Chap 04 Marlin 2002

7/29/2019 Chap 04 Marlin 2002

1/44

CHAPTER 4 : MODELLING &

ANALYSIS FOR PROCESS CONTROL

When I complete this chapter, I want to be

able to do the following.

Analytically solve linear dynamic modelsof first and second order

Express dynamic models as transfer

functions

Predict important features of dynamic

behavior from model without solving

7/29/2019 Chap 04 Marlin 2002

2/44

Outline of the lesson.

Laplace transform

Solve linear dynamic models

Transfer function model structure

Qualitative features directly from model

Frequency response

Workshop



7/29/2019 Chap 04 Marlin 2002

3/44

WHY WE NEED MORE DYNAMIC MODELLING

T

A

I can model this;

what more do

I need?

T

A

I would like to

model elements

individually

combine as needed

determine key

dynamic features

w/o solving

7/29/2019 Chap 04 Marlin 2002

4/44


T

A

I would like to

model elements

individually

This will be a

transfer function

Now, I can combine

elements to model

many process

structures

7/29/2019 Chap 04 Marlin 2002

5/44


Now, I can combine

elements to model

many processstructures

Even more amazing,

I can combine toderive a simplified

model!

7/29/2019 Chap 04 Marlin 2002

6/44

THE FIRST STEP: LAPLACE TRANSFORM

==

0

dtetfsftfL st)()())((

s

Ces

CdtCeCLt

t

stst ====

=

00

)(:Constant

Step Change at t=0: Same as constant for t=0 to t=

t=0

f(t)=0

7/29/2019 Chap 04 Marlin 2002

7/44


==

0

dtetfsftfL st)()())((

+==

000

11 dteedtedte)e())e((L st/tstst/t/t

s/1=

/s

e

/s

dtet)s/(t)s/(

10

1

10

1 11

+

=

+

==

+

+

)s(sss/ss 11111

11 +=+=+=

We have seen

this term

often! Its the

step response toa first order

dynamic system.

7/29/2019 Chap 04 Marlin 2002

8/44


Lets consider plug flow through a pipe. Plug flow has no

backmixing; we can think of this a a hockey puck

traveling in a pipe.

What is the dynamic response of the outlet fluid property

(e.g., concentration) to a step change in the inlet fluid

property?

Lets learn a new

dynamic response

& its Laplace

Transform

7/29/2019 Chap 04 Marlin 2002

9/44


Lets learn a new

dynamic response

& its Laplace

Transform

time

Xin

Xout

= dead time

What is the value of

dead time for

plug flow?

7/29/2019 Chap 04 Marlin 2002

10/44

0 1 2 3 4 5 6 7 8 9 10-0.5

0

0.5

1

time

Y,outletfromd

eadtime

0 1 2 3 4 5 6 7 8 9 10-0.5

0

0.5

1

time

X,

inlettodeadtime


Lets learn a new

dynamic response

& its Laplace

Transform Is this a

dead time?

What is thevalue?

7/29/2019 Chap 04 Marlin 2002

11/44


Lets learn a new

dynamic response

& its Laplace

Transform

The dynamic model for dead time is

)t(X)t(X inout =

The Laplace transform for a variable after dead time is

)())(())(( sXetXLtXL ins

inout ==

Our plants have

pipes. We will

use this a lot!

7/29/2019 Chap 04 Marlin 2002

12/44


We need the Laplace transform of

derivatives for solving dynamic models.

First

derivative:

General:

0==

t

)t(f)s(sf

dt

)t(dfL

constant

+++=

=

=

=

0

1

1

0

1

0

1

t

n

n

t

n

t

nn

n

n

dt

)t(fd....

dt

)t(dfs)t(fs)s(fs

dt

)t(fdL

constant

I am in desperate

need of examples!

7/29/2019 Chap 04 Marlin 2002

13/44

SOLVING MODELS USING THE LAPLACE

TRANSFORM

Textbook Example 3.1: The CSTR (or mixing tank)

experiences a step in feed composition with all other

variables are constant. Determine the dynamic response.

(Well solve this in class.)

F

CA0VCA

AAAA VkC')C'F(C'

dt

dC'V = 0

AA kCr

BA

=

kVF

FKand

kVF

Vwith''

'

+=

+==+ 0AA

A KCCdt

dC

I hope we get the sameanswer as with the

integrating factor!

