressure of hexane may 

be calculated from the Antoine equation as follows: 

phexane exp( ) 5.019 151.28 mmHg 298.15 48.78

2697.55

exp 15.8366 ⎟ = = ⎠

⎞ ⎜

⎝

⎛

− = −

therefore pN 2 = PT − phexane = 760 mmHg −151.28 mmHg = 608.719 mmHg

Ideal Gas Law: lb moles

C

mmHg ft

RT

p V

E hexane

n hexane = − ° +

⋅ − = = 0.244

(998.9)(25 273.15)

151.281 481.28 3
Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models 

EIIP Volume II 16.3-3 

 lb moles mmHg ft

RT

p V

E N

n N = − ⋅ − = = 0.984

(998.9)(298.15)

608.719 481.28 3

2

2

Emission level: ( ) lb

lb mole

lb E lb moles Wt hexane 0.244 86.17 ⎟ = 21.03 ⎠

⎞ ⎜

⎝

⎛

− − = −

 

( ) lb

lb mole

lb E lb moles Wt N 0.984 28.01 27.562 2 ⎟ =

⎠

⎞ ⎜

⎝

⎛

− − = −

Step 3. Calculate the emission rate based on the 1 hour addition. 

Emission Rates: 

hr

lb

hr

lb ER hexane 21.03

1.0

21.03 − = =

 hr

lb

hr

lb ER N 27.56

1.0

27.56

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) 

PT = 1.0 atm = 760 mm Hg (Total system pressure) 

Vdisp = 1,500 gal = 200.53 ft3 (Displacement volume) 

Time = 5 min (Time for event) 

Constants and Relationships: 

Universal Gas Constant: 

lb mole K

mmHg ft R − ⋅°

⋅ =

3

998.9

Antoine Vapor Pressure Equation: 

 

Pi = exp a − b

T + c

⎛

⎝ ⎞

⎠

Gas Law: 

RT

PV

n = , also 

RT

p V

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
Chapter 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 contains nitrogen as the remaining gas component. Pure 

component vapor pressures for the liquid components may be estimated using the Antoine equation. 

 

 

Pheptane = exp 15.8737 − 2911.32

293.15 − 56.51

⎛

⎝ ⎞ ⎠ = exp(3.571) = 35.55mmHg

 

 

Ptoluene = exp 16.0137 − 3096.52

293.15 −53.67

⎛

⎝ ⎞ ⎠ = exp(3.0835) = 21.84mmHg

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: 

 

lb moles mmHg ft

RT

p V

E

sys

heptane

n-heptane = − ⋅ = = 0.0102

(998.9)(293.15)

14.93 200.53 3

lb moles mmHg ft

RT

p V

n

sys

toluene

n toluene = − ⋅ − = = 0.0087

(998.9)(293.15)

12.67 200.53 3

n

p V

RT

mmHg ft lb moles nitrogen

air

sys

= = ⋅ = − 732 40 20053

998 9 29315

05015

3 . .

( . )( . ) .

Emission level: 

( ) lbs

lb mole

lbs E lb moles Wt-heptane 0.0102 100.205 ⎟ =1.02 ⎠

⎞ ⎜

⎝

⎛

− = −

( ) lbs

lb mole

lbs E lb moles Wt toluene 0.0087 92.13 ⎟ =0.80 ⎠

⎞ ⎜

⎝

⎛

− − = −

( ) lbs

lb mole

lbs E lb moles Wt nitrogen 0.5015 28.0134 ⎟ = 14.05 ⎠

⎞ ⎜

⎝

⎛

− − = −

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 

lbs hr

min

min/hr E lbs R heptane 2.04 /

30

60 1.02 =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ − =

lbs hr

min

min/hr E lbs R toluene 1.6 /

30

60 0.80 =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ − =

lbs hr

min

min/hr E lbs R nitrogen 28.1 /

30

60 14.05 =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ − =

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 batch will also change in accordance to the batch 

composition. 

At any point in the filling operation one may calculate the batch composition from the initial vessel 

contents and the amount of material that has been added. Let nA represent the moles of inlet mixture that 

are to be added to the vessel and let NB represent the total number of moles of mixture that are initially 

contained in the vessel regardless of composition. For example, if the inlet stream contained ethanol, 

water, and methanol then nA would represent the total moles of ethanol, water, and methanol that have 

been charged at any point in the operation. 

 

A B

A

A

n N

n

+

φ = Eq. 3-3 

Where φ A is the degree of dilution of the inlet stream mixture at any point during the addition, 

 nA are the moles of inlet steam mixture charged to the vessel, and 

 NB are the moles of mixture that were initially contained in the vessel prior to the 

addition. 

The average dilution φ A of the inlet stream A from being mixed with the contents of the vessel may be 

calculated by integrating a differential expression for A dφ 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

A B

A

A

A dn

n N

n

N

A

∫ + =

0

1

φ Eq. 3-4 

 ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+ = +

A B

B

A

B

A N N

N

N

N

φ 1 ln Eq. 3-5 

A similar calculation may be made for the average dilution factor of mixture B (the initial vessel contents) 

that exists during the filling process. 

 ( ) A

N

A A B

B

B dn

N n N

N A

∫ + =

0

1

φ Eq. 3-6 
Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models 

EIIP Volume II 16.3-6 

 ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+ = −

A B

B

A

B

B N N

N

N

N

φ ln Eq. 3-7 

Once φ A and φ B 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 calculate the average vessel headspace vapor composition. 

φ A and φ B 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 addition 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 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 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: 

lb mole K

mmHg ft R − ⋅°

⋅ =

3

998.9

Antoine Vapor Pressure Equation: 

 

Pi = exp a − b

T + c

⎛

⎝ ⎞

⎠

Gas Law: 

RT

PV

n = , also 

RT

p V

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
Chapter 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 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 

Calculate the inlet stream dilution factor: 

From Eq 3.5: ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+ = +

A B

B

A

B

A N N

N

N

N

φ 1 ln 

⎟

⎠

⎞ ⎜

⎝

⎛

+ = +33.90 101.37

101.37 ln

33.90

101.37

φ A 1

= 1+ 2.99ln(0.7494) = 0.14 φ A

Calculate the dilution factor for the initial vessel contents: 

From Eq 3.7 ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+ = −

A B

B

A

B

B N N

N

N

N

φ ln

⎟

⎠

⎞ ⎜

⎝

⎛

+ = − 33.90 101.37

101.37 ln

33.90

101.37

φ B

= −2.99ln(0.7494) = 0.86 φ B

Step 3. Calculate the average batch and vapor-phase compositions. 

VOC Xi φA, φB Xi

Pi 

(mm Hg) 

pi 

(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 1.00 760.00 
Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models 

EIIP Volume II 16.3-8 

Ideal Gas Law: lb moles mmHg ft

RT

p V E

sys

acetone

n-acetone = − ⋅ = = 0.00354

(998.9)(293.15)

25.87 40.1 3

lb moles mmHg ft

RT

p V

E

sys

heptane

n-heptane = − ⋅ = = 0.00176

(998.9)(293.15)

12.83 40.1 3

lb moles mmHg ft

RT

p V E

sys

toluene

n toluene = − ⋅ − = = 0.0015

(998.9)(293.15)

10.92 40.1 3

lb moles mmHg ft

RT

p V

E

sys

nitrogen

n nitrogen = − ⋅ − = = 0.0973

(998.9)(293.15)

710.41 40.1 3

Emission level: ( ) lbs

lb mole

lbs E lb moles wt-acetone 0.00354 58.08 ⎟ = 0.21 ⎠

⎞ ⎜

⎝

⎛

− = −

( ) lbs

lb mole

lbs E lb moles wt-heptane 0.00176 100.205 ⎟ = 0.18 ⎠

⎞ ⎜

⎝

⎛

− = −

( ) lbs

lb mole

lbs E lb moles wt toluene 0.0015 92.13 ⎟ =0.14 ⎠

⎞ ⎜

⎝

⎛

− − = −

( ) lbs

lb mole

lbs E lb moles wt nitrogen 0.0973 28.0134 ⎟ = 2.73 ⎠

⎞ ⎜

⎝

⎛

− − = −

Step 4. Calculate the emission rate in lb/hr units. 

lbs hr

hr

lbs ER acetone 0.42 /

0.5

0.21 − = =

lbs hr

hr

lbs ER heptane 0.36 /

0.5

0.18 − = =

lbs hr

hr

lbs ER toluene 0.56 /

0.5

0.28 − = =

lbs hr

hr

lbs ER nitrogen 5.46 /

0,5

2.73 − = =

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 added to the vessel featured in Illustration 3-2 by way 

of above surface addition. For this problem, the initial 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 headspace 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 φ A 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) 

Vdisp = 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: 

lb mole K

mmHg ft R − ⋅°

⋅ =

3

998.9

Antoine Vapor Pressure Equation: 

 

Pi = exp a − b

T + c

⎛

⎝ ⎞

⎠

Gas Law: 

RT

PV

n = also 

RT

p V

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 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 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 

Calculate the dilution factor for the initial vessel contents: 

From Eq 3-7 ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+ = −

A B

B

A

B

B N N

N

N

N

φ ln

⎟

⎠

⎞ ⎜

⎝

⎛

+ = − 33.90 101.37

101.37 ln

33.90

101.37

φ B

= −2.99ln( ) 0.7494 = 0.86 φ B
Chapter 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 φA, φB Xi

Pi 

(mm Hg) 

pi 

(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: lb moles mmHg ft

RT

p V E

sys

acetone

n-acetone = − ⋅ = = 0.0.0253

(998.9)(293.15)

184.80 40.1 3

lb moles mmHg ft

RT

p V

E

sys

heptane

n-heptane = − ⋅ = = 0.00176

(998.9)(293.15)

12.83 40.1 3

lb moles mmHg ft

RT

p V E

sys

toluene

n toluene = − ⋅ − = = 0.0015

(998.9)(293.15)

10.92 40.1 3

lb moles mmHg ft

RT

p V

E

sys

nitrogen

n nitrogen = − ⋅ − = = 0.0755

(998.9)(293.15)

551.48 40.1 3

Emission level: ( ) lbs

lb mole

lbs E lb moles wt-acetone 0.0253 58.08 ⎟ =1.47 ⎠

⎞ ⎜

⎝

⎛

− = −

( ) lbs

lb mole

lbs E lb moles wt-heptane 0.00176 100.205 ⎟ = 0.18 ⎠

⎞ ⎜

⎝

⎛

− = −

( ) lbs

lb mole

lbs E lb moles wt toluene 0.0015 92.13 ⎟ =0.14 ⎠

⎞ ⎜

⎝

⎛

− − = −

( ) lbs

lb mole

lbs E lb moles wt nitrogen 0.0755 28.0134 ⎟ = 2.12 ⎠

⎞ ⎜

⎝

⎛

− − = −

Step 4. Calculate the emission rate in lb/hr units. 

lbs hr

hr

lbs ER acetone 2.94 /

0.5

1.47 − = =

lbs hr

hr

lbs ER heptane 0.36 /

0.5

0.18 − = =

lbs hr

hr

lbs ER toluene 0.56 /

0.5

0.28 − = =

lbs hr

hr

lbs ER nitrogen 4.24 /

0,5

2.12 − = =
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) 

PT = 1.0 atm = 760 mm Hg (Total system pressure) 

Vdisp = 750 gal = 100.2 ft3 (Displacement volume) 

Time = 30 min (Time for event) 

Constants and Relationships: 

Universal Gas Constant: 

lb mole K

mmHg ft R − ⋅°

⋅ =

3

998.9

Antoine Vapor Pressure Equation: 

 

Pi = exp a − b

T + c

⎛

⎝ ⎞

⎠

Gas Law: 

RT

PV

n = , also 

RT

p V

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 Xi

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 Xi

Water 18.02 8.33 500 4,165 231.13 1.00 

Total 231.13 1.00 
Chapter 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 Xi

Pi 

(mm Hg) 

Toluene 1.00 21.84 

Water 1.00 17.35 

Nitrogen 720.81 

Total 1.00 760.00 

Ideal Gas Law: lb moles mmHg ft

RT

p V E

sys

toluene

n-toluene = − ⋅ = = 0.0.00747

(998.9)(293.15)

21.84 100.2 3

lb moles mmHg ft

RT

p V E

sys

water

n-water = − ⋅ = = 0.00594

(998.9)(293.15)

17.35 100.2 3

lb moles mmHg ft

RT

p V

E

sys

nitrogen

n nitrogen = − ⋅ − = = 0.24665

(998.9)(293.15)

720.81 100.2 3

Emission level: ( ) lbs

lb mole

lbs E lb moles wt toluene 0.00747 92.13 ⎟ =0.69 ⎠

⎞ ⎜

⎝

⎛

− − = −

( ) lbs

lb mole

lbs E lb moles wt water 0.00594 18.02 ⎟ = 0.11 ⎠

⎞ ⎜

⎝

⎛

− − = −

( ) lbs

lb mole

lbs E lb moles wt nitrogen 0.24665 28.0134 ⎟ = 6.91 ⎠

⎞ ⎜

⎝

⎛

− − = −

Step 4. Calculate the emission rate in lb/hr units. 

lbs hr

hr

lbs ER toluene 1.38 /

0.5

0.69 − = =

lbs hr

hr

lbs ER water 0.22 /

0.5

0.11 − = =

lbs hr

hr

lbs ER nitrogen 13.82 /

0,5

6.91 − = =

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 determined by the following expression. 

 ⎟

⎠

⎞ ⎜

⎝

⎛ = − −

− V Ft in i e

RT

p V

E 1 ,1

Where En-i are the moles of component i that are emitted due to vapor displacement, 

pi,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 gal

gal ft = = ft

2000

7 48

267 3

3

. /

n

F t

V

acf

ft = ⋅ = = 1073267

4 02 3 .

Step 2. Calculate the acetone loss from the vessel by conducting a material balance around the 

vessel. 

Where Nacetone is the amount of acetone displaced from the vessel: 

⎟

⎠

⎞ ⎜

⎝

⎛ = − −

− V Ft in i e

RT

p V

E 1 ,1

( ) 4.02 1

(998.9)(293.15)

(184.86)(267) − E − = − e n i

E ( e ) lb moles n acetone = − = = − −

− (0.1686) 1 (0.1686)(0.9821) 0.166 4.02

E lb moles lb lb mole lb wt acetone = (0.166 − )(58.08 / − ) = 9.58 −

Step 3. Calculate the amount of nitrogen emitted from the vessel. 

Before the empty vessel purge began, the gas space was saturated with acetone vapors at a partial pressure 

of 184 mm Hg. At the completion of the purge operation, 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. 

