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Tier II
Worked Examples
Module 5 – Controllability Analysis
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Tier II Statement of intent
The goal of this tier is to demonstrate various concepts and
tools of Controllability Analysis using real examples. Some
examples will be given, focusing mainly on Controllability
Analysis tools. At the end of Tier II, the student should have
a general idea of what is:
Relative Gain Array
Niederlinski Index
Controller design
Design of multivariable controllers
Steady State Decoupling (Singular Value
Decomposition)
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2.0 Inverse of a matrix
To obtain the Relative Gain Array (RGA) of a transfer function matrix,
first of all it is necessary to review how to obtain the inverse of a
matrix.
The inverse of a matrix does not exist for all matrices, it exists
only if:
The matrix is square, and
Its determinant is not zero (non-singular matrix).
Now given the 3x3 matrix A, it is desired to obtain the inverse
of the matrix A.
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1. Determinant of matrix A.
The determinant of matrix A is obtained using the cofactors based
on any row. In this case the first row is selected.
1 4 0 4 0 1
 3
A  2
 1



5
2
2
3
5
3


 
 
2 1 3
A  0 1 4
5 2 3
 2(3  8)  1(0  20)  3(0  5)  5
As it can be seen, the determinant is not equal to zero
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In a 3x3 matrix there is no problem to calculate the determinant.
Nevertheless it is important to note that the second cofactor is being
multiplied by the factor of (-1).
This is because the cofactor of each matrix element must be multiplied
by the following term:
(1)i j
where i is the row number and j is the column number of the element
for which the cofactor is calculated.
It means that for matrix A, the cofactors which are multiplied by (-1)
are:
C(12), C(21), C(23) and C(32)
2 1 3
A  0 1 4
5 2 3
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2. Matrix of cofactors (C).
Now it is necessary to calculate the cofactors matrix (C) for
each element of matrix A.
C11 
1 4
2 3
 5
0 4
C12  
5 3
 20
1 3
C21  
3
2 3
2 3
C22 
 9
5 3
1 3
C31 
1
1 4
2 3
C32  
0 4
 8
2 1 3 


A  0 1 4 
0 1
5 2 3 
C13 
 5
5 2
2 1
C23  
1
5 2
2 1
C33 
2
0 1
And this way the cofactor matrix has been obtained:
Module 5 – Controllability Analysis
 5 20  5
C   3  9 1 
 1  8 2 
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3. Adjoint of (A).
The Adjoint of A is the transpose of matrix C.
 5 20  5
C   3  9 1 
 1  8 2 
1
 5 3
adj( A)  CT   20  9  8
 5 1
2 
4. Now the inverse of A is obtained using the determinant and the
adjoint of A.
 1 3

5
3
1


5

1
CT
1 

1
Α 
adj( A) 

20  9  8   4 9

5
Α
Α
5

 5 1
2   1  1
5

1 
5
8 
5 
2 
5
5. Check now that by matrix multiplication that the identity matrix is
obtained multiplying A and A-1.
3
2 1 3  1  5
A  A 1  0 1 4   4 9
5

5 2 3  1  1
5

Module 5 – Controllability Analysis
 1  1 0 0
5
8   0 1 0   

5  
 2  0 0 1
5
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As can be seen, the amount of work is extensive, just to calculate the
inverse of a matrix!.
Therefore, it will now be shown how to obtain the same inverse
matrix A-1 of matrix A now using Excel.
1. Determinant of matrix A, using Excel.
Fill the numbers of the matrix in an excel sheet, using a cell for
each element of the matrix.
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2.
Calculate the matrix determinant, this will be done using cofactors of
the first row.
Each cofactor must be calculated using the appropriate formula as shown
in the Excel formula bar. Pay attention on cofactor C(12) because it must be
negative, as it has been shown previously.
-
+
Elements C(11), C(12) and C(13) are the cofactors of each element of the
first row, which are elements a11, a12 and a13 according to matrix A:
a11
a12
a13
A  a21
a31
a22
a32
a23
a33
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3. Calculate the matrix determinant, using the formula shown on the
excel sheet.
Remember that the cofactor of element a12 must be
negative!
The determinant of this matrix is not equal to zero. For this reason it
is possible to obtain its inverse matrix.
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4. Calculate the cofactors of matrix A for the second and third rows.
Now remember, the cofactors for the elements C(21), C(23), and C(32), are:
(select one)
POSITIVE
NEGATIVE
As you can see in the Excel sheet, it is necessary to include the
sign for each cofactor, and it arithmetically appears on the cell that
calculates the matrix determinant (red one). Do not confuse this
sign with the one that has been written in front of the matrix in the
excel sheet (blue one). The latter one is just to show the sign of
the cofactor.
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5. Create the cofactors matrix C placing each cofactor in the
corresponding place of the element of which it has been
calculated, as the excel formula bar shows.
Do the same with the second and third rows of matrix C.
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6. Calculate the adjoint of matrix A, this will be done just transposing
the matrix C (matrix of cofactors).
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6. Calculate the inverse of matrix A (A-1). To do this, divide the
Adjoint matrix by the determinant of matrix A, already calculated.
And do the same for the remaining elements of matrix A-1
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7. Check by matrix multiplication that matrix A multiplied by A-1
gives the Identity Matrix.
And do the same for the rest of elements of matrix A-1
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As can be seen, even using excel to obtain the inverse of a matrix is
still hard work.
But there exist some functions in Excel that allow you to obtain the
inverse of a matrix rapidly. This will be shown next.
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Now it will be shown how to use some Excel functions to
manipulate matrices.
1. Determinant of matrix A, using an Excel functions.
Fill the numbers, as before, of the matrix in an Excel sheet, using a
cell for each element of the matrix. Select the cell where the
determinant is to appear. Select function/insert and choose the
function mdeterm.
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2. Next, it is necessary to select the range of the matrix data.
This way, the determinant of A is easily calculated.
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3. Now the transpose matrix of A will be calculated.
Select the cell where the first element a11T is to appear. Select
function/insert and choose the function transpose.
Next, select the range of the matrix, as the next slide shows.
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4. Select the range of the matrix, as shown below.
Once the range of the matrix has been selected press OK.
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5. No values appear in the cell because it is necessary to introduce
the formula as an array. To do that, select the range that the
transpose matrix will occupy and press F2. Then, press
shift+ctrl+enter. The Excel sheet should look like the one below.
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6. It is possible to obtain directly the inverse of a matrix. Select the
cell where the first element a11-1 is to appear. Select
function/insert and choose the function minverse.
Then, select the range of the matrix, as shown on the next slide.
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7. Select the range of the matrix, as shown below.
Once the range of the matrix has been selected press OK.
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8. Again no values appear in the cell because it is necessary to
introduce the formula as an array. To do that, select the range
that the inverse matrix will occupy and press F2. Next, press
shift+ctrl+enter. The Excel sheet must look like the one below.
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The determinant, transpose and inverse of matrix A can easily be obtained.
To verify if the Identity matrix is obtained, the function MMULT of excel is
used. Matrix A can be multiplied by Matrix A-1.
9. Multiplying matrices. Select the cell where the element I11 is to
appear. Select function/insert and select the function mmult.
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10. Next, select the range of each matrix to be multiplied, as shown
below.
Once the ranges of the matrices have been selected, press OK.
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11. No values appear in the cell because it is necessary to introduce the
formula as an array. To do that, select the range that the identity
matrix will occupy and press F2. Next, press shift+ctrl+enter. The
Excel sheet should look like the one below.
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As it was seen, Excel functions are very helpful to obtain the
determinant, the transpose and the inverse of a matrix, even
for matrix multiplication.
Of course there are several software packages with the
availability to work with matrices, but Excel has been chosen
because it is available in almost every computer that students
have access.
These Excel functions are the main tools that will serve to
obtain the Relative Gain Array, as it will be shown shortly.
Few examples will be covered.
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Despite the great help that Excel can provide, some limitations
must be specified. These limitations are:
Determinant.
The size of the array must not exceed 73 rows by 73
columns.
Multiplication.
The size of the resulting array must not be equal or greater
than a total of 5 461 cells.
Inverse.
The size of the array must not exceed 52 columns by 52
rows
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2.1 Relative Gain Array
2.1.1 Obtain the RGA for the linear model of a distillation column used
in separating methanol and water as reported in [1] (see next
slide). It is a system with two output variables, two input
variables, and one disturbance variable. All variables are defined in
terms on deviation variables:
y1= overhead mole fraction methanol
y2= bottoms mole fraction methanol
d = column feed flowrate
u1= overhead reflux flowrate
u2= bottoms steam flowrate
The 2x2 transfer function matrix is:
 12.8e  s

