Chapter 6

Dynamic Behavior of Ideal Systems

Overall Course Objectives

• Develop the skills necessary to function as an industrial process control engineer.– Skills

• Tuning loops

• Control loop design

• Control loop troubleshooting

• Command of the terminology

– Fundamental understanding• Process dynamics

• Feedback control

Ideal Dynamic Behavior

• Idealized dynamic behavior can be effectively used to qualitatively describe the behavior of industrial processes.

• Certain aspects of second order dynamics (e.g., decay ratio, settling time) are used as criteria for tuning feedback control loops.

Inputs

A A

t

a

A

P

First Order Process

• Differential equation

• Transfer function

• Note that gain and time constant define the behavior of a first order process.

)()()(

tuKtydt

tdypp

1)(

s

KsG

p

pp

First Order Process

u

y (t )

0.95 AK p

0.63 AK p

0 p 3

p

t

y

u

Determine the Process Gain and Process Time Constant from Gp(s)

85.0

directlydeterminedbecanandThen

15.0

8)(G

formstandardtoRearrange

2

16)(

p

p

p

pp

p

K

K

ss

ssG

Estimate of First-Order Model from Process Response

4timesettling

p

p u

yK

In-Class Exercise

• By observing a process, an operator indicates that an increase of 1,000 lb/h of feed (input) to a tank produces a 8% increase in a self-regulating tank level (output). In addition, when a change in the feed rate is made, it takes approximately 20 minutes for the full effect on the tank to be observed. Using this process information, develop a first-order model for this process.

Second Order Process

)()()(

2)(

2

22 tuKty

dt

tdy

dt

tydppp

• Differential equation

• Transfer function

• Note that the gain, time constant, and the damping factor define the dynamic behavior of 2nd order process.

12)(

22

ss

KsG

pp

pp

Underdamped vs Overdamped

Effect of on Overdamped Response

0

0.2

0.4

0.6

0.8

1

0 4 8 12t/ p

y(t)

/AK

p

=1

=3

=2

Effect of on Underdamped Response

0

0.5

1

1.5

2

0 4 8 12t/ p

y(t)

/AK

p

=1.0

=0.1

0.4

0.7

Effect of on Underdamped Response

-2

-1

0

1

2

3

4

0 4 8 12t /n

y( t

)/A

Kp

z=-0.1

z=0

Characteristics of an Underdamped Response

y(t)

trise

D

B C

T

±5%

trt

Time

• Rise time• Overshoot (B)• Decay ratio

(C/B)• Settling or

response time• Period (T)

Example of a 2nd Order Process

PT

PC

Vent

P sp

C.W.

• The closed loop performance of a process with a PI controller can behave as a second order process.

• When the aggressiveness of the controller is very low, the response will be overdamped.

• As the aggressiveness of the controller is increased, the response will become underdamped.

Determining the Parameters of a 2nd Order System from its Gp(s)

75.02

3

224

Then

134

2)(

formstandardtheintogRearrangin

5.05.12

1)(

2

2

p

pp

p

p

K

sssG

sssG

Second-Order Model Parameters from Process Response

p

Data: PI controller with 20% overshoot and

with a period of oscillation equal to 5 min.

Solution: PI controller yields K 1. With Equation 5.15

20% overshoot yields ζ 0.456. Then, Equation 5.17

with the period of oscillation yields 0.p

2

708 min

1( )

0.0502 1.29 1pG ss s

High Order Processes

Time

y

n=15

n=3

n=5

• The larger n, the more sluggish the process response (i.e., the larger the effective deadtime)

• Transfer function:

np

pp

s

KsG

1)(

Example of Overdamped Process

AT

LC

LC

AT

DL

B

V

• Distillation columns are made-up of a large number of trays stacked on top of each other.

• The order of the process is approximately equal to the number of trays in the column

Integrating Processes

0 20 40 60 80 100Time (seconds)

Fout

Ls

• In flow and out flow are set independent of level

• Non-self-regulating process

• Example: Level in a tank.

• Transfer function:

sAsG

cp

1)(

Deadtime

F

C A 0

F

FT

FCF spec

AT

L

• Transport delay from reactor to analyzer:

• Transfer function:

FALtCtC cs /where)()(

sp esG )(

FOPDT Model

• High order processes are well represented by FOPDT models. As a result, FOPDT models do a better job of approximating industrial processes than other idealized dynamic models.

Time

FOPDT Model

5th OrderProcess

Determining FOPDT Parameters

Timet1/3 t2/3

1/3 y

2/3 y

y

0

• Determine time to one-third of total change and time to two-thirds of total change after an input change.

• FOPDT parameters:

u

yKt

ttpppp

4.0

7.0 3/13/13/2

Determination of t1/3 and t2/3

616

615

314414

112313

4212

2011

6000

3/2

3/1

3/2

3/1

t

t

y

y

y

yut

In-Class Exercise

• Determine a FOPDT model for the data given in Problem 5.51 page 208 of the text.

Inverse Acting Processes

Time

y(t)

u(t)

• Results from competing factors.

• Example: Thermometer• Example of two first

order factors:

pppp

p

p

p

pp

andKK

s

K

s

KsG

11

)(

Lead-Lag Element

1

1)(

lg

s

ssG ld

Time

y(t

)

ld> lg

ld< lg

1.0

0.0

Recycle Processes

Feed

Product T f

T o T r

• Recycle processes recycle mass and/or energy.

• Recycle results in larger time constants and larger process gains.

• Recycles (process integration) are used more today in order to improve the economics of process designs.Energy Recycle

Mass Recycle Example

Steam

TT

C Product

LC

LC

LC

TT

Fresh BFeed

Fresh AFeed

PT

Steam

Overview

• It is important to understand terms such as:– Overdamped and underdamped response– Decay ratio and settling time– Rectangular pulse and ramp input– FOPDT model– Inverse acting process– Lead-Lag element– Process integration and recycle processes