E (out) N (in) N (accum.) n−nitrogen = nitrogen − nitrogen

RT

p p V

RT

PV

E out purge i i vessel

n nitrogen

( ) ( ) ,1 − ,2

− = −

E out lb moles n nitrogen = − − − = − 2.62

(998.9)(293.15)

(184.86 3.32)(267)

(998.9)(293.15)

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

EIIP Volume II 16.3-14 

E lb moles lb lb mole lb wt nitrogen = (2.62 − )(28 / − ) = 73.4 −

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 criteria. 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 Si 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 Si

implies that the exit vent gas is at 25% saturation level with respect to the solvent vapors in the tank, 

while a Si 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

R i R nc p

S p E − = E − Eq. 3-8

Where ER- = moles of volatile component i emitted per unit time, 

 Si = saturation level of the exit vent gas stream, 

 Enc = moles of non-condensable gas emitted per unit time (= inlet purge rate), 

sat pi = partial pressure of component i at saturated conditions, 

sat pnc = 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 function of the evaporation mass transfer coefficient Ki, the 

liquid surface area in the vessel, and the solvent partial pressure in the vessel headspace. When the exit 

emission rate for a vessel is set equal to the evaporation rate in the vessel then the following expression 

results for the saturation level Si. 

Saturation level SI: sat

i nc i i

i

i

i

Sat

i

i

i K A F S F

K A

K A F

K A

p

p S

+ + = + = = Eq. 3-9 

 Where 3

1

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

i

o

i o M

M K K , Eq. 3-10 

 and ( ) sat

sys i

sat

i

sat nc

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, 

Ki = mass transfer coefficient i, 

Ko = mass transfer coefficient of a reference compound, 

Mi = molecular weight of compound i, 
Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models 

EIIP Volume II 16.3-15 

Mo = molecular weight of the reference compound o, 

A = surface area of the liquid, 

Fnc = volumetric flow rate of the non-condensable gas (i.e. air, nitrogen), 

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

sat pi = saturated vapor pressure of component i, 

sat pnc = partial pressure of the noncondensable gas (i.e. air, nitrogen) at saturated solvent 

pressure conditions, 

Psys = 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 partial pressure of the non-condensable at saturated 

conditions. 

Si may be solved using the standard quadratic solution. Although the standard quadratic equation contains 

two roots, only the one solution shown in Eq. 3-12 produces a realistic value since Si must be a positive 

number between 0 and 1.0. 

Quadratic solution: sat

i

i

sat

i nc i nc i

i F

K A F K A F F K A

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 Si , where sat pi = Si pi . 

 RT

M S F P

E sys

sat

i i i

i = Eq. 3-13 

For multi-component liquid mixtures, Eq. 3-9 may be expanded to provide partial volumetric flow levels 

for each volatile component in the liquid. 

 

sat

n n

sat

j j

sat

i nc i i

i

i K A F S F S F S F

K A 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 iteration. Finally, when the saturation level S of each 

compound remains the same for subsequent iterations then the calculation process is stopped. 

Emission rates for each solvent compound are then calculated based on the partial volumetric flow rate as 

calculated from the saturation level and the saturated volumetric flow rate sat Fi = SiFi . 

Solvent emission rate 

RT

M S F P

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

0.92

30.48 min

60 0.4685

100.2

18.02 0.83 3

1 3

1

ft

cm

s ft

s

cm

s

cm

M

M

K K

i

o

i o = ⋅ ⋅

⋅ ⋅ ⎟ = ×

⎠

⎞ ⎜

⎝

⎛ = ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

45.86 . P tan mmHg sat

hep e = P 760 45.86 714.14 mmHg. satnitrogen = − =

10.92 / min

273.15

298.15 10 3 ft KK Fnitrogen scfm = °° =

0.70 / min

714.14

45.86 10.92 3

tan ft mmHg

mmHg F sat

hep e = =

2

2 2

28.26

4

3.14 36

4 ft d ft A = ∗ = = π

2 0.70

(0.92 28.26 10.92) (0.92 28.26 10.92) 4 0.70 0.92 28.26 2

tan ∗

− ∗ + + ∗ + + ∗ ∗ ∗ Shep e =

0.69

1.40

0.97

1.40

36.92 1363.0 72.79 tan = = − + + Shep e =

( )( )( )( )

( ) ft atm lb mol R ( ) R

lb lb mole ft mmHg

RT

M S F P

E

L

sys

sat

i i i

m − ° °

− = = 554.98 / 537100.2 / 0.69 0.70 / min 760 3

3

E lb lb hr R m = 0.12 / min = 7.2 / −

As an illustration on how the saturation factor for heptane could have been calculated using the iteration 

technique Eq 3-14 can be applied directly. 

i i

sat

i nc i i

i

i K A F S F S S

K A S

+ ∗ = ∗ + +

∗ = + + = 36.92 0.7

25.99

0.92 28.26 10.92 0.70

0.92 28.26

Starting with an initial guess for Si to equal 1.0, the convergence occurs rapidly with the final result 0.69 

being equal to earlier results using the quadratic equation. 

 Iteration 0 Si= 1.00 

 Iteration 1 Si = f(1.00) = 0.69 

 Iteration 2 Si = 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 contained a solvent mixture consisting of 20% heptane, 70% 

toluene, and 10% methanol. Assuming that the composition is specified in mole percents, calculate the 
Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models 

EIIP Volume II 16.3-17 

saturation factor for each component using Eq. 3-14. Before 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 substituted into Eq. 3-14 for an iterative trial and error 

procedure. 

 sat

n n

sat

j j

sat

i nc i i

i

i K A F S F S F S F

K A S

+ + + + + + = ... 1 Eq. 3-14 

Table 3-2. Iterative Trial and Error Results for Si using Eq. 3-14 

Compound KiA FiSat 

Si 

(iteration 0)

Si 

(iteration 1)

Si 

(iteration 2)

Si 

(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 component may be calculated from the relationship. 

 RT

M S F P

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 Si FiSat Qi (lb/min) Qi (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 distillation 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 temperature, 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 applied 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 nitrogen 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 the vent emissions are calculated for the system as 

being from the distillate receiver based on the condensate 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 that 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

n i n nc p

p E − = E − Eq. 3-15

Where En-i = moles of volatile component i emitted from the process, 

 En-nc = total moles of noncondensable gas emitted from the process, 

pi = partial pressure of volatile component i, 

pnc = partial pressure of the noncondensable gas (i.e. air, nitrogen) at saturated 

solvent pressure conditions. 

En-nc represents the total moles of non-condensable gas component that are removed from the system by 

the vacuum pump. 

 En−nc = En−nc−leakage + En−nc−displacement + En−nc−gas sweep Eq. 3-16

Where En-nc-leakage are the moles of air due to leakage into the system, 

 En-nc-displacement are the moles of air that are displaced by the condensate, 

 En-nc-gas sweep are the moles of air or nitrogen admitted as a gas sweep. 

If the distillation phase has been completed and the 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 (ft3

/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 (ft3

/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 mixture under vacuum conditions in 2.5 hours. The 

equipment consists of a 1000 gallon still, condenser, and 1000 gallon receiver. 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 the 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 collected 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: Pnc = 100 mm Hg – 12.43 mm Hg. = 87.57 mm Hg. 

Displacement volume in ft3

: 

3

3 53.44

7.485 /

400 ft gal ft

gal V = =

lb moles

K

mmHg ft

RT

p V E

sys

nc

n nc displacement = − °

⋅ − − = = 0.25

(998.9)(293.15 )

87.57 53.44 3

Leakage volume in ft3

: 3 V (10 acfm)(2.5hr)(60min/ hr) 1560 ft leakage = =

lb moles

K

mmHg ft

RT

p VT E

K

nc m

n nc leakage = − °

⋅ = =

°

− − 0.53

(998.9)(293.15 )

100 1560 3

Therefore 

En−nc = En−nc−leakage + En−nc−displacement + En−nc−gas sweep

E lb moles n−nc = 0.53n−nc−leakage + 0.25n−nc−displacement + 0n−nc−gas sweep = 0.78 −

Finally 

lb moles

p

p E E

nc

i

n−toluene = n−nc = = 0.11 − 87.57

12.47 0.78

Ewt−toluene = En−tolueneMwtoluene = (0.11lb − moles)(92.13 lb/lb - mole) =10.23 lb

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 ft3

 tray drier, condenser, and 250 gallon receiver. 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. Although 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 leak 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 V (1 acfm)(60min) 60 ft leakage = =

lb moles

K

mmHg ft

RT

p VT E

K

nc m

n nc leakage = − °

⋅ = =

°

− − 0.0054

(998.9)(280.15 )

25 60 3

Gas sweep (nitrogen): 

(1 )(60min) 60 ( ) 3 Vgas sweep = scfm = ft stp

lb moles

K

mmHg ft

RT

P V

E

stp

stp

n nc gas sweep = − °

⋅ − − = = 0.17

(998.9)(273.15 )

760 60 3

Therefore 

En−nc = En−nc−leakage + En−nc−displacement + En−nc−gas sweep

E lb moles n−nc = 0.0054n−nc−leakage + 0n−nc−displacement + 0.17n−nc−gas sweep = 0.18 −

Finally 

p P p mmHg nc sys toluene = − = 25.0 −10.4 =14.6

lb moles

p

p E E

nc

i

n−toluene = n−nc = = 0.13 − 14.60

10.40 0.18

= = (0.13 − )(92.13 lb/lb - mole) =11.98 lb E − E − Mw lb moles wt toluene n toluene toluene

3.4 Gas Evolution 

Certain processes generate off gases as function of the 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 components in the batch. The partial pressure of each 

component is calculated based on the pure component vapor 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

n i n rxn p

p E − = E − Eq. 3-17

where: En-i = moles of volatile component i emitted from the process, 

 En-rxn = total moles of reaction off gas emitted from the process, 

pi = partial pressure of volatile component i, 

prxn = 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 determined 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 soluble 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. Eight 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. 

lb mole

lb lb mole

lbs

MWt

Wt E

HCl

HCl

n HCl = − − − = = 0.219

36.461 /

8

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

Ptoluene C exp( ) 4.52298 92.11 mmHg 323.15 53.67

3096.52 ,50 exp 16.0137 ⎟ = = ⎠

⎞ ⎜

⎝

⎛

− ° = −

ptoluene,50°C = 0.95× 92.11 mmHg = 87.50 mmHg

Step 3. Calculate the toluene emission rate by the ratio of vapor pressures. 

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ − =

HCl

toluene

n toluene HCl p

p E E

lb moles

mmHg mmHg

mmHg E lb moles n toluene = − ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− − = − 0.0285

760 87.50

87.50 0.219

E lb moles lb lb mole wt−toluene = 0.0285 − ×92.13 / −

E lbs wt toluene = 2.63 −

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. Eight pounds of hydrogen 

chloride along with an equal molar quantity of sulfur dioxide is generated 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 and sulfur dioxide that is discharged from the vessel. 

The molar amount of HCl is calculated from the molecular weight and the quantity of HCL emitted. The 

molar amount of SO2 is set equal to the calculated molar amount of HCl. 

lb mole

lb lb mole

lbs

MWt

Wt E

HCl

HCl

n HCl = − − − = = 0.219

36.461 /

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

EIIP Volume II 16.3-22 

E n lb moles n−SO = HCl = 0.219 − 2

0.219 64.063 lb/lb - mole 14.03 lbs 2 Ewt−SO = lb − moles × =

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

Ptoluene C exp( ) 4.52298 92.11 mmHg 323.15 53.67

3096.52 ,50 exp 16.0137 ⎟ = = ⎠

⎞ ⎜

⎝

⎛

− ° = −

ptoluene,50°C = 0.95×92.11 mmHg = 87.50 mmHg

Step 3. Calculate the toluene emission rate by the ratio of vapor pressures. 

( ) ( ) 2 2

2

HCl SO

sys toluene

toluene

HCl SO

HCl SO

toluene

n toluene n n

P p

p n n

p p

p E + ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− + = ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+ − =

lb mole lb moles

mmHg mmHg

mmHg En toluene − = − ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− − = 0.438 0.057760 87.50

87.50

E lb moles lb lb mole wt−toluene = 0.057 − × 92.13 / −

E lbs wt−toluene = 5.26

E lbs wt−HCl = 8.0

E lbs wt SO 14.03 2 − =

Illustration 3-13: Reaction involving multicomponent gas evolution and nitrogen purge. 

A reaction takes place over a 1.0 hour period in a vessel at 50°C with toluene as the primary solvent. 

Eight pounds of hydrogen chloride along with an equal molar quantity of sulfur dioxide is generated 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 and a nitrogen purge is being applied at 30 SCFH. Also, 

assume that the batch consist of 95% toluene (mole/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 SO2 is set equal 

to the calculated molar amount of HCl. 

lb mole

lb lb mole

lbs

MWt

Wt E

HCl

HCl

n HCl = − − − = = 0.219

36.461 /

8

E n lb moles n−SO = HCl = 0.219 − 2

E lb moles lb lb moles lbs wt SO 0.219 64.063 / 14.03 2 − = − × − =

The molar amount of N2 is calculated based on the 30 SCFH (standard 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 

lb mole

SCF lb mole

SCFH HR En N = − −× − = 0.084

359.046 /

30.0 1.0

2

E lb moles lbs wt N 0.084 28.013 lb/lb - moles 2.35 2 − = − × =

The total moles of non-condensable compounds are calculated by summing the molar amounts of HCl, 

SO2, and N2. 

0.219 0.219 0.084 2 2 En−nc = En−NCl + En−SO + En−N = + +

E lb moles n−nc = 0.522 −

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

Ptoluene C exp( ) 4.52298 92.11 mmHg 323.15 53.67

3096.52 ,50 exp 16.0137 ⎟ = = ⎠

⎞ ⎜

⎝

⎛

− ° = −

ptoluene,50°C = 0.95×92.11 mmHg = 87.50 mmHg

Step 3, Calculate the toluene emission rate by the ratio of vapor pressures. 

n nc

sys toluene

toluene

n nc

nc

toluene

n toluene E

P p

p E

p

p E − − − ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− = ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

=

∑

lb mole lb moles

mmHg mmHg

mmHg En toluene − = − ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− − = 0.522 0.13

760 87.50

87.50

E lb moles lb lb mole wt−toluene = 0.13 − × 92.13 / −

E lbs wt−toluene = 11.98

E lbs wt−HCl = 8.0

E lbs wt SO 14.03 2 − =

E lbs wt N 2.35 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 certain assumptions and approximations be made: 

 The system pressure is decreased linearly over time. 

 Air leakage into the vessel during the operation is negligible. 

