Introduction to Steady State Simulation

Introduction to Steady State Simulation
NAMP Program for North American Mobility In Higher Education Module 9 Introduction to Steady State Simulation Welcome to module number 8: Introduction to Steady State Simulation. This module is a part of a series of modules sponsored by the North American Mobility Program (NAMP). Introducing Process integration for Environmental Control in Engineering Curricula PIECE Module 9 – Steady state simulation

PIECE Process integration for Environmental Control in Engineering Curricula Paprican École Polytechnique de Montréal Universidad Autónoma de San Luis Potosí University of Ottawa Universidad de Guanajuato North Carolina State University The North American Mobility Program is a joint education venture between 6 universities and 2 technical institutes across North America. Their locations are listed here. The universities are: (list universities and locations of countries) Instituto Mexicano del Petróleo Texas A&M University Program for North American Mobility in Higher Education NAMP Module 9 – Steady state simulation

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

Participating institutions
Project Summary 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 The objectives of this project are to: Create web-based modules to assist universities to address the introduction of process integration into engineering curricula. Make the modules widely available to universities and companies in all three countries. As mentioned earlier, there are six universities and two research institutes participating in the program. Each university has financially sponsored seven exchange students during the period of the grant. Partially funded by the countries’ governments, a student from one university is sent to another of his/her choice to develop a specific module specified by the lead professor of the program at that university. Module 9 – Steady state simulation

What is the structure of this module?
Structure of Module 9 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. Each module is divided into three tiers. Tier 1 is intended to provide background information that will provide students with a solid foundation of the material in the module. Tier 2 is intended to present examples that demonstrate the module’s concepts, which students are encouraged to follow along. The third module presents a design problem that students will work through using the various concepts studied in the module. The tiers are to be completed in that order, and students may be quizzed at various points in the first tier to measure their level of understanding of the background information Module 9 – Steady state simulation

What is the purpose of this module?
Purpose of Module 9 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. The intention of this module is to cover the basic aspects of steady state simulation and discuss the subject in a way that will be understandable to the student. This module is planned to be a prerequisite for other modules that will focus on the learning of steady state simulation. As a student, it is expected that you be familiar with mass and energy balances and have some knowledge of thermodynamics and transport processes. Module 9 – Steady state simulation

Tier I Background Information
Module 9 – Steady state simulation

Statement 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 In Tier 1, you will review basic chemical concepts, such as mass and energy balances, heat and mass transfer, thermodynamic laws, and analytical methodology; you will understand why steady-state simulation is important in industry today; learn how to develop a steady-state based model in general, and discuss problem solving techniques. Module 9 – Steady state simulation

Input + Generation – Output – Consumption = 0
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 time Accumulation term in mass balance set to zero Input + Generation – Output – Consumption = 0 In definition, chemical processes can be defined as steady state or transient, and either batch, continuous, or semibatch. We’ll talk about the latter three processes later. If the values of all the variables in a process (i.e. temperature, pressure, volume, flowrates, etc…) do not change with time , the process is operating at steady state. However, if any of the process variables change with time, transient or unsteady-state operation is said to occur. We can use steady-state prcoesses Module 9 – Steady state simulation

Three types of processes:
Batch Continuous Semi-batch A process where inputs and outputs flow continuously through duration of process. A process where a set amount of input enters a process, where it is removed from the process at a later time. As stated earlier, processes can be classified as steady-state or transient: batch, continuous, or semibatch. Let’s discuss the latter of these comparisons. In a batch process, a set amount of input (heat, mass) enters the process, where it is removed from the process at a later time. No mass crosses the system boundaries between the time the feed is charged and the time the product is removed. A continuous process is one in which inputs and outputs flow continuously throughout duration of process. A semibatch process can be a combination of both batch and continuous processes or neither of them. By their nature, batch and semibatch process are non steady state, so in this module, we’ll be looking at continuous processes. Neither batch or continuous, may be combination of both. In steady-state processes, we will be looking at continuous processes. Module 9 – Steady state simulation

Batch Example Ammonia is produced from nitrogen and hydrogen. At time t = t0, nitrogen and hydrogen are added to the reactor. No ammonia leaves the reactor between t = t0 and t = tf. At tf, nf moles of ammonia are released. H2 N2 NH3 Batch processing is commonly used with small quantities of a product that are to be produced on a single occasion. Here is an example of a batch process. We know that the reaction of hydrogen and nitrogen gas makes ammonia. At time t=t0, a specific amount of hydrogen and nitrogen is dumped into the reactor. No ammonia leaves the reactor between t0 and tf. At tf, nf moles of ammonia are released from the reactor. t = to t = tf Module 9 – Steady state simulation

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. Escaping gas from a pressurized gas tank is an example of a semibatch process. It is partially continuous because the gas continuously diffuses from the tank. It is partially batch because a specific amount of helium was inputted into the tank. There is no continuous input of helium entering the tank. Module 9 – Steady state simulation

50% molar CH3OH 50% molar H2O Mostly H2O Mostly CH3OH Continuous 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. A distillation column such as this one above provides a good example of a continuous process. A constant feed of methanol and water enters the column. Because methanol is the more volatile component (it has a lower boiling point than water) most of the methanol boils off and exits the column on the top. The water exits from the bottom of the column. Both the methanol and water continuously exit the column as the feed enters. As stated earlier, studies in steady state simulation will involve these types of processes. Module 9 – Steady state simulation

Classify the following processes as batch, continuous, or semibatch.
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. Module 9 – Steady state simulation

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: Carbon (C) Air (79% N2, 21% O2) It is important that when you are trying to model a continuous process, you need to develop a simple diagram to help you picture the process. You may remember the importance of block diagrams in solving material balances; they help to define the look of the process and you can indicate information about the process, such as flow rates, species compositions, temperatures, and pressures. Reactor Separator Module 9 – Steady state simulation

Degree Of Freedom Analysis
Analysis done to determine if there is enough information to solve a given problem. Draw and completely label a flowchart Count the unknown variables, then the independent equations relating them, Subtract the number of equations from the number of variables. This gives ndf, or the number of degrees of freedom in the process. Once you have defined the process and have drawn a diagram to depict the process pictorially, you need to see if the given information is enough to solve the problem. This is where a degree of freedom analysis is important in order to answer this very question. So how do we go about doing this? First, draw and label a flowchart (block diagram). Label all necessary variables (known and unknown). Second, count the number of unknown variables and the equations relating them. Sources of equations include material balances, energy balances, process specifications, physical properties and laws, physical constraints, and stoichometric relations. Then subtract the number of equations from the number of variables, which will get you the number of degrees of freedom in the process. Module 9 – Steady state simulation

Degree of Freedom Analysis
If ndf = 0 there are n independent equations in n unknowns and the problem can be solved If ndf >0, there are more unknowns than independent equations relating them, and at least ndf additional variable values must be specified. If ndf <0, there are more independent equations than unknowns. The flowchart is incompletely labeled or inconsistent and redundant relations exist. If the difference is equal to zero, then the problem can be solved. If the difference is greater than zero, than there are more unknowns than equations relating them, and additional variable values must be specified. If the difference is less than zero, there are more independent equations than unknowns. The problem is incompletely labeled or there are redundant and possibly inconsistent relations that exist. Either way, solving the problem would be a waste of time if the difference is not zero. Module 9 – Steady state simulation

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 Let’s now look at the principles that balances are based on. You most likely are familiar with the law of conservation of mass, which states that mass can never be created or destroyed. The general material balance is shown above, with the rate of mass that accumulates in the system equal to the rate mass enters through system, minus the rate of mass leaving the system. We mark out the production and consumed terms, as mass cannot be created or destroyed. Module 9 – Steady state simulation

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 Note that because mass flow continuously throughout a steady state process, there is no mass accumulation. Accumulation is a time dependent term, and since steady-state processes are independent of time, we can eliminate the accumulation term. Therefore, the mass balance simply is the mass coming in equals the mass coming out. The PDF file attached describes how to go about solving material balances. Study it if you need some review. Module 9 – Steady state simulation

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 Now let’s look at the flow of energy through a steady-state process.. A system can be defined as open or closed according to whether or not mass crosses the system boundary during the period of time covered by the energy balance. Batch processes are by definition closed, and continuous and semibatch systems are by definition open. The first law of thermodynamics for a steady state open system has the form (input = output). Input signifies the total rate of transport of kinetic, potential, and internal energy by all output streams plus the energy transferred as heat, and the output is the total rate of energy transport by the output streams plus the energy transferred out as work. The balance above states that the net rate at which energy is transferred to a system as heat and/or shaft work equals the difference between the rates at which the quantity (enthalpy + kinetic energy + potential energy) is transported in and out of the system. Module 9 – Steady state simulation

Quiz #2 Explain the Degree of Freedom analysis.
What term goes to zero in a steady-state process? Is a continuous process closed or open? How about a batch process? Module 9 – Steady state simulation

Heat Transfer Also classified as energy transfer
Three types of heat transfer modes: Conduction Convection Radiation Let’s discuss transport properties. Many operations carried out in chemical engineering involve either the production or absorption of energy in the form of heat. Heat transfer occurs when two objects at different temperatures are brought into contact. Heat will flow from the object at the higher temperature to the object at the lower temperature. The net flow is always in the direction of the temperature decrease. Let’s discuss three types of heat transfer: conduction, convection and radiation. Module 9 – Steady state simulation

