Chapter_2 (Mathematika Models)

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Chapter

2

Mathematical Modeling of

Chemical Processes

Mathematical Model (Eykhoff, 1974)“a representation of the essential aspects of an existing

system (or a system to be constructed) which

represents knowledge of that system in a usable form”

Everything should be made as simple as possible, butno simpler.



General Modeling Principles

• The model equations are at best an approximation to the real

process.• Adage: “All models are wrong, but some are useful.”

• Modeling inherently involves a compromise between model

accuracy and complexity on one hand, and the cost and effort

required to develop the model, on the other hand.

• Process modeling is both an art and a science. Creativity is

required to make simplifying assumptions that result in an

appropriate model.

• Dynamic models of chemical processes consist of ordinary

differential equations (ODE) and/or partial differential

equations (PDE), plus related algebraic equations.

Chapter

2



Table 2.1. A Systematic Approach for

Developing Dynamic Models

1. State the modeling objectives and the end use of the model.They determine the required levels of model detail and model

accuracy.

2. Draw a schematic diagram of the process and label all process

variables.3. List all of the assumptions that are involved in developing the

model. Try for parsimony; the model should be no more

complicated than necessary to meet the modeling objectives.

4. Determine whether spatial variations of process variables areimportant. If so, a partial differential equation model will be

required.

5. Write appropriate conservation equations (mass, component,

energy, and so forth).

Chapter

2



6. Introduce equilibrium relations and other algebraic

equations (from thermodynamics, transport phenomena,chemical kinetics, equipment geometry, etc.).

7. Perform a degrees of freedom analysis (Section 2.3) to

ensure that the model equations can be solved.

8. Simplify the model. It is often possible to arrange the

equations so that the dependent variables (outputs) appear

on the left side and the independent variables (inputs)

appear on the right side. This model form is convenient

for computer simulation and subsequent analysis.

9. Classify inputs as disturbance variables or as manipulated

variables.

Table 2.1. (continued)

Chapter

2



Chapter

2

Modeling Approaches Physical/chemical (fundamental, global)

• Model structure by theoretical analysis Material/energy balances Heat, mass, and momentum transfer Thermodynamics, chemical kinetics

Physical property relationships• Model complexity must be determined

(assumptions)• Can be computationally expensive (not real-

time)

• May be expensive/time-consuming to obtain• Good for extrapolation, scale-up• Does not require experimental data to obtain

(data required for validation and fitting)



• Conservation Laws

Theoretical models of chemical processes are based on

conservation laws.

Conservation of Mass

rate of mass rate of mass rate of mass(2-6)

accumulation in out

= −

Conservation of Component i

rate of component i rate of component i

accumulation in

rate of component i rate of component i(2-7)

out produced

=

− +

Chapter

2



Conservation of Energy

The general law of energy conservation is also called the First

Law of Thermodynamics. It can be expressed as:

= −

+ +

rate of energy rate of energy in rate of energy out

accumulation by convection by convection

net rate of heat addition net rate of work

to the system from performed on the sys

the surroundings

tem (2-8)

by the surroundings

The total energy of a thermodynamic system, U tot , is the sum of itsinternal energy, kinetic energy, and potential energy:

int (2-9)tot KE PE U U U U = + +

Chapter

2



Chapter

2

Black box (empirical)

•Large number of unknown parameters

•Can be obtained quickly (e.g., linear regression)

•Model structure is subjective

•Dangerous to extrapolate Semi-empirical

•Compromise of first two approaches

•Model structure may be simpler •Typically 2 to 10 physical parameters

estimated

(nonlinear regression)

•Good versatility, can be extrapolated



Chapter

2

• linear regression

• nonlinear regression

• number of parameters affects accuracy of model,

but confidence limits on the parameters fitted mustbe evaluated

• objective function for data fitting – minimize sum of squares of errors between data points and modelpredictions (use optimization code to fit

parameters)