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

 The vapor space of the vessel remains in equilibrium with the volatile 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 depressurization operation then 

the equilibrium vapor pressure of the vessel liquid contents remains constant as well. The moles of 

solvent vapor that exist within the vessel headspace during the depressurization remain constant for this 

reason. However, the volatile solvent vapor occupies a greater fraction of the vessel headspace and exit 

vent gas as the depressurization takes place since the system pressure is being reduced toward the solvent 

vapor pressure level. As the depressurization unfolds then more and more solvent must evaporate in order 

to maintain the equilibrium vapor pressure. Therefore, the solvent emissions that occur during 

depressurization are equal to the net solvent evaporation within the vessel, based on the assumptions for 

this model. 

Eq. 3-18 is based on a material balance for the 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 integrated as shown in Eq. 3-19 to result in Eq. 3-20. . 

[Hatfield, 1998b] 

 nc

nc

i

i dp

p

p

RT

V dn out ⎟

⎠

⎞ ⎜

⎝

⎛ ( ) = − Eq. 3-18 

 ∫ ⎟ ∫ ⎟

⎠

⎞

⎜

⎜

⎝

⎛ = −

2

1

2

1 ,

( )

p

p nc

nc

T V

i

p

p

i p

dp

RT

Vp dn out Eq. 3-19 

 ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

,2

,1

, ln

nc

i nc

i out p

p

RT

Vp

n Eq. 3-20 

Where: ni,out = moles of volatile component i leaving the vessel 

V = vessel headspace volume 

pi = partial pressure of the volatile component 

R = Universal gas constant 

T = system temperature 

pnc,1 = partial pressure of the non-condensable component at initial conditions 

pnc,2 = partial pressure of the non-condensable 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 onto 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 (Pnc,1) and final conditions (Pnc,2). 

p mmHg acetone C , . exp . ( ) .

. . exp . . 26 7 16 6513 2940 46299 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. 

lb moles

p

p

RT

Vp E

nc

i nc

n acetone ⎟ = − ⎠

⎞ ⎜

⎝

⎛ =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− = 0.0827

512.89

2322.89 ln

(998.9)(300)

(66.843)(247.11) ln

,2

,1

Ewt-acetone = (0.0827 lb-moles) ( 58.08 lb/lb-mole) = 4.8 lbs 

Illustration 3-15: Vessel depressurization involving a solvent mixture.

A 1,200 gallon process vessel contains 700 gallons of solvent 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 depressurization 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: 

lb mole K

mmHg ft R − ⋅°

⋅ =

3

998.9

Gas Law: n

PV

RT = , also n

pV

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

Xi

Pi=Xi*Pi 

(mm Hg) 

Acetone 184.86 0.2 36.97 

Methanol 97.30 0.3 29.19 

Toluene 21.84 0.5 10.92 

Total 77.08 
Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models 

EIIP Volume II 16.3-26 

lb moles

p

p

RT

Vp E

nc

i nc

n acetone ⎟ = − ⎠

⎞ ⎜

⎝

⎛ =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− = 0.0286

22.92

682.92 ln

(998.9)(293.15)

(66.843)(36.97) ln

,2

,1

lb moles

p

p

RT

Vp E

nc

i nc

n methanol ⎟ = − ⎠

⎞ ⎜

⎝

⎛ =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− = 0.0226

22.92

682.92 ln

(998.9)(293.15)

(66.843)(29.19) ln

,2

,1

lb moles

p

p

RT

Vp E

nc

i nc

n toluene ⎟ = − ⎠

⎞ ⎜

⎝

⎛ =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− = 0.0085

22.92

682.92 ln

(998.9)(293.15)

(66.843)(10.92) ln

,2

,1

3.6 Heating 

Heat-up losses that occur during the operation of reactors, 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. The 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 amount of the non-condensable component (air, nitrogen, 

etc.) that is displaced from the vessel is determined by the initial and final gas space composition. If a 

nitrogen purge or sweep is placed on the vessel during the heating step, then the amount of noncondensable component that is displaced from the vessel 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 from the vapor space. This model assumes that the 

average molar headspace volume 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

n i out avg n n

p

p E N ,2 ,1

,2

,1

, ln − − ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− =

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

EIIP Volume II 16.3-27 

 Where: ( ) 1 2 2

1 Navg = n + n Eq. 3-22 

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

Navg = average gas space molar volume during the heating process 

pnc1 = partial pressure of non-condensable in the vessel headspace at temperature T1

pnc2 = partial pressure of non-condensable in the vessel headspace at temperature T2

ni,1 = moles of volatile component i in the vessel headspace at temperature T1

ni,2 = moles of volatile component i in the vessel headspace at temperature T2 

n1 = total moles of gas in the vessel headspace at temperature T1 

n2 = total moles of gas in the vessel headspace at temperature T2 

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 the heat up. How much toluene will be emitted? 

Step 1. Calculate the average molar volume of the vessel headspace. 

Ti = 20°C = 293.15°K (Initial temperature) 

Tf = 70°C = 343.15°K (Final temperature) 

PT = 1.0 atm = 760 mm Hg (Total system pressure) 

Vgas = 500 gallons = 66.843 ft3

 (Gas space volume) 

R = 998.9 (mmHg-ft3

)/(lb-mole-oK) (Universal gas constant) 

Gas Law: n

PV

RT = , also n

pV

RT i

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

( ) 1 2 2

1 Navg = n + n

lb moles

RT

PV

RT

PV Navg ⎟ = − ⎠

⎞ ⎜

⎝

⎛ +

×

×

⎥ =

⎦

⎤ ⎢

⎣

⎡ ⎟

⎠

⎞ ⎜

⎝

⎛ ⎟ + ⎠⎞ ⎜

⎝

⎛ = 0.1608

343.15

1

293.15

1

2 998.9

760 66.843

2

1

1 2

Step 2. Calculate the initial and final partial pressures of nitrogen. 

Use the Antoine equation to calculate the partial pressure of toluene: 

 

Ptoluene, 20o C = exp 16.0137 − 3096.52

293.15 −53.67

⎛

⎝ ⎞ ⎠ = exp(3.0835) = 21.84mmHg

 

Ptoluene,70 o C = exp 16.0137 − 3096.52

343.15 −53.67

⎛

⎝ ⎞ ⎠ = exp(5.317) = 203.74mmHg

The partial pressure of nitrogen is the difference between the total system pressure and the partial pressure 

of toluene: 

pnc,1 = 760 − ptoluene,1 = 760 − 21.84 = 738.16mmHg

pnc,2 = 760 − ptoluene,2 = 760 − 203.74 = 556.26mmHg
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

p V

n toluene

toluene 0.00499

998.9 293.15

21.84 66.843

1

,1

,1 = ×

× =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

moles

RT

p V

n

toluene

toluene 0.0397

998.9 343.15

203.74 66.843

2

,2

,2 = ×

× =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

Step 4. Calculate the toluene emission using Eq. 3-21. 

The moles of toluene that are displaced 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

n i avg n n

p

p E N ,2 ,1

,2

,1 ln − − ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− =

E ( ) ( ) lb moles n toluene ⎟ − − = − ⎠

⎞ ⎜

⎝

⎛ − = 0.0397 0.00499 0.0108

556.26

738.16 0.1608 ln

E ( )( ) lb moles lb lb mole lbs wt toluene = 0.0108 − 92.13 / − = 0.995 −

Illustration 3-17: Heatup losses from a vessel containing a volatile mixture. 

A 2000 gallon reactor contains 1,500 gallons of a solvent 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 

Pi @ 20C

(mm Hg) 

XiPi (20C) 

(mm Hg) 

Pi @ 70C 

(mm Hg) 

XiPi (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. 

( ) 1 2 2

1 Navg = n + n
Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models 

EIIP Volume II 16.3-29 

lb moles

RT

PV

RT

PV Navg ⎟ = − ⎠

⎞ ⎜

⎝

⎛

⎟ + ⎠

⎞ ⎜

⎝

⎛

×

×

⎥ =

⎦

⎤ ⎢

⎣

⎡ ⎟

⎠

⎞ ⎜

⎝

⎛ ⎟ + ⎠

⎞ ⎜

⎝

⎛ = 0.1608

343.15

1

293.15

1

2 998.9

760 66.843

2

1

1 2

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

p V

n i

i 0.01623

998.9 293.15

71.12 66.843

1

,1

,1 = ×

× =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

moles

RT

p V

n i

i 0.09544

998.9 343.15

489.43 66.843

2

,2

,2 = ×

× =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

( )vessel i i

nc

nc

n i avg n n

p

p E N ,2 ,1

,2

,1 ln − − ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− =

E ( ) ( ) lb moles n i ⎟ − − = − ⎠

⎞ ⎜

⎝

⎛ − = 0.09544 0.01623 0.07106

270.57

688.873 0.1608 ln

Table 3-5. Final Calculated Results for This Problem 

Compound 

avg pi 

(mm Hg)

Fraction 

To total

ni 

(lb-moles) MWti 

wti 

(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 solution 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. Assuming a 25% vapor saturation of the gas sweep vapors, 

how much toluene will be emitted? 

This problem differs from the prior heating illustration because we wish to take into account air or 

nitrogen that is entering the head space of the vessel during 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. 

Ti = 20°C = 293.15°K (Initial temperature) 

Tf = 70°C = 343.15°K (Final temperature) 

PT = 1.0 atm = 760 mm Hg (Total system pressure) 

Vgas = 500 gal = 66.843 ft3

 (Gas space volume) 

R = 998.9 (mmHg-ft3

)/(lb-mole-oK) (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

pV

RT i

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

( ) 1 2 2

1 Navg = n + n

lb moles

RT

PV

RT

PV Navg ⎟ = − ⎠

⎞ ⎜

⎝

⎛ +

×

×

⎥ =

⎦

⎤ ⎢

⎣

⎡ ⎟

⎠

⎞ ⎜

⎝

⎛ ⎟ + ⎠

⎞ ⎜

⎝

⎛ = 0.1608

343.15

1

293.15

1

2 998.9

760 66.843

2

1

1 2

Step 2. Calculate the initial and final partial pressures of nitrogen. 

Use the Antoine equation to calculate the partial pressure of toluene: 

 

Ptoluene, 20o C = exp 16.0137 − 3096.52

293.15 −53.67

⎛

⎝ ⎞ ⎠ = exp(3.0835) = 21.84mmHg

 

Ptoluene,70 o C = exp 16.0137 − 3096.52

343.15 −53.67

⎛

⎝ ⎞ ⎠ = exp(5.317) = 203.74mmHg

The partial pressure of nitrogen is the difference between the total system pressure and the partial pressure 

of toluene: 

pnc,1 = 760 − ptoluene,1 = 760 − 21.84 = 738.16mmHg

p p mmHg nc toluene 760 760 203.74 556.26 ,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 the moles of toluene and air: 

moles

RT

p V

n toluene

toluene 0.00499

998.9 293.15

21.84 66.843

1

,1

,1 = ×

× =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

moles

RT

p V

n nc

nc 0.1685

998.9 293.15

738.16 66.843

1

,1

,1 = ×

× =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

moles

RT

p V

n toluene

toluene 0.0397

998.9 343.15

203.74 66.843

2

,2

,2 = ×

× =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

moles

RT

p V

n nc

nc 0.1085

998.9 343.15

556.26 66.843

2

,2

,2 = ×

× =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

Step 4. Calculate the toluene emission using Eq. 3-21. 

The number of moles of toluene displaced from the vessel is equal to the moles of nitrogen that are 

displaced from the vessel during the heating operation multiplied by the average molar ratio. 

( )vessel i i

nc

nc

n i avg n n

p

p E N ,2 ,1

,2

,1 ln − − ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

− =

E ( ) ( ) lb moles n toluene ⎟ − − = − ⎠

⎞ ⎜

⎝

⎛ − = 0.0397 0.00499 0.0108

556.26

738.16 0.1608 ln
Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models 

EIIP Volume II 16.3-31 

E ( )( ) lb moles lb lb mole lbs wt toluene = 0.0108 − 92.13 / − = 0.995 −

Step 5. Calculate the emission of air without the gas sweep present. 

E n n moles n−nc = nc,1 − nc,2 = 0.1685 − 0.1085 = 0.06

Step 6. Calculate the moles of gas sweep. 

( ) moles air scfm En gas sweep 0.501

998.9 273.15

760 3 60min =⎥

⎦

⎤ ⎢

⎣

⎡

×

× × − =

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

saturation level. 

( ) E lb moles n toluene .0108 3.08 0.0108 0.0333

0.06

0.06 0.25 0.501 ⎟× = × = ⎠

⎞ ⎜

⎝

⎛ + × − =

E ( )( ) lb moles lb lb mole lbs wt toluene = 0.0333 − 92.13 / − = 3.068 −

E lb moles n−nc = 0.06 + 0.501 = 0.561 −

E ( )( ) lb moles lb lb mole lbs wt nc = 0.561 − 28.97 / − =16.25 −

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

n i RT

M K A P P E − − = Eq. 3-23 

Where En is the evaporation rate (mass/time). 

Mi is the molecular weight of the volatile substance, 

Ki is a mass transfer coefficient (length/time), 

A is the evaporation surface area, sat Pi is the saturated solvent vapor pressure, 

Pi is the actual vapor pressure near the liquid surface, 

R is the ideal gas constant, and 

TL 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

M K AP 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 between the compound of interest K and reference compound 

Ko 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

i o M

M K K

 Eq. 3-27 

Water is commonly used as a base reference for estimating 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 contains 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

hr cm

s ft

s

cm

s

cm

M

M K K

i

o

i o 55.33

30.48

3600 0.4685

100.2

18.02 0.83 3

1 3

1

= ⋅ ⋅

⋅ ⋅ ⎟ = × ⎠

⎞ ⎜

⎝

⎛ = ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

45.86 . P tan mmHg sat

hep e =

2

2 2

28.26

4

3.14 36

4 ft d ft A = ∗ = = π

( )( )( )( )

( ) ft atm lb mol K ( ) K

lb lb mole ft hr ft mmHg

RT

M K AP E

L

sat

i i i

n i − ° °

− − = = 998.9 / 298.15100.2 / 55.33 / 26.26 45.86 3

2

E lb hr wt i = 24.42 / −

Illustration 3-20: Evaporation losses from a spill. 

Toluene is spilled onto the ground outside of a building. 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 ft2. 

The molecular weight of toluene is 92.13 lb/lb-mole. 

The vapor pressure of toluene is 28.445 mm Hg. 

hr

ft

hr cm

s ft

s

cm

s

cm

M

M

K K

i

o

i o 68.50

30.48

3600 0.58

92.13

18.02 0.83 3

1 3

1

= ⋅ ⋅

⋅ ⋅ ⎟ = ×

⎠

⎞ ⎜

⎝

⎛ = ∗ ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

( )( )( )( )

( ) ft mmHg lb mol K ( ) R

lb lb mole ft hr ft mmHg

RT

M K A p E

L

sat

i i

n i − ° °

− − = = 998.9 / 298.1592.13 / 68.50 / 100.0 28.44 3

2

E lb hr n i = 60 / −

3.8 Emission Model for Liquid Material Storage 

The preferred method for calculating emissions from storage 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 TANKS 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 Wastewater Collection and Treatment Facilities, of this 

volume. These documents, as well as models such 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. However, collection and analysis of air samples from 

manufacturing facilities can be very expensive and especially complicated for chemical manufacturing 

facilities where a variety of VOCs are emitted and where most of the emissions may be fugitive in nature. 

