PIECENAMP Module 9 – Steady state simulation 1 Program for North American Mobility In Higher Education NAMP Introducing Process integration for Environmental

Module 9 – Steady state simulation 11

PIECENAMPProgram for North American Mobility In Higher EducationProgram for North American Mobility In Higher Education

NAMPNAMP

Introducing Process integration for Environmental Control in Engineering CurriculaIntroducing Process integration for Environmental Control in Engineering Curricula

Introduction to Steady Introduction to Steady State SimulationState Simulation

Module 9Module 9

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Program for North American Mobility in Higher EducationProgram for North American Mobility in Higher Education NAMPNAMP

Process integration for Environmental Control in Engineering CurriculaProcess integration for Environmental Control in Engineering Curricula

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University of University of OttawaOttawa

École École Polytechnique Polytechnique de Montréalde Montréal

Instituto Instituto Mexicano del Mexicano del

PetrPetróóleoleo

North Carolina North Carolina State State

UniversityUniversity

PapricanPaprican

Universidad Universidad AutAutóónoma de noma de

San Luis PotosSan Luis Potosíí

Texas A&M Texas A&M UniversityUniversity

Universidad de Universidad de GuanajuatoGuanajuato


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Module 9Module 9

This module was This module was created by:created by:

Amy WestgateAmy Westgate

Richard EzikeRichard Ezike

Host Host UniversityUniversity

FroFromm


UniversityUniversity University of University of OttawaOttawa

University of University of OttawaOttawa


UniversityUniversity


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ObjectivesObjectives Create web-based modules to assist universities to Create web-based modules to assist universities to address the introduction to Process Integration into address the introduction to Process Integration into engineering curriculaengineering curriculaMake these modules widely available in each of the Make these modules widely available in each of the participating countriesparticipating countries

Participating institutionsParticipating institutions Six universities in three countries (Canada, Mexico and Six universities in three countries (Canada, Mexico and the USA)the USA)Two research institutes in different industry sectors: Two research institutes in different industry sectors: petroleum (Mexico) and pulp and paper (Canada)petroleum (Mexico) and pulp and paper (Canada)Each of the six universities has sponsored 7 exchange Each of the six universities has sponsored 7 exchange students during the period of the grant subsidised in students during the period of the grant subsidised in part by each of the three countries’ governmentspart by each of the three countries’ governments

Project SummaryProject Summary


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What is the structure of this module?What is the structure of this module?

All modules are divided into 3 tiers, each with a All modules are divided into 3 tiers, each with a specific goal:specific goal:

Tier I: Background InformationTier I: Background InformationTier II: Case Study ApplicationsTier II: Case Study ApplicationsTier III: Open-Ended Design ProblemTier III: Open-Ended Design Problem

These tiers are intended to be completed in that These tiers are intended to be completed in that particular order. In the first tier, students are quizzed particular order. In the first tier, students are quizzed at various points to measure their degree of at various points to measure their degree of understanding, before proceeding to the next two understanding, before proceeding to the next two tiers.tiers.

Structure of Module 9Structure of Module 9


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What is the purpose of this module?What is the purpose of this module?

It is the intent of this module to cover the basic It is the intent of this module to cover the basic aspects of aspects of Steady State SimulationSteady State Simulation. It is identified . It is identified as a pre-requisite for other modules related to the as a pre-requisite for other modules related to the learning of learning of Steady State Simulation.Steady State Simulation.

This module is intended for students familiar with This module is intended for students familiar with basic mass and energy balances and may have had basic mass and energy balances and may have had some training with thermodynamics and transport some training with thermodynamics and transport processes. processes.

Purpose of Module 9Purpose of Module 9


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Tier IBackground Information


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• Statement of IntentStatement of Intent

– Review basic chemical engineering Review basic chemical engineering concepts employed in steady state concepts employed in steady state simulationsimulation

– Understand the purpose of steady-Understand the purpose of steady-state simulationstate simulation

– Learn how to develop models of Learn how to develop models of processes in steady-stateprocesses in steady-state

– Discuss problem solving techniquesDiscuss problem solving techniques


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Steady State ProcessSteady State Process

We can use steady-state processes to We can use steady-state processes to determine the optimum operation conditions determine the optimum operation conditions for a process that can be limited by safety, for a process that can be limited by safety, equipment performance, and product quality equipment performance, and product quality constraints.constraints.

• Concentration does not change with Concentration does not change with respect to respect to timetime• Accumulation term in mass balance set to Accumulation term in mass balance set to zerozero

Input + Generation – Output – Consumption Input + Generation – Output – Consumption = 0= 0


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• Three types of processes:Three types of processes:

Batch ContinuousBatch Continuous

Semi-batchSemi-batch

•A process where a set amount of input enters a process, where it is removed from the process at a later time.

•A process where inputs and outputs flow continuously through duration of process.

•Neither batch or continuous, may be combination of both.

In steady-state processes, we will be looking at continuous processes.In steady-state processes, we will be looking at continuous processes.


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Batch ExampleBatch Example• Ammonia is produced from nitrogen and hydrogen. At Ammonia is produced from nitrogen and hydrogen. At

time t = ttime t = t00, nitrogen and hydrogen are added to the , nitrogen and hydrogen are added to the reactor. No ammonia leaves the reactor between t = reactor. No ammonia leaves the reactor between t = tt00 and t = t and t = tff. At t. At tff, n, nff moles of ammonia are released. moles of ammonia are released.

t = to

H2

N2

t = tf

NH32 2 3N + 3H 2NH


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Semibatch ExampleSemibatch Example

• Helium is Helium is pressurized in pressurized in large tanks for large tanks for storage. When the storage. When the tank valve is open, tank valve is open, the gas diffuses the gas diffuses out due to the out due to the difference in difference in pressure.pressure.


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•Continuous ExampleContinuous Example– Pump a methanol/water Pump a methanol/water

mixture into a mixture into a distillation column and distillation column and withdraw the more withdraw the more volatile component volatile component (methanol) from the top (methanol) from the top of the column and the of the column and the less volatile component less volatile component (water) from the bottom (water) from the bottom of the column.of the column.

50% molar CH3OH50% molar H2O

Mostly H2O

Mostly CH3OH


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Quiz #1

Classify the following processes as batch, continuous, or semibatch.

• A balloon is filled with air at a steady rate of 2 m3/min.

• Pump a mixture of liquids into a distillation column at a constant rate and steadily withdraw product streams from the top and bottom.

• Slowly blend several liquids in a tank from which nothing is being withdrawn.


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Block DiagramsBlock Diagrams• When solving a problem, it is helpful to

develop a block diagram, such as the one below, that defines what the process looks like as well as to indicate all information about the process such as flow rates and species compositions.

Process:

ReactoReactorr

SeparatorSeparator

Carbon (C)Carbon (C)Air (79% NAir (79% N22, 21% , 21% OO22))


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Degree Of Freedom AnalysisDegree Of Freedom Analysis

• Analysis done to determine if there is enough Analysis done to determine if there is enough information to solve a given problem.information to solve a given problem.

1.1. Draw and completely label a flowchartDraw and completely label a flowchart

2.2. Count the unknown variables, then the Count the unknown variables, then the independent equations relating them, independent equations relating them,

3.3. Subtract the number of equations from the Subtract the number of equations from the number of variables. This gives nnumber of variables. This gives ndfdf, or the , or the number of degrees of freedom in the process.number of degrees of freedom in the process.


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Degree of Freedom AnalysisDegree of Freedom Analysis

• If nIf ndfdf = 0 there are n independent equations in n = 0 there are n independent equations in n unknowns and the problem can be solvedunknowns and the problem can be solved

• If nIf ndfdf >0, there are more unknowns than >0, there are more unknowns than independent equations relating them, and at least independent equations relating them, and at least nndfdf additional variable values must be specified. additional variable values must be specified.

• If nIf ndfdf <0, there are more independent equations <0, there are more independent equations than unknowns. The flowchart is incompletely than unknowns. The flowchart is incompletely labeled or inconsistent and redundant relations labeled or inconsistent and redundant relations exist.exist.


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Mass (Material) BalanceMass (Material) Balance

A mass (material) balance is an essential calculation that accounts for the mass that enters

and leaves a particular process.

Accumulation of mass = Mass flow rate in – Mass flow rate out


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Mass (Material) Balance (continued)Mass (Material) Balance (continued)

In the case of a steady-state process we are able to set the accumulation term to zero since it is a time dependent term. Since

steady-state does not depend on time as it is constant, we are able to eliminate this

term:

Mass Flow Rate In = Mass Flow Rate Out

Material Balance Procedure


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First Law of Thermodynamics (Energy First Law of Thermodynamics (Energy Balance) for a Steady State Open Balance) for a Steady State Open

System System

The net rate at which energy is transferred to The net rate at which energy is transferred to system as heat and/or shaft work equals the system as heat and/or shaft work equals the

difference between rates at which (enthalpy+ difference between rates at which (enthalpy+ kinetic energy + potential energy) is kinetic energy + potential energy) is

transported into and out of the systemtransported into and out of the system

k p sΔH + ΔE + ΔE = Q - W


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Quiz #21. Explain the Degree of Freedom analysis.