Conduction Accomplished in two ways Molecular interaction
“Free electrons” Conduction equation called Fourier’s Law qy = heat transfer area in y direction (W) A = area normal to direction flow (m2) dT/dy = temperature gradient (oC/m) k = thermal conductivity (W/moC) Heat can flow unaccompanied by any motion of matter if a temperature gradient exists in some continuous surface. This kind of heat transfer is termed conduction. Fourier’s Law for one-dimensional flow, which is listed on the slide, states that the heat flux (the left side of the equation) is proportional to the temperature gradient and opposite in sign to it. In gases, conduction results from random interaction between molecules, while in metals, conduction results from the motion of free electrons. Module 9 – Steady state simulation

Convection Accomplished in two ways Natural convection
Forced convection Convection equation called Newton’s Law qy = rate of convective heat transfer (W) A = area normal to direction flow (m2) ΔT = temperature gradient (oC) h = convective heat transfer coefficient (W/m2 oC) Convection can refer to the flow of heat associated with the movement of a fluid. Some examples include hot air from a furnace entering a room, or the transfer of heat from a hot surface to a flowing liquid. The convective flux is proportional to the difference between the surface temperature and the temperature of the fluid. A modified form of this relationship is shown in the equation above, called Newton’s Law of Cooling. When currents in a fluid result from buoyant forces created by a density gradient, and the density gradient is a product of temperature gradients in the fluid, the action is called natural convection. When the currents are due to some action of a mechanical device like a pump, the flow is independent of a density gradient and is termed forced convection. Module 9 – Steady state simulation

Radiation (Thermal) Exhibits same optical properties as optical light
May be absorbed, reflected, or transmitted Total radiation for unit area of opaque body of area A1, emissivity ε1, and absolute temperature T1, and a universal constant σ Radiation is a term given to the transfer of energy through space by electromagnetic waves. When radiation passes through matter, it will be absorbed by the matter, reflected off, or transmitted to another location. Only the absorbed energy appears as heat, and this transformation is quantitative. The equation above describes the total radiation for a unity area of an opaque body. Module 9 – Steady state simulation

Radiation Between Surfaces
Simplest type occurs where each surface can see only the other and where both surfaces are black Energy emitted by first plane is σT14; the second plane emits σT24 if T1 > T2, then net loss energy per unit area by first plane and net gain by second are σT14- σT24, or σ(T14-T24) The simplest type of radiation occurs between two surfaces where each surface can see only the other, for example, when both surfaces are parallel planes and both surfaces are black. Each plane emits different energy, and if one surface is hotter than the other, then the cold surface will gain as much energy as the hot surface loses. However, note that this type of radiation transfer is only valid in ideal cases. Because the two surfaces most likely have different reflective properties, emissivities must be considered. Cold surface Note this is only in ideal cases: no surface is exactly black, and emissivities must be considered Hot surface Module 9 – Steady state simulation

Mass Transfer The transport of one constituent from a region of higher concentration to a region of lower concentration Molecular mass transfer Random molecular motion in quiescent fluid Convective mass transfer From a surface into a moving fluid or vice-versa Mass transfer is the movement of mass from a region of higher concentration to a region of lower concentration. The driving force behind mass transfer is a concentration difference or a difference in activity. Such techniques that employ mass transfer include distillation, liquid extraction, and gas absorption. In general, there are two types of mass transfer. There is molecular mass transfer, in which mass travels by random molecular motion in a quiescent fluid. And there is convective mass transfer, in which mass moves from a surface into a moving fluid. Module 9 – Steady state simulation

Fick rate equation (restricted to isothermal/isobaric systems)
Flux = - (overall density)*(diffusion coefficient)*(concentration gradient) Fick rate equation (restricted to isothermal/isobaric systems) de Groot equation is more general Units: JA – mol A/m2s CA- mol A/m3 DAB- m2/s A mass flux is a product of the density of the fluid, the diffusion coefficient, and the change in concentration within the fluid. Fick’s First Law provides the basic relations for molecular diffusion. Note that the Fick equation is only restricted to systems at constant temperature (isothermal) or isobaric (constant pressure). The de Groot equation applies to more general systems. Dab is the diffusion coefficient. This value is dependent on temperature, pressure and the composition of the fluid. For binary gas mixtures at low pressures, it is practically independent of composition. However, for most systems, the coefficient is proportional to the inverse of pressure and directly to temperature. Module 9 – Steady state simulation

Molar flux of species A in binary system (A + B)
c = concentration DAB = diffusivity of species A in B = change of molar species in y with respect to a specified direction NA, NB = molar fluxes of components If we have a binary system, and we want to explain the molar flux of one species in this system, we can do that using the equation above. The molar flux of species A in a binary system is explained by the product of the total concentration of the system, the diffusion coefficient, the concentration gradient, and the sum of the concentration of A plus the sum of the individual molar fluxes. Module 9 – Steady state simulation

Quiz #3 What are two ways in which conduction occurs?
Define natural and forced convection. What is the restriction to the use of Fick’s Law? Module 9 – Steady state simulation

Modeling What is Modeling? Steady-State vs. Dynamic Modeling
Empirical vs. Mechanistic Modeling Derivation of a Steady State Model Modeling and Process Design Implications Now that we skimmed over some basic concepts, let’s begin our studies into modeling. First, we’ll provide a broad definition of what modeling is. We’ll compare steady-state versus dynamic modeling and empirical versus mechanistic modeling. We’ll then discuss the steps to derive a steady-state model, and then discuss the implications that modeling and process design have in the industry. Module 9 – Steady state simulation

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 Selection Comparison of Different Process Configurations Evaluation of Process Performance Against Limitations Optimization So what is a model? The definition of a model is simply a pictorial depiction of a process operation that is used to design, change, improve, or control a process. Models are employed when designing, sizing, and selecting equipment for a plant, comparison of difference configurations for a process, evaluating process performance against process limitations, and optimizing a process itself to allow it to perform at peak efficiency. There are many other uses for models in addition to these ones as well. Module 9 – Steady state simulation

Models vary by: Phenomena represented Energy, phase changes
Level of details Assumptions (perfect mixing, heat loss) Inputs required Functions performed (satisfaction of constraints, optimization) Outputs generated No one model is going to be exactly the same as another model. Of course, those distinctions will depend on the type of process that you are trying to model and desired process parameters. Models can vary by the phenomena that is depicted,. For example, one model could describe the compression of synthesis gas (hydrogen/carbon monoxide mixture) in a compressor, but another model in the same plant would describe the heating of the gas to a higher temperature for later use The level of detail may be different as well. In some models, flow rates may be the only significant parameter to note, while another model, temperature and pressure are significant as well. You also have to take assumptions into account when comparing models. Models also vary by the inputs required to the process. The feed into one model may be different from the feed into another model. In addition, the functions performed within the model’s calculation window vary. For example, you may only need to do a mass balance in one model to make sure the amount of synthesis gas flowing through the plant is sufficient, while in another model, you would have to employ an energy balance to verify that the temperature of the compressed synthesis gas is sufficient for your process. Vice versa, the outputs generated vary as well. One output could be condensed water from a heat exchanger; another model may have compressed gas as an output. You have to take these changes into account as your modeling a system progresses. Module 9 – Steady state simulation

Requirements of a good model
Accuracy: the model should be close to the target description. 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. Computational efficiency: models should be calculable using reasonable amounts of time and computing resources. Anybody can develop a model, but if that model is not thorough, it will have little use to the development of your plant. The development of strong models is a crucial skill in your work as an engineer. Good models must be accurate. It is important to have your model looked over and you be fully confident that the numbers you calculated are correct. The model should be close to describing the target goals of your process. Models must also be valid. They have to be developed off sound chemical engineering and mathematic principles and have the ability to be easily justified. Thirdly, models should be easy to understand and lastly, they should calculable using a reasonable amount of time and computing resources. Module 9 – Steady state simulation

Level of Knowledge-based Modeling
Time-based Modeling Steady State Dynamic Model The concept of modeling can be divided into two primary divisions; time-based and level-of-knowledge based. Time-based modeling can be separated into steady-state modeling (constant with time) and dynamic modeling (not constant of time). Level-of-knowledge based modeling can be divided into empirical modeling, mechanistic modeling, and hybrid (a combination of both). Empirical Mechanistic Hybrid Level of Knowledge-based Modeling Module 9 – Steady state simulation

Dynamic 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 In steady state modeling, the process is usually studied at equilibrium conditions, since time is not a factor in the process. All unit operations will be studied under equilibrium conditions as well. Because we are at equilibrium, the factor of equipment sizing is not a very important consideration. Also, knowing that we are at equilibrium does not require as much process information as a dynamic-influenced model. Dynamic models are dependent on time, so you cannot safely assume that unit operations function at equilibrium conditions. This uncertainty of equilibrium requires that in the design of a process, equipment sizes need to be taken into account. Naturally, more data is needed to develop an accurate model of the dynamic system. Module 9 – Steady state simulation

Continuous Stirred Tank Reactor (CSTR)
Steady State Example Continuous Stirred Tank Reactor (CSTR) Concentration profile at one point in reactor does not change with time Continuous stirred tank reactor (CSTR) are examples of steady state model. In most undergraduate reaction and kinetics classes, problems will be based on reactors being of this nature. In CSTRs, it is assumed that the concentration profile at one point in the reactor does not change with time. Ca t Module 9 – Steady state simulation

Concentration profile at one point in reactor does change with time
Dynamic Example Batch Reactor Concentration profile at one point in reactor does change with time ca In contrast, a batch reactor is a good example of a dynamic model. Recall earlier in the presentation about the batch example. If you discharge nitrogen and hydrogen into a reactor and let it react, ammonia will be formed. Now pick any point within the reactor and monitor the concentration of any of the three components. Because you are not putting a continuous amount of feed into the reactor, the concentration profile at that specified point is going to change over time – decreasing for hydrogen and nitrogen, and increasing for ammonia – until equilibrium is reached. t Module 9 – Steady state simulation

