Lecture 4:

Important structures of simple systems

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Outline of the lesson.

• Important structures of simple systems- Series - Parallel- Recycle

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When I complete this chapter, I want to be able to do the following.

• Derive the dynamics for important structures of simple dynamic systems

• Recognize the strong effects on process dynamics caused by process structures

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3.4.3. Interacting Tanks-in-Series

• The following tanks are interacting:

• The output of the second tank (level h2) affects the flow between tanks and hence affects the input of the same tank. Hence, this is called an interacting series.

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3.4.3. Non-Interacting Tanks-in-Series

• The output from an element does not influence the input to the same element

• Common example is tanks in series with pumped flow in between.

• Block diagram as shown

T

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

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

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STRUCTURES OF PROCESS SYSTEMS

NON-INTERACTING SERIES

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

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

)()(

)(sG

sX

sYi

n

i1

In general:

With each element a first order system:

)()(

)(

11 s

K

sX

sY

i

in

i

• overall gain is product of gains

• no longer first order system (becomes of

higher-order )

• slower than any single element

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3.4. Higher-Order Processes are Slower

• The more tanks we have in a series, the longer we have to wait until the last tank “sees” the changes that we have made in the first one.

• Hence, the more tanks in the series, the more sluggish the response of the overall process.

• Processes that are products of first-order functions are also called multicapacity processes.

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Numerical illustration

• Numerical calculation is used to illustrate that the resulting response becomes more sluggish.

• If we assume that all stages have the same time constant, then the whole system can be modeled as

• This particular case is not common in reality, but is a useful textbook illustration.

• Let us simulate this system for n=1,2,3,…

8

nn

ssU

sY

)1(

1

)(

)(

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tau = 3; % Just an arbitrary time constantG = tf(1,[tau 1]);step(G); % First-order function unit-step responseholdstep(G*G); % Second-order responsestep(G*G*G); % Third-order responsestep(G*G*G*G); % Fourth-order responsestep(G*G*G*G*G); % Fifth-order response

Approximation of high-order processes with First-Order Process plus Dead Time (FOPDT)

• It is clear that, as n increases, the response becomes slower. If we ignore the “data” at small times, it looks like that some dead time occurs.

• For this reason, high-order processes can be usually approximated with first-order process plus dead-time (FOPDT) of the following form

10

,1)(

)(

s

Ke

sU

sY Ls

Example 3.4

• Find a simple first-order approximation of the following transfer function

• In this example, the dominant pole is at − 1 / 3, corresponding to the largest time constant at 3 (time unit).

• Accordingly, we may approximate the full-order function as

• where 1.6 is the sum of the smaller time constants 0.1, 0.5, and 1.

11

,13

3

)(

)( 6.1

s

e

sU

sY s

.)13)(1)(15.0)(11.0(

3

)(

)(

sssssU

sY

Example 3.4

• The unit-step response of the full-order function and that of the FOPDT approximation are shown below.

• The approximation is reasonable when time is large enough when the pole at − 1 / 3 can indeed be considered as dominant.

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STRUCTURES OF PROCESS SYSTEMS

Class Exercise: Sketch the step response for the system below.

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STRUCTURES OF PROCESS SYSTEMS

Class Exercise: Sketch the step response for the system below.

0 5 10 15 20 250

1

2

3

4

5DYNAMIC SIMULATION

Time

Con

trol

led

Var

iabl

e

0 5 10 15 20 250

1

2

3

4

5

Time

Man

ipul

ated

Var

iabl

e

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STRUCTURES OF PROCESS SYSTEMS

Class Exercise: Sketch the step response for each of the systems below and compare the results.

Case 1

= 2 = 2 = 2 = 2

= 2 = 1 = 1 = 2 & = 2

Case 2

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0 2 4 6 8 10 12 14 16 18 200

1

2

3

4

time

case

1 r

esp

on

ses

0 2 4 6 8 10 12 14 16 18 200

1

2

3

4

time

case

2 r

esp

on

ses

Two plants can have different intermediate variables and have the same input-output behavior!

Step

Case1

Case2

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3.5. Effect of Zeros in Time Response

• As we know, the inherent dynamics is governed by the poles, but the zeros can impart finer “fingerprint” features by modifying the coefficients of each term in the time-domain solution.

• One common illustrations on the effects of zeros is the sum of two functions in parallel.

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3.5.2. Transfer Functions in Parallel

• PARALLEL STRUCTURES result from more than one causal path, with different time constants, between the input and output. This can be a flow split, but it can be from other process relationships.

G1(s)

G2(s)

U(s) Y(s)

A B C

Example process systems Block diagram

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3.5.2. Transfer Functions in Parallel

• Assume that both elements are first order, then the overall model is

• We can combine the two terms to give the second-order function with a zero

• where

1

1

2

2

1

1

s

K

s

K

U

Y

.,

,)1)(1(

)1(

21

122121

21

KK

KKKKK

ss

sK

U

Y

z

z

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Example

• We will compare the response of three transfer functions:

zeropositivess

ssG

zeronegativess

ssG

zerosnoss

sG

.)12)(1(

13 )(

,)12)(1(

13 )(

,)12)(1(

1 )(

3

2

1

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Matlab code

G1 = tf(1,conv([1 1],[2 1]));

G2 = tf([3 1],conv([1 1],[2 1]));

G3 = tf([-3 1],conv([1 1],[2 1]));

step(G1)

hold on

step(G2)

step(G3)

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Which wouldbe difficult/easy

to control?

Conclusion on Parallel Structures

PARALLEL STRUCTURES can experience complex dynamics due to the presence of zeros in the transfer function and this may be sometimes difficult to control.

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STRUCTURES OF PROCESS SYSTEMS: RECYCLE STRUCTURES

RECYCLE STRUCTURES result from recovery of material and energy. They are essential for profitable operation, but they strongly affect dynamics.

RECYCLE can be considered analogous to a positive feedback mechanism. Hence, systems with recycle tends to have longer response times (large time constants) and also may cause instability.

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RECYCLE STRUCTURES

H1(s) G(s)

H2(s)

Y0(s) Y(s)

)()(1

)()(

)(

)(

2

1

0 sHsG

sHsG

sY

sY

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Class exercise: Determine the effect of recycle on the dynamics of the given chemical reactor (faster or slower?).

110

3)(

30.0)(

40.0)(

2

1

ssG

sH

sH

OVERVIEW OF PROCESS SYSTEMS

Even simple elements can yield complex dynamics when combined in typical process structures.

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Mason’s gain formula

• It is a powerful method to find the transfer function between two variables in a way that is easier than block diagram reduction.

• Mason’s gain formula gives the transfer function between two variables as

• Where the different terms are defined in the next slide.

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f

iiiF

sG 1)(

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Example

• Find the transfer function C/R in the following system.

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