NAMP for North American Mobility In Higher Education
Program
PIECE
NAMP
Module 9
Introduction to Steady
State Simulation
Introducing
Process
integration
for Environmental Control in Engineering Curricula
Module
9 – Steady
state
simulation
PIECE
1
PIECE
NAMP integration for Environmental Control in Engineering Curricula
Process
Paprican
PIECE
École
Polytechnique
de Montréal
Universidad
Autónoma de
San Luis Potosí
University of
Ottawa
Universidad de
Guanajuato
North Carolina
State University
Instituto
Mexicano del
Petróleo
Program
North
American
Mobility in Higher Education
Module
9 –for
Steady
state
simulation
Texas A&M
University
NAMP
2
NAMP
PIECE
Module 9
This module was created by:
Amy Westgate
North Carolina
State University
University of
Ottawa
From
Host University
North Carolina
State University
Richard Ezike
Module 9 – Steady state simulation
University of
Ottawa
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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
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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.
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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 prerequisite 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.
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Tier I
Background Information
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• 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
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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
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• Three types of processes:
Batch
• A process where a set
amount of input enters a
process, where it is
removed from the process
at a later time.
Continuous
• A process where inputs
and outputs flow
continuously through
duration of process.
Semi-batch
• Neither batch or continuous,
may be combination of both.
In steady-state processes, we will be looking at continuous processes.
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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
N 2 + 3H 2  2N H 3
t = to
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t = tf
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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.
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• 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.
Mostly CH3OH
50% molar CH3OH
50% molar H2O
Mostly H2O
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Quiz #1
Classify the following processes as batch,
continuous, or semibatch.
•
A balloon is filled with air at a steady rate of 2
m3/min.
•
Pump a mixture of liquids into a distillation
column at a constant rate and steadily withdraw
product streams from the top and bottom.
•
Slowly blend several liquids in a tank from which
nothing is being withdrawn.
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Block Diagrams
• When solving a problem, it is helpful to develop a
Process:
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.
Carbon (C)
Air (79% N2, 21% O2)
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Reactor
Separator
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Degree Of Freedom Analysis
•
Analysis done to determine if there is enough information
to solve a given problem.
1. Draw and completely label a flowchart
2. Count the unknown variables, then the independent
equations relating them,
3. Subtract the number of equations from the number of
variables. This gives ndf, or the number of degrees of
freedom in the process.
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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.
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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
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Mass (Material) Balance (continued)
In the case of a steady-state process we are able
to set the accumulation term to zero since it is a
time dependent term. Since steady-state does
not depend on time as it is constant, we are able
to eliminate this term:
Mass Flow Rate In = Mass Flow Rate Out
Material Balance Procedure
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First Law of Thermodynamics (Energy
Balance) for a Steady State Open
System
ΔH + ΔEk + ΔEp = Q - Ws
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
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Quiz #2
1. Explain the Degree of Freedom analysis.
2. What term goes to zero in a steady-state process?
3. Is a continuous process closed or open? How
about a batch process?
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Heat Transfer
• Also classified as energy transfer
• Three types of heat transfer modes:
– Conduction
– Convection
– Radiation
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Conduction
• Accomplished in two ways
qy
dT
– Molecular interaction
= - k
A
dy
– “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)
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Convection
• Accomplished in two ways
q =  hx ΔTdA = hAΔT
–Natural convection
A
–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)
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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
q
A1
=   1T
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1
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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)
Cold surface
Note this is only in ideal cases: no surface is
exactly black, and emissivities must be
considered
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Hot surface
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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
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• Flux
= - (overall density)*(diffusion
coefficient)*(concentration gradient)
• Fick rate equation (restricted to
isothermal/isobaric systems)
JA = - D AB
JA – mol A/m2s
dC A
CA- mol A/m3
dz
DAB- m2/s
• de Groot equation is more general
J A = - cD A B
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Units:
dy A
dz
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• Molar flux of species A in binary system (A + B)
N A   cD
AB
 y A  y A (N A  N B )
c = concentration
DAB = diffusivity of species A in B
yA =
change of molar species in y with respect to a
specified direction
NA, NB = molar fluxes of components
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Quiz #3
1. What are two ways in which conduction occurs?
2. Define natural and forced convection.
3. What is the restriction to the use of Fick’s Law?
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Modeling
• What is Modeling?
• Steady-State vs. Dynamic Modeling
• Empirical vs. Mechanistic Modeling
• Derivation of a Steady State Model
• Modeling and Process Design Implications
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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
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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
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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.
