part one che621 adv pdc

8/11/2019 Part One Che621 Adv Pdc

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CHE 621

Advanced Process Dynamics and Control

1

1. Quick Overview of Chemical Process Control

2. Mathematical Modeling of Chemical Processes

Dr. Waheed Afzal

Associate Professor of Chemical Engineering

Institute of Chemical Engineering and Technology

University of the Punjab, Lahore

[email protected]


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Course Quick Revision of PDC

Mathematical Modeling of Chemical Processes

Linearization of Non-Linear Models

Development of Transfer Functions and Input-Output Models

Dynamic Behavior of First, Second (and Higher) Order Systems Analysis and design of feedback-controlled processes

Analysis and design of advanced control systems

Plant-wide process control

Computer applications in PDC

2


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3

Donald R. Coughanowr and Steven E. LeBlanc. Process SystemsAnalysis and Control. McGraw-Hill, 2008

Dale E. Seborg, Thomas F. Edgar, and Duncan A. Mellichamp. Process

Dynamics and Control. 2nd Edition, Wiley, 2004.

Carlos A. Smith, and Armando Corripio. Principles and Practice ofAutomatic Process Control. 3rd ed. John Wiley & Sons, Inc., 2006.

William L Luyben. Process Modeling, Simulation and Control for

Chemical Engineers. 2nd Edition, McGraw-Hill, 1996

George Stephanopoulos. Chemical process control. Englewood Cliffs,New Jersey: Prentice-Hall, 1984

Recommended Books


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Take-Home Assignment

Assignment # 1 (Compulsory)

Mathematical Modeling

Part I. Solve all of the problems at the end of Chapters 4

and 5 of Stephanopoulos (1984).

Part II.Solve allof the problems at the end of Chapter 3

of Luyben (1996).

4

Fun Assignment

Introduction to Chemical Process Control

Part I. Prepare short answers to things to think about

(Stephanopoulos, 1984) pages 33-35 and 78-79.

Part II. Solve the problems for Part I (page 36-41) PI.1 to 1.10 of

Stephanopoulos (1984)


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Test yourself (and Define):

Dynamics (of openloop

and closedloop) systems

Manipulated Variables

Controlled/ Uncontrolled

Variables

Load/Disturbances

Feedback, Feedforward

and Inferential controls

Error

Offset (steady-statevalue of error)

Set-point

5

Stability

Block diagram Transducer

Final control element

Mathematical model

Input-out model,transfer function

Deterministic and

stochastic models

Optimization Types of Feedback

Controllers (P, PI, PID)


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Need of a ControlSafety:

Equipment and PersonnelProduction Specifications:

Quality and Quantity

Environmental Regulations:Effluents

Operational Constraints:

Distillation columns (flooding, weeping); Tanks

(overflow, drying), Catalytic reactor (maximumtemperature, pressure)

Economics:

Minimum operating cost, maximum profits6


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Requirements from a control

7

Suppressing External Disturbances

Ensure the Stabilityof a Process

Optimizationof the Performance

of a Process

Control system is an agent ofmanagement of process variabilityin

which the impact of disturbance is

shifted to a benign location in a

process or plant.


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Process Control in a Chemical Plant

Identify Sources of Disturbances

8

Luyben (1996)


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10

Classification of Variables

Input variables(sometime called as load variables or LV)Further classified as disturbances and manipulated or control

variables)

Output variables

Further classified into measured and unmeasured variables

Often, manipulated variable effects (measured) output

variable; controlled variable is a measured variable

When an output variable is chosen as a manipulated variable,it becomes an input variable.

A manipulated variable is always an input variable.


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11

Design Elements in a Control

Objective: h= hs (Controlled Variable or CV)

Scenario Contrd.

Variable

Manip.

Variable

Input

Variable

Output

Variable

1 (shown) h F Fi h

2 h Fi F, h

Define Control Objective:what are the operational objectives of a control

system

Select Measurements: what variables must be measured to monitor the

performance of a chemical plant

Select Manipulated Variables:what are the manipulated variables to be used

to control a chemical process

Select the Control Configuration: information structure for measured and

controlled variables. Configurations include feedback control, inferential

control, feedforward control

Fh

A


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12

Input variables

Fi, Fst, Ti, (F)

Output variables

F, T, h

Control Objective

(a) T = Ts(b) h = hs

F, T

Fst

h A

F, T

h A

Fst

Temperature and level control in a stirred

tank heater (Stephanopoulos, 1984)

Design Elements in a Control


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13

Control Configurations in a Distillation Column

Define Control Objective:

95 % top product

Select Measurements:

composition of Distillate

Select Manipulated variables:

Reflux ratio

Select the Control Configuration:

feedback control

(Stephanopoulos, 1984)


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14

Feed-forward Control Configuration in a Distillation

Column


ControlxD


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15

Inferential Control in a Distillation Column


Control Objective:xD

Unmeasured input =f(secondary measurements)


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The process (chemical or physical)

Measuring instruments and sensors (inputs, outputs)

what are the sensors for measuring T, P, F, h, x, etc?

