chapter_2 (mathematika models)
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Chapter
2
Mathematical Modeling of
Chemical Processes
Mathematical Model (Eykhoff, 1974)“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, butno simpler.
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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.
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Table 2.1. A Systematic Approach for
Developing Dynamic Models
1. State the modeling objectives and the end use of the model.They determine the required levels of model detail and model
accuracy.
2. Draw a schematic diagram of the process and label all process
variables.3. 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.
4. Determine whether spatial variations of process variables areimportant. If so, a partial differential equation model will be
required.
5. Write appropriate conservation equations (mass, component,
energy, and so forth).
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6. Introduce equilibrium relations and other algebraic
equations (from thermodynamics, transport phenomena,chemical kinetics, equipment geometry, etc.).
7. Perform a degrees of freedom analysis (Section 2.3) to
ensure that the model equations can be solved.
8. 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.
9. Classify inputs as disturbance variables or as manipulated
variables.
Table 2.1. (continued)
Chapter
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Chapter
2
Modeling Approaches Physical/chemical (fundamental, global)
• Model structure by theoretical analysis Material/energy balances Heat, mass, and momentum transfer Thermodynamics, chemical kinetics
Physical property relationships• Model complexity must be determined
(assumptions)• Can be computationally expensive (not real-
time)
• May be expensive/time-consuming to obtain• Good for extrapolation, scale-up• Does not require experimental data to obtain
(data required for validation and fitting)
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• 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)
accumulation in out
= −
Conservation of Component i
rate of component i rate of component i
accumulation in
rate of component i rate of component i(2-7)
out produced
=
− +
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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 heat addition net rate of work
to the system from performed on the sys
the surroundings
tem (2-8)
by the surroundings
The total energy of a thermodynamic system, U tot , is the sum of itsinternal energy, kinetic energy, and potential energy:
int (2-9)tot KE PE U U U U = + +
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Chapter
2
Black box (empirical)
•Large number of unknown parameters
•Can be obtained quickly (e.g., linear regression)
•Model structure is subjective
•Dangerous to extrapolate Semi-empirical
•Compromise of first two approaches
•Model structure may be simpler •Typically 2 to 10 physical parameters
estimated
(nonlinear regression)
•Good versatility, can be extrapolated
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Chapter
2
• linear regression
• nonlinear regression
• number of parameters affects accuracy of model,
but confidence limits on the parameters fitted mustbe evaluated
• objective function for data fitting – minimize sum of squares of errors between data points and modelpredictions (use optimization code to fit
parameters)
• nonlinear models such as neural nets arebecoming popular (automatic modeling)
2
210 xc xcc y ++=
( )τ /1 t e K y −−=
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Chapter
2
Number of sightings of storks
Number of
births (West
Germany)
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
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• 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 (2-1)
of mass in the tank mass in mass out
= −
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• The unsteady-state component balance is:
( )1 1 2 2
ρ(2-3)
d V xw x w x wx
dt = + −
The corresponding steady-state model was derived in Ch. 1 (cf.
Eqs. 1-1 and 1-2).
1 2
1 1 2 2
0 (2-4)
0 (2-5)
w w w
w x w x wx
= + −
= + −
or
where w1, w2, and w are mass flow rates.
( )1 2
ρ(2-2)
d V w w w
dt = + −
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The Blending Process Revisited
For constant , Eqs. 2-2 and 2-3 become: ρ
1 2 (2-12)dV w w wdt
ρ = + −
( )1 1 2 2 (2-13)
d Vxw x w x wx
dt
ρ = + −C
hapter
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Equation 2-13 can be simplified by expanding the accumulation
term using the “chain rule” for differentiation of a product:
( )(2-14)
d Vx dx dV V x
dt dt dt ρ ρ ρ = +
Substitution of (2-14) into (2-13) gives:
1 1 2 2 (2-15)dx dV
V x w x w x wxdt dt
ρ ρ + = + −
Substitution of the mass balance in (2-12) for in (2-15)
gives:
/dV dt ρ
( )1 2 1 1 2 2 (2-16)dx
V x w w w w x w x wxdt
ρ + + − = + −
After canceling common terms and rearranging (2-12) and (2-16),
a more convenient model form is obtained:
( )
( ) ( )
1 2
1 21 2
1(2-17)
(2-18)
dV w w w
dt
w wdx x x x xdt V V
ρ
ρ ρ
= + −
= − + −
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Stirred-Tank Heating Process (cont’d.)
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 outletflow rates are equal.
