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CHEE 321 CHEMICAL REACTION ENGINEERING

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at t=0, CA = CA,initial = 2 mol/L. Reaction: A B. Rate Kinetics: (-rA) ... Applying the initial condition (at t=0, CA= CA, initial) to the above equation, we get ... – PowerPoint PPT presentation

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Title: CHEE 321 CHEMICAL REACTION ENGINEERING


1
CHEE 222 CHEMICAL PROCESS DYNAMICS AND NUMERICAL
METHODS Module 3 Dynamic Lumped Parameter
System
2
Topics to be covered in this module
  • Origin of Ordinary Differential Equations
  • Transient balance for lumped parameter systems
  • Steady state balance for distributed parameter
    system
  • Numerical Methods
  • Numerical Integration of Ordinary Differential
    Equation
  • Eulers Method
  • Explicit and Implicit Methods Numerical
    Stability of Explicit Method
  • Runge-Kutta Method
  • Second- and Fourth-Order Runge-Kutta Methods
  • Computer-Based ODE Solution Techniques
  • Matlab Integration Routines
  • Ode23 and Ode45Study of transient behavior
  • Process Dynamics
  • Case study CSTR with first-order reaction
    kinetics

3
Types of System and Resulting Equations
Distributed Parameter System
Lumped Parameter System
Dynamic
Dynamic
Steady State
Steady State
X
Algebraic Equations
Differential Equations
Differential Equations
Partial Differential Equations
Linear
Non-Linear
Single Variable
Multi- Variable
4
Transient Lumped Parameter System
  • For a constant density (r) fluid of density being
    flowed in and out of a storage tank of uniform
    cross-section area (A), the transient mass
    balance (derived earlier in the class) takes the
    following form

h
Note The density parameter does not appear
because it cancels out
5
Steady State Distributed Parameter System
  • Consider viscous flow in a pipe.
  • The differential equation to describe the local
    pressure drop takes the following form (youll
    learn more in CHEE 223)

D
6
Numerical Methods for SolvingOrdinary
Differential Equations
7
Eulers Method
8
Numerical Solutions for ODEs The Euler Method
  • Let us consider a single-variable ODE by the
    following equation
  • with the initial
    condition
  • Objective To find the solution x(t).
  • Strategy Approximate or over a
    finite time length or
  • time-step (Dt)
  • The basic idea is to approximate by a
    difference scheme
  • The value of expression f(x) is evaluated and is
    then used to estimate the value of x (tDt)

Graphical illustration will be provided in class
9
Explicit Euler Methods
  • In this method, the value of f(x) is evaluated at
    known value of x or at xk. The term dx/dt is
    represented by the following difference scheme
  • The derivative is then equated to f(x)
  • The value of x at ttDt is then calculated by
    the following equation
  • Note that the RHS of the above equation contains
    all known values, therefore, the equation can be
    solved explicitly

10
Eulers IMPLICIT Method
11
Implicit Euler Methods
  • This method differs from the implicit method in
    that the value of f(x) is evaluated at unknown
    value of x or at xk1. The dx/dt is represented
    by a difference scheme
  • The derivative is equated to f(xk1) and we get
  • The equation can be rearranged as follows
  • Clearly, the equation is implicit in the unknown
    xk1 . Depending on the non-linearity of f(x), an
    iterative solution may be required

12
Stability of Explicit Method
13
Explicit Eulers Method Stability
  • Tutorial5 problem illustrates that explicit
    Eulers Method may result in a solution that
    generates oscillatory solution.
  • The oscillatory behavior is due to numerical
    approximation.
  • The solution can also be unstable.
  • For Eulers explicit method, both the instability
    and oscillation are the result of step-size
    chosen for integration.
  • From mathematical theory, the conditions for
    stability can be found. Here, we provide you with
    that condition (the derivation is beyond the
    scope of this course).

14
Explicit Eulers Method Stability
  • The general form of solution for Explicit Euler
    Method is as follows
  • Let us denote the RHS of the above equation as a
    function g(xk). The requirement for stable
    solution is as follows
  • where, g?(xk) is the derivative of g(xk) with
    respect to xk
  • In lecture, we will apply this criterion to the
    Tutorial6 problem

15
Eulers Method Closing Remarks
  • Eulers method is rarely used in advanced
    computational routines. Nonetheless, it is very
    useful in demonstrating the general concept on
    numerical integration
  • Explicit Euler Method is easy to implement due to
    explicit nature of the equation but numerical
    stability can be a problem
  • Stability problems can be overcome with an
    appropriate choice of step-size.
  • Smaller steps sizes are always desirable because
    they lead to increased accuracy. However, small
    step-size come at the cost of increased
    computation time.
  • Implicit methods are inherently stable.
  • Implicit Euler Method for non-linear equations
    require an iterative solution such as Newtons
    method.
  • Finally, stable does not mean accurate.

16
Runge-Kutta Method
17
Runge-Kutta Methods
  • Runge-Kutta methods are variations of explicit
    methods.
  • Runge-Kutta methods are used extensively in
    computational packages.
  • MATLAB uses 2nd and 4th order methods ode23
    and ode45 solvers
  • The basic idea is to estimate an improved guess
    for the derivative and, in turn, a better
    estimate for the integrated variable.

