Transcript Mathematical Modeling of Chemical Processes Mathematical Model “a representation of the essential aspects of an existing system (or a system to be constructed) which.

Mathematical Modeling of
Chemical Processes
Mathematical Model
“a representation of the essential aspects of an
existing system (or a system to be constructed)
which represents knowledge of that system in a
usable form”
Everything should be made as simple as
possible, but no simpler.
Uses of Mathematical Modeling
to improve understanding of the process
to optimize process design/operating conditions
to design a control strategy for the process
to train operating personnel
General Modeling Principles
The model equations are at best an approximation to the
real process.
Adage: “All models are wrong, but some are useful.”
Modeling inherently involves a compromise between
model accuracy and complexity on one hand, and the
cost and effort required to develop the model, on the
other hand.
Process modeling is both an art and a science. Creativity
is required to make simplifying assumptions that result in
an appropriate model.
Dynamic models of chemical processes consist of
ordinary differential equations (ODE) and/or partial
differential equations (PDE), plus related algebraic
equations.
A Systematic Approach for Developing
Dynamic Models
1.
2.
3.
4.
5.
State the modeling objectives and the end use of the
model. They 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 that are involved in
developing the model. Try for parsimony; the model
should be no more complicated than necessary to
meet the modeling objectives.
Determine whether spatial variations of process
variables are important. If so, a partial differential
equation model will be required.
Write appropriate conservation equations (mass,
component, energy, and so forth).
A Systematic Approach for Developing
Dynamic Models
6.
7.
8.
9.
Introduce equilibrium relations and other algebraic
equations (from thermodynamics, transport
phenomena, chemical kinetics, equipment geometry,
etc.).
Perform a degrees of freedom analysis to ensure that
the model equations can be solved.
Simplify the model. It is often possible to arrange the
equations so that the dependent variables (outputs)
appear on the left side and the independent variables
(inputs) appear on the right side. This model form is
convenient for computer simulation and subsequent
analysis.
Classify inputs as disturbance variables or as
manipulated variables.
Conservation Laws
Theoretical models of chemical processes are based on
conservation laws.
Conservation of Mass
 rate of mass  rate of mass  rate of mass 



 (2-6)
in
out
accumulation  
 

Conservation of Component i
rate of component i  rate of component i 



in
 accumulation  

rate of component i  rate of component i 



out
produced

 

(2-7)
Conservation of Energy
The general law of energy conservation is also called the
First Law of Thermodynamics. It can be expressed as:
rate of energy  rate of energy in  rate of energy out 




accumulation
by
convection
by
convection

 
 

net rate of work
net rate of heat addition  


 

  to the system from   performed on the system 
 the surroundings
  by the surroundings 

 

(2-8)
The total energy of a thermodynamic system, Utot, is the
sum of its internal energy, kinetic energy, and potential
energy:
U tot  U int  U KE  U PE
(2-9)
Example
Simple tank Problem
Chapter 2
Degrees of Freedom Analysis
1. List all quantities in the model that are known constants
(or parameters that can be specified) on the basis of
equipment dimensions, known physical properties, etc.
2. 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.
3. Calculate the number of degrees of freedom, NF = NV NE.
4. Identify the NE output variables that will be obtained by
solving the process model.
5. 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.
Chapter 2
Stirred-Tank Heating Process
Stirred-tank heating process with constant holdup, V.
Stirred-Tank Heating Process (cont’d.)
Chapter 2
Assumptions:
1. Perfect mixing; thus, the exit temperature T is also the
temperature of the tank contents.
2. The liquid holdup V is constant because the inlet and
outlet flow rates are equal.
3. The density r and heat capacity C of the liquid are
assumed to be constant. Thus, their temperature
dependence is neglected.
4. Heat losses are negligible.
Degrees of Freedom Analysis for the Stirred-Tank
Model:
3 parameters:
V , r,C
T , Ti , w, Q
dT
1 equation:
V rC
 wC Ti  T   Q
dt
Thus the degrees of freedom are NF = 4 – 1 = 3. The process
variables are classified as:
Chapter 2
4 variables:
1 output variable:
T
3 input variables:
Ti, w, Q
For temperature control purposes, it is reasonable to classify the
three inputs as:
2 disturbance variables:
Ti, w
1 manipulated variable:
Q
Degrees of Freedom Analysis
Degrees of Freedom Analysis
System comprises of only 2 chemical species A
and B
Can write only 2 independent mass balances
write for species A and species B
write overall balance & one component balance
(either for species A or B)
Degrees of Freedom Analysis
Degrees of Freedom Analysis
Focus on the control volume (A ∆z) over the time interval t to t + ∆t
Degrees of Freedom Analysis
Dimensional Analysis
A conceptual tool often applied to understand physical
situations involving a mix of different kinds of physical
quantities.
It is routinely used by physical scientists and engineers
to check the plausibility of derived equations.
Only like dimensioned quantities may be added,
subtracted, compared, or equated.
When unlike dimensioned quantities appear opposite of
the "+" or "−" or "=" sign, that physical equation is not
plausible, which might prompt one to correct errors
before proceeding to use it.
When like dimensioned quantities or unlike dimensioned
quantities are multiplied or divided, their dimensions are
likewise multiplied or divided.
Dimensional Analysis
Dimensions of a physical quantity is associated
with symbols, such as M, L, T which represent
mass, length and time
Assume to determine the power required to
drive a house fan. Torque is chosen as the
dependent variable and the following are known
physical variables
Fan diameter (d)
Fan design (R)
Air density (r)
Rotative speed (n)
Dimensions
Quantity
Symbol
M
L
T
Torque
t
1
2
-2
Fan diameter
D
0
1
0
Fan design
R
0
0
0
Air density
r
1
-3
0
Relative speed
n
0
0
-1
Dividing torque by density gives
Symbol
M
L
T
t/r
0
5
-2
D
0
1
0
n
0
0
-1
t/r divided by D5n2 gives:
 t 
 rD 5 n 2   1