1
G ( s )  16.7 s
7s
 6.6e
10.9s  1
 18.9e 3s 

21.0s  1 
 19.4e 3s 
14.4s  1 
Module 5 – Controllability Analysis
 3.8e 8.1s 


14
.
9
s

1

G d ( s)  
3.4 s
 4.9e

 13.2s  1 
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Distillation column used in separating methanol
and water
Overhead reflux
flow rate (u1)
Feed
flow
rate (d)
Overhead mole fraction
methanol (y1)
Bottoms steam
flow rate (u2)
Bottoms mole
fraction methanol
(y2)
It is easy to identify both manipulated and controlled variables.
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Problem description.
From the transfer function matrix G(s), it is possible to obtain the
steady-state gain matrix.
The steady-state gain matrix is:
12.8  18.9
K  G(0)  

6
.
6

19
.
4


Based on this matrix, the Excel functions seen previously can be
used to obtain the RGA as the next slide shows.
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It is possible to use any Excel sheet shown before, or you can
start a new one as the next figure shows.
R=
Matrix K
Matrix R
Be careful, because the multiplication of matrices K and R is a
multiplication term by term.
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The RGA has therefore been obtained very easily:
 2.01  1.0


 1.0 2.01
The pairing rules recommend pairing 1-1/2-2, which means to
use the overhead reflux flowrate to control the overhead mole
fraction of methanol, and the bottoms steam flowrate to
control the bottoms mole fraction of methanol.
The final coupling is shown in the next slide.
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Final coupling suggested for RGA from a distillation
column used to separate methanol and water
LC
CC
Feed
flow
rate (d)
Overhead reflux
flow rate (u1)
Overhead mole fraction
methanol (y1)
Bottoms steam
flow rate (u2)
LC
CC
Bottoms mole
fraction methanol
(y2)
Next, the RGA for a 3x3 system will be calculated.
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2.1.2 Obtain the RGA for pilot scale binary distillation column used to
separate ethanol and water for which the transfer function matrix
is given below [2]. The process variables are (in terms of
deviations from their respective steady state values):
y1 = overhead mole fraction ethanol
u1 = overhead reflux flowrate
y 2 = side stream ethanol mole fraction
u2 = side stream draw-off rate
y3 = Temperature on Tray #19
u3 = reboiler steam pressure
 0.66e-2.6s

6.7s+1

y
 1
-6.5s
 y  =  1.11e
 2   3.5s+1
 y 3  
 -33.68e-9.2s

 8.15s+1
-0.61e-3.5s
8.64s+1
-2.3e-3s
5s+1
46.2e-9.4s
10.9s+1
Module 5 – Controllability Analysis

-0.0049e-s

9.06s+1
 u 
  1
-0.012e-1.2s
 u2 
7.09s+1
 u 
0.87 11.61s+1 e-s   3 

 3.89s+118.8s+  
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Distillation column used in separating ethanol and
water
Temperature on
tray #19 (y3)
Overhead
reflux flow
rate (u1)
Feed flow
rate (d)
Overhead mole fraction ethanol (y1)
Side stream draw-off rate (u2)
Mole fraction of ethanol in the side stream (y2)
Reboiler steam pressure (u3)
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Problem description.
From the transfer function model, the steady-state gain matrix is:
 0.66 -0.61 -0.0049 
K = G(0) =  1.11
-2.3 -0.012 
 -33.68 46.2
0.87 
R=
Matrix K
Module 5 – Controllability Analysis
Matrix R
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Again, remember that the multiplication of matrices K and R is
a multiplication term by term.
The RGA for the 3x3 system has been obtained easily:
 1.95 -0.67 -0.27 
 = -0.66 1.90 -0.23 
-0.28 -0.23 1.51 
The pairing rules recommend pairing 1-1/2-2/3-3, which
means that the overhead mole fraction of ethanol can be
controlled using the overhead reflux flowrate. In the same way,
the mole fraction of ethanol in the side stream can be
controlled using the side stream draw-off rate and the
temperature on tray #19 can be controlled using the reboiler
steam pressure.
The next slide shows the final coupling.
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Final coupling suggested for RGA from a distillation
column used in separating ethanol and water
Overhead mole fraction ethanol (y1)
Temperature
on tray #19
(y3)
CC
Overhead reflux flow rate (u1)
Mole fraction of ethanol in the side stream (y2)
TC
CC
Feed flow
rate (d)
Side stream draw-off rate (u2)
Reboiler steam pressure (u3)
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2.2 Niederlinski Index
Up to now the utility of the RGA has been used to find
the appropriate pairing for the process variables.
Despite the utility of the RGA, sometimes it is necessary
to use the RGA with another important tool such as the
NIEDERLINSKI INDEX (NI).
The NI is very useful because it allows to identify
structurally unstable pairings and as a result to avoid
them.
Next the use of NI is shown.
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2.2.1 Determine the best pairing using the RGA and the NI for the
system with three output variables and three input variables, the
steady state matrix is:
1  0.1
1
K   2  3
1 
 0.1 2
 1 
u1 u2 u3
y1
y2
y3
Problem description.
First of all, since the steady sate matrix is know, the RGA is
obtained as before and the next slide shows just the RGA..
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Then, the RGA obtained is:
  1.89 3.58  0.70
RGA   3.02  5.60 3.58 
 0.13 3.02  1.89
Click to
Interchange
Row 2 and 3
According to the RGA, the only feasible pairing has to involve a negative
RGA element, so it is possible to interchange rows 2 and 3.
Rows 2 and 3 of RGA have been interchanged.
Now the steady state matrix is:
1  0.1 y1
1
K   0.1 2
 1  y3
 2  3
1  y2
u1 u2 u3
Interchanging the rows was necessary to calculate the Niederlinski
Index, because all the pairing elements must be on the diagonal of
the K matrix, as next slide shows.
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Now the NI shows that the system is not structurally unstable even
pairing a negative element.
Niederlinski Index
Index
Niederlinski
1. Obtain the steady
state matrix lKl and
its determinant.
2. Obtain the
diagonal matrix of
lKl and its
determinant.
3. Obtain the NI
ratio:
NI 
G(0)
n

i 1
gii (0)
Module 5 – Controllability Analysis
K
0.53


 0.27
Diag(K )
2
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According to these calculations, for the pairing 1-1/2-3/3-2, the system is
not unstable, despite the rules of RGA. However if the first loop is
opened (y1– u1) or not included in the process model, the resulting
subsystem is unstable as will be shown.
o
First loop of matrix K open (K):
1  0.1 y1
1

 2  1 y3

K   0.1 2 K  1  y3
 3 1  y2
 2  3
1  y2 
u1 u2 u3u2 u3
o
And the NI for matrix K is:

NI K  
2  (3)
1

2
2
Reminder: A negative NI indicates that the system is structurally
UNSTABLE.
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2.3 Nonlinear Systems
Despite that many chemical processes can be adequately
represented by linear systems, via linear transfer function models,
the majority of chemical processes are inherently nonlinear and
sometimes need to use nonlinear models in order to be valid in a
wider range of operation.
8
6
4
2
0
-2
-4
-6
5
10
5
0
-5
-5
0
-8
30
20
It is therefore necessary to see how the pairing of input and output
variables of nonlinear systems is performed.
-10
-10
10
0
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5
10
15
20
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The same information used for the RGA of steady state systems is
used for nonlinear systems. This feature can appear for someone a
"disadvantage", but it is precisely this "disadvantage" involving only
steady state values that can be used to handle nonlinear systems.
Disadvantage
or advantage???
Take a decision
PLEASE !!!!!
Next, an example shows that according to the steady state values, the
pairing of both manipulated and controlled variables is selected.
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2.3.1 Obtain the RGA for the multivariable system, a stirred mixing tank,
consisting in a hot stream and a cold stream which are used to control
the liquid level and the tank temperature, and use it to recommend
which of the manipulated variables should be used to control the liquid
level and the tank temperature. The transfer function matrix is shown
below.
y1= Liquid level
u1= Hot stream flowrate
y2= Tank temperature
u2= Cold stream temperature
1




k
 A  s+

 C  2A C hs 


G  s = 

 TH -Ts 


k 

A
h
s+
 C s  Ac h 
s 


1




 


k
k
 A  s+
A C  s+

 
 2A h  
 C  2A C hs 
C
s 



 G  s = 

 d
 Tds -Ts 
 TC -Ts 





k 

k

A Chs  s +

 A Chs  s + A h 


A C hs  
C
s 



1






Fds


k 
A Chs  s+

 A h 
C
s 

0
Where k is a constant (see next slide) and Ac is the cross-section area
for the tank.
Module 5 – Controllability Analysis
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Problem description.
A diagram of the tank is shown below.
COLD STREAM
FLOWRATE (u2)
TANK TEMPERATURE (y2)
DISTURBANCE (Td, Fd)
HOT STREAM
FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
h
Output Flow rate (F, T)
F=k(h)½
Next it is necessary to obtain the steady state gain matrix, as next
slide shows.
Module 5 – Controllability Analysis
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 2 hs

k

K = G 0 =
  TH -Ts 

 k hs
 2 hs
1
K =   TH -Ts 
k
 hs
2 hs 

k 
 TC -Ts  

k hs 
2 hs 

 TC -Ts  
hs 
Now with the steady-state gain matrix, it is possible to
obtain the RGA, using the full matrix method or, since it is a
2x2 matrix, in a more direct way as seen before.
Module 5 – Controllability Analysis
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To calculate the RGA of this 2x2 system it is possible to obtain l11
using the First Principles.
In Tier 1 has seen that:
1
l11 
 l22
1 
and,
K12 K 21