18
2nd Order Runge-Kutta (RK2) Method
  • Objective To numerically integrate the following
    ordinary differential equation.
  • Integration Procedure
  • Apply Eulers method to estimate the slope or
    derivative at xk
  • Estimate the value of state variable at a
    mid-point, i.e. x(tDt/2).
  • Calculate the slope or derivative at this
    mid-point
  • Estimate the value of state variable at next time
    step, i.e. xk1 using slope at mid-point
  • Illustrative examples to be discussed in lecture

19
4th Order Runge-Kutta Method (RK4 Method)
  • Integration Procedure
  • Apply Eulers method to estimate the slope (m1)
    or derivative at xk
  • Estimate the value of state variable at a
    mid-point, i.e. x(tDt/2).
  • Calculate the slope or derivative (m2) at this
    mid-point to estimate new mid-point state
    variable x?k1/2
  • Find corrected mid-point slope (m3) at this new
    mid-point x?k1/2
  • A final slope (m4)is evaluated at the end of the
    step

20
4th Order Runge-Kutta Method (Cont.)
  • The new state variable at the integration step is
    calculated by a weighted average of three slopes

21
Multi-variable RK method
22
RK Methods Closing Remarks
  • 2nd order Runge-Kutta Method is more accurate
    than Eulers explicit method
  • 4th order Runge-Kutta Method is preferred when
    higher accuracy is required.
  • Selection of step-size is crucial for accuracy
    and usually class problems are solved using a
    fixed step-size. However, Most of the integration
    routines use a variable step-size.

23
Numerical Integration in Matlab
  • So far, we have learned about Runge-Kutta methods
    for integration of ordinary differential
    equation.
  • Runge-Kutta methods are used extensively in
    computational packages.
  • MATLAB uses 2nd and 4th order methods ode23
    and ode45 solvers

24
Study of Transient Behavior of a Chemical
ProcessCSTR with 1st-order Rate Kinetics
25
CSTR Steady State and Transient Behavior
  • CSTR Behavior and Transience
  • CSTRs are designed to operate at steady-state
    and, as such, transient effects arise from the
    following situations
  • start-ups or shut-down of reactor
  • upsets or disturbances (fluctuations) of state
    and/or design variables during operation
  • transience from one steady-state to the other.
  • Role of a process engineer operating a CSTR.
  • The primary interest is to analyze the steady
    state conditions. As well, to predict how the
    system behaves during transient period.
  • How do we undertake this study of steady state
    and transient behavior?
  • Derivation of Total Mass Balance
  • Derivation of Mole Balance on Reactant A
  • Calculation of Steady-State Concentration
  • Determination of analytical solution for the
    problem, i.e. C(t) ?

26
Case Study CSTR Transient Behavior
27
Study of SS Transient Behavior of a CSTR
  • Objective Apply differential mass and mole
    balance to a CSTR to study transient effects,
    i.e. to investigate the process dynamics of the
    system.
  • Problem description Consider the following CSTR
    system wherein a chemical reaction following
    first-order rate kinetics is being carried out.

Initial Condition at t0, CA CA,initial 2
mol/L
Reaction A ? B Rate Kinetics (-rA) k?CA
where, k0.005 s-1
V 10 (L)
28
Fundamental Balance Equations for a CSTR
  • 1. Differential Mass Balance Equation
  • Accumulation Input Rate Output Rate
  • ? for constant density system (e.g. liquids)
  • 2. Differential Mole Balance on Reactant A
  • Accumulation Input Rate Output Rate Gen.
    Rate Cons.Rate

Rate of consumption is given by a constitutive
relationship reaction kinetics. Reaction
kinetics describes how reaction rate is related
to CA via a reaction rate constant
29
Fundamental Balance Equations for a CSTR (cont.)
  • 2. Differential Mole Balance on Reactant A
    (Cont.)
  • For constant volume system, the following
    equation
  • reduces to
  • and can be rearranged as follows
  • Introducing a new variable (tp,) below, the
    above equation can be re-written as

30
Steady State Behavior of the CSTR
  • 3. Calculation of the Steady-State Concentration
  • The steady-state (SS) concentration (CA,s) can be
    calculated by setting dCA/dt term equal to zero
    in the following equation
  • Thus, we have
  • ?
  • tp 1/(0.15/10 .005)-1 seconds 50 s
  • CA,s 50 s ? (.15/10) s-1? 10 mol/L 7.5 mol/L

31
Transient Behavior of the CSTR
  • 4. Analytical Solution for Transient Mole Balance
    Equation
  • The mole balance equation below represents a
    first-order ordinary differential equation.
  • ?
  • The equation can be re-written as follows
  • ?
  • The above equation can be solved to yield CA (t)
    by simple integration
  • Applying the initial condition (at t0, CA CA,
    initial) to the above equation, we get

32
Transient Behavior of the CSTR
  • The solution can therefore can be derived as
    follows
  • Dividing by -tp
  • Rearranging, we get
  • Finally,

DCA
33
Case Study Graphical Representation of the
Solution
0.865 DCA
0.993 DCA
0.633 DCA
DCA
t 2tp
t 5tp
t tp
Can you see the relevance and utility of tp ?
34
Where do we go from here ?
35
Transient Effects of a CSTR
  • Let us explore the dynamic behavior of the CSTR
    subjected to the following WHAT IF scenarios
  • What if the initial concentration was varied.
  • What if there was upset in feed concentration
    after SS operation was established.
  • What if there was upset in feed volumetric rate
    after SS operation was established.
  • The above scenarios will be explored via MATLAB
    simulations (to be demonstrated in class)

This exercise aims to demonstrate application of
computer-aided balance equations.
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