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CHAPTER II PROCESS DYNAMICS AND MATHEMATICAL MODELING

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... stirred tank blending system where two input systems are blended to produce an ... a more general version of the blending system where stream 2 is not pure ... – PowerPoint PPT presentation

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Title: CHAPTER II PROCESS DYNAMICS AND MATHEMATICAL MODELING


1
CHAPTER II PROCESS DYNAMICS AND MATHEMATICAL
MODELING
Process Control
2
  • Dynamic Models are also referred as unsteady
    state models.
  • These models can be used for
  • Improve understanding of the process
  • Train plant operating personnel
  • Develop a control strategy for a new process
  • Optimize process operating conditions.

3
  • Theoretical Models are developed using the
    principles of chemistry, physics and biology.
  • Advantages
  • they provide physical insight into process
    behavior.
  • they are applicable over a wide range of
    conditions.
  • Disadvantages
  • they tend to be time consuming and expensive to
    develop.
  • theoretical models of complex processes include
    some model parameters that are not readily
    available.

4
  • 2. Empirical Models are obtained by fitting
    experimental data.
  • Advantages
  • easier to develop
  • Disadvantages
  • do not extrapolate well
  • the range of data is usually quite small compared
    to the full range of operating conditions.

5
  • 3. Semi-Empirical Models are a combination of
    theoretical and empirical models where the
    numerical values of parameters in a theoretical
    model are calculated from experimental data.
  • Advantages
  • they incorporate theoretical knowledge
  • they can be extrapolated over a large range of
    conditions
  • they require less development of effort than
    theoretical models.

6
A Systematical Approach for Modeling
  • State the modeling objectives and the end use of
    the model. Then determine the required levels of
    model detail and model accuracy.
  • Draw a schematic diagram of the process and label
    all process variables.
  • List all of the assumptions involved in
    developing the model. The model should not be no
    more complicated than necessary to meet the
    modeling objectives.

7
  • 4. Determine if spatial variations are important.
    If so, a partial differential equation model will
    be required.
  • 5. Write appropriate conservation equations
    (mass, component, energy etc)
  • 6. Introduce constitutive equations, which are
    equilibrium relations and other algebraic
    equations
  • 7. Perform a degrees of freedom analysis to
    ensure that the model equations can be solved.

8
  • 8. Simplify the model.
  • output f (inputs)
  • This model form is convenient for computer
    simulation and subsequent analysis.
  • 9. Classify inputs as disturbance variables or as
    manipulated variables.

9
Degrees of freedom
  • To simulate a process, model equations should be
    solvable set of relations. In order for a model
    to have a unique solution, number of degrees of
    freedom should be zero.

Number of degrees of freedom
Number of process variables
Number of independent equations
10
  • For Degrees of Freedom Analysis
  • List all quantities in the model that are known
    constants (or parameters).
  • Determine the number of equations NE and the
    number of process variables NV. Note that time t
    is not considered to be a process variable
    because it is neither a process input nor a
    process output.
  • Calculate the number of degrees of freedom.
  • Identify the NE output variables that will be
    obtained by solving the process model.
  • Identify the NF input variables that must be
    specified as either disturbance variables or
    manipulated variables, in order to utilize the NF
    degrees of freedom.

11
  • Summary
  • if DOF0 The system is exactly specified
  • if DOFlt0 The system is over specified, and in
    general no solution to model exists. The likely
    cause is either (1) improperly designating a
    variable(s) as a parameter or external variable
    or (2) including an extra, dependent equation(s)
    in the model.
  • if DOFgt0 The system is underspecified, and an
    infinite number of solutions to the model exists.
    The likely cause is either (1) improperly
    designating a parameter or external variable as a
    variable or (2) not including in the model all
    equations that determine the systems behavior.

12
Example
  • Consider a continuous stirred tank blending
    system where two input systems are blended to
    produce an outlet stream that has the desired
    composition.
  • Stream 1 is a mixture of two species A and B. We
    assume that its mass flow rate is constant but
    the mass fraction of A (x1) varies with time.
    Stream 2 consists of pure A and thus x21. The
    mass flow rate of stream 2 (w2) can be
    manipulated using a control valve. The mass
    fraction of A in the exit stream is denoted by x
    and the desired value by xsp.

