ChE 6303 – Advanced Process Controlteacher.buet.ac.bd/shoukat/ChE6303_Handout1.pdf · Review of...

© Dr. M. A. A. Shoukat Choudhury

ChE 6303 – Advanced Process Control Teacher: Dr. M. A. A. Shoukat Choudhury, Email: [email protected]

Syllabus:

1. SISO control systems: Review of the concepts of process dynamics and control, process models, Laplace transform, transfer functions,Poles and zeros,state-space models, feedback controllers, controller design – direct synthesis and IMC rules, controllertuning, Concept of stability, feedforward and ratio control,cascade control, time delay compensation,inferential control, adaptive control, selective control/override systems

2. MIMO control systems: Control loop interactions, RGA analysis, , pairing control loops,decoupler design

mailto:[email protected]


ChE 6303 – Advanced Process Control3. Process Monitoring:Traditional process monitoring, multivariate statistical monitoring

4. Process Faults:

Control valve problems, Valve stiction, data-based methods for detection of valve problems

5. Troubleshooting Plantwide Oscillations

Marks Distribution:

1. Assignments 20%2. Project 30%3. Final exam 50%


Process Dynamics and Control

Concepts of Process Dynamics and Control:

- Dynamics is concerned with the transient state ( steady state too) behaviour of the process.

- Control is concerned with the manipulation of process behaviour- make processes operate closer to the operating

conditions- regulate the process well in the presence of disturbances


Why study process dynamics and control

- Increased emphasis on efficient plant operation- Continued impact of energy conservation measures

(Energy integration)- Tighter integration of plant design (Mass integration)- Emphasis on increased plant/human safety- Stringer environmental regulationsso on…..


Process Dynamics – where and why?- Refers to unsteady-state or transient behavior.- ChE curriculum emphasizes steady-state or equilibrium

situations: Examples: ChE 111, 201, 303, 205- Continuous processes: Examples of transient behavior:

i. Start up & shutdownii. Polymer grade changesiii. Disturbances, especially major disturbances, e.g., refinery

during stormy or hurricane conditions, seasonal variationiv. Equipment or instrument failure (e.g., pump failure)v. Process degradation, catalyst poisoning, heat exchanger

fouling


Process Dynamics (cont’d)- Batch processes

i. Inherently unsteady-state operation

ii. Example: Batch reactor

1. Composition changes with time

2. Other variables such as temperature could be constant.


Process Control- Ubiquitous, everywhere in life

- starting from very basic household equipments to large chemical processes

- Household equipments: Refrigerators, ACs, TVs

- Large scale, continuous processes:

i. Oil refinery, ethylene plant, pulp mill. Typically, several thousands process variables are measured. Examples: flow rate, T, P, liquid level, composition

ii. To control the process variables, one needs manipulated variables such as feed rate, cooling rate, product flow rate

Question: How do we control processes?

• We will consider an illustrative example


Process Control

Question:

How do we control a process or a variable?


Illustrative Example – A Blending system

Notation:• w1, w2 and w are mass flow rates• x1, x2 and x are mass fractions of component A


A Blending System (cont’d)Assumptions:

1. w1 is constant

2. x2 = constant = 1 (stream 2 is pure A)

3. Perfect mixing in the tankControl Objective:

Keep x at a desired value (or “set point”) xsp, despite variations in x1(t). Flow rate w2 can be adjusted for this purpose.Terminology:

• Controlled variable (or “output variable”): x

• Manipulated variable (or “input variable”): w2

• Disturbance variable (or “load variable”): x1


A Blending System (cont’d)Design Question. What value of is required to have 2w

?SPx x=Overall balance:

1 20 (1-1)w w w= + −

Component A balance:

1 1 2 2 0 (1-2)w x w x wx+ − =

(The overbars denote nominal steady-state design values.)

• At the design conditions, . Substitute Eq. 1-2, and , then solve Eq. 1-2 for :

SPx x= SPx x=

2 1x = 2w

2 1 (1-3)1 SP

w wx

=−

1SPx x−


A Blending System (cont’d)• Equation 1-3 is the design equation for the blending

system.

• If our assumptions are correct, then this value of will keep at . But what if conditions change?

xSPx

2w

Control Question. Suppose that the inlet concentration x1changes with time. How can we ensure that x remains at or near the set point ?

