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Transcript of MMCE Chap1 Mm
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Mathematical Modeling of Chemical Processes
Mathematical Model:
“a representation of the essential aspects of an existing system (or a system to beconstructed) which represents knowledge of that system in a usable form”
Everything should be made as simple as possible, but no simpler.
General Modeling Principles
1. The model equations are at best an approximation to the real process.
2. Adage: “All models are wrong, but some are useful.”3. Modeling inherently involves a compromise between model accuracy and
complexity on one hand, and the cost and effort required to develop the model, onthe other hand.
4.
Process modeling is both an art and a science. Creativity is required to makesimplifying assumptions that result in an appropriate model.
5. 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. 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 themodeling objectives.
4. Determine whether spatial variations of process variables are important. If so, a partial differential equation model will be required.
5. Write appropriate conservation equations (mass, component, energy, and soforth).
6. Introduce equilibrium relations and other algebraic equations (fromthermodynamics, 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 independentvariables (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.
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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)• Conservation Laws
Theoretical models of chemical processes are based on conservation laws.
Conservation of Mass
Conservation of Component i
Conservation of Energy
The general law of energy conservation is also called the First Law of
Thermodynamics. It can be expressed as:
rate of mass rate of mass rate of mass(2-6)
accumulation in out
! " ! " ! "= #$ % $ % $ %
& ' & ' & '
rate of component i rate of component i
accumulation in
rate of component i rate of component i(2-7)
out produced
! " ! "=
# $ # $% & % &
! " ! "' +# $ # $
% & % &
! " ! " ! "= #$ % $ % $ %
& ' & ' & '
! "( (
+ +$ %( (& '
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
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The total energy of a thermodynamic system, U tot , is the sum of its internal energy,
kinetic energy, and potential energy:
! 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• Can be run in real-time
• linear regression• nonlinear regression
• number of parameters affects accuracy of model, but confidence limits onthe parameters fitted must be evaluated
• objective function for data fitting – minimize sum of squares of errors between data points and model predictions (use optimization code to fit
parameters)• nonlinear models such as neural nets are becoming popular (automatic
modeling)
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
int (2-9)tot KE PE
U U U U = + +
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1. Blending Process
An unsteady-state mass balance for the blending system:
or
where w1, w2, and w are mass flow rates.
• The unsteady-state component balance is:
At steady state.
The Blending Process Revisited
For constant !, Esq. 2-2 and 2-3 become:
rate of accumulation rate of rate of (2-1)
of mass in the tank mass in mass out
! " ! " ! "= #$ % $ % $ %
& ' & ' & '
( )1 2
!(2-2)
d V w w w
dt = + !
( )1 1 2 2
!(2-3)
d V xw x w x wx
dt = + !
1 2
1 1 2 2
0 (2-4)
0 (2-5)
w w w
w x w x wx
= + !
= + !
1 2 (2-12)dV
w w wdt
! = + "
( )1 1 2 2 (2-13)
d Vxw x w x wx
dt
! = + "
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Equation 2-13 can be simplified by expanding the accumulation term using the “chainrule” for differentiation of a product:
Substitution of (2-14) into (2-13) gives:
Substitution of the mass balance in (2-12) for /dV dt ! in (2-15) gives:
After canceling common terms and rearranging (2-12) and (2-16), a more convenient
model form is obtained:
2. Stirred Tank Heater
We consider the liquid tank of the last example but at non-isothermal conditions.
The liquid enters the tank with a flow rate F f (m3 /s), density ! f (kg/m3) and temperature T f
(K). It is heated with an external heat supply of temperature T st (K), assumed constant.
The effluent stream is of flow rate F o (m3 /s), density !o (kg/m
3) and temperature T (K)
(Fig. 2.2). Our objective is to model both the variation of liquid level and its temperature.
As in the previous example we carry out a macroscopic model over the whole system.
Assuming that the variations of temperature are not as large as to affect the density then
the mass balance of Eq. 2.8 remains valid.
To describe the variations of the temperature we need to write an energy balance
equation. In the following we develop the energy balance for any macroscopic system
(Fig. 2.3) and then we apply it to our example of stirred tank heater.
