Eng. Zaid Abed Aljasim Muhasain                   Eng. saif Abed Aljasim Muhasain
Control Eng, Al- Technology University.            Computer and software Eng, AL-Nahrain University
e-mail:zaid_logic@yahoo.com.                            E-mail:safe_aljibouri@yahoo.com  





                     Z. A. Muhasain      S.A . Muhasain     


Abstract:
This paper deals with process control, generalized definition and modeling design of flow systems and control it based on compare to constant as principle of operation approach. After reviewing the operation of the plant, a compare system between water level and desire level is formulated to control flow rate. The scheme is then designed and tested by Matlab and Simulink   Toolbox. Simulation is presented and shown that on/off switch with constraint is better and more versatile compare with conventional approach.
Keywords:    Process; controller; flow.
1. Introduction
The liquid level control systems have been of good interest to engineers for along time. In the development this system has increased because of studies expansion, methods multitude and easier to implement this system in digital computer.  These experiments have been design and set up for the laboratory .In the simple model of this system consist the valve, tank or more tank, float ,transducer and controller. The water will going to the two tanks interactive (in the simple model) and its level will  be measured by float or any sensor and transducer in the conventional state  and then compare with desired level to control input signal(inflow) as shown  fig bellow:

Figure (1) Two tank fluid system using transducer to measure level.

 In this paper, we design a new scheme of two tanks system. The purpose of it to present a new approach is similar to the previous without using any float, sensor and transducer if we know the parameter of pipe resistance (Resistivity) and cross section are of the two tanks. We use a generalized definition for modeling and block diagram in design it. It can be designed and implemented by using Matlab program simulation. Show that this method is better and easier compared with conventional approach. This approach simulation by Matlab is allow us to solve more complicate problem and developing  the measuring states to make better modeling and analyzing of system and control allow more in the designs.
 The design of a tow tanks fluid system can control it according to the level without use any transducer. And the effect of Resistivity is observed on the level of the two tanks and controller designs, and show that the controller affects to the system performance. All simulation is performed using Matlab / simulink. These results are achieved for research and future study.


2. Modeling of the Liquid System
The fluid flow system are very common in the process control .It can be modeled by using generalized definition and modeling of hydraulic system.

2.1 Single Tank
For a single tank liquid level system is illustrated in fig. (2). It must be know that, how to simplify the design in the equation. This type of the system consists of source inflow rate qi into the tank. A drain valve in this to regulate the water out flow rate (qo): 


Figure (2) Single tank liquid level system


R is the Resistivity which represent a parameter due to pipe resistance, s/m2
Q is the Volume flow rate, m3/s
C is the Cross sectional area of the tank, m2
qi is the inflow rate,   m3/s
qo is the outflow rate,  m3/s
H   is the water level in the tank, m
For the liquid level system in Fig.(2). It can be obtained a Mathematical Model to get the relationship between qi and qo  or between qi and H.
The turbulent flow resistance can be determining by the following relationship:


R =                     in the steady state
          
We will obtain this equation

R=        if the relationship is linear
The system is modeled by using physical principles .The physical equation governing the change in liquid store (ACC)

Change in the liquid store (ACC) = input - output


Cross section area (C) =   


C =                 

C is constant

So that .Applied the balance equation

Acc= I/P - O/P =  


2.2 Two Tanks
Now we design a hydraulic system with two tanks as interaction network and use this equation in the two tanks .It consists of source, two tanks and pipe between them. qi may be represent a water from motor to tank 1 as show in fig (3) :



Figure (3) Liquid system in the two tanks as interaction network
In the tank 1
Acc1=In – Out = q - q1                                         (1)

R =                         In the steady state
          
R1 =                        q1=             (2)

The Laplace domain can be used to solve this equation:
 Q1(S) =                                                 (3)

Modeling of system requires us to determine the relationship between (h1-h2 ) and divided by R1 as show in the block diagram fig. (4):





Figure (4)


Cross section area (C) =                                   Acc = 
From equation 1 we obtain this equation

Acc1=In – Out = q- q1                            

q - q1 =                                    
We   take Laplace domain   so we get               

Q (S) – Q1(S) =C1SH1                              (4)




From the previous equation draw the second block diagram by subtract Q – Q1 and divided by C1S as shown in fig. (5).


Figure (5)

The liquid inflow q(s) and q1 as the flow out from tank1 to tank2

In tank2

R2 =         in the steady state
R 2=        

The third block diagram from dividing H2 by R2

Acc2 = in – Out = q1-q2

Cross section area of tank 2 (C2) =                     = C2. 
In Laplace domain Acc2=C2SH2 
Applying these equations obtain

Q1(S) – Q2(S) =C2SH2                                          (5)
H2 =      
This equation can be modeled by block diagram by subtract Q1(S) – Q2(S) and divided by C¬2s and we get h2(s) as o/p



Figure (6)

It can obtain the produce block diagram by getting the component block diagram from the previous shape, and putting all form pieces to make the final block as show in fig. (7)  Diagram. It must determine what the input and output for the block


Figure (7) Final block diagram

In this work, the output (level) is used to control the input flow rate and maintained the water with the desired level so as not to overflow the water.


