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

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Presentation on theme: "PIECENAMP Module 9 – Steady state simulation 1 Program for North American Mobility In Higher Education NAMP Introducing Process integration for Environmental."— Presentation transcript:

1 PIECENAMP Module 9 – Steady state simulation 1 Program for North American Mobility In Higher Education NAMP Introducing Process integration for Environmental Control in Engineering Curricula Introduction to Steady State Simulation Module 9 PIECE

2 PIECENAMP Module 9 – Steady state simulation 2 Program for North American Mobility in Higher Education NAMP Process integration for Environmental Control in Engineering Curricula PIECE University of Ottawa École Polytechnique de Montréal Instituto Mexicano del Petróleo North Carolina State University Paprican Universidad Autónoma de San Luis Potosí Texas A&M University Universidad de Guanajuato

3 PIECENAMP Module 9 – Steady state simulation 3 Module 9 This module was created by: Amy Westgate Richard Ezike Host University From North Carolina State University University of Ottawa North Carolina State University

4 PIECENAMP Module 9 – Steady state simulation 4 Objectives Create web-based modules to assist universities to address the introduction to Process Integration into engineering curricula Make these modules widely available in each of the participating countries Participating institutions Six universities in three countries (Canada, Mexico and the USA) Two research institutes in different industry sectors: petroleum (Mexico) and pulp and paper (Canada) Each of the six universities has sponsored 7 exchange students during the period of the grant subsidised in part by each of the three countries’ governments Project Summary

5 PIECENAMP Module 9 – Steady state simulation 5 What is the structure of this module? All modules are divided into 3 tiers, each with a specific goal: Tier I: Background Information Tier II: Case Study Applications Tier III: Open-Ended Design Problem These tiers are intended to be completed in that particular order. In the first tier, students are quizzed at various points to measure their degree of understanding, before proceeding to the next two tiers. Structure of Module 9

6 PIECENAMP Module 9 – Steady state simulation 6 What is the purpose of this module? It is the intent of this module to cover the basic aspects of Steady State Simulation. It is identified as a pre- requisite for other modules related to the learning of Steady State Simulation. This module is intended for students familiar with basic mass and energy balances and may have had some training with thermodynamics and transport processes. Purpose of Module 9

7 PIECENAMP Module 9 – Steady state simulation 7 Tier I Background Information

8 PIECENAMP Module 9 – Steady state simulation 8 Statement of IntentStatement of Intent –Review basic chemical engineering concepts employed in steady state simulation –Understand the purpose of steady-state simulation –Learn how to develop models of processes in steady-state –Discuss problem solving techniques

9 PIECENAMP Module 9 – Steady state simulation 9 Steady State Process We can use steady-state processes to determine the optimum operation conditions for a process that can be limited by safety, equipment performance, and product quality constraints. Concentration does not change with respect to Concentration does not change with respect to time time Accumulation term in mass balance set to zero Accumulation term in mass balance set to zero Input + Generation – Output – Consumption = 0

10 PIECENAMP Module 9 – Steady state simulation 10 Three types of processes:Three types of processes: Batch Continuous Batch Continuous Semi-batch Semi-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.

11 PIECENAMP Module 9 – Steady state simulation 11 Batch Example Ammonia is produced from nitrogen and hydrogen. At time t = t 0, nitrogen and hydrogen are added to the reactor. No ammonia leaves the reactor between t = t 0 and t = t f. At t f, n f moles of ammonia are released.Ammonia is produced from nitrogen and hydrogen. At time t = t 0, nitrogen and hydrogen are added to the reactor. No ammonia leaves the reactor between t = t 0 and t = t f. At t f, n f moles of ammonia are released. t = t o H2N2H2N2 t = t f NH 3

12 PIECENAMP Module 9 – Steady state simulation 12 Semibatch Example Helium is pressurized in large tanks for storage. When the tank valve is open, the gas diffuses out due to the difference in pressure.Helium is pressurized in large tanks for storage. When the tank valve is open, the gas diffuses out due to the difference in pressure.

13 PIECENAMP Module 9 – Steady state simulation 13 Continuous ExampleContinuous Example –Pump a methanol/water mixture into a distillation column and withdraw the more volatile component (methanol) from the top of the column and the less volatile component (water) from the bottom of the column. 50% molar CH 3 OH 50% molar H 2 O Mostly H 2 O Mostly CH 3 OH

14 PIECENAMP Module 9 – Steady state simulation 14 Quiz #1 Classify the following processes as batch, continuous, or semibatch. A balloon is filled with air at a steady rate of 2 m 3 /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.

15 PIECENAMP Module 9 – Steady state simulation 15 Block 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: Reactor Separator Carbon (C) Air (79% N 2, 21% O 2 )

16 PIECENAMP Module 9 – Steady state simulation 16 Degree Of Freedom Analysis Analysis done to determine if there is enough information to solve a given problem. Analysis done to determine if there is enough information to solve a given problem. 1.Draw and completely label a flowchart 2.Count the unknown variables, then the independent equations relating them, 3.Subtract the number of equations from the number of variables. This gives n df, or the number of degrees of freedom in the process.

17 PIECENAMP Module 9 – Steady state simulation 17 Degree of Freedom Analysis If n df = 0 there are n independent equations in n unknowns and the problem can be solvedIf n df = 0 there are n independent equations in n unknowns and the problem can be solved If n df >0, there are more unknowns than independent equations relating them, and at least n df additional variable values must be specified.If n df >0, there are more unknowns than independent equations relating them, and at least n df additional variable values must be specified. If n df <0, there are more independent equations than unknowns. The flowchart is incompletely labeled or inconsistent and redundant relations exist.If n df <0, there are more independent equations than unknowns. The flowchart is incompletely labeled or inconsistent and redundant relations exist.

18 PIECENAMP Module 9 – Steady state simulation 18 Mass (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

19 PIECENAMP Module 9 – Steady state simulation 19 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

20 PIECENAMP Module 9 – Steady state simulation 20 First Law of Thermodynamics (Energy Balance) for a Steady State Open System The net rate at which energy is transferred to system as heat and/or shaft work equals the difference between rates at which (enthalpy+ kinetic energy + potential energy) is transported into and out of the system

21 PIECENAMP Module 9 – Steady state simulation 21 Quiz #2 1.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?

22 PIECENAMP Module 9 – Steady state simulation 22 Heat Transfer Also classified as energy transferAlso classified as energy transfer Three types of heat transfer modes:Three types of heat transfer modes: –Conduction –Convection –Radiation

23 PIECENAMP Module 9 – Steady state simulation 23 Conduction Accomplished in two waysAccomplished in two ways –Molecular interaction –“Free electrons” Conduction equation called Fourier’s LawConduction equation called Fourier’s Law –q y = heat transfer area in y direction (W) –A = area normal to direction flow (m 2 ) –dT/dy = temperature gradient ( o C/m) –k = thermal conductivity (W/m o C)