The energy E ( J ) of any system of (Fig. 2.3) is the sum of its internal U ( J ), kinetic K ( J )
and potential energy "(J):
( )(2-14)
d Vx dx dV V x
dt dt dt ! ! ! = +
1 1 2 2 (2-15)dx dV
V x w x w x wxdt dt
! ! + = + "
( )1 2 1 1 2 2 (2-16)dx
V x w w w w x w x wxdt
! + + " = + "
( )
( ) ( )
1 2
1 21 2
1(2-17)
(2-18)
dV w w w
dt
w wdx x x x x
dt V V
!
! !
= + "
= " + "
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E = U + K + " (2.14)
Consequently, the flow of energy into the system is:
! f F f ( f f f K U !++~
~~
) (2.15)
where the ( •~ ) denotes the specific energy ( J/kg ).
Q
Heat supply
T st
F f , T f ,9 f
L
F o , T o ,9o
Figure 2-1 Stirred Tank Heater
System
F1
Inlets
F2
Fn
outlets
F1
F2
F3
Ws
Q
Figure 2-2 General Macroscopic System
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The flow of energy out of the system is:
!o F o( ooo K U !++~
~~
) (2.16)
The rate of accumulation of energy is:
dt
K U V d )~
~~( !++"
(2.17)
As for the rate of generation of energy, it was mentioned in Section 1.8.4, that the
energy exchanged between the system and the surroundings may include heat of reaction
Qr ( J/s), heat exchanged with surroundings Qe ( J/s) and the rate of work done against
pressure forces (flow work) W pv ( J/s), in addition to any other work W o.
The flow of work W pv done by the system is given by:
f f oo pv P F P F W != (2.18)
where P o and P f are the inlet and outlet pressure, respectively.
In this case, the rate of energy generation is:
)( f f ooor e P F P F W QQ !+!+ (2.19)
Substituting all these terms in the general balance equation (Eq. 1.7) yields:
( ) ( )
)(
~
~~
~
~~
)~
~~(
f f ooor e
ooooo f f f f f
P F P F W QQ
K U F K U F dt
K U V d
!+!++
"++#!"++#="++#
(2.20)
We can check that all terms of this equation have the SI unit of (J/s). Equation (2.20) can
be also written as:
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( ) ( )
f
f
f f o
o
ooor e
ooooo f f f f f
P F
P F W QQ
K U F K U F dt
K U V d
!!+
!!""++
#++!"#++!=#++! ~
~~
~
~~
)~
~~(
(2.21)
The term != /1~V is the specific volume (m3/kg ). Thus Eq. 2.21 can be written as:
( ) ( )
or e
ooooooo f f f f f f f
W QQ
K V P U F K V P U F dt
K U V d
!++
+++!+++=++
" # " # " # ~
~~~
~
~~~
)~
~~(
(2.22)
The term V P U ~~
+ that appears in the equation is the specific enthalpy h~
. Therefore, the
general energy balance equation for a macroscopic system can be written as:
( ) ( ) ( ) or eooooo f f f f f W QQ K h F K h F dt
K U V d !++"++#!"++#=
"++# ~
~~
~
~~
)~
~~(
(2.23)
We return now to the liquid stirred tank heater. A number of simplifying assumptions can be introduced:
• We can neglect kinetic energy unless the flow velocities are high.
• We can neglect the potential energy unless the flow difference between the inlet
and outlet elevation is large.
• All the work other than flow work is neglected, i.e. W o = 0.
• There is no reaction involved, i.e. Qr = 0.
The energy balance (Eq. 2.23) is reduced to:
eooo f f f Qh F h F dt
U V d +!"!=
! ~
~
~
(2.24)
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Here Qe is the heat ( J/s) supplied by the external source. Furthermore, as mentioned in
Section 1.12, the internal energy U ~
for liquids can be approximated by enthalpy, h~
. The
enthalpy is generally a function of temperature, pressure and composition. However, it
can be safely estimated from heat capacity relations as follows:
)(~~
ref T T pC h != (2.25)
where pC ~
is the average heat capacity.
Furthermore since the tank is well mixed the effluent temperature T o is equal to process
temperature T . The energy balance equation can be written, assuming constant density ! f
= !o = ! , as follows:
eref poref f p f ref
p QT T C F T T C F dt
T T V d C +!"!!"=
!" )(
~ )(
~
)(~
(2.26)
Taking T ref = 0 for simplicity and since V = AL result in:
( )e po f p f p QT C F T C F
dt
LT d AC +!"!=!