3. The density and heat capacity C of the liquid are assumed to
be constant. Thus, their temperature dependence is neglected.
4. Heat losses are negligible.
ρ
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For the processes and examples considered in this book, it
is appropriate to make two assumptions:
1. Changes in potential energy and kinetic energy can be
neglected because they are small in comparison with changesin internal energy.
2. The net rate of work can be neglected because it is small
compared to the rates of heat transfer and convection.
For these reasonable assumptions, the energy balance in
Eq. 2-8 can be written as
( )int (2-10)dU
wH Qdt
= −∆ +)
int the internal energy of
the system
enthalpy per unit mass
mass flow rate
rate of heat transfer to the system
U
H
w
Q
=
==
=
)
( )
denotes the difference between outlet and inlet
conditions of the flowing
streams; therefore
-Δ wH = rate of enthalpy of the inlet
stream(s) - the enthalpyof the outlet stream(s)
∆ =
)
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For a pure liquid at low or moderate pressures, the internal energy
is approximately equal to the enthalpy, U int , and H depends
only on temperature. Consequently, in the subsequent
development, we assume that U int = H and where the
caret (^) means per unit mass. As shown in Appendix B, a
differential change in temperature, dT , produces a correspondingchange in the internal energy per unit mass,
H ≈
ˆ ˆint U H =
ˆ ,int dU
intˆ ˆ (2-29)dU dH CdT = =
where C is the constant pressure heat capacity (assumed to be
constant). The total internal energy of the liquid in the tank is:
int int
ˆ(2-30)U VU ρ =
Model Development - I
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An expression for the rate of internal energy accumulation can be
derived from Eqs. (2-29) and (2-30):
int (2-31)dU dT
VC dt dt
ρ =
Note that this term appears in the general energy balance of Eq. 2-
10.
Suppose that the liquid in the tank is at a temperature T and has an
enthalpy, . Integrating Eq. 2-29 from a reference temperature
T ref to T gives,
ˆ H
( )ˆ ˆ
(2-32)ref ref H H C T T − = −where is the value of at T ref . Without loss of generality, we
assume that (see Appendix B). Thus, (2-32) can be
written as:
ˆref H ˆ H
ˆ 0ref H =
( )ˆ (2-33)ref H C T T = −
Model Development - II
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( )ˆ (2-34)i i ref H C T T = −
Substituting (2-33) and (2-34) into the convection term of (2-10)
gives:
( ) ( ) ( )ˆ (2-35)i ref ref wH w C T T w C T T −∆ = − − −
Finally, substitution of (2-31) and (2-35) into (2-10)
( ) (2-36)i
dT V C wC T T Q
dt ρ = − +
Model Development - III
For the inlet stream
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steam-heating: = ∆ s vQ w H
( ) (1)= − + ∆i s v
dT
V C wC T T w H dt ρ
0 ( ) (2)= − + ∆ si vwC T T w H
subtract (2) from (1)
( ) ( )= − + − ∆ s s v
dT V C wC T T w w H
dt ρ
divide by wC
( )∆
= − + −v s s
H V dT T T w w
w dt wC
ρ
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Define deviation variables (from set point)
1
1
is desired operating point( ) from steady state
Vnote that and
w
dynote when 0dt
General linear ordinary differential equation solution:
s s s
v v p
p
p
y T T T u w w w T
H H V dy y u K
w dt wC wC
y K u
dy y K u
dt
ρ ρ τ
τ
= −= −
∆ ∆= − + = =
= =
= − +
1
sum of exponential(s)
Suppose u 1 (unit step response)
( ) 1
t
p y t K e τ −
=
= −
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Example 1:Example 1:
C0=TC,90=TC,40=T o
i
oo
i′
wC(T - T ) = w Hi s v∆6
s
v
o
4
3 3
3
4
w =0.83 10 g hr
H =600 cal g
C=l cal g C
w=10 kg hr =10 kg m
V=20 m
2 10 kgV
ρ
ρ
×
∆
= ×4
4V 2 10 kg 2hr
w 10 kg hr ρ τ ×= = =
dynamic model2dy
dt= -y + 6 10 u-5×
y T T
u w w s s
= −
= −
s.s. balance:
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Step 1: t=0 double ws
final
T(0) = T y(0) = 0
hr g10+0.83=u 6×
2 dydt
= -y + 50
( )y = 50 l - e-0.5t
T = y + T = 50 + 90 = 140 Css
o
Step 2: maintain
Step 3: then set
hr 24/C140=T o
u = 0, w = 833 kg hr s
2
dy
dt = -y + 6 10 u, y(0) = 50
-5
×Solve for u = 0
y = 50e ty 0
-0.5t → ∞→ (self-regulating, but slow)
how long to reach y = 0.5 ?