K11K 22
Substituting the values of the matrix K:
 2 hs

 k
 = 
 2 hs

 k

   TH -Ts  


 k h 
TH -Ts 

s 

=
   TC -Ts  
 TC -Ts 


 k h 
s 

Module 5 – Controllability Analysis
TC -Ts 

1
l11 =
=
TH -Ts 
TC -TH 


1 TC -Ts 
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In a similar way it is possible to obtain l12 using:

l12 
 l21
1 
According to this the value of l12 is:
l12 =
TH -Ts 

 TC -Ts 
TH -Ts 

1 TC -Ts 
=
-  TH -Ts 
 TC -TH 
And the RGA for this systems is:
Module 5 – Controllability Analysis
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 TC -Ts

TC -TH

Λ=
 -  T -T 
 H s
 TC -TH
-  TH -Ts  

TC -TH 
TC -Ts 

TC -TH 
Here, it is due to mention that the RGA depends only of values of
hot and cold streams, and also for the values of the steady state
values involved in the steady-sate gain matrix.
The values given to hot and cold stream are:
TH = 65ºC
TC = 15ºC
These values are fixed and just the value of Ts will be changed
according to different scenarios.
Module 5 – Controllability Analysis
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Five different values will given to Ts :
Ts > (TH+TC)/2;
Ts < (TH+TC)/2;
Ts = 55ºC
Ts = 25ºC
Ts = (TH+TC)/2 ;
Ts = 40ºC
Ts = TH:
Ts = TC;
Ts = 65ºC
Ts = 15ºC
According to this values the system of the stirred mixing tank is:
COLD STREAM FLOWRATE (u2)
TC = 15ºC
TANK TEMPERATURE (y2)
TH = 65ºC
DISTURBANCE (Td, Fd)
HOT STREAM FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
h
F=k(h)½
Output Flow rate (F, T)
Module 5 – Controllability Analysis
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Case 1.
Ts > (TH+TC)/2;
TC
COLD STREAM FLOWRATE (u2)
TC= 15ºC
Ts = 55ºC
TANK TEMPERATURE (y2)
DISTURBANCE (Td,Fd)
TH = 65ºC
HOT STREAM FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
LLC
h
Output Flow rate
(F, T)
F=k(h)½
From the RGA, the suggested pairing is 1-1/2-2.
The physical meaning of this pairing is: since the temperature
of cold stream (TC) is farther away from the steady state
Click to
operating tank temperature, small
changes in the cold stream
Pairing
produce noticeable changes in1-1/2-2
the tank temperature, whereas
the temperature of the hot stream (TH) is closer to the
operating steady state temperature, it can be used to control
the level without causing significant changes in the tank
temperature.
Module 5 – Controllability Analysis
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Case 2.
Ts < (TH+TC)/2;
Ts = 25ºC
TC
COLD STREAM FLOWRATE (u2)
TANK TEMPERATURE (y2)
TC = 15ºC
DISTURBANCE (Td,Fd)
TH = 65ºC
HOT STREAM FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
F=k(h)½
LLC
h
Output Flow rate
(F, T)
From the RGA, the suggested pairing is 1-2/2-1.
Again, the physical meaning of this pairing is: since the
temperature of the hot stream (TH) is farther away from the
Click to
steady state operating tank temperature,
small changes in the
Pairing
hot stream produce noticeable1-2/2-1
changes in the tank
temperature, whereas the temperature of cold stream (TC) is
closer to the operating steady state temperature, it can be
used to control the level without causing significant changes in
the tank temperature.
Module 5 – Controllability Analysis
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Case 3.
Ts = (TH+TC)/2;
COLD STREAM FLOWRATE (u2)
Ts = 40ºC
TANK TEMPERATURE (y2)
TC = 15ºC
DISTURBANCE (Td,Fd)
TH = 65ºC
HOT STREAM FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
F=k(h)½
h
Output Flow rate
(F, T)
Here the values of the RGA are all equal to 0.5 For
this reason it is equally bad to pair 1-1/2-2 than 12/2-1, because the operating temperature is exactly
equidistant from both the cold stream temperature
and the hot stream temperature.
A poor control of the process under this undesirable
special condition is obtained.
Module 5 – Controllability Analysis
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Case 4.
Ts = TH;
Ts = 65ºC
TC
COLD STREAM FLOWRATE (u2)
TC = 15ºC
TANK TEMPERATURE (y2)
DISTURBANCE (Td,Fd)
TH = 65ºC
HOT STREAM FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
F=k(h)½
LLC
h
Output Flow rate
(F, T)
From the RGA, the suggested pairing is 1-1/2-2. It is
possible to achieve a perfect control of the level tank,
without interacting with the temperature, using the hot
stream.
Here, the temperature of the hot stream (TH) is the same that
Click
to
the steady state operating tank
temperature.
For that reason
Pairing
this stream is used to control the level of the tank, whereas the
1-1/2-2
temperature of the cold stream (TC) is used to control the
temperature because a small change of cold stream cause
significant changes in the tank temperature.
Module 5 – Controllability Analysis
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Case 5.
Ts = TC;
Ts = 15ºC
TC
COLD STREAM FLOWRATE (u2)
TANK TEMPERATURE (y2)
TC = 15ºC
DISTURBANCE (Td,Fd)
TH = 65ºC
HOT STREAM FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
F=k(h)½
LLC
h
Output Flow rate
(F, T)
From the RGA, the suggested pairing is 1-2/2-1. It is
possible to achieve a perfect control of the tank level,
without affecting the temperature, using the cold stream.
Here, the temperature of the hot stream (TC) is the same than
Click to
the steady state operating tank
temperature. For that reason
Pairing
this stream is used to control 1-2/2-1
the level of the tank, whereas the
temperature of hot stream (TH) is used to control the
temperature because a small change of hot stream cause
significant changes in the tank temperature.
Module 5 – Controllability Analysis
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These five different analyses have demonstrated that the RGA
can indeed be used for nonlinear as well as for linear systems.
10
Pay attention to the fact that the RGA suggests different pairings at
different operating conditions. This because even that the analysis
has been based on approximate linearized models, this property
characteristic of the nonlinear systems is not lost.
STEADY STATE
5
0
2
-5
1
-10
30
0
25
20
-1
20
15
10
10
0
5
0
-2
2
1
0
-1
-2
-2
-1
0
2
1
It is as if the nonlinear system was analyzed on different sections, or
slides around `fixed` points, or in this case around steady states.
Module 5 – Controllability Analysis
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2.4 Non Square Systems
This section discusses another important point about RGA, the
selection of variables for Underdefined and Overdefined systems.
To do this task, it is first of all absolutely necessary to manipulate
the non square system in order to obtain a square system. This is
done according to the type of non square systems.
The objective in non square systems is pairing, as before, the
process variables to minimize the interaction between them.
Next slides show how to obtain a square system from a
UNDERDEFINED (therefore non square) system.
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2.4.1 Obtain the RGA of a pilot scale binary distillation column used to
separate ethanol and water for which the transfer function is given
below [2]. In addition, consider the side stream draw-off rate set at
a fixed amount and it cannot be changed. Use the same process
variables, that in Ex. 2.1.2.
Problem description.
Now , the side stream draw off rate is not a controlled variable,
because it is fixed. For this reason the process model is:
y1= overhead mole fraction ethanol
y2= ethanol mole fraction in side stream
y3= Temperature on Tray #19
 0.66e-2.6s