~
~
~
(2.27)
or equivalently:
( )
p
eo f f
C
QT F T F
dt
LT d A
~
!+"=
(2.28)
Since
( ) ( ) ( )dt
T d ALdt
Ld AT dt
LT d A +=
(2.29)
and using the mass balance (Eq. 2.8) we get:
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p
eo f f o f
C
QT F T F F F T
dt
dT AL ~)(
! +"="+
(2.30)
or equivalently:
p
e f f
C
QT T F
dt
dT AL ~)(
!+"=
(2.31)
The stirred tank heater is modeled, then by the following coupled ODE's:
o f F F
dt
dL A !=
(2.32)
p
e f f
C
QT T F
dt
dT AL ~)(
!+"=
(2.33)
This system of ODE's can be solved if it is exactly specified and if conditions at initial
time are known,
L(t i) = Li and T (t i) = T i (2.34) Degree of freedoms analysis
For this system we can make the following simple analysis:
• Parameter of constant values: A, ! and C p
• (Forced variable): F f and T f
• Remaining variables: L, F o, T , Qe
• Number of equations: 2 (Eq. 2.32 and Eq. 2.33)
The degree of freedom is therefore, 4 " 2 = 2. We still need two relations for our problem
to be exactly specified. Similarly to the previous example, if the system is operated
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without control then F o is related to L through (Eq. 2.10). One additional relation is
obtained from the heat transfer relation that specifies the amount of heat supplied:
Qe = UA H (T st "T ) (2.35)
U and A H are heat transfer coefficient and heat transfer area. The source temperature T st
was assumed to be known. If on the other hand both the height and temperature are under
control, i.e. kept constant at desired values of L s and T s then there are two control laws
that relate respectively F o to L and L s and Qe to T and T s:
F o = F o( L, L s), and Qe = Qe (T, T s) (2.36)
3. Isothermal CSTR
We revisit the perfectly mixed tank of the first example but where a liquid phase
chemical reactions taking place:
B A k !" ! (2.37)
The reaction is assumed to be irreversible and of first order. As shown in figure 2.4, the
feed enters the reactor with volumetric rate F f (m3/ s), density ! f (kg/m
3) and concentration
C Af (mole/m3). The output comes out of the reactor at volumetric rate F o, density !0 and
concentration C Ao (mole/m3) and C Bo (mole/m
3). We assume isothermal conditions.
Our objective is to develop a model for the variation of the volume of the reactor
and the concentration of species A and B. The assumptions of example 2.1.1 still hold andthe total mass balance equation (Eq. 2.6) is therefore unchanged
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V
F f
9 f
C Af
C Bf
F o9o
C Ao
C Bo
Figure 2.4 Isothermal CSTR
The component balance on species A is obtained by the application of (Eq. 1.3) to
the number of moles (n A = C AV ). Since the system is well mixed the effluent
concentration C Ao and C Bo are equal to the process concentration C A and C B.
Flow of moles of A in:
F f C Af (2.38)
Flow of moles of A out:
F o C Ao (2.39)
Rate of accumulation:
dt
VC d
dt
dn A)(=
(2.40)
Rate of generation: -rV
where r (moles/m3 s) is the rate of reaction.
Substituting these terms in the general equation (Eq. 1.3) yields:
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rV C F C F dt
VC d Ao Af f
A!!=
)(
(2.41)
We can check that all terms in the equation have the unit (mole/s).
We could write a similar component balance on species B but it is not needed
since it will not represent an independent equation. In fact, as a general rule, a system of
n species is exactly specified by n independent equations. We can write either the total
mass balance along with (n "1) component balance equations, or we can write n
component balance equations.
Using the differential principles, equation (2.41) can be written as follows:
rV C F C F dt
V d C
dt
C d V
dt
VC d Ao Af f A
A A!!=+=
)()()(
(2.42)
Substituting Equation (2.6) into (2.42) and with some algebraic manipulations we obtain:
rV C C F dt
C d
V A Af f
A!!=
)(
)(
(2.43)
In order to fully define the model, we need to define the reaction rate which is for a first-
order irreversible reaction:
r = k C A (2.44)
Equations 2.6 and 2.43 define the dynamic behavior of the reactor. They can be solved ifthe system is exactly specified and if the initial conditions are given:
V (t i) = V i and C A(t i) = C Ai (2.45)
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Degrees of freedom analysis
• Parameter of constant values: A
• (Forced variable): F f and C Af
• Remaining variables: V , F o, and C A
• Number of equations: 2 (Eq. 2.6 and Eq. 2.43)
The degree of freedom is therefore 3 " 2 =1. The extra relation is obtained by the relation
between the effluent flow F o and the level in open loop operation (Eq. 2.10) or in closed
loop operation (Eq. 2.11).