2. What term goes to zero in a steady-state process?

3. Is a continuous process closed or open? How about a batch process?


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Heat TransferHeat Transfer

• Also classified as energy transferAlso classified as energy transfer

• Three types of heat transfer modes:Three types of heat transfer modes:– ConductionConduction– ConvectionConvection– RadiationRadiation


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ConductionConduction• Accomplished in two waysAccomplished in two ways

– Molecular interactionMolecular interaction– ““Free electrons”Free electrons”

• Conduction equation called Fourier’s LawConduction equation called Fourier’s Law

– qqyy = heat transfer area in y direction (W) = heat transfer area in y direction (W)

– A = area normal to direction flow (mA = area normal to direction flow (m22))– dT/dy = temperature gradient (dT/dy = temperature gradient (ooC/m)C/m)– k = thermal conductivity (W/mk = thermal conductivity (W/mooC)C)

yq dT = - k

A dy


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ConvectionConvection• Accomplished in two waysAccomplished in two ways

–Natural convectionNatural convection–Forced convectionForced convection

• Convection equation called Newton’s Law Convection equation called Newton’s Law

–qqyy = rate of convective heat transfer (W) = rate of convective heat transfer (W)

–A = area normal to direction flow (mA = area normal to direction flow (m22))–ΔΔT = temperature gradient (T = temperature gradient (ooC)C)–h = convective heat transfer coefficient (W/mh = convective heat transfer coefficient (W/m2 2

ooC)C)

xq = h ΔTdA = hAΔTA


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Radiation (Thermal)Radiation (Thermal)

Exhibits same optical properties as optical Exhibits same optical properties as optical lightlight

•May be absorbed, reflected, or May be absorbed, reflected, or transmittedtransmittedTotal radiation for unit area of

opaque body of area A1, emissivity ε1, and absolute temperature T1,

and a universal constant σ

41 1

1

q =

AT


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Radiation Between SurfacesRadiation Between Surfaces

Cold Cold surfacesurface

Hot surfaceHot surface

• Simplest type occurs where each surface can see only Simplest type occurs where each surface can see only the other and where both surfaces are blackthe other and where both surfaces are black

• Energy emitted by first plane is Energy emitted by first plane is σσTT1144; the second plane ; the second plane

emits emits σσTT2244

• if Tif T1 1 > T> T22, then net loss energy per unit area by first , then net loss energy per unit area by first plane and net gain by second are plane and net gain by second are σσTT11

44- - σσTT2244, or , or σσ(T(T11

44-T-T2244))

Note this is only in ideal cases: no Note this is only in ideal cases: no surface is exactly black, and surface is exactly black, and emissivities must be consideredemissivities must be considered


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Mass TransferMass Transfer

• The transport of one constituent from a The transport of one constituent from a region of higher concentration to a region region of higher concentration to a region of lower concentrationof lower concentration

• Molecular mass transferMolecular mass transfer– Random molecular motion in quiescent fluidRandom molecular motion in quiescent fluid

• Convective mass transferConvective mass transfer– From a surface into a moving fluid or vice-From a surface into a moving fluid or vice-

versaversa


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• FluxFlux= = - (overall density)*(diffusion - (overall density)*(diffusion

coefficient)*(concentration gradient)coefficient)*(concentration gradient)

• Fick rate equation (restricted to Fick rate equation (restricted to isothermal/isobaric systems)isothermal/isobaric systems)

• de Groot equation is more general de Groot equation is more general

AA AB

dCJ = - D

dz

AA AB

dyJ = - cD

dz

Units:

JA – mol A/m2s

CA- mol A/m3

DAB- m2/s


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• Molar flux of species A in binary system (A + Molar flux of species A in binary system (A + B) B)

)N(NyycDN BAAAABA

c = concentrationc = concentration

DDABAB = diffusivity of species A in B = diffusivity of species A in B

= change of molar species in y with respect = change of molar species in y with respect to a specified directionto a specified direction

NNAA, N, NBB = molar fluxes of components = molar fluxes of components

Ay


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Quiz #31. What are two ways in which conduction

occurs?

2. Define natural and forced convection.

3. What is the restriction to the use of Fick’s Law?


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ModelingModeling

• What is Modeling?What is Modeling?• Steady-State vs. Dynamic ModelingSteady-State vs. Dynamic Modeling• Empirical vs. Mechanistic ModelingEmpirical vs. Mechanistic Modeling• Derivation of a Steady State ModelDerivation of a Steady State Model• Modeling and Process Design Modeling and Process Design

ImplicationsImplications


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What is a Model?What is a Model?

A model is an depiction of a process operation A model is an depiction of a process operation

used to design, change, improve or control a used to design, change, improve or control a processprocess..

Uses of ModelUses of Model• Equipment Design, Size and SelectionEquipment Design, Size and Selection• Comparison of Different Process Comparison of Different Process

ConfigurationsConfigurations• Evaluation of Process Performance Against Evaluation of Process Performance Against

LimitationsLimitations• OptimizationOptimization


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Models vary by:Models vary by:

– Phenomena representedPhenomena represented• Energy, phase changesEnergy, phase changes

– Level of detailsLevel of details– Assumptions (perfect mixing, heat loss)Assumptions (perfect mixing, heat loss)– Inputs requiredInputs required– Functions performed (satisfaction of Functions performed (satisfaction of

constraints, optimization)constraints, optimization)– Outputs generatedOutputs generated


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Requirements of a good modelRequirements of a good model

• AccuracyAccuracy: the model should be close to the : the model should be close to the target description.target description.

• ValidityValidity: model must have a solid : model must have a solid foundation and ability to be easily foundation and ability to be easily justified.justified.

• ComplexityComplexity: the level of the model should : the level of the model should be considered and easy to understand.be considered and easy to understand.

• Computational efficiencyComputational efficiency: models should be : models should be calculable using reasonable amounts of calculable using reasonable amounts of time and computing resources.time and computing resources.


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Model

Steady State

Dynamic

Empirical

Mechanistic

Hybrid

Level of Knowledge-based Level of Knowledge-based ModelingModeling

Time-based ModelingTime-based Modeling


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Steady – StateSteady – State

Balance at equilibrium Balance at equilibrium conditioncondition

Time dependent resultsTime dependent results

Equilibrium results for all Equilibrium results for all unit operationsunit operations

Equilibrium conditions not Equilibrium conditions not assumed for all unitsassumed for all units

Equipment sizes not neededEquipment sizes not needed Equipment sizes neededEquipment sizes needed

Amount of information Amount of information required: small to mediumrequired: small to medium

Amount of information Amount of information required: medium to largerequired: medium to large

DynamicDynamic


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Steady State Steady State ExampleExample

Continuous Stirred Tank Reactor Continuous Stirred Tank Reactor (CSTR)(CSTR)

Concentration profile at one point in reactor does not change with time

t

Ca


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Dynamic ExampleDynamic Example

Batch Reactor Batch Reactor

Concentration profile at one

point in reactor does change with

timet

ca


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Empirical ModelingEmpirical Modeling

• Definition:Definition:– a model that is based on data whether a model that is based on data whether

it has been collected from a process or it has been collected from a process or some other sourcesome other source..

• Key NotesKey Notes– Derived from observation– Often simple– May or may not have theoretical

foundation– Valid only within range of observation


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• Procedure – Empirical ModelingProcedure – Empirical Modeling1.1. Obtain data from process you wish to model.Obtain data from process you wish to model.

– Temperature, pressure, flow, etc…Temperature, pressure, flow, etc…

2.2. Perform appropriate statistical analysis and Perform appropriate statistical analysis and develop accurate correlations from data.develop accurate correlations from data.

3.3. Develop mathematical equations to accurately Develop mathematical equations to accurately represent the data and the correlations found in represent the data and the correlations found in step 2, and determine which equations are useful step 2, and determine which equations are useful in the development of the model.in the development of the model.

4.4. Check for correctness in your analysis and Check for correctness in your analysis and equations, and determine if the model is equations, and determine if the model is satisfactory. satisfactory.

