Math Modellling

44 Modeling the Dynamic and Static Behavior of Chemical Processes Part II

2. To describe a methodology for the modeling of a chemical proc ess using the balance equations and provide examples of its implementation

3. To determine the scope and the difficulties of the mathematicalmodeling for process control purposes

It should be noted that the subsequent chapters do not constitute a complete treatment of all the aspects on mathematical modeling, but it is limited to those of interest for process control.

Development 4of a Mathematical Model

Consider a general processing system with its associated variables as shown in Figure 2.1. To investigate how the behavior of a chemical process (i.e., its outputs) changes with time under the influence of changes in the external disturbances and manipulated variables and consequently design an appropriate controller, we can use two different approaches:

1. Experimental approach: In this case the physical equipment(s) of the chemical process is available to us. Consequently, we change deliberately the values of various inputs (disturbances, manipulated variables) and through appropriate measuring dev ices we observe how the outputs (temperatures, pressures, flow rates, concentrations) of the chemical process change with time. Such a procedure is time and effort consuming and it is usually quite costly because a large number of such experiments must be performed.

2. Theoretical approach: It is quite often the case that we have todesign the control system for a chemical process before the process has been constructed. In such a case we cannot rely on the experimental procedure, and we need a different representation ofthe chemical process in order to study its dynamic behavior. This representation is usually given in terms of a set of mathematical equations (differential, algebraic) whose solution yields thedynamic or static behavior of the chemical process we examine.

46 Modeling the Dynamic and Static Beha'lipf pf P:h~'lmicaPl rocesses Part II

In this text we discuss both approaches fpr .the development of a model for a chemical process. Initially, we will ~kamine the theoretical approach, leaving the experimental for suq~~qutnt chapters (Chapters16 and 31).

Chap. 4 Development of a Mathematical Model 47

Disturbance

4.1 Why Do We Need MathematicalModeling for Process Control'?

Manipulated variable

ProcessOutput

Let us repeat that our goal is to develop a cpnt(g! system for a chemical process which will guarantee that the OR~mH{)naol bjectives of our process are satisfied in the presence of ~V~r':ehangingdisturbances. Then, why do we need to develop a mat~~!Dat~(?adl escription (model) for the process we want to control?

In the introductory paragraphs earlier Yff( n~'!Mthat often the physi

cal equipment of the chemical process we wliUlll0control have not been constructed. Consequently, we cannot e'lcperim~ntto determine how theprocess reacts to various inputs and therefore; we cannot design the appropriate control system. But even if the prp~ @ : s e s quipment is available for experimentation, the procedure ~~~~HH\n}vfery costly. Therefore, we need a simple description of how the protess reacts to various inputs, and this is what the mathematical mf.lUelscan provide to the control designer. .

Let us demonstrate now in terms of some ~~~mples the need for the development of a mathematical model ~~fRr¢ we design the control system for a chemical process.

Example 4.1: Design an Integral Contfqllef..i('r a Stirred TankHeater .

Consider the problem of controlling the temR~rature of a liquid in a tank using integral control (Example 2.12),FrAm Fi~gure2.8 we notice thatthe quality of the control depends on the vf\IU~gfthe parameter a. Butthe question is: How does Ct' affect the quality qf~ontrol, and what is itsbest value? To answer this question we need tQ l!;~gwhow the value of the liquid temperature T is affected by chll-n~~~in the value of the inlet temperature T, or the integral action of the q)!'!tri~BerT. his is given byeq.(2.7), which constitutes the mathematical model ~tthe tank with integralcontrol.

Example 4.2: Design a Feedforward Cont,.t!/~~f'"or a Process

In the feedforward control arrangement ~hQ"'tlin Figure 4.1 we mea sure the value of the disturbance and we II-nfi~lmntw: hat its effect will be on the output of the process that we want to ~q!·!t~;:(I)nl.order to keep the

mailto:prp~@:sesquipment

Figure 4.1 Feedforward control configuration.

value of this output at the desired level, we need to change the value of the manipulated variable by such an amount as to eliminate the impact that the disturbance would have on the output. The question is: By how much should we change the manipulated variable in order to cancel the effect of the disturbance? To answer this question we must know the following two relationships:

output = fl(disturbance)

output = h(manipulated variable)

which are provided by a mathematical model of the process. Indeed, if the output is to remain the same, the manipulated variable must take such a value that

fl(disturbance) - h(manipulated variable) = 0

This example demonstrates very vividly how important mathematical modeling is for the design of a feedforward control system. In fact, with out good and accurate mathematical modeling we cannot design efficient feedforward control systems.

Example 4.3: Design of an Inferential Control SystemIn the inferential control scheme shown in Figure 4.2 we measure the

measured output and try to regulate the value of the unmeasured control objective at a desired value. Since the control objective is not measured directly, it can only be estimated from the value of the measured output if a relationship such as the following is available:

control objective = f(measured output)

Such a relationship in turn is not possible if we do not have a mathemati cal representation of the process (mathematical model). Once the value of the control objective can be estimated from a relationship such as the above, it can be compared to the desired value (set point) and the control ler can be activated for appropriate action as in feedback control.

We notice, therefore, that the availability of a good mathematical model for the process is indispensable for the design of good inferential control systems.

48 49Modeling the Dynamic and Static Behavior of Chemical Processes

Part II Chap. 4 Development of a Mathematical Model

Disturbance The principle of conservation of a quantity S states that:

Manipulated Controlled acc~~ulation of SJ flow of S ] flow of S ]variable variable

Process[ within a system [ in the system [ out of the system

'------....------' (u nrne asured) time period time period time period (4.1)

Measured output

Figure 4.2 Inferential control configuration.

The three examples above indicate very clearly that mathematical modeling of a process is of paramount importance for the design of good and efficient control systems for a chemical process. In the follow ing sections we develop a methodology for the concise modeling of chemical processes.

4.2 State Variablesand State Equations for a Chemical Process

In order to characterize a processing system (tank heater, batch reactor, distillation column, heat exchanger, etc.) and its behavior we need:

1. A set of fundamental dependent quantities whose values will describe the natural state of a given system

2. A set of equations in the variables above which will describe how the natural state of the given system changes with time

For most of the processing systems of interest to a chemical engineer there are only three such fundamental quantities: mass, energy, and momentum. Quite often, though, the fundamental dependent variables cannot be measured directly and conveniently. In such cases we select other variables which can be measured conveniently, and when grouped appropriately they determine the value of the fundamental variables. Thus mass, energy, and momentum can be characterized by variables such as density, concentration, temperature, pressure, and flow rate. These characterizing variables are called state variables and their values define the state of a processing system.

The equations that relate the state variables (dependent variables) to the various independent variables are derived from application of the conservation principle on the fundamental quantities and are called state equations.