The steady state behavior can be simply obtained by setting the accumulation terms to
zero. Equation 2.6 and 2.43 become:
F 0 = F f (2.46)
rV C C F A Af f =! )( (2.47)
More complex situations can also be modeled in the same fashion. Consider the
catalytic hydrogenation of ethylene:
A + B # P (2.48)
where A represents hydrogen, B represents ethylene and P is the product (ethane). The
reaction takes place in the CSTR shown in figure 2.5. Two streams are feeding the
reactor. One concentrated feed with flow rate F 1 (m3/ s) and concentration C B1 (mole/m
3)
and another dilute stream with flow rate F 2 (m
3
/ s) and concentration C B2 (mole/m
3
). Theeffluent has flow rate F o (m
3/ s) and concentration C B (mole/m
3). The reactant A is
assumed to be in excess.
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V
Fo, CB
F1, CB1 F2, CB2
Figure 2-5 Reaction in a CSTR
The reaction rate is assumed to be:
)./()1(
3
2
2
1 smmole
C k
C k r
B
B
+
= (2.49)
where k 1 is the reaction rate constant and k 2 is the adsorption equilibrium constant.
Assuming the operation to be isothermal and the density is constant, and following the
same procedure of the previous example we get the following model:
Total mass balance:
o F F F dt
dL A !+= 21
(2.50)
Component B balance:
rV C C F C C F dt
C d V B B B B
A!!+!= )()(
)(2211
(2.51)
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Degrees of freedom analysis
• Parameter of constant values: A, k 1 and k 1
• (Forced variable): F 1 F 2 C B1 and C B2
• Remaining variables: V , F o, and C B
• Number of equations: 2 (Eq. 2.50 and Eq. 2.51)
The degree of freedom is therefore 3 " 2 =1. The extra relation is between the effluent
flow F o and the level L as in the previous example.
4. Heat Exchanger
Consider the shell and tube heat exchanger shown in figure 2.10. Liquid A of
density ! A is flowing through the inner tube and is being heated from temperature T A1 to
T A2 by liquid B of density ! B flowing counter-currently around the tube. Liquid B sees its
temperature decreasing from T B1 to T B2. Clearly the temperature of both liquids varies not
only with time but also along the tubes (i.e. axial direction) and possibly with the radial
direction too. Tubular heat exchangers are therefore typical examples of distributed
parameters systems. A rigorous model would require writing a microscopic balance
around a differential element of the system. This would lead to a set of partial
differential equations. However, in many practical situations we would like to model the
tubular heat exchanger using simple ordinary differential equations. This can be possible
if we think about the heat exchanger within the unit as being an exchanger between two
perfect mixed tanks. Each one of them contains a liquid.
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Liquid, A
T A1 T A2
Liquid, B
T B1
Liquid, B
T B2
T w
Figure 2-10 Heat Exchanger
For the time being we neglect the thermal capacity of the metal wall separating the two
liquids. This means that the dynamics of the metal wall are not included in the model. We
will also assume constant densities and constant average heat capacities.
One way to model the heat exchanger is to take as state variable the exit temperatures T A2
and T B2 of each liquid. A better way would be to take as state variable not the exit
temperature but the average temperature between the inlet and outlet:
2
21 A A
A
T T T
+=
(2.93)
2
21 B B
B
T T T
+=
(2.94)
For liquid A, a macroscopic energy balance yields:
QT T C F dt
dT V C A A p A A
A A p A A A
+!= )( 21 " " (2.95)
where Q ( J/s) is the rate of heat gained by liquid A. Similarly for liquid B:
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QT T C F dt
dT V C B B p B B
B B p B B B
!!= )( 21 " " (2.96)
The amount of heat Q exchanged is:
Q = UA H (T B – T A) (2.97)
Or using the log mean temperature difference:
Q = UA H #T lm (2.98)
where
)(
)(ln
)()(
21
12
2112
B A
B A
B A B A
lm
T T
T T
T T T T T
!