Test data from one specific process may not be representative 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 occur 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 fitting. Additionally, these vessels may have top mounted 

agitators that can be raised or lowered depending upon mixing elevation requirements. A gas tight agitator 

seal may not be practical for this type of process vessel. In other cases, the mixing vessel my be 

completely open top. Quantifying the gas sweep rate and volatile vapor saturation level for the overall 
Chapter 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 document may not be reliable in this case. 

US EPA Method 204D (Fugitive VOCs from Temporary Total Enclosure) may be used as a means of 

quantifying air emissions from specific equipment 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 measurement 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 complete production operations. 

Illustration 3-21: Using emission measurements to represent production operations. 

A large mixing tank is used as part of a solvent blending operation. The vessel has a flat lid that contains 

several smaller hinged covers for hand holes and charge 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 monitored for air flow rate and solvent concentration 

using standardized instrumental analysis. During the time of testing the atmospheric pressure is measured 

to be 760 mm Hg (1.0 atm). The flow rate and temperature of the exhaust gas are measured to be 10 

ft3

/min (or 600 ft3/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 heptane emission rate during this process operation. 

Using the partial pressure for each compound in conjunction with the ideal gas law and exhaust gas flow 

rate the following expression can be used to calculate the emission rate for a given pollutant. 

 RT

M p F Q i i

i =

where Mi = molecular weight of the pollutant, 

pi = partial pressure of the pollutant in mm Hg, 

F = exhaust gas flow rate in ft3

/hr, 

T = exhaust gas temperature in °K, 

R = universal gas constant for mm Hg, ft3

, and °K. 

For Toluene: 

hr

lbs

RT

M p F Q toluene toluene

toluene 5.2

998.9 298.15

92.13 2.8 600 = ∗∗ ∗ = =

For Heptane: 

hr

lbs

RT

M p F

Q hep e hep e

hep e 7.5

998.9 298.15

tan tan 100.21 3.7 600 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 measurement 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 used to correlate with conventional batch modeling 

formulas. Process variables such as vessel 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, surface area, and other known variables. [Crowl & Louvar, 

2002] 

 ( )i

Sat

m Pi p RT

MKA Q = − Eq. 3-28 

where Qm = evaporation rate (lb/min) 

M = molecular weight of compound i 

K = mass transfer coefficient (ft/min) 

A = surface area (ft2

) 

R = ideal gas constant 

T = temperature of liquid 

PSat = saturated vapor pressure of compound i 

pi = 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

T i

v = =

 Eq. 3-29 

where Qv = emissions from vessel vent (lb/min) 

F = exit gas flow rate (ft3

/min) 

PT = overall system pressure 

R = ideal gas constant 

T = temperature of liquid 

pi = 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 Si as follows: 

Qm = Qv

Substitute basic expressions ( ) RT

MFp P p RT

MKA i

i

Sat

i − =

Cancel out common terms ( )i i

Sat KA Pi − p = Fp

Expand expression i i

Sat KAPi − KAp = Fp
Chapter 16 – Chemical Manufacturing 3.0 – Basic Air Emission Models 

EIIP Volume II 16.3-36 

Rearrange terms i i i

Sat KAPi = Fp + KAp = (F + KA) p

Divide by (F+FA) F KA

KAP

p

Sat

i

i

+ =

Rearrange to obtain saturation. F KA

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 (ft2

), and F (ft3/min) then Si 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 becomes 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 component the exit gas sweep rate (F) and the evaporation 

rate (KA). However, Eq 3.30 represents a worst case condition because it assumes 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 greater than KA then the saturation level is greater 

than 50% and when F is smaller than KA then 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 ft2

. Assuming that the gas space in the vessel is 

mixed then the following calculations can be made. 

min

0.949

30.48 min

60 0.482

92.13

18.02 0.83 3

1 3

1

ft

cm

s ft

s

cm

s

cm

M

M

K K

i

o

i o = ⋅ ⋅

⋅ ⋅ ⎟ = × ⎠

⎞ ⎜

⎝

⎛ = ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

18.59

18.59

(0.949)(19.6)

(0.949)(19.6)

+ = + = + = F K A F F

K A S

i

i

i

Saturation Si 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 estimated emission rate for 

compound i can be plotted as a function of F as by substituting Sat pi = Si pi . 

 RT

MFS p

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 cover 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 while 
Chapter 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 the established gas sweep and 

saturation level can be used in support of a re-evaluation of the revised operation. In this case the vapor 

pressure and other physical properties of the new process 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 standard 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 where the equipment is open to the atmosphere and/or 

does not fit typical process vessel designs. 

In such cases, the quantity of initial 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 such as oil or other materials are contained in the 

residual spent solvent then the material would need 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 spent toluene is assayed using loss on drying (LOD) analysis 

to be 98.8% toluene by weight. Calculate the 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 emissions 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 contains 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: Condensers, 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 condensate while a vacuum pump or steam jet would be 

used if vacuum was needed at the 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 discharged along with the remaining vapor. In the case 

of a vacuum steam jet or a liquid ring vacuum pump, working fluid (steam, water, or other fluid) enters 

the device and is allowed to mix with the process stream as it passes through the unit. A liquid phase is 

discharged with any remaining vapors similarly to the condenser. Simple diagrams of each type of device 

are presented in Figure 4-1. 

For calculation purposes, one tends to consider that the 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 device 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

Fp

Fp

Fb

Ff

V

L

Ff

Fb

Fp 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 = Fp + Fb + Ff Eq. 4-1 

Where Fp is the inlet process rate, Fb is any air or nitrogen bleed, and Ff 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. 

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

EIIP Volume II 16.4-2 

 ziF = xiL + yiV 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 stream. 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. 

 zncF = yncV where xnc = 0 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 component contained in the exit vapor stream V. 

Additionally, Eq. 4-4 provides a way to calculate the 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

i nc 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 exit liquid stream L and the exit vapor stream V are 

needed in order to relate the relative amount of component 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 Pi pi x γ = Eq. 4-6 

 

sys

i i i

sys

i

i P

x P

P

p

y

γ = = Eq. 4-7 

Where x and y are mole fractions of component i in streams L and V. Pi and pi are the pure vapor 

pressure and partial pressure of component i. Psys 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 component (air, nitrogen, or other gases) then it’s 

important to check to see if condensation will likely occur before proceeding to solve the condenser 

problem. This is not normally an issue with calculations 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 contains a large amount of noncondensable 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 first 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 temperature in order for the partial pressure to be the 
Chapter 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 xi’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 xi’s is less than 1.0 then one can assume that 

condensation does not take place and that all of the process exits the condenser as a vapor stream. 

 i iPi pi x γ = Eq. 4-6 

 

i i

i

i P

p

x

γ = Eq. 4-8 

Where Pi is the pure vapor pressure at condenser temperature, 

 pi is the partial pressure at the condenser inlet. 

∑ ≥ 1.0 i x ∑ < 1.0 i x

Condensation takes place at Tcondenser. Condensation does not take place at Tcondenser

4.1.4 General Calculation Approach for Solving Single Stage Vent Device Problems 

If the inlet process stream to the single stage device 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, pi is 

equal to Pi 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 condensable 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 equilibrium relationships that exist between exit streams L and V. The 

general solution to the problem is found by iterating 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 iteration. Iterations continue until all equations are 

satisfied within an acceptable tolerance. 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 (Tb = 110°C) chemically produces methyl chloride (Tb = -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 process stream is in the vapor state then a partial 

condenser could be used at a temperature that is above 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 tend 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 small 
Chapter 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 condenser 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 industry 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 still 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 solvent vapor that entered the condenser was condensed and 

collected in the receiving vessel and none of the solvent left as a vapor. A condenser operated in this 

manner would be referred to as a total condenser since the entire inlet process vapor was converted into a 

liquid condensate. Emissions would not occur from the condenser directly since the process vapor does 

not contain a non condensable gas component. However, 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 result of non condensable gases (air, nitrogen, etc) that 

are leaving the process vessel which carry solvent 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 temperature 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 solvent vapors from a distillation operation then they 

typically operate like a vent condenser at other times 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 material balance correction method. 

4.1.8 Liquid Ring Vacuum Pumps 

Liquid ring vacuum pumps are used to supply vacuum for certain process activities such as vacuum 

distillations, solids drying, and other operations. Liquid enters the pump and operates 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 discharged 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 single process stream. The final discharge temperature 

and pressure are used for the equilibrium calculations. 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 vacuum 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 special nozzle where it is released and allowed to flow through a chamber 

with an increasing cross sectional area. As the chamber 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 and mix with the working steam. The combined steam 

and the process vapor stream leave the vacuum chamber and pass through an internal condenser where the 

steam and other vapors are condensed and the remaining vapor discharged to the atmosphere. 

Vacuum steam jets problems can be solved using the same 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 discharge temperature and pressure are used for the 

equilibrium calculations. A worked illustration is presented which applies the material balance correction 

method to solve the emission calculation. 

4.2 Illustrations 

Several worked illustrations are presented as a general guide for performing 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 normally 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 involves only a single condensable component then the 

calculation of the exit vapor stream is straight forward. 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 inlet 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 the outlet side of the vacuum pump. Each of these 

examples uses Eq. 4-5 to calculate the amount of the condensable component leaving in the condenser 

outlet vapor stream. The fraction of the inlet condensable 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 heptane vapor at a 69% level is vented from a process 

vessel (see Illustration 3-7 in section 3.2.2 of this document) into a condenser. The nitrogen gas 

constituent has a flow rate of 600.0 SCFH. The gas stream 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 

Pheptane 45.69 mmHg @ 25°C 

Pheptane 15.43 mmHg @ 5°C 

Tinlet 25°C, 298.15°K Condenser gas inlet 

Texit 5°C, 278.15°K Condenser gas exit 

Psys 760 mm Hg System pressure 

Time 1 hr Time for event 

Universal Gas Constant: 

lb mole K

mmHg ft R − ⋅°

⋅ =

3

998.9

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 ft3

 (at 273.15°C & 760 mm Hg). 

 lb moles

K

mmHg ft

RT

P V

n N

N = − °

⋅ = = 1.671

(998.9)(273.15 )

760 600 3

2

2

 EN 1.671* 28.013 MWt 46.817 lb 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 heptane saturation level of 69%. In order to calculate 

the heptane content in the inlet gas stream we must first know the pure component vapor pressure for 

heptane at the 25°C stream temperature. Then the partial pressure of the heptane can be calculated for the 

69% saturation level. The moles of heptane in the inlet gas stream can be calculated from the ratio of the 

partial pressure of heptane to the partial pressure of nitrogen and the molar rate of nitrogen. 

 0.69× 45.69 mmHg = 31.53 mmHg
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. 

 pN Psys pH 760 mmHg 31.53 mmHg 728.47 mmHg 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 N2

H

N

H

p

p

n

n = therefore 2

2

N

N

H

H n

p

p n =

lb moles lb moles

mmHg

mmHg nH 1.671 0.07233

728.47

31.53 = =

F lb moles MWt lb H in 0.07233 100.205 7.247 , = × =

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 pressure of the nitrogen component at the condenser exit 

can be calculated by subtracting the heptane vapor pressure from the total system pressure. 

pN ,5 C Psys pH 760 mmHg 15.43 mmHg 744.57 mmHg 2 ° = − = − =

Step 5. Calculate the heptane emissions from the condenser over the one hour period. 

The final heptane emission level EH,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 

EH,exit is known then LH,exit may be calculated by subtracting EH,exit from FH,inlet. 

2 N2

H

N

H

p

p

n

n = therefore 2

2

N

N

H

H n

p

p n =

lb moles lb moles

mmHg

mmHg nH C 1.671 0.03463

744.57

15.43 ,5° = =

E ( lb moles)( ) lb H exit 0.03463 100.205 MWt 3.47 , = =

L lb lb lb H exit 7.25 3.47 3.78 , = − =

47.86%

7.25

3.47 % 100% * tan = = lblb E hep e
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-8 

Contdensed %hep tan e = 100% − 47.86% = 52.14%

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 vented 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 operate 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: 

lb mole K

mmHg ft R − ⋅°⋅ =

3

998.9

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 the inlet stream can be calculated from the respective 

partial pressures. The same calculation can be made at the condenser exit. The percent toluene emitted 
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-9 

from the condenser may then be calculated by dividing the molar ratio of toluene to nitrogen at the exit by 

the molar ratio at the inlet. 

 ( ) mole nitrogen

mole toluene

mmHg

mmHg

p

p

N

T 1.059

175.0 90.0

90.0

2

= − =

 ( ) mole nitrogen

mole toluene

mmHg

mmHg

p

p

N

T 0.076

175.0 -12.43

12.43

2

= =

 *100% 7.18%

1.059

0.076 ET % = =

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 the condenser in the gas phase since little if any would 

condense with the toluene. The rate that nitrogen flows through the system over 10.0 lb/hr. 

0.357 lb - moles hr

28.013

10.0 2 = = MWtlb hr

nN

hr

mmHg

mmHg

p

p

n n

N

T

T N 0.027 lb - moles

(175 12.43)

12.43 * 0.357 *

2

2 = − = =

hr

lb 0.027 *92.13 = 2.49 −

− = lb moleslb

hr

lb moles ET

Illustration 4-3: Condenser with toluene vapor and nitrogen at atmospheric pressure 

If the condenser described in Illustration 4-2 were located at the atmospheric discharge of the vacuum 

pump instead of on the vacuum inlet side what would the condenser performance been? The vacuum 

pump is a dry pump and the discharge vapor temperature 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 operate 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: 

lb mole K

mmHg ft R − ⋅°⋅ =

3

998.9

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 temperature that would result in a 391 mm Hg vapor pressure 

for toluene. 

Antoine Equation:

( 48.78) 16.0127 ln(391)

3096.52

ln( ) − − − − = − = c

a p

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 the 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 the molar ratio of toluene to nitrogen at the exit by 

the molar ratio at the inlet. 

( ) mole nitrogen

mole toluene

mmHg

mmHg

p

p

N

T 1.060

760.0 391.0

391.0

2

= − =

( ) mole nitrogen

mole toluene

mmHg

mmHg

p

p

N

T 0.017

760.0 -12.43

12.43

2

= =

*100% 1.60%

1.060

0.017 ET % = =

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 the condenser in the gas phase since little if any would 

condense with the toluene. The rate that nitrogen flows through the system is 10.0 lb/hr. 