[ ge:::~e: :~t~in] [co:::e: :~t~in ]the system the system

+ -------''-----time period time period

The quantity S can be any of the following fundamental quantities:

Total massMass of individual componentsTotal energyMomentum

Remark. It should be remembered that for the physical and chemical processes we will be studying, the total mass and total energy cannot be generated from nothing; neither do they disappear.

Let us review now the forms used most often for the balance equations. Consider the system shown in Figure 4.3. We have:

Q

Inlets

2 Outlets

w,

Figure 4.3 A general system and its interactions with the external world.

50 Modeling the Dynamic and Static Behavior of Chemical Processes

Part II 51Chap. 4 Development of a Mathematical Model

L

L

Total mass balance:

d(pV)---= PiFj - pjF} (4.1a)

as defined by eqs. (4.1) will yield a set of differential equations with the fundamental quantities as the dependent variables and time as the independent variable. The solution of the differential equations will

dt j:inlet j:outlet determine how the fundamental quantities, or equivalently, the stateMass balance on component A:

denA) = d(CA V)~.....:..:.=~ CAjFj - CAjF} ± rV

dt dt j:inlet j:outlet

Total energy balance:

(4.1b)

variables, change with time; that is, it will determine the dynamic behavior of the process.

If the state variables do not change with time, we say that the process is at steady state. In this case, the rate of accumulation of a fundamental quantity S per unit of time is zero, and the resulting balances yield a set of algebraic equations.

dE d(U+K+P)dt dt L pjFjhj - L pjFjhj ± Q ± Ws

j:inlet j:outlet(4.1c)

The variables appearing in the equations above have the following meaning:

p = density of the material in the systemp, = density of the material in the ith inlet stream p, = density of the material in the jth outlet stream V = total volume of the systemF, = volumetric flow rate of the ith inlet streamF} = volumetric flow rate of the jth outlet streamn A = number of moles of component A in the systemCA = molar concentration (moles/volume) of A in the system

CAj = molar concentration of A in the ith inletCAj = molar concentration of A in the jth outlet

r = reaction rate per unit volume for component A in the system

h, = specific enthalpy of the material in the ith inlet streamhj = specific enthalpy of the material in the jth outlet stream

U, K, P = internal, kinetic, and potential energies of the system, respectively

Q = amount of heat exchanged between the system and its surroundings per unit time

Ws = shaft work exchanged between the system and its surroundings per unit time

Example 4.4: State Variables and State Equations for a StirredTank Heater; Its Static and Dynamic Behavior

Consider the stirred tank heater of Example 1.1 (Figure 1.1). The fundamental quantities whose values provide the information about the heater are:

(a) The total mass of the liquid in the tank(b) The total energy of the material in the tank(c) Its momentum

The momentum of the heater remains constant even when the disturbances change value and will not be considered further.

Let us now identify the state variables for the tank heater.Total mass in the tank:

total mass = p V = pAh (4.2)

where p the density of liquid, V the volume of liquid, A the crosssectional area of the tank, and h the height of the liquid level.

Total energy of the liquid in the tank:

E=U+K+P

but since the tank does not move, dK/dt = dr/dt = 0 and dli/dt = dll/dt .For liquid systems,

dU dHdt dt

where H is the total enthalpy of the liquid in the tank. Furthermore,

By convention, a quantity is considered positive if it flows in the where cp = heat capacity of the liquid in the tank

(4.3)

system and negative if it flows out.The state equations with the associated state variables constitute the

mathematical model of a process, which yields the dynamic or static

behavior of the process. The application of the conservation principle


Part II 51Chap. 4 Development of a Mathematical Model

T,« = reference temperature where the specific enthalpy of the liquid is assumed to be zero.

From eqs. (4.2) and (4.3) we conclude that the state variables for the stirred tank heater are the following;


Part II Chap. 4 Development of a Mathematical Model 53

State variables: hand T

while the

Constant parameters: p, A, cp, Tref

are characteristic of the tank system.

Additional algebraic manipulations On eq. (4.5a) yield

A d(hT) = Ah dT + AT dh = Ah dT + T(F; - F) = F;T; - FT + _Qdt dt dt dt pc,

or

Note. It has been assumed that the density p is independent of the temperature.

Let us now proceed to develop the state equations for the stirred tank heater. We will apply the conservation principle on the two fundamental

Ah dT = F;( T; - T) + _Qdt pc,

Summarizing the modeling steps above, we have:State equations:

(4.5b)

quantities: the total mass and the total energy.Total mass balance:

A dh = F - F (4.4a)dt I

accumulation Of] input Of] output Of] Ah dT = F;(T;- T)+_Q (4.5b)[ total mass [ total mass [ total mass

dt pc,The variables in eqs. (4.4a) and (4.5b) can be classified as follows (see also

time

or

time

d(pAh) = pF; _ pFdt

time

(4.4)

Section 2.1):

State variables: h , TOutput variables: h, T (both measured)Input variables

where F; and F are the volumetric flow rates [i.e., volume per unit oftime (fe/min, or m3/min)] for the inlet and outlet streams, respectively. Assuming constant density (independent of temperature), eq. (4.4) becomes

dh

Disturbances: T;, F;Manipulated variables: Q, F (for feedback control)

F; (for feedforward control)Parameters: A, p, cp


A-=F;-Fdt

(4.4a)The state equations (4.4a) and (4.5b), with the state variables, the inputs, and the parameters, constitute the mathematical model of the stirred tank heater. We need only solve them in order to find the tank's dynamic

accumulation Of] input of ] output of ] or steady-state behavior.[ total energy [ total energy [ total energy Let us now study the dynamic and static behavior of the stirred tank

heater using the state equations (4.4a) and (4.5b). We will assume thattime time

energy sUPPlied]

time initially the tank heater is at steady state (i.e., nothing is changing). This situation is described by the state equations if the rate of accumulation

+ [ _ ::[left-hand sides of (4.4a) and (4.5b)] is set to zero:

:::....

or

:b:_=.y_:s:..:_te_:a_:mtime F;" - F, = 0

F;.,(T;., - T,) + Q, = 0pc,

d[pAh cp( T - T ref)] = pF;cp( T; _ Tref) _ pFcp( T - Trer) + Q (4.5)

dt

where Q is the amount of heat supplied by the steam per unit of time. The equation above can take the following simpler form (assume that T ref = 0):

The subscript s denotes the steady-state value of the corresponding variable.