6.7s+1
 y1  
-6.5s
 y  =  1.11e
 2   3.5s+1
 y 3  
 -33.68e-9.2s

 8.15s+1
Module 5 – Controllability Analysis
u1= overhead reflux flowrate
u2= reboiler steam pressure
d = column feed flowrate

-0.0049e-s

9.06s+1

  u1 
-0.012e-1.2s
 
7.09s+1
 u2 
0.87 11.61s+1 e-s 

 3.89s+118.8s+  
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Distillation column used in
separating ethanol and water
Overhead mole
fraction ethanol (y1)
Temperature on tray #19 (y3)
Feed flow rate (d)
Overhead reflux
flow rate (u1)
Mole fraction of ethanol in
the side stream (y2)
Reboiler steam
pressure (u2)
Module 5 – Controllability Analysis
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It is impossible to control all three output (ys) variables with only two
input variables (us).
For that reason it is necessary to select the two most important
variables to be controlled, in this case the variables selected have
been y1 and y3, as next diagram shows.
Overhead mole
fraction ethanol (y1)
Temperature on
tray #19 (y3)
Feed flow
rate (d)
Overhead reflux
flow rate (u1)
Mole fraction of
ethanol in the side
stream (y2)
Reboiler
steam
pressure (u2)
Module 5 – Controllability Analysis
Here the side stream
mole fraction of
ethanol is taken as
the less important of
the output variables.
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Once it has been decided to leave the control of the side stream
composition out of control scheme, the control model is now:
y1= overhead mole fraction ethanol
y3= Temperature on Tray #19
 0.66e-2.6s

6.7s+1
y
 1

 y  =  -33.68e-9.2s
 3

 8.15s+1
u1= overhead reflux flowrate
u2= reboiler steam pressure

-0.0049e -s

9.06s+1
  u1 
0.87 11.61s+1 e-s  u2 

3.89s+1
18.8s+


 
This is a square (modified) subsystem. Therefore, now it is possible to
perform the RGA analysis and also to obtain the additional relation:
1.11e-6.5s
-0.012e-1.2s
y2 =
u1 +
u2
3.5s+1
7.09s+1
Module 5 – Controllability Analysis
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From the subsystem, the steady state gain matrix and the RGA obtained is:
According to
these values of
RGA, a 1-1/2-2
pairing is
recommended.
It means to use the overhead reflux (u1) to control the overhead
composition (y1), and use the reboiler steam pressure (u2) to control
Tray #19 temperature (y2). This makes sense.
Module 5 – Controllability Analysis
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It must notice that according to the relation:
1.11e-6.5s
-0.012e-1.2s
y2 =
u1 +
u2
3.5s+1
7.09s+1
The side stream composition will drift according to the values of the
overhead reflux (u1) and the reboiler steam pressure (u2).
This is the nature of UNDERDEFINED systems. The previous system
showed that it is only possible to achieve arbitrarily good control of two
[overhead mole fraction ethanol (y1) and temperature on Tray #19 (y3)]
of the three output variables and accept the drift of the third one
(composition on the side stream).
The strategy to work with an UNDERDEFINED system is to choose a
square subsystem by dropping off the excess number of output
variables on the basis of economic importance; the subsequent analysis
is the same as for square systems.
Module 5 – Controllability Analysis
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Next will be show how to deal with Overdefined
systems.
And this is the real
challenge of non square
systems, so you must
put all your attention…
Module 5 – Controllability Analysis
…and follow the
instructions
given in the next
example.
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NAMP
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2.4.2 According to a certain system with two outputs (y1 and y2) to be
controlled using two of three available inputs (u1, u2, and u3),
which loop pairing is expected to give the best control?. Through
pulse testing, the following transfer function model was obtained.
 0.5e-0.2s
 y1   3s+1
y  = 
-0.5s

0.004e
 2
 1.5s+1
0.07e-0.3s
2.5s+1
-0.003e-0.2s
s+1
0.04e-0.03s 

2.8s+1 
-0.001e-0.4s 
1.6s+1 
 u1 
u 
 2
u3 
Problem description.
This is a 2x3 system, this implies that only two of the three candidate
input variables will be used for control, while the third input variable will
have to be set at a fixed value and will therefore be redundant.
To determine which variables should be active and which ones should be
redundant, first of all possible 2x2 subsystems must be obtained.
Module 5 – Controllability Analysis
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Number of subsystems =
u!
n! u - n!
Where: u are the number of input variables and n the number of output
variables.
According to the previous system: u=3 and n=2
3  21

3!
Number of subsystems =
=
=3
2!  3 - 2 !
 211
Subsystem 1.
Utilizing u1 and u2 for control:
 0.5e-0.2s
 y1   3s+1
y  = 
-0.5s

0.004e
 2
 1.5s+1
Module 5 – Controllability Analysis
0.07e-0.3s 

2.5s+1   u1 
u 
-0.2s
  2
-0.003e

s+1
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From the subsystem 1, the steady state matrix and the RGA are:
Subsystem 2.
Utilizing u1 and u2 for control:
 0.5e-0.2s

 y1 
3s+1

=
y 
 0.004e-0.5s
 2
 1.5s+1
Module 5 – Controllability Analysis
0.04e-0.03s 

2.8s+1   u1 
u 
-0.4s
  3
-0.001e
1.6s+1 
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From the subsystem 2, the steady state matrix and the RGA are:
Subsystem 3.
Utilizing u2 and u3 for control:
 0.07e-0.3s

 y1 
2.5s+1

=
y 
-0.2s

-0.003e
 2

s+1
Module 5 – Controllability Analysis
0.04e-0.03s 

2.8s+1   u1 
u 
-0.4s
  3
-0.001e
1.6s+1 
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From the subsystem 3, the steady state matrix and the RGA are:
Next slide shows the three RGA values obtained for each subsystem.
Module 5 – Controllability Analysis
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RGA for subsystems 1 to 3.
 0.843 0.157 
Λ1  

0.157
0.843


0.758 0.242
Λ2  

0.242
0.758


 1.4 2.4 
Λ3  

2.4

1.4


According to these values of RGA for each subsystem, the
best possible control is the subsystem 1, because it is closest
to subsystem
the ideal situation;
is somewhat
subsystem
This
involves it
pairing
u1 with better
y1 and than
y2 with
u2, and 2this
andimplies
far superior
3.
also
that u3than
is tosubsystem
be redundant.
Module 5 – Controllability Analysis
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2.5 Factors Influencing the Loop Pairing
As seen in TIER I, there are some factors that affect how the
variables are paired. Some of those are:
Constraints in the input variable,
Time delay,
Inverse response,
Slow dynamics in the best RGA paring,
Timescale decoupling of loop dynamics
Next slides show how to pair the process variables according
to these factors.
Module 5 – Controllability Analysis
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2.5.1 Take again Ex. 2.3.1, but now, an in-tank heater was added to the
stirred mixing tank to control the temperature with the heater power
Q. Obtain the RGA for this system if Ts=(TH+TC)/2. The transfer
function is:
1
1


0






k
k
 A  s+

 A c  s+

c
 2A h 
 2A h 


c
s 
c
s 




G s = 

1
ρCp
 TH -Ts 
 TC -Ts 









k
k
k
 A h  s+
 A chs  s+
 A c hs  s+

c s
 A h 
 A h 
 A h 

c
s 
c
s 
c
s 




u1= Hot stream flowrate
y = Liquid level
1
y2= Tank temperature
u2= Cold stream temperature
u3= Heater power
Where k is the same constant as in Ex. 2.3.1 and Ac is the cross-section
area for the tank.
Module 5 – Controllability Analysis
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Problem description.
A diagram of the tank with the Heater is show below.
COLD STREAM
FLOWRATE (u2)
HOT STREAM
FLOWRATE (u1)
HEAT POWER (u3)
TANK TEMPERATURE (y2)
DISTURBANCE (Td, Fd)
h
TANK LIQUID LEVEL (y1)
Output Flow rate (F, T)
F=k(h)½
Now it is an overdetermined system with more than one subsystem
to pair. First the RGA for each subsystem will be obtained.
Module 5 – Controllability Analysis
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In a similar way as Ex. 2.4.2, there are three different subsystems:
Subsystem 1. Utilizing u1 (Hot Stream) and u2 (Cold stream) for control.
The steady state gain for this subsystem is the same that ex. 2.3.1:
 2 hs

k

K=
  TH -Ts 

 k hs
2 hs 

k 
 TC -Ts  

k hs 
And the RGA is the same as obtained in ex. 2.3.1:
 TC -Ts

TC -TH

Λ1 =
 -  T -T 
 H s
 TC -TH
Module 5 – Controllability Analysis
-  TH -Ts  

TC -TH 
TC -Ts 

TC -TH 
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NAMP
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Subsystem 2. Utilizing u1 (Hot Stream) and u3 (Heater) for control.
The transfer function matrix for this subsystem is:
1




k
 A  s+

c



2A
h
c
s



G(s) = 
 TH -Ts 




 A h  s+ k 
 c s A h 
c
s 


And the steady state gain matrix is:
 2 hs

 k
K= 
  TH -Ts 

 k hs
Module 5 – Controllability Analysis






1
ρCp




k
A chs  s+

 A h 
c
s 

0

0 


1
ρCp 

k hs 
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To calculate the RGA of this 2x2 system it is possible to obtain l11 using the
First Principles.
Again:
1
l11 
 l22
1 
and,
K12 K 21

