L6-9 Chapter 2

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CCB3013 - Chemical Process Dynamics, Instrumentation

and Control 16/10/2014

Chapter 2

Mathematical Modeling of

Chemical Processes

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Chapter Objectives

End of this chapter, you should be able to:

1. Develop unsteady-state models (dynamic models) of

chemical processes from physical and chemical principles

2. Explain the rationale for dynamic models

3. Explain a general strategy for deriving dynamic models

4. Carry out simulation studies (Solution of the dynamic

models that consist of sets of ODE’s and AE’s)

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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,

but no simpler

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Uses of mathematical models

Design:

• Exploring the sizing and arrangement of processing

equipment for dynamic performance

• Studying the interactions of various parts of the

process, particularly when material recycle or heat

integration is used

• Evaluating alternate process and control structures

and strategies

• Simulating start-up, shutdown, and emergency

simulations and procedures.

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Uses of mathematical models

Plant operation:

• Troubleshooting and processing problems

• Aiding in start-up and operator training

• Studying the effects of and the requirements for

expansion (bottle-neck removal) projects

• Optimizing plant operation

• It is usually much cheaper, safer, and faster to

conduct the kinds of studies listed above on a

mathematical model than experimentally on an

operating unit.

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Principles of Model Formulation

• The model equations are at best an

approximation to the real process.

• Adage: “All models are wrong, but some are

useful”

• Inherently involves a compromise between

– model accuracy and complexity

– the and effort required to develop the

model

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Process modeling

• Both an art and a science.

• Requires creativity to make simplifying

assumptions that result in an appropriate

model

• Dynamic models of chemical processes

consist of ODEs and/or PDEs, plus related

algebraic equations

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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.

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Systematic Approach

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 so forth).

6. Introduce equilibrium relations and other algebraic

equations (from thermodynamics, transport

phenomena, chemical kinetics, equipment

geometry, etc.).

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A Systematic Approach…

6. Perform a degrees of freedom analysis (Section 2.3) to ensure that the model equations can be solved.

7. Simplify the model.

• 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.

8. Classify inputs as disturbance variables or as manipulated variables.

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Modeling Approaches

Physical/chemical (fundamental, global)

• Model structure by theoretical analysis

– Material/energy balances

– Heat, mass, and momentum transfer

– Thermodynamics, chemical kinetics

– Physical property relationships

• Model complexity must be determined (assumptions)

• Can be computationally expensive (not real-time)

• May be expensive/time-consuming to obtain

• Good for extrapolation, scale-up

• Does not require experimental data to obtain (data

required for validation and fitting)

• Theoretical models of chemical processes are based on

conservation laws.

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Mass Balance

Conservation of Component i

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

accumulation in out

(2.1)

rate of component i rate of component i

accumulation in

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

out produced

(2.2)

Conservation of Mass

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Energy Balance

Conservation of Energy

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 (2.3)

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Energy Balance …

• Total energy of a system, Utot

int (2-9)tot KE PEU U U U (2.4)

• For the processes and examples considered in

this course, it is appropriate to make two

assumptions:

– Changes in potential energy and kinetic

energy can be neglected

– The net rate of work can be neglected

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Energy Balance …

• For these reasonable assumptions, the energy balance in Eq.

2-3 can be written as

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 enthalpy

of the outlet stream(s)

int (2-10)dU

wH Qdt

(2.5)

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Black box (empirical)

– Large number of unknown parameters

– Can be obtained quickly (e.g., linear

regression)

– Model structure is subjective

– Dangerous to extrapolate

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• 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

Semi-empirical

2

210 xcxccy

/1 teKy

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Semi-empirical

• Number of parameters affects accuracy of model,

but confidence limits on the 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)

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Development of Dynamic Models

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

1 2

ρ(2-2)

d Vw w w

dt

(2.1)

(2.6)

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

or

Illustrative Example: A Blending Process

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Blending Process

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

(2.8)

(2.7)

1 1 2 2

ρ(2-3)

d V xw x w x wx

dt

The unsteady-state component balance is:

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Equation 2-10 can be simplified by expanding the

accumulation term using the “chain rule” for

differentiation of a product:

The Blending Process Revisited

1 2 (2-12)dV

w w wdt

(2.9)

1 1 2 2 (2-13)

d Vxw x w x wx

dt

(2.10)

For constant ρ, Eqs. 2-6 and 2-7 become:

Substitution of (2-11) into (2-10) gives:

(2.12)1 1 2 2 (2-15)

dx dVV x w x w x wx

dt dt

(2.11) (2-14)

d Vx dx dVV x

dt dt dt

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Blending Process Revisited …

Substitution of the mass balance in (2-9) for in (2-12) gives:

(2.13) 1 2 1 1 2 2 (2-16)dx

V x w w w w x w x wxdt

After canceling common terms and rearranging (2-9) and

(2-13), a more convenient model form is obtained:

(2.14)

(2.15)

1 2

1 21 2

1(2-17)

(2-18)

dVw w w

dt

w wdxx x x x

dt V V

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• The dynamic model in Eqs. 2-14 and 2-15 is in a

convenient form for subsequent investigation based on

analytical or numerical techniques

• Specify the inlet compositions (x1 and x2) and the flow

rates (w1, w2 and w) as functions of time

• Specify the initial conditions for the dependent variables,

V(0) and x(0)

• Determine the transient responses, V(t) and x(t)

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Dynamic models of representative

processes

• Stirred-Tank Heating Process

Figure 2.3 Stirred-tank heating process with constant holdup, V.

