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.



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



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 = + +



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

! = + "



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

!

! !

= + "

= " + "



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



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:



( ) ( )

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)



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:



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



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



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:



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)



Degrees of freedom analysis

• Parameter of constant values: A

• (Forced variable): F f and C Af

• Remaining variables: V , F o, and C A


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.



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)




• 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


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.



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:



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).


• 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.



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.



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)



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.


• 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.



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



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



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)



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:



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 )



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


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)



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



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:



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.



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 =



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 =



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

=

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