• nonlinear models such as neural nets arebecoming popular (automatic modeling)

2

210 xc xcc y ++=

( )τ /1 t e K y −−=



Chapter

2

Number of sightings of storks

Number of

births (West

Germany)

Uses of Mathematical Modeling

• to improve understanding of the process

• to optimize process design/operating conditions

• to design a control strategy for the process

• to train operating personnel



• Development of Dynamic Models

• Illustrative Example: A Blending Process

An unsteady-state mass balance for the blending system:

rate of accumulation rate of rate of (2-1)

of mass in the tank mass in mass out

= −

Chapter

2



• The unsteady-state component balance is:

( )1 1 2 2

ρ(2-3)

d V xw x w x wx

dt = + −

The corresponding steady-state model was derived in Ch. 1 (cf.

Eqs. 1-1 and 1-2).

1 2

1 1 2 2

0 (2-4)

0 (2-5)

w w w

w x w x wx

= + −

= + −

or

where w1, w2, and w are mass flow rates.

( )1 2

ρ(2-2)

d V w w w

dt = + −

Chapter

2



The Blending Process Revisited

For constant , Eqs. 2-2 and 2-3 become: ρ

1 2 (2-12)dV w w wdt

ρ = + −

( )1 1 2 2 (2-13)

d Vxw x w x wx

dt

ρ = + −C

hapter

2



Equation 2-13 can be simplified by expanding the accumulation

term using the “chain rule” for differentiation of a product:

( )(2-14)

d Vx dx dV V x

dt dt dt ρ ρ ρ = +

Substitution of (2-14) into (2-13) gives:

1 1 2 2 (2-15)dx dV

V x w x w x wxdt dt

ρ ρ + = + −

Substitution of the mass balance in (2-12) for in (2-15)

gives:

/dV dt ρ

( )1 2 1 1 2 2 (2-16)dx

V x w w w w x w x wxdt

ρ + + − = + −

After canceling common terms and rearranging (2-12) and (2-16),

a more convenient model form is obtained:

( )

( ) ( )

1 2

1 21 2

1(2-17)

(2-18)

dV w w w

dt

w wdx x x x xdt V V

ρ

ρ ρ

= + −

= − + −

Chapter

2



Chapter

2



Stirred-Tank Heating Process (cont’d.)

Assumptions:

1. Perfect mixing; thus, the exit temperature T is also the

temperature of the tank contents.

2. The liquid holdup V is constant because the inlet and outletflow rates are equal.

3. The density and heat capacity C of the liquid are assumed to

be constant. Thus, their temperature dependence is neglected.

4. Heat losses are negligible.

ρ

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2



For the processes and examples considered in this book, it

is appropriate to make two assumptions:

1. Changes in potential energy and kinetic energy can be

neglected because they are small in comparison with changesin internal energy.

2. The net rate of work can be neglected because it is small

compared to the rates of heat transfer and convection.

For these reasonable assumptions, the energy balance in

Eq. 2-8 can be written as

( )int (2-10)dU

wH Qdt

= −∆ +)

int the internal energy of

the system

enthalpy per unit mass

mass flow rate

rate of heat transfer to the system

U

H

w

Q

=

==

=

)

( )

denotes the difference between outlet and inlet

conditions of the flowing

streams; therefore

-Δ wH = rate of enthalpy of the inlet

stream(s) - the enthalpyof the outlet stream(s)

∆ =

)

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For a pure liquid at low or moderate pressures, the internal energy

is approximately equal to the enthalpy, U int , and H depends

only on temperature. Consequently, in the subsequent

development, we assume that U int = H and where the

caret (^) means per unit mass. As shown in Appendix B, a

differential change in temperature, dT , produces a correspondingchange in the internal energy per unit mass,

H ≈

ˆ ˆint U H =

ˆ ,int dU

intˆ ˆ (2-29)dU dH CdT = =

where C is the constant pressure heat capacity (assumed to be

constant). The total internal energy of the liquid in the tank is:

int int

ˆ(2-30)U VU ρ =

Model Development - I

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2



An expression for the rate of internal energy accumulation can be

derived from Eqs. (2-29) and (2-30):

int (2-31)dU dT

VC dt dt

ρ =

Note that this term appears in the general energy balance of Eq. 2-

10.