Empirical Modeling Definition: Key Notes
a model that is based on data whether it has been collected from a process or some other source. Key Notes Derived from observation Often simple May or may not have theoretical foundation Valid only within range of observation Let’s now discuss empirical modeling. A empirical model is completely based on collected data. That data could have come from another process. A model of this type is derived from observation, meaning the data was collected from observation of the process itself. These models are often simple to depict, and may or may not have foundations based on chemical engineering theory. Most importantly, these models are only valid within the range of observation – you cannot derive an accurate empirical model with data that has not been observed in the process. Module 9 – Steady state simulation

Procedure – Empirical Modeling
Obtain data from process you wish to model. Temperature, pressure, flow, etc… Perform appropriate statistical analysis and develop accurate correlations from data. 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. Check for correctness in your analysis and equations, and determine if the model is satisfactory. Statistical Analysis with Excel Developing an empirical model is fairly straightforward. First observe and collect the data that you wish model. This may include temperatures, pressures, and flows. Next, perform the appropriate statistical analysis to see if the data is relevant and develop accurate correlations from data. Once the relevant data has been sorted, develop mathematical equations that will accurately represent the data and the correlations, and determine which equations are useful in the development of the model. Finally, check for accuracy in your equations and overall analysis, and determine if your model is satisfactory. The hyperlink shows some Excel formulas of basic statistical functions that can be used. Module 9 – Steady state simulation

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 Th1, leaves at Th2 and a cool stream coming in at Tc1 and leaving at Tc2. If the physical properties of the fluids are the same, then the temperature difference describes the amount of energy transferred. Let’s take an example of an ideal heat exchanger to describe empirical modeling. Heat exchangers are used in process industries to transfer energy (in the form of heat) from a hotter stream to a cooler stream. Remember the theory of heat transfer. Let’s say we have a hot stream of fluid at temperature Th1 going into the shell inlet and comes out cooler at Th2 from the shell outlet, while a cool stream comes into the tube inlet and Tc1 and comes out hotter at Tc2 from the tube outlet. Note that we don’t know Tc2; that is the value we are trying to find. If the physical properties of the fluid are the same (note that the fluids are not the same, but they have similar physical properties), then the temperature difference describes the amount of energy transferred. So we’ve observed and collected our data (Th1, Th2, Tc1). Module 9 – Steady state simulation

We do not know Tc2, but we can take various measurements of Th1, Th2 and Tc1 to find Tc2 . Using certain statistical procedures, it can be determined that Tc2 is related to the other three temperatures by this equation: Tc2 = Tc1 + a(Th2-Th1) 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. As stated earlier, we don’t know Tc2, but we can take multiple measurements of Th1, Th2, and Tc1. If we employ statistical analysis (we won’t get into how to do this, it’s not relevant to this presentation), it can be determined that Tc2 is related to the other three temperatures by the equation above. (With some rearranging, you can see that the energy transferred to the cold stream is directly proportional to the energy transferred from the hot stream). However, we don’t know the formula for the proportionality constant a – we can determine a value for it, but for only the specific fluids tested. If we knew the heat capacities and mass flow rates of the fluids through the heat exchanger, we can use them to determine a model that will relate the four temperatures for any combination of two fluids. This type of modeling is called mechanistic. Module 9 – Steady state simulation

Mechanistic Modeling Definition Key Notes
a model that is derived from fundamental physical laws or basic principles Key Notes Model construction – time-consuming and costly Most reliable, but often not enough data available Mechanistic models are ones that are derived from fundamental physical laws (thermodynamics, conservation) or basic principles. These models can be time-consuming and costly to construct, because of the amount of known data needed to derive a model. Although a mechanistic model is more reliable than an empirical model, they are hard to construct because sometimes, there is simply not enough information available. Module 9 – Steady state simulation

Procedure – Mechanistic Modeling
Know physical and chemical properties of the process. Determine the appropriate process model using mass and/or heat balance. Determine appropriate model run conditions and parameters Complete runs and use output data to compare against the predicted model results 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. In developing a mechanistic model, you first have to know the chemical and physical properties of the constituents within the process. (example: heat capacities of fluids, specific enthalpies, reaction stoichometry). Next, develop the appropriate model by employing balances, such as heat and/or mass. Next, determine appropriate run parameters and limitations that the model must use. Once you have developed the preliminary model and the run conditions, conduct the runs and use the output to compare against the predicted model results. If there is a discrepancy between the predicted model results and the actual results, re-examine the assumptions and the collected data and repeat steps 2-4. Module 9 – Steady state simulation

Let us go back to the heat exchanger
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: Now let us go back to the heat exchanger. Remember that we were trying to determine a relationship between Tc2 and the other three temperature readings. We had determined a relationship between the four temperatures that involved the empiricism a. However, we noted that we could find a specific value of a only for a specific pair of fluids. In our mechanistic modeling technique, we can determine that this empiricism a is a function of the mass flow and heat capacity of any two fluids. Knowing this allows us to model a heat exchanger for any two fluids. The model is determined to be the following equation (with some rearrangement). Module 9 – Steady state simulation

Steady state model derivation
1. Define Goals. a) Specific design decisions. b) Numerical values. c) Functional relationships. d) Required accuracy. We’ve discussed the various forms of modeling – now let’s employ those techniques to develop a steady state model. First off, you should define your goals on what you want your model to look like. Take into account specific design decisions you make, the numerical values of your data, relationships and correlations between your data, and the required accuracy that the model needs to employ. Secondly, collect and prepare the information for the model. Sketch the process and identify the variables needed to accurately explain the system. Also, do not forget to state assumptions and collect relevant data that is known in order to simplify the model. 2. Prepare information. a) Sketch process. b) Identify variables of interest. c) State assumptions and data. Module 9 – Steady state simulation

3. Formulate model a) Conservation balances. b) Constitutive equations. c) Rationalize (combine equations and collect terms). d) Check degrees of freedom. Next, begin to formulate your model. Employ conservation balances (mass, energy), constitutive equations (vapor-liquid equilibrium, etc…) to develop a preliminary description. Then perform the degree of freedom analysis to see if the problem is solvable. If it is, determine the solution to your model. You may have to employ analytical techniques, numerical techniques, or a combination of both to solve the model. 4. Determine solution a) Analytical b) Numerical Module 9 – Steady state simulation

5. Analyze results a) Check results for correctness Accuracy of numerical/analytical methods Plot solution Relate results to data and assumptions Answer “what if questions” Compare with experimental results Once a solution has been found, check it for accuracy. Check your calculations – did you make any mistakes?. Plot the solution if possible and see if the resulting curve(s) agree with the behavior of the model. Relate the resulting solution to the predicted model solution. Do they match? Also ask and answer what if questions. What if the mass flow was increased two fold? What would happen? What if we changed the fluid? How would the transfer properties of the model change? Module 9 – Steady state simulation

Process insights resulting from modeling
Identification: If we know the input (I) and output (O) parameters, we can determine the structure (R) of the model. What can we learn from the process about modeling? One, we can determine the structure of the model if we are given the experimental input and output of the model. This is called identification, because we are trying to identify the behavior of the model. I O R? Module 9 – Steady state simulation

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. We can simulate what the output would be if we know the structure of the model and the inputs entering the model. If we want to change the output, we can modify the input and (less likely) the structure of the model, since we have control of those two. I O? R Module 9 – Steady state simulation

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. If we know the structure of the model and want to obtain a desired output value, we can manipulate the input to optimize the process. This is called control or optimization. In your chemical engineering education, you will take a course completely focusing on this topic. I? O R Module 9 – Steady state simulation

Quiz #4 What are some uses of modeling?
Name and explain three requirements of a good model. What distinguishes a steady-state model and a dynamic model? Review the procedures for developing a mechanistic and empirical model. What are some differences between the two procedures? Discuss the control/optimization insight of modeling. Module 9 – Steady state simulation

Solving Problems Analytical Methods Process Design Methods
Spreadsheets Simulation Software Solution Determination Let’s talk about problem solving now. We’ve talked about various chemical engineering principles and deriving models. Now how do we go about solving those models? We’ll discuss a couple of analytical methods you may be familiar with, discuss the design of a process from a model, talk about different methods to solve problems (spreadsheets and simulation packages) and ways to determine the accurate solution of a problem. Module 9 – Steady state simulation

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 error Precision error Systematic error Calculation error Error propagation Curve Fitting Example You are probably most likely familiar with fitting curves on a plot. You employ this method when you generate a set of data that you can represent on a plot. You may collect data that follows a seemingly random pattern, but by fitting a curve, you can get a general idea of the behavior of the data; therefore, you can get a general idea of the behavior of the process. You 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 uniform. However, you can encounter errors, including measurement error, precision error, systematic error, calculation errors, and propagating errors from a combination of the above given. An example of curve fitting techniques in Excel is given in the hyperlink. Module 9 – Steady state simulation

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) Least Squares Method The Least squares method allows you to plot an accurate curve by minimizing the sum of the squares of the residuals of the data. A spreadsheet depicting the least squares method in Excel is accessible from the hyperlink. Module 9 – Steady state simulation

Process Design Let’s focus on how we design processes.