As a specific example, if and , then x > xSP.

SPx

1 1x x> 2 2w w=Some Possible Control Strategies:

Method 1. Measure x and adjust w2.

• Intuitively, if x is too high, we should reduce w2;


A Blending System (cont’d)

• Manual control vs. automatic control

• Proportional feedback control law,

( ) ( )2 2 (1-4)c SPw t w K x x t = + − - =

1. where Kc is called the controller gain.

2. w2(t) and x(t) denote variables that change with time t.

3. The change in the flow rate, is proportional to the deviation from the set point, xSP – x(t).

( )2 2,w t w−


How to implement the feedback?


Method 2 - Feedforward

Method 2. Measure x1 and adjust w2.

• Thus, if x1 is greater than , we would decrease w2 so that

• One approach: Consider Eq. (1-3) and replace and with x1(t) and w2(t) to get a control law:

1x2 2;w w<

1x 2w

( ) ( )12 1 (1-5)

1SP

SP

x x tw t w

x−

=−


Method 2 – Feedforward (cont’d)

• Because Eq. (1-3) applies only at steady state, it is not clear how effective the control law in (1-5) will be for transient conditions.


Other methods

Method 3. Measure x1 and x, adjust w2.

• This approach is a combination of Methods 1 and 2.

Method 4. Use a larger tank.

• If a larger tank is used, fluctuations in x1 will tend to be damped out due to the larger capacitance of the tank contents.

• However, a larger tank means an increased capital cost.


Control Strategies

Table. 1.1 Control Strategies for the Blending System

Design change--4

FF/FBw2x1 and x3

FFw2x12

FBw2x1

CategoryManipulated Variable

Measured VariableMethod


Feedback Control

Distinguishing feature: measure the controlled variable

• Advantages:

Corrective action is taken regardless of the source of the disturbance.

Reduces sensitivity of the controlled variable to disturbances and changes in the process (shown later).

• Disadvantages:

No corrective action occurs until after the disturbance has upset the process, that is, until after x differs from xsp.

Very oscillatory responses, or even instability.


Feedforward Control• Features

Distinguishing feature: measure a disturbance variable

• Advantage:

Correct for disturbance before it upsets the process.

• Disadvantage:

Must be able to measure the disturbance.

No corrective action for unmeasured disturbances.


Hierarchy of process control activities

1. Measurement and Actuation

2. Safety, Environment and Equipment Protection

3a. Regulatory Control

4. Real-Time Optimization

5. Planning and Scheduling

Process

3b. Multivariable and Constraint Control

(days-months )

(< 1 second )

(< 1 second )

(seconds-minutes )

(minutes-hours )

(hours-days )

Figure 1.7 Hierarchy of process control activities.


Major Steps in control system development

Figure 1.9 Major steps in control system development


MODELS

MODELS


What are Models?A model is a mathematical abstraction of a process A model can be formulated on the basis of a physio-chemical or a

mechanistic knowledge of the processA model can capture the transient and/or steady state of the processSteady state is a special case of transient states

ProcessInputs Outputs

Mathematically,

G ( . )Output Space

Y (.)

Input Space

U (.)


Type of Models

1. Unsteady vs. Steady state models2. First principle vs. empirical models/black-box

models3. Semi-empirical/gray box models

Advantages and Disadvantages of these models

Example: Model for a simple cylindrical tank- mass balance- linearization (if necessary)- deviation variable- time constant equivalent to residence time- gain of the process


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.


Developing Dynamic ModelsA 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 are important. If so, a partial differential equation model will berequired.

5. Write appropriate conservation equations (mass, component, energy, and so forth).


Developing Dynamic Models (cont’d)

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.


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.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.