0.357 lb - moles hr

28.013

10.0 2 = = MWtlb hr

nN

hr

mmHg

mmHg

p

p

n n

N

T

T N 0.00594 lb - moles

(760 12.43)

12.43 * 0.357 *

2

2 = − = =

hr

lb 0.00594 *92.13 = 0.55 −

− = lb moleslb

hr

lb moles ET
Chapter 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

K V F

K z

F f V 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 γi can be set equal to 1.0 and the value for Ki becomes a 

constant for all compositions at the fixed system temperature and pressure. 

 

sys

i

i P

P K = , were γ i = 1.0 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/F

f(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][Rachford & 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. 

 V = F (V F) Eq. 4-11 

 L = F – V Eq. 4-12 

 1+ ( )( ) −1 =

i

i

i V F K

z

x Eq. 4-13 

 i iKi y = x 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 fN(V/F) is expressed in Eq. 4-16. 

 ( ) ( ) ( )

( )k

k

k k f V F

f V F

V F V F ′ = − +1 Eq. 4-15 

 ( ) ( )

[ ] ( ) ∑= + −

− ′ =

C

i i k

i i

k K V F

K z 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. 

 ( ) ( )

( ) tolerance level

V F

V F V F

k

k k < − + / 1 Eq. 4-17 

The acceptable tolerance level may vary depending on the 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 ideal 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 discharge 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 Fi (lb-moles/hr) and mole fraction zi 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. Ki 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 

Fi 

lb/hr MWt 

Fi 

lb-moles/hr 

zi

mole frac. 

Pure VPi 

(mm Hg) 

45°C Ki=Pi/Psys

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 Ki and zi 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

K V F

K z

F f V 

( )

( )

( )

( )

( )

( ) K V F

K z

K V F

K z

K V F

K z

t

t t

h

h h

mc

mc mc

1 1 /

1

1 1 /

1

1 1 /

1 0

+ −

− +

+ −

− +

+ −

− =

( )

( )

( )

( )

( )

V F V F 1 ( ) 0.097 1 V / F

0.097 1 0.45

1 0.445 1 /

0.445 1 0.301

1 9.696 1 /

9.696 1 0.249 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 i k

i i

k

K V F

K z f V F

1

2

2

1 1 ( )

1 

 ( ) ( )

( ) ( )

( )

( ) ( )

( )

( ) ( ) 2

2

2

2

2

2

1 0.097 1 /

0.097 1 0.45

1 0.445 1 /

0.445 1 0.301

1 9.696 1 /

9.696 1 0.249

V F V F V F

f V F k

+ −

− +

+ −

− +

+ −

− ′ = Eq. 4-19 
Chapter 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 guess 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 fN(V/F) of 0.567262 and 

2.053191, respectively. The next estimate of V/F was 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 Results for Illustration 4-4 

Iteration ( ) F k

V ( ) F f V ( ) F f ′V ( ) F k +1V ( ) ( )

( )k

k k

V F

V F +1 − V F

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 showing 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 shown 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 = 23.494 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

K V F

K z

F f V

1

1

1 1

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

EIIP Volume II 16.4-15 

Table 4-3. Product Outlet Stream Analysis 

Compound xi

Li 

(lb-moles/hr) yi

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 condensable components and also a noncondensable gas 

component (air, nitrogen, etc.) then the Rachford-Rice 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 iteratively. Alternatively, two other iterative approaches may be used. 

Both of these alternatives involve making an initial guess of the condensate composition. This 

information is used to solve material balances or equilibrium 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 alternatives 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 achieve 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 for a condenser with the same inlet stream. These 

illustrations use an inlet stream with three condensable 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 xi = 0, and Ki becomes infinity. When Ki equals infinity then the standard 

Rachford-Rice model becomes indeterminate since Ki appears in the numerator and the denominator. This 

problem may be solved by dividing both numerator and denominator by (Ki-1) as shown. 

 ( )

( ) ∑= + −

− ⎟ =

⎠

⎞ ⎜

⎝

⎛ C

i i

i i

K V F

K z

F

V f

1 1 1

1

 Eq. 4-9 

 ( )

( )

( )

( ) ∑=

⎥

⎦

⎤ ⎢

⎣

⎡

−

− ×

+ −

− ⎟ =

⎠

⎞ ⎜

⎝

⎛ C

i i

i

i

i i

K

K

K V F

K z

F

V f

1 1 1

1 1

1 1

1

 Eq. 4-20 

 ( ) ∑= − +

⎟ =

⎠

⎞ ⎜

⎝

⎛ C

i i

i

K V F

z

F

V f

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

EIIP Volume II 16.4-16 

For the case when component i is a noncondensable gas (nitrogen, air, etc) where xi = 0 then: 

 

i

i

i

x

y K = Eq. 4-10 

 [ ] ⎥ = ∞

⎦

⎤ ⎢

⎣

⎡ = → → i

i

x i x x

y K i 0 i 0

lim lim Eq. 4-22 

The modified Rachford-Rice expression for the noncondensable component may now be evaluated as 

shown. 

 ( ) nc

i

nc

i

nc i

i

K V F

z

V F

z

K V F

z

i ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ = ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

∞ + ⎥ =

⎦

⎤ ⎢

⎣

⎡

→∞ 1 −1 + 1

lim Eq. 4-23 

Combining Eq. 4-21 (for the condensable components) with Eq. 4-23 (for the noncondensable 

components) leads to a modified Rachford-Rice equation that can be used to solve condenser problems 

when non-condensable components such as air or nitrogen are present in the process stream. 

 ( ) ∑ ∑ = =

+ − +

⎟ =

⎠

⎞ ⎜

⎝

⎛ c Cnc

j

j

C

i i

i

V F

z

K V F

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. 

 ( ) ( )

( ) [ ] ( ) [ ] ∑ ∑ = =

− + − +

− ′ = = c Cnc

j

j

C

i i

i

V F

z

K V F

z

d V F

f V F

F f V

1

2

1

2 1 1

 Eq. 4-25 

 ( ) ( ) ( )

f ( ) V F

f V F V F k V F k ' //

1 = − + Eq. 4-15 

After V/F is determined, the composition of each exit stream may be calculated using Eq. 4-11 through 

Eq. 4-14. 

 V = F (V F) Eq. 4-11 

 L = F - V Eq. 4-12 

 1+ ( )( ) −1 =

i

i

i V F K

z

x Eq. 4-13 

 i iKi y = x Eq. 4-14 
Chapter 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 cooled 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 condenser 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 emissions 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 condenser from the receiver is shown in Table 4-4. 

Table 4-4. Analysis of the Condenser Inlet Feed Stream 

Compound 

Fi 

lb/hr MWt 

Fi 

lb-moles/hr 

zi 

Mole frac. 

Partial 

pressure 

pi 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 whether condensation will occur 

Before proceeding to calculate the material balance for the vent condenser it’s a good idea to perform a 

quick test to see if condensation is likely to occur 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 xi for each condensable component in hypothetical condensate at the 

condenser temperature (2°C in this example). If the sum of the calculated xi 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 xi 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 Pi is the pure component vapor pressure at 2°C, 

pi is the component partial pressure at the condenser inlet conditions (30°C), and 

i γ is 1.0 for the ideal mixture. 

The partial pressure pi for each component in the inlet stream to the condenser was established in step 1. 

The pure component vapor pressure (Pi) for each component in the feed stream at 2°C has been calculated 

based on the Antoine Equation, and the resulting values are shown in Table 4-5. The hypothetical 
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-18 

condensate mole fraction for each component xi has been calculated using Eq. 4-26 because the problem 

stated that the process mixture could be treated as ideal. 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) Pi @ 2°C 

xi 

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 zi are needed for all four components; these values were established in step 1. Values for Ki 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 Ki 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 Ki values. 

Table 4-6. Preparation for Applying the Modified Rachford-Rice Method 

Compound 

Fi 

lb-moles/hr 

zi 

Mole frac. 

Pi 

(mm Hg) 2°C Ki=Pi/Psys

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 Ki and zi from Table 4-6. Note that a 

separate term is needed for each component in the inlet stream. 

( ) ∑ ∑ = =

+ − +

⎟ =

⎠

⎞ ⎜

⎝

⎛ c Cnc

j

j

C

i i

i

V F

z

K V F

z

F

V f

1 1 1 1

( ) ( ) ( ) V F

z

K V F

z

K V F

z

K V F

z n

t

t

h

h

mc

mc 2

1 1 / 1 1 / 1 1 /

0 + − +

+ − + − + =

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

0.354

1/ 0.011 1 /

0.0276

1/ 0.066 1 /

0.0854

1/ 2.749 1 /

0.532 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. 

 ( ) ( )

( ) [ ] ( ) [ ] ∑ ∑ = =

− + − +

− ′ = = c Cnc

j

j

C

i i

i

V F

z

K V F

z

d V F

f V F

F f V

1

2

1

2 1 1

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 /

0.354

1/ 0.011 1 /

0.45

1/ 0.066 1 /

0.301

1/ 2.749 1 /

0.249

V F V F V F V F F f V − + − +

− + − +− + − +

− ′ =

Step 6. Solve for V/F using Newton’s method. 

An initial guess of V/F is needed to start the iterative process of solving the modified Rachford-Rice 

equation. Rather than picking a value arbitrarily, sometimes it’s helpful to graphically plot f(V/F) as a 

function of V/F as shown in Figure 4-4 for this problem. 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 ensure that the method will converge on a valid solution. Newton’s method 

relies on the slope of f(V/F) to help locate the solution. At points where the slope of the function is 

sufficiently low the method has been known to extrapolate 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 Rachford-Rice equation for this condenser problem is 

close to 0.9. 

Figure 4-4. Rachford-Rice Plot showing solution at V/F = 0.90076. 

Based on Figure 4-4, an initial guess for V/F was assumed 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 fN(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 1.0
Chapter 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 was repeated for four iterations. At that point the 

value of V/F had converged within the specified tolerance level. 

Table 4-7. Iteration results for Illustration 4-5 

Iteration ( ) F k

V ( ) F f V ( ) F f ′V ( ) F k +1V ( ) ( )

( )k

k k

V F

V F +1 − V F

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 shown 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. 

 V = F (V F) 

 = (0.47126)(0.90076)

 = 0.42454lbmoles/ hr

 L = F −V

 = 0.47126 − 0.42454

 = 0.0467lb moles/ hr

The mole fractions of components in each exit stream are calculated using Eq. 4-13 and Eq. 4-14. For 

example, the mole fractions for methyl chloride are calculated as follows: 

 1 + ( )( ) −1 =

mc

mc

mc V F K

z

x 

 1 ( )( ) 0.90076 2.749 1

.532

+ − =

 = 0.20676

 i iKi y = x

 = (.20676)(2.749)

 = 0.56831
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-21 

The molar flow rates of each component are calculated by multiplying the total moles by the applicable 

mole fractions. For example, the molar flow rates of methyl chloride are calculated as follows: 

 n ( )( ) lbmoles hr mc L .20676 .0467 0.00967 / , = =

 n ( )( ) lbmoles hr mc V .56831 .42454 0.24127 / , = =

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) 

Fi

lb-moles/hr Zi Ki

Li

lb-moles/hr Xi

Vi

lb-moles/hr Yi

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 MWt

Fi

Inlet Stream 

lb/hr 

Li

Condensate 

lb/hr 

Vi

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 Regarding 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 solution. In some cases a damping factor may be used to 

improve the tendency for Newton’s method to continually move toward the desired solution. 

Additionally, other numerical methods exist including the False Position, Bisection, and Fibonacci 

Methods that can be applied to either the standard Rachford-Rice or the modified Rachford-Rice solution 

model. If it is desired to use standard spreadsheet software 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 problem when two or more condensable components are 

present in the system involves making an initial guesstimate of the condensate composition and then 

placing the balance of the process in the vapor phase. Component vapor pressures are then calculated for 
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-22 

the condensate and compared to the component partial 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 partial pressures are the same for each component (within 

an acceptable tolerance). Material balance accounting 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: 

 ni,F = ni,L + ni,V 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 extended with activity coefficients. 

 L

n P

p x P i L i i

i L i i i

γ

γ ,

, = = Eq. 4-28 

The partial pressure of each condensable component in the exit gas phase may be calculated using 

Dalton’s Law. 

 sys i sys

i V

i V P y P

V

n

p = = ,

, Eq. 4-29 

At equilibrium both partial pressures should be the same for each condensable component. 

 pi,L = pi,V Eq. 4-30 

However, since we are making only an initial guesstimate, the two partial pressures will likely not be the 

same. An error of estimate can be calculated by subtracting one from the other. 

 ni pi L pi V f , , ( ) = − Eq. 4-31 

Newton’s method is applied to all condensable components simultaneously until the two partial pressures 

(from the condensate and the gas phase) agree within acceptable tolerance for each component. 

 ( )

( ) i k

i k

i L k i L k f n

f n

n n

,

,

, , 1 , , ' + = − Eq. 4-15 

'( ) ni f may be approximated using the calculus of finite differences where only condensable component 

i is varied and the other components are held constant. [Constantinides] 

 ( ) ( ) ( )

T P j i

i

T P j

i L k i L k

i L k

n

f n f n p f n

, , , ,

, , , ,

, , 2 ' ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

Δ

Δ = ⎥

⎦

⎤ ⎢

⎣

⎡ + − − ≅

δ

δ δ

 Eq. 4-32 

If the process contains only two condensable components then Eq. 4-15 would be required for both 

components. If the process contains 5 condensable compounds then five separate equations would be 
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-23 

required and so on. For each equation f(ni,L,k) and fN(ni,L.k) are calculated with only ni being varied while all 

other compound quantities held constant. Once f(ni,L,k) and fN(ni,L,k) have been calculated ni,L,k+1 is 

calculated for each condensable compound in the system as shown. 

Component 1 

( )

( ) 1, , , , 1

1, ,

1, , 1 1, , ' ≠

+ = −

T P j L k

L k

L k L k f n

f n

n n

Component 2 

( )

( ) 2, , , , 2

2, ,

2, , 1 2, , ' ≠

+ = −

T P j L k

L k

L k L k f n

f n

n n

· 

· 

· 

•

•

•

Component n 

( )

( ) n L k T P j n

n L k

n L k n L k f n

f n

n n

≠

+ = −

, , , ,

, ,

, , 1 , , '

Illustration 4-6: Partial Pressure Correction Method for Vent Condenser 

500 gallons of the condensate from Illustration 4-4 was cooled to 30°C and is being held in a 1,000 gallon 

receiver. A 1.0 scfm nitrogen gas sweep enters that vessel 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 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 condenser 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 

Fi 

lb/hr MWt 

Fi 

lb-moles/hr 

zi 

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 whether condensation will occur. 