!
!!!="
(2.99)
with U ( J/m2 s) and A H (m
2) being respectively the overall heat transfer coefficient and
heat transfer area. The heat exchanger is therefore describe by the two simple ODE's (Eq.
2.95) and (Eq. 2.96) and the algebraic equation (Eq. 2.97).
Degrees of freedom analysis
• Parameter of constant values: !$,Cp A , V A , !% , Cp B , V B, U, A H
• (Forced variable): T A1, T B1, F A, F B
• Remaining variables: T A2, T B2, Q
•
Number of equations: 3 (Eq. 2.95, 2.96, 2.97)
The degree of freedom is 5 " 3 = 2. The two extra relations are obtained by noting that
the flows F A and F B are generally regulated through valves to avoid fluctuations in their
values.
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So far we have neglected the thermal capacity of the metal wall separating the two
liquids. A more elaborated model would include the energy balance on the metal wall as
well. We assume that the metal wall is of volume V w, density !w and constant heat
capacity Cpw. We also assume that the wall is at constant temperature T w, not a bad
assumption if the metal is assumed to have large conductivity and if the metal is not very
thick. The heat transfer depends on the heat transfer coefficient ho,t on the outside and on
the heat transfer coefficient hi,t on the inside. Writing the energy balance for liquid B
yields:
)()( ,,21 W Bt ot o B B p B B B
B p B T T AhT T C F dt
dT V C
B B!!!= " "
(2.100)
where Ao,t is the outside heat transfer area. The energy balance for the metal yields:
)()( ,,,, Awt it iw Bt ot ow
w pw T T AhT T Ahdt
dT V C
w!!!= "
(2.101)
where Ai,t is the inside heat transfer area. . The energy balance for liquid A yields:
)()( ,,21 Awt it i A A p A A A
A p A T T AhT T C F dt
dT V C
A A!+!= " "
(2.102)
Note that the introduction of equation (Eq. 2.101) does not change the degree of freedom
of the system.
Heat Exchanger with Steam
A common case in heat exchange is when a liquid L is heated with steam (Figure 2.11). If
the pressure of the steam changes then we need to write both mass and energy balance
equations on the steam side.
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Liquid, L
T L1 T L2
Steam
T s(t )
condensate, T s
T w
Figure 2-11 Heat Exchanger with Heating Steam
The energy balance on the tube side gives:
s L L p L L L
L p L QT T C F dt
dT V C
L L+!"=" )( 21
(2.103)
where
2
21 L L
L
T T T
+=
(2.104)
Q s = UA s (T s – T L) (2.105)
The steam saturated temperature T s is also related to the pressure P s:
T s = T s ( P ) (2.106)
Assuming ideal gas law, then the mass flow of steam is:
s
s s s s
RT
V P M m =
(2.107)
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where M s is the molecular weight and R is the ideal gas constant. The mass balance for
the steam yields:
cc s s s
s s
F F dt
dP
RT
V M
!"!=
(2.108)
where F c and !c are the condensate flow rate and density. The heat losses at the steam
side are related to the flow of the condensate by:
Q s = F c & s (2.109)
Where & s is the latent heat.
Degrees of freedom analysis
• Parameter of constant values: ! L,Cp L , M s , A s , U , M s , R
• (Forced variable): T L1
• Remaining variables: T L2, F L, T s, F s, P s, Qs, F c
•
Number of equations: 5 (Eq. 2.103, 2.105, 2.106, 2.108, 2.109)
The degrees of freedom is therefore 7 – 5 = 2. The extra relations are given by the
relation between the steam flow rate F s with the pressure P s either in open-loop or closed-
loop operations. The liquid flow rate F 1 is usually regulated by a valve.
5. Multi-component Distillation Column
Distillation columns are important units in petrochemical industries. These units
process their feed, which is a mixture of many components, into two valuable fractions
namely the top product which rich in the light components and bottom product which is
rich in the heavier components. A typical distillation column is shown in Figure 2.18.
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The column consists of n trays excluding the re-boiler and the total condenser. The
convention is to number the stages from the bottom upward starting with the re-boiler as
the 0 stage and the condenser as the n+1 stage.