K11K 22
Substituting the values of the matrix K of this subsystem:
  TH -Ts  
(0) 

 k h 
s 

=
=0
1
 2 hs  ρCp


 k  k h
s


Module 5 – Controllability Analysis
Therefore l11 is:
1
l11 
1
1 0
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And since, l12:

l12 
 l21
1 
Finally the value of l12 is:
And the RGA for subsystem 2 is :
l12  0
 1 0
Λ2 = 

0
1


Next the RGA for subsystem 3, will be obtained.
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Subsystem 3. Utilizing u2 (Cold Stream) y u3 (Heater) for control.
The transfer function matrix for this subsystem is:
1




k
 A  s+

 c  2A c hs 



G(s) = 
 TC -Ts 




 A h  s+ k 
 c s A h 
c
s 


And the steady state gain matrix is:
 2 hs

 k
K= 
  TC -Ts 

 k hs
Module 5 – Controllability Analysis






1
ρCp




k
A chs  s+

 A h 
c
s 

0

0 


1
ρCp 

k hs 
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To calculate the RGA of this 2x2 system it is possible to obtain l11 using the
First Principles.
Again:
1
l11 
 l22
1 
and,
K12 K 21

K11K 22
Substituting the values of the matrix K of this subsystem:
  TC -Ts  
(0) 

 k h 
s 

 =
=0
1
 2 hs  ρCp


 k  k h
s


Module 5 – Controllability Analysis
Therefore l11 is:
1
l11 
1
1 0
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And since, l12:

l12 
 l21
1 
Finally the value of l12 is:
l12  0
 1 0
And the RGA for subsystem 3 is : Λ 3 = 

0
1


You must noted that the RGA for subsystem 2 and 3 is the same and
both are independent of Ts.
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RGA for subsystems 1 to 3.
 TC -Ts

TC -TH

Λ1 =
 -  T -T 
 H s
 TC -TH
-  TH -Ts  

TC -TH 
TC -Ts 

TC -TH 
 1 0
Λ2 = 

0
1


 1 0
Λ3 = 

0
1


Taking the case where Ts=(TH+TC)/2, the RGA for each subsystem is:
Subsystem 1.
Subsystem 2.
Subsystem 3.
0.5 0.5 
Λ1 = 

0.5 0.5 
 1 0
Λ2 = 

0
1


 1 0
Λ3 = 

0
1


Again, note that the RGA for subsystem 1 was obtained in Ex. 2.3.1.
Subsystems 2 and 3 are the same and both are independent of Ts.
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According to this analysis, the pairing in subsystem 2 involves to use
the Hot stream temperature (u1) to control the liquid level (y1) and
use the in tank heater (u3) to control the tank temperature (y2):
Subsystem 2.
COLD STREAM FLOWRATE (u2)
TANK TEMPERATURE (y2)
DISTURBANCE (Td, Fd)
HOT STREAM FLOWRATE (u1)
Subsystem 3.
LLC
h TANK LIQUID LEVEL (y1)
HEAT POWER (u3)
TC
Click to
Pairing
Subsystem 2
Module 5 – Controllability Analysis
F=k(h)½
Output Flow rate (F, T)
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For subsystem 3 the pairing involves to use the Cold stream
temperature (u2) to control the liquid level (y1) and use the in-tank
heater (u3) to control the tank temperature (y2):
Subsystem 3.
TANK TEMPERATURE (y2)
COLD STREAM FLOWRATE (u2)
HOT STREAM FLOWRATE (u1)
DISTURBANCE (Td, Fd)
Subsystem 3.
LLC
h
HEAT POWER (u3)
TC
Click to
Pairing
Subsystem 3
Module 5 – Controllability Analysis
TANK LIQUID LEVEL (y1)
F=k(h)½
Output Flow rate (F, T)
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But both pairings of these subsystems can become an
undesirable pairing control as it will be discussed next.
If the in-tank heater can barely achieve the steady
state, Ts, at maximum power, there is a major problem.
Thus, this subsystem would not be desirable for the regulatory
temperature control because, following variations of the other
process variable (hot or cold stream), the IN-TANK HEATER
has no more power to supply (or extract) heat to keep
the new steady state temperature.
Module 5 – Controllability Analysis
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Now to overcome the power limitation, a much larger
heater is installed in the tank, but as a consequence of this,
there is a VERY LARGE TIME DELAY, between the
control signal and the actual power delivery.
And because of this sluggish closed-loop response
in the heater, the best choice for pairing the
process variables could be the poor RGA of
subsystem 1.
Next will be show another factor to considerer in the loop
pairing of process variables.
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2.5.2 Now considering a system with a transfer function given as below,
obtain the RGA for this system and analyze a unit set point change in
(y1) and a diagonal PI controller (Kc1= 4,t I1=0.5; Kc2=-4, t I2=0.3)
using the resulting pairing.
2 
 2
 10s+1 s+1 
G s = 

-4 
 1
 s+1 10s+1
Problem description.
First of all it is necessary to obtain the steady state gain matrix, as
it is shows below.
2 2 
G  0  =K= 

1

4


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Now, the RGA obtained as before is:
According to the RGA, the recommended pairing is y1-u1 and y2-u2.
Next step is to analyze a change in the set point.
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Since the dynamic simulation of the analyzed
system is beyond the scope of this module,
only the result of the change in the set point
will be display.
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As mentioned in last slide, the next graphic shows the closed loop response
for a unit set point change in y1 using the pairing suggested for the RGA
and a diagonal PI controller.
Pairing 1-1/2-2
y1
y2
Set point y1
The performance of this
Despite this “not too
pairing is not too bad
bad” performance,
considering that the open
the inverse pairing
loop time constants on
will be analyzed for
the diagonal are 10
the same set point
minutes
change.
point
Inverse loop pairing involves to take the value ofSet
l=0.2
inythe
2
pairing, but it has been mentioned as a situation to avoid !!!!!!!!.
Module 5 – Controllability Analysis
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Different pairing also implies to use a different PI controller, for that
reason the inverse pairing analysis of a unit set point change in (y1), the
new diagonal PI controller is (Kc1= 10,t I1=0.3; Kc2= 20, t I2=0.3).
Pairing 1-2/2-1
Set point y1
y1
The reason is that
The performance in this case
is
the control
loops are
dramatically better than able
the to respond so
recommended pairing byrapidly
the
that the
RGA, because the open loop
interactions that
time constants on the diagonal
appear more slowly
are only 1 minute.
are easily dealt with.
Set point y2
y
2
Finally in this example, the best loop pairing was obtained using the
inverse pairing, than the suggested by the RGA.
Module 5 – Controllability Analysis
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After this example, do you fell like this ?...
You
should
not
fell
like
any
of
The purpose of this example is not to confuse you
about how to this,
select a because…
loop pairing, the purpose is
to show you that RGA provides only a guideline
to steady state interactions, for that reason, all other
engineering considerations must be used together in
choosing the loop pairing.
Module 5 – Controllability Analysis
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2.3 Nonlinear Systems
Despite that many chemical processes can be adequately
represented by linear systems, via linear transfer function models,
the majority of chemical processes are inherently nonlinear and
sometimes need to use nonlinear models in order to be valid in a
wider range of operation.
8
6
4
2
0
-2
-4
-6
5
10
5
0
-5
-5
0
-8
30
20
It is therefore necessary to see how the pairing of input and output
variables of nonlinear systems is performed.
-10
-10
10
0
Module 5 – Controllability Analysis
0
5
10
15
20
96
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The same information used for the RGA of steady state systems is
used for nonlinear systems. This feature can appear for someone a
"disadvantage", but it is precisely this "disadvantage" involving only
steady state values that can be used to handle nonlinear systems.
Disadvantage
or advantage???
Take a decision
PLEASE !!!!!
Next, an example shows that according to the steady state values, the
pairing of both manipulated and controlled variables is selected.
Module 5 – Controllability Analysis
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2.3.1 Obtain the RGA for the multivariable system, a stirred mixing tank,
consisting in a hot stream and a cold stream which are used to control
the liquid level and the tank temperature, and use it to recommend
which of the manipulated variables should be used to control the liquid
level and the tank temperature. The transfer function matrix is shown
below.
y1= Liquid level
u1= Hot stream flowrate
y2= Tank temperature
u2= Cold stream temperature
1