This procedure is presented in illustration 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 method, 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. Since 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 pressure of methyl chloride (2089 mm Hg) at 2°C is 

greater than the inlet partial pressure of methyl chloride (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 mixture where the activity coefficient equals 1.0 for 

each component in the condensate, then an initial guess for the concentration of methyl chloride can be 

estimated using Eq. 4-8, where xi 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

1* 2089

405

2 = = = °C

i i

feed

i

i P

p x

γ

 Eq. 4-8 

Once xi is known, then the moles of methyl chloride that should be allocated to the liquid condensate can 

be calculated by the ratio of the methyl chloride mole fraction and the non methyl chloride portion of the 

condensate as shown in Eq. 4-33, where j represents n-hexane and toluene. 

 j L

i

i

i L n

x

x

n , , 1∗ − ≈ Eq. 4-33 

 0.0533

1 0.19

0.19

∗ − =

 = 0.01249lb moles/ hr

The resulting initial estimate of the condensate is presented in Table 4-11. 

Table 4-11. Initial Estimate of Condensate 

Compound 

Fi

lb-moles/hr 

pi at Feed 

mm Hg 

Pi at 2°C 

mm Hg 

Li, 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 0.06575 
Chapter 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 liquid composition and the pure component 

vapor pressure at 2°C is calculated using Eq. 4-28. For example, the calculation is as follows for toluene: 

 ( )( )( )

mmHg

Pi L xi iPi

1.6

0.2 1 8

,

=

=

= 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 outlet 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 methyl chloride in the outlet vapor stream. The partial 

pressures are then estimated using Eq. 4-29. For example, the calculation is as follows for methyl chloride 

in the first iteration: 

 ( )( )

mmHg

pi V yiPsys

447

0.59 760

,

=

=

=

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

chloride: 

 

( )

mmHg

f ni pi L pi V

50

397 447

, ,

= −

= −

= −

The derivative of the function is estimated using finite differences as described 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 the 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 ni results in f(ni) 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 condensate for the second iteration is estimated using Eq. 

4-15. For example, the next second estimate for methyl chloride is estimated as follows: 
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-26 

 

( )

( )

lb moles hr

f n

f n

n n

i k

i k

i L k i L k

0.0144 /

26,507

50 0.01249

,

,

, , 1 , ,

=

− = −

′ + = −

This process is continued until the difference in the partial pressures estimated based on the liquid and 

vapor moles fractions for all components agree within 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 for all compounds. One can calculate the percent 

agreement for each component by dividing Δpi by the average of both partial pressures. For Iteration 7 

the % agreement is shown in Table 4-12. 

Iteration 1 

Compound ni.L,k xi pi,L Vi yi pi,V ∆pi ∆pi/∆ni ni,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 ni.L,k xi pi,L Vi yi pi,V ∆pi ∆pi/∆ni ni,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 ni.L,k xi pi,L Vi yi pi,V ∆pi ∆pi/∆ni ni,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 0.0119 
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-27 

Iteration 4 

Compound ni.L,k xi pi,L Vi yi pi,V ∆pi ∆pi/∆ni ni,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 ni.L,k xi pi,L Vi yi pi,V ∆pi ∆pi/∆ni ni,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 ni.L,k xi pi,L Vi yi pi,V ∆pi ∆pi/∆ni ni,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 ni.L,k xi pi,L Vi yi pi,V ∆pi ∆pi/∆ni ni,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 

pi avg. 

mm Hg 

∆pi

mm Hg 

∆pi/pi

% 

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 Table 4-13 with the material balance displayed in pound 

moles and in pounds for the exit vapor and liquid streams. The composition of the exit vapor streams is 

also listed by the average partial pressure 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 

Fi

lb/hr 

Fi

lbmoles/

hr 

Li

lbmoles/

hr 

Li

lb/hr 

Vi

lbmoles/hr Vi

lb/hr 

Pi

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 problem when two or more condensable components are 

present involves making an initial guesstimate of the liquid condensate composition. Component vapor 

pressures are then calculated for the condensate and used in conjunction 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 these 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 liquid streams is calculated to be 20% greater than 

the moles of component i in the incoming feed stream then the amount of i in the condensate is divided by 

1.2 for the next iteration. This iterative process for correcting 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 the exit vapor and the liquid condensate at steady state 

operation. 

Material Balance Considerations 

Overall material balance: F = L +V 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: ziF = xiL + yi

V Eq. 4-35 

 ni,F = ni,L + ni,V Eq. 4-36 

Assuming that all of the noncondensable gases (air, nitrogen, etc) exit the condenser in the vapor stream 

then Eq. 4-35 and Eq. 4-36 can be easily simplified. The moles of noncondensable nnc in the exit gas 

stream are equal to the moles that are contained in the inlet stream. 

Noncondensable component: nnc,V = zncF = yncV Eq. 4-37 
Chapter 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 iPi p = x 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 moles of the noncondensable component as shown. 

 

nc

i

nc V

i V

p

p

n

n =

,

, Eq. 4-39 

 

nc

i L i i

nc

nc

i i i

nc

nc

i

i V nc Lp

n P

n

p

x P

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 and Equilibrium Considerations 

If the process contains two or more condensable components 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

i F i L i V i L nc p

p

n , = n , + n , = n , + n Eq. 4-41 

 Eq. 4-41 ties the material balance and the equilibrium 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 educated guess of the condensate liquid is made. Eq. 4-41 

is used to calculate the molar inventory of component i from the condensate and calculated vapor phase. 

Eq. 4-42 is then used to calculate the ratio Фi 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 Фi must 

equal 1.0. 

 i

i F

i L i V

n

n n = φ

+

,

, , Eq. 4-42 

 i F

nc

nc i i

i L i L

i

n

p L

n P

n , n , ,

1 ≈⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+ γ

φ Eq. 4-43 
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-30 

 i F

nc

nc i i

i

i L

n

p L

n n P ,

, 1 ≈⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+ γ

φ Eq. 4-44 

ni,L,k+1 is estimated by dividing ni,L,k by Фi as shown in Eq. 4-45 and applied to the next iteration until Фi

becomes close to 1.0 within an acceptable 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. 

 

i k

i L k

i L k

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 fallen 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 will 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 ni,L in the condensate liquid are calculated by multiplying the moles of ni,F in the 

feed by (1-ν i ) since (1-ν i ) would be a fraction of a compound to remain in the liquid phase. 

 ( ) ( ) ni,L k = 1−ν i

* ni,F Eq. 4-47 

Also, pnc,k is set equal to the partial pressure of the noncondensable as contained in the feed stream pnc,F . 

This is because we know that any condensation to occur will lead to value of pnc that is greater than what 

we started with. So the noncondensable component partial pressure at the inlet feed is considered to be 

lowest starting value possible and can only become larger as the convergence takes place. 

*********************** 

Step 1. Calculate L, xi, and pi for each component in the liquid phase. 

 = ∑ i L L n , , L

n

x

i L

i

, = , i i i Pi p = x γ

Step 2. Calculate ni,L,k+1 for each condensable component in the vapor phase and calculate Vk+1

from the sum of the moles of condensable components in the vapor phase plus the 

noncondensable moles (which remain constant). 

 

nc k

i k

i V k nc k

p

p

n n

,

,

, , +1 = , nc

C

j

k j V k V = ∑n + n =

+ +

1

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

EIIP Volume II 16.4-31 

Step 3. Calculate pnc,k. 

 sys

k

nc

nc k P

V

n

p

1

, 1

+

+ = 

Step 4. Calculate Фi for each condensable component using Eq. 4-39 substituted in Eq. 4-42. 

 , 1

,

, 1

,

+

+

=

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+

i k

i F

nc k

i

i L nc

n

p

p

n n

φ

Step 5. Calculate the next iteration value for (ni,L)k+1. 

 ( ) ( )

1

,

, 1

+

+ =

k

L k i

L k 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 cooled to 30°C and is being held in a 1,000 gallon 

receiver. A 1.0 scfm nitrogen gas sweep enters that vessel 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 

Fi 

lb/hr MWt 

Fi 

lb-moles/hr 

zi 

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 the 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 vi 1 - vi n,i,F n,i,L xi

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 

ni,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 Pnc,k+1 = 375.56 

Iteration 2 

Components 

Fi 

lb-m 

ni,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 Pnc,k+1 = 329.06 

Iteration 3 

Components 

Fi 

lb-m 

ni,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 Pnc,k+1 = 290.65 
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-33 

Iteration 20 

Components 

Fi 

lb-m 

ni,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 Pnc,k+1 = 298.76 

Iteration 21 

Components 

Fi 

lb-m 

ni,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 Pnc,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 undergoes 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 vapor and liquid streams. The composition of the exit 

vapor streams is also listed by the average partial pressure for each component being emitted. 

Table 4-17. Material Balance for Condenser 

Compound MWt 

ni,F 

lb-m/hr 

Fi 

lb/hr 

ni,L 

lb-m/hr 

Li 

lb/hr 

ni,V 

lb-m/hr 

Vi 

lb/hr 

pi 

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 2.03 
Chapter 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 Modified 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 behavior 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 pressure and other conditions: 

Pethyl acetate 66.19 mm Hg 18.7°C 

Pethanol 36.36 mm Hg 18.7°C 

Psystem 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 analyzed using standard gas property relationships: 
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-35 

 

i

i

i F MWt

F

n , =

∑

=

F

i F

i

n

n

y ,

i iPi p = y

Table 4-18. Condenser Inlet Stream Analysis for 1.0 hour of Operation 

Compound 

Fi 

lb/hr 

Molecular 

Weight 

ni, F 

lb-mole 

yi 

mole fraction 

pi 

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 Estimating Activity Coefficients 

If azeotrope data is available for a two component liquid 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 γi can be calculated since xi = yi for each component at the azeotrope. 

 

i

sat vp

i i

i sat vp

i P

P

x P

y P

γ = = Eq. 4-50 

From the azeotrope data that has been supplied, the following calculations can be made: 

1.16

66.19

77.4

1 = = mmHg

mmHg γ 2.13

36.36

77.4

2 = = mmHg

mmHg γ

ethyl acetate ethanol 

 ( )

( ) ln( ) 1.16 1.202

0.734ln 1.16

0.266ln 2.13 1

2

= ⎥

⎦

⎤ ⎢

⎣

⎡

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

EIIP Volume II 16.4-36 

 ( )

( ) ln( ) 2.13 1.797

0.266ln 2.13

0.734ln 1.16 1

2

= ⎥

⎦

⎤ ⎢

⎣

⎡ β = +

2

2

1

1

1.797

1.202 1

1.202 ln

⎥

⎦

⎤ ⎢

⎣

⎡

+

=

x

x

γ 2

1

2

2

1.202

1.797 1

1.797 ln

⎥

⎦

⎤ ⎢

⎣

⎡

+

=

x

x

γ Eq. 4-51 

Where 

( ) 1 1 γ = exp lnγ ( ) 2 2 γ = exp lnγ 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 vent condenser has been analyzed and is shown in 

Table 4-18. A van Laar equation has been developed 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 for ethyl acetate and ethanol are not known since the 

composition of the liquid condensate is not initially known. However, if condensation is to occur then the 

first drop of condensate formed would be at equilibrium 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 equilibrium 
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-37 

condensate composition assuming that the dew point for the process vapor stream is reached at the 

18.7°C. The vapor composition for the ethyl acetate-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 condensate at dew point conditions for the inlet 

condenser stream. 

The partial pressures of ethyl acetate and ethanol at the condenser inlet conditions are used to calculate 

the relative mole fractions of ethyl acetate and ethanol that would exist at dew point conditions. These 

calculations are shown in Table 4-19 where the relative 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 

pi 

mm Hg 

yi 

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 initial condensate composition at the vapor dew point. By 

inspection, this value is between 7% and 8% ethyl acetate. Alternatively, it is possible to use a goal seek 

routine in a spread sheet to back calculate the 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 completed using the following relationships. 
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-38 

liq phase

n

i i i

i

n

i

i P

p x ⎥

⎦

⎤ ⎢

⎣

⎡

∑ = ∑ < = =

1.0

1 1 γ liq phase

n

i i i

i

n

i

i P

p x ⎥

⎦

⎤ ⎢

⎣

⎡

∑ = ∑ ≥ = =

1.0

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 total 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 condensate are corrected by the resulting ratio to obtain the 

corrected value to use for the next iteration, k+1. 

 i

i F

i L i V

n

n n = φ

+

,

, , Eq. 4-53 

 

i k

i L k

i L k

n

n

,

, ,

, , 1 φ + = Eq. 4-54 

 Eq. 4-53 and Eq. 4-54 are repeated until the error of estimate 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 composition 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 ni xi γi pi ni,V ni,L+V ni,F Φ-1 ni,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 ni xi γi pi ni,V ni,L+V ni,F Φ-1 ni,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 ni xi γi pi ni,V ni,L+V ni,F Φ-1 ni,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 ni xi γi pi ni,V ni,L+V ni,F Φ-1 ni,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 ni xi γi pi ni,V ni,L+V ni,F Φ-1 ni,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 ni xi γi pi ni,V ni,L+V ni,F Φ-1 ni,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 Fi, lb 

ni,L 

lb-mole Li, lb

ni,V 

lb-mole Vi, lb Li + Vi

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 pi, mm Hg M. Wt. Lb-Moles Vi

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 can be solved algebraically because the two liquid 

phases must be in equilibrium. This situation is examined in illustration 4-9. The problem is more 

complicated with additional liquid phases. The modified Rachford 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/hr). The condenser temperature is such that the exit 

streams are at 20°C. 

Using the simplified condensate analysis approach, calculate the emissions (Vi) and the material balance 

for this condenser. 

Data Provided 

Pt 21.827 mm Hg 20°C 

Pw 17.351 mm Hg 20°C 

Psys 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 and the rate in lb-mole/hr, mole fraction, and partial 

pressures are calculated using the standard equations. 

i

i

i F MWt

F

n , =

∑

=

F

i F

i

n

n

y ,

i iPi p = y

Table 4-24. Inlet Stream Analysis over 1.0 Hour 

Compound 

Fi 

Rate (lb/hr) 

Molecular 

Weight 

ni,F 

(lb-mole) Mole Fraction 

Partial Pressure 

(mm Hg) 

Nitrogen 5.0 lb/hr 28.013 0.1785 0.188 142.72 
Chapter 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 

pi for component i in the liquid. 

 i iPi pi 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, 

Pi is the pure component vapor pressure of component i at the mixture temperature, and 

For two liquid phases in equilibrium, the vapor pressure pi of component i is the same for each phase. 

 pi,1 = pi,2 Eq. 4-56 

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

i γ is the activity coefficient of component i, 

Pi is the pure component vapor pressure of component i at the mixture temperature. 