3. Simulation and Controller
3.1 Simulation
After all block diagram was completed for the liquid level system. It can now be implemented this diagram and system in a simulation model. This block is used to represent the two tanks system. Matlab program was used to simulate this system. A step input in the simulink is referred   to the process input as an inflow rate (qi). The simulation in the system consists of a function block of the cross section area(C) and parameter resistivity (R) in a transfer function; we can obtain it from a block transfer function. A scope is used to display output (q2) and scope1 to display the water level. In the two tanks (h1, h2) this figure illustrates a simulation of model of this system:



Figure (8) Simulation represents block diagram of the
 Liquid level system with two tanks

In the simulink, simulate inflow (Q=2 m3/s) as a step input is 2 and step time is zero. Let cross section area in tank 1 (C1) equals 0.785 m2 and in tank 2 (C2) is equal 0.635 m2 (C1 and C2 is constant) in this design. Firstly assume the Resistivity parameters equal (R1=100 s/m2, R2= 100 s/m2). We must change a stop time 1000 (remember that simulink used second as time unit). From simulink scope monitors (q out), scope1 monitors water level in the two tanks h1 and h2. This figure below illustrates the trajectory of these scopes.


Figure (9) response of liquid level in the two tanks with time

To know that the level in both tanks is less or more than the desire level, A (Compare To Constant) block from simulink is used for implementing this and connect to the level (h1, h2). In this design assume that the height (in both tanks) is 13m and the water level must be less than 10 m. It can implement by using a (Compare To Constant) block from simulink and go to the logic and bit operation. The constant value is compared with h1, h2. If the level of the water equals or less than 10m, The (Compare To Constant) will give an o/p equals 1, else of this (the water level is more than 10m) the o/p is zero. The figure illustrates this process.


Figure (10) show the detection the level in the tank by using compare constant

From this design (Compare1) is shown from scope1. Firstly Compare To Constant 1   equals 1 depending on h1. If h1 reaches to 10m, constant compare will being to zero. Also compare 2 is used for h2 if h2 reaches to 10 m compare2 drops to zero, it illustrates by scope1 in fig. (11)


Figure (11) changing in the Compare to constant corresponding with the level in the two tanks

3.2 Controller Design
After a (Compare To Constant) block is used in the previous to detect if the water is less or more than ten meters. A control system can be used to open or close the i/p depending on (Compare To Constant) block. The output signal from a (Compare To Constant) block is back to the input and multiplying by using Product from simulink for implementing this controller as show in fig. (12):


Figure (12) controller design by used product

This control will make the input is close if water in tank 1 is more than 10m , also input is close if water level in tank 2 is more than 10m . The table blow illustrates this process with
Q = i/p * Compare To Constant 1 * Compare To Constant 1

Water level in
Tank1(h1)       Water level in
Tank2(h2)       Compare To Constant 1         Compare To Constant 2         q
Less or equal 10 m     Less or equal 10 m     1          1          i/p(open)
More than  10 m         Less or equal 10 m     0          1          0(close)
Less or equal 10 m     More than  10 m         1          0          0(close)
More than  10 m         More than  10 m         0          0          0(close)

Table (1) controller effect in the i/p
If h1 < = 10 and   h2 <= 10    q is open else q is zero
We conclude from this if the level of the water accesses more than ten meter, the inflow will closed depending on the controller so as no to overflow any tank of this system .The next simulation show in fig. (13)  Illustrates   this control in the system.


Figure (13) Simulation of the system with controller design

4. Implementation and Result
From fig.(13), after run the system the scope1 shows that the level in both tanks (h1, h2) will increase with the time. The research finds another way to control. The controller is close the input and prevent any increase of the level of the water so as not to overflow any tank. The simulink specification provides information about the level with time and time of open and closes the input by the controller. The system adjusted to control the water level. Firstly assume that the resistivity parameters is R1=100 and R2=100 and stopping time is 50 seconds. A scope1 shows this process as show in the figure:
  

Figure (14) result of control of liquid level and inflow using controller design R1=100 R2=100.

From fig (14) one can notice that, the level in tank1 doesn’t increase of ten meters and the input will be close from 2 to zero when the water level will reach to 10m at 4 sec .A series of tests using in the system by changing the parameters R1, R2 and show the comparative result and efficiency of system and controller which describe the output response.
In case 2 .We try to change this parameter by increasing R2 and parameters will be R1 =100, R2=200 to show the ability and effect in the controller as show in fig. (15):
 
Figure (15) System behavior and controller by changing Parameters to R1=100 R2=200.

From fig (15), When a scope1 trajectory is finished the change in the resistivity parameter (increasing R2) from 100 to 200 in the system will cause increasing of height in tank2 (R2) from 4 to 4.5 m. The tank 1 stills in the same height.
In case3, Change the resistivity parameters (by decreasing R1) R1= 50,   R2 = 200 in the system and the test results are showed by scope1 as show in figure below:
 

Figure (16) System behavior and efficiency by changing parameters to R1=50 R2=200.
From fig.(16) (case3), The changing in the parameters will cause change in the level in tank2 which reach to 7 m and input will close and open more than time.
In the case 4 .Change the resistivity  by decreasing R1 , R1= 10 ,R2 = 200 and change  stop time to 1000 to show more effect of controller with the time .This leads to the steady state as show in the fig.(17).
:

Figure (17) Show results in the level in the two tanks and controller efficiency.

And it notice that the level in both tanks will reach almost to ten meter and doesn’t access this level. The change in the   parameters can be adjusted the water level.



5. Conclusion:
The purpose of this paper is to present a controller of the liquid system with two tanks without using any float, sensor, transducer if we know the parameters of pipe resistance. The problem used to control level in the tanks by adjusting flow rate of the liquid entering the tank. After reviewing the operations of the system, simulation are present and are showed that the level of the water doesn’t access the desire level although of a little error of a few accessing of the water from desire level because of (Compare to Constant). In this work will be shown that high efficiency, fast response and accuracy in the controller. It can be conclude that the controller yield is better performance than the other controller. The results of modeling can be present an important means to measure and control more instruments and support the studies in the scientific laboratories. In the experiments installation in Matlab program, we must interface computer with electrical instruments that give the facilities understanding.
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