k
 A  s+

 C  2A C hs 


G  s = 

 TH -Ts 


k 

A
h
s+
 C s  Ac h 
s 


1




 


k
k
 A  s+
A C  s+

 
 2A h  
 C  2A C hs 
C
s 



 G  s = 

 d
 Tds -Ts 
 TC -Ts 





k 

k

A Chs  s +

 A Chs  s + A h 


A C hs  
C
s 



1






Fds


k 
A Chs  s+

 A h 
C
s 

0
Where k is a constant (see next slide) and Ac is the cross-section area
for the tank.
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Problem description.
A diagram of the tank is shown below.
COLD STREAM
FLOWRATE (u2)
TANK TEMPERATURE (y2)
DISTURBANCE (Td, Fd)
HOT STREAM
FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
h
Output Flow rate (F, T)
F=k(h)½
Next it is necessary to obtain the steady state gain matrix, as next
slide shows.
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 2 hs

k

K = G 0 =
  TH -Ts 

 k hs
 2 hs
1
K =   TH -Ts 
k
 hs
2 hs 

k 
 TC -Ts  

k hs 
2 hs 

 TC -Ts  
hs 
Now with the steady-state gain matrix, it is possible to
obtain the RGA, using the full matrix method or, since it is a
2x2 matrix, in a more direct way as seen before.
Module 5 – Controllability Analysis
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To calculate the RGA of this 2x2 system it is possible to obtain l11
using the First Principles.
In Tier 1 has seen that:
1
l11 
 l22
1 
and,
K12 K 21

K11K 22
Substituting the values of the matrix K:
 2 hs

 k
 = 
 2 hs

 k

   TH -Ts  


 k h 
TH -Ts 

s 

=
   TC -Ts  
 TC -Ts 


 k h 
s 

Module 5 – Controllability Analysis
TC -Ts 

1
l11 =
=
TH -Ts 
TC -TH 


1 TC -Ts 
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In a similar way it is possible to obtain l12 using:

l12 
 l21
1 
According to this the value of l12 is:
l12 =
TH -Ts 

 TC -Ts 
TH -Ts 

1 TC -Ts 
=
-  TH -Ts 
 TC -TH 
And the RGA for this systems is:
Module 5 – Controllability Analysis
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 TC -Ts

TC -TH

Λ=
 -  T -T 
 H s
 TC -TH
-  TH -Ts  

TC -TH 
TC -Ts 

TC -TH 
Here, it is due to mention that the RGA depends only of values of
hot and cold streams, and also for the values of the steady state
values involved in the steady-sate gain matrix.
The values given to hot and cold stream are:
TH = 65ºC
TC = 15ºC
These values are fixed and just the value of Ts will be changed
according to different scenarios.
Module 5 – Controllability Analysis
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Five different values will given to Ts :
Ts > (TH+TC)/2;
Ts < (TH+TC)/2;
Ts = 55ºC
Ts = 25ºC
Ts = (TH+TC)/2 ;
Ts = 40ºC
Ts = TH:
Ts = TC;
Ts = 65ºC
Ts = 15ºC
According to this values the system of the stirred mixing tank is:
COLD STREAM FLOWRATE (u2)
TC = 15ºC
TANK TEMPERATURE (y2)
TH = 65ºC
DISTURBANCE (Td, Fd)
HOT STREAM FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
h
F=k(h)½
Output Flow rate (F, T)
Module 5 – Controllability Analysis
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Case 1.
Ts > (TH+TC)/2;
TC
COLD STREAM FLOWRATE (u2)
TC= 15ºC
Ts = 55ºC
TANK TEMPERATURE (y2)
DISTURBANCE (Td,Fd)
TH = 65ºC
HOT STREAM FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
LLC
h
Output Flow rate
(F, T)
F=k(h)½
From the RGA, the suggested pairing is 1-1/2-2.
The physical meaning of this pairing is: since the temperature
of cold stream (TC) is farther away from the steady state
Click to
operating tank temperature, small
changes in the cold stream
Pairing
produce noticeable changes in1-1/2-2
the tank temperature, whereas
the temperature of the hot stream (TH) is closer to the
operating steady state temperature, it can be used to control
the level without causing significant changes in the tank
temperature.
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Case 2.
Ts < (TH+TC)/2;
Ts = 25ºC
TC
COLD STREAM FLOWRATE (u2)
TANK TEMPERATURE (y2)
TC = 15ºC
DISTURBANCE (Td,Fd)
TH = 65ºC
HOT STREAM FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
F=k(h)½
LLC
h
Output Flow rate
(F, T)
From the RGA, the suggested pairing is 1-2/2-1.
Again, the physical meaning of this pairing is: since the
temperature of the hot stream (TH) is farther away from the
Click to
steady state operating tank temperature,
small changes in the
Pairing
hot stream produce noticeable1-2/2-1
changes in the tank
temperature, whereas the temperature of cold stream (TC) is
closer to the operating steady state temperature, it can be
used to control the level without causing significant changes in
the tank temperature.
Module 5 – Controllability Analysis
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Case 3.
Ts = (TH+TC)/2;
COLD STREAM FLOWRATE (u2)
Ts = 40ºC
TANK TEMPERATURE (y2)
TC = 15ºC
DISTURBANCE (Td,Fd)
TH = 65ºC
HOT STREAM FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
F=k(h)½
h
Output Flow rate
(F, T)
Here the values of the RGA are all equal to 0.5 For
this reason it is equally bad to pair 1-1/2-2 than 12/2-1, because the operating temperature is exactly
equidistant from both the cold stream temperature
and the hot stream temperature.
A poor control of the process under this undesirable
special condition is obtained.
Module 5 – Controllability Analysis
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Case 4.
Ts = TH;
Ts = 65ºC
TC
COLD STREAM FLOWRATE (u2)
TC = 15ºC
TANK TEMPERATURE (y2)
DISTURBANCE (Td,Fd)
TH = 65ºC
HOT STREAM FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
F=k(h)½
LLC
h
Output Flow rate
(F, T)
From the RGA, the suggested pairing is 1-1/2-2. It is
possible to achieve a perfect control of the level tank,
without interacting with the temperature, using the hot
stream.
Here, the temperature of the hot stream (TH) is the same that
Click
to
the steady state operating tank
temperature.
For that reason
Pairing
this stream is used to control the level of the tank, whereas the
1-1/2-2
temperature of the cold stream (TC) is used to control the
temperature because a small change of cold stream cause
significant changes in the tank temperature.
Module 5 – Controllability Analysis
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Case 5.
Ts = TC;
Ts = 15ºC
TC
COLD STREAM FLOWRATE (u2)
TANK TEMPERATURE (y2)
TC = 15ºC
DISTURBANCE (Td,Fd)
TH = 65ºC
HOT STREAM FLOWRATE (u1)
TANK LIQUID LEVEL (y1)
F=k(h)½
LLC
h
Output Flow rate
(F, T)
From the RGA, the suggested pairing is 1-2/2-1. It is
possible to achieve a perfect control of the tank level,
without affecting the temperature, using the cold stream.
Here, the temperature of the hot stream (TC) is the same than
Click to
the steady state operating tank
temperature. For that reason
Pairing
this stream is used to control 1-2/2-1
the level of the tank, whereas the
temperature of hot stream (TH) is used to control the
temperature because a small change of hot stream cause
significant changes in the tank temperature.
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These five different analyses have demonstrated that the RGA
can indeed be used for nonlinear as well as for linear systems.
10
Pay attention to the fact that the RGA suggests different pairings at
different operating conditions. This because even that the analysis
has been based on approximate linearized models, this property
characteristic of the nonlinear systems is not lost.
STEADY STATE
5
0
2
-5
1
-10
30
0
25
20
-1
20
15
10
10
0
5
0
-2
2
1
0
-1
-2
-2
-1
0
1
It is as if the nonlinear system was analyzed on different sections, or
slides around `fixed` points, or in this case around steady states.
Module 5 – Controllability Analysis
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2.4 Non Square Systems
This section discusses another important point about RGA, the
selection of variables for Underdefined and Overdefined systems.
To do this task, it is first of all absolutely necessary to manipulate
the non square system in order to obtain a square system. This is
done according to the type of non square systems.
The objective in non square systems is pairing, as before, the
process variables to minimize the interaction between them.
Next slides show how to obtain a square system from a
UNDERDEFINED (therefore non square) system.
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2.4.1 Obtain the RGA of a pilot scale binary distillation column used to
separate ethanol and water for which the transfer function is given
below [2]. In addition, consider the side stream draw-off rate set at
a fixed amount and it cannot be changed. Use the same process
variables, that in Ex. 2.1.2.
Problem description.
Now , the side stream draw off rate is not a controlled variable,
because it is fixed. For this reason the process model is:
y1= overhead mole fraction ethanol
y2= ethanol mole fraction in side stream
y3= Temperature on Tray #19
 0.66e-2.6s