Substituting the 

 i i i i i Pi x P x ,1 ,1 ,2 ,2 γ = γ Eq. 4-57 

 i,1 i,1 i,2 i,2 x γ = x γ Eq. 4-58 

Therefore, if a two phase liquid exists of known composition and the activity coefficient i,1 γ is known 

then i,2 γ may be estimated as shown. 

 

,1

,2 ,2

,1

,2 ,2

,1

i

i i

i i

i i i

i

x

x

x P

x γ P 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) 
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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-. 

 

i i j j

i i

i wt MWt wt MWt

wt MWt

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 coefficients 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 reasonable 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. 

 ( )( )

( ) 9974.5

0.0001

0.99745 1.0

γ t,1 = = Eq. 4-62 

Calculate the activity coefficient for water in the non-aqueous phase (phase #2) using 

 Eq. 4-61. 

 ( )( )

( ) 392.1

0.00255

0.9999 1.0

γ w,2 = = Eq. 4-63 

Table 4-27. Estimated Activity Coefficients 

Non-Aqueous Phase Aqueous Phase 

Toluene 1.0 9974.5 

Water 392.1 1.0 
Chapter 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 through the condenser (air leakage, toluene vapors, and 

water vapors) has been analyzed and is shown in Table 4-24. Now that the activity coefficients for toluene 

and water have been estimated and are shown in Table 4-27 for a two liquid phase mixture, it is important 

to evaluate whether condensation will occur in the condenser 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 xi 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 component partial pressures in the inlet stream and 

the pure component vapor pressures at the condenser temperature 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 the 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 iPi pi x γ = Eq. 4-64 

 

i i

i

i P

p

x

γ = Eq. 4-65 

liquid phase

n

i i i

i

n

i

i P

p

x ⎥

⎦

⎤ ⎢

⎣

⎡

∑ = ∑ = = =

1.0

1 1 γ

 Eq. 4-66 

The feasibility that condensate will form in the condenser is determined for an aqueous liquid phase and a 

non-aqueous liquid phase using Eq. 4-66. Results from these calculations 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 fractions 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 fractions for a non-aqueous phase is 8.02 and also greater 

then 1.0. Therefore, a condensate mixture from this condenser will likely have a non-aqueous phase as 

well as an aqueous liquid phase. 

During the course of iterating to determine the condenser solution, it is possible for the sum of the xi

values might fall below 1.0 for one of the liquid phases, in which case that phase would no longer be 

considered as feasible. This is because, as iterations 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 support 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, the 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 VPi Partial VPi

Component Xi γi Xi γi Pi pi

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 

 pN = 0 − pt − pw 760. 2 Eq. 4-67 

 pN 2 = 760.0 − 21.772 −17.349 = 720.879 mmHg
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-45 

 lb moles

p

p

n n

N

t

t V = N = = 0.0054 − 720.879

21.772 0.1785

2

, 2

 lb moles

p

p

n n

N

w

w V = N = = 0.0043 − 720.879

17.349 0.1785

2

, 2

Exit Liquid Analysis 

The moles of condensate can be determined by subtracting the moles of toluene and water that exit in the 

vapor phase from the moles of toluene and water that exit in the two liquid phases. 

 n n n lb moles t,1&2 = t,F − t,V = 0.2171− 0.0054 = 0.2117 − Eq. 4-68 

 n n n lb moles w,1&2 = w,F − w,V = 0.5549 − 0.0043 = 0.5506 − Eq. 4-69 

nT,1&2 and nW,1&2 are the moles of toluene and water that are contained in the combined two liquid phases. 

Equations can be written that relate the composition of the aqueous phase and the nonaqueous phase to 

the total moles of toluene and water as liquid. 

 t,1 1 t,2L2 nt,1&2 x L + x = Eq. 4-70 

 w,1 1 w,2L2 nw,1&2 x L + x = Eq. 4-71 

L1 are the moles of nonaqueous phase and L2 are the moles of aqueous phase in the condensate. The 

objective is to now solve Eq. 4-70 and Eq. 4-71 for L1 and L2 and then calculate the moles of toluene and 

water that are contained in each phase based on the known solubility data. This can be accomplished 

through algebraic manipulation of Eq. 4-70 and Eq. 4-71 as shown: 

 w,2 t,1 1 w,2 t,2 2 w,2 nt,1&2 x x L + x x L = x Eq. 4-72 

 t,2 w,1 1 t,2 w,2 2 t,2 nw,1&2 x x L + x x L = x Eq. 4-73 

 w,2 t,1 1 t,2 w,1 1 w,2 t,1&2 t,2 nw,1&2 x x L − x x L = x n − x Eq. 4-74 

 

,2 ,1 ,2 ,1

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

1

w t t w

w t t w

x x x x

x n x n

L −

− = Eq. 4-75 

L1 and L2 can finally be calculated using Eq. 4-76 and Eq. 4-77. Molar quantities of toluene and water 

can then be calculated once L1 and L2 are known. 

L = = lb − moles

∗ − ∗

∗ − ∗ = 0.2117

0.9998

0.2116

0.9999 0.9975 0.0001 0.0025

0.9999 0.2117 0.0001 0.5506

1 Eq. 4-76 

 L = L − L = 0.7623 - 0.2067 = 0.5506 lb − moles 2 1&2 1 Eq. 4-77 
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-46 

Nonaqueous Phase 

 n x L lb moles t,1 = t,1 ∗ 1 = 0.9975∗ 0.2117 = 0.2111 −

 n x L x lb moles w = w ∗ = ∗ = − -4

,1 ,1 1 0.0025 0.2117 5.292 10

Aqueous Phase 

 n x L lb moles t = t ∗ = ∗ = − -5

,2 ,2 2 0.0001 0.5506 5.506 x10

 n x L lb moles w,2 = w,2 ∗ 2 = 0.9999 ∗ 0.5506 = 0.5506 −

Table 4-31. Exit Liquid Stream Analysis 

Nonaqueous Aqueous 

Xi n1 (lb-moles) Xi n2 (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

ni,1 lb ni,2 lb ni,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 with 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 equation, the partial pressure correction method, or the 
Chapter 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 maintain a distillation operation at 100 mm Hg absolute 

pressure. The inlet process vapor stream at the vacuum pump consists of air (0.95 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 water 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 gas and liquid streams that exit the vacuum pump are 

considered to be in equilibrium. Table 4-34 provides 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 

Fi 

lb/hr MWt 

ni,F 

lb-m/hr zi

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 Pressure Correction method, we must first arrive at an 

initial guess for the composition of the liquid that will discharge from the vacuum pump. Upon inspection 

of Table 4-34, it becomes apparent that the bulk of the inlet stream (97.8%) consists of water (480 lb/hr) 

and that the remainder of the components (including 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 and Initial Guess of the Exit Liquid Composition 

Component 

Pi @ 25°C 

mm Hg 

Feed from Table 4-34 

ni,F lb-moles/hr 

Initial Guess 

ni,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 ( ) pi,L − pi,V = pi,L

since pi,V 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 iteration. Only three iterations were required to 

calculate a solution to the desired tolerance as shown: 

Iteration 1 

Components ni,L,k xi pi,L ni,V yi pi,V pi,L – pi,V Δni,L ni,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 ni,L,k xi pi,L ni,V yi pi,V pi,L – pi,V Δni,L ni,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 ni,L,k xi pi,L ni,V yi pi,V pi,L – pi,V Δni,L ni,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 26.6332 
Chapter 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 met. 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 calculated 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 minimize 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 work with. The Modified Rachford-Rice approach adjusts V/F until the 

complete based equation is solved. However, since 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 the convergence of each individual component in a 

concurrent fashion. So regardless whether the component occupied as much as 97.8% (as in the case of 

water) of the process or as little as 0.0026% (as in the 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 process 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 stream 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 problem 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 absolute 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: 

PT 113.8 mm Hg 55°C 

PW 118.0 mm Hg 55°C 

Psys 760 mm Hg discharge pressure 

Time 1 hr Time for event 

Solubility data at 55°C: 

XT,ph-1 0.000168 mole fraction aqueous phase 

XW, 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 equations and relationships: 

 

i

i

i F MWt

F

n , =

∑

=

F

i F

i

n

n

y ,

i iPi p = y
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 

Fi 

lb/hr 

Molecular 

Weight 

ni,F 

lb-mole 

yi 

mole Fraction pi, 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 Estimating 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 water can be estimated for each liquid phase using Eq. 4-78. For 

the aqueous phase it is reasonable to assume that water 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 (phase #1) is calculated as follows: 

 ( )( )

( ) 5903

0.000168

0.993697 1.0

γ t,1 = =

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

 ( )( )

( ) 159

0.006303

0.999832 1.0

γ w,2 = = Eq. 4-79 

Table 4-40. Estimated Activity Coefficients 

Aqueous Phase Non-Aqueous Phase 

Toluene 5903.0 1.0 

Water 1.0 159.0 
Chapter 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 Table 4-38. Now that the activity coefficients for toluene 

and water have been estimated and are shown in Table 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 fractions for an aqueous phase condensate is 6.34. 

Therefore, we can assume that the discharge process stream 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

i F

i L i V

n

n n = φ

+

,

, , Eq. 4-80 

 

i k

i L k

i L k

n

n

,

, ,

, , 1 φ + = Eq. 4-81 
Chapter 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 fallen 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 first iteration. The partial pressures of toluene and water 

in the outlet vapor, assuming equilibrium with the condensate 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 and 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 for 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 Steam Jet Problem with Aqueous Phase Condensate 

Iteration 1 ni,L,k xi pi ni,V ni,L+V ni,F Φi,k ni,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 ni,L,k xi pi ni,V ni,L+V ni,F Φi,k ni,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 ni,L,k xi pi ni,V ni,L+V ni,F Φi,k ni,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 ni,L,k xi pi ni,V ni,L+V ni,F Φi,k ni,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 ni,L,k xi pi ni,V ni,L+V ni,F Φi,k ni,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 0.00094

 
Chapter 16 – Chemical Manufacturing 4.0 – Single Stage Vent Devices 

EIIP Volume II 16.4-54 

Iteration 6 ni,L,k xi pi ni,V ni,L+V ni,F Φi,k ni,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) ni,1 (lb) ni,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 1.44 
Chapter 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, waste 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 raised to the solvent boiling point temperature. 

3. The recovery phase when purified distillate is collected in the receiving vessel. 

4. At the completion of the distillation process any remaining 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 charged 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 has 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 the 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: P exp(2.975) 19.582 mmHg 291.15 53.67

3096.52

exp 16.0137 toluene ⎟ = = ⎠

⎞ ⎜

⎝

⎛

− = −

Raoult’s Law: ptoluene = 0.985 Ptoluene = (0.985)(19.582 mm Hg) = 19.288 mm Hg. 

Displacement Vol. V (ft3

) = 100 gallons * 0.13368 (ft3/gal) = 13.368 ft3 

Ideal Gas Constant R = 998.9 mmHg• ft3 / lb-mole•°K 

Moles of toluene emitted per batch: 0.000887

998.9 291.15

19.288 13.368 = ⋅

⋅ ni = lb-moles 

Weight of toluene emitted per batch: wti = (0.000887 lb-moles)(92.13 lb/lb-mole) = 0.082 lbs 
Chapter 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 condenser operating at 20°C then we can assume that the 

exit vent gas caused by heating will leave the equipment 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 than the 20°C saturation level. However, this error is 

considered to be in significant when compared to the total emissions that occur while heating to the final 

111.2°C boiling point temperature prior to distillation. 

Ti = 18°C = 291.15°K (Initial temperature) 

PT = 1.0 atm = 760 mm Hg (Total system pressure) 

Vgas = 120 gallons = 16.04 ft3

 (Gas space volume) 

R = 998.9 (mmHg-ft3

)/(lb-mole-oK) (Universal gas constant) 

Ptoluene, 18°C = 19.906 mm Hg. (Antoine Equation) 

The partial pressure of nitrogen at 18°C is the difference between the total system pressure and the partial 

pressure of toluene: p p mmHg nc toluene 760 760 19.6 740.4 ,1 = − ,1 = − =

Gas Law: n

PV

RT = , also n

pV

RT i

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

Calculate the total amount of nitrogen that is displaced from the still during the heating activity using the 

ideal gas law at the initial conditions. When the waste 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: 0.041

998.9 291.15

2 740.4 16.04 = ∗

∗ = = RT

p V

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 toluene 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: lb moles

mmHg

mmHg ntoluene = − ⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ = 0.041 0.00121

738.2

21.8

Final toluene emission:

wt ( )( ) lb moles lb lb mole lbs toluene = 0.00121 − 92.13 / − = 0.0012

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

EIIP Volume II 16.5-3 

toluene to be recovered will be 97 gallons. The emissions that are associated with the actual distillation 

operation are calculated using the filling or charging model 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/ft3

) = 12.967 ft3

Toluene emitted per batch: 0.001048

998.9 291.15

21.85 12.967 = ∗

∗ = = RT

PV

n i

i lb-moles 

Toluene emitted per batch: wt ( lb moles)( lb lb mole) lbs toluene = 0.001048 − 92.13 / − = 0.097

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. Emissions for this operation are 

calculated as a simple filling operation. 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: 

0.001048

998.9 291.15

21.85 12.967 = ∗

∗ = = RT

PV

n i

i lb-moles 

Toluene emitted per batch: wt ( ) lb moles ( lb lb mole) lbs toluene = 0.001048 − 92.13 / − = 0.097

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 receiver 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 when 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. Different 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-101
Chapter 16 – Chemical Manufacturing 5.0 – Basic Process Operations 

EIIP Volume II 16.5-4 

Examples of batch filtration might include processing when crystalline product that 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

Slurry Receiver

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 emissions that occur during this processing step. Since the 

filter is connected directly to the filtrate receiving vessel then all emissions exit through the receiver 

process vent. In this case the emissions from the operation 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 filter 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 model 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 separate the solids portion of a 

process slurry from the liquid portion. Most centrifuge 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 include a feed step (when slurry is feed to the 

centrifuge so that the internal basket can be loaded), a spin step (when filtrate is permitted to exit the 

centrifuge cake), a washing step (when water or process solvent 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 material 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 receiver from the centrifuge. The charging or filling with 

gas sweep model is used to calculate the process vent 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 plant data) is used for the gas sweep portion of the 

centrifuge emission calculation. If the centrifuge is tightly connected to other ancillary equipment such as 

an enclosed bottom hopper then the gas sweep rate may be negligible. 