Statistical Analysis with ExcelStatistical Analysis with Excel


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Example:Example:The figure below depicts a heat exchanger. Heat exchangers function as a The figure below depicts a heat exchanger. Heat exchangers function as a medium to transfer energy (in the form of heat) from a hotter stream to a medium to transfer energy (in the form of heat) from a hotter stream to a cooler stream. Let’s say we have a hot stream of fluid coming into the cooler stream. Let’s say we have a hot stream of fluid coming into the exchanger at Texchanger at Th1h1, leaves at T, leaves at Th2h2 and a cool stream coming in at T and a cool stream coming in at Tc1c1 and and leaving at Tleaving at Tc2c2. If the physical properties of the fluids are the same, then the . If the physical properties of the fluids are the same, then the temperature difference describes the amount of energy transferred.temperature difference describes the amount of energy transferred.


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• We do not know TWe do not know Tc2c2, but we can , but we can take various measurements of take various measurements of TTh1h1, T, Th2h2 and T and Tc1c1 to find T to find Tc2c2 . . Using certain statistical Using certain statistical procedures, it can be procedures, it can be determined that Tdetermined that Tc2c2 is related is related to the other three to the other three temperatures by this equation:temperatures by this equation:

TTc2c2 = T = Tc1c1 + + aa(T(Th2h2-T-Th1h1))

• We have empirically We have empirically determined a value for determined a value for aa, but , but only for the specific fluids and only for the specific fluids and conditions tested. conditions tested.

• If we knew, say, the If we knew, say, the mass flow rates and mass flow rates and heat capacities of the heat capacities of the two fluids, we can use two fluids, we can use them to determine the them to determine the mechanistic model mechanistic model that relates the four that relates the four temperatures for any temperatures for any combination of two combination of two fluids. fluids.


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Mechanistic ModelingMechanistic Modeling• DefinitionDefinition

– a model that is derived from fundamental a model that is derived from fundamental physical laws or basic principlesphysical laws or basic principles

• Key NotesKey Notes– Model construction – time-consuming and Model construction – time-consuming and

costlycostly– Most reliable, but often not enough data Most reliable, but often not enough data

availableavailable


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• Procedure – Mechanistic ModelingProcedure – Mechanistic Modeling1.1. Know physical and chemical properties of the Know physical and chemical properties of the

process.process.

2.2. Determine the appropriate process model using Determine the appropriate process model using mass and/or heat balance.mass and/or heat balance.

3.3. Determine appropriate model run conditions and Determine appropriate model run conditions and parameters parameters

4.4. Complete runs and use output data to compare Complete runs and use output data to compare against the predicted model resultsagainst the predicted model results

5.5. Develop an acceptable conclusion for the model. Develop an acceptable conclusion for the model. Should the conclusion not be acceptable, re-Should the conclusion not be acceptable, re-examine the assumptions, process and the examine the assumptions, process and the physical and chemical properties made in Step 1. physical and chemical properties made in Step 1. Make appropriate modifications and repeat Steps Make appropriate modifications and repeat Steps 2-4.2-4.


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Let us go back to the heat exchanger. Now we Let us go back to the heat exchanger. Now we know that the empiricism know that the empiricism aa that we determined that we determined earlier is related to the mass flow and heat earlier is related to the mass flow and heat capacity of the two fluids. This knowledge capacity of the two fluids. This knowledge allows us to model a heat exchanger for any allows us to model a heat exchanger for any two fluids. The model is determined to be:two fluids. The model is determined to be:

H 1 2 C 2 1H p H H C p C Cm C (T - T ) = m C (T - T )


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Steady state model derivationSteady state model derivation

1. Define Goals.1. Define Goals.a) a) Specific design decisions.Specific design decisions.b)b) Numerical values. Numerical values.c)c) Functional relationships. Functional relationships.d)d) Required accuracy. Required accuracy.

2. Prepare information.a) Sketch process.a) Sketch process.b) Identify variables of interest.b) Identify variables of interest.c) State assumptions and data.c) State assumptions and data.


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Steady state model derivationSteady state model derivation

3. Formulate model3. Formulate modela)a) Conservation balances.Conservation balances.b) Constitutive equations.b) Constitutive equations.c) Rationalize (combine equations and collect c) Rationalize (combine equations and collect

terms).terms).d) Check degrees of freedom.d) Check degrees of freedom.

4. Determine solution4. Determine solutiona)a) Analytical Analyticalb)b) Numerical Numerical


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Steady state model derivationSteady state model derivation5. Analyze results5. Analyze results

a)a) Check results for correctnessCheck results for correctnessAccuracy of numerical/analytical Accuracy of numerical/analytical

methodsmethodsPlot solutionPlot solutionRelate results to data and assumptions Relate results to data and assumptions Answer “what if questions”Answer “what if questions” Compare with experimental results Compare with experimental results


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Process insights resulting from Process insights resulting from modelingmodeling

1.1. Identification: If we know the Identification: If we know the input (I) and output (O) input (I) and output (O) parameters, we can determine parameters, we can determine the structure (R) of the model.the structure (R) of the model.

I OR?


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2. 2. Simulation: If we know the Simulation: If we know the structure of the model, we can structure of the model, we can simulate what the output of the simulate what the output of the process will be for a given input.process will be for a given input.

I O?R


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3. 3. Control/Optimization: If we know Control/Optimization: If we know the desired output (O) and the the desired output (O) and the structure (R) of the model, we structure (R) of the model, we can determine what the input (I) can determine what the input (I) should be to optimize the should be to optimize the process.process.

I? OR


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Quiz #41. What are some uses of modeling?

2. Name and explain three requirements of a good model.

3. What distinguishes a steady-state model and a dynamic model?

4. Review the procedures for developing a mechanistic and empirical model. What are some differences between the two procedures?

5. Discuss the control/optimization insight of modeling.


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Solving ProblemsSolving Problems

– Analytical MethodsAnalytical Methods– Process DesignProcess Design

– MethodsMethods• SpreadsheetsSpreadsheets• Simulation SoftwareSimulation Software

– Solution DeterminationSolution Determination


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Curve fittingCurve fitting– Try to find the best fit of a curve through the Try to find the best fit of a curve through the

data such that the distribution of the data data such that the distribution of the data points on either side of the line is equalpoints on either side of the line is equal

– Possible errorsPossible errors• Measurement errorMeasurement error• Precision errorPrecision error• Systematic errorSystematic error• Calculation errorCalculation error• Error propagationError propagation• Curve Fitting ExampleCurve Fitting Example


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Least SquaresLeast Squares• The best curve through the data is The best curve through the data is

the one that minimizes the sum of the one that minimizes the sum of the squares of the residuals the squares of the residuals (differences between predicted and (differences between predicted and experimental values)experimental values)

Least Squares MethodLeast Squares Method


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Process Process DesignDesign


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Process designProcess design

The design of chemical products begins with the The design of chemical products begins with the identification and creation of potential identification and creation of potential opportunities to satisfy societal needs and to opportunities to satisfy societal needs and to generate profit. The scope of chemical product generate profit. The scope of chemical product is extremely broad. They can be roughly is extremely broad. They can be roughly classified as:classified as:

1.1. Basic chemical products.Basic chemical products.

2.2. Industrial products.Industrial products.

3.3. Consumer products.Consumer products.


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Process designProcess design

ManufacturingProcess

NaturalResources

Basic chemicalProducts

ManufacturingProcess

Basic ChemicalProducts

IndustrialProducts

ManufacturingProcessIndustrial Products

ConsumerProducts


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Motivation for Process DesignMotivation for Process Design

1.1. Desires of customers for chemicals with Desires of customers for chemicals with improved properties for many applications.improved properties for many applications.

2.2. Discovery of a new inexpensive source of a Discovery of a new inexpensive source of a raw material with comparable physical and raw material with comparable physical and chemical properties to the old source.chemical properties to the old source.

3.3. New markets are discovered.New markets are discovered.


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Steps in a Process DesignSteps in a Process Design

1.1. Process Design – Questions to AnswerProcess Design – Questions to Answer

Is the chemical structure known?Is the chemical structure known? Is a process required to produce the Is a process required to produce the

chemicals?chemicals? Is the gross profit favorable?Is the gross profit favorable? Is the process still promising after Is the process still promising after

further elaboration?further elaboration? Is the process and/or product feasible?Is the process and/or product feasible?


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Steps in a Process DesignSteps in a Process Design

Develop objective(s).Develop objective(s). Find inputs that have the desired properties Find inputs that have the desired properties

and performance.and performance. Create process.Create process. Develop a base case for which to conduct Develop a base case for which to conduct

initial testing on process.initial testing on process. (does it stay stable at steady state?)(does it stay stable at steady state?)

Improve/maintain processImprove/maintain process

2.2. Process Design – StepsProcess Design – Steps


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Stability of the processStability of the processWhen a process is disturbed from an initial When a process is disturbed from an initial

steady state, it will generally respond in steady state, it will generally respond in one of 3 ways.one of 3 ways.

a)a) Proceed to a steady state and remain Proceed to a steady state and remain there.there.