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: Basic chemical products. Industrial products. Consumer products. So how is a product designed? Most products begin with the identification and creation of a potential opportunity to satisfy a societal need and generate profit. There are three classifications that products can be roughly identified in – basic chemical products (such as stock chemicals, like ethanol and propanediol), which are converted into industrial products (such as polymer and plastics), which are processed into consumer products (such as carpet and plastic containers). Module 9 – Steady state simulation

Process design Manufacturing Process Natural Resources Basic chemical
Products Basic Chemical Industrial Industrial Products Consumer Natural resources/raw materials are converted into basic chemical products through a manufacturing process. Another process is employed to convert these basic chemical products to industrial products that the company can sell. Then these industrial products are converted into products the consumer can use through another manufacturing process. Module 9 – Steady state simulation

Motivation for Process Design
Desires of customers for chemicals with improved properties for many applications. Discovery of a new inexpensive source of a raw material with comparable physical and chemical properties to the old source. New markets are discovered. Why are products designed? For one thing, customers will have ever changing desires for new chemicals that have improved properties for many applications. For example, in the early days of the computer, customers wanted efficient chips to power the large machines. Copper was the metal of choice. As the technology progressed and computers got smaller, the benefits of silicon became present, and customers desired to have smaller silicon-based chips. Another motivation is that a new inexpensive source of raw material with new physical and chemical properties is discovered. For example, with petroleum becoming increasingly expensive, some car companies are looking into biological chemicals, such as ethanol, to power cars. Also, the discovery of a new market will lead to a new process. For example, soft drinks were once bottled in glass. But with an increasing need for a more inexpensive product to produce, plastic was introduced as a means of bottling soft drinks Module 9 – Steady state simulation

Steps in a Process Design
Process Design – Questions to Answer Is the chemical structure known? Is a process required to produce the chemicals? Is the gross profit favorable? Is the process still promising after further elaboration? Is the process and/or product feasible? Now let’s talk about the steps in designing a process. First, we must ask ourselves questions. Do we know the chemical structure of the product we are about to make? Do we need to process the chemicals or can they be found as raw materials? Can we make a profit out of a new process? How about long term implications - if the product is to be made for many years, will the process be of a strong enough design to efficiently function for many years and still create a profit? Can we even make the process and resulting product feasibly? Module 9 – Steady state simulation

Steps in a Process Design
Process Design – Steps Develop objective(s). Find inputs that have the desired properties and performance. Create process. Develop a base case for which to conduct initial testing on process. (does it stay stable at steady state?) Improve/maintain process Once we have answered these questions (and concluded that the process and resulting products are feasible to make), we begin designing the process. First, decide the objectives of the process. What is going to be the resulting product and how efficiently should we make it? What safety protocols must be implemented in the process? How much of the product will we want to make in a fiscal year? How much do we want to spend on cost, labor, raw materials? Once you’ve answered these questions and developed other relevant objectives for your process, decide what inputs you want to enter your process. Whether it may be raw materials from natural resources, chemical/chemical mixtures, or industrial products to make consumer products, you have to take into account cost, safety, and environmental effects. Develop your process once these inputs have been decided. Monitor initial outputs for a certain amount of time to develop a base case for which to conduct initial testing on process. Do you find that the output composition is what it is expected to be? Do you find that the temperature of the product is too high? Is the process showing signs of instability and not behaving like a steady-state system as you expected? Finally, after running tests from your base case, continue to add improvements and perform maintenance on your process. The earlier and more timely that you do this, the better chance that the process will function according to its parameters for a long time. Module 9 – Steady state simulation

Stability of the process
When a process is disturbed from an initial steady state, it will generally respond in one of 3 ways. Proceed to a steady state and remain there. Stability was mentioned in the last slide as a way to test the functionality of your process. Let’s talk about stability in a little more detail. It is important to note that even though you may design a process to mimic a steady-state system, it will be disturbed. Disturbances can occur from many causes, such as a spike in temperatures or flows, a malfunctioning unit operation downstream in the process, changes in the weather, etc… When this disturbance occurs, the process will respond in three ways. 1. The process will return back to steady-state operation and remain there, absorbing the disturbance quickly. Module 9 – Steady state simulation

Fail to attain to a steady state condition because its output grows indefinitely. The system is unstable. 2. Fail to attain steady state because the output increase indefinitely. This causes the process to become unstable and could lead to disastrous results if the output is not controlled immediately. Module 9 – Steady state simulation

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. 3. The process will fail to attain a steady-state condition because the output of the process oscillates indefinitely. Because attaining ideal steady-state behavior is all but impossible in reality, most processes will behave like this. Therefore, engineers will design processes to output between two limits. This is called the limit of stability. The output can fluctuate within the limit interval without causing any alarm. Module 9 – Steady state simulation

Quiz #5 What are some errors that may arise when attempting to fit a curve? What are the three products developed from process design? Provide an example of each product. What happens to an unstable system over time? Module 9 – Steady state simulation

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. You are familiar with the use of spreadsheets. Almost everyday, engineers interact with spreadsheet programs to store and sort data, plot graphs, and in a limited manner, perform simulations. Most people use Microsoft Excel as their spreadsheet program, although there are other programs available. Spreadsheets are used to store and calculate information in a structured array of data cells. By defining relationships between information in cells, a user can observe the effects of data changes on other data in parts of the spreadsheet. Spreadsheets are beneficial in that they provide an easy, efficient method for solving sets of equations and data that may be too difficult to solve simply by the use of a hand calculator. Module 9 – Steady state simulation

Columns are designated by letters, rows by numbers
Here is an example of a spreadsheet. Columns are designated by letters, rows by numbers. Module 9 – Steady state simulation

Goalseek Under Tools Menu
want to know input value formula needs to determine result Excel varies value in cell specified until dependent formula returns value you want Goalseek is part of Excel’s analysis tool. When you know the desired result of a single formula, but not the input value the formula needs to determine the result, you can use the Goal Seek feature available by clicking Tools, then Goalseek. When goal seeking, (which is a method to find a specific value for a cell by adjusting the value of one other cell), Excel varies the value in a cell that you specify until a formula that's dependent on that cell returns the result you want. (From Microsoft Office Online) Module 9 – Steady state simulation

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. Excel can handle large amounts of data but there is a point where Excel may have difficulty in solving a system of equations. Spreadsheets do have drawbacks. You have the ability to enter the equations yourself in spreadsheet cells to determine formula values. However, sometimes you can make a mistake in an equation that can mess up your entire calculation. If you don’t catch this mistake early, it can be difficult to find the mistake, especially if you are performing multiple calculations. Also, spreadsheet programs like Excel can handle large amounts of data but there is a point where Excel may have difficulty solving a system of equations. This is where simulation programs are helpful. Module 9 – Steady state simulation

Simulation Predicts behavior of a process by solving mathematical relationships that describe the behavior of the process components. Involves performance of experiments with a process model Simulation programs predict the behavior of a process by solving the mathematical relationships that describe the behavior of the process and its components. To do this, experiments are performed with a process model to determine the predicted output and compared to your desired output. Module 9 – Steady state simulation

Simulation Software – Why use it?
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 So why should we use simulation? Simulation software provides an economical way for engineers to construct/modify a process before doing a test in reality. Engineers can use the software to determine optimum operation conditions (for example, we can quantify how much equipment and sizing needed and raw materials required) determine process problems (for example, accurate changes in process can be made without sacrificing safety or money), determine the composition of streams, and simplify the design of complex unit operations. Module 9 – Steady state simulation

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 Simulation software allows for many things. 1. We can manipulate and compare previous data as well as for research. 2. We can manipulate the process to produce a certain output, and modify the process until that output is achieved. 3. We can easily explain complex process much better than what spreadsheets could do. 4. We can easily modify the process conditions and see how the output is changed through equipment behavior. Module 9 – Steady state simulation

Simulation 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 Even with simulation software, we must consider a few things. Many programs have built-in assumptions integrated into their databases – for example, the molecular weights of triglycerides – and since the assumptions may be incorrect, you have to take those into account when designing your process. Human error is introduced when mistakes are made in calculations, so it is safe to do mass balances over the process for checking purposes Also take into account the number of variables involved,the physical and chemical properties of the streams ,and the size of the process being simulated. Module 9 – Steady state simulation

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. Fresh feed Heater Reactor Flash Distillation Product Now that we have discussed what simulation is, let’s talk about some diagrams you’ll see in simulation programs. Remember when we talked about block diagrams in solving material balances earlier in the presentation? Well, what you see on this slide is also a block diagram, but developed in a simulation program. This type of process flowsheet is defined as a collections of icons to represent multiple process units and arrows to represent the flow of materials through the units. So in essence, this is a block diagram, just built on the simulation package. And you can do much more on developing this block diagram. steam Module 9 – Steady state simulation

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. When you add unit operations into your process and the necessary data for your inputs, the simulation package will likely calculate the information around each operation one at a time. In the overall process, the calculation order will most likely progress from beginning to end, depending on the specifications of the process itself. 1 2 3 4 Module 9 – Steady state simulation

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. When a simulation is attempting to calculate process parameters, there may be times where there is a missing parameter that prevents the simulation from completely simulating the process. This is where information recycle loops come in. If there are unknown variables which prevent the system from being solved completely, the simulation will return back to the beginning of the process to attempt to calculate the missing variables. For this situation, a solution technique is needed to solve the equations for all the units in the recycle loop. 1 2 3 4 For these processes, a solution technique is needed to solve the equations for all the units in the recycle loop. Module 9 – Steady state simulation

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 This solution technique is called iteration. In iteration, an initial guess is taken at the input and a preliminary solution is determined for the missing variables. The guess is taken back to the beginning of the calculation after being compared to the initial output, and the simulation makes a more educated guess and the system is solved based on the initial solution. The iteration continues until solution converges to a certain value that is in agreement with the output. Module 9 – Steady state simulation