6.7s+1
 y1  
-6.5s
 y  =  1.11e
 2   3.5s+1
 y 3  
 -33.68e-9.2s

 8.15s+1
Module 5 – Controllability Analysis
u1= overhead reflux flowrate
u2= reboiler steam pressure
d = column feed flowrate

-0.0049e-s

9.06s+1

  u1 
-0.012e-1.2s
 
7.09s+1
 u2 
0.87 11.61s+1 e-s 

 3.89s+118.8s+  
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Distillation column used in
separating ethanol and water
Overhead mole
fraction ethanol (y1)
Temperature on tray #19 (y3)
Feed flow rate (d)
Overhead reflux
flow rate (u1)
Mole fraction of ethanol in
the side stream (y2)
Reboiler steam
pressure (u2)
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It is impossible to control all three output (ys) variables with only two
input variables (us).
For that reason it is necessary to select the two most important
variables to be controlled, in this case the variables selected have
been y1 and y3, as next diagram shows.
Overhead mole
fraction ethanol (y1)
Temperature on
tray #19 (y3)
Feed flow
rate (d)
Overhead reflux
flow rate (u1)
Mole fraction of
ethanol in the side
stream (y2)
Reboiler
steam
pressure (u2)
Module 5 – Controllability Analysis
Here the side stream
mole fraction of
ethanol is taken as
the less important of
the output variables.
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Once it has been decided to leave the control of the side stream
composition out of control scheme, the control model is now:
y1= overhead mole fraction ethanol
y3= Temperature on Tray #19
 0.66e-2.6s

6.7s+1
y
 1

 y  =  -33.68e-9.2s
 3

 8.15s+1
u1= overhead reflux flowrate
u2= reboiler steam pressure

-0.0049e -s

9.06s+1
  u1 
0.87 11.61s+1 e-s  u2 

3.89s+1
18.8s+


 
This is a square (modified) subsystem. Therefore, now it is possible to
perform the RGA analysis and also to obtain the additional relation:
1.11e-6.5s
-0.012e-1.2s
y2 =
u1 +
u2
3.5s+1
7.09s+1
Module 5 – Controllability Analysis
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From the subsystem, the steady state gain matrix and the RGA obtained is:
According to
these values of
RGA, a 1-1/2-2
pairing is
recommended.
It means to use the overhead reflux (u1) to control the overhead
composition (y1), and use the reboiler steam pressure (u2) to control
Tray #19 temperature (y2). This makes sense.
Module 5 – Controllability Analysis
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It must notice that according to the relation:
1.11e-6.5s
-0.012e-1.2s
y2 =
u1 +
u2
3.5s+1
7.09s+1
The side stream composition will drift according to the values of the
overhead reflux (u1) and the reboiler steam pressure (u2).
This is the nature of UNDERDEFINED systems. The previous system
showed that it is only possible to achieve arbitrarily good control of two
[overhead mole fraction ethanol (y1) and temperature on Tray #19 (y3)]
of the three output variables and accept the drift of the third one
(composition on the side stream).
The strategy to work with an UNDERDEFINED system is to choose a
square subsystem by dropping off the excess number of output
variables on the basis of economic importance; the subsequent analysis
is the same as for square systems.
Module 5 – Controllability Analysis
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Next will be show how to deal with Overdefined
systems.
And this is the real
challenge of non square
systems, so you must
put all your attention…
Module 5 – Controllability Analysis
…and follow the
instructions
given in the next
example.
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2.4.2 According to a certain system with two outputs (y1 and y2) to be
controlled using two of three available inputs (u1, u2, and u3),
which loop pairing is expected to give the best control?. Through
pulse testing, the following transfer function model was obtained.
 0.5e-0.2s
 y1   3s+1
y  = 
-0.5s

0.004e
 2
 1.5s+1
0.07e-0.3s
2.5s+1
-0.003e-0.2s
s+1
0.04e-0.03s 

2.8s+1 
-0.001e-0.4s 
1.6s+1 
 u1 
u 
 2
u3 
Problem description.
This is a 2x3 system, this implies that only two of the three candidate
input variables will be used for control, while the third input variable will
have to be set at a fixed value and will therefore be redundant.
To determine which variables should be active and which ones should be
redundant, first of all possible 2x2 subsystems must be obtained.
Module 5 – Controllability Analysis
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Number of subsystems =
u!
n! u - n!
Where: u are the number of input variables and n the number of output
variables.
According to the previous system: u=3 and n=2
3  21

3!
Number of subsystems =
=
=3
2!  3 - 2 !
 211
Subsystem 1.
Utilizing u1 and u2 for control:
 0.5e-0.2s
 y1   3s+1
y  = 
-0.5s

0.004e
 2
 1.5s+1
Module 5 – Controllability Analysis
0.07e-0.3s 

2.5s+1   u1 
u 
-0.2s
  2
-0.003e

s+1
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From the subsystem 1, the steady state matrix and the RGA are:
Subsystem 2.
Utilizing u1 and u2 for control:
 0.5e-0.2s

 y1 
3s+1

=
y 
 0.004e-0.5s
 2
 1.5s+1
Module 5 – Controllability Analysis
0.04e-0.03s 

2.8s+1   u1 
u 
-0.4s
  3
-0.001e
1.6s+1 
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From the subsystem 2, the steady state matrix and the RGA are:
Subsystem 3.
Utilizing u2 and u3 for control:
 0.07e-0.3s

 y1 
2.5s+1

=
y 
-0.2s

-0.003e
 2

s+1
Module 5 – Controllability Analysis
0.04e-0.03s 

2.8s+1   u1 
u 
-0.4s
  3
-0.001e
1.6s+1 
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From the subsystem 3, the steady state matrix and the RGA are:
Next slide shows the three RGA values obtained for each subsystem.
Module 5 – Controllability Analysis
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RGA for subsystems 1 to 3.
 0.843 0.157 
Λ1  

0.157
0.843


0.758 0.242
Λ2  

0.242
0.758


 1.4 2.4 
Λ3  

2.4

1.4


According to these values of RGA for each subsystem, the
best possible control is the subsystem 1, because it is closest
to subsystem
the ideal situation;
is somewhat
subsystem
This
involves it
pairing
u1 with better
y1 and than
y2 with
u2, and 2this
andimplies
far superior
3.
also
that u3than
is tosubsystem
be redundant.
Module 5 – Controllability Analysis
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2.5 Factors Influencing the Loop Pairing
As seen in TIER I, there are some factors that affect how the
variables are paired. Some of those are:
Constraints in the input variable,
Time delay,
Inverse response,
Slow dynamics in the best RGA paring,
Timescale decoupling of loop dynamics
Next slides show how to pair the process variables according
to these factors.
Module 5 – Controllability Analysis
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2.5.1 Take again Ex. 2.3.1, but now, an in-tank heater was added to the
stirred mixing tank to control the temperature with the heater power
Q. Obtain the RGA for this system if Ts=(TH+TC)/2. The transfer
function is:
1
1