24 PIECENAMP Module 9 – Steady state simulation 24 Convection Accomplished in two waysAccomplished in two ways –Natural convection –Forced convection Convection equation called Newton’s LawConvection equation called Newton’s Law –q y = rate of convective heat transfer (W) –A = area normal to direction flow (m 2 ) –ΔT = temperature gradient ( o C) –h = convective heat transfer coefficient (W/m 2 o C)

25 PIECENAMP Module 9 – Steady state simulation 25 Radiation (Thermal) Exhibits same optical properties as optical light May be absorbed, reflected, or transmittedMay be absorbed, reflected, or transmitted Total radiation for unit area of opaque body of area A 1, emissivity ε 1, and absolute temperature T 1, and a universal constant σ

26 PIECENAMP Module 9 – Steady state simulation 26 Radiation Between Surfaces Cold surface Hot surface Simplest type occurs where each surface can see only the other and where both surfaces are black Simplest type occurs where each surface can see only the other and where both surfaces are black Energy emitted by first plane is σT 1 4 ; the second plane emits σT 2 4 Energy emitted by first plane is σT 1 4 ; the second plane emits σT 2 4 if T 1 > T 2, then net loss energy per unit area by first plane and net gain by second are σT σT 2 4, or σ(T 1 4 -T 2 4 ) if T 1 > T 2, then net loss energy per unit area by first plane and net gain by second are σT σT 2 4, or σ(T 1 4 -T 2 4 ) Note this is only in ideal cases: no surface is exactly black, and emissivities must be considered

27 PIECENAMP Module 9 – Steady state simulation 27 Mass Transfer The transport of one constituent from a region of higher concentration to a region of lower concentrationThe transport of one constituent from a region of higher concentration to a region of lower concentration Molecular mass transferMolecular mass transfer –Random molecular motion in quiescent fluid Convective mass transferConvective mass transfer –From a surface into a moving fluid or vice-versa

28 PIECENAMP Module 9 – Steady state simulation 28 FluxFlux = - (overall density)*(diffusion coefficient)*(concentration gradient) Fick rate equation (restricted to isothermal/isobaric systems)Fick rate equation (restricted to isothermal/isobaric systems) de Groot equation is more generalde Groot equation is more general Units: J A – mol A/m 2 s C A - mol A/m 3 D AB - m 2 /s

29 PIECENAMP Module 9 – Steady state simulation 29 Molar flux of species A in binary system (A + B)Molar flux of species A in binary system (A + B) c = concentration D AB = diffusivity of species A in B = change of molar species in y with respect to a specified direction = change of molar species in y with respect to a specified direction N A, N B = molar fluxes of components

30 PIECENAMP Module 9 – Steady state simulation 30 Quiz #3 1.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?

31 PIECENAMP Module 9 – Steady state simulation 31 Modeling 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 ImplicationsModeling and Process Design Implications

32 PIECENAMP Module 9 – Steady state simulation 32 What is a Model? A model is an depiction of a process operation used to design, change, improve or control a process. Uses of Model Equipment Design, Size and SelectionEquipment Design, Size and Selection Comparison of Different Process ConfigurationsComparison of Different Process Configurations Evaluation of Process Performance Against LimitationsEvaluation of Process Performance Against Limitations OptimizationOptimization

33 PIECENAMP Module 9 – Steady state simulation 33 Models vary by: –Phenomena represented Energy, phase changesEnergy, phase changes –Level of details –Assumptions (perfect mixing, heat loss) –Inputs required –Functions performed (satisfaction of constraints, optimization) –Outputs generated

34 PIECENAMP Module 9 – Steady state simulation 34 Requirements of a good model Accuracy: the model should be close to the target description.Accuracy: the model should be close to the target description. Validity: model must have a solid foundation and ability to be easily justified.Validity: model must have a solid foundation and ability to be easily justified. Complexity: the level of the model should be considered and easy to understand.Complexity: the level of the model should be considered and easy to understand. Computational efficiency: models should be calculable using reasonable amounts of time and computing resources.Computational efficiency: models should be calculable using reasonable amounts of time and computing resources.

35 PIECENAMP Module 9 – Steady state simulation 35 Model Steady State Dynamic EmpiricalMechanisticHybrid Level of Knowledge-based Modeling Time-based Modeling

36 PIECENAMP Module 9 – Steady state simulation 36 Steady – State Balance at equilibrium condition Time dependent results Equilibrium results for all unit operations Equilibrium conditions not assumed for all units Equipment sizes not needed Equipment sizes needed Amount of information required: small to medium Amount of information required: medium to large Dynamic

37 PIECENAMP Module 9 – Steady state simulation 37 Steady State Example Continuous Stirred Tank Reactor (CSTR) Concentration profile at one point in reactor does not change with time t CaCa

38 PIECENAMP Module 9 – Steady state simulation 38 Dynamic Example Batch Reactor Concentration profile at one point in reactor does change with time t caca

39 PIECENAMP Module 9 – Steady state simulation 39 Empirical Modeling Definition:Definition: –a model that is based on data whether it has been collected from a process or some other source. Key NotesKey Notes – –Derived from observation – –Often simple – –May or may not have theoretical foundation – –Valid only within range of observation

40 PIECENAMP Module 9 – Steady state simulation 40 Procedure – Empirical ModelingProcedure – Empirical Modeling 1.Obtain data from process you wish to model. –Temperature, pressure, flow, etc… 2.Perform appropriate statistical analysis and develop accurate correlations from data. 3.Develop mathematical equations to accurately represent the data and the correlations found in step 2, and determine which equations are useful in the development of the model. 4.Check for correctness in your analysis and equations, and determine if the model is satisfactory. Statistical Analysis with Excel Statistical Analysis with Excel

41 PIECENAMP Module 9 – Steady state simulation 41 Example: 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 cooler stream. Let’s say we have a hot stream of fluid coming into the exchanger at T h1, leaves at T h2 and a cool stream coming in at T c1 and leaving at T c2. If the physical properties of the fluids are the same, then the temperature difference describes the amount of energy transferred.

42 PIECENAMP Module 9 – Steady state simulation 42 We do not know T c2, but we can take various measurements of T h1, T h2 and T c1 to find T c2. Using certain statistical procedures, it can be determined that T c2 is related to the other three temperatures by this equation:We do not know T c2, but we can take various measurements of T h1, T h2 and T c1 to find T c2. Using certain statistical procedures, it can be determined that T c2 is related to the other three temperatures by this equation: T c2 = T c1 + a(T h2 -T h1 ) T c2 = T c1 + a(T h2 -T h1 ) We have empirically determined a value for a, but only for the specific fluids and conditions tested.We have empirically determined a value for a, but only for the specific fluids and conditions tested. If we knew, say, the mass flow rates and heat capacities of the two fluids, we can use them to determine the mechanistic model that relates the four temperatures for any combination of two fluids.If we knew, say, the mass flow rates and heat capacities of the two fluids, we can use them to determine the mechanistic model that relates the four temperatures for any combination of two fluids.