The system will be disturbed from the steady-state situation if any of the input variables changes value. Let us examine the following two situations:

A d(hT) = F;T; - FT + _Q dt (4.5a)


Part II Chap. 4 Development of a Mathematical Model 531. Consider that the inlet temperature T;

decreases by 10% from its steady-state value. The liquid level will remain the same at the

54 Modeling the Dynamic and Static Behavior of Chemical Processes Part II IOld steady state I


T

T,

New steady state

o Time

Figure 4.4 Temperature response of a stirred heater to a step decrease in inlet temperature.

steady-state value hs, since T; does not influence the total mass in the tank [see also eq. (4.4a»). On the contrary, the temperature of the liquid will start decreasing with time. How the temperature T changes with time will be determined from the solution of eq. (4.5b) using as initial condition the steady-state value of T:

T(t = 0) = T,

Figure 4.4 indicates the static and dynamic behavior of the tank for this case. We observe that after a certain time the tank heater again reaches steady-state conditions.

2. Consider that initially the tank heater is at steady state. Then, at time t = 0, the inlet flow rate decreases by 10%.It is clear that both the level and the temperature of the liquid in the tank will startchanging [notice that F; is present in both state equations (4.4a) and (4.5b»). How hand T change with time will be given from the solution of eqs. (4.4a) and (4.5b) using as initial conditions

h(t = 0) = hs and T(t = 0) = T,

Figure 4.5 summarizes the static and dynamic behavior of the tank heater for this case.

Remark. It is worth noting that after F; has changed, the level handthe temperature T reach their new steady states with different speeds. In particular, the level, h, achieves its new steady state faster than thetemperature. In a subsequent chapter we will analyze the reasons for such behavior.

4.3 Additional Elements of the Mathematical

Models

In addition to the balance equations, we need other relationships to express thermodynamic equilibria, reaction rates, transport rates for heat, mass, momentum, and so on. Such additional relationships needed to complete the mathematical modeling of various chemical and/or physical processes can be classified as follows:

Transport rate equations

They are needed to describe the rate of mass, energy, and momen tum transfer between a system and its surroundings. These equations are developed in courses on transport phenomena.

Example 4.5The amount of heat Q supplied by the steam to the liquid ofthe tank

heater (Example 4.4) is given by the following heat transfer rate equation:

Q = UAt(Ts1 - T)

where U = overall heat transfer coefficientAt = total area of heat transfer

TSI = temperature of the steam

h T

Old steady state

New steady state

New steady state

Old steady state

Kinetic rate equations

They are needed to describe the rates of chemical reactions taking place in a system. Such equations are developed in a course on chemical kinetics.

Example 4.6o Time o Time

Figure 4.5 Dynamic response of a stirred tank heater to a step decrease in inlet flow rate.

The reaction rate of a first-order reaction taking place in a CSTR is given by

= =

P

Modeling the Dynamic and Static Behavior of Chemical Processes Part II

dI.1.;'rhere ko = preexponential kinetic constantshi] E = activation energy for the reaction

I R = ideal gas constant

T,CA = temperature and concentration of the reacting fluid.

l~eaction and phase equilibria relationships

H'hese are needed to describe the equilibrium situations reached ling a chemical reaction or by two or more phases. These relation A)sare developed in courses on thermodynamics..~I

l'!xample 4.7I

Consider a liquid stream composed of two components A and B at aiigh pressure PI and temperature TI. If the pressure PI is larger than the

)ubble-point pressure of the liquid at temperature Tf, no vapor phase willee present. The liquid stream passes through a restriction (valve) and is'flashed" in a drum; that is, its pressure is reduced from PI to P (Figure J.6). This abrupt expansion takes place under constant enthalpy. If the pressure P in the drum is smaller than the bubble-point pressure of the

.liquid stream at the temperature TI,the liquid will partially vaporize and.wo phases at equilibrium with each other will be present in the flash

.drum.I The thermodynamic equilibrium between the vapor and liquid phases


2. Pressure of liquid phase = pressure of vapor phase3. Chemical potential of component i in the liquid phase = chemical

potential of component i in the vapor phase

The equilibrium relationships introduce additional equations among the state variables of a system. Care must be exercised so that all the equilibrium relationships are accounted for.

Equations of state

Equations of state are needed to describe the relationship among the intensive variables describing the thermodynamic state of a system. The ideal gas law and the van der Waals equation are two typical equations of state for gaseous systems.

Example 4.8Let us return to the flash drum system discussed above in Example

4.7. For the vapor phase, from the ideal gas law, we have

p Vvapor = (moles of A + moles of B) . RT

(4.6) But

Ilmposes certain restrictions on the state variables of the system, andI must be included in the mathematical model of the flash drum if it is to be consistent and correct. These equilibrium relationships, as known from chemical thermodynamics, are:

mo Ies 0fA + mo Ies 0fB =

Therefore, from eq. (4.6) we have

mass of A + mass ofB----------average molecular weight

I

1. Temperature of liquid phase = temperature of vapor phasemass of A + mass ofB P .

Pvapor (average molecular weight)Vvapor RT

Considering that

average molecular weight = YAM A + yBM Bwe have

Pvapor = - [YAM A + yBM B]

RT

(4.6a)

F where YA and YB are the molar fractions of components A and Band M A p and M B are the molecular weights of A and B. Equation (4.6a) indicates a

relationship among the state variables of the flash drum and must beincluded in the mathematical model of the flash drum.

56 Starting from an appropriate equation of thermodynamic state for the liquid phase of the flash drum, we can develop an expression for its density of the form

Figure 4.6 Flash drum unit.

Pliquid = p( T, x A)

where x A is the molar fraction of component A in the liquid.

58 Modeling the Dynamic and Static Behavior of Chemical Processes Part "

4.4 Dead Time

In all of the modeling examples discussed in earlier sections it was assumed that whenever a change takes place in one of the input vari ables (disturbances, manipulated variables), its effect is instantaneously observed in the state variables and the outputs. Thus whenever the feed composition, c Ai' or the feed temperature, Ti; or the coolant tempera


clear that the temperature of the outlet, Tout, will remain the same until the change reaches the ~nd of the pipe. Then we will observe the tempera ture of the outlet changing, as shown by curve B in Figure 4.7b. We notice that the chang~ of the outlet temperature follows the same pattern as the c.hangeof the Inlet t.emperature with a delay of td seconds. t« is the dead time and from physical considerations it is easy to see that

td = volume of the pipe = A· L = ~ture, TCi' change in the CSTR of Figure 1.7, the effect of the change isfelt immediately and the temperature T or concentration CA of the

volumetric flow rate A· o: u; seconds

outlet stream start changing.The oversimplified picture given above is contrary to our physical

experience, which dictates that whenever an input variable of a system changes, there is a time interval (short or long) during which no effect is observed on the outputs of the system. This time interval is called dead time, or transportation lag, or pure delay, or distance-velocity lag.