Convergence Is the process to compare the guessed value with the computed value until a value is found within the tolerance range. Guessed value No Yes Guessed value – calculated value < Tolerance So what is convergence? Convergence is the process to compare the guessed value with the computed value until the guessed value is found within the tolerance range. The tolerance range is usually decided by the engineer designing the process, or market factors; for example, producing methanol with no more than 0.5% impurity. Let’s say you’re monitoring a certain concentration, and you want to keep the impurity level at 1% of the total concentration of product or less. If the process you design initially outputs product with 1.5% impurity, the simulation will recycle back to the initial calculation point of the process (which could be a concentration of a feedstock) and modify that value. It will keep changing that value until the impurity level reaches 1% or less. This value is called the tolerance. When the criterion is achieved, no more iteration needs to be done. Convergence When the criterion is achieved, the solution is found and no more iteration needs to be done. Module 9 – Steady state simulation

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. 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. Design to target: we design and specify unit operations to operate according to the desired target operation of the process. Let’s discuss the methods that drive process design. Providing a total account of an process is an obvious methodology, for here the engineer plans the process with multiple alternative designs. The designs are compared against each other for accuracy, and the best one is chosen. Evolution of design follows from the generation of the selected design. Then if the design needs improvement or modification, designers can make many small changes to improve the design incrementally. Doing this prevents any large changes from upsetting the process. Employing structured decision making provides that all the alternatives have been considered and a sound decision was made. We want to also design the process to achieve the target concentration, or temperature, etc…, so we design unit operations (columns, pumps, compressors, etc…) to operate according to the desired target operation of the process. Module 9 – Steady state simulation

Solution Determination
Sequential Solution Work 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 time Generally requires simulation software Now that the parameters of the process have been determined, how do we come to a solution? We can come up with the solution sequentially, where we work backwards from one point in order, solving one equation at a time. The iterative method is a key component of sequential solution determination, although it can be very time consuming if the problem is complicated. Simulation software employs simultaneous solution determination, in which multiple equations with multiple variables are solved at the same time. The ability of software to perform many calculations at once makes process design much easier to handle for the engineer. Module 9 – Steady state simulation

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. If you are an initial designer within the company, and you have faced your first time with simulation or have not touched any software in a while, you can do a lot to help you in your cause. Talk with trained professionals in the field (engineers, chemists). They will provide input to help you design the process better. Secondly, as mentioned earlier in the presentation, watch out for built-in assumptions and estimated parameters when designing various alternatives. Doing this may ultimately skew results and cause you to make an invalid decision when choosing a process. Also, visit the plant and talk with operators. Their insight and your knowledge of modeling can be beneficial in helping to solve problems. Module 9 – Steady state simulation

With a simulator, one day of process operation can be simulated in just seconds, and make as many changes as you want. Fresh Feed Steam Heater Reactor Flash Distillation Product Change in Reactor Properties Change composition in feed So let’s add closure to our talk of simulation software. A simulator is a very useful tool in that an entire process can be simulated and modified quickly and efficiently. Change in Heat Duty Change in Column Properties Module 9 – Steady state simulation

Commercial Simulation Software Packages
There are many of them, some of them are: Excel (spreadsheet) Excel Tutorial Matlab MATLAB homepage Fortran and C++ (programming languages) Aspen AspenTech HYSYS HYSYS WinGEMS WinGEMS SuperPro Designer SuperPro Designer IDEAS (Simons) There are many simulation programs; we include Excel as well. The hyperlinks given will transfer you to homepages of the programs and/or tutorials. Module 9 – Steady state simulation

Final Quiz What is a drawback of using spreadsheets?
What are two functions that simulation allows for? How are units calculated within a simulation process? Explain how iteration works and why you should use it. 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? Module 9 – Steady state simulation

Tier II Worked Examples
Now that we have talked about some background information in steady state simulation, let’s employ these concepts in a couple of worked examples. We’ll be discussing a single effect evaporator and dehydrogenation of isobutane. Both examples will be done in Excel. Module 9 – Steady state simulation

Statement 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 Tier 2 is specifically designed to take you step-by-step through two examples using steady-state simulation theories and Excel to help you to understand how to understand and develop a solution to a steady-state based problem. You will employ these problem solving techniques in the open-ended design problem in the third tier Module 9 – Steady state simulation

First Example: A Single Effect Evaporator
(to be done in Excel) First we’ll introduce what evaporation is, then discuss the problem and go step by step through the solution. Module 9 – Steady state simulation

Function is to concentrate solution
Evaporation Function is to concentrate solution What affects evaporation? Rate at which heat is transferred to the liquid Quantity of heat required to evaporate mass of water Maximum allowable temperature of liquid Pressure which evaporation takes place The main purpose of evaporation is to concentrate a solution to a higher solid content. Evaporation is affected by the rate at which heat is transferred to the liquid, the quantity of heat required to evaporate the desired mass of water, the maximum allowable temperature of the liquid, and the pressure at which the evaporation takes place. Module 9 – Steady state simulation

Single Effect Vertical Evaporator
Three functional sections Heat exchanger Evaporation section liquid boils and evaporates Separator vapor leaves liquid and passes off to other equipment A single effect vertical evaporator has three functional sections: a heat exchanger (also called a calandria), an evaporation section, where the liquid boils and evaporates, and a separator, in which the vapor leaves the liquid and passes off to other equipment. All three of these sections are contained in a vertical alignment. Three sections contained in a vertical cylinder Module 9 – Steady state simulation

In 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 stack As evaporation proceeds, the remaining liquors become more concentrated In the heat exchanger section, the steam condenses in the outer jacket, then the liquid resulting from the condensed steam boils on the insides of the tubes and in the space above the upper tube plate. As evaporation proceeds, the remaining liquors become more concentrated. Module 9 – Steady state simulation

Diagram of Single Effect Evaporator
Vapor V Tv, yv, Hv, ṁV Tf, xf, hf, ṁf U = J/m2 s oC Feed F P = kPa Ts, Hs, ṁs Here is a diagram of a single effect evaporator. You have a feed stream with a temperature, solids concentration, and enthalpy entering the evaporator. You also have steam entering at a temperature and enthalpy. The steam condenses and exits as condensate. The steam transfers its energy to the water in the feed, causing it to boil off and exit as vapor. The feed now has less water, but yet the same amount of solids, and exits as a concentrated liquid. The evaporator has a specific heat transfer area and is at a specific pressure. The heat transfer coefficient U describes the behavior of energy transfer between the steam and the feed. A = ? m2 Condensate S Ts, hs, ṁs Steam S Concentrated liquid L TL, xL, hL, ṁL Module 9 – Steady state simulation

Material and Heat Balances
q = UAΔT ΔT = Ts – TL Heat given off by vapor λ = Hs – hs ṁFhF + ṁsHs = ṁLhL + ṁVHV+ ṁshs ṁFhF + ṁsλ = ṁLhL + ṁVHV q = ṁs(Hs-hs) = ṁsλ ṁsλ – ideal heat transferred in evaporator ṁF = ṁL + ṁV ṁFxF = ṁLxL + ṁVyV The material and heat balances are shown above (material in red, heat in blue). The mass flow rate of feed (F) is equal to the sum of the mass flow rates of liquid (L) and vapor (V). Also, the product of the mass flow and the mole fraction of solids in the feed is equal to the sum of similar quantities within the liquid and vapor stream. The total heat passed in the reactor is equal to the product of the heat transfer coefficient, heat transfer area, and temperature gradient (which is equal to the difference of temperature between the steam and the concentrated liquid). The heat given off by the vapor is the difference in enthalpy between the steam and condensate. The sum of the product of the mass flow rate and respective enthalpy of the inputs (steam, feed) is equal to the same quantity for the outputs (liquid, condensate, vapor). We can also describe the total heat transferred as the product of the heat given off by vapor and the flow rate of steam. Module 9 – Steady state simulation

Finding the Latent Heat of Evaporation of Solution and the Enthalpies
Using the temperature of the boiling solution TL, the latent heat of evaporation can be found; The heat capacities of the liquid feed (CpF) and product (CpL) are used to calculate the enthalpies of the solution. Most likely, when designing an evaporator, we will not know the latent heat of vaporization or the enthalpies of the streams off-head. The temperature of the boiling solution (water, methanol, etc…) is used to find the latent heat. The enthalpies are calculated from the heat capacities of the feed and product. Reference to Tables B.1 and B.2, in Felder and Rousseau, Elementary Principles of Chemical Engineering, to see how to calculate heat capacities and therefore enthalpies. Module 9 – Steady state simulation

Property Effects on the Evaporator
Feed Temperature Large effect Preheating can reduce heat transfer area requirements Pressure Reduction Reduction in boiling point of solution Increased temperature gradient Lower heating surface area requirements Effect of Steam Pressure Increased temperature gradient when higher pressure steam is used. As mentioned earlier in the presentation, various factors can affect the performance of the evaporator. The feed temperature can have a fairly large effect on the evaporator’s performance, as preheating of the feed can reduce the heat transfer area. A reduction in pressure within the evaporator can reduce the boiling point in solution, resulting in a greater temperature difference and less heating surface area required. An increase in steam pressure will also result in a increased temperature difference. Module 9 – Steady state simulation

Boiling-Point Rise of Solutions
Increase in boiling point over that of water is known as the boiling point elevation (BPE) of solution BPE 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 For a given pressure the boiling point of an aqueous solution is higher than that of pure water. The increase in boiling point over that of water is known as the boiling point elevation of solution. For strong solutions, the BPE is best found using a rule known as Duhring’s rule, which states that the boiling point of a given solution is a linear function of the boiling point of pure water at the same pressure. Thus, if the boiling point of the solution is plotted against that of water at the same pressure, a straight line will result. Module 9 – Steady state simulation