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Time-based Modeling
Steady
State
Dynamic
Model
Empirical
Mechanistic
Hybrid
Level of Knowledge-based Modeling
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Steady – State
Dynamic
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: Amount of information required:
small to medium
medium to large
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Steady State Example
Continuous Stirred Tank Reactor (CSTR)
Concentration
profile at one
point in reactor
does not
change with
time
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Ca
t
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Dynamic Example
Batch Reactor
Concentration
profile at one
point in reactor
does change
with time
ca
t
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Empirical Modeling
•
Definition:
– 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
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Procedure – Empirical Modeling
1. Obtain data from process you wish to model.
–
Temperature, pressure, flow, etc…
2. Perform appropriate statistical analysis and develop
accurate correlations from data.
3. Develop mathematical equations to accurately
represent the data and the correlations found in step 2,
and determine which equations are useful in the
development of the model.
4. Check for correctness in your analysis and equations,
and determine if the model is satisfactory.
Statistical Analysis with Excel
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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.
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• 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)
• 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.
• We have empirically
determined a value for a, but
only for the specific fluids
and conditions tested.
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Mechanistic Modeling
• Definition
– 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
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Procedure – Mechanistic Modeling
1. Know physical and chemical properties of the process.
2. Determine the appropriate process model using mass
and/or heat balance.
3. Determine appropriate model run conditions and
parameters
4. Complete runs and use output data to compare
against the predicted model results
5. Develop an acceptable conclusion for the model.
Should the conclusion not be acceptable, re-examine
the assumptions, process and the physical and
chemical properties made in Step 1. Make appropriate
modifications and repeat Steps 2-4.
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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:
m H C p H (TH1 - TH 2 ) = m C C p C (TC 2 - TC 1 )
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Steady state model derivation
1. Define Goals.
a) Specific design decisions.
b) Numerical values.
c) Functional relationships.
d) Required accuracy.
2. Prepare information.
a) Sketch process.
b) Identify variables of interest.
c) State assumptions and data.
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Steady state model derivation
3. Formulate model
a) Conservation balances.
b) Constitutive equations.
c) Rationalize (combine equations and collect terms).
d) Check degrees of freedom.
4. Determine solution
a) Analytical
b) Numerical
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Steady state model derivation
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
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Process insights resulting from modeling
1.
Identification: If we know the input
(I) and output (O) parameters, we can
determine the structure (R) of the
model.
I
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R?
O
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Process insights resulting from modeling
2.
Simulation: If we know the structure
of the model, we can simulate what
the output of the process will be for a
given input.
I
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R
O?
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Process insights resulting from modeling
3.
Control/Optimization: If we know the
desired output (O) and the structure
(R) of the model, we can determine
what the input (I) should be to
optimize the process.
I?
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R
O
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Quiz #4
1. What are some uses of modeling?
2. Name and explain three requirements of a good
model.
3. What distinguishes a steady-state model and a
dynamic model?
4. Review the procedures for developing a
mechanistic and empirical model. What are some
differences between the two procedures?
5. Discuss the control/optimization insight of
modeling.
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Solving Problems
– Analytical Methods
– Process Design
– Methods
• Spreadsheets
• Simulation Software
– Solution Determination
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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
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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
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Process Design
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Process design
The design of chemical products begins with the
identification and creation of potential opportunities to
satisfy societal needs and to generate profit. The scope
of chemical product is extremely broad. They can be
roughly classified as:
1. Basic chemical products.
2. Industrial products.
3. Consumer products.
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Process design
Natural
Resources
Manufacturing
Process
Basic chemical
Products
Basic Chemical
Products
Manufacturing
Process
Industrial
Products
Industrial Products
Manufacturing
Process
Consumer
Products
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Motivation for Process Design
1.
Desires of customers for chemicals with
improved properties for many applications.
2.
Discovery of a new inexpensive source of a
raw material with comparable physical and
chemical properties to the old source.
3.
New markets are discovered.
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Steps in a Process Design
1.
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?



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Steps in a Process Design
2. 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
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Stability of the process
When a process is disturbed from an initial steady
state, it will generally respond in one of 3 ways.
a) Proceed to a steady state and remain there.
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Stability of the process
b) Fail to attain to a steady state condition
because its output grows indefinitely. The
system is unstable.
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Stability of the process
c)
Fail to attain a steady state condition
because the output of the process oscillates
indefinitely with a constant amplitude. The
system is at the limit of stability.