Description of the process:
The feed containing nc components is fed at specific location known as the feed tray
(labeled f ) where it mixes with the vapor and liquid in that tray. The vapor produced from
the re-boiler flows upward. While flowing up, the vapor gains more fraction of the light
component and loses fraction of the heavy components. The vapor leaves the column at
the top where it condenses and is split into the product (distillate) and reflux which
returned into the column as liquid. The liquid flows down gaining more fraction of the
heavy component and loses fraction of the light components. The liquid leaves the
column at the bottom where it is evaporated in the re-boiler. Part of the liquid is drawn as
bottom product and the rest is recycled to the column. The loss and gain of materials
occur at each stage where the two phases are brought into intimate phase equilibrium.
B
xb
F
z
D
xd
C w
steam
Figure 2-18 Distillation Column
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Modeling the unit :
We are interested in developing the unsteady state model for the unit using the flowing
assumptions:
• 100% tray efficiency
• Well mixed condenser drum and re-boiler.
• Liquids are well mixed in each tray.
• Negligible vapor holdups.
• liquid-vapor thermal equilibrium
Since the vapor-phase has negligible holdups, then conservation laws will only be written
for the liquid phase as follows:
Stage n+1 (Condenser), Figure 2.19a:
Total mass balance:
)( D RV dt
dM n
D+!=
(2.161)
Component balance:
1,1)()(
,,
,!=+!= nc j x D R yV
dt
x M d j D jnn
j D D (2.162)
Energy balance:
c Dnn D D Qh D RhV
dt
h M d !+!= )(
)(
(2.163)
Note that R = Ln+1 and the subscript D denotes n+1
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Stage n, Figure fig2.19b
Total Mass balance:
nnnn
L RV V dt
dM !+!=
!1
(2.164)
Component balance:
1,1)(
,,,,11
,!=!+!=
!!
nc j x L Rx yV yV dt
x M d jnn j D jnn jnn
jnn
(2.165)
Energy balance:
nn Dnnnn
nnh L Rh H V H V
dt
h M d !+!=
!! 11
)(
(2.166)
Stage i, Figure 2.19c
Total Mass balance:
iiiii
L LV V dt
dM !+!=
+! 11
(2.167)
Component balance:
1,1)(
,,11,,11
,!=!+!=
++!!
nc j x L x L yV yV dt
x M d jii jii jii jii
jii (2.168)
Energy balance:
iiiiiiii
iih Lh L H V H V
dt
h M d !+!=
++!! 1111
)(
(2.169)
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Stage f (Feed stage), Figure 2.19d
Total Mass balance:
)())1(( 11 qF L L F qV V dt
dM f f f f
f +!+!+!=
+!
(2.170)
Component balance:
1,1
)())1(()(
,,11,,11
,
!=
+!+!+!=++!!
nc j
qFz x L x L Fz q yV yV
dt
x M d j j f f j f f j j f f j f f
j f f
(2.171)
Energy balance:
)())1(()(
1111 f f f f f f f f f f
f f qFhh Lh L Fhq H V H V
dt
h M d +!+!+!=
++!!