43 PIECENAMP Module 9 – Steady state simulation 43 Mechanistic Modeling DefinitionDefinition –a model that is derived from fundamental physical laws or basic principles Key NotesKey Notes –Model construction – time-consuming and costly –Most reliable, but often not enough data available

44 PIECENAMP Module 9 – Steady state simulation 44 Procedure – Mechanistic ModelingProcedure – Mechanistic Modeling 1.Know physical and chemical properties of the process. 2.Determine the appropriate process model using mass and/or heat balance. 3.Determine appropriate model run conditions and parameters 4.Complete runs and use output data to compare against the predicted model results 5.Develop an acceptable conclusion for the model. Should the conclusion not be acceptable, re-examine the assumptions, process and the physical and chemical properties made in Step 1. Make appropriate modifications and repeat Steps 2-4.

45 PIECENAMP Module 9 – Steady state simulation 45 Let us go back to the heat exchanger. Now we know that the empiricism a that we determined earlier is related to the mass flow and heat capacity of the two fluids. This knowledge allows us to model a heat exchanger for any two fluids. The model is determined to be:

46 PIECENAMP Module 9 – Steady state simulation 46 Steady state model derivation 1. Define Goals. 1. Define Goals. a) Specific design decisions. b) Numerical values. c) Functional relationships. d) Required accuracy. 2. Prepare information. a) Sketch process. b) Identify variables of interest. c) State assumptions and data.

47 PIECENAMP Module 9 – Steady state simulation 47 Steady state model derivation 3. Formulate model a) Conservation balances. b) Constitutive equations. c) Rationalize (combine equations and collect terms). d) Check degrees of freedom. 4. Determine solution 4. Determine solution a) Analytical b) Numerical

48 PIECENAMP Module 9 – Steady state simulation 48 Steady state model derivation 5. Analyze results a) Check results for correctness Accuracy of numerical/analytical methods Accuracy of numerical/analytical methods Plot solution Plot solution Relate 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

49 PIECENAMP Module 9 – Steady state simulation 49 Process insights resulting from modeling 1.Identification: If we know the input (I) and output (O) parameters, we can determine the structure (R) of the model. IO R?

50 PIECENAMP Module 9 – Steady state simulation 50 Process insights resulting from modeling 2. Simulation: If we know the structure of the model, we can simulate what the output of the process will be for a given input. IO? R

51 PIECENAMP Module 9 – Steady state simulation 51 Process insights resulting from modeling 3. Control/Optimization: If we know the desired output (O) and the structure (R) of the model, we can determine what the input (I) should be to optimize the process. I?O R

52 PIECENAMP Module 9 – Steady state simulation 52 Quiz #4 1.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.

53 PIECENAMP Module 9 – Steady state simulation 53 Solving Problems –Analytical Methods –Process Design –Methods SpreadsheetsSpreadsheets Simulation SoftwareSimulation Software –Solution Determination

54 PIECENAMP Module 9 – Steady state simulation 54 Curve fitting –Try to find the best fit of a curve through the data such that the distribution of the data points on either side of the line is equal –Possible errors Measurement errorMeasurement error Precision errorPrecision error Systematic errorSystematic error Calculation errorCalculation error Error propagationError propagation Curve Fitting ExampleCurve Fitting ExampleCurve Fitting ExampleCurve Fitting Example

55 PIECENAMP Module 9 – Steady state simulation 55 Least Squares The best curve through the data is the one that minimizes the sum of the squares of the residuals (differences between predicted and experimental values)The best curve through the data is the one that minimizes the sum of the squares of the residuals (differences between predicted and experimental values) Least Squares Method Least Squares Method

56 PIECENAMP Module 9 – Steady state simulation 56 Process Design

57 PIECENAMP Module 9 – Steady state simulation 57 Process design The design of chemical products begins with the identification and creation of potential opportunities to satisfy societal needs and to generate profit. The scope of chemical product is extremely broad. They can be roughly classified as: 1.Basic chemical products. 2.Industrial products. 3.Consumer products.

58 PIECENAMP Module 9 – Steady state simulation 58 Process design Manufacturing Process Natural Resources Basic chemical Products Manufacturing Process Basic Chemical Products Industrial Products Manufacturing Process Industrial Products Consumer Products

59 PIECENAMP Module 9 – Steady state simulation 59 Motivation for Process Design 1.Desires of customers for chemicals with improved properties for many applications. 2.Discovery of a new inexpensive source of a raw material with comparable physical and chemical properties to the old source. 3.New markets are discovered.

60 PIECENAMP Module 9 – Steady state simulation 60 Steps in a Process Design 1.Process Design – Questions to Answer Is the chemical structure known? Is the chemical structure known? Is a process required to produce the chemicals? Is a process required to produce the chemicals? Is the gross profit favorable? Is the gross profit favorable? Is the process still promising after further elaboration? Is the process still promising after further elaboration? Is the process and/or product feasible? Is the process and/or product feasible?

61 PIECENAMP Module 9 – Steady state simulation 61 Steps in a Process Design Develop objective(s). Develop objective(s). Find inputs that have the desired properties and performance. Find inputs that have the desired properties and performance. Create process. Create process. Develop a base case for which to conduct initial testing on process. Develop a base case for which to conduct initial testing on process. (does it stay stable at steady state?) (does it stay stable at steady state?) Improve/maintain process Improve/maintain process 2.Process Design – Steps

62 PIECENAMP Module 9 – Steady state simulation 62 Stability of the process When a process is disturbed from an initial steady state, it will generally respond in one of 3 ways. a)Proceed to a steady state and remain there.

63 PIECENAMP Module 9 – Steady state simulation 63 Stability of the process b)Fail to attain to a steady state condition because its output grows indefinitely. The system is unstable.

64 PIECENAMP Module 9 – Steady state simulation 64 Stability of the process c)Fail to attain a steady state condition because the output of the process oscillates indefinitely with a constant amplitude. The system is at the limit of stability.

65 PIECENAMP Module 9 – Steady state simulation 65 Quiz #5 1.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?

66 PIECENAMP Module 9 – Steady state simulation 66 Spreadsheet – –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.

67 PIECENAMP Module 9 – Steady state simulation 67 –Columns are designated by letters, rows by numbers

68 PIECENAMP Module 9 – Steady state simulation 68 Goalseek Under Tools Menu Under Tools Menu want to know input value formula needs to determine result want to know input value formula needs to determine result Excel varies value in cell specified until dependent formula returns value you want Excel varies value in cell specified until dependent formula returns value you want

69 PIECENAMP Module 9 – Steady state simulation 69 Spreadsheet Drawbacks Entering the equations yourself could lead to false answers as you can make a mistake. Mistakes can become unmanageable very quickly causing debugging to be difficult.Entering the equations yourself could lead to false answers as you can make a mistake. Mistakes can become unmanageable very quickly causing debugging to be difficult. Excel can handle large amounts of data but there is a point where Excel may have difficulty in solving a system of equations.Excel can handle large amounts of data but there is a point where Excel may have difficulty in solving a system of equations.

