lecture 4: important structures of simple systems 1
TRANSCRIPT
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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
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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
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,1)(
)(
s
Ke
sU
sY Ls
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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
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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?
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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
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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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