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Step 4: How can we speed up the return from 140°C to 90°C?
ws = 0 vs. ws = 0.83× 106
g/hr at s.s ws =0
y →-50°C T →40°C
Process DynamicsProcess control is inherently concerned with unsteady
state behavior (i.e., "transient response", "process
dynamics")
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Stirred tank heater : assume a "lag" between heating
element temperature Te, and process fluid temp, T.
heat transfer limitation = heA(Te – T)
Energy balances
Tank:
Chest:
At s.s.
Specify Q →calc. T, Te
2 first order equations ⇒1 second order equation in T
Relate T to Q (Te is an intermediate variable)
i e e
dTwCT +h A(T -T)-wCT=mC
dt
dt
dTCm=T)-A(Th-Q e
eeee
0dt
dT 0,
dt
dT e ==
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fixed T Q-Q=u T-T=y i
Note Ce →0 yields 1st order ODE (simpler model)
Fig. 2.2
hR
1=q
v
Rv: line resistance
R v
uwC
1
ydt
dy
w
m
wC
Cm
Ah
Cm
dt
yd
Awh
Cmm ee
ee
ee
2
2
ee
ee
=+
+++
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linear ODE
If
nonlinear ODE
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57)-(2 1
h
R
q
dt
dh A
v
i −=
*(2-61)i v i v
dh A q Cρgh q C h
dt
= − = −
gh p P ρ +=
pressureambient:P P-P=q aa
*
vC ghP ρ += p
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Table 2.2. 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 equipmentdimensions, known physical properties, etc.
2. Determine the number of equations N E and the number of
process variables, N V . 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, N F = N V - N E .
4. Identify the N E output variables that will be obtained by solving
the process model.
5. Identify the N F input variables that must be specified as either
disturbance variables or manipulated variables, in order to
utilize the N F degrees of freedom.
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D f F d A l i f th Sti d T k
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Thus the degrees of freedom are N F = 4 – 1 = 3. The process
variables are classified as:
1 output variable: T
3 input variables: T i , w, Q
For temperature control purposes, it is reasonable to classify the
three inputs as:
2 disturbance variables: T i , w
1 manipulated variable: Q
Degrees of Freedom Analysis for the Stirred-Tank
Model:
3 parameters:
4 variables:
1 equation: Eq. 2-36
, ,V C ρ
, , ,iT T w Q
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Biological Reactions
• Biological reactions that involve micro-organisms and enzyme catalysts
are pervasive and play a crucial role in the natural world.
• Without such bioreactions, plant and animal life, as we know it, simply
could not exist.
• Bioreactions also provide the basis for production of a wide variety of
pharmaceuticals and healthcare and food products.
• Important industrial processes that involve bioreactions include
fermentation and wastewater treatment.
• Chemical engineers are heavily involved with biochemical and
biomedical processes.
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Bioreactions
• Are typically performed in a batch or fed-batch reactor.
• Fed-batch is a synonym for semi-batch.
• Fed-batch reactors are widely used in the pharmaceutical
and other process industries.
• Bioreactions:
• Yield Coefficients:
substrate more cells + products (2-90)cells
→
/ (2-92) P S
mass of product formed Y
mass of substrate consumed to form product
=
/ (2-91) X S mass of new cells formed Y mass of substrate consumed to form new cells
=
d h i
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Fed-Batch Bioreactor
Figure 2.11. Fed-batch reactor
for a bioreaction.
Monod Equation
Specific Growth Rate
(2-93) g r X µ =
max (2-94) s
S
K S µ µ =
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1. The exponential cell growth stage is of interest.
2. The fed-batch reactor is perfectly mixed.3. Heat effects are small so that isothermal reactor operation can
be assumed.
4. The liquid density is constant.
5. The broth in the bioreactor consists of liquid plus solidmaterial, the mass of cells. This heterogenous mixture can be
approximated as a homogenous liquid.
6. The rate of cell growth r g is given by the Monod equation in (2-
93) and (2-94).
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( )(2-98) g
d XV V r
dt
=
1 1(2-100) f g P
X / S P / S
d( SV ) F S V r V r
dt Y Y −= −
( )(2-101)
d V F
dt =
( )(2-99) p
d PV Vr
dt =
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