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Stability of the processStability of the process

b)b) Fail to attain to a steady state condition Fail to attain to a steady state condition because its output grows indefinitely. The because its output grows indefinitely. The system is unstable.system is unstable.


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Stability of the processStability of the process

c)c) Fail to attain a steady state condition because Fail to attain a steady state condition because the output of the process oscillates indefinitely the output of the process oscillates indefinitely with a constant amplitude. The system is at the with a constant amplitude. The system is at the limit of stability.limit of stability.


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Quiz #51. What are some errors that may arise when

attempting to fit a curve?

2. What are the three products developed from process design? Provide an example of each product.

3. What happens to an unstable system over time?


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SpreadsheetSpreadsheet– A computer program (Microsoft Excel) used to

store and calculate information in a structured array of data cells. By defining relationships between information in cells, a user can see the effects of certain data changes on other data in other parts of the spreadsheet.

– Provides an easy, efficient method for solving sets of equations and other forms of data that are not too numerous but complex enough that it would be difficult to solve by hand.


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– Columns are designated by letters, rows by numbersColumns are designated by letters, rows by numbers

http://www.instrunet.com/images/Direct%20To%20Excel%20Spreadsheet.png


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GoalseekGoalseek• Under Tools MenuUnder Tools Menu

• want to know input value formula needs to want to know input value formula needs to determine resultdetermine result

• Excel varies value in cell specified until Excel varies value in cell specified until dependent formula returns value you wantdependent formula returns value you want


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Spreadsheet DrawbacksSpreadsheet Drawbacks

• Entering the equations yourself could lead Entering the equations yourself could lead to false answers as you can make a mistake. to false answers as you can make a mistake. Mistakes can become unmanageable very Mistakes can become unmanageable very quickly causing debugging to be difficult.quickly causing debugging to be difficult.

• Excel can handle large amounts of data but Excel can handle large amounts of data but there is a point where Excel may have there is a point where Excel may have difficulty in solving a system of equations.difficulty in solving a system of equations.


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SimulationSimulation

• Predicts behavior of a process by solving Predicts behavior of a process by solving mathematical relationships that describe the mathematical relationships that describe the behavior of the process components.behavior of the process components.

• Involves performance of experiments with a Involves performance of experiments with a process modelprocess model


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• Simulation Software – Why use it?Simulation Software – Why use it?– economical way for engineers to construct economical way for engineers to construct

or modify a process before doing a test in or modify a process before doing a test in realityreality.• Can determine optimum operating conditions

– Quantify equipment, raw materials required with accuracy

• Can discover process problems– Make accurate changes in process without

sacrificing money or safety• Determine composition of streams and simplify

complex unit operations


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• Simulation Software – What does it allow?Simulation Software – What does it allow?

– Manipulation and comparison of previous data as well as for research

– Manipulation of a process until a desired target is reached

– Allows complex processes to be easily calculated

– Can easily change conditions and see how the output is changed and the equipment behaves


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• Simulation Issues and ConsiderationsSimulation Issues and Considerations– Built-in assumptions in programs – must Built-in assumptions in programs – must

be taken into account and validatedbe taken into account and validated– Can make mistakes in calculations – do Can make mistakes in calculations – do

mass balances over process as a check mass balances over process as a check overover

– Number of variables involvedNumber of variables involved– Physical properties of streamsPhysical properties of streams– Size of process being simulatedSize of process being simulated


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Process Flowsheet (Block Diagram)Process Flowsheet (Block Diagram)A process flowsheet is a collection of icons A process flowsheet is a collection of icons to represent process units and arrows to to represent process units and arrows to represent the flow of materials to and from represent the flow of materials to and from the units.the units.

Heater

Reactor

Flash

Distillation

Product

steam

Fresh feed


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Calculation OrderCalculation Order

In most process simulators, the units are In most process simulators, the units are computed one at a time. The calculation order is computed one at a time. The calculation order is automatically computed to be consistent with automatically computed to be consistent with the flow of information in the simulation the flow of information in the simulation flowsheet, where the information flow depends flowsheet, where the information flow depends on the specifications for the chemical process. on the specifications for the chemical process.

1 2 3 4


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Recycle FlowsRecycle FlowsA simulation flowsheet usually contains information A simulation flowsheet usually contains information recycle loops. That is, there are variables that are not recycle loops. That is, there are variables that are not known which prevent the equations in the process model known which prevent the equations in the process model from being solved completely. These variables are from being solved completely. These variables are recycled back to the initial calculation point. recycled back to the initial calculation point.

1 2 3 4

For these processes, a solution technique is needed to solve the equations for all the units in the recycle loop.


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IterationIteration

– Initial guess is taken at the input and a solution is determined for the system

– Second, a more educated guess is made and the system is solved based on initial solution

– Iterations continue until solution converges to one value


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ConvergenceConvergenceIs the process to compare the guessed value Is the process to compare the guessed value with the computed value until a value is found with the computed value until a value is found within the tolerance range.within the tolerance range.

Guessed value – calculated value < Tolerance

Guessed value

YesNo

Convergence

When the criterion is achieved, the solution is When the criterion is achieved, the solution is found and no more iteration needs to be done. found and no more iteration needs to be done.


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Process synthesis Process synthesis methodologiesmethodologies

Total account of an explicit process:Total account of an explicit process: is the most is the most obvious. Here we generate and evaluate every alternative obvious. Here we generate and evaluate every alternative design. We locate the better alternative by directly design. We locate the better alternative by directly comparing the evaluations.comparing the evaluations.

Evolution of design:Evolution of design: follow from the generation of a good follow from the generation of a good base case design. Designers can then make many small base case design. Designers can then make many small changes, a few at a time, to improve the design changes, a few at a time, to improve the design incrementally.incrementally.

Structured Decision Making:Structured Decision Making: following a plan that following a plan that contains all the alternatives.contains all the alternatives.

Design to target:Design to target: we design and specify unit operations to we design and specify unit operations to operate according to the desired target operation of the operate according to the desired target operation of the process.process.


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Solution DeterminationSolution Determination

– Sequential SolutionSequential Solution• Work backwards from one point in a sequential Work backwards from one point in a sequential

order solving one equation at a timeorder solving one equation at a time– Iterative MethodIterative Method

– Simultaneous SolutionSimultaneous Solution• Have to solve multiple equations with multiple Have to solve multiple equations with multiple

variables all at same timevariables all at same time• Generally requires simulation softwareGenerally requires simulation software


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Some advice when running a Some advice when running a simulationsimulation

1. Talk with trained professionals1. Talk with trained professionals (chemists, (chemists, vendors, other engineers in the field).vendors, other engineers in the field).

2.2. Beware of using estimated parameters and Beware of using estimated parameters and interaction parameters when screening interaction parameters when screening process alternatives.process alternatives.

3.3. Go see the plant. Plant personnel are usually Go see the plant. Plant personnel are usually helpful. Their insight and your knowledge of helpful. Their insight and your knowledge of modeling can help solve problems efficiently.modeling can help solve problems efficiently.


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Fresh Feed

Steam

Heater

Reactor Flash

Distillation

Product

Change inHeat Duty

Change inReactor Properties

Change inColumn Properties

Changecomposition

in feed

With a simulator, one day of process With a simulator, one day of process operation can be simulated in just seconds, operation can be simulated in just seconds, and make as many changes as you want.and make as many changes as you want.


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Commercial Simulation Software Commercial Simulation Software PackagesPackages

There are many of them, some of them are:There are many of them, some of them are:

Excel (spreadsheet) Excel (spreadsheet) Excel TutorialExcel Tutorial

Matlab Matlab MATLAB MATLAB homepagehomepage Fortran and CFortran and C++++ (programming languages) (programming languages) Aspen Aspen AspenTechAspenTech HYSYS HYSYS HYSYSHYSYS WinGEMS WinGEMS WinGEMSWinGEMS SuperPro Designer SuperPro SuperPro Designer SuperPro DesignerDesigner IDEAS (Simons)IDEAS (Simons)


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Final Quiz1. What is a drawback of using spreadsheets?

2. What are two functions that simulation allows for?

3. How are units calculated within a simulation process?

4. Explain how iteration works and why you should use it.

5. You are an engineer who has been tabbed to design a new chemical process for a company. What are some steps you can take to help you in your design?


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Tier IIWorked Examples


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– Review basic chemical engineering Review basic chemical engineering concepts employed in steady state concepts employed in steady state simulation through examplessimulation through examples

– Understand how to develop a steady-Understand how to develop a steady-state simulation problem in Excelstate simulation problem in Excel


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First Example: A First Example: A Single Effect Single Effect EvaporatorEvaporator

(to be done in Excel)(to be done in Excel)


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EvaporationEvaporationFunction is to concentrate solution

What affects evaporation?What affects evaporation?