70 PIECENAMP Module 9 – Steady state simulation 70 Simulation Predicts behavior of a process by solving mathematical relationships that describe the behavior of the process components.Predicts behavior of a process by solving mathematical relationships that describe the behavior of the process components. Involves performance of experiments with a process modelInvolves performance of experiments with a process model

71 PIECENAMP Module 9 – Steady state simulation 71 Simulation Software – Why use it?Simulation Software – Why use it? –economical way for engineers to construct or modify a process before doing a test in reality –economical way for engineers to construct or modify a process before doing a test in reality. 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

72 PIECENAMP Module 9 – Steady state simulation 72 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

73 PIECENAMP Module 9 – Steady state simulation 73 Simulation Issues and ConsiderationsSimulation Issues and Considerations –Built-in assumptions in programs – must be taken into account and validated –Can make mistakes in calculations – do mass balances over process as a check over –Number of variables involved –Physical properties of streams –Size of process being simulated

74 PIECENAMP Module 9 – Steady state simulation 74 Process Flowsheet (Block Diagram) A process flowsheet is a collection of icons to represent process units and arrows to represent the flow of materials to and from the units. Heater Reactor Flash Distillation Product steam Fresh feed

75 PIECENAMP Module 9 – Steady state simulation 75 Calculation Order In most process simulators, the units are computed one at a time. The calculation order is automatically computed to be consistent with the flow of information in the simulation flowsheet, where the information flow depends on the specifications for the chemical process. 1234

76 PIECENAMP Module 9 – Steady state simulation 76 Recycle Flows A simulation flowsheet usually contains information recycle loops. That is, there are variables that are not known which prevent the equations in the process model from being solved completely. These variables are recycled back to the initial calculation point For these processes, a solution technique is needed to solve the equations for all the units in the recycle loop.

77 PIECENAMP Module 9 – Steady state simulation 77 Iteration – –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

78 PIECENAMP Module 9 – Steady state simulation 78 Convergence Is the process to compare the guessed value with the computed value until a value is found within the tolerance range. Guessed value – calculated value < Tolerance Guessed value Yes No Convergence When the criterion is achieved, the solution is found and no more iteration needs to be done.

79 PIECENAMP Module 9 – Steady state simulation 79 Process synthesis methodologies Total account of an explicit process: is the most obvious. Here we generate and evaluate every alternative design. We locate the better alternative by directly comparing the evaluations. Total account of an explicit process: is the most obvious. Here we generate and evaluate every alternative design. We locate the better alternative by directly comparing the evaluations. Evolution of design: follow from the generation of a good base case design. Designers can then make many small changes, a few at a time, to improve the design incrementally. Evolution of design: follow from the generation of a good base case design. Designers can then make many small changes, a few at a time, to improve the design incrementally. Structured Decision Making: following a plan that contains all the alternatives. Structured Decision Making: following a plan that contains all the alternatives. Design to target: we design and specify unit operations to operate according to the desired target operation of the process. Design to target: we design and specify unit operations to operate according to the desired target operation of the process.

80 PIECENAMP Module 9 – Steady state simulation 80 Solution Determination –Sequential Solution Work backwards from one point in a sequential order solving one equation at a timeWork backwards from one point in a sequential order solving one equation at a time –Iterative Method –Simultaneous Solution Have to solve multiple equations with multiple variables all at same timeHave to solve multiple equations with multiple variables all at same time Generally requires simulation softwareGenerally requires simulation software

81 PIECENAMP Module 9 – Steady state simulation 81 Some advice when running a simulation 1. Talk with trained professionals (chemists, vendors, other engineers in the field). 2.Beware of using estimated parameters and interaction parameters when screening process alternatives. 3.Go see the plant. Plant personnel are usually helpful. Their insight and your knowledge of modeling can help solve problems efficiently.

82 PIECENAMP Module 9 – Steady state simulation 82 Fresh Feed Steam Heater Reactor Flash Distillation Product Change in Heat Duty Change in Reactor Properties Change in Column Properties Change composition in feed With a simulator, one day of process operation can be simulated in just seconds, and make as many changes as you want.

83 PIECENAMP Module 9 – Steady state simulation 83 Commercial Simulation Software Packages There are many of them, some of them are: Excel (spreadsheet) Excel Tutorial Excel (spreadsheet) Excel Tutorial Excel Tutorial Excel Tutorial Matlab MATLAB homepage Matlab MATLAB homepage MATLAB homepage MATLAB homepage Fortran and C ++ (programming languages) Fortran and C ++ (programming languages) Aspen AspenTech Aspen AspenTech AspenTech HYSYS HYSYS HYSYS HYSYS HYSYS WinGEMS WinGEMS WinGEMS WinGEMS WinGEMS SuperPro Designer SuperPro Designer SuperPro Designer SuperPro DesignerSuperPro DesignerSuperPro Designer IDEAS (Simons) IDEAS (Simons)

84 PIECENAMP Module 9 – Steady state simulation 84 Final Quiz 1.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?

85 PIECENAMP Module 9 – Steady state simulation 85 Tier II Worked Examples

86 PIECENAMP Module 9 – Steady state simulation 86 Statement of IntentStatement of Intent –Review basic chemical engineering concepts employed in steady state simulation through examples –Understand how to develop a steady-state simulation problem in Excel

87 PIECENAMP Module 9 – Steady state simulation 87 First Example: A Single Effect Evaporator (to be done in Excel)

88 PIECENAMP Module 9 – Steady state simulation 88 Evaporation Function is to concentrate solution What affects evaporation? Rate at which heat is transferred to the liquid Rate at which heat is transferred to the liquid Quantity of heat required to evaporate mass of water Quantity of heat required to evaporate mass of water Maximum allowable temperature of liquid Maximum allowable temperature of liquid Pressure which evaporation takes place Pressure which evaporation takes place

89 PIECENAMP Module 9 – Steady state simulation 89 Single Effect Vertical Evaporator Three functional sections Heat exchanger Heat exchanger Evaporation section Evaporation section liquid boils and evaporates liquid boils and evaporates Separator Separator vapor leaves liquid and passes off to other equipment vapor leaves liquid and passes off to other equipment Three sections contained in a vertical cylinder

90 PIECENAMP Module 9 – Steady state simulation 90 In the heat exchanger section (calandria), steam condenses in the outer jacketIn the heat exchanger section (calandria), steam condenses in the outer jacket Liquid being evaporated boils on inside of the tubes and in the space above the upper tube stackLiquid being evaporated boils on inside of the tubes and in the space above the upper tube stack As evaporation proceeds, the remaining liquors become more concentratedAs evaporation proceeds, the remaining liquors become more concentrated

91 PIECENAMP Module 9 – Steady state simulation 91 T f, x f, h f, ṁ f T L, x L, h L, ṁ L T s, H s, ṁ s T v, y v, H v, ṁ V T s, h s, ṁ s P = kPa U = J/m 2 s o C A = ? m 2 Condensate S Vapor V Concentrated liquid L Steam S Feed F Diagram of Single Effect Evaporator