0






k
k
 A  s+

 A c  s+

c
 2A h 
 2A h 


c
s 
c
s 




G s = 

1
ρCp
 TH -Ts 
 TC -Ts 









k
k
k
 A h  s+
 A chs  s+
 A c hs  s+

c s
 A h 
 A h 
 A h 

c
s 
c
s 
c
s 




u1= Hot stream flowrate
y = Liquid level
1
y2= Tank temperature
u2= Cold stream temperature
u3= Heater power
Where k is the same constant as in Ex. 2.3.1 and Ac is the cross-section
area for the tank.
Module 5 – Controllability Analysis
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Problem description.
A diagram of the tank with the Heater is show below.
COLD STREAM
FLOWRATE (u2)
HOT STREAM
FLOWRATE (u1)
HEAT POWER (u3)
TANK TEMPERATURE (y2)
DISTURBANCE (Td, Fd)
h
TANK LIQUID LEVEL (y1)
Output Flow rate (F, T)
F=k(h)½
Now it is an overdetermined system with more than one subsystem
to pair. First the RGA for each subsystem will be obtained.
Module 5 – Controllability Analysis
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In a similar way as Ex. 2.4.2, there are three different subsystems:
Subsystem 1. Utilizing u1 (Hot Stream) and u2 (Cold stream) for control.
The steady state gain for this subsystem is the same that ex. 2.3.1:
 2 hs

k

K=
  TH -Ts 

 k hs
2 hs 

k 
 TC -Ts  

k hs 
And the RGA is the same as obtained in ex. 2.3.1:
 TC -Ts

TC -TH

Λ1 =
 -  T -T 
 H s
 TC -TH
Module 5 – Controllability Analysis
-  TH -Ts  

TC -TH 
TC -Ts 

TC -TH 
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NAMP
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Subsystem 2. Utilizing u1 (Hot Stream) and u3 (Heater) for control.
The transfer function matrix for this subsystem is:
1




k
 A  s+

c



2A
h
c
s



G(s) = 
 TH -Ts 




 A h  s+ k 
 c s A h 
c
s 


And the steady state gain matrix is:
 2 hs

 k
K= 
  TH -Ts 

 k hs
Module 5 – Controllability Analysis






1
ρCp




k
A chs  s+

 A h 
c
s 

0

0 


1
ρCp 

k hs 
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To calculate the RGA of this 2x2 system it is possible to obtain l11 using the
First Principles.
Again:
1
l11 
 l22
1 
and,
K12 K 21

K11K 22
Substituting the values of the matrix K of this subsystem:
  TH -Ts  
(0) 

 k h 
s 

=
=0
1
 2 hs  ρCp


 k  k h
s


Module 5 – Controllability Analysis
Therefore l11 is:
1
l11 
1
1 0
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And since, l12:

l12 
 l21
1 
Finally the value of l12 is:
And the RGA for subsystem 2 is :
l12  0
 1 0
Λ2 = 

0
1


Next the RGA for subsystem 3, will be obtained.
Module 5 – Controllability Analysis
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Subsystem 3. Utilizing u2 (Cold Stream) y u3 (Heater) for control.
The transfer function matrix for this subsystem is:
1




k
 A  s+

 c  2A c hs 



G(s) = 
 TC -Ts 




 A h  s+ k 
 c s A h 
c
s 


And the steady state gain matrix is:
 2 hs

 k
K= 
  TC -Ts 

 k hs
Module 5 – Controllability Analysis






1
ρCp




k
A chs  s+

 A h 
c
s 

0

0 


1
ρCp 

k hs 
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To calculate the RGA of this 2x2 system it is possible to obtain l11 using the
First Principles.
Again:
1
l11 
 l22
1 
and,
K12 K 21

K11K 22
Substituting the values of the matrix K of this subsystem:
  TC -Ts  
(0) 

 k h 
s 

 =
=0
1
 2 hs  ρCp


 k  k h
s


Module 5 – Controllability Analysis
Therefore l11 is:
1
l11 
1
1 0
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And since, l12:

l12 
 l21
1 
Finally the value of l12 is:
l12  0
 1 0
And the RGA for subsystem 3 is : Λ 3 = 

0
1


You must noted that the RGA for subsystem 2 and 3 is the same and
both are independent of Ts.
Module 5 – Controllability Analysis
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RGA for subsystems 1 to 3.
 TC -Ts

TC -TH

Λ1 =
 -  T -T 
 H s
 TC -TH
-  TH -Ts  

TC -TH 
TC -Ts 

TC -TH 
 1 0
Λ2 = 

0
1


 1 0
Λ3 = 

0
1


Taking the case where Ts=(TH+TC)/2, the RGA for each subsystem is:
Subsystem 1.
Subsystem 2.
Subsystem 3.
0.5 0.5 
Λ1 = 

0.5 0.5 
 1 0
Λ2 = 

0
1


 1 0
Λ3 = 

0
1


Again, note that the RGA for subsystem 1 was obtained in Ex. 2.3.1.
Subsystems 2 and 3 are the same and both are independent of Ts.
Module 5 – Controllability Analysis
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According to this analysis, the pairing in subsystem 2 involves to use
the Hot stream temperature (u1) to control the liquid level (y1) and
use the in tank heater (u3) to control the tank temperature (y2):
Subsystem 2.
COLD STREAM FLOWRATE (u2)
TANK TEMPERATURE (y2)
DISTURBANCE (Td, Fd)
HOT STREAM FLOWRATE (u1)
Subsystem 3.
LLC
h TANK LIQUID LEVEL (y1)
HEAT POWER (u3)
TC
Click to
Pairing
Subsystem 2
Module 5 – Controllability Analysis
F=k(h)½
Output Flow rate (F, T)
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For subsystem 3 the pairing involves to use the Cold stream
temperature (u2) to control the liquid level (y1) and use the in-tank
heater (u3) to control the tank temperature (y2):
Subsystem 3.
TANK TEMPERATURE (y2)
COLD STREAM FLOWRATE (u2)
HOT STREAM FLOWRATE (u1)
DISTURBANCE (Td, Fd)
Subsystem 3.
LLC
h
HEAT POWER (u3)
TC
Click to
Pairing
Subsystem 3
Module 5 – Controllability Analysis
TANK LIQUID LEVEL (y1)
F=k(h)½
Output Flow rate (F, T)
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But both pairings of these subsystems can become an
undesirable pairing control as it will be discussed next.
If the in-tank heater can barely achieve the steady
state, Ts, at maximum power, there is a major problem.
Thus, this subsystem would not be desirable for the regulatory
temperature control because, following variations of the other
process variable (hot or cold stream), the IN-TANK HEATER
has no more power to supply (or extract) heat to keep
the new steady state temperature.
Module 5 – Controllability Analysis
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Now to overcome the power limitation, a much larger
heater is installed in the tank, but as a consequence of this,
there is a VERY LARGE TIME DELAY, between the
control signal and the actual power delivery.
And because of this sluggish closed-loop response
in the heater, the best choice for pairing the
process variables could be the poor RGA of
subsystem 1.
Next will be show another factor to considerer in the loop
pairing of process variables.
Module 5 – Controllability Analysis
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2.5.2 Now considering a system with a transfer function given as below,
obtain the RGA for this system and analyze a unit set point change in
(y1) and a diagonal PI controller (Kc1= 4,t I1=0.5; Kc2=-4, t I2=0.3)
using the resulting pairing.
2 
 2
 10s+1 s+1 
G s = 

-4 
 1
 s+1 10s+1
Problem description.
First of all it is necessary to obtain the steady state gain matrix, as
it is shows below.
2 2 
G  0  =K= 

1

4


Module 5 – Controllability Analysis
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Now, the RGA obtained as before is:
According to the RGA, the recommended pairing is y1-u1 and y2-u2.
Next step is to analyze a change in the set point.
Module 5 – Controllability Analysis
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Since the dynamic simulation of the analyzed
system is beyond the scope of this module,
only the result of the change in the set point
will be display.
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As mentioned in last slide, the next graphic shows the closed loop response
for a unit set point change in y1 using the pairing suggested for the RGA
and a diagonal PI controller.
Pairing 1-1/2-2
y1
y2
Set point y1
The performance of this
Despite this “not too
pairing is not too bad
bad” performance,
considering that the open
the inverse pairing
loop time constants on
will be analyzed for
the diagonal are 10
the same set point
minutes
change.
point
Inverse loop pairing involves to take the value ofSet
l=0.2
inythe
2
pairing, but it has been mentioned as a situation to avoid !!!!!!!!.
Module 5 – Controllability Analysis
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Different pairing also implies to use a different PI controller, for that
reason the inverse pairing analysis of a unit set point change in (y1), the
new diagonal PI controller is (Kc1= 10,t I1=0.3; Kc2= 20, t I2=0.3).
Pairing 1-2/2-1
Set point y1
y1
The reason is that
The performance in this case
is
the control
loops are
dramatically better than able
the to respond so
recommended pairing byrapidly
the
that the
RGA, because the open loop
interactions that
time constants on the diagonal
appear more slowly
are only 1 minute.
are easily dealt with.
Set point y2
y
2
Finally in this example, the best loop pairing was obtained using the
inverse pairing, than the suggested by the RGA.
Module 5 – Controllability Analysis
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After this example, do you fell like this ?...
You
should
not
fell
like
any
of
The purpose of this example is not to confuse you
about how to this,
select a because…
loop pairing, the purpose is
to show you that RGA provides only a guideline
to steady state interactions, for that reason, all other
engineering considerations must be used together in
choosing the loop pairing.
Module 5 – Controllability Analysis
145