Transducers (converts measurements to current/ voltage)

Transmission lines/ amplifier

The controller (intelligence)

The final control element

Recording/ display

elements

RecallProcess

Instrumentation

16

Hardware for a Process Control System



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Mathematical representation of a process (chemical

or physical) intended to promote qualitative and

quantitative understanding

Set of equations

Steady state, unsteady state (transient) behavior

Model should be in good agreement with

experiments

17

Mathematical Modeling

Experimental Setup

Set of Equations

(process model)

Inputs Outputs

Outputs

Compare


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1. Determine objectives, end-use, required details andaccuracy

2. Draw schematic diagram and label all variables, parameters

3. Develop basis and list all assumptions; simplicity Vs reality

4. If spatial variables are important (partial or ordinary DEs)

5. Write conservation equations, introduce auxiliary equations

6. Never forget dimensional analysis while developing

equations

7. Perform degree of freedom analysis to ensure solution

8. Simplify model by re-arranging equations

9. Classify variables (disturbances, controlled and manipulated

variables, etc.) 18

Systematic Approach for Modelling(Seborg et al 2004)


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To understandthe transient behavior, how inputs

influence outputs, effects of recycles, bottlenecks To trainthe operating personnel (what will happen

if, emergency situations, no/smaller thanrequired reflux in distillation column, pump is not

providing feed, etc.) Selection of control pairs (controlled v. /

manipulated v.) and control configurations(process-based models)

To troubleshoot

Optimizingprocess conditions (most profitablescenarios)

19

Need of a Mathematical Model


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Theoretical Models

based on principal of conservation- mass, energy,

momentum and auxiliary relationships,, enthalpy,

cp, phase equilibria, Arrhenius equation, etc)

Empirical modelbased on large quantity of experimental data)

Semi-empirical model(combination of theoretical

and empirical models)Any available combination of theoretical principles

and empirical correlations

20

Classification of Process Modelsbased on how they are developed


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Theoretical Models Physical insight into the process

Applicable over a wide range of conditions

Time consuming (actual models consist of large

number of equations)

Availability of model parameters e.g. reaction rate

coefficient, over-all heart transfer coefficient, etc.

Empirical model

Easier to develop but needs experimental data

Applicable to narrow range of conditions

21

Advantages of Different Models


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State variables describe natural state of a process Fundamental quantities (mass, energy, momentum)

are readily measurable in a process are described by

measurable variables (T, P,x, F, V)

State equations are derived from conservationprinciple (relates state variables with other variables)

(Rate of accumulation) = (rate of input)(rate ofoutput) + (rate of generation) - (rate of consumption)

22

State Variables and State Equations


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Basis

Flow rates are volumetric

Compositions are molar

A B, exothermic, first order

Assumptions

Perfect mixing

, cPare constant

Perfect insulation

Coolant is perfectly mixed

No thermal resistance of jacket

23

Modeling Examples

Jacketed CSTR

Coolant

Fi, CAi, Ti

F, CA, T


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Overall Mass Balance

(Rate of accumulation) = (rate of input)

(rate of output)

Component Mass Balance

(Rate of accumulation of A) = (rate of

input of A)(rate of output of A) + (rate

of generation of A)(rate of

consumption of A)

24

Modeling of a Jacketed CSTR (Contd.)

Coolant

Fci,Tci

V

CA

T

Fi, CAi, Ti

F, CA, T

Coolant

Fco,Tco

Energy Balance

(Rate of energy accumulation) = (rate of energy input)(rate of

energy output) - (rate of energy removal by coolant) + (rate of

energy added by the exothermic reaction)


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=

Component Mass Balance

=

0/

Energy Balance

=

0/

= ( )25

Modeling of a Jacketed CSTR (Contd.)