7/29/2019 Chap 04 Marlin 2002

14/44


TRANSFORM

AA kCr

BA

=

F

CA0V1CA1

V2CA2

Two isothermal CSTRs are initially at steady state and

experience a step change to the feed composition to the

first tank. Formulate the model for CA2.


222212

2

111101

1

AAAA

AAAA

C'kV)C'F(C'dt

dC'V

C'kV)C'F(C'dt

dC'V

=

=

'''

'''

1222

2

0111

1

AAA

AAA

CKCdt

dC

CKCdt

dC

=+

=+

Much easier than

integrating factor!

7/29/2019 Chap 04 Marlin 2002

15/44


TRANSFORM

Textbook Example 3.5: The feed composition experiences a

step. All other variables are constant. Determine the

dynamic response of CA.


2AA kCr

BA

=

F

CA0VCA

Non-linear!

7/29/2019 Chap 04 Marlin 2002

16/44

TRANSFER FUNCTIONS: MODELS VALID

FOR ANY INPUT FUNCTION

Lets rearrange the Laplace transform of a dynamic model

Y(s)X(s) G(s)Y(s) = G(s) X(s)

A TRANSFER FUNCTION is the output variable, Y(s),

divided by the input variable, X(s), with all initial

conditions zero.

G(s) = Y(s)/X(s)

7/29/2019 Chap 04 Marlin 2002

17/44



Y(s)X(s) G(s)G(s) = Y(s)/ X(s)

How do we achieve zero initial

conditions for every model?

We dont have primes on the

variables; why?

Is this restricted to a step input?

What about non-linear models?

How many inputs and outputs?

7/29/2019 Chap 04 Marlin 2002

18/44




Some examples:

?)()(

)(:CSTRsTwo

?)()(

)(:tankMixing

==

==

sGsC

sC

sGsC

sC

A

A

A

A

0

2

0

7/29/2019 Chap 04 Marlin 2002

19/44




Why are we doing this?

To torture students.

We have individual models that we can

combine easily - algebraically.

We can determine lots of information

about the system without solving the

dynamic model.

I chose the

first answer!

7/29/2019 Chap 04 Marlin 2002

20/44



T

open

sm

sv

sFsG

valve %

/10.)(

)()(

30 ==

1250

2.1

)(

)()(

0

1

+

==

ssF

sTsG /sm

K 3tank1

1300

01

1

2

+==

s

KK

sT

sTsG

/.

)(

)()(tank2 110

01

2

+=

=

s

KK

sT

sT

sGmeasured

sensor

/.

)(

)(

)(

(Time in seconds)

Lets see how to

combine models

7/29/2019 Chap 04 Marlin 2002

21/44



The BLOCK DIAGRAM

Gvalve(s) Gtank2(s)Gtank1(s) Gsensor(s)

v(s) F0(s) T1(s) T2(s) Tmeas(s)

Its a picture of the model equations!

Individual models can be replaced easily

Helpful visualization

Cause-effect by arrows

7/29/2019 Chap 04 Marlin 2002

22/44


FOR ANY INPUT FUNCTIONCombine using BLOCK DIAGRAM ALGEBRA

Gvalve(s) Gtank2(s)Gtank1(s) Gsensor(s)

v(s) F0(s) T1(s) T2(s) Tmeas(s)

)()()()(

)(

)(

)(

)(

)(

)(

)(

)()(

)(

)(

sGsGsGsG

sv

sF

sF

sT

sT

sT

sT

sTsG

sv

sT

vTTs

measmeas

12

0

0

1

1

2

2

=

==

G(s)v(s) Tmeas(s)

7/29/2019 Chap 04 Marlin 2002

23/44


FOR ANY INPUT FUNCTIONKey rules for BLOCK DIAGRAM ALGEBRA

7/29/2019 Chap 04 Marlin 2002

24/44

QUALITATIVE FEATURES W/O SOLVING

FINAL VALUE THEOREM: Evaluate the final valve of

the output of a dynamic model without solving for the

entire transient response.

sY(s)lim)(

=st

tY

Example for first order system

pA

pA

stA KC

ss

KCstC 0

0

0

)1(

lim|)( =+

=

7/29/2019 Chap 04 Marlin 2002

25/44

...)]sin()cos([...

..)(...)(

+++

++++++++=t

ttt

q

p

etCtC

etBtBBeAeAAtY

21

2210210

21

What about dynamics

can we determine

without solving?


We can use partial fraction

expansion to prove the following

key result.