Example 4.9Consider the flow of an incompressible, nonreacting liquid through a

pipe (Figure 4.7a). If the pipe is completely thermally insulated and the heat generated by the friction of the flowing fluid is negligible, it is easy to see that at steady state the temperature Tout of the outlet stream will be equal to that of the inlet, Tin. Assume now that starting at t = 0, the temperature of the inlet changes as shown by curve A in Figure 4.7b. It is

-:: -O~~======::::::::(1=j==~(()f----I ~ -- L-----+-I·I

(a)

T

(b)

Figure 4.7 (a) Pipe flow of Example 4.9; (b) delayed response of exit

temperature to inlet temperature change.

where ~av is the average velocity of the fluid over the cross-sectional areaof the pipe, assumed to be constant. Functionally, we can relate Tin andTout as follows:

(4.7) The ~ead time is an important ~leme.ntin the mathematical modeling

of chemical processes and has a serIOUSImpact on the design of effectivecontroll~rs. As we.~ill see in Chapter 19, the presence of dead time can very easily destabilize the dynamic behavior of a controlled system.

4.5 Additional Examples of MathematicalModeling

In this section we examine some typical chemical processes and develop their mathematical models.

Example 4.10: Mathematical Model 0/ a Continuous Stirred TankReactor (CSTR)

Consider the continuous stirred tank reactor system discussed inExample 1.2 (Figure 1.7). A simple exothermic reaction A ...B takes place~n the reactor, which is in turn cooled by a coolant that flows through aJacket around the reactor.

The fundamental dependent quantities for the reactor are:(a) Total mass of the reacting mixture in tank(b) Mass of chemical A in the reacting mixture(c) Total energy ofthe reacting mixture in the tank

Remarks

1. The mass of component B can be found from the total mass and the mass of component A. Therefore, it is not an independent fundamental quantity.

2. The ~?mentum of the CSTR does not change under any operating conditions for the reactor and will be neglected.

-=F;-

60 Modeling the Dynamic and Static Behavior of Chemical Processes Part

II

Let us apply the conservation principle on the three fundamental quantities:

Total mass balance:

aCCUmulation] [input Of] [output Of] [total mass generated] [ of total mass total mass _ total mass + or consumed

time = time time - time

or


dE = d( U + K + P) dUdt dt dt

Since the system is a liquid system, we can make the following approximation:

accumulation OftotalJ laccumUlation OftotalJenergy of the dU dH enthalpy of the material in the CSTR = dt ~dt = material in the CSTR

d(pV) = p.F, - pF ± 0dt

(4.8) l per unit time per unit time

where Pi,P = densities of the inlet and ou~let streams J .F. F = volumetric flow rates of the Inlet and outlet streams, ft /mIn

" or mJ/minV = volume of the reacting mixture in the tank

Mass balance on component A:

aCCUmulation] [input] [output] [disappearanc~ of A] [ of A of A of A due to reactIOn

time = time - time - time

Furthermore,

input of total energy with feed per unit time = p.F, h;( Ti)

and

output of total energy with the outlet stream per unit time = pF h (T)

where hi is the specific enthalpy (enthalpy per unit mass) of the feed stream and h is the specific enthalpy of the outlet stream. Consequently, the total energy balance leads to the equation

or

d(nA) = d(CA V) = cA;F; _ cAF - rV (4.9)dH = PiFihi( Ti) - pFh( T) _ Qdt (4.10)

dt dt

where r = rate of reaction per unit volume ..CA;,CA= molar concentrations (moles/volume) of A In the Inlet and

outlet streams andn A= number of moles of A in the reacting mixture


accumulation Of] [ total energy

time

where Q is the amount of heat removed by the coolant per unit time.Equations (4.8), (4.9) and (4.10)are not in their final and most conve

nient form for process control design studies. To bring them to such formwe need to identify the appropriate state variables.

Characterize Total Mass. We need the density of the reacting mixture, p, and its volume, V. The density will be a function of the concentration C A

and CB and of the temperature T. Quite often the dependence of p on C A, CB, and T is weak and the density can be considered constant as the reaction proceeds. Therefore, the left-hand side ofeq. (4.8) yields

input of total ] [ output of total ] [energy removed] [ energy with feed energy with outlet _ by coolant

d(pV) dV--=p-dt dt

while p; =p

= time - time time

In the balance above we have neglected the shaft work done by ~he impeller of the stirring mechanism. The total energy of the reacting mixture is

Under the assumption above, V is the only state variable that is needed to characterize the total mass. Then eq. (4.8) becomes

dVdt

(4.8a)

E=U+K+P

where U is the internal energy, K the kinetic energy, ~nd P the poten tial energy of the reacting mixture. Therefore, assuming that. the reac tor does not move (i.e., dK/dt = dr/dt = 0), the left-hand Side of the total energy balance yields

Characterize the Mass a/Component A. This is simple. From eq. (4.9) we realize that the state variables needed are C A and V. Algebraic manipula tions on eq. (4.9) lead to

d(CA V) dV V de A . k -E/RT---=CA-+ -=cAF;-cAF- oe cAV

dt dt dt I

c

c

-: --'. (T - T) + J.k e-


or

and finally,

Chap. 4 Development ot a Mathematical Model 63Let us now notice that

FiPih,(Ti): F;£Pihi(T) +PiCPi(Ti - T)]: F,[CAiHA(T) +PiCPi(T, - T)]and

Fph(T): F[CAH A(T) + cBHB(T)]

-d:C- A Fi ( CA·-CA ) - dt V '

k oe-E/RTCA

(4.9a) Consequently, eq. (4.lOa) becomes

pVcp ~~ :-~+~+HArV~-HBrv[Note. We have introduced r : koe-E/RTcA.]

Characterize the Total Energy. We know from thermodynamics that the enthalpy of a liquid system is a function of the temperature and its

+ ~A + FiPiCp,(Ti - T) - ~ _ ~_ Qor

composition:

H: H(T, nA, nB)P

v.dTFdt: ,p,cp,(T, - T)

+

~(H A -~HB)rV - Q

where n A and n B are the moles of A and B in the CSTR. Differentiating the expression above, we take

dH aH dT aH dn s; aH dn«

Finally, since (HA-HB):(-AH,): heat of reaction at temperature Tand P : Pi, Cp : Cp" '

V dT _ . (-AH,)rV Q-:--+--+--dt aT dt anA dt anB dt

(4.11) dt - F,(T, - T) +pcp

- - (4.10b)pcp

But

aH-:pV p

aTaH ~-: HB(T)anB

From eq'.(4.l0b) we conclude that temperature T is the state variable that charactenzes the total energy of the system.

Summarizing all the steps above in the mathematical modeling of aCSTR, we have the following:

where Cp is the specific heat capacity of the reacting mixture and H A and H B are the partial molar enthalpies of A and B. Furthermore, from eq. (4.9),

and a similar balance on component B,

dnB: d(cBV): 0 _ cBF + rVdt dt

Substitute the quantities above in eq. (4.11)and take

dH dT ~ ~

State variables: V.CA, T State equations:

dV-:F,-Fdt

dCA F,-: - (CA· - cAl - koe-EIRTCdt V ' A

dT F Qdt V' 0 A---

pCpVwhere J: (-AH,)/pcp.