• Rate at which heat is transferred to the liquidRate at which heat is transferred to the liquid

• Quantity of heat required to evaporate mass of waterQuantity of heat required to evaporate mass of water

• Maximum allowable temperature of liquidMaximum allowable temperature of liquid

• Pressure which evaporation takes placePressure which evaporation takes place


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Single Effect Vertical EvaporatorSingle Effect Vertical Evaporator

Three functional sectionsThree functional sections

• Heat exchangerHeat exchanger

• Evaporation sectionEvaporation section

• liquid boils and evaporatesliquid boils and evaporates

• SeparatorSeparator

• vapor leaves liquid and passes vapor leaves liquid and passes off to other equipmentoff to other equipment

Three sections contained in a vertical Three sections contained in a vertical cylindercylinder


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• In the heat exchanger section (calandria), steam In the heat exchanger section (calandria), steam condenses in the outer jacketcondenses in the outer jacket

• Liquid being evaporated boils on inside of the Liquid being evaporated boils on inside of the tubes and in the space above the upper tube tubes and in the space above the upper tube stackstack

• As evaporation proceeds, the remaining liquors As evaporation proceeds, the remaining liquors become more concentratedbecome more concentrated


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Tf, xf, hf, ṁf

TL, xL, hL, ṁL

Ts, Hs, ṁs

Tv, yv, Hv,

ṁV

Ts, hs, ṁs

P = kPa

U = J/m2 s oC

A = ? m2

Condensate S

Vapor V

Concentrated liquid L

Steam S

Feed F

Diagram of Single Effect Diagram of Single Effect Evaporator Evaporator


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Material and Heat BalancesMaterial and Heat Balances

ṁF = ṁL + ṁV

ṁFxF = ṁLxL + ṁVyV

q = UAq = UAΔΔTT

ΔT = Ts – TL

Heat given off by vaporλ = Hs – hs

ṁFhhFF + + ṁsHHss = = ṁLhhLL + + ṁVHHVV+ + ṁshhss

ṁFhhFF + + ṁsλλ = = ṁLhhLL + + ṁVHHVV

q = q = ṁs(H(Hss-h-hss) = ) = ṁsλλ

ṁsλλ – ideal heat – ideal heat transferred in transferred in evaporatorevaporator


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Finding the Latent Heat of Evaporation of Finding the Latent Heat of Evaporation of Solution and the EnthalpiesSolution and the Enthalpies

• Using the temperature of the boiling solution Using the temperature of the boiling solution TTLL, the latent heat of evaporation can be , the latent heat of evaporation can be found;found;

• The heat capacities of the liquid feed (CpThe heat capacities of the liquid feed (CpFF) ) and product (Cpand product (CpLL) are used to calculate the ) are used to calculate the enthalpies of the solution.enthalpies of the solution.


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Property Effects on the EvaporatorProperty Effects on the Evaporator• Feed TemperatureFeed Temperature

– Large effectLarge effect– Preheating can reduce heat transfer area Preheating can reduce heat transfer area

requirementsrequirements

• PressurePressure– ReductionReduction

• Reduction in boiling point of solutionReduction in boiling point of solution• Increased temperature gradientIncreased temperature gradient• Lower heating surface area requirementsLower heating surface area requirements

• Effect of Steam PressureEffect of Steam Pressure– Increased temperature gradient when higher Increased temperature gradient when higher

pressure steam is used.pressure steam is used.


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Boiling-Point Rise of SolutionsBoiling-Point Rise of Solutions

• Increase in boiling point over that of water Increase in boiling point over that of water is known as the is known as the boiling point elevation boiling point elevation (BPE)(BPE) of solutionof solution

• BPE is found using BPE is found using Duhring’s RuleDuhring’s Rule– Boiling point of a given solution is a linear Boiling point of a given solution is a linear

function of the boiling point of pure water function of the boiling point of pure water at the same pressureat the same pressure


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Duhring lines (sodium chloride)Duhring lines (sodium chloride)

http://www.nzifst.org.nz/unitoperations/evaporation4.htm


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Problem StatementProblem Statement(McCabe 16.1 modified)(McCabe 16.1 modified)

A single-effect evaporator is used to concentrate A single-effect evaporator is used to concentrate 9070 kg/h of a 5% solution of sodium chloride to 9070 kg/h of a 5% solution of sodium chloride to 20% solids. The gauge pressure of the steam is 20% solids. The gauge pressure of the steam is 1.37 atm; the absolute pressure in the vapor 1.37 atm; the absolute pressure in the vapor space is 100 mm Hg. The overall heat transfer space is 100 mm Hg. The overall heat transfer coefficient is estimated to be 1400 W/mcoefficient is estimated to be 1400 W/m2 o2 oC. The C. The feed temperature is 0feed temperature is 0ooC. Calculate the amount of C. Calculate the amount of steam consumed, the economy, and required steam consumed, the economy, and required heating surface. heating surface. First Example Excel

Spreadsheet


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1. Draw Diagram and Label Streams1. Draw Diagram and Label Streams

9070 kg/h feed, 0oC, 5%

solids, hF

TL, 20% solids, hL, ṁL

Ts, Hs, 1.37 atm gauge, ṁs

Tv, 0% solids, Hv, ṁv

Ts, hs, ṁs

P= 100 mm Hg

U = 1400 W/m2 oC

A=?

Condensate S

Vapor V

Liquor L

Steam S

Feed F

q=?


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2. Perform Mass Balances2. Perform Mass BalancesṁF = ṁL + ṁV

[9070 kg/h = ṁL kg/h+ ṁV kg/h]

ṁFxF = ṁLxL + ṁVyV (note that yv is zero because only

vapor is present, no solids)

[0.05 * 9070 kg/h = 0.2 * ṁL kg/h + 0]

• Can solve for ṁv and ṁL

ṁV = 6802.5 kg/h, ṁL = 2267.5 kg/h


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3. Perform Heat Balances to find the 3. Perform Heat Balances to find the EconomyEconomy

ṁFhhFF + + ṁSHHSS = = ṁLhhLL + + ṁVHHVV+ + ṁShhSS

ṁFhhFF + + ṁSλλ = = ṁLhhLL + + ṁVHHVV

q = q = ṁS(H(HSS- h- hSS) = ) = ṁSλλ

The economy is defined as the mass of water The economy is defined as the mass of water evaporated per mass of steam supplied.evaporated per mass of steam supplied.


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Needed DataNeeded Data• Boiling point of water at 100 mm Hg = Boiling point of water at 100 mm Hg = 5151ooCC (from steam (from steam tables)tables)

www.nzifst.org.nz/unitoperations/appendix8.htm

• Boiling point of solution = Boiling point of solution = 8888ooCC (from Duhring lines)(from Duhring lines)

http://www.nzifst.org.nz/unitoperations/evaporation4.htm

• Boiling point elevation = 88 – 51 = Boiling point elevation = 88 – 51 = 3737ooCC

• Enthalpy of vapor leaving evaporator (enthalpy of Enthalpy of vapor leaving evaporator (enthalpy of superheated vapor at 88superheated vapor at 88ooC and 100 mm Hg [.133 bar])C and 100 mm Hg [.133 bar]) == 2664 kJ/kg2664 kJ/kg (F&R, p.650) – (F&R, p.650) – also called the also called the latent heat of latent heat of evaporationevaporation

• Heat of vaporization of steam (HHeat of vaporization of steam (Hss-h-hss = = λλ ) at 1.37 atm ) at 1.37 atm gauge [20 lbgauge [20 lbff/in/in22] = 939 Btu/lb = ] = 939 Btu/lb = 2182 kJ/kg 2182 kJ/kg (McCabe, App.7, (McCabe, App.7, p.1073)p.1073)


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Finding the enthalpy of the Finding the enthalpy of the feed feed

1.1. Find the heat capacity of the liquid feedFind the heat capacity of the liquid feed

feed is 5% sodium chloride, 95% waterfeed is 5% sodium chloride, 95% water

p, i pimixall mixturecomponents

C = x C

yyNaClNaCl=0.05=0.05

yywaterwater=0.95=0.95

CCp,waterp,water=4.18 kJ/kg=4.18 kJ/kgooCC

CCp,NaClp,NaCl=0.85 kJ/kg=0.85 kJ/kgooCC

(C(Cpp))FF = .05*0.85 + .95*4.18 = = .05*0.85 + .95*4.18 = 4.01 kJ/kg4.01 kJ/kgooCC

2. Calculate Enthalpy (neglecting heats of 2. Calculate Enthalpy (neglecting heats of dilution)dilution)

F p F ref,Fh = C (T - T )

hF = 4.01 kJ/kgoC (0 - 0 oC) = 0 kJ/kg


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Finding the enthalpy of the liquor Finding the enthalpy of the liquor