Coolant

Fci,Tci

V

CA

T

Fi, CAi, Ti

F, CA, T

Coolant

Fco,Tco

Input variables: CAi, Fi, Ti, Q, (F)

Output variables: V, CA, T


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Nf= Nv- NE

Case (1): Nf= 0 i.e. Nv= NE(exactly specified system)

We can solve the model without difficulty

Case (2): f> 0 i.e. Nv> NE(under specified system), infinite

number of solutions because Nfprocess variables can be fixed

arbitrarily. either specify variables (by measuring disturbances)

or add controller equation/s

Case (3): Nf< 0 i.e. Nv< NE(over specified system) set ofequations has no solution

remove Nfequation/s

We must achieve Nf= 0 in order to simulate (solve) the model26

Degrees of Freedom (Nf) Analysis


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Basis/ Assumptions Perfectly mixed, Perfectly insulated

, cPare constant

27

Stirred Tank Heater: Modeling and

Degree of Freedom Analysis

Steam

Fst

A


=

Energy Balance

=


Independent Equations: 2 Variables: 6 (h, Fi, F, Ti, T, Q)

Nf= 6-2 (= 4) Underspecified


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Nf= 4 Specifyload variables (or disturbance)

Measure Fi, Ti (Nf= 4 - 2 = 2)

Include controller equations (not

studied yet); specify CV-MV pairs:

28

Stirred Tank Heater: Modeling and


Steam

Fst

A

=

=

CV MV

h F

T Q

=

=

Nf

= 2 - 2 = 0Can you draw these control loops?


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F

Z

mB

mD

VB

D

xD

B

xB

R xD

Reboiler

Condenser

Reflux Drum

(Stephanopoulos, 1984)29

Basis/ Assumptions

1. Saturated feed

2. Perfect insulation of column

3. Trays are ideal

4. Vapor hold-up is negligible

5. Molar heats of vaporization of A

and B are similar6. Perfect mixing on each tray

7. Relative volatility () is constant

8. Liquid holdup follows Francis weir

formulae

9. Condenser and Reboiler dynamics

are neglected

10. Total 20 trays, feed at 10

2, 4, 5 V1= V2= V3= VN

(not valid for high-pressure columns)

Modeling an Ideal Binary Distillation Column

= 20


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31

NthStage

mN

vN LN+1

LN vN-1

NthStage (stages 19 to 11 and 9 to 2)

Overall()

= +

Component()

= ++

. simplify!

Modeling Distillation Column

Feed Stage

(10th)

mN

v10 L11

L10 v9

FZ

Feed Stage (10th)

Overall(

)

= 9

Component()

= 99

. simplify!


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VBL1

V1 L2

1stStage

Modeling Distillation Column1stStage

Overall

()

=

Component

()

= simplify!

VB

L1

Column

Base

mB

B

Column Base

Overall(

)

=

Component()

=

. simplify!

VB


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33


Equilibrium relationships (to determine y)

Mass balance (total and component) around 6 segments of a

distillation column: reflux drum, top tray, Nthtray, feed tray, 1st

tray and column base.

Solution of ODE for total mass balance gives liquid holdups (mN)

Solution of ODE for component mass balance gives liquid

compositions (xN)

V1= V2= = VN= VB(vapor holdups)

How to calculate y(vapor composition) and L(liquid flow rate)

Recall ijis constant throughout the column

Use ij = ki/kj, xi+ xj=1,yi+ yj= 1, and k = y/xto prove

=

+( ) Phase-equilibrium relationship (recall

thermodynamics)


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34

Liquid flow rate can be calculated using well-known Francisweir hydraulic relationship; simple form of this equation is

linearized version:

=

LNis flow rate of liquid coming from Nthstage

LN0is reference value of flow rate LN

mNis liquid holdup at Nthstage

mN0is reference value of liquid holdup mN is hydraulic time constant (typically 3 to 6 seconds)


Hydraulic relationships (to determine L)


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35



Total number of independent equations:

Equilibrium relationships (y1, y2, yN, yB) N+1 (21)

Hydraulic relationships (L1, L2, LN) N (20)

(does not work for liquid flow rates D and B)

Total mass balances (1 for each tray, reflux drum andcolumn base) N+2 (22)

Total component mass balances (1 for each tray, reflux

drum and column base) N+2 (22)

Total Number of equations NE= 4N + 5 (85)

44 differential and 41 algebraic equations

Note the size of modeleven for a simple system with several

simplifying assumptions!