Y(s) = G(s)X(s) = [N(s)/D(s)]X(s) = C1/(s- 1) + C2/(s- 2) + ...

With i the solution to the denominator of the transfer

function being zero, D(s) = 0.

...)]sin()cos([...

..)(...)(

+++

++++++++=t

ttt

q

p

etCtC

etBtBBeAeAAtY

21

2210210

21

...)]sin()cos([...

..)(...)(

+++

++++++++=t

ttt

q

p

etCtC

etBtBBeAeAAtY

21

2210210

21

Real, distinct iReal, repeated i

Complex i

q is Re( i)

7/29/2019 Chap 04 Marlin 2002

26/44


With i the solutions to D(s) = 0, which is a polynomial.

...)]sin()cos([.....)(...)(

+++

++++++++=t

ttt

q

p

etCtCetBtBBeAeAAtY

21

2

210210

21

1. If all i are ???, Y(t) is stable

If any one i is ???, Y(t) is unstable

2. If all i are ???, Y(t) is overdamped(does not oscillate)

If one pair ofi

are ???, Y(t) is

underdamped

Complete statements

based on equation.

7/29/2019 Chap 04 Marlin 2002

27/44


With i the solutions to D(s) = 0, which is a polynomial.

...)]sin()cos([.....)(...)(

+++

++++++++=t

ttt

q

p

etCtCetBtBBeAeAAtY

21

2

210210

21

1. If all real [ i] are < 0, Y(t) is stable

If any one real [ i] is 0, Y(t) is unstable

2. If all i are real, Y(t) is overdamped (does notoscillate)

If one pair ofi

are complex, Y(t) is underdamped

7/29/2019 Chap 04 Marlin 2002

28/44

AA kCr

BA

=

F

CA0

V1C

A1V2CA2


'''

'''

1222

2

0111

1

AAA

AAA

CKCdt

dC

CKCdt

dC

=+

=+


1. Is this system stable?

2. Is this system over- or underdamped?

3. What is the order of the system?

(Order = the number of derivatives

between the input and output variables)

4. What is the steady-state gain?

Without

solving!

7/29/2019 Chap 04 Marlin 2002

29/44


FREQUENCY RESPONSE:The response to a sine input of

the output variable is of great practical importance. Why?

Sine inputs almost never occur. However, many

periodic disturbances occur and other inputs can be

represented by a combination of sines.

For a process without control, we want a sine input tohave a small effect on the output.

7/29/2019 Chap 04 Marlin 2002

30/44


0 1 2 3 4 5 6-0.4

-0.2

0

0.2

0.4

time

Y,outletfroms

ystem

0 1 2 3 4 5 6-1

-0.5

0

0.5

1

time

X,inlettosystem

input

output B

A

P

P

Amplitude ratio = |Y(t)| max / |X(t)| max

Phase angle = phase difference between

input and output

7/29/2019 Chap 04 Marlin 2002

31/44


Amplitude ratio = |Y(t)| max / |X(t)| max

Phase angle = phase difference between

input and output

For linear systems, we can evaluate directly using transfer function!

Set s = j , with = frequency and j = complex variable.

===

+===

))(Re(

))(Im(tan)(anglePhase

))(Im())(Re()(RatioAmp.

jG

jGjG

jGjGjGAR

1

22

These calculations are tedious by hand but easily performed in

standard programming languages.

7/29/2019 Chap 04 Marlin 2002

32/44


Example 4.15 Frequency response of mixing tank.

Time-domain

behavior.

Bode Plot - Shows

frequency response for

a range of frequencies

Log (AR) vs log( )

Phase angle vs log( )

7/29/2019 Chap 04 Marlin 2002

33/44


F

CA0V1CA1

V2CA2Sine disturbance with

amplitude = 1 mol/m3

frequency = 0.20 rad/min

Must have

fluctuations

< 0.050 mol/m3

CA2

Using equations for the frequency response amplitude ratio

0500120120011

1

2

2202

220

2

..).)(.(||)(

||||

)(|)(|

||||

>== +

=

+==

A

p

AA

p

A

A

C

KCC

KjGCC

Not acceptable. We need

to reduce the variability.

How about feedback

control?

Data from 2 CSTRs

7/29/2019 Chap 04 Marlin 2002

34/44

OVERVIEW OF ANALYSIS METHODS

We can determine

individual modelsand combine

1. System order

2. Final Value

3. Stability

4. Damping

5. Frequency response

We can determine

these features without

solving for theentire transient!!