Duhring lines (sodium chloride)
This is a plot of the Duhring lines for sodium chloride. As you can see, different lines are plotted for different concentrations, and over a moderate range of pressure the lines are nearly straight. Let’s describe how to use this plot. If for example, if the pressure over a 25% solution of sodium chloride is such that water boils at 80oC, by reading up from the x axis at 80oC to the line at 25% solution and then horizontally to the y axis, it is found that the boiling point of the solution at this pressure is approximately 88oC. The BPE for this solution at this pressure is 8oC. Module 9 – Steady state simulation

First Example Excel Spreadsheet
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/m2 oC. The feed temperature is 0oC. Calculate the amount of steam consumed, the economy, and required heating surface. The problem statement comes from Example 16.1, McCabe, Smith and Harriott (Unit Operations of Chemical Engineering, Sixth Edition). The hyperlink will take you to a spreadsheet of the problem, where you can interact and see the calculations being performed. First Example Excel Spreadsheet Module 9 – Steady state simulation

1. Draw Diagram and Label Streams
9070 kg/h feed, 0oC, 5% solids, hF Vapor V Tv, 0% solids, Hv, ṁv U = 1400 W/m2 oC Feed F P= 100 mm Hg First, draw diagram and label streams. We’ve labeled the feed, vapor, steam, condensate, and liquor streams with relevant information (including temperatures, mass flows, and enthalpies), and labeled the evaporator with some of its characteristics, including the heat transfer coefficient within the evaporator and the pressure within the evaporator. There is a lot of information to sort through, so drawing the diagram will help to place the information in a carefully, neat manner. Ts, Hs, 1.37 atm gauge, ṁs q=? Condensate S Ts, hs, ṁs Steam S A=? Liquor L TL, 20% solids, hL, ṁL Module 9 – Steady state simulation

[9070 kg/h = ṁL kg/h+ ṁV kg/h]
2. Perform Mass Balances ṁF = 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 = kg/h, ṁL = kg/h Second, perform mass balances around the evaporator. They are shown above (explain). Yv is zero because there is no solids present in the vapor stream. Module 9 – Steady state simulation

3. Perform Heat Balances to find the Economy
The economy is defined as the mass of water evaporated per mass of steam supplied. ṁFhF + ṁSHS = ṁLhL + ṁVHV+ ṁShS ṁFhF + ṁSλ = ṁLhL + ṁVHV q = ṁS(HS- hS) = ṁSλ After we perform the mass balances, let’s focus on performing the heat balances to find the economy. The economy is defined as the mass of water evaporated per mass of steam supplied. The heat balances shown were explained at the beginning of the example – they are the balances that describe the flow of heat through the evaporator. Module 9 – Steady state simulation

Needed Data Boiling point of water at 100 mm Hg = 51oC (from steam tables) Boiling point of solution = 88oC (from Duhring lines) Boiling point elevation = 88 – 51 = 37oC Enthalpy of vapor leaving evaporator (enthalpy of superheated vapor at 88oC 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 (Hs-hs = λ ) at 1.37 atm gauge [20 lbf/in2] = 939 Btu/lb = 2182 kJ/kg (McCabe, App.7, p.1073) To perform the heat balances, we need some data. First, we need to find the boiling point of water at the pressure of the evaporator. From steam tables, the pressure of the evaporator is at 100 mm Hg is 51oC. From Duhring lines, we can find that the boiling point of the solution is at 88oC. Therefore, the BPE is the difference between the boiling point of water and the solution, which is 37oC. The enthalpy of the vapor leaving the evaporator, or the enthalpy of the superheated vapor at the boiling point of solution and at the pressure of the evaporator, is found to be 2664 kJ/kg. The heat of vaporization of steam at the pressure of steam (1.37 atm gauge) is found to be 939 Btu/lb, which is converted 2182 k-Joules/k-gram (to keep units in metric terms) Module 9 – Steady state simulation

Finding the enthalpy of the feed
yNaCl=0.05 ywater=0.95 Cp,water=4.18 kJ/kgoC Cp,NaCl=0.85 kJ/kgoC Find the heat capacity of the liquid feed feed is 5% sodium chloride, 95% water (Cp)F = .05* *4.18 = 4.01 kJ/kgoC Now we need to find the enthalpies of the streams. Let’s first focus on the feed. First, we need to find the heat capacity of the liquid feed. The feed is a mixture of 5% sodium chloride and 95% water. The equation is given above on how to calculate the heat capacity of a mixture (it is the product of the mole fraction of the component and its heat capacity at the specified temperature. The feed is at a temperature of 0oC. The heat capacity of the feed is calculated to by 4.01 kJ/kgoC. Then calculate the enthalpy, assuming that there is no heat given or absorbed during dilution of the solution. The enthalpy is the product of the heat capacity of the mixture and the difference between the feed temperature and the reference temperature, which is given as 0oC. The enthalpy is 0 kJ/kg 2. Calculate Enthalpy (neglecting heats of dilution) hF = 4.01 kJ/kgoC (0 - 0 oC) = 0 kJ/kg Module 9 – Steady state simulation

Finding the enthalpy of the liquor
yNaCl=0.20 ywater=0.80 Cp,water=4.18 kJ/kgoC Cp,NaCl=0.85 kJ/kgoC Find the heat capacity of the liquor feed is 20% sodium chloride, 80% water Cp,L = .20* *4.18 = 3.51 kJ/kgoC The heat capacity of the liquor is solved in the same way. 2. Calculate Enthalpy (neglecting heats of dilution) hL = 3.51 kJ/kgoC (88-0 oC) = 309 kJ/kg Module 9 – Steady state simulation

q = 8626.5 kg/h*2182 kJ/kg = 1.88x107 kJ/h = 5228621 W = 5.23 MW
Heat Balances ṁLhL + ṁVHV - ṁFhF = ṁSHS - ṁShS = ṁS(HS- hS) = ṁSλ λ = (HS-hS) = 2182 kJ/kg ( kg/h * kJ/kg) + ( kg/h * 2664 kJ/kg) – (0) = ṁS (HS-hS) q = ṁS (2182 kJ/kg) Now we use the heat balances to find the transferred heat q and the flow rate of steam/condensate ṁs. ṁs= kg/h q = kg/h*2182 kJ/kg = 1.88x107 kJ/h = W = 5.23 MW Module 9 – Steady state simulation

Find the Economy = ṁV/ṁS
To find the economy, we simply divide the mass flow rate of the vapor by the mass flow rate of the steam/condensate. Module 9 – Steady state simulation

Condensing temperature of steam (1.37 atm gauge = 126.1oC
4. Calculate Required Heating Surface Condensing temperature of steam (1.37 atm gauge = 126.1oC q = UAΔT A = q/UΔT To calculate the required area, we need to first find the condensing temperature of steam at 1.37 atm gauge (2.37 atm absolute). This value can be found in the saturated steam tables. If we look for the temperature reading at 2.37 atm pressure, we extrapolate the temperature to be approximately 126.1oC. Then we just rearrange the heat balance to solve for the area, substitute in the value for the transferred heat, the heat transfer coefficient, and the value for the temperature gradient between the liquor and the condensate, and we find the heat transfer area. Module 9 – Steady state simulation

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

Second Example: Simulation of Cyclic Process (Felder and Rousseau, Example 10.2-3, pp. 516-519)
(to be done in Excel) The second example comes from example from Elementary Principles of Chemical Engineering, Felder and Rousseau. Like the first example, this will be done in Excel, and there are hyperlinks at the beginning and end of the section which will link you to a spreadsheet with which you can modify numbers and control the simulation. Module 9 – Steady state simulation

Problem Statement C4H10 C4H8 + H2
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. Here is the problem statement and the reaction (read) C4H C4H8 + H2 Module 9 – Steady state simulation

Second Example Cyclic Process
Diagram of Process Second Example Cyclic Process Here is a diagram of the process. Click on the hyperlink to see an Excel spreadsheet of the process. Module 9 – Steady state simulation

Notes A 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 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 Some important notes: Through the duration of this problem, A will denote isobutane, B will denote isobutene, and C will denote hydrogen. All streams are gases, Qr is the required rate of heat transfer to the reactor, and Qs is the net rate of heat transfer to the separation process. When solving for the enthalpies, they are for the gaseous species at the stream temperatures relative to 25oC. Module 9 – Steady state simulation

1. Perform Degree of Freedom Analysis
The first step in this problem is to perform the degree of freedom analysis In this problem, because we have so many unit operations, we are going to do the DOF analysis for each unit (the reactor, separator, and mixing point, which could be a continuously stirred tank). Then we will do the net degrees of freedom for the entire process. Module 9 – Steady state simulation

Review – Degrees of Freedom
Draw and completely label a flowchart Count the unknown variables, then the independent equations relating them, Subtract the number of equations from the number of variables. This gives ndf, or the number of degrees of freedom in the process. First, draw and label a flowchart (block diagram). Label all necessary variables (known and unknown). Second, count the number of unknown variables and the equations relating them. Sources of equations include material balances, energy balances, process specifications, physical properties and laws, physical constraints, and stoichometric relations. Then subtract the number of equations from the number of variables, which will get you the number of degrees of freedom in the process. Module 9 – Steady state simulation