Final analysis
 t


f
, R   0
5 2
 rD n

The torque for a given design R is proportional
to the dimensionless product
 t 
 rD5 n 2 


Buckingham p theorem
every physically meaningful equation involving n
variables can be equivalently rewritten as an
equation of n – m dimensionless parameters,
where m is the number of fundamental
dimensions used
it provides a method for computing these
dimensionless parameters from the given
variables, even if the form of the equation is still
unknown
Buckingham p theorem
In mathematical terms, if we have a physically
meaningful equation such as
where the qi are the n physical variables, and they are
expressed in terms of k independent physical units, then
the above equation can be restated as
where the πi are dimensionless parameters constructed
from the qi by p = n − k equations of the form
where the exponents mi are constants.
Example
If a moving fluid meets an object, it exerts a
force on the object, according to a complicated
(and not completely understood) law. We might
suppose that the variables involved under some
conditions to be the speed, density and viscosity
of the fluid, the size of the body (expressed in
terms of its frontal area A), and the drag force.
Example
Buckingham p theorem states that there
will be two such groups
Development of Dynamic Models
Illustrative Example: A Blending Process
An unsteady-state mass balance for the blending system:
rate of accumulation   rate of   rate of 




 of mass in the tank  mass in  mass out 
(2-1)
or
d Vρ 
dt
 w1  w2  w
(2-2)
where w1, w2, and w are mass flow rates.
The unsteady-state component balance is:
d Vρx 
dt
 w1x1  w2 x2  wx
(2-3)
The corresponding steady-state model was derived in Ch. 1 (cf.
Eqs. 1-1 and 1-2).
0  w1  w2  w
(2-4)
0  w1x1  w2 x2  wx
(2-5)
The Blending Process Revisited
For constant r , Eqs. 2-2 and 2-3 become:
dV
r
 w1  w2  w
dt
r d Vx 
dt
 w1x1  w2 x2  wx
(2-12)
(2-13)
Equation 2-13 can be simplified by expanding the accumulation
term using the “chain rule” for differentiation of a product:
d Vx 
dx
dV
 rx
(2-14)
dt
dt
dt
Substitution of (2-14) into (2-13) gives:
dx
dV
rV  r x
 w1x1  w2 x2  wx
(2-15)
dt
dt
Substitution of the mass balance in (2-12) for r dV/dt in (2-15)
gives:
dx
rV  x  w1  w2  w   w1x1  w2 x2  wx
(2-16)
dt
After canceling common terms and rearranging (2-12) and (2-16),
a more convenient model form is obtained:
dV 1
  w1  w2  w 
(2-17)
dt r
w2
dx w1

(2-18)
 x1  x    x2  x 
dt V r
Vr
r
 rV