Output variables: V. CA, T

(4.8a)

(4.9a)

(4.lOb)

- :pVcp - + H A[CA F, - cAF - rV] + HB[-cBF + rV]dt dt '

Substitute dH /dt by its equal from the total energy balance Ceq.(4.l0))and take

pVcp- dT

Input variables: CA" Fi, T" Q, F (when feedback control is used)

Among the input variables the most common disturbances are:

Disturbances' CA· F T·• I' " I

dt : -H A[CAiFi - cAF - rV] - HB[-cBF + rV] + p.Fih, - pFh - Q

(4.10a) while the usual manipulated variables are:

Manipulated variables: Q, F (occasionally r, or T;)


The remaining variables are parameters characteristic of the reactor system:

Constant parameters: P, cP' (-tlHr), k», E (activation energy), R

In the presence of changes in the input variables, the state variables change. Integration of eqs. (4.8a), (4.9a), and (4.10b) yields V(t), CA(t), and T(t) as functions of time.

The steady-state behavior of the CSTR is given by eqs. (4.8a), (4.9a), and (4.lOb) if their left-hand sides are set equal to zero.

Example 4.11: Mathematical Model of a Mixing ProcessTwo streams 1 and 2 are being mixed in a well-stirred tank, producing

a product stream 3 (Figure 4.8). Each of the two feed streams is composed of two components, A and B, with molar concentrations CAl' CBI and CA2' CB2' respectively. Also let FI and F2 be the volumetric flow rates of the two streams (fe/min, m3/min) and TI and T2 their corresponding tem peratures. Finally, let CA3' CB3' F3, and T3 be the concentrations, flow rate, and temperature of the product stream. A coil is also immersed in the liquid of the tank and it is used to supply heat to the system with steam, or remove heat with cooling water.

The fundamental quantities needed to describe the mixing process are:

(a) Total mass in the tank(b) Amounts of components A and B in the tank(c) Total energy(d) Momentum of the material in the tank

Remarks


Consider now the balances on the fundamental quantities:Total mass balance:

accumu~ation of total] [inp~t of total ] [ output of total ] [ mass In the tank = mass In the tank mass from the tanktime time time

or

(4.12)

~here PI, P2, and P3 are the densities of streams 1, 2, and 3, respectively. SInce the content of the tank is well mixed, the density of the product str~am P3 is equal to the density of the material in the tank, P (i.e., P3 = p). V IS the volume of the material in the tank and is equal to the product of the cross-sectional area of the tank, A, and the height, h, of the liquidlevel:

V=A ·h

In general, the densities, p, PI, and P2 depend on the corresponding concentrations and temperatures:

p=P3=f(CA3,CB3' T3) PI=f(CAI'CBI' TI) P2=f(CA2,CB2' T2)Usually (but not always) the dependencies above are weak and we assume that the densities are independent of the concentrations and tempera tures. Therefore, we assume that

PI =P2 =P3 =PThis transforms eq. (4.12) to the following:

dV dh

1. The momentum does not change under any operating conditions and it will be neglected in further treatment.

2. We only need to consider two of the following three quantities: total

- = A - = (F I + F 2) - F 3dt dt

Balance on component A:

(4.12a)

mass, mass of A, mass of B. The third can be computed from the other two.

Stream 1

~

I < nn))n )(W4-_- Q (Heat added or removed)

accumulation ?f] [total input Of]component A In component A[

the tank in the tank [

time time or

or

total output Of]component Afrom the tank

time

(4.13)

Figure 4.8 Mixing process. Substituting dV/dt by its equal from eq. (4.12a), we take

--


and since C A = C A3 due to the well-stirred assumption,

Part II Chap. 4 Development of a Mathematical Model 67

The question now is how to characterize h h h 2, and h 3 in terms of other variables (i.e., temperatures, concentrations, etc.). We know that

h3(T3) = h3(To) + cp3(T3 - To) (4.l5a)

h,(T,) = hl(To) + cp,(TI - To)=0 (4.l5b)

V dCA3 = (CAl - CA3)FI + (CA2 - CA3)F2 - (~ CA3)F3 (4.13a)

h2(T2) h2(To) + c (T2 - To) (4.l5c)

dt '''A3~ = P2

Total energy balance:where To is the reference temperature. At this temperature

ph3(To) = cAli A + cBlIB + CA3t:dis3(To) (4.l6a)accumulation Of]

[ total energy [

time

input of total energy]with feed streams

time

output of total energy] [ with product stream

time

ph I(To) = cAli A + cB/IB + CAl t1Hs,(To)

ph2\ To) = CA2HA + CB2HB + CA2 t1H S2( To)

(4.l6b)

(4.l6c)

heat added or removed] [ with the coil±=-----------------~time

The total energy of the material in the tank is

E = U (internal) + K (kinetic) + P (potential)

Since the tank is not moving, dK/dt = dP/dt = O. Thus dE/dt = dllt dtand for liquid systems,

dU dHdt dt

where H is the total enthalpy of the material in the tank. Furthermore,

input of total energy)

where H A and HB are the molar enthalpies (enthalpy per mole) of com ponents A and B at temperature To. t1H Sl' t1H S2' and t1H S3 are the heats of solution for streams I, 2, and 3 per mole of A at temperature To. Substituting eqs. (4.l5a,b,c) and (4.l6a,b,c) into the total energy balance eq. (4.14), we take

d[ V(c A3H A + CB3HB + C A3 t1H S3) + pVc p]( T 3 -

To)1 dt

= FI(CAIH A + CBIHB + CAl t1H s) + pF1cp,(TI - To)

+ F2(CA2H A + CB2HB + CA2 t1HS2) + pF2cP2(T2 - To)

- F3(CA3H A + CB3HB + CA3 t1HS3) - pF3cP3(T3 - To) ± Q

or

and

( with fee~ st~eams =p(Flh, + F2h2)

per unit time

output of total energy)

( with prod~ct.stream =pF3h3

= 0 (balance on A)

pCp3 d[V(T;t- TO)]+HA[d(d;A3 -CAl F2+CA3F3]

+HB~d( 3 _1<'. rCB1-2<F'-2"'+/CB3F3]+t1HS3d(VCA3)L

per unit time

where h c; h i, and h3 are the specific enthalpies (enthalpy per unit mass) of streams I, 2, and 3. Due to the perfect stirring assumption, the specific enthalpy of the material in stream 3 is the same as the specific enthalpy of the material in the tank. Thus

dt .. "~ dt

= 0 (balance on B)

= FlcAI t1HS, + pFlcp,(T, - To) + F2CA2 t1HS2 + pF2cpiT2 - To)