Suppose that the liquid in the tank is at a temperature T and has an

enthalpy, . Integrating Eq. 2-29 from a reference temperature

T ref to T gives,

ˆ H

( )ˆ ˆ

(2-32)ref ref H H C T T − = −where is the value of at T ref . Without loss of generality, we

assume that (see Appendix B). Thus, (2-32) can be

written as:

ˆref H ˆ H

ˆ 0ref H =

( )ˆ (2-33)ref H C T T = −

Model Development - II

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( )ˆ (2-34)i i ref H C T T = −

Substituting (2-33) and (2-34) into the convection term of (2-10)

gives:

( ) ( ) ( )ˆ (2-35)i ref ref wH w C T T w C T T −∆ = − − −

Finally, substitution of (2-31) and (2-35) into (2-10)

( ) (2-36)i

dT V C wC T T Q

dt ρ = − +

Model Development - III

For the inlet stream

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2



steam-heating: = ∆ s vQ w H

( ) (1)= − + ∆i s v

dT

V C wC T T w H dt ρ

0 ( ) (2)= − + ∆ si vwC T T w H

subtract (2) from (1)

( ) ( )= − + − ∆ s s v

dT V C wC T T w w H

dt ρ

divide by wC

( )∆

= − + −v s s

H V dT T T w w

w dt wC

ρ



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2

Define deviation variables (from set point)

1

1

is desired operating point( ) from steady state

Vnote that and

w

dynote when 0dt

General linear ordinary differential equation solution:

s s s

v v p

p

p

y T T T u w w w T

H H V dy y u K

w dt wC wC

y K u

dy y K u

dt

ρ ρ τ

τ

= −= −

∆ ∆= − + = =

= =

= − +

1

sum of exponential(s)

Suppose u 1 (unit step response)

( ) 1

t

p y t K e τ −

=

= −



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Example 1:Example 1:

C0=TC,90=TC,40=T o

i

oo

i′

wC(T - T ) = w Hi s v∆6

s

v

o

4

3 3

3

4

w =0.83 10 g hr

H =600 cal g

C=l cal g C

w=10 kg hr =10 kg m

V=20 m

2 10 kgV

ρ

ρ

×

∆

= ×4

4V 2 10 kg 2hr

w 10 kg hr ρ τ ×= = =

dynamic model2dy

dt= -y + 6 10 u-5×

y T T

u w w s s

= −

= −

s.s. balance:



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Step 1: t=0 double ws

final

T(0) = T y(0) = 0

hr g10+0.83=u 6×

2 dydt

= -y + 50

( )y = 50 l - e-0.5t

T = y + T = 50 + 90 = 140 Css

o

Step 2: maintain

Step 3: then set

hr 24/C140=T o

u = 0, w = 833 kg hr s

2

dy

dt = -y + 6 10 u, y(0) = 50

-5

×Solve for u = 0

y = 50e ty 0

-0.5t → ∞→ (self-regulating, but slow)

how long to reach y = 0.5 ?



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2

Step 4: How can we speed up the return from 140°C to 90°C?

ws = 0 vs. ws = 0.83× 106

g/hr at s.s ws =0

y →-50°C T →40°C

Process DynamicsProcess control is inherently concerned with unsteady

state behavior (i.e., "transient response", "process

dynamics")



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Stirred tank heater : assume a "lag" between heating

element temperature Te, and process fluid temp, T.

heat transfer limitation = heA(Te – T)

Energy balances

Tank:

Chest:

At s.s.