92 PIECENAMP Module 9 – Steady state simulation 92 Material and Heat Balances ṁ F = ṁ L + ṁ V ṁ F x F = ṁ L x L + ṁ V y V q = UAΔT q = UAΔT ΔT = T s – T L Heat given off by vapor λ = H s – h s h F + H s = h L + H V + h s ṁ F h F + ṁ s H s = ṁ L h L + ṁ V H V + ṁ s h s h F + λ = h L + H V ṁ F h F + ṁ s λ = ṁ L h L + ṁ V H V q = (H s -h s ) = λ q = ṁ s (H s -h s ) = ṁ s λ λ – ideal heat transferred in evaporator ṁ s λ – ideal heat transferred in evaporator

93 PIECENAMP Module 9 – Steady state simulation 93 Finding the Latent Heat of Evaporation of Solution and the Enthalpies Using the temperature of the boiling solution T L, the latent heat of evaporation can be found;Using the temperature of the boiling solution T L, the latent heat of evaporation can be found; The heat capacities of the liquid feed (Cp F ) and product (Cp L ) are used to calculate the enthalpies of the solution.The heat capacities of the liquid feed (Cp F ) and product (Cp L ) are used to calculate the enthalpies of the solution.

94 PIECENAMP Module 9 – Steady state simulation 94 Property Effects on the Evaporator Feed TemperatureFeed Temperature –Large effect –Preheating can reduce heat transfer area requirements PressurePressure –Reduction 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 pressure steam is used.

95 PIECENAMP Module 9 – Steady state simulation 95 Boiling-Point Rise of Solutions Increase in boiling point over that of water is known as the boiling point elevation (BPE) of solutionIncrease in boiling point over that of water is known as the boiling point elevation (BPE) of solution BPE is found using Duhring’s RuleBPE is found using Duhring’s Rule –Boiling point of a given solution is a linear function of the boiling point of pure water at the same pressure

96 PIECENAMP Module 9 – Steady state simulation 96 Duhring lines (sodium chloride)

97 PIECENAMP Module 9 – Steady state simulation 97 Problem Statement (McCabe 16.1 modified) A single-effect evaporator is used to concentrate 9070 kg/h of a 5% solution of sodium chloride to 20% solids. The gauge pressure of the steam is 1.37 atm; the absolute pressure in the vapor space is 100 mm Hg. The overall heat transfer coefficient is estimated to be 1400 W/m 2 o C. The feed temperature is 0 o C. Calculate the amount of steam consumed, the economy, and required heating surface. First Example Excel Spreadsheet

98 PIECENAMP Module 9 – Steady state simulation Draw Diagram and Label Streams 9070 kg/h feed, 0 o C, 5% solids, h F T L, 20% solids, h L, ṁ L T s, H s, 1.37 atm gauge, ṁ s T v, 0% solids, H v, ṁ v T s, h s, ṁ s P= 100 mm Hg U = 1400 W/m 2 o C A=? Condensate S Vapor V Liquor L Steam S Feed F q=?

99 PIECENAMP Module 9 – Steady state simulation Perform Mass Balances ṁ F = ṁ L + ṁ V [9070 kg/h = ṁ L kg/h+ ṁ V kg/h] ṁ F x F = ṁ L x L + ṁ V y V (note that y v is zero because only vapor is present, no solids) [0.05 * 9070 kg/h = 0.2 * ṁ L kg/h + 0] Can solve for ṁ v and ṁ L ṁ V = kg/h, ṁ L = kg/h

100 PIECENAMP Module 9 – Steady state simulation Perform Heat Balances to find the Economy h F + H S = h L + H V + h S ṁ F h F + ṁ S H S = ṁ L h L + ṁ V H V + ṁ S h S h F + λ = h L + H V ṁ F h F + ṁ S λ = ṁ L h L + ṁ V H V q = (H S - h S ) = λ q = ṁ S (H S - h S ) = ṁ S λ The economy is defined as the mass of water evaporated per mass of steam supplied.

101 PIECENAMP Module 9 – Steady state simulation 101 Needed Data Boiling point of water at 100 mm Hg = 51 o C (from steam tables) Boiling point of water at 100 mm Hg = 51 o C (from steam tables) Boiling point of solution = 88 o C (from Duhring lines) Boiling point of solution = 88 o C (from Duhring lines) Boiling point elevation = 88 – 51 = 37 o C Boiling point elevation = 88 – 51 = 37 o C Enthalpy of vapor leaving evaporator (enthalpy of superheated vapor at 88 o C and 100 mm Hg [.133 bar]) = 2664 kJ/kg (F&R, p.650) – also called the latent heat of evaporation Enthalpy of vapor leaving evaporator (enthalpy of superheated vapor at 88 o C and 100 mm Hg [.133 bar]) = 2664 kJ/kg (F&R, p.650) – also called the latent heat of evaporation Heat of vaporization of steam (H s -h s = λ ) at 1.37 atm gauge [20 lb f /in 2 ] = 939 Btu/lb = 2182 kJ/kg (McCabe, App.7, p.1073) Heat of vaporization of steam (H s -h s = λ ) at 1.37 atm gauge [20 lb f /in 2 ] = 939 Btu/lb = 2182 kJ/kg (McCabe, App.7, p.1073)

102 PIECENAMP Module 9 – Steady state simulation 102 Finding the enthalpy of the feed 1.Find the heat capacity of the liquid feed feed is 5% sodium chloride, 95% water y NaCl =0.05 y water =0.95 C p,water =4.18 kJ/kg o C C p,water =4.18 kJ/kg o C C p,NaCl =0.85 kJ/kg o C C p,NaCl =0.85 kJ/kg o C (C p ) F =.05* *4.18 = 4.01 kJ/kg o C 2. Calculate Enthalpy (neglecting heats of dilution) h F = 4.01 kJ/kg o C (0 - 0 o C) = 0 kJ/kg

103 PIECENAMP Module 9 – Steady state simulation 103 Finding the enthalpy of the liquor 1.Find the heat capacity of the liquor feed is 20% sodium chloride, 80% water y NaCl =0.20 y water =0.80 C p,water =4.18 kJ/kg o C C p,NaCl =0.85 kJ/kg o C C p,L =.20* *4.18 = 3.51 kJ/kg o C 2. Calculate Enthalpy (neglecting heats of dilution) h L = 3.51 kJ/kg o C (88-0 o C) = 309 kJ/kg

104 PIECENAMP Module 9 – Steady state simulation 104 ṁ L h L + ṁ V H V - ṁ F h F ṁ S H S - ṁ S h S = ṁ S (H S - h S ) = ṁ S λ ṁ L h L + ṁ V H V - ṁ F h F = ṁ S H S - ṁ S h S = ṁ S (H S - h S ) = ṁ S λ λ = (H S -h S ) = 2182 kJ/kg ( kg/h * kJ/kg) + ( kg/h * 2664 kJ/kg) – (0) = ṁ S (H S -h S ) ( kg/h * kJ/kg) + ( kg/h * 2664 kJ/kg) – (0) = ṁ S (H S -h S ) q = ṁ S (2182 kJ/kg) q = ṁ S (2182 kJ/kg) Heat Balances ṁ s = kg/h q = kg/h*2182 kJ/kg = 1.88x10 7 kJ/h = W = 5.23 MW