- F3CA3 t1HS3 - pF3c P3(T3 - To) ± Q

or

H =pVh3pCp3 V -dT3 + pCp3(T3 - To) -

dV+ t1H-S3[CAIFI +

1

(4.14

F F CA2 2 - CA3 3

Consequently, the total energy balance yields

d(pVh3) =p(Flh, + F2h2) - pF3h3 ± Qdt

dt dt

= FICAI t1Hs, + pF,cp,(TI - To) + F2CA2 t1HS2 + pF2cP2(T2 - To)

- F3CA3 t1HS3 - pF3Cp](T3 - To) ± Q


Chap. 4

Development of a Mathematical Model 69

d

l

1.1 SI S3 2 3 . h

and finally,dT ~ ~ ~ ~

Example 4.12: Mathematical Model 0/ a Tubular Heat Exchangerpc V_J == CA Fl[AHsl - AHsJ] + CA2F2[AHs2 - AHsJ]

Pl I

+ pF,[cp,(T, - To) - cPJ(Tl - To)]

+ pF2[cP2(T2 - To) - cPJ(T3 - To)] ± Q

If we assume that cPI == CP2 = CPJ == cp, we have

+ pFlcp(T, - T3) + pF2Cp(T2 - T3) ± Q (4.14a)

Summarizing the steps above, we have:

State variables: V. C AJ. T 3State equations:

Consider the shell-and-tube heat exchanger shown in Figure 4.9. Aliquid flows through the inner tube and it is being heated by steam thatflows countercurrently around the tube. The temperature of the liquid not only changes with time but also changes along the axial direction z from the value TI at the entrance to the value T2 at the exit. We will assumethat the temperature does not change along the radius of the pipe. Consequently, we have two independent variables, z and t. The state variable of interest for the heat exchanger is the temperature T of the heated liquid. Therefore, we need the energy balance for the characterization of the temperature. To perform this balance, consider the element of length Az defined in Figure 4.9 by the dashed lines. For this system and over aperiod of time At, we have:

Energy balance:

pc.A Az [( T) I 1+ AI - (T)t] = pc, vA (T) I z At ldV =(FI+F2)-F3

V dCAJ= (CAl - CA3)FI + (CA2 - CA3)F2

(4.12a)

(4.l3a)

accumulation OfJenthalpy during lflOWin of llflOW out of

dt

dt

(4.14a)

Input variables: F" CAl' TI, F2, CA2' T2, F3 (for feedback control)

the time periodAt enthalpy during enthalpy during

the time period the time periodAt At

+ Q At (nD Az) (4.17)

enthalpy tranSferred] from the steam to the liquid, through the

Output variables: V (or equivalently the height of liquid level, h),CA3,T3 ~ ~ ~ Parameters (constant): p, Cp, AHsl, AHs2, AHs3

Remarks

3. Usually, a mixing tank is equipped wit? a cooling or heating.coil or jacket through which flows a coolant (If h~at ISrel~ased ~unng th.e mixing of the two solutions) or a heating medlU~ .(If .heat IS absorbed during mixing) in an attempt to keep the mixmg Isother-

[ wall, during the time period At

where Q == amount of heat transferred from the steam to the liquid per unit of time and unit of heat transfer area

A = cross-sectional area of the inner tubev = average (assumed constant) velocity of the liquid

D = external diameter of the inner tubeDi viding both sides of eq. (4.17) by Az At and letting Az ...0 and At ...0,

'_-Z

4. mal. heats of solution are stro~g functions of . (iI.e., if IIf the concentra~l?n I I

[AH~ - AH~ ] and [Ails - AHs] are not small quantities), then I


Chap. 4

Development of a Mathematical Model 69from the total energy balance, eq. (4.14a), we notice t at tempera- TIture T 3 depends strongly on the concentrations of ~e feed~streams and their temperatures. If on the other hand, [AHs! - AHs3] and [Ails2 - Ails3] are nearly zero, then T3 depends basically only on TI and T2•

Liquid ----"'1 : TI, + a,

-.::).Z-

Figure 4.9 Tubular heat exchanger.


Part II Chap. 4

Development of a Mathematical Model 71we take

aT aTpc.A -at + pCpvA -az '" 1CDQ (4.18) r----------- -II

---------,I

In eq. (4.18) we can substitute Q by its equal,

Q", Ut.T« - T)

and take

aT aTpc.A -at + pcp vA -az '" 1CDU(Tst - T) (4.19)

I I I I IIII

Feed I

IJICooling waterI I IIDistillate

~---().----+-- ...F.D•xDproduct

This is the equation of state that models the behavior of liquid's tempera ture (state variable) along the length of the exchanger. Since eq. (4.19)is a partial differential equation we say that the exchanger has been modeled as a distributed parameter system. Note that U is the overall heat transfer coefficient between steam and the liquid in the tube, and T« is the temperature of saturated steam.

Example 4.13: Mathematical Model of an Ideal Binary DistillationJII .. ~---_-~&~t~to~m~sI

I I I I I I

ISteam

I I I I

ColumnConsider a binary mixture of components A and B, to be separated

into two product streams using conventional distillation. The mixture is fed in the column as a saturated liquid (i.e., at its bubble point), onto the

IL _ ------- ..J

product FB,XB

feed tray f (Figure 4.10), with a molar flow rate (mol/min) Ff and a molarfraction of component A, cj. The overhead vapor stream is cooled andcompletely condensed, and then it flows into the reflux drum. The cooling of the overhead vapor is accomplished with cooling water. The liquid from the reflux drum is partly pumped back in the column (top tray, N)with a molar flow rate FR (reflux stream) and is partly removed as the distillate product with a molar flow rate FD• Let us call MRD the liquid holdup in the reflux drum and XD the molar fraction of component A in the liquid of the reflux drum. It is clear that XD is the composition for both the reflux and distillate streams.

At the base of the distillation column, a liquid product stream (the bottoms product) is removed with a flow rate FB and a composition XB (molar fraction of A). A liquid stream with a molar flow rate V is also drawn from the bottom of the column and after it has been heated with steam, it returns to the base of the column. The composition of the recirculating back to column stream is XB. Let MB be the liquid holdup at the base of the column.

The column contains N trays numbered from the bottom of the column to the top. Let M, be the liquid holdup on the ith tray. The vapor holdup on each tray will be assumed to be negligible.

In Figure 4.1la we see the material flows in and out of the feed tray. Similarly, Figure 4.1lb and c show the material flows for the top (Nth) and

Figure 4.10 Binary distillation column.

I. Vapor holdup on each tray will be neglected.2. The m~lar heats of vaporization of both components A and Bare

approximately equal. This means that 1 mol of condensing vapor releases enough heat to vaporize 1 mol of liquid

3. The heat ~o~sesfrom the column to the surroundings are assumed to be negligible,

4. The relative volatility 0: of the two components remains constant throughout the column.

5. Each tray. is. assu~~d ~o be 100%efficient (i.e., the vapor leaving each tray IS In equilibrium with the liquid on the tray).