Transfer function and block diagram

7/29/2019 Chap 04 Marlin 2002

35/44

Flowchart of Modeling Method

Goal: Assumptions: Data:

Variable(s): related to goals

System: volume within which variables are independent of position

Fundamental Balance: e.g. material, energy

Check

D.O.F.

Is model linear? Expand in Taylor Series

DOF = 0 Anothe r equa tion:-Fundamental balance

-Constitutive equations

DOF 0

No

Express in deviation variables

Group parameters to evalua te [gains (K), time-constants (), dead-times()]

Take Laplace transform

Substitute specific input, e.g.,

step, and solve for output

Ana lytical solution

(step)

Numerical solution

Analyze the model for:

- causality- order- stability

- damping

Yes

Combine several models into

integrated sys tem

We can use astandard modelling

procedure to

focus our

creativity!

Combining Chapters 3

and 4

7/29/2019 Chap 04 Marlin 2002

36/44

Too small to read here - check it out in the textbook!

7/29/2019 Chap 04 Marlin 2002

37/44

CHAPTER 4: MODELLING & ANALYSIS WORKSHOP 1

Example 3.6 The tank with a drain has a continuous flow in

and out. It has achieved initial steady state when a step

decrease occurs to the flow in. Determine the level as afunction of time.

Solve the linearized model using Laplace transforms

7/29/2019 Chap 04 Marlin 2002

38/44


1. System order

2. Final Value

3. Stability

4. Damping

5. Frequency response

The dynamic model for a non-

isothermal CSTR is derived in

Appendix C. A specific example

has the following transferfunction.

Determine the features

in the table for this

system.

)..()..(

)()(

80357918345076

2 ++=ss

ssFsT

c

T

A

7/29/2019 Chap 04 Marlin 2002

39/44


F

CA0V1CA1

V2CA2

Answer the following using the MATLAB program

S_LOOP.

Using the transfer function derived in Example 4.9,determine the frequency response for CA0 CA2. Check

one point on the plot by hand calculation.

7/29/2019 Chap 04 Marlin 2002

40/44


Feed

Vaporproduct

Liquid

productProcessfluid

Steam

F1

F2 F3

T1 T2

T3

T5

T4

T6 P1

L1

A1

L. Key

We often measure pressure for process monitoring and

control. Explain three principles for pressure sensors, select

one for P1 and explain your choice.

7/29/2019 Chap 04 Marlin 2002

41/44



When I complete this chapter, I want to be

able to do the following.

Analytically solve linear dynamic models of first

and second order

Express dynamic models as transfer functions

Predict important features of dynamic behavior

from model without solving

Lots of improvement, but we need some more study!

Read the textbook

Review the notes, especially learning goals and workshop

Try out the self-study suggestions Naturally, well have an assignment!

7/29/2019 Chap 04 Marlin 2002

42/44

LEARNING RESOURCES

SITE PC-EDUCATION WEB

- Instrumentation Notes

- Interactive Learning Module (Chapter 4)

- Tutorials (Chapter 14) Software Laboratory

- S_LOOP program

Other textbooks on Process Control (see course

outline)

7/29/2019 Chap 04 Marlin 2002

43/44

SUGGESTIONS FOR SELF-STUDY

1. Why are variables expressed as deviation variables

when we develop transfer functions?

2. Discuss the difference between a second order reaction

and a second order dynamic model.

3. For a sine input to a process, is the output a sine for aa. Linear plant?

b. Non-linear plant?

4. Is the amplitude ratio of a plant always equal to or

greater than the steady-state gain?

7/29/2019 Chap 04 Marlin 2002

44/44

SUGGESTIONS FOR SELF-STUDY

5. Calculate the frequency response for the model in

Workshop 2 using S_LOOP. Discuss the results.

6. Decide whether a linearized model should be used for

the fired heater for

FT

1

FT

2

PT

1

PIC

1

AT

1

TI

1

TI

2

TI

3

TI

4

PI

2

PI

3

PI

4

TI

5

TI

6

TI7

TI

8

TI

9

FI

3

TI

10

TI

11

PI

5

PI

6

a. A 3% increase in thefuel flow rate.

b. A 2% change in the

feed flow rate.

c. Start up from ambienttemperature.

d. Emergency stoppage

of fuel flow to 0.0.

fuel

feed

air

Chap 04 Marlin 2002

Documents

Transcript of Chap 04 Marlin 2002