WELCOME TO:

SACCITO’S DAY

2

4 COUPLED TANKS CONTROL

Goals:

• Apply diverse control methods to our 4 tanks model, such as PID, and GPC(Generalized Predictive Control).

• Analyze the interaction of our system.

• Create a simulation model in SIMULINK.MATLAB in order to manipulate and analyze our 4 tanks system.

• Learn to read and interpret the results of our simulation and optimize it.

• Predict and control the future behavior of our model and prevent the instability.

APPLICATION

• Sewage Treatment

• Textile Industry

• Dairy Industry

Sewage Treatment

Coupled tanks are used in the sewage treatment industry to apply the different phases of cleaning to the water.The sewage goes amongst various specialized tanks as the homogenization tank, selector tank, biodigestor, bioreactor, etc.

Textile Industry

• Level control is very important in the Textile Industry, they use it to control the steam between the process tanks, that is essential to the processes like: ironing, gumming, bleaching, washing and dyeing of the clothing they produce.

Dairy Industry

• In the big industries, the process of making ice cream uses coupled tanks, they separate the ingredients like water, milk, flavors in tanks that later will be given to the blender machine and get the different types of ice cream.

¿What is a model?• A model is a simplified representation of a system that relate

the outputs variables with the entry variables. Models are designed to understand, analyze and control the systems behavior.

¿What is a system?• A system is a set of elements with a common objective, where

those elements are interrelated.

We are reviewing now the SISO and MIMO systems.

SISO: Simple Input Simple Output• SISO systems have only one input and one output, we can see

this in the control of one tank, where the process simulates the tank.

• The process is being affected by the controller, and is giving an output, that is our measured variable y, in this case the caudal.

MIMO: Multiple Input Multiple Output• MIMO systems have 2 or more inputs and 2 or more outputs.

In our project we have a MIMO system , with 4 entrys that are the caudal given by the pumps, and 4 outputs that are the caudal of each tank.

The realization of the model• In order to find out the model of our system, we use two

principles Conservation of the mass:

Conservation of the energy:

• We assume the entrance to the tanks as zeroFrom the filling and emptying of tanks equations we have:• =

• =

Where =Constriction factorKa=Converter factorA=Entry areaa=Output areaQ=Caudal

• Then, based on the system, and applying the respective equations we got the non-linear model of the tanks:

T1

T2

T3

T4

• To simplify the model we realize a Taylor linearization:T1

T2

T3

T4)

• Now the converter factor is applied to get the outputs on centimeters:

T1

T2

T3

T4

Implementation in Simulink

Assume the following values in some variables, such as:• u1 = 10 (V)• u2 = 8 (V)• g = 9.81 m / s2; X = 1; A = 200 cm2• a1 = 0.5 cm2; a2 = 1.5 cm2; a3 = 1 cm2; a4 = 2 cm2; a13 = 2cm2

Non-linear model of the tanks

Tank 1

Tank 2

Tank 3

Tank 4

Show the levels of tanks voltages indicated for each pump:

In this simulation, we can obtain the values of the heights at steady, where we have the following values:

10 3.935h cm

20 5.84h cm

30 3.77h cm

40 2.473h cm

Simulink Implementation of Linear System

We have to assign the values given for the pipes, tank area and values of the pumps.All valves are open for all tanks, including tank valve 1 and 3.

One observation that can be made of all tanks is that the linear system now depends on only one input and one output, ie, no longer is a MIMO system, but of a SISO system.

Variable Valor Unidades

u1o 10 Volts

u2o 8 Volts

h1o 3.935 Cm.

h2o 5.84 Cm.

h3o 3.77 Cm.

h4o 2.473 Cm.

A 200 Cm2.

•Values for each variable

a1 0.5 Cm2.

a2 1.5 Cm2.

a3 1 Cm2.

a4 2 Cm2.

a13 2 Cm2.

0.4 No units

0.7 No units

0.3 No units

0.6 No units

g 9.81 Cm./seg2.

Tank 1

Tank 2

Tank 3

Tank 4

In the figure below, we see the 4 tanks system coupled to the linear system

NO LINEAL LINEAL

h2o 5.84 cm. 5.1635 cm.

h4o 2.473 cm. 2.8075cm.

State Variables Representation .(t) A*X(t) B*u(t)

Y(t) C*X(t) D*u(t)

X

1

s*X(s) A*X(s) B*u(s) (sI A)*X(s) B*u(s)

X(s) (sI A) * * ( ).....( )B u s

Y(s) C*X(s) D*u(s)......( )

1Y(s) *(sI A) * * ( ) D*u(s)C B u s

1 1Y(s)*u(s) *(sI A) * DC B

1( ) *(sI A) *T S C B

Laplace transform

Finally

Finding the values of the matrix A and B:

0.573 0 0.545 0

0.0279 0.0687 0 0

0.545 0 0.602 0

0 0 0.06 0.141

A

0 0.033

0.058 0

0.025 0

0 0.05

B

The matrices A and B are:

0100

0001C

The matrix C are:

11 12

21 22

T TT

T T

3 2 5

11 4 3 2 5

0.058 0.06889 0.005663 8.273*10

1.256 0.1727 0.007234 6.706*10

s s sT

s s s s

12 3 2

0.0015 0.0007755

1.26 0.172 0.002003

sT

s s s

21 3 2

0.0009207 0.0005543

1.188 0.09108 0.0009762

sT

s s s

3 2

22 4 3 2

0.05 0.063 0.009678 0.0002523

1.401 0.3496 0.02625 0.0002825

s s sT

s s s s

1.2337 0.5678

0.3871 0.8932T

In steady state, when s=0:

Matrix RGA:

RGA=Tf0.*inv(Tf0)'eval (RGA)

1.2491 0.2491

0.2491 1.2491RGA

From RGA see that the best combination, since it has less interaction

1 2u y

2 1u y

Condition Number

cond (Tf0)eval (ans) CN = 2.809

CN is a low value whats means by my system has few interactions.

Conclusion:From the values of CN and RGA matrix, we can deduce that for tuning our system, simply take the values of T11 and T22 to find in this way our PI’s, due to little interaction having my system.

Tuning of PID's

For T11:3 2 5

11 4 3 2 5

0.058 0.06889 0.005663 8.273*10

1.256 0.1727 0.007234 6.706*10

s s sT

s s s s

Replacing when S = wj

2 5 3

11 4 2 5 3

( 0.06889 8.273*10 ) ( 0.058 0.005663 )

( 0.1727 6.706*10 ) ( 1.256 0.007234 ) j

w w w jT

w w w w

31.256 0.007234 0w w When:

Parameter for T11 PID:

K=0.5*Kc = 0.8714Ti= 0.8*Tc =66.136 seg Td=0.125*Tc=10.33 seg

2 3

22 4 2 3

( 0.063 0.0002523) ( 0.05 0.009678 )

( 0.3496 0.0002825) ( 1.401 0.02625 )

w w w jT

w w w w j

For T22:

31.401 0.02625 0w w

When:

Parameter for T22 PID :

K=0.5*Kc = 1.563Ti= 0.8*Tc =36.717 seg Td=0.125*Tc=5.737 seg