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Quiz #5
1. What are some errors that may arise when
attempting to fit a curve?
2. What are the three products developed from
process design? Provide an example of each
product.
3. What happens to an unstable system over time?
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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.
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– Columns are designated by letters, rows by numbers
http://www.instrunet.com/images/Direct%20To%20Excel%20Spreadsheet.png
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• Under Tools Menu
PIECE
Goalseek
• want to know input value formula needs to
determine result
• Excel varies value in cell specified until dependent
formula returns value you want
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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.
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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
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• 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
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• Simulation Software – What does it allow?
– Manipulation and comparison of previous data as
well as for research
– Manipulation of a process until a desired target is
reached
– Allows complex processes to be easily calculated
– Can easily change conditions and see how the output
is changed and the equipment behaves
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• Simulation Issues and 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
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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
Distillation
steam
Reactor
Heater
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Flash
Product
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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.
1
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2
3
4
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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.
1
2
3
4
For these processes, a solution technique is needed to
solve the equations for all the units in the recycle loop.
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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
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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
Guessed value – calculated value < Tolerance
Yes
Convergence
When the criterion is achieved, the solution is found
and no more iteration needs to be done.
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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.
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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
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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.
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With a simulator, one day of process operation
can be simulated in just seconds, and make as
many changes as you want.
Fresh Feed
Change
composition
in feed
Change in
Reactor Properties
Distillation
Reactor
Steam
Flash
Product
Heater
Change in
Heat Duty
Module 9 – Steady state simulation
Change in
Column Properties
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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)
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Final Quiz
1. What is a drawback of using spreadsheets?
2. What are two functions that simulation allows for?
3. How are units calculated within a simulation
process?
4. Explain how iteration works and why you should
use it.
5. You are an engineer who has been tabbed to
design a new chemical process for a company.
What are some steps you can take to help you in
your design?
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Tier II
Worked Examples
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• 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
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First Example: A
Single Effect
Evaporator
(to be done in Excel)
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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
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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
Three sections contained in a vertical cylinder
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• 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
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Diagram of Single Effect Evaporator
Tf, xf, hf, ṁf
Feed F
Vapor V
Tv, yv, Hv, ṁV
U = J/m2 s oC
P = kPa
Ts, Hs, ṁs
A = ? m2
Condensate S
Ts, hs, ṁs
Steam S
Concentrated
liquid L
Module 9 – Steady state simulation
TL, xL, hL, ṁL
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Material and Heat Balances
q = UAΔT
ΔT = Ts – TL
ṁF = ṁL + ṁV
Heat given off by vapor
λ = H s – hs
ṁFxF = ṁLxL + ṁVyV
ṁFhF + ṁsHs = ṁLhL + ṁVHV+ ṁshs
ṁFhF + ṁsλ = ṁLhL + ṁVHV
q = ṁs(Hs-hs) = ṁsλ
ṁsλ – ideal heat
transferred in evaporator
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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.
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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.
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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
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Duhring lines (sodium chloride)
http://www.nzifst.org.nz/unitoperations/evaporation4.htm
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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.
First Example Excel Spreadsheet
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1. Draw Diagram and Label Streams
9070 kg/h feed,
0oC, 5% solids,
hF
Feed F
Ts, Hs, 1.37 atm
gauge, ṁs
Steam S
Vapor V Tv, 0% solids,
Hv, ṁv
U = 1400
W/m2 oC
P= 100
mm Hg
q=?
Condensate S
Ts, hs, ṁs
A=?