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36



Total number of independent variables:

Liquid composition (x1, x2, xN, xD,xB) N+2

Liquid holdup (m1, m2, mN, mD,mB) N+2

Vapor composition (y1, y2, yN, yB) N+1

Liquid flow rates (L1, L2, LN)

N Additional variables 6

(Feed: F,Z; Reflux: D, R; Bottom: B, VB)

Total Number of independent variables NV= 4N + 11

Degree of Freedom = (4N + 11) (4N + 5)

= 6

System is underspecified


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38

Feedback Control on a Binary Distillation ColumnCV MV loop

xD R 1

xB VB 2

mD D 3

mB B 4R


M d li CSTR i S i


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39

Modeling CSTRs in Seriesconstant holdup, isothermal

F0

F1CA1

F2CA2 F3

CA3

V1K1T1

V2K2T2

V3K3T3

Basis and Assumptions

A B (first order reaction)Compositions are molar and flow rates are volumetric

Constant V,, T


= 0 1= 0i.e. at constant V, F3=F2=F1=F0 F

So overall mass balance is not required!

Luyben (1996)

M d li CSTR i S i


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40


Component A mass balance on each tank (A is chosen arbitrarily)

1

= 1

0 1 11

2

=

21 2 22

3

= 3

2 3 33

kndepends upon temperature kn= k0e-E/RTn where n= 1, 2, 3

Apply degree of freedom analysis!

Parameters/ Constants (to be known): V1, V2, V3, k1, k2, k3

Specified variables (orforcing functions): Fand CA0(known but not

constant) . Unknown variables are 3(CA1, CA2, CA3) for 3ODEs

Simplify the above ODEs for constant V, T and putting = V/F


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41


If throughput F, temperature Tand holdup Vare same inall tanks, then for = V/F (note its dimension is time)

1

1 1

=

1

0

2

2 1

=

1

1

3

3

1

=

1

2

In this way, only forcing function (variable to be specified)

is CA0.

Modeling CSTRs in Series


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Modeling CSTRs in SeriesVariable Holdups, nthorder

Mass Balances (Reactor 1)

1 = 0 1

(

)

= 00 1111(1)

n

Mass Balances (Reactor 2)

2

= 1 2

(

)

= 11 2222(2)

n

Mass Balances (Reactor 1)3

= 2 3

(

)

= 22 3333(3)

n

Changes from previous case:

Vof reactors (and F) varies

with time,

reaction is nthorder

Parameters to be known:

k1, k2, k3, n

Disturbances to be specified:CA0, F0

Unknown variables:

CA1, CA2, CA3, V1, V2, V3, F1, F2, F3

CV MV IncludeController eqns

V1(or h1) F1 F1= f(V1)

V2(or h2) F2 F2= f(V2)

V3(or h3) F3 F3= f(V3)


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H1H2

H3

Qin or out

43

Modeling a Mixing Process

Overall Mass Balance()

= 1 2 3

()

= ( ) Component Mass Balance

( )

= (1 2) 3

cAis concentration of A in CSTR; hence cA= cA3

Basis and Assumptions

F(volumetric), CA(molar); Well Stirred

Feed (1, 2) consists of components A and B

Enthalpy of mixing is significant

Process includes heating/ cooling

Stephanopoulos (1984)


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H1H2

H3

44

Modeling a Mixing ProcessConservation of energy

(recall first law of thermodynamics) =

(for constant / liquid system, is zero)

Energy Balance

enthalpybalance (h is energy/mass)

()

= ( )

We were familiar with energy ; how to characterize h(specific enthalpy) into familiar quantities (T, CA, parameters, )

His enthalpy, his specific enthalpy; CPis heat capacity, cPis specific

heat capacity .


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45


Since enthalpy depends upon temperatureso lets replace hwith h(T)

1 1 = 0 1 1 0

2 2 = (0) 2 2 0

3 3 = (0) 3 3 0 enthalpy associated with T was easy to obtain, how to obtain h(T0)

0 = 1 1 11(0)

0 = 2 2 22(0)

0 =3 3 33(0)

and are molar enthalpy of component A and B and is heat of

solution for stream iat T0.

()

= ( )

Put values of hin overall energy balance


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Re-arranging (and using component mass balance equations)

3

3

= 11 1

3 22 2 3

1 1 1 0 3 3 0

2[2 2 0 3 3 0 ]

If we assume cP1= cP2 = cP3 = cP

3

= 111

3 22 2 3

1(

1

3) c

p

2(

2

3)

If heats of solutions are strong functions of concentrations

then 1 3 and 2

3 are significant

Mixing process is generally kept isothermal (how?)

part one che621 adv pdc

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