105 PIECENAMP Module 9 – Steady state simulation 105 Find the Economy = ṁ V / ṁ S

106 PIECENAMP Module 9 – Steady state simulation Calculate Required Heating Surface Condensing temperature of steam (1.37 atm gauge = o C q = UAΔT A = q/UΔT

107 PIECENAMP Module 9 – Steady state simulation 107 Click on the Hyperlink and click on the “Final Solution” tab to see the final answer for the system. First Example Final Solution

108 PIECENAMP Module 9 – Steady state simulation 108 Second Example: Simulation of Cyclic Process (Felder and Rousseau, Example , pp ) (to be done in Excel)

109 PIECENAMP Module 9 – Steady state simulation 109 Problem Statement The 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. C 4 H 10 C 4 H 8 + H 2

110 PIECENAMP Module 9 – Steady state simulation 110 Diagram of Process Second Example Cyclic Process

111 PIECENAMP Module 9 – Steady state simulation 111 Notes A will denote isobutane, B denotes isobutene, C denotes hydrogenA will denote isobutane, B denotes isobutene, C denotes hydrogen All streams are gases, is the required rate of heat transfer to the reactor and is the net rate of heat transfer to the separation processAll streams are gases, is the required rate of heat transfer to the reactor and is the net rate of heat transfer to the separation process Specific enthalpies are for the gaseous species at the stream temperatures relative to 25 o C - Heats of formation are taken from Table B.1, and heat capacity formulas are taken from Table B.2 in Felder and Rousseau

112 PIECENAMP Module 9 – Steady state simulation Perform Degree of Freedom Analysis

113 PIECENAMP Module 9 – Steady state simulation 113 Review – Degrees of Freedom 1.Draw and completely label a flowchart 2.Count the unknown variables, then the independent equations relating them, 3.Subtract the number of equations from the number of variables. This gives n df, or the number of degrees of freedom in the process.

114 PIECENAMP Module 9 – Steady state simulation 114 Degree of Freedom Analysis If n df = 0 there are n independent equations in n unknowns and the problem can be solvedIf n df = 0 there are n independent equations in n unknowns and the problem can be solved If n df >0, there are more unknowns than independent equations relating them, and at least n df additional variable values must be specified.If n df >0, there are more unknowns than independent equations relating them, and at least n df additional variable values must be specified. If n df <0, there are more independent equations than unknowns. The flowchart is incompletely labeled or inconsistent and redundant relations exist.If n df <0, there are more independent equations than unknowns. The flowchart is incompletely labeled or inconsistent and redundant relations exist.

115 PIECENAMP Module 9 – Steady state simulation 115 Degree of Freedom Analysis – Mixing Point 4 unknowns (ṅ A1, ṅ B1, ṅ 4,T 1 ) - 3 balances (2 material balances, 1 energy balance) = 1 local degree of freedom

116 PIECENAMP Module 9 – Steady state simulation 116 Degree of Freedom Analysis – Reactor 7 unknowns (ṅ A1, ṅ B1, ṅ A2, ṅ B2, ṅ C2, T 1, ) -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

117 PIECENAMP Module 9 – Steady state simulation 117 Degree of Freedom Analysis – Separator 8 unknowns (ṅ A2, ṅ B2, ṅ C2, ṅ A3, ṅ B3, ṅ C3, ṅ 4, ) -4 balances (3 material balances, 1 energy balance) -1 additional relation (isobutane split) = 3 local degrees of freedom

118 PIECENAMP Module 9 – Steady state simulation 118 Net Degree of Freedom Analysis – Overall Process 7 local degrees of freedoms (1+3+3) -7 ties ( ṅ A1, ṅ B1, ṅ A2, ṅ B2, ṅ C2, ṅ 4, and T 1 were counted twice) = 0 net degrees of freedom The problem can be solved for all labeled variables.

119 PIECENAMP Module 9 – Steady state simulation Equation Based Solution Process

120 PIECENAMP Module 9 – Steady state simulation 120 Tearing the Cycle Can’t solve system in a unit-to-unit manner without trial and errorCan’t solve system in a unit-to-unit manner without trial and error “tear” between two units“tear” between two units -Purpose is to have the least number of variables that have to be determined by trial and error We tear between separation process and mixing unit We tear between separation process and mixing unit - Only have to determine ṅ 4 by trial and error.

121 PIECENAMP Module 9 – Steady state simulation 121 Solution Process Assume value of recycle flow rate (ṅ 4A = 100 mol/s)Assume value of recycle flow rate (ṅ 4A = 100 mol/s) Assume mixing point outlet temperature (T 1 = 50 o C)Assume mixing point outlet temperature (T 1 = 50 o C) Vary ṅ 4A until calculated recycle flow rate (ṅ 4C* ) equals assumed value in ṅ 4AVary ṅ 4A until calculated recycle flow rate (ṅ 4C* ) equals assumed value in ṅ 4A - Will be done by driving (ṅ 4A - ṅ 4C* ) using Goalseek Mixing point temperature (T 1 ) will be varied to determine the value that drives Δ Ḣ mix to zero (remember, the mixer is adiabatic)Mixing point temperature (T 1 ) will be varied to determine the value that drives Δ Ḣ mix to zero (remember, the mixer is adiabatic)

122 PIECENAMP Module 9 – Steady state simulation 122 Known Values X A = 0.35 (fractional conversion of A) 100 mol/s (basis of calculation) Feed temperature – 20 o C Reactor Effluent Temperature – 90 o C Product Stream Temperature – 30 o C Guess for recycle stream flow rate (ṅ A4 ) = 100 mol/s Mole fraction of A in recycle stream = 0.9 Mole fraction of B in recycle stream = 0.1 Temperature of recycle stream – 85 o C Initial guess for combined stream temperature – 50 o C

123 PIECENAMP Module 9 – Steady state simulation 123 Mass Balances (based on initial guesses) ṅ A1 = 100 mol/s feed + (100 mol/s recycle * 0.9 mol fraction = 190 mol/s) ṅ B1 = 100 mol/s recycle * 0.1 mol fraction = 10 mol/s ṅ A2 = ṅ A1 * (1-X A ) = mol/s ṅ B2 = ṅ B1 + (ṅ A1 *X A )= 76.5 mol/s ṅ C2 = ṅ A1 * X A = 66.5 mol/s ṅ A3 = 0.01* ṅ A2 = 1.24 mol/s ṅ C4 = (ṅ A2 - ṅ A3 )/0.9 mol fraction = mol/s ṅ B3 = ṅ B2 – (0.1 mol fraction * ṅ C4 )= 62.9 mol/s ṅ C3 = ṅ C2 = 66.5 mol/s