Degree of Freedom Analysis
If ndf = 0 there are n independent equations in n unknowns and the problem can be solved If ndf >0, there are more unknowns than independent equations relating them, and at least ndf additional variable values must be specified. If ndf <0, there are more independent equations than unknowns. The flowchart is incompletely labeled or inconsistent and redundant relations exist. If the difference is equal to zero, then the problem can be solved. If the difference is greater than zero, than there are more unknowns than equations relating them, and additional variable values must be specified. If the difference is less than zero, there are more independent equations than unknowns. The problem is incompletely labeled or there are redundant and possibly inconsistent relations that exist. Module 9 – Steady state simulation

Degree of Freedom Analysis – Mixing Point
4 unknowns (ṅA1, ṅB1, ṅ4,T1) - 3 balances (2 material balances, 1 energy balance) = 1 local degree of freedom Around the mixing point (where the recycle stream enters the feed stream, which is highlighted in green and indicated with a red arrow) we have 4 unknown variables, 3 balances (two material for A and B, energy balance) = 1 local degree of freedom. Module 9 – Steady state simulation

Degree of Freedom Analysis – Reactor
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 Around the reactor, we have 7 unknown variables (list), 4 balances (A, B, and C material, 1 energy balance), an additional relation of the single pass conversion of 35%, and a chemical reaction. This equals 3 local degrees of freedom Module 9 – Steady state simulation

Degree of Freedom Analysis – Separator
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 Around the separator, we have 8 unknown variables, 4 material balances, and 1 additional relation (the isobutane split to the output and recycle), which equals 3 local degrees of freedom. Module 9 – Steady state simulation

Net Degree of Freedom Analysis – Overall Process
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 Now we do the overall balance. We add all the local degrees of freedom from the three “units,” subtract the number of variables that were counted twice (called ties) and find the difference. Because this difference is zero, this problem can be solved for all labeled variables. The problem can be solved for all labeled variables. Module 9 – Steady state simulation

2. Equation Based Solution Process
We will employ an equation based solution process to solve this problem. In this approach, the equations for all the units are collected and solve simultaneously. Programs such as Maple and Matlab are available to do this type of equation solving. Make sure to refer to the Excel spreadsheet to follow the model. Module 9 – Steady state simulation

Tearing the Cycle Can’t solve system in a unit-to-unit manner without trial and error “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 - Only have to determine ṅ4 by trial and error. If we tried to solve this system in a unit-to-unit manner (solve for variables in the mixing point, then the reactor, then the separator), we would have to employ trial and error guessing. This would dramatically increase the time it would take to solve the problem. Therefore, we “tear” the cycle between two units. “Tearing” is like adding another stream to the process. The purpose is to have the least number of variables that have to be determined by trial and error. If we tear between the separation process and the mixing unit, we would only have to solve for n4 by trial and error. If we did this between the mixing point and the reactor, we would have to determine two variables by trial and error – and if we did this between the reactor and the separator, we would have to determine three variables by trial and error. Module 9 – Steady state simulation

Solution Process Assume value of recycle flow rate (ṅ4A = 100 mol/s)
Assume mixing point outlet temperature (T1 = 50oC) Vary ṅ4A until calculated recycle flow rate (ṅ4C*) equals assumed value in ṅ4A - Will be done by driving (ṅ4A - ṅ4C*) using Goalseek Mixing point temperature (T1) will be varied to determine the value that drives ΔḢmix to zero (remember, the mixer is adiabatic) In the solution, we will assume that we have a recycle flow rate of 100 mol/s. This is just an initial guess for this value. We will also make a guess for the mixing point outlet temperature of 50oC. We will vary the assumed recycle flow rate until the calculated recycle flow rate equals the assumed value of the recycle flow rate. This will determine the true recycle flow rate of the process. This will be done by driving the difference between the calculated flow rate and the assumed flow rate using the Goalseek function, for which was explained earlier in the presentation. To find the mixing point temperature, we will vary that value until the change in enthalpy of the mixture is driven to zero (the reactor is adiabatic). Module 9 – Steady state simulation

Known Values XA = 0.35 (fractional conversion of A)
100 mol/s (basis of calculation) Feed temperature – 20oC Reactor Effluent Temperature – 90oC Product Stream Temperature – 30oC Guess for recycle stream flow rate (ṅA4) = 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 – 85oC Initial guess for combined stream temperature – 50oC These are the known values from the problem statement, including the basis of calculation (we said 100 mol/s, any value can be put in here depending on how much isobutene you want to simulate) and the initial guesses for the mixing point temperature and the recycle stream flow rate. Module 9 – Steady state simulation

Mass Balances (based on initial 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/s ṅA2 = ṅA1 * (1-XA) = 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 = mol/s ṅB3 = ṅB2 – (0.1 mol fraction * ṅC4 )= 62.9 mol/s ṅC3 = ṅC2 = 66.5 mol/s The mass balances are listed here. Because you are familiar with how to do these balances, I won’t go into detail here. Refer to the Excel spreadsheet to see the mass balances being done and also reference to the previous slide, which had the known values. Module 9 – Steady state simulation

- (heats of formation) are located in Table B.1 of F&R
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 To calculate the specific enthalpies of the species in the problem, we will need to refer to tables B.1 and B.2 on pages of Elementary Principles of Chemical Engineering by Felder and Rousseau, Third Edition. The formula for finding the specific enthalpy of a species is listed above. It is equal to the sum of the heat of formation of the species and the integral of its heat capacity over a specified temperature range. The heats of formation are listed in Table B:1. A (isobutane [g]) = kJ/mol B (isobutene [g]) = 1.17 kJ/mol C (hydrogen [g])= 0 kJ/mol Module 9 – Steady state simulation

Calculation of Specific Enthalpies (Tables B. 1 and B
Calculation of Specific Enthalpies (Tables B.1 and B.2, Felder and Rousseau) - heat capacity of component i (kJ/moloC) = a+ bT + cT-2 + dT-3 , where T is temperature in oC Chemicals A* 103 B* 105 C* 108 D* 1012 isobutane 89.46 30.13 -18.91 49.87 isobutene 82.88 25.64 -17.27 50.50 hydrogen 28.84 0.3288 To find the heat capacity of a component, we use the formula given in Table B.2, which is written in green above. The heat capacity is a function of temperature, and the values to find each heat capacity for the species is given in the table above. Module 9 – Steady state simulation

Heat Balances (based on initial guesses)
ΔḢmix = ṅA1*ĤA1 + ṅB1*ĤB1 – 100 mol/s*ĤA0 - (ṅA4*0.9 mol A/mol *ĤA4) - (ṅA4*0.1 mol fraction*ĤB4) = kJ/mol = ṅA2*ĤA2 + ṅB2*ĤB2 + ṅC2*ĤC2 - ṅA1*ĤA1 – ṅB1*ĤB1 = kJ/s Based on the initial guesses, we can solve for the enthalpy of mixing, Qr and Qs. Of course, when we drive the enthalpy of mixing to zero and the mixing temperature to zero, Qr and Qs will change to reflect of the actual process. = ṅA3*ĤA3 + ṅB3*ĤB3 + ṅC3*ĤC3+(ṅA4*0.900 mol fraction*ĤA4)+(ṅB4*0.100 mol fraction*ĤB4) – ṅA2*ĤA2 - ṅB2*ĤB2 - ṅC2*ĤC2 = kJ/s Module 9 – Steady state simulation

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

Tier III Open-ended problem
Approach to open-ended problem Case Study. In this section an open – ended design problem will be presented. An open–ended problem is one which can have many solutions depending on problem conditions. Open-ended problems are similar to most real-world processes and will help to develop and use your creative thought process to determine a myriad of possible solutions for any process design you may encounter. First, we’ll talk how to approach open ended problems, then provide a case study for you to study and solve. Module 9 – Steady state simulation

Statement of Intent Learn how to approach open-ended design problems
Solve a problem on your own In Tier 3, we will explain briefly how to tackle an open-ended design problem. Sometimes, these types of problem will have one specific set of answers; sometimes they will not. You will then have the opportunity to solve a problem on your own. You will be able to practice developing spreadsheet models, employing degree of freedom analyses, and solving mass and energy balances equations. Module 9 – Steady state simulation

How to approach open–ended problems
State the problem clearly, including goals, constraints, and data requirements. Define the trade-offs necessary. Define the criteria for a valid solution. Develop a set of cases to simulate possible solutions. Perform the simulation and evaluate results against solution criteria. Evaluate solutions against environmental, safety and financial considerations. To tackle open-ended problems, first state the problem clearly, including goals, constraints and data requirements. Next, define what trade-offs you may have to make. For example, you may have to reduce your initial target production rate because one of the reactors that you would have to get to achieve the rate is too expensive for the company to purchase. Next, develop the criteria required to obtain a valid solution, and use that criteria to develop a set of cases to simulate possible solutions. Perform the simulation and evaluate the pending results against the solution criteria that you initially established. When you have determined a number of possible solutions, compare and evaluate those solutions against environmental, safety, and financial considerations. Module 9 – Steady state simulation

The Use of Limestone Slurry Scrubbing to Remove Sulfur Dioxide from Power Plant Flue Gases
This case study is called The Use of Limestone Slurry Scrubbing to Remove Sulfur Dioxide from Power Plant Flue Gases. 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 Module 9 – Steady state simulation

About Coal 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 Predominately carbon, but contains other elements and hydrocarbon volatile matter Chemical engineers have a responsibility to contribute to the improvement of the environment, and they can do this through their knowledge of process development. It is often necessary to employ engineering practices to remedy existing environmental problems. We’ll look at a situation in which a coal-burning power plant is designed to minimize its impact on the environment. Coal is an abundant source of energy and a source of raw materials in production. Coal is predominately carbon, but it contains other elements, including metals, nitrogen compounds, and sulfur, and other volatile matter that can be burned to produce heat, water, and carbon dioxide. Module 9 – Steady state simulation