1.1. Find the heat capacity of the liquorFind the heat capacity of the liquor

feed is 20% sodium chloride, 80% waterfeed is 20% sodium chloride, 80% water

yyNaClNaCl=0.20=0.20

yywaterwater=0.80=0.80

CCp,waterp,water=4.18 =4.18 kJ/kgkJ/kgooCC

CCp,NaClp,NaCl=0.85 kJ/kg=0.85 kJ/kgooCC

CCp,Lp,L = .20*0.85 + .80*4.18 = = .20*0.85 + .80*4.18 = 3.51 kJ/kg3.51 kJ/kgooCC

2. Calculate Enthalpy (neglecting heats of dilution)2. Calculate Enthalpy (neglecting heats of dilution)

L p, L refLh = C (T - T )

hL = 3.51 kJ/kgoC (88-0 oC) = 309 kJ/kg

p, i pimixall mixturecomponents

C = x C


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ṁṁLLhhLL + ṁ + ṁVVHHVV - ṁ - ṁFFhhFF = ṁ ṁSSHHSS - ṁ - ṁSShhSS = ṁ = ṁSS(H(HSS- h- hSS) = ) = ṁṁSSλλ

λλ = (H = (HSS-h-hSS)) = = 2182 kJ/kg2182 kJ/kg

(2267.5 kg/h *309.23 kJ/kg) + (6802.5 kg/h * 2664 (2267.5 kg/h *309.23 kJ/kg) + (6802.5 kg/h * 2664 kJ/kg) – (0) = kJ/kg) – (0) = ṁṁS S (H(HSS-h-hSS))

q = ṁq = ṁSS (2182 kJ/kg) (2182 kJ/kg)

Heat BalancesHeat Balances

ṁṁss=8626.5 kg/h=8626.5 kg/h

q = q = 8626.5 kg/h*2182 kJ/kg8626.5 kg/h*2182 kJ/kg = 1.88x10= 1.88x1077 kJ/h = kJ/h = 5228621 W = 5.23 MW5228621 W = 5.23 MW


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Find the EconomyFind the Economy

= = ṁṁVV//ṁṁSS

6802.5 kg/hEconomy = = 0.788

8626.5 kg/h


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4. Calculate Required Heating Surface4. Calculate Required Heating Surface

Condensing temperature of steam (1.37 atm Condensing temperature of steam (1.37 atm gauge = 126.1gauge = 126.1ooCC

q = UAq = UAΔΔTT

A = q/UA = q/UΔΔTT

2

o2 o

5228621 WA = = 98.02 m

W1400 (126.1 - 88) C

m C


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Click on the Hyperlink and click on the “Final Solution” tab to see the

final answer for the system.

First Example Final Solution


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Second Example: Second Example: Simulation of Cyclic Simulation of Cyclic

Process Process (Felder and (Felder and Rousseau, Example 10.2-3, Rousseau, Example 10.2-3,

pp. 516-519)pp. 516-519)(to be done in Excel)(to be done in Excel)


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Problem StatementProblem StatementThe gas-phase dehydrogenation of isobutane (A) to isobutene (B) is carried out in a continuous reactor. A stream of pure isobutane (the fresh feed to the process) is mixed adiabatically with a recycle stream containing 90% mole isobutane and the balance isobutene, and the combined stream goes to a catalytic reactor. The effluent from this process goes through a multistage separation process; one product stream containing all the hydrogen (C) and 10% of the isobutane leaving the reactor as well as some isobutene is sent to another part of the plant for additional processing, and the other product stream is the recycle to the reactor. The conversion of isobutane in the reactor is 35%. Assume a fresh feed of 100 mol isobutane. Simulate the process using a spreadsheet to find the desired process variables.

C4H10 C4H8 + H2


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Diagram of ProcessDiagram of Process

Second Example Cyclic Process


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NotesNotes

• A will denote isobutane, B denotes isobutene, C A will denote isobutane, B denotes isobutene, C denotes hydrogendenotes hydrogen

• All streams are gases, is the required rate All streams are gases, is the required rate of heat transfer to the reactor and is the of heat transfer to the reactor and is the net rate of heat transfer to the separation net rate of heat transfer to the separation process process

•Specific enthalpies are for the gaseous species at the stream temperatures relative to 25oC

- Heats of formation are taken from Table B.1, and heat capacity formulas are taken from Table B.2 in Felder and Rousseau

rQ

sQ


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1. Perform Degree 1. Perform Degree of Freedom Analysisof Freedom Analysis


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Review – Degrees of FreedomReview – Degrees of Freedom

1.1. Draw and completely label a flowchartDraw and completely label a flowchart

2.2. Count the unknown variables, then the Count the unknown variables, then the independent equations relating them, independent equations relating them,

3.3. Subtract the number of equations from the Subtract the number of equations from the number of variables. This gives nnumber of variables. This gives ndfdf, or the , or the number of degrees of freedom in the number of degrees of freedom in the process.process.


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Degree of Freedom AnalysisDegree of Freedom Analysis

• If nIf ndfdf = 0 there are n independent equations in = 0 there are n independent equations in n unknowns and the problem can be solvedn unknowns and the problem can be solved

• If nIf ndfdf >0, there are more unknowns than >0, there are more unknowns than independent equations relating them, and at independent equations relating them, and at least nleast ndfdf additional variable values must be additional variable values must be specified.specified.

• If nIf ndfdf <0, there are more independent <0, there are more independent equations than unknowns. The flowchart is equations than unknowns. The flowchart is incompletely labeled or inconsistent and incompletely labeled or inconsistent and redundant relations exist.redundant relations exist.


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Degree of Freedom Analysis – Mixing Degree of Freedom Analysis – Mixing PointPoint 4 unknowns (ṅA1, ṅB1, ṅ4,T1)

- 3 balances (2 material balances, 1 energy balance)

= 1 local degree of freedom


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Degree of Freedom Analysis – Degree of Freedom Analysis – ReactorReactor

7 unknowns (ṅA1, ṅB1, ṅA2, ṅB2, ṅC2, T1, )

- 4 balances (3 molecular species balances, 1 energy balance)

- 1 additional relation (35% conversion of isobutane)

+ 1 independent chemical reaction

= 3 local degrees of freedom

rQ


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Degree of Freedom Analysis – Degree of Freedom Analysis – SeparatorSeparator

8 unknowns (ṅA2, ṅB2, ṅC2, ṅA3, ṅB3, ṅC3, ṅ4, )

- 4 balances (3 material balances, 1 energy balance)

- 1 additional relation (isobutane split)

= 3 local degrees of freedom

sQ


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Net Degree of Freedom Analysis – Overall Net Degree of Freedom Analysis – Overall ProcessProcess

7 local degrees of freedoms (1+3+3)

- 7 ties (ṅA1, ṅB1, ṅA2, ṅB2, ṅC2, ṅ4, and T1

were counted twice)

= 0 net degrees of freedom

The problem can be solved for all labeled The problem can be solved for all labeled variables. variables.


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2. Equation Based 2. Equation Based Solution ProcessSolution Process


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Tearing the CycleTearing the Cycle

• Can’t solve system in a unit-to-unit manner Can’t solve system in a unit-to-unit manner without trial and errorwithout trial and error

• ““tear” between two units tear” between two units

- Purpose is to have the least number of Purpose is to have the least number of variables that have to be determined by trial variables that have to be determined by trial and errorand error

• We tear between separation process and We tear between separation process and mixing unitmixing unit

- Only have to determine ṅ- Only have to determine ṅ44 by trial and error. by trial and error.


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Solution ProcessSolution Process

• Assume value of recycle flow rate (ṅAssume value of recycle flow rate (ṅ4A4A = 100 = 100 mol/s)mol/s)

• Assume mixing point outlet temperature (TAssume mixing point outlet temperature (T11 = = 5050ooC)C)

• Vary ṅVary ṅ4A4A until calculated recycle flow rate (ṅ until calculated recycle flow rate (ṅ4C*4C*) ) equals assumed value in ṅequals assumed value in ṅ4A4A

- Will be done by driving (ṅ- Will be done by driving (ṅ4A4A - ṅ - ṅ4C*4C*) using ) using GoalseekGoalseek

• Mixing point temperature (TMixing point temperature (T11) will be varied to ) will be varied to determine the value that drives determine the value that drives ΔΔḢḢmixmix to zero to zero (remember, the mixer is adiabatic)(remember, the mixer is adiabatic)


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Known ValuesKnown ValuesXXAA = 0.35 (fractional conversion of A) = 0.35 (fractional conversion of A)

100 mol/s (basis of calculation)100 mol/s (basis of calculation)Feed temperature – 20Feed temperature – 20ooCC

Reactor Effluent Temperature – 90Reactor Effluent Temperature – 90ooCCProduct Stream Temperature – 30Product Stream Temperature – 30ooCC