124 PIECENAMP Module 9 – Steady state simulation 124 Calculation of Specific Enthalpies (Tables B.1 and B.2, Felder and Rousseau) - (heats of formation) are located in Table B.1 of F&R A (isobutane [g]) = kJ/mol B (isobutene [g]) = 1.17 kJ/mol C (hydrogen [g])= 0 kJ/mol

125 PIECENAMP Module 9 – Steady state simulation 125 Calculation of Specific Enthalpies (Tables B.1 and B.2, Felder and Rousseau) - heat capacity of component i (kJ/mol o C) = a+ bT + cT -2 + dT -3, where T is temperature in o CChemicals A* 10 3 B* 10 5 C* 10 8 D* isobutane isobutene hydrogen

126 PIECENAMP Module 9 – Steady state simulation 126 Heat Balances (based on initial guesses) ΔḢ mix = ṅ A1 * Ĥ A1 + ṅ B1 *Ĥ B1 – 100 mol/s*Ĥ A0 - (ṅ A4 *0.9 mol A/mol *Ĥ A4 ) - (ṅ A4 *0.1 mol fraction*Ĥ B4 ) = kJ/mol = ṅ A2 *Ĥ A2 + ṅ B2 *Ĥ B2 + ṅ C2 *Ĥ C2 - ṅ A1 *Ĥ A1 – ṅ B1 *Ĥ B1 = ṅ A2 *Ĥ A2 + ṅ B2 *Ĥ B2 + ṅ C2 *Ĥ C2 - ṅ A1 *Ĥ A1 – ṅ B1 *Ĥ B1 = kJ/s = kJ/s = ṅ A3 *Ĥ A3 + ṅ B3 *Ĥ B3 + ṅ C3 *Ĥ C3 +(ṅ A4 *0.900 mol fraction*Ĥ A4 )+(ṅ B4 *0.100 mol fraction*Ĥ B4 ) – fraction*Ĥ A4 )+(ṅ B4 *0.100 mol fraction*Ĥ B4 ) – ṅ A2 *Ĥ A2 - ṅ B2 *Ĥ B2 - ṅ C2 *Ĥ C2 ṅ A2 *Ĥ A2 - ṅ B2 *Ĥ B2 - ṅ C2 *Ĥ C2 = kJ/s = kJ/s

127 PIECENAMP Module 9 – Steady state simulation 127 Click on the Hyperlink and click on the “Final Solution” tab to see the final answer for the system. Second Example Final Solution

128 PIECENAMP Module 9 – Steady state simulation 128 Tier III Open-ended problem  Approach to open-ended problem  Case Study.

129 PIECENAMP Module 9 – Steady state simulation 129 Statement of IntentStatement of Intent –Learn how to approach open-ended design problems –Solve a problem on your own

130 PIECENAMP Module 9 – Steady state simulation 130 How to approach open–ended problems State the problem clearly, including goals, constraints, and data requirements. State the problem clearly, including goals, 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 solutions. Develop a set of cases to simulate possible solutions. Perform the simulation and evaluate results against solution criteria. Perform the simulation and evaluate results against solution criteria. Evaluate solutions against environmental, safety and financial considerations. Evaluate solutions against environmental, safety and financial considerations.

131 PIECENAMP Module 9 – Steady state simulation 131 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 Slurry Scrubbing to Remove Sulfur Dioxide from Power Plant Flue Gases

132 PIECENAMP Module 9 – Steady state simulation 132 Protection of environment through process development is an important responsibility for chemical engineers Protection of environment through process development is an important responsibility for chemical engineers Coal is an abundant source of energy and source of raw materials in production Coal is an abundant source of energy and source of raw materials in production Predominately carbon, but contains other elements and hydrocarbon volatile matter Predominately carbon, but contains other elements and hydrocarbon volatile matter About Coal

133 PIECENAMP Module 9 – Steady state simulation 133 burned in many of world’s power plants to produce electricity burned in many of world’s power plants to produce electricity can produce a lot of pollution if gases not treated, like soot and ash can produce a lot of pollution if gases not treated, like soot and ash sulfur dioxide emissions regulated in the U.S. by the Environmental Protection Agency sulfur dioxide emissions regulated in the U.S. by the Environmental Protection Agency current regulations are no more than 520 ng SO 2 per joule of heating value of the fuel fed to the furnace current regulations are no more than 520 ng SO 2 per joule of heating value of the fuel fed to the furnace plants must remove 90% of SO 2 released when coal-burning plants must remove 90% of SO 2 released when coal-burning

134 PIECENAMP Module 9 – Steady state simulation 134 About Commercial Processing SO 2 removal is classified as regenerative or throwaway SO 2 removal is classified as regenerative or throwaway throwaway processing can be modified to produce gypsum throwaway processing can be modified to produce gypsum throwaway processing uses separating agent to remove SO 2 from stack gases followed by disposal of SO 2 innocuously (CaSO 3 * ½ H 2 O) and a slurried separating agent of calcium carbonate throwaway processing uses separating agent to remove SO 2 from stack gases followed by disposal of SO 2 innocuously (CaSO 3 * ½ H 2 O) and a slurried separating agent of calcium carbonate

135 PIECENAMP Module 9 – Steady state simulation 135 Process Description

136 PIECENAMP Module 9 – Steady state simulation 136 want to produce 500 MWe (megawatts of electricity) want to produce 500 MWe (megawatts of electricity) properties of coal given in table on next slide properties of coal given in table on next slide coal fed at 25 o C to furnace, burned with 15% excess air coal fed at 25 o C to furnace, burned with 15% excess air sulfur reacts to form SO 2 and negligible SO 3 sulfur reacts to form SO 2 and negligible SO 3 carbon, hydrogen oxidized completely to CO 2 and water carbon, hydrogen oxidized completely to CO 2 and water nitrogen in coal leaves furnace as N 2 nitrogen in coal leaves furnace as N 2 ash in coal leaves furnace in two streams ash in coal leaves furnace in two streams 80% leaves as fly ash in furnace flue gas 80% leaves as fly ash in furnace flue gas remainder as bottom ash at 900 o C remainder as bottom ash at 900 o C

137 PIECENAMP Module 9 – Steady state simulation 137 Component Dry Weight % Carbon75.2 Hydrogen5.0 Nitrogen1.6 Sulfur3.5 Oxygen7.5 Ash7.2 Moisture 4.8 kg/100 kg dry coal HHV KJ/kg dry coal C p dry coal kJ/(kg o C) C p ash KJ/(kg o C)

138 PIECENAMP Module 9 – Steady state simulation 138 combustion air brought into process at 25 o C, 50% RH combustion air brought into process at 25 o C, 50% RH air sent to heat exchanger, temperature increased to 315 o C air sent to heat exchanger, temperature increased to 315 o C air then fed to boiler, reacts with coal air then fed to boiler, reacts with coal flue gas leaves furnace at 330 o C, goes to electrostatic precipitator flue gas leaves furnace at 330 o C, goes to electrostatic precipitator 99.9% of particulate material removed 99.9% of particulate material removed goes to air preheater, exchanges heat with combustion air goes to air preheater, exchanges heat with combustion air leaves air preheater and split into two equal streams leaves air preheater and split into two equal streams each stream is feed to one of two identical scrubber trains each stream is feed to one of two identical scrubber trains trains sized to process 60% of flue gas trains sized to process 60% of flue gas