Liquor L
Module 9 – Steady state simulation
TL, 20% solids,
hL, ṁL
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ṁF = ṁL + ṁV
2. Perform Mass Balances
[9070 kg/h = ṁL kg/h+ ṁV kg/h]
ṁFxF = ṁLxL + ṁVyV
vapor is present, no solids)
(note that yv is zero because only
[0.05 * 9070 kg/h = 0.2 * ṁL kg/h + 0]
• Can solve for ṁv and ṁL
ṁV = 6802.5 kg/h, ṁL = 2267.5 kg/h
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3. Perform Heat Balances to find the 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
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Needed Data
• Boiling point of water at 100 mm Hg = 51oC (from steam tables)
www.nzifst.org.nz/unitoperations/appendix8.htm
• Boiling point of solution = 88oC (from Duhring lines)
http://www.nzifst.org.nz/unitoperations/evaporation4.htm
• 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)
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Finding the enthalpy of the feed
1. Find the heat capacity of the liquid feed
yNaCl=0.05
feed is 5% sodium chloride, 95% water
ywater=0.95
C p ,m ix =

x iC p i
a ll m ixtu re
c o m p o n e n ts
Cp,water=4.18 kJ/kgoC
Cp,NaCl=0.85 kJ/kgoC
(Cp)F = .05*0.85 + .95*4.18 = 4.01 kJ/kgoC
2. Calculate Enthalpy (neglecting heats of dilution)
h F = C p ,F (TF - Tre f )
hF = 4.01 kJ/kgoC (0 - 0 oC) = 0 kJ/kg
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Finding the enthalpy of the liquor
yNaCl=0.20
1. Find the heat capacity of the liquor
feed is 20% sodium chloride, 80% water
C p ,m ix =

x iC p i
a ll m ixtu re
c o m p o n e n ts
ywater=0.80
Cp,water=4.18 kJ/kgoC
Cp,NaCl=0.85 kJ/kgoC
Cp,L = .20*0.85 + .80*4.18 = 3.51 kJ/kgoC
2. Calculate Enthalpy (neglecting heats of dilution)
h L = C p ,L (TL - Tre f )
hL = 3.51 kJ/kgoC (88-0 oC) = 309 kJ/kg
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Heat Balances
ṁLhL + ṁVHV - ṁFhF = ṁSHS - ṁShS = ṁS(HS- hS) = ṁSλ
λ = (HS-hS) = 2182 kJ/kg
(2267.5 kg/h *309.23 kJ/kg) + (6802.5 kg/h * 2664
kJ/kg) – (0) = ṁS (HS-hS)
q = ṁS (2182 kJ/kg)
ṁs=8626.5 kg/h
q = 8626.5 kg/h*2182 kJ/kg = 1.88x107 kJ/h =
5228621 W = 5.23 MW
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Find the Economy
= ṁV/ṁS
E conom y =
6802.5 k g/h
= 0.788
8626.5 k g/h
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4. Calculate Required Heating Surface
Condensing temperature of steam (1.37 atm gauge =
126.1oC
q = UAΔT
A = q/UΔT
5228621
A =
1400
W
m
2 o
Module 9 – Steady state simulation
W
= 9 8 .0 2 m
2
o
C
(1 2 6 .1 - 8 8 ) C
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Click on the Hyperlink and click on the
“Final Solution” tab to see the final
answer for the system.
First Example Final Solution
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Second Example:
Simulation of Cyclic
Process (Felder and
Rousseau, Example 10.2-3, pp.
516-519)
(to be done in Excel)
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Problem Statement
The gas-phase dehydrogenation of isobutane (A) to isobutene (B)
is carried out in a continuous reactor. A stream of pure isobutane
(the fresh feed to the process) is mixed adiabatically with a recycle
stream containing 90% mole isobutane and the balance isobutene,
and the combined stream goes to a catalytic reactor. The effluent
from this process goes through a multistage separation process; one
product stream containing all the hydrogen (C) and 10% of the
isobutane leaving the reactor as well as some isobutene is sent to
another part of the plant for additional processing, and the other
product stream is the recycle to the reactor. The conversion of
isobutane in the reactor is 35%. Assume a fresh feed of 100 mol
isobutane. Simulate the process using a spreadsheet to find the
desired process variables.
Module 9 – Steady state simulation
C4H10
C4H8 + H2
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Diagram of Process
Second Example Cyclic Process
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Notes
• A will denote isobutane, B denotes isobutene, C
denotes hydrogen
• All streams are gases, Q r is the required rate of heat
transfer to the reactor and Q s 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
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1. Perform Degree of
Freedom Analysis
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Review – Degrees of Freedom
1. Draw and completely label a flowchart
2. Count the unknown variables, then the
independent equations relating them,
3. Subtract the number of equations from the
number of variables. This gives ndf, or the
number of degrees of freedom in the process.
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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.
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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
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Degree of Freedom Analysis – Reactor
7 unknowns (ṅA1, ṅB1, ṅA2, ṅB2, ṅC2, T1,
Qr
)
- 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
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Degree of Freedom Analysis – Separator
8 unknowns (ṅA2, ṅB2, ṅC2, ṅA3, ṅB3, ṅC3, ṅ4, Q s )
- 4 balances (3 material balances, 1 energy
balance)
- 1 additional relation (isobutane split)
= 3 local degrees of freedom
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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
The problem can be solved for all labeled variables.
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2. Equation Based
Solution Process
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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.