13
X1, w1
X2, w2
X, w
Control Question Suppose that inlet
concentration x1 varies with time. How can we
ensure that the outlet composition x remains at
or near its desired value.
14
  • Method 1.
  • Measure x and adjust w2.
  • if x is high, w2 should be reduced
  • if x is low, w2 should be increased
  • (Feedback Control)

X2, w2
X1, w1
X, w
AT
(Analyzer-Transmitter)
15
  • Method 2
  • Measure x1 and adjust w2.
  • (Feed forward Control)

AT
X2, w2
X1, w1
X, w
16
  • Design Question If the nominal value of x1 is
    x1,s what nominal flow rate w2 is required to
    produce the desired outlet concentration xsp.
  • With a st-st material balance,
  • w1 w2 w 0 ( overall balance)
  • w1x1,s w2x2,s wxsp 0 (component A
    balance)
  • w1x1,sw2(1.0)-(w1w2)xsp0

17
X2, w2
X1, w1
  • Consider a more general version of the blending
    system where stream 2 is not pure and volume of
    the tank may vary with time.
  • (! Not an overflow system any more but a draining
    system)

X, w
Objective is again to keep x at the desired value
18
  • A unsteady state mass balance gives
  • The mass of liquid in the tank can be expressed
    as product of the liquid and the density.

rate of mass in
rate of mass out
rate of accumulation of mass in the tank
19
  • system liquid in the tank
  • assumptions
  • tank is well mixed
  • density of liquid is not changing with
    composition change
  • Total material balance
  • rate of mass in - rate of mass out rate
    of accumulation of mass

20
  • The rate expression in real form is,
  • Dividing by ?t and taking limit as ?t ?0 gives

21
  • Component balance
  • Considering the constant density assumption
    equations become

Equation 1
Equation 2
22
  • Replacing Equation (1) into Equation (2) gives
  • With these two equations system behavior is
    mathematically defined.

23
  • Degrees of Freedom Analysis
  • Parameter(s) ?
  • variables V, x1, w1, x2, w2, x, w
  • equations (dV/dt and dx/dt)
  • D.O.F 7-2 5
  • outputs V, x
  • inputs x1, w1, x2, w2, w
  • disturbances x1, w1, x2
  • manipulated variables w2, w (x vs w2 and V vs
    w as the control structures)

24
Example
  • Goal The dynamic response of temperature of the
    liquid in the tank is to be determined.
  • System The liquid in the tank.
  • Assumptions
  • tank is well mixed
  • Physical properties of the system are not
    changing during the process.

Fi, Ti
F, T
25
  • Total mass balance
  • with constant ? and cross-sectional area A
  • flow rates are given in units of volumetric flow
    rates

26
  • Total energy balance
  • with constant ?, Cp and assuming Tref 0 gives

27
  • D.O.F Analysis
  • Parameter(s) ?,cp
  • variables V, T
  • equations (dh/dt and dT/dt)
  • D.O.F 2-2 0
  • outputs V, T
  • inputs Fi, Ti, Fst
  • disturbances Fi, Ti,
  • manipulated variables no control structures

28
Example
Consider the typical liquid storage process shown
in the figure, where qi and q are volumetric flow
rates. Assuming constant density and cross
sectional area A a mass balance gives
qi

h
q
29
  • There are three important variations in the
    liquid storage processes
  • The inlet or outlet flow rates might be constant.
    In that case the exit flow rate is independent of
    the liquid level over a wide range of conditions.
    Consequently qinqout at the steady state
    conditions.

30
  • The tank exit line may function simply as a
    resistance to flow from the tank or it may
    contain a valve that provides significant
    resistance to flow at a single point. In the
    simplest case, the flow may be assumed to be
    linearly related to the driving force, the
    liquid level.

31
  • A more realistic expression for flow rate q can
    be obtained when a fixed valve has been placed in
    the exit line and turbulent flow can be assumed.
    The driving force for flow through the valve is
    the pressure drop ?P, ?PP-Pa where P is pressure
    at the bottom of the tank and Pa is pressure at
    the end of the exit line.
  • Cv is the valve constant

32
Example

Fi, Ti, CAi
Consider the Continuous Stirred Tank Reactor
(CSTR) in which a simple liquid phase
irreversible chemical reaction takes place. A
B rkCA r rate of reaction k reaction
rate constant, kk0exp(-E/RT) CA molar
concentration
F, T, CA
Cooling medium
33
  • system liquid in the tank
  • assumptions
  • CSTR is perfectly mixed
  • Mass densities of feed and product are equal and
    constant
  • Liquid volume is kept constant by an overflow
    line.
  • The thermal capacitances of the coolant and the
    cooling coil wall are negligible compared to the
    thermal capacitance of the liquid.

34
  • Coolant temperature is constant. (change in the
    tank is negligible)
  • Rate of heat transfer to coolant is given by,
    QUA(Tc-T), where U,A are parameters.
  • Heat of mixing is negligible compared to the heat
    of reaction.
  • Shaft work and heat losses are negligible.

35
  • Total mass balance
  • With constant ? and V, FiF
  • Component balance for species A (in molar units)
  • Energy balance
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