139 PIECENAMP Module 9 – Steady state simulation 139 divided gas stream fed to scrubber, contacts aqueous slurry of limestone, undergoes adiabatic cooling to 53 o C. divided gas stream fed to scrubber, contacts aqueous slurry of limestone, undergoes adiabatic cooling to 53 o C. sulfur dioxide absorbed in the slurry and reacts with the limestone: sulfur dioxide absorbed in the slurry and reacts with the limestone: CaCO 3 + SO 2 + ½ H 2 O CaSO 3 · ½ H 2 O + CO 2 CaCO 3 + SO 2 + ½ H 2 O CaSO 3 · ½ H 2 O + CO 2 solid/liquid slurry enters scrubber at 50 o C solid/liquid slurry enters scrubber at 50 o C liquid slurry flows at 15.2 kg liquid/kg inlet gas liquid slurry flows at 15.2 kg liquid/kg inlet gas solid to liquid ratio in the slurry is 1:9 by weight solid to liquid ratio in the slurry is 1:9 by weight liquid saturated with CaCO 3 and CaSO 3 liquid saturated with CaCO 3 and CaSO 3 cleaned flue gas cleaned flue gas meets EPA SO 2 requirements meets EPA SO 2 requirements leaves scrubber with saturated water at 53 o C leaves scrubber with saturated water at 53 o C

140 PIECENAMP Module 9 – Steady state simulation 140 cleaned flue gas contains CO 2 generated in scrubbing but no fly ash cleaned flue gas contains CO 2 generated in scrubbing but no fly ash cleaned flue gas reheated to 80 o C, blended with clean flue gas stream from other train, and sent to be released to atmosphere cleaned flue gas reheated to 80 o C, blended with clean flue gas stream from other train, and sent to be released to atmosphere solids in spent aqueous slurry solids in spent aqueous slurry unreacted CaCO 3, flyash from flue gas, inert materials, CaSO 3 unreacted CaCO 3, flyash from flue gas, inert materials, CaSO 3 liquid portion of slurry saturated with CaCO 3, CaSO 3 liquid portion of slurry saturated with CaCO 3, CaSO 3 specific gravity of specific gravity of spent slurry split in two spent slurry split in two one stream sent to a blending tank, mixed with freshly ground limestone, makeup water, and recycle stream one stream sent to a blending tank, mixed with freshly ground limestone, makeup water, and recycle stream fresh slurry stream from blending tank fed to top of scrubber fresh slurry stream from blending tank fed to top of scrubber

141 PIECENAMP Module 9 – Steady state simulation 141 second stream sent to filter where wet solids containing fly ash, inert materials, CaSO 3 and CaCO 3 are separated from filtrate second stream sent to filter where wet solids containing fly ash, inert materials, CaSO 3 and CaCO 3 are separated from filtrate filtrate saturated with CaSO 3, CaCO 3, and is the recycle stream fed to the blending tank filtrate saturated with CaSO 3, CaCO 3, and is the recycle stream fed to the blending tank wet solids contain 50.2% liquid that has similar composition to filtrate wet solids contain 50.2% liquid that has similar composition to filtrate fresh ground limestone fed to blending tank at rate of 5.2% excess of that is required to react with SO 2 absorbed from flue gas fresh ground limestone fed to blending tank at rate of 5.2% excess of that is required to react with SO 2 absorbed from flue gas limestone – 92.1% CaCO 3 and rest is insoluble inert material limestone – 92.1% CaCO 3 and rest is insoluble inert material

142 PIECENAMP Module 9 – Steady state simulation 142 Boiler generates steam at supercritical conditions Boiler generates steam at supercritical conditions 540 o C and 24.1 MPa absolute 540 o C and 24.1 MPa absolute mechanical work derived by expanding steam through a power- generating system of turbines mechanical work derived by expanding steam through a power- generating system of turbines low pressure steam extracted from power system contains 27.5% liquid water at 6.55 kPa absolute low pressure steam extracted from power system contains 27.5% liquid water at 6.55 kPa absolute heat removed from wet low pressure steam in a condenser by cooling water heat removed from wet low pressure steam in a condenser by cooling water cooling water enters condenser at 25 o C and leaves at 28 o C cooling water enters condenser at 25 o C and leaves at 28 o C saturated condensate at 38 o C is produced by condenser and pumped back to boiler saturated condensate at 38 o C is produced by condenser and pumped back to boiler

143 PIECENAMP Module 9 – Steady state simulation 143 Assume a basis of 100 kg dry coal/min fed to the furnace. 1.Construct a flowchart of the process and completely label the streams. Show the details of only one train in the scrubber operation. Do this in Excel. 2.Estimate the molar flow rate (kmol/min) of each element in the coal (other than those in the ash). 3.Determine the feed rate (kmol/min) of O 2 required for complete combustion of the coal.

144 PIECENAMP Module 9 – Steady state simulation If 15% excess oxygen is fed to combustion furnace, estimate the following: a.The oxygen and nitrogen feed rates (kmol/min) b.The mole fraction of water in the wet air, the average molecular weight, and the molar flow rate of water in the air stream (kmol/min) c.The air feed rate (kmol/min, m 3 /min) 5.Estimate flow rate (kmol/min, kg/min) of each component and composition (mole frac) of furnace flue gas (ignore fly ash). At what rate (kg/min) is fly ash removed from flue gas by the electrostatic precipitator?

145 PIECENAMP Module 9 – Steady state simulation If system is assumed to meet standard 90% SO 2 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.

146 PIECENAMP Module 9 – Steady state simulation (continued) e. What are flow rates (kg/min) of inerts, CaSO 3, CaCO 3, fly ash, and water, in the wet solids removed from the filter? f. Estimate rate (kg/min, L/min) at which filtrate is recycled to blending tank. At what rate (kg/min, L/min) is makeup water added to blending tank? 7. At what rate is heat removed from the furnace? Estimate the rate of steam generation in the power cycle, assuming all the heat removed from the furnace is used to make steam.

147 PIECENAMP Module 9 – Steady state simulation 147 References: Felder, R.F. and Rousseau, R.W. Elementary Principles of Chemical Processes, Third Edition. New York, John Wiley and Sons, Smith, J.C. and Harriott, Peter. Unit Operations of Chemical Engineering, Sixth Edition. Boston, McGraw Hill, Earle, R.L. Unit Operations in Food Processing, Second Edition. Thibault, Jules. Notes, CHE 4311: Unit Operations. University of Ottawa, August Genzer, Jan. Notes, CHE 225: Chemical Process Systems. North Carolina State University, August 2002.

148 PIECENAMP Module 9 – Steady state simulation 148 Source on pictures for slides 13, 41,Source on pictures for slides 13, 41,


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