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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)
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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
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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) = 123.5 mol/s
ṅB2 = ṅB1+ (ṅA1*XA)= 76.5 mol/s
ṅC2 = ṅA1 * XA = 66.5 mol/s
ṅA3 = 0.01* ṅA2 = 1.24 mol/s
ṅC4 = (ṅA2- ṅA3)/0.9 mol fraction = 135.85 mol/s
ṅB3 = ṅB2 – (0.1 mol fraction * ṅC4 )= 62.9 mol/s
ṅC3 = ṅC2 = 66.5 mol/s
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Calculation of Specific Enthalpies (Tables B.1 and
B.2, Felder and Rousseau)
o
ˆ
ˆ
H i = (Δ H f ) i +
o
ˆ
(Δ H f ) i
C
pi
dT
- (heats of formation) are located in Table B.1 of F&R
A (isobutane [g]) = -134.5 kJ/mol
B (isobutene [g]) = 1.17 kJ/mol
C (hydrogen [g])= 0 kJ/mol
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Calculation of Specific Enthalpies (Tables
B.1 and B.2, Felder and Rousseau)
oC)
heat
capacity
of
component
i
(kJ/mol
C
d
T
p
i

= 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
0.00765
0.3288
-0.8698
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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) = -78.64
kJ/mol
Q r = ṅA2*ĤA2 + ṅB2*ĤB2 + ṅC2*ĤC2 - ṅA1*ĤA1 – ṅB1*ĤB1
= 9980.4 kJ/s
Q s = ṅA3*ĤA3 + ṅB3*ĤB3 + ṅC3*ĤC3+(ṅA4*0.900 mol
fraction*ĤA4)+(ṅB4*0.100 mol fraction*ĤB4) –
ṅA2*ĤA2 - ṅB2*ĤB2 - ṅC2*ĤC2
= -568.4 kJ/s
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Click on the Hyperlink and click on the
“Final Solution” tab to see the final
answer for the system.
Second Example Final Solution
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Tier III
Open-ended problem
 Approach to open-ended problem
 Case Study.
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• Statement of Intent
– Learn how to approach open-ended design
problems
– Solve a problem on your own
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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.
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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
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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
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• 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
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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
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Process
Description
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• 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
Module 9 – Steady state simulation
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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)
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• 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
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• 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
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• 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
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• 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
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• Boiler generates steam at supercritical conditions
• 540oC and 24.1 MPa absolute
• mechanical work derived by expanding steam through a powergenerating 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
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Assume a basis of 100 kg dry coal/min fed to the
furnace.
1. Construct a flowchart of the process and
completely label the streams. Show the details of
only one train in the scrubber operation. Do this in
Excel.
2. Estimate the molar flow rate (kmol/min) of each
element in the coal (other than those in the ash).
3. Determine the feed rate (kmol/min) of O2
required for complete combustion of the coal.
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4. If 15% excess oxygen is fed to combustion furnace,
estimate the following:
a. The oxygen and nitrogen feed rates (kmol/min)
b. The mole fraction of water in the wet air, the average
molecular weight, and the molar flow rate of water in
the air stream (kmol/min)
c. The air feed rate (kmol/min, m3/min)
5. Estimate flow rate (kmol/min, kg/min) of each component
and composition (mole frac) of furnace flue gas (ignore fly
ash). At what rate (kg/min) is fly ash removed from flue gas
by the electrostatic precipitator?
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6. If system is assumed to meet standard 90% SO2 removal
released upon combustion:
a. Determine flow rate (kg/min and kmol/min) of each
component in the flue gas leaving scrubber
b. Determine flow rate (kg/min) of slurry entering
scrubber
c. Estimate solid-to-liquid mass ratio in slurry leaving
scrubber.
d. Estimate feed rate (kg/min) of fresh ground
limestone to the blending tank.
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6. (continued)
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.
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References:
• Felder, R.F. and Rousseau, R.W. Elementary Principles of Chemical
Processes, Third Edition. New York, John Wiley and Sons, 2000.
• Smith, J.C. and Harriott, Peter. Unit Operations of Chemical Engineering,
Sixth Edition. Boston, McGraw Hill, 2001.
• Earle, R.L. Unit Operations in Food Processing, Second Edition.
http://www.nzifst.org.nz/unitoperations/index.htm
• Thibault, Jules. Notes, CHE 4311: Unit Operations. University of Ottawa,
August 2002.
• Genzer, Jan. Notes, CHE 225: Chemical Process Systems. North Carolina
State University, August 2002.
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• Source on pictures for slides 13, 41,
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