The first three assumptions imply that

V", VI '" V2", ••• '" VN

and there is no need for e~erg~ balance around each tray.. The la~t tw~ assumptions Imply that a simple vapor-liquid equilib

~um relatI.onship ~an be used to relate the molar fraction of A in the ap~r leaving the lth tray (y;) with the molar fraction of A in the liquid

Ieavmg the same tray (Xi):

bottom (first) trays, while Figure 4.11drefers to any other tray. ~'" o:~1 + (0: - l)xi

(42. 0)


Part II Chap. 4

Development of a Mathematical Model 71To simplify the system, we will make the following assumptions: where 0: is the relative volatility of the two components A and B.

Part II Chap. 4

Development of a Mathematical Model72

Modeling the Dynamic and Static Behavior of Chemical Processes

73

-

-

-

The final assumptions that we will make are the following:

6. Neglect the dynamics of the condenser an~ the reboile~. It is clear that these two units (heat exchangers) constitute pr~cessmg ~ystems on their own right and as such they have a dynamic behavior (see Example 4.12). Therefore, accurate mode.ling sho.uld include the state equations, which describe the dynamic behavior of condenser and reboiler.

7. Neglect the momentum balance for each tray and. assume that themolar flow rate of the liquid leaving each tray IS related to theliquid holdup of the tray through the Francis weir formula:

L;=f(M;) i=I,2, ... ,/, ... ,N (4.21)

the boundary of the system of interest is outlined by dashed lines in Figure 4.10. Such a boundary clearly identifies the inputs and outputs of practical significance for the overall system. It is also evident that unless we can describe how the concentrations and liquid holdups on each tray change with time, we cannot find how the variables of practical signifi cance, such as Xo and XB, change with time. Therefore, we are forced to consider the balances around each tray. Thus we have (see also Figure4.11):

Feed tray (i =f):

Total mass: d(MJ) = Ff+ Lf+1 + Vf-I - Lr Vf= Ff+ Lf+1 - Lf (4.22a)dt

Let us now develop the state equations that will describ~ ~hedynamic behavior of a distillation column. The fundamental quantities are total mass and mass of component A. But the question is: What is .the sys~em around which we will make the balances? From a practical point of VIew,

d(MfXf)Component A: dtTop tray (i = N):

= Ffcf+ Lf+IXf+1+ Vf-1Yf-1- Lfxf- VrYf (4.22b)

Total mass: d(MN) = FR + VN-1- LN - VN = FR - LN (4.23a)dtd(MNXN)

-Lf+1 t I t LN t

vN

Component A: dtBottom tray (i = I):

= FRxD + VN-1YN_ LNxN - VNYN (4.23b)

Ff t I

Cf

Lf tVf t t VN_1 Total mass: d(M,) = L2 - LI + V - VI = L2 - LI

dt

(4.24a)

t r. -I t

- t r -

FD'xO d(M,x,)Component A: dt = L2x2 + VYB- L,x, -

V1YI(4.24b)

Feed tray Nth tray ith tray (i = 2, ... , N -I and i * f):(a) (b) Total mass:

d(M;)= L;+I - L; + V;_I - V; = L;+I - L;

(4.25a)

- t -

dt

Component A: d(M;x;)

= L;+lx;+, + V;-IYi-I - Lix, - V;y;dt (4.25b)

rL;+I tv; ~; Reflux drum:

L;_I tV;-11

t I

ith tray

(d)

_I..-

Column base:

Total mass: d(MRO) = VN - FR - FD (4.26a)dt

Component A: d(MRDXO)

= VNYN- (FR + Fotx» (4.26b)dt

(c)

Part II Chap. 4

Development of a Mathematical Model72


73

Figure4.11 Modeling details of the binary distillation column: (a) f~edTotal mass: d(MB) = L 1 - V - FB

dt (4.27a)

section; (b) top section and overhead accumulator; (c) bottom section and reboiler; (d) general ith tray.

Component A: d(MBXB)

dt

= Lix , - VYB- FBxB

(4.27b)

75

Part II Chap. 4 Development of a Mathematical Model


74

All the equations above are state equations and describe the dynamic behavior of the distillation column. The state variables ofthe model are:

Liquid holdups: MI, M2, ... , MI, .. ·' MN; MRD and MB

Liquid concentrations: XI, X2, .. ·, XI,.··, XN; XD and XB

To complete the modeling of the column, in addition to the state equations, we need the following relationships:

1. Equilibrium relationships:

ax,

and the concentration CA. How do we decide that this dependence is weak (so that we can use constant values as in the example) or strong (in which case the modeling becomes very complicated)? The same questions arise for the densities p and p, and the heat of reaction (-dB,).

3. During the operation of the CSTR, scaling, fouling, and so on, willalter the value of the overall heat transfer coefficient. How can we account for this effect in the mathematical model?

4. We have considered first-order kinetics to describe the reactionrate. Is this correct?

Yj=-----1 + (a - I)x j

i = 1,2, ... ,f, ... , N, B (4.20)We can classify the difficulties encountered during the mathematical

2. Hydraulic relationships (Francis weir formula):

L, =f(Mj) i = I, 2, ... , f, ... , N (4.21)

When all the modeling equations above are solved, we find how the flow rates and concentrations of the two product streams (distillate, bottom) change with time, in the presence of changes in the various inputvariables.

The modeling steps outlined above indicate that the overall proceduremay be tedious and full of simplifying assumptions. At times the result~ng model is overwhelming in size and the solution of the corresponding equations may be cumbersome. For the binary distillation column wehave to solve a system of

2N + 4 nonlinear differential equations (state equations)

and2N + 1 algebraic equations (equilibrium, and hydraulic relationships)

4.6 Modeling Difficulties

The modeling examples discussed in previous sections of this chapter should have alerted the reader to a series of difficulties that we may encounter in our efforts to develop a meaningful and realistic mathe matical description of a chemical process.

Example 4.14: Difficulties in the Modeling of a CSTRConsidering the mathematical modeling of a CSTR (Example 4.10),

the following difficulties arise:1. Determine with the desired accuracy the values of various parame

ters such as the preexponential kinetic constant ko, the activation energy E, and the overall heat transfer coefficient U.