Guess for recycle stream flow rate (ṅGuess for recycle stream flow rate (ṅA4A4) = 100 ) = 100 mol/smol/s

Mole fraction of A in recycle stream = 0.9Mole fraction of A in recycle stream = 0.9Mole fraction of B in recycle stream = 0.1Mole fraction of B in recycle stream = 0.1

Temperature of recycle stream – 85Temperature of recycle stream – 85ooCCInitial guess for combined stream temperature – Initial guess for combined stream temperature –

5050ooCC


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Mass Balances (based on initial Mass Balances (based on initial guesses)guesses)

ṅA1 = 100 mol/s feed + (100 mol/s recycle * 0.9 mol fraction = 190 mol/s)ṅB1 = 100 mol/s recycle * 0.1 mol fraction = 10 mol/sṅA2 = ṅA1 * (1-XA) = 123.5 mol/s

ṅB2 = ṅB1+ (ṅA1*XA)= 76.5 mol/s

ṅC2 = ṅA1 * XA = 66.5 mol/s

ṅA3 = 0.01* ṅA2 = 1.24 mol/s

ṅC4 = (ṅA2- ṅA3)/0.9 mol fraction = 135.85 mol/s

ṅB3 = ṅB2 – (0.1 mol fraction * ṅC4 )= 62.9 mol/s

ṅC3 = ṅC2 = 66.5 mol/s


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Calculation of Specific Enthalpies (Tables Calculation of Specific Enthalpies (Tables B.1 and B.2, Felder and Rousseau)B.1 and B.2, Felder and Rousseau)

oi f i pi

ˆ ˆH = (ΔH ) + C dT

of i

ˆ(ΔH ) - (heats of formation) are located in Table B.1 of F&R

A (isobutane [g]) = -134.5 kJ/mol

B (isobutene [g]) = 1.17 kJ/mol

C (hydrogen [g])= 0 kJ/mol


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piC dT

Calculation of Specific Enthalpies Calculation of Specific Enthalpies (Tables B.1 and B.2, Felder and (Tables B.1 and B.2, Felder and

Rousseau)Rousseau)- heat capacity of component i (kJ/moloC)

= a+ bT + cT-2 + dT-3 , where T is temperature in oC

ChemicalsChemicals A* 10A* 1033 B* 10B* 1055 C* 10C* 1088 D* 10D* 101212

isobutaneisobutane 89.4689.46 30.1330.13 -18.91-18.91 49.8749.87

isobuteneisobutene 82.8882.88 25.6425.64 -17.27-17.27 50.5050.50

hydrogenhydrogen 28.8428.84 0.007650.00765 0.32880.3288 -0.8698-0.8698


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Heat Balances (based on initial Heat Balances (based on initial guesses)guesses)

ΔΔḢḢmixmix = ṅ = ṅA1A1**ĤĤA1A1 + ṅ + ṅB1B1*Ĥ*ĤB1B1 – 100 mol/s*Ĥ – 100 mol/s*ĤA0A0 - (ṅ - (ṅA4A4*0.9*0.9mol A/mol *Ĥmol A/mol *ĤA4A4) - (ṅ) - (ṅA4A4*0.1 mol fraction*Ĥ*0.1 mol fraction*ĤB4B4) = ) = --78.64 kJ/mol78.64 kJ/mol = = ṅṅA2A2*Ĥ*ĤA2A2 + + ṅṅB2B2*Ĥ*ĤB2B2 + + ṅṅC2C2*Ĥ*ĤC2C2 - ṅ - ṅA1A1*Ĥ*ĤA1A1 – – ṅṅB1B1*Ĥ*ĤB1B1

= 9980.4 kJ/s= 9980.4 kJ/s

sQ

rQ

= ṅ= ṅA3A3*Ĥ*ĤA3A3 + ṅ + ṅB3B3*Ĥ*ĤB3B3 + + ṅṅC3C3*Ĥ*ĤC3C3+(ṅ+(ṅA4A4*0.900 mol *0.900 mol fraction*Ĥfraction*ĤA4A4)+(ṅ)+(ṅB4B4*0.100 mol fraction*Ĥ*0.100 mol fraction*ĤB4B4) – ) – ṅṅA2A2*Ĥ*ĤA2A2 - ṅ - ṅB2B2*Ĥ*ĤB2B2 - - ṅṅC2C2*Ĥ*ĤC2C2

= = -568.4 kJ/s-568.4 kJ/s


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Click on the Hyperlink and click on the “Final Solution” tab to see the

final answer for the system.

Second Example Final Solution


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Tier IIIOpen-ended problem

Approach to open-ended problemApproach to open-ended problem

Case Study.Case Study.


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– Learn how to approach open-ended Learn how to approach open-ended design problemsdesign problems

– Solve a problem on your ownSolve a problem on your own


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How to approach open–ended How to approach open–ended problemsproblems

State the problem clearly, including goals, State the problem clearly, including goals, constraints, and data requirements.constraints, and data requirements.

Define the trade-offs necessary.Define the trade-offs necessary. Define the criteria for a valid solution.Define the criteria for a valid solution. Develop a set of cases to simulate possible Develop a set of cases to simulate possible

solutions.solutions. Perform the simulation and evaluate results Perform the simulation and evaluate results

against solution criteria.against solution criteria. Evaluate solutions against environmental, safety Evaluate solutions against environmental, safety

and financial considerations. and financial considerations.


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Prepared by Ronald W. Rousseau and Jack Winnick, Georgia Tech Department of Chemical Engineering, and Norman

Kaplan, National Risk Management Research Laboratory, United States EPA

The Use of Limestone The Use of Limestone Slurry Scrubbing to Slurry Scrubbing to

Remove Sulfur Dioxide Remove Sulfur Dioxide from Power Plant Flue from Power Plant Flue

GasesGases


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• Protection of environment through Protection of environment through process development is an important process development is an important responsibility for chemical engineersresponsibility for chemical engineers

• Coal is an abundant source of energy Coal is an abundant source of energy and source of raw materials in and source of raw materials in productionproduction

• Predominately carbon, but contains Predominately carbon, but contains other elements and hydrocarbon other elements and hydrocarbon volatile mattervolatile matter

About CoalAbout Coal


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• burned in many of world’s power plants to burned in many of world’s power plants to produce electricityproduce electricity

• can produce a lot of pollution if gases not can produce a lot of pollution if gases not treated, like soot and ashtreated, like soot and ash

• sulfur dioxide emissions regulated in the U.S. sulfur dioxide emissions regulated in the U.S. by the Environmental Protection Agencyby the Environmental Protection Agency

• current regulations are no more than 520 ng current regulations are no more than 520 ng SOSO22 per joule of heating value of the fuel fed to per joule of heating value of the fuel fed to the furnacethe furnace

• plants must remove 90% of SOplants must remove 90% of SO22 released released when coal-burningwhen coal-burning


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About Commercial ProcessingAbout Commercial Processing

• SOSO22 removal is classified as regenerative or removal is classified as regenerative or throwawaythrowaway

• throwaway processing can be modified to throwaway processing can be modified to produce gypsum produce gypsum

• throwaway processing uses separating throwaway processing uses separating agent to remove SOagent to remove SO22 from stack gases from stack gases followed by disposal of SOfollowed by disposal of SO22 innocuously innocuously (CaSO(CaSO33 * ½ H * ½ H22O) and a slurried separating O) and a slurried separating agent of calcium carbonateagent of calcium carbonate


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Process Process DescriptionDescription


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• want to produce 500 MWe (megawatts of want to produce 500 MWe (megawatts of electricity)electricity)

• properties of coal given in table on next slideproperties of coal given in table on next slide

• coal fed at 25coal fed at 25ooC to furnace, burned with 15% C to furnace, burned with 15% excess air excess air

• sulfur reacts to form SOsulfur reacts to form SO22 and negligible SO and negligible SO33

• carbon, hydrogen oxidized completely to COcarbon, hydrogen oxidized completely to CO22 and and waterwater

• nitrogen in coal leaves furnace as Nnitrogen in coal leaves furnace as N22

• ash in coal leaves furnace in two streamsash in coal leaves furnace in two streams

• 80% leaves as fly ash in furnace flue gas80% leaves as fly ash in furnace flue gas

• remainder as bottom ash at 900remainder as bottom ash at 900ooCC


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ComponentComponent Dry Weight %Dry Weight %

CarbonCarbon 75.275.2

HydrogenHydrogen 5.05.0

NitrogenNitrogen 1.61.6

SulfurSulfur 3.53.5

OxygenOxygen 7.57.5

AshAsh 7.27.2

MoistureMoisture 4.8 kg/100 kg dry 4.8 kg/100 kg dry coalcoal

HHVHHV 30780 KJ/kg dry 30780 KJ/kg dry coalcoal

CCpp dry coal dry coal 1.046 kJ/(kg1.046 kJ/(kgooC)C)

CCpp ash ash 0.921 KJ/(kg0.921 KJ/(kgooC)C)


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• combustion air brought into process at 25combustion air brought into process at 25ooC, 50% RHC, 50% RH

• air sent to heat exchanger, temperature increased to air sent to heat exchanger, temperature increased to 315315ooCC

• air then fed to boiler, reacts with coalair then fed to boiler, reacts with coal

• flue gas leaves furnace at 330flue gas leaves furnace at 330ooC, goes to electrostatic C, goes to electrostatic precipitator precipitator

• 99.9% of particulate material removed99.9% of particulate material removed

• goes to air preheater, exchanges heat with goes to air preheater, exchanges heat with combustion aircombustion air

• leaves air preheater and split into two equal streamsleaves air preheater and split into two equal streams

• each stream is feed to one of two identical each stream is feed to one of two identical scrubber trainsscrubber trains

• trains sized to process 60% of flue gastrains sized to process 60% of flue gas


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• divided gas stream fed to scrubber, contacts aqueous slurry of divided gas stream fed to scrubber, contacts aqueous slurry of limestone, undergoes adiabatic cooling to 53limestone, undergoes adiabatic cooling to 53ooC. C.