Similarly, the volume of filtrate that enters the filtrate 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 phase 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 centrifuge 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 receiver in conjunction with any gas sweep that also 

might be present forms the basis for the emission 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 final 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 pressure to the design level (3) heating the batch for 

evaporation to take place, and (4) collecting 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 vacuum 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 from wet solids because each process is different and the 

material being processed varies with respect to particle size, solvent content, exposure condition, and 

other variables. A conservative approach to estimating 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

M K A P P Q − = Eq. 5-1 

where Qm 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 Pi is the saturated solvent vapor pressure, 

Pi 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 

TL 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 existing as a continuous liquid with a fixed surface 

area. As evaporation takes place the amount of solvent 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 Pi 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 representative 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 calculated 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 calculated between the initial temperature conditions and 

within 2°C of the boiling point temperature for the 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 drain into the distillate receiver. The distillate receiver continues to 

fill with solvent as long as solvent is being removed from the product solids. Air emissions that occur 

while the vacuum drying process is ongoing originate 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 emissions during this part of the drying process. The 

composition and quantity of the distillate being collected is the same as the original amount of solvent 

that was contained in the wet cake. The fill volume is equal 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 Properties 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 calculate the total number of moles in a gas space from known variables, 

such as pressure, volume, and temperature. 

 PV = nRT Eq. 6-4 

where: n = moles of gas 

P = pressure (absolute) 

V = volume 

T = temperature (absolute) 

R = Universal Gas Constant 

The following properties exist for 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/in2 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 Vd = 550 gal = 0.133680555 ft3/gal * 550 gal = 73.52 ft3 

Temperature T = 35°C = 35°C + 273.15°K = 308.15°K 

Moles 0.182 lb - moles

998.99 308.15

760 73.52 ⎟ =

⎠

⎞ ⎜

⎝

⎛

∗

∗ ⎟ =

⎠

⎞ ⎜

⎝

⎛ = RT

PV

n

Part B (Moles of gas retained in the vessel headspace). 

Gas Space Volume Vg = 1200 gal – 550 gal = 650 gal. 

Volume conversion Vg = 0.133680555 ft3/gal * 650 gal = 86.89 ft3 

Moles 0.215 lb - moles

998.99 308.15

760 86.89 ⎟ =

⎠

⎞ ⎜

⎝

⎛

⋅

⋅ ⎟ =

⎠

⎞ ⎜

⎝

⎛ = 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 determine the number of moles of a specific compound in the 

vapor space of the vessel. 

 Dalton’s Law

RT

pV

n i

i = Eq. 6-5 

 Amagat’s Law 

ni = Pvi

RT Eq. 6-6 

where: P = pressure 

V = volume 
Chapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships 

EIIP Volume II 16.6-3 

T = temperature 

R = Universal Gas Constant 

ni = moles of i 

pi = partial pressure of i 

vi = partial volume of I

Illustration 6-2: Calculating molar quantities for gas mixtures. 

The headspace of a vessel contains the vapors of water (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

pV

n i

i =

Moles of water 0.00393 lb - moles

998.99 308.15

20 60.56 ⎟ =

⎠

⎞ ⎜

⎝

⎛

∗

∗ ⎟ =

⎠

⎞ ⎜

⎝

⎛ = RT

p V

n i

water

Moles of methanol 0.00885 lb - moles

998.99 308.15

45 60.56 ⎟ =

⎠

⎞ ⎜

⎝

⎛

∗

∗ ⎟ =

⎠

⎞ ⎜

⎝

⎛ = RT

p V

n i

methanol

Moles of ethanol 0.00787 lb - moles

998.99 308.15

40 60.56 ⎟ =

⎠

⎞ ⎜

⎝

⎛

∗

∗ ⎟ =

⎠

⎞ ⎜

⎝

⎛ = RT

p V

n i

ethanol

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

Moles of nitrogen 0.00885 lb - moles

998.99 308.15

45 60.56 ⎟ =

⎠

⎞ ⎜

⎝

⎛

∗

∗ ⎟ =

⎠

⎞ ⎜

⎝

⎛ = RT

p V

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, Xi. The liquid 

mole fraction of a compound is used later for calculating the vapor pressure for the same compound using 

Raoult’s Law. 

Mole fraction 

total

i

i

n

n

x = Eq. 6-7 

where: xi = mole fraction for component i

ni = moles of i in a single liquid phase 

ntotal = total moles in a single liquid phase 
Chapter 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 (Xi) 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: 

44.79 lb - moles

32.04

1,435 = = mwtlb

n

moles of isopropyl alcohol: 

9.09 lb - moles

60.096

546 = = mwtlb

n

moles of acetone: 

10.06 lb - moles

58.08

584 = = mwt

lb

n

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

mole fraction of methanol: 

9.09lb- moles

60.096

546 = = mwtlb

n

mole fraction of isopropyl alcohol: 

0.142

69.93

9.09 = −

− = lb moles

lb moles Xi

mole fraction of acetone: 

0.157

69.93

10.06 = −

− = lb moles

lb moles Xi

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 have been formulated over the years that enable the pure component 

vapor pressure of a compound to be accurately estimated 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 ln(Pi) = A − Eq. 6-8 
Chapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships 

EIIP Volume II 16.6-5 

where Pi = pure component vapor pressure 

T = temperature, degrees Kelvin 

A = empirical constant 

B = ΔHv RΔZv (ΔHv - heat of vaporization, ΔZv - 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 ln(P) = A −

(a) Solve for B ln( ) ln( ) ( )

1

1

1

2 1 2

− − P − P = −B T − T

 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

2 1 ln( ) ln( )

T T

P P B

4372.7371

0.0026059 0.0032792

6.6333184 3.6888795 =⎥

⎦

⎤ ⎢

⎣

⎡

−

− B = −

Solve for A 18.0281

383.75

4372.7371 ln( ) 6.6333184

2

= 2 + = + = T

B A P

Calculate the vapor pressure of toluene at 59.9 C (325.05 K) 

4.575567

325.05

4372.7371 ln( ) = − = 18.0281− = T

B P A

 P51.9 C=97.1 mm Hg. 

Note that the vapor pressure of toluene from literature 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 Clapeyron 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(Pi) = ( )

i

i

i T c

b

a

+ − Eq. 6-9 

where: Pi = pure component pressure of compound i 

T = absolute temperature 

ai, bi, ci = Antoine constants 

Coefficients ai, bi, and ci are published in several literature sources for many compounds. It is important to 

note the specific temperature and pressure units that are associated with the Antoine coefficients that are 

listed in the literature. 

The reader should keep in mind that the Antoine equation is a general mathematical relationship and can 

be used with temperatures in either Centigrade, Kelvin, 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 toluene 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 T c

b

a

+ −

Temperature (35°C) T = 35°C = 35.0°C + 273.15°K = 308.15°K 

3.8457

308.15 53.67

3096.52 ln(P ) = (16.0137 ) i = − −

46.79 mm Hg. 3.8457 P35°C = e =

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

5.3169

343.15 53.67

3096.52 ln(P ) = (16.0137 ) i = − −

203.74 mm Hg. 5.3169 P70°C = e =

6.3.3 Other Vapor-Pressure Equation Forms 

Over the years researchers have published many different 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 CT DT

T

B Pvp = A + + + Eq. 6-10 

Rankine-Kirchhoff Equation: C T

T

B ln Pvp = A + + ln Eq. 6-11 

Riedel Equation: 6 ln C lnT DT

T

B Pvp = A + + + Eq. 6-12 

Coefficients that apply to each equation are normally provided through commercially available databases. 

The DIPPER Database (AIChE) provides modeling coefficients for an equation of the general form: 

 E

vp C T DT

T

B ln P = A + + ln + Eq. 6-13 

Where Pvp = 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? 

T = 25°C + 273.15 = 298.15°K

2 7.098ln 0.0000062237 5599.6 ln 69.006 T T

T

Pvp = − − +

 

ln Pvp = 69.006 −18.781− 40.444 + 0.553 = 10.334

mmHg Pa

mmHg Pvp Pa 230.697

133.35

1 3.074 104 = × × =

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 molecules 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 equal 
Chapter 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 pi = XiPi

where: pi = effective vapor pressure of i 

Pi = pure component pressure i 

Xi = mole fraction of component i 

Illustration 6-7: Estimating component vapor pressures using Raloult’s Law. 

A solution contains 20% toluene (wt/wt), 50% heptane (wt/wt), 30% acetone. The solution temperature is 

35°C. The pure component vapor pressures for toluene, 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 component in the solution on a 100 lb basis. 

Moles of toluene 0.22 lb - moles

92.13

20 ⎟ =

⎠

⎞ ⎜

⎝

⎛ =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ = lb

MWt

wt

n

toluene

toluene

toluene

Moles of heptane 0.50 lb - moles

100.205

50

tan

tan

tan ⎟ =

⎠

⎞ ⎜

⎝

⎛ =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

= lb

MWt

wt

n

hep e

hep e

hep e

Moles of acetone 0.52 lb - moles

58.08

30 ⎟ =

⎠

⎞ ⎜

⎝

⎛ =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ = lb

MWt

wt

n

acetone

acetone

acetone

Sum of moles ntotal = 0.22 + 0.50 + 0.52 = 1.23 lb-moles 

Mole fraction of toluene 0.176

1.23

0.22 ⎟ =

⎠

⎞ ⎜

⎝

⎛ =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ = total moles

moles X toluene

toluene

Mole fraction of heptane 0.405

1.23

tan 0.50 tan ⎟ =

⎠

⎞ ⎜

⎝

⎛ =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ = total moles

moles

X hep e

hep e

Mole fraction of acetone 0.419

1.23

0.52 ⎟ =

⎠

⎞ ⎜

⎝

⎛ =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ = total moles

moles X acetone

acetone

Calculate the equilibrium vapor pressure for each component. 

Vapor pressure for of toluene ptoluene = 0.176 ∗ 46.79 = 8.24 mm Hg

Vapor pressure for of heptane phep tan e = 0.405∗ 74.04 = 29.97 mm Hg

Vapor pressure for of acetone pacetone = 0.419 ∗347.1 = 145.46 mm Hg
Chapter 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 determine 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 waste 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 

6.675

385.25 53.67

3096.52 ln(P ) = (16.0137 ) i = − −

792.362 mm Hg. 6.675 P112.1°C = e =

Raoult’s Law pi = xiPi

Molar concentration: 100% 95.9%

792.362

760.0

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. The contents contain 850 lbs of toluene, 525 lbs of 

heptane, and 2,500 lbs of other non-volatile materials. 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 9.22 lb - moles

92.13

850 ⎟ =

⎠

⎞ ⎜

⎝

⎛ =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ = lb

MWt

wt

n

toluene

toluene

toluene

Moles of heptane 5.24 lb - moles

100.205

525

tan

tan

tan ⎟ =

⎠

⎞ ⎜

⎝

⎛ =⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

= lb

MWt

wt

n

hep e

hep e

hep e

A sample of the process mixture is placed in a vacuum flask equip with a temperature probe, reflux 

condenser, vacuum pump, vacuum controller, and pressure 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 identified to consist entirely of toluene and heptane. 

However, the mole fraction of these two components cannot be determined because the molecular weight 
Chapter 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

n P

y Eq. 6-14 

Finally, the liquid mole fraction for each volatile component may be determined from Raoult’s Law. 

 i

i T

i

i

i P

y P

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 0.521

(9.22 45.79) (5.24 74.04)

(9.22 45.79) = × + ×

× = + =

i i j j

i i

i n P n P

n P

y

For heptane 0.479

(9.22 45.79) (5.24 74.04)

(5.24 74.04) = × + ×

×

y j =

Finally, 

For toluene 0.455

45.79

0.521 40.0 = × = =

i

i T

i P

y P

x

For heptane 0.258

74.04

0.479 40.0 = × = =

j

j T

j P

y P

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 nature 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 hydrogen chloride being orders of magnitude lower than 

expected by Raoult’s Law. Other examples of solutions with non ideal vapor pressure behavior would 

include systems with azeotropic properties such as acetone – hexane. 

Non ideal equilibrium systems should be calculated using an activity coefficient (or correction factor) as 

part of the basic equilibrium calculation. The equilibrium vapor pressure equation for ideal solutions has 

been modified to include the activity coefficient as part of the calculation. 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 may be estimated from known vapor pressure data. Eq. 6-16
Chapter 16 – Chemical Manufacturing 6.0 – Physical Property Relationships 

EIIP Volume II 16.6-11 

shows how the vapor pressure of a component in solution 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: pi = effective vapor pressure of i 

Pi = pure component pressure i 

Xi = mole fraction of component i 

γi = activity coefficient for component i 

T = absolute temperature 

ai, bi, ci = Antoine constants 

Illustration 6-10: Estimating activity coefficients from solution measurements. 

Using Raoult’s Law the pure component vapor pressure of ammonia over a 19.1% (wt/wt) aqueous 

solution of ammonia acid at 21.1°C is 1308 mm Hg and 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 0.16917

1308

221.3

3 = = mmHg

mmHg

NH γ

Activity Coefficient 0.972

14.9

14.47 = = mmHg

mmHg

Water γ

Assuming that the activity coefficients remain constant between 10°C and 30°C, what would the vapor 

pressure of hydrogen chloride and water be for the same solution at 25°C? 

Based on 100 lbs of solution, 

Mole fraction of NH3: 0.20

19.1 17.031 79.9 18.02

19.1 17.031 = +

Mole fraction of H2O: 0.80

79.9 18.02 19.1 17.031

79.9 18.02 = +

At 25 C the calculated pure vapor pressure for NH3 and H2O are 7,370 mm Hg and 23.58 mm Hg, 

respectively. 

Vapor pressure of NH3: pNH 3 = pNH 3γ NH 3PNH 3 = 0.20 * 0.16917 * 7370 = 249.3 mm Hg

Vapor pressure of H2O: pH 2O = X H 2Oγ H 2O PH 2O = 0.80 ∗ 0.972 ∗ 23.58 = 18.35 mm Hg

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 the 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 and 
Chapter 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, respectively. Note that the liquid and vapor compositions 

for an azeotropic system are the same at the boiling point. 

For this problem the composition of heptane and methanol 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 heptane and methanol under these conditions. 

γheptane 3.672

47.9

0.231 760 tan = ∗ = mmHgmmHg

hep e γ

γmethanol 1.243

470

0.769 760 = ∗ = mmHgmmHg 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 

pheptane = X hep tan eγ hep tan ePhep tan e = 0.231∗3.672 ∗58.54 = 50 mmHg

pmethanol = X methanolγ methanolPmethanol = 0.769 ∗1.243∗163.8 = 157 mmHg

pnitrogen = 760 - 50 -157 = 553 mm Hg

Vapor space molal percentage compositions yi: 

100% 6.6%

760

50

yhep tan e = ∗ =

100% 20.5%

760

156

ymethanol = ∗ =

100% 72.9%

760

553

ynitrogen = ∗ =
Chapter 16 – Chemical Manufacturing 7.0 – References 

EIIP Volume II 16.7-1 

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Chapter 16 – Chemical Manufacturing 7.0 – References 

EIIP Volume II 16.7-2 

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