2. Although the specific heat capacities, cp and CPj, have been considered constant, they are in general functions of the temperature T

modeling of a process in three categories:

1. Those arising from poorly understood chemical or physical phenomena

2. Those caused from inaccurate values of various parameters3. Those caused from the size and the complexity of the resulting

model

Poorly understood processes

To understand completely the physical and chemical phenomena occurring in a chemical process is virtually impossible. Even an accept able degree of knowledge is at times very difficult. Typical examples include:

Multicomponent reaction systems with poorly known interactions among the various components and imprecisely known kinetics Vapor-liquid or liquid-liquid thermodynamic equilibria for multicomponent systemsHeat and mass transfer interactions in distillation columns with nonideal multicomponent mixtures, azeotropic mixtures, and so on.

Example 4.15Consider the fluidized catalytic cracking process shown in Figure 4.12.

An oil feed composed of heavy hydrocarbon molecules is mixed with catalyst and enters a fluidized bed reactor. The long molecules react on the surface of the catalyst and are cracked into lighter product molecules (such as gasoline) which leave the reactor from the top. While cracking is taking place, carbon and other heavy uncracked organic materials are deposited on the surface of the catalyst, leading to its deactivation. The catalyst is then taken into a regenerator, where the material deposited on its surface is burned with air. Then, the regenerated catalyst returns to the reactor after it has been mixed with fresh feed.

76 Modeling the Dynamic and Static Behavior of Chemical Processes Part

II

~ Product (to the separators)gases

Regenerator Reactor

Air Regenerated catalyst

Heavy oil feed

Figure 4.12 Fluid catalytic cracking (FCC) system.

To model the two units, the following information must be available:

1. The reaction rate ofthe cracking process2. The rate with which carbon and heavy material are deposited on

the catalyst (this will determine the rate of catalyst deactivation)3. The dependence of the two rates above on the temperature of the

reactor and the quality of the feed (light or heavy)4. The rate with which carbonaceous material deposited on the cata

lyst is burned off in the regenerator, and its dependence on temper-ature

All of the foregoing information is not only difficult to acquire, but at times it leads to contradicting contentions. For example, in Figure 4.13 we see two models that describe the effect of the heavy oil feed rate on the reactor temperature. We notice that the qualitative behavior predicted bythe two models is quite different.

Finally, the two units (reactor, regenerator) are fluidized beds and it iswell known how poorly understood the fluid mechanical characteristicsof such units are.

Reactor temperature


Imprecisely known parameters

The availability of accurate values for the parameters of a model is indispensable for any quantitative analysis of the behavior ofa process. Unfortunately, this is not always possible. Typical examples include the preexponential constant of a kinetic rate expression.

It should also be pointed out that the values of the parameters do not remain constant over long periods of time. Therefore, for effectivemode.lin~we need not only accurate values but also some quantitative description of ho~ the parametric values change with time. Typical examples of changing parameters are the activity of a catalyst and the overall heat transfer coefficient of heat transfer systems (heat exchangers, jacketed reactors, etc.).. The. dead time is also a critical parameter whose value is usually imprecisely known and ~arying. As we will see in a later section, poor knowledge of the dead time can lead to serious stability problems for the process.

When no reliable values for the parameters are available we resortto experiments on the real process in an effort to estimate some "good" values for them. The experimental procedures will be discussed further in Chapter 31.

Size and complexity of a model

In an effort to develop as accurate and precise a mathematical model as possible, its size and complexity increase significantly.

Example 4.16Consider a distillation column with 20 trays, a reboiler, and a con

denser. The feed is a two-component mixture. Then, as we have seen in Example 4.13, the mathematical model is composed of

2N + 4 '"'2(20) + 4 = 44 differential equations and

2N + 1 '"'2(20) + 1 = 41 algebraic equations

T~e size of the model for such a simple system is already prohibitive. SInce the common distillation systems include feeds with more than two components and possess larger numbers of trays, it is clear that such an extensive modeling would lead to cumbersome and hard-to-use models.

Figure 4.13 Two different models to describe the effect of heavy oil feed rate on reactor temperature for the FCC unit.

Therefore, care must be exercised that the size and complexity of a model do n~t exceed certain manageable levels, beyond which the model loses its value and becomes less attractive.


THINGS TO THINK ABOUT

1. What is a mathematical model ofa physical process, and what do we mean when we talk about mathematical modeling?

2. In Figure 4.13 we see two different curves tha~ relate ~he te~pe.rature a~d the feed rate of the reactor for the fluid catalytic cracking unit discussed m Example 4.15. Is the term "model" appropriate for each of these curves?

3. Let us recall that the steam tables give the temperature at which water liquid and water vapor are at equilibrium for a given pressure. The~ al~o give the specific values for enthalpy, entropy, and volume of both liquid and vapor phases. Do these tables of values constitute a mathematical model?

4. Consider the graphs shown in Figure Q4.1.These graphs were produced by measuring the concentration of B in the reaction A ....B, over time, and at various temperatures. Do these graphs represent a mathematical model?

Concentration of B

Temperature TI


a multicomponent liquid at temperature T and pressure p, with known composition for the N components.

10. Repeat question 9, but with a gas instead of a liquid.

11. Consider the flash drum of Examples 4.7 and 4.8. Develop an expression for the density of the vapor phase, using the van der Waals equation of state. State also an expression for the density of the liquid phase.

12. When is a system at steady state?

13. What is the main reason for the presence of dead time in a process?

14. Do you know of any systems that do not possess dead time?

15. How would you find the dead time of a system?

16. In Figure Q4.2 we see the behavior of the concentration at the outlet of two processes after concentration at the inlets and at time t '" 0 was increased by 10%.Which process possesses dead time?

Concentration

FigureQ4.1

Time

Figure Q4.2

Time

5. Why do you need to develop the mathematical model of a process you want to control?

6. What are the state variables, and what are the state equations? What are they used for?

7. How many state variables do you need to describe a system that is composed of M phases and N components?

8. We know that when two phases are at thermodynamic equilibrium, the chemical potential /1-;,( of every component (i) in phase I is equal to the chemical potential /1-;,n of the same component in phase II:

/1-;,( = /1-;,n, i = 1,2, ... , N

Express the equilibrium relationship above in terms of the mole,concen~ra tions of the N components in the two phases. The answer to this question will demonstrate to you that we do not need the concentrations of the N components in both phases in order to describe the system.

9. Write a relationship that will give you the molar or the specific enthalpy of

17. What are the assumptions leading to equimolar vapor flow rates (i.e.,VI'" V2", ' •• '" VN '" V) for a binary distillation column?

18. Why have we neglected the energy balances for the binary ideal distillation column of Example 4.13?

19. What are the assumptions leading to the equilibrium relationship (4.20), and how is it derived?

20. Could you have dead time between the overhead vapor and the distillate product? If yes, why?

21. Consider again Example 4.9. Show that the dead time can be computed from the following equation

(' F(8)d8", volume of the pipeJ'-'dwhere F(8) is the volumetric flow rate of the liquid through the pipe as a function of time. The above equation is more general than that of Example

4.9, where the volumetric flow rate was assumed to be constant.

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