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 sulfur dioxide emissions regulated in the U.S. by the Environmental Protection Agency current regulations are no more than 520 ng SO2 per joule of heating value of the fuel fed to the furnace plants must remove 90% of SO2 released when coal-burning Coal is burned in many of the world’s power plants to produce electricity; however, the burning of coal can produce a lot of pollution if the gases are not treated before being released into the atmosphere. Untreated gases can contain soot, nitrogen oxides, ash, and sulfur dioxide. Sulfur dioxide emissions are regulated in the United States by the Environmental Protection Agency and appropriate state and local agencies. Current regulations require that gases released to the atmosphere from new and existing power plants contain no more than 520 nanograms of sulfur dioxide per joule of higher heating value of the fuel fed to the furnace. In addition, plants must remove 90% of the sulfur dioxide released when coal is burned. Module 9 – Steady state simulation

About Commercial Processing
SO2 removal is classified as regenerative or throwaway throwaway processing can be modified to produce gypsum throwaway processing uses separating agent to remove SO2 from stack gases followed by disposal of SO2 innocuously (CaSO3 * ½ H2O) and a slurried separating agent of calcium carbonate Commercial processing of sulfur dioxide removal are classified as regenerative or throwaway, depending on whether or not the agent used to remove the sulfur dioxide is reusable. Some throwaway processing can be modified to produce gypsum, a saleable product, without regeneration of the absorbent material. Throwaway processing utilizes a separating agent to remove sulfur dioxide from the stack gases follow by disposal of both the sulfur dioxide in an innocuous form and the slurried separating agent of calcium carbonate. The following process will employ the use of limestone-slurry scrubbing. Module 9 – Steady state simulation

Process Description Module 9 – Steady state simulation

want to produce 500 MWe (megawatts of electricity)
properties of coal given in table on next slide coal fed at 25oC to furnace, burned with 15% excess air sulfur reacts to form SO2 and negligible SO3 carbon, hydrogen oxidized completely to CO2 and water nitrogen in coal leaves furnace as N2 ash in coal leaves furnace in two streams 80% leaves as fly ash in furnace flue gas remainder as bottom ash at 900oC This process is to produce 500 megawatts of electricity in the present facility. Coal with properties given on the table in the next slide is fed to 25oC to a furnace and burned with 15% excess air. During combustion of the coal, the sulfur reacts to form sulfur dioxide and a negligible amount of sulfur trioxide. The carbon and the hydrogen are oxidized completely to carbon dioxide and water. All the nitrogen in the coal leaves the furnace as N2. The ash in the coal leaves the furnace in two streams: 80% leaves as fly ash in the furnace flue gas, and the remainder leaves the furnace as bottom ash at 900oC. Module 9 – Steady state simulation

Component Dry Weight % Carbon 75.2 Hydrogen 5.0 Nitrogen 1.6 Sulfur
3.5 Oxygen 7.5 Ash 7.2 Moisture 4.8 kg/100 kg dry coal HHV 30780 KJ/kg dry coal Cp dry coal 1.046 kJ/(kgoC) Cp ash 0.921 KJ/(kgoC) A table depicting the properties of coal to be burned. Module 9 – Steady state simulation

combustion air brought into process at 25oC, 50% RH
air sent to heat exchanger, temperature increased to 315oC air then fed to boiler, reacts with coal flue gas leaves furnace at 330oC, goes to electrostatic precipitator 99.9% of particulate material removed goes to air preheater, exchanges heat with combustion air leaves air preheater and split into two equal streams each stream is feed to one of two identical scrubber trains trains sized to process 60% of flue gas Combustion air is brought into the process at 25oC and 50% relative humidity. The air is sent to an heat exchanger, where its temperature is increased to 315oC by exchanging heat with the furnace flue gas. The air is then fed to the boiler and reacts with the coal. The flue gas leaves the furnace at 330oC, goes to an electrostatic precipitator where 99.9% of the particulate material is removed, and then to the air preheater, where it exchanges heat with the combustion air. The flue gas leaves the air preheater and is split into two equal streams. Each stream is feed to one of two identical scrubber trains. Each of the trains is sized to process 60% of the flue gas. Module 9 – Steady state simulation

sulfur dioxide absorbed in the slurry and reacts with the limestone:
divided gas stream fed to scrubber, contacts aqueous slurry of limestone, undergoes adiabatic cooling to 53oC. sulfur dioxide absorbed in the slurry and reacts with the limestone: CaCO3 + SO2 + ½ H2O CaSO3 · ½ H2O + CO2 solid/liquid slurry enters scrubber at 50oC liquid slurry flows at 15.2 kg liquid/kg inlet gas solid to liquid ratio in the slurry is 1:9 by weight liquid saturated with CaCO3 and CaSO3 cleaned flue gas meets EPA SO2 requirements leaves scrubber with saturated water at 53oC In each of the scrubber trains, the divided-off gas stream is fed to the scrubber, where it is contacted with an aqueous slurry of limestone, and undergoes adiabatic cooling to 53oC. Sulfur dioxide is absorbed in the slurry and reacts with the limestone according to the reaction above. The solid-liquid limestone enters the scrubber at 50oC; the liquid portion of the slurry flows at a rate of 15.2 kg liquid/kg inlet gas and the solid to liquid ratio in the slurry is 1 to 9 by weight. The liquid is saturated with calcium carbonate and calcium sulfate. The cleaned flue gas meets the EPA sulfur dioxide requirements for emissions. The sulfur dioxide leaves the scrubber with saturated water at 53oC, containing only the carbon dioxide generated in scrubbing but no fly ash. Module 9 – Steady state simulation

cleaned flue gas contains CO2 generated in scrubbing but no fly ash
cleaned flue gas reheated to 80oC, blended with clean flue gas stream from other train, and sent to be released to atmosphere solids in spent aqueous slurry unreacted CaCO3, flyash from flue gas, inert materials, CaSO3 liquid portion of slurry saturated with CaCO3, CaSO3 specific gravity of 0.988 spent slurry split in two 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 The cleaned flue gas contains carbon dioxide generated in scrubbing, but no fly ash. The cleaned flue gas is reheated to 80oC, blended with a clean flue gas stream from the other train, and is sent to be released to the atmosphere. The solids in the spent aqueous slurry contain unreacted calcium carbonate, fly ash from the flue gas, inert materials, and calcium sulfate. The liquid portion of the slurry that is saturated with calcium carbonate and calcium sulfate has a specific gravity of The spent slurry is split in two; one stream is mixed with freshly ground limestone and makeup water and sent to a blending tank – the other is a recycle stream that will be described shortly. A fresh slurry stream from the blending tank is fed to the top of the scrubber. Module 9 – Steady state simulation

limestone – 92.1% CaCO3 and rest is insoluble inert material
second stream sent to filter where wet solids containing fly ash, inert materials, CaSO3 and CaCO3 are separated from filtrate filtrate saturated with CaSO3, CaCO3, and is the recycle stream fed to the blending tank 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 SO2 absorbed from flue gas limestone – 92.1% CaCO3 and rest is insoluble inert material The second stream is sent to the filter, where wet solids that contain fly ash, inert materials, calcium sulfate, and calcium carbonate are separated from the filtrate. The filtrate is saturated with calcium sulfate and calcium carbonate and is the recycle stream fed to the blending tank. The wet solids contain 50.2% liquid that has the same composition as the filtrate. The fresh ground limestone is fed to the blending tank at the rate of 5.2% in excess of that required to react with the sulfur dioxide absorbed from the flue gas. The limestone contains 92.1% calcium carbonate and the rest is insoluble inert material. Module 9 – Steady state simulation

Boiler generates steam at supercritical conditions
540oC and 24.1 MPa absolute 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 heat removed from wet low pressure steam in a condenser by cooling water cooling water enters condenser at 25oC and leaves at 28oC saturated condensate at 38oC is produced by condenser and pumped back to boiler The boiler generates steam at supercritical conditions; 540oC at 24.1 megaPascals absolute pressure. Mechanical work is derived by expanding the steam through a power-generated system of turbines. The low pressure stream extracted from power system contains 27.5% liquid water at 6.55 kiloPascals absolute pressure. Heat is removed from the wet low pressure steam in a condenser by cooling water that enters the condenser at 25oC and leaves at 28oC. Saturated condensate at 38oC is produced by the condenser and pumped back to the boiler. Module 9 – Steady state simulation

Assume a basis of 100 kg dry coal/min fed to the furnace.
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. Estimate the molar flow rate (kmol/min) of each element in the coal (other than those in the ash). Determine the feed rate (kmol/min) of O2 required for complete combustion of the coal. Assume a basic of 100 kilograms dry coal/min fed to the furnace and answer the questions on the next three slides. Module 9 – Steady state simulation

The oxygen and nitrogen feed rates (kmol/min)
If 15% excess oxygen is fed to combustion furnace, estimate the following: The oxygen and nitrogen feed rates (kmol/min) 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) The air feed rate (kmol/min, m3/min) 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? Module 9 – Steady state simulation

b. Determine flow rate (kg/min) of slurry entering scrubber
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. Module 9 – Steady state simulation

6. (continued) e. What are flow rates (kg/min) of inerts, CaSO3, CaCO3, 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. Module 9 – Steady state simulation

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. 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. Module 9 – Steady state simulation

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Introduction to Steady State Simulation

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