• sulfur dioxide absorbed in the slurry and reacts with the sulfur dioxide absorbed in the slurry and reacts with the limestone: limestone:

• CaCOCaCO33 + SO + SO22 + ½ H + ½ H22O CaSOO CaSO33 · ½ H· ½ H22O + COO + CO22

• solid/liquid slurry enters scrubber at 50solid/liquid slurry enters scrubber at 50ooCC

• liquid slurry flows at 15.2 kg liquid/kg inlet gasliquid slurry flows at 15.2 kg liquid/kg inlet gas

• solid to liquid ratio in the slurry is 1:9 by weightsolid to liquid ratio in the slurry is 1:9 by weight

• liquid saturated with CaCOliquid saturated with CaCO33 and CaSO and CaSO33

• cleaned flue gascleaned flue gas

• meets EPA SOmeets EPA SO22 requirements requirements

• leaves scrubber with saturated water at 53leaves scrubber with saturated water at 53ooCC


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• cleaned flue gas contains COcleaned flue gas contains CO22 generated in scrubbing but no generated in scrubbing but no fly ashfly ash

• cleaned flue gas reheated to 80cleaned flue gas reheated to 80ooC, blended with clean flue C, blended with clean flue gas stream gas stream from other train, and sent to be released from other train, and sent to be released to atmosphere to atmosphere

• solids in spent aqueous slurrysolids in spent aqueous slurry

• unreacted CaCOunreacted CaCO33, flyash from flue gas, inert materials, , flyash from flue gas, inert materials, CaSOCaSO33

• liquid portion of slurry saturated with CaCOliquid portion of slurry saturated with CaCO33, CaSO, CaSO33

• specific gravity of 0.988specific gravity of 0.988

• spent slurry split in twospent slurry split in two

• one stream sent to a blending tank, mixed with freshly one stream sent to a blending tank, mixed with freshly ground limestone, makeup water, and recycle streamground limestone, makeup water, and recycle stream

• fresh slurry stream from blending tank fed to top of fresh slurry stream from blending tank fed to top of scrubberscrubber


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• second stream sent to filter where wet solids containing second stream sent to filter where wet solids containing fly ash, inert materials, CaSOfly ash, inert materials, CaSO33 and CaCO and CaCO33 are separated are separated from filtratefrom filtrate

• filtrate saturated with CaSOfiltrate saturated with CaSO33, CaCO, CaCO33, and is the recycle , and is the recycle stream fed to the blending tankstream fed to the blending tank

• wet solids contain 50.2% liquid that has similar wet solids contain 50.2% liquid that has similar composition to filtratecomposition to filtrate

• fresh ground limestone fed to blending tank at rate of fresh ground limestone fed to blending tank at rate of 5.2% excess of that is required to react with SO5.2% excess of that is required to react with SO22 absorbed absorbed from flue gasfrom flue gas

• limestone – 92.1% CaCOlimestone – 92.1% CaCO33 and rest is insoluble inert and rest is insoluble inert materialmaterial


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• Boiler generates steam at supercritical conditionsBoiler generates steam at supercritical conditions

• 540540ooC and 24.1 MPa absoluteC and 24.1 MPa absolute

• mechanical work derived by expanding steam through a mechanical work derived by expanding steam through a power-power- generating system of turbinesgenerating system of turbines

• low pressure steam extracted from power system contains low pressure steam extracted from power system contains 27.5% liquid 27.5% liquid water at 6.55 kPa absolutewater at 6.55 kPa absolute

• heat removed from wet low pressure steam in a condenser by heat removed from wet low pressure steam in a condenser by cooling cooling waterwater

• cooling water enters condenser at 25cooling water enters condenser at 25ooC and leaves at 28C and leaves at 28ooCC

• saturated condensate at 38saturated condensate at 38ooC is produced by condenser C is produced by condenser and pumped and pumped back to boilerback to boiler


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Assume a basis of 100 kg dry coal/min fed to Assume a basis of 100 kg dry coal/min fed to the furnace. the furnace.

1.1. Construct a flowchart of the process and Construct a flowchart of the process and completely label the streams. Show the completely label the streams. Show the details of only one train in the scrubber details of only one train in the scrubber operation. Do this in Excel.operation. Do this in Excel.

2.2. Estimate the molar flow rate (kmol/min) of Estimate the molar flow rate (kmol/min) of each element in the coal (other than those each element in the coal (other than those in the ash).in the ash).

3.3. Determine the feed rate (kmol/min) of ODetermine the feed rate (kmol/min) of O22 required for complete combustion of the required for complete combustion of the coal.coal.


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4.4. If 15% excess oxygen is fed to combustion furnace, If 15% excess oxygen is fed to combustion furnace, estimate the following:estimate the following:

a.a. The oxygen and nitrogen feed rates (kmol/min)The oxygen and nitrogen feed rates (kmol/min)

b.b. The mole fraction of water in the wet air, the The mole fraction of water in the wet air, the average molecular weight, and the molar flow rate average molecular weight, and the molar flow rate of water in the air stream (kmol/min)of water in the air stream (kmol/min)

c.c. The air feed rate (kmol/min, mThe air feed rate (kmol/min, m33/min) /min)

5.5. Estimate flow rate (kmol/min, kg/min) of each Estimate flow rate (kmol/min, kg/min) of each component and composition (mole frac) of furnace component and composition (mole frac) of furnace flue gas (ignore fly ash). At what rate (kg/min) is fly flue gas (ignore fly ash). At what rate (kg/min) is fly ash removed from flue gas by the electrostatic ash removed from flue gas by the electrostatic precipitator?precipitator?


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6. If system is assumed to meet standard 90% SO2 removal released upon combustion:

a. Determine flow rate (kg/min and kmol/min) of each component in the flue gas leaving scrubber

b. Determine flow rate (kg/min) of slurry entering scrubber

c. Estimate solid-to-liquid mass ratio in slurry leaving scrubber.

d. Estimate feed rate (kg/min) of fresh ground limestone to the blending tank.


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6. (continued)6. (continued)

e. What are flow rates (kg/min) of inerts, e. What are flow rates (kg/min) of inerts, CaSOCaSO33, , CaCOCaCO33, fly ash, and water, in the wet , fly ash, and water, in the wet solids solids removed from the filter?removed from the filter?

f. Estimate rate (kg/min, L/min) at which f. Estimate rate (kg/min, L/min) at which filtrate is filtrate is recycled to blending tank. At what rate recycled to blending tank. At what rate (kg/min, (kg/min, L/min) is makeup water added to L/min) is makeup water added to blending tank?blending tank?

7. At what rate is heat removed from the furnace? 7. At what rate is heat removed from the furnace? Estimate the rate of steam generation in the power Estimate the rate of steam generation in the power cycle, assuming all the heat removed from the cycle, assuming all the heat removed from the furnace is used to make steam. furnace is used to make steam.


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References:References:

• Felder, R.F. and Rousseau, R.W. Elementary Principles of Chemical Processes, Third Edition. New York, John Wiley and Sons, 2000.

• Smith, J.C. and Harriott, Peter. Unit Operations of Chemical Engineering, Sixth Edition. Boston, McGraw Hill, 2001.

• Earle, R.L. Unit Operations in Food Processing, Second Edition. http://www.nzifst.org.nz/unitoperations/index.htm

• Thibault, Jules. Notes, CHE 4311: Unit Operations. University of Ottawa, August 2002.

• Genzer, Jan. Notes, CHE 225: Chemical Process Systems. North Carolina State University, August 2002.


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• Source on pictures for slides 13, 41, Source on pictures for slides 13, 41,

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