Conservation LawsTheoretical 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 irate of component i rate of component i

accumulation in

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

out produced

=

− +


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 outaccumulation by convection by convection

net rate of heat addition net rate of workto the system from performed on the systhe surroundings

tem (2-8)by the surroundings

The total energy of a thermodynamic system, Utot, is the sum of its internal energy, kinetic energy, and potential energy:

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


Conservation of Energy

For the processes and examples considered in this course, itis appropriate to make two assumptions:

1. Changes in potential energy and kinetic energy can be neglected because they are small in comparison with changes in 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 inEq. 2-8 can be written as

( )int (2-10)dU wH Qdt

= −∆ +)

int the internal energy of the system

enthalpy per unit massmass flow raterate of heat transfer to the system

U

HwQ

=

===

)

( )

denotes the differencebetween outlet and inletconditions of the flowingstreams; therefore

-∆ wH = rate of enthalpy of the inletstream(s) - the enthalpyof the outlet stream(s)

∆ =

)


Development of Dynamic Models – An Example

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

= −


Example (cont’d)or

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

( )1 2

ρ(2-2)

d Vw w w

dt= + −

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 ww x w x wx

= + −= + −


Example (cont’d)

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

dtρ

= + −

Equation 2-13 can be simplified by expanding the accumulation term using the “chain rule” for differentiation of a product:


Example (cont’d)( ) (2-14)

d Vx dx dVV xdt dt dt

ρ ρ ρ= +

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

1 1 2 2 (2-15)dx dVV x w x w x wxdt dt

ρ ρ+ = + −

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

/dV dtρdx ( )1 2 1 1 2 2 (2-16)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 wdt

w wdx x x x xdt V V

ρ

ρ ρ

= + −

= − + −


Continuous Stirred Tank Reactor (CSTR)

Fig. 2.6. Schematic diagram of a CSTR.


CSTR: Model Development

[ ]A

A

(2-62) where = moles of A reacted per unit time, per unit volume, is the concentration of A (mol

r = kc r c

es per unit volume), and is the rate constant (units of reciprocal time).7. The rate constant has an Arrhenius temperature dependence:

exp(- ) 0

k

k = k E/RT (2-63)

where is the frequency factor, is the activation energy, and is the the gas constant.

0

k ER

Assumptions:1. Single, irreversible reaction, A → B.2. Perfect mixing.3. The liquid volume V is kept constant by an overflow line.4. The mass densities of the feed and product streams are equal

and constant. They are denoted by ρ.5. Heat losses are negligible.6. The reaction rate for the disappearance of A, r, is given by,


8.

9.

10.

11.

12.

CSTR: Model Development (cont’d)


CSTR: Model Development (cont’d)

• Unsteady-state mass balance

• Unsteady-state component balance

Because ρ and V are constant, . Thus, the mass balance is not required.

.


CSTR Model: Some Extensions

• How would the dynamic model change for:

1. Multiple reactions (e.g., A → B → C) ?2. Different kinetics, e.g., 2nd order reaction?3. Significant thermal capacity of the coolant liquid?4. Liquid volume V is not constant (e.g., no overflow line)?5. Heat losses are not negligible?6. Perfect mixing cannot be assumed (e.g., for a very

viscous liquid)?


LAPLACE TRANSOFRM – A Review

LAPLACE TRANSOFRM - Review


Pierre – Simon Laplace

Pierre – Simon LaplaceBorn: 23 March 1749 in Beaumont-en-Auge, Normandy, FranceDied: 5 March 1827 in Paris, France


Laplace Transforms

• Important analytical method for solving linear ordinarydifferential equations.

- Application to nonlinear ODEs? Must linearize first.

• Laplace transforms play a key role in important process control concepts and techniques.

- Examples:

• Transfer functions

• Frequency response

• Control system design

• Stability analysis


Definition

The Laplace transform of a function, f(t), is defined as

[ ] ( )0

( ) ( ) (3-1)stF s f t f t e dt∞ −= = ∫L

where F(s) is the symbol for the Laplace transform, L is the Laplace transform operator, and f(t) is some function of time, t.

Note: The L operator transforms a time domain function f(t) into an s domain function, F(s). s is a complex variable: s = a + bj, 1j −=


Inverse Laplace Transform, L-1

By definition, the inverse Laplace transform operator, L-1, converts an s-domain function back to the corresponding time domain function:

( ) ( )1f t F s− = L

Important Properties:

Both L and L-1 are linear operators. Thus,

( ) ( ) ( ) ( )( ) ( ) (3-3)

ax t by t a x t b y t

aX s bY s

+ = + = +

L L L


LT Propertieswhere:

- x(t) and y(t) are arbitrary functions

- a and b are constants

- ( ) ( ) ( ) ( )X s x t Y s y t L Land= =

Similarly,

( ) ( ) ( ) ( )1 aX s bY s ax t b y t− + = + L


Laplace Transforms of Common Functions

1. Constant Function

Let f(t) = a (a constant). Then from the definition of the Laplace transform in (3-1),