(2.172)
Stage 1, Figure 2.19e
Total Mass balance:
121
1 L LV V
dt
dM B
!+!= (2.173)
Component balance:
1,1)(
,11,22,11,
,11!=!+!= nc j x L x L yV yV
dt
x M d j j j j B B
j (2.174)
Energy balance:
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11221111 )(
h Lh L H V H V dt
h M d B B
!+!= (2.175)
Stage 0 (Re-boiler), Figure 2.19f
Total Mass balance:
B LV dt
dM B
B!+!=
1
(2.176)
Component balance:
1,1)(
,,11,
,!=!+!= nc j Bx x L yV
dt
x M d j B j j B B
j B B
(2.177)
Energy balance:
r B B B B B Q Bhh L H V
dt
h M d +!+!= 11
)(
(2.178)
Note that L0 = B and B denotes the subscript 0
Additional given relations:
Phase equilibrium: y j = f ( x j, T,P )
Liquid holdup: M i = f ( Li)
Enthalpies: H i = f (T i, yi,j), hi = f (T i, xi,j)
Vapor rates: V i = f ( P )
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Notation:
Li , V i Liquid and vapor molar rates
H i , hi Vapor and liquid specific enthalpies
xi , yi Liquid and vapor molar fractions
M i Liquid holdup
Q Liquid fraction of the feed
Z Molar fractions of the feed
F Feed molar rate
Degrees of freedom analysis
Variables
M i n
M B , M D 2
Li n
B,R,D 3
xi,j n(nc " 1)
x B,j ,x D,j 2(nc " 1)
yi,j n(nc " 1)
y B,j nc " 1
hi n
h B , h D 2
H i n
H B 1
V i n
V B 1
T i n
T D , T B 2
Total 11+6n+2n(nc"1)+3(nc"1)
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Equations:
Total Mass n + 2
Energy n + 2
Component (n + 2)(nc " 1)
Equilibrium n(nc " 1)
Liquid holdup n
Enthalpies 2n+2
Vapor rate n
h B = h1 1
y B = x B (nc " 1)
Total 7+6n+2n(nc-1)+3(nc-1)
Constants: P , F , Z
Therefore; the degree of freedom is 4
To well define the model for solution we include four relations imported from inclusion
of four feedback control loops as follows:
• Use B, and D to control the liquid level in the condenser drum and in the re-boiler.
• Use V B and R to control the end compositions i.e., x B, x D
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R, xd
Ln, xn
Vn, yn
Vn-1, yn-1
stage n
Li, xi Vi-1, yi-1
stage i
Li+1, xi+1 Vi, yi
(a)(b)
Lf+1, xf+1
Lf , xf Vf-1, yf-1
stage f
Vf , yf
(c)
L2, x2
L1, x1
V1, y1
stage 1
(d)
VB, yB
VB, yB
L1, x1
B, xB
(e)
Vn, yn
R, xd D, xd
Qc
Qr
(f)
Figure 2-19 Distillation Column Stages
Simplified Model
One can further simplify the foregoing model by the following assumptions:
(a) Equi-molar flow rates, i.e. whenever one mole of liquid vaporizes a tantamount of
vapor condenses. This occur when the molar heat of vaporization of all
components are about the same. This assumption leads to further idealization that
implies constant temperature over the entire column, thus neglecting the energy
balance. In addition, the vapor rate through the column is constant and equal to:
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V B = V 1 = V 2 =… = V n (2.179)
(b) Constant relative volatility, thus a simpler formula for the phase equilibrium can be
used:
y j = ' j x j/(1+(' j " 1) x j) (2.180)
Degrees of Freedom:
Variables:
M i , M B , M D n + 2
Li , B,R,D n + 3
xi ,x B ,x D (n + 2)(nc " 1)
y j, y B (n + 1)(nc " 1)
V 1
Total 2 + 2n + (2n + 3)(nc " 1)
Equations:
Total Mass n + 2
Component (n + 2)(nc " 1)Equilibrium n(nc " 1)
Liquid holdup n
y B = x B 1
Total 2+2n+(2n+3)(nc-1)
It is obvious that the degrees of freedom is still 4.
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6. 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 notexist.
• 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 andwastewater treatment.
• Chemical engineers are heavily involved with biochemical and biomedical processes.
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 pharmaceuticaland other process industries.
• Bioreactions:
• Yield Coefficients:
Fed-Batch Bioreactor
Figure 2.11. Fed-batch reactor for a bioreaction.
substrate more cells + products (2-90)cells
!
/ (2-91) X S
mass of new cells formed Y
mass of substrate consumed to form new cells
=
/ (2-92) P S
mass of product formed Y
mass of substrate consumed to form product =
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Monod Equation:
Specific Growth Rate
• Modeling Assumptions
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 solid material, 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).
7. The rate of product formation per unit volume r p can be expressed as
where the product yield coefficient Y P/X is defined as:
8. The feed stream is sterile and thus contains no cells.
(2-93) g r X µ =
max (2-94) s
S
K S µ µ =
+
/ (2-95) p P X g r Y r =
/ (2-96) P X
mass of product formed Y
mass of new cells formed =
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General Form of Each Balance
•
Individual Component Balances
• Cells:
• Product:
• Substrate:
• Overall Mass Balance
• Mass:
{ } { } { } (2-97) Rate of accumulation rate in rate of formation= +
( )(2-98) g
d XV V r
dt =
( )(2-99) p
d PV Vr
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
=