Specify Q →calc. T, Te

2 first order equations ⇒1 second order equation in T

Relate T to Q (Te is an intermediate variable)

i e e

dTwCT +h A(T -T)-wCT=mC

dt

dt

dTCm=T)-A(Th-Q e

eeee

0dt

dT 0,

dt

dT e ==



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2

fixed T Q-Q=u T-T=y i

Note Ce →0 yields 1st order ODE (simpler model)

Fig. 2.2

hR

1=q

v

Rv: line resistance

R v

uwC

1

ydt

dy

w

m

wC

Cm

Ah

Cm

dt

yd

Awh

Cmm ee

ee

ee

2

2

ee

ee

=+

+++



linear ODE

If

nonlinear ODE

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2

57)-(2 1

h

R

q

dt

dh A

v

i −=

*(2-61)i v i v

dh A q Cρgh q C h

dt

= − = −

gh p P ρ +=

pressureambient:P P-P=q aa

*

vC ghP ρ += p



Table 2.2. Degrees of Freedom Analysis

1. List all quantities in the model that are known constants (or

parameters that can be specified) on the basis of equipmentdimensions, known physical properties, etc.

2. Determine the number of equations N E and the number of

process variables, N V . Note that time t is not considered to be a

process variable because it is neither a process input nor a process output.

3. Calculate the number of degrees of freedom, N F = N V - N E .

4. Identify the N E output variables that will be obtained by solving

the process model.

5. Identify the N F input variables that must be specified as either

disturbance variables or manipulated variables, in order to

utilize the N F degrees of freedom.

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2

D f F d A l i f th Sti d T k



Thus the degrees of freedom are N F = 4 – 1 = 3. The process

variables are classified as:

1 output variable: T

3 input variables: T i , w, Q

For temperature control purposes, it is reasonable to classify the

three inputs as:

2 disturbance variables: T i , w

1 manipulated variable: Q

Degrees of Freedom Analysis for the Stirred-Tank

Model:

3 parameters:

4 variables:

1 equation: Eq. 2-36

, ,V C ρ

, , ,iT T w Q

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Biological Reactions

• Biological reactions that involve micro-organisms and enzyme catalysts

are pervasive and play a crucial role in the natural world.

• Without such bioreactions, plant and animal life, as we know it, simply

could not exist.

• Bioreactions also provide the basis for production of a wide variety of

pharmaceuticals and healthcare and food products.

• Important industrial processes that involve bioreactions include

fermentation and wastewater treatment.

• Chemical engineers are heavily involved with biochemical and

biomedical processes.

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i i



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2

Bioreactions

• Are typically performed in a batch or fed-batch reactor.

• Fed-batch is a synonym for semi-batch.

• Fed-batch reactors are widely used in the pharmaceutical

and other process industries.

• Bioreactions:

• Yield Coefficients:

substrate more cells + products (2-90)cells

→

/ (2-92) P S

mass of product formed Y

mass of substrate consumed to form product

=

/ (2-91) X S mass of new cells formed Y mass of substrate consumed to form new cells

=

d h i



Fed-Batch Bioreactor

Figure 2.11. Fed-batch reactor

for a bioreaction.

Monod Equation

Specific Growth Rate

(2-93) g r X µ =

max (2-94) s

S

K S µ µ =

+Chapt

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1. The exponential cell growth stage is of interest.

2. The fed-batch reactor is perfectly mixed.3. Heat effects are small so that isothermal reactor operation can

be assumed.

4. The liquid density is constant.

5. The broth in the bioreactor consists of liquid plus solidmaterial, the mass of cells. This heterogenous mixture can be

approximated as a homogenous liquid.

6. The rate of cell growth r g is given by the Monod equation in (2-

93) and (2-94).

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( )(2-98) g

d XV V r

dt

=

1 1(2-100) f g P

X / S P / S

d( SV ) F S V r V r

dt Y Y −= −

( )(2-101)

d V F

dt =

( )(2-99) p

d PV Vr

dt =

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Chapter_2 (Mathematika Models)

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