( )0

0

0 (3-4)st sta a aa ae dt es s s

∞∞ − − = = − = − − =

∫L


Step Function2. Step Function

The unit step function is widely used in the analysis of processcontrol problems. It is defined as:

( ) 0 for 0(3-5)

1 for 0t

S tt

< ≥

=

Because the step function is a special case of a “constant”, it follows from (3-4) that

( ) 1 (3-6)S ts

= L


Derivatives3. Derivatives

This is a very important transform because derivatives appear in the ODEs we wish to solve. In the text (p.53), it is shown that

( ) ( )0 (3-9)df sF s fdt

= − L

initial condition at t = 0Similarly, for higher order derivatives:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

11 2

2 1

0 0

... 0 0 (3-14)

nn n n

n

n n

d f s F s s f s fdt

sf f

− −

− −

= − − −

− − −

L


Derivatives (cont’d)where:

- n is an arbitrary positive integer

- ( ) ( )0

0k

kk

t

d ffdt =

=

Special Case: All Initial Conditions are Zero

Suppose Then

In process control problems, we usually assume zero initial conditions.

( ) ( ) ( ) ( ) ( )1 10 0 ... 0 .nf f f −= = =

( )n

nn

d f s F sdt

=

L


Exponential and Pulse Functions4. Exponential Functions

Consider where b > 0. Then, ( ) btf t e−=

( )

( )

0 0

0

1 1 (3-16)

b s tbt bt st

b s t

e e e dt e dt

eb s s b

∞ ∞ − +− − −

∞− +

= =

= − = + +

∫ ∫L

5. Rectangular Pulse Function

It is defined by:

( )0 for 0

for 0 (3-20)0 for

w

w

tf t h t t

t t

<= ≤ < ≥


h

( )f t

wtTime, t

The Laplace transform of the rectangular pulse is given by

( ) ( )1 (3-22)wt shF s es

−= −


Impulse function

6. Impulse Function (or Dirac Delta Function)The impulse function is obtained by taking the limit of therectangular pulse as its width, tw, goes to zero but holdingthe area under the pulse constant at one. (i.e., let )

Let,

Then,

1

wh

t=

( )tδ impulse function

( ) 1tδ = L

=


Laplace Table


Solution of ODEs by Laplace Transforms

Procedure:

1. Take the L of both sides of the ODE.

2. Rearrange the resulting algebraic equation in the s domain to solve for the L of the output variable, e.g., Y(s).

3. Perform a partial fraction expansion.

4. Use the L-1 to find y(t) from the expression for Y(s).


Transfer Functions

• Convenient representation of a linear, dynamic model.

• A transfer function (TF) relates one input and one output:

( )( )

( )( )system

x t y tX s Y s

→ →

The following terminology is used:

y

output

response

“effect”

x

input

forcing function

“cause”


Definition of TFLet G(s) denote the transfer function between an input, x, and an output, y. Then, by definition

( ) ( )( )

Y sG s

X s=

where:

( ) ( )( ) ( )

Y s y t

X s x t

L

L

=

=

Example: Model for a simple cylindrical tank- mass balance- linearization (if necessary)- deviation variable- time constant equivalent to residence time- gain of the process


First Order SystemThe standard form for a first-order TF is:

Cha

pter

5 where:

Consider the response of this system to a step of magnitude, M:

Substitute into (5-16) and rearrange,

( )( ) (5-16)

τ 1Y s KU s s

=+

steady-state gainτ time constantK

( ) ( )for 0 MU t M t U ss

= ≥ ⇒ =

=

=

( ) ( ) (5-17)τ 1KMY s

s s=

+


First Order SystemTake L-1 (cf. Table 3.1),

( ) ( )/ τ1 (5-18)ty t KM e−= −

Let steady-state value of y(t). From (5-18), y∞ .y KM∞ ==

t ___0 0

0.6320.8650.9500.9820.993

yy∞

τ2τ3τ4τ5τ

Cha

pter

5

00

21 543

1.0

0.5y

y∞

τt

Note: Large means a slow response.τ

ChE 6303 – Advanced Process Controlteacher.buet.ac.bd/shoukat/ChE6303_Handout1.pdf · Review of...

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