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Assumptions:

• Perfect mixing

The exit temperature T is also the temperature of

the tank contents

• The liquid holdup V is constant because the inlet and

outlet flow rates are equal

• The density ρ and heat capacity C of the liquid are

assumed to be constant

Temperature dependence is neglected

• Heat losses are negligible

Stirred-Tank Heating Process

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intˆ ˆ (2-29)dU dH CdT (2.16)

Model Development

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

energy, Uint, is approximately equal to the enthalpy, H

• H depends only on temperature

• We assume that Uint = H and where the caret (^) means per

unit mass.

A differential change in temperature, dT, produces a

corresponding change in the internal energy per unit mass,

where C is the constant pressure heat capacity.

(2.17)int intˆ (2-30)U VU

The total internal energy of the liquid in the tank is:

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An expression for the rate of internal energy accumulation

can be derived from Eqs. (2-16) and (2-17):

(2.18)int (2-31)dU dT

VCdt dt

• Note that this term appears in the general energy

balance of Eq. 2-5.

(2.19) ˆ ˆ (2-32)ref refH H C T T

Suppose that the liquid in the tank is at a temperature T

and has an enthalpy, . Integrating Eq. 2-16 from a

reference temperature Tref to T gives,

H

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Without loss of generality, we assume that .

Thus, (2-19) can be written as:

0ˆ refH

(2.20) ˆ (2-33)refH C T T

(2.21)For the inlet stream: ˆ (2-34)i i refH C T T

Substituting (2-20) and (2-21) into the convection term

of (2-5) gives:

(2.22) ˆ (2-35)i ref refwH w C T T w C T T

Finally, substitution of (2-18) and (2-22) into (2-5)

(2.23) (2-36)i

dTV C wC T T Q

dt

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1

1

is desired operating point

( ) from steady state

Vnote that and

w

dynote when 0

dt

General linear ordinary differential equation solution:

s ss

v vp

p

p

y T T T

u w w w T

H HV dyy u K

w dt wC wC

y K u

dyy K u

dt

1

sum of exponential(s)

Suppose u 1 (unit step response)

( ) 1

t

py t K e

Define deviation variables (from desired operating/set point)

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Example 1:

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

4

4

V 2 10 kg2hr

w 10 kg hr

C0= C,90= C,40= o

i

oo

i TTT

The dynamic model is

u106+-y=dt

dy2 5-

ss wwu

TTy

Steady-state balance:

vsi ΔHw)=-TTwC(

hr/kg) ( 106

) (cal/g 1 (g/hr) 1010

)cal/g(600

o5

o34

C

CwC

HK v

p

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Example 1:

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

Step 1: at t=0 double ws

hrg10+0.83=u 6

50+-y=dt

dy2

C140=90+50=T+y=T o

ssFinal

-0.5te-l50=y

t = 0

ws

sw

50

1083.0106 65

uK p

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Example 1:

hr 24 / C140 = oTStep 2: Maintain New steady

state

Step 3: Then set hrkg830 = 0, = swu

50=(0) ,106+-=2 5- yuydt

dy

Solve for u = 0

0 , as

50= -0.5

yt

ey t

How long to reach y = 0.5 ?

Self-regulating, but slow

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Example 1:

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

90°C?

Now, find the time to reach y = 0.5?

ws = 0 g/hr u = -0.83 x 106 g/hr

50--=2 ydt

dy tey -0.5-l50- =

At steady state with ws =0, y -50°C T 40°C

Definitely much quicker

than the case u = 0

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Process Dynamics• Process control is inherently concerned with unsteady

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

dynamics"

Stirred tank heater: Assume a "lag" between heating element

temperature Te, and process fluid temp, T.

Heat transfer limitation = he A(Te – T)

Energy balances

Tank:

Heating element:dt

dTC-T)=mA(TQ-h e

eeee

0 0, dt

dT

dt

dT eAt steady state :

dt

dTmCwCTTTAhwCT eei )(

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Process Dynamics

Specify Q Calculate T, Te

2 first order equations 1 second order equation in T

Relate T to Q (Te is an intermediate variable)

fixed i TQ- u=QTy=T-

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

uwC

ydt

dy

w

m

wC

Cm

Ah

Cm

dt

yd

Ah

Cmm ee

ee

ee

ee

ee 1 2

2

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Liquid storage systems

h R

1=q

v

Rv: line resistance

57)-(2 1

hR

qdt

dhA

v

i

linear ODE

pressureambient : aav P P-Pq=C

ghpP If

* (2-61)i v i v

dhA q C ρgh q C h

dt

Non-linear

ODE

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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

equipment dimensions, known physical properties, etc.

2. Determine the number of equations NE and the number

of process variables, NV. 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, NF = NV -

NE.

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Degrees of Freedom

4. Identify the NE output variables that will be obtained

by solving the process model.

5. Identify the NF input variables that must be specified as

either disturbance variables or manipulated variables, in

order to utilize the NF degrees of freedom.

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Degrees of Freedom Analysis

for the Stirred-Tank Model

, ,V CParameters (3): , , ,iT T w QVariables (4):

Equation (1) : Eq. 2-23 NF = 4 – 1 = 3

The process variables are classified as:

Output variable: T Input variables : Ti, w, Q

For temperature control purposes, it is reasonable to classify

the three inputs as:

Disturbance variables : Ti, w

Manipulated variable : Q

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Conclusion!

• Need for mathematical modeling

• Types of mathematical models

• A systematic approach

• Blending example – revisited

• Tank heater example – model developed

• Second-order system

• Linear and nonlinear systems

• Degrees of freedom

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For Chapter 3

• Go through Laplace Transform

• Xerox Laplace Transform Table 3.1

page 54-55, Seborg Book

L6-9 Chapter 2

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