Saturday, 5 October 2013

EXPERIMENT ( LAB 1 )

Course Code : JM507
Course          : Control System
Program        : DEM

LEVEL CONTROL DEMONSTRATION PANEL

Control technology is now an integral part of nearly all areas of engineering. Accordingly, a description of its basic principles is a standard feature of technical training programs.

The demonstration models of the RT-6x4 series from G.U.N.T. make it possible to ascertain relationships between control parameters in practical experiments and demonstrate these relationships so that they are clear and easily memorizable.

Every model comprises a fully functional system of processes with its own control circuit. An extensive use of modern industrial components makes the models as realistic as possible. As the students, we thus not only obtain knowledge of basic control principles but also an overview of the control elements's design, functionality and application.

The models have a desktop design and require very little maintenance. They are ideal as training aids for laboratory experiments at technical colleges and universities and intended exclusively for educational purposes.

1. Devices description

  • The RT 614 demonstration model is a desktop device to control filling levels.
  • Water is used as the operating medium here. Filling levels are controlled by an electronic industrial unit which can be configured as a P.PI or PID controller.
  • Experiments with the demonstration model involve modification and adaption to system control parameters.
2. Device layout
  • The RT 614 demonstration model for filling level control has following layout:





















3. Process scheme













4. The controller 
  • The digital universal controller is equipped with a microprocessor with digitally processes input signals and converts them back to analog variable prior to output.
5. Question
  • Draw block diagram for the control system
  • Explain the system base on each block diagram?
  • Draw a conclusion base on the system?
Solution
  • Draw block diagram for the control system

  • Explain the system base on each block diagram?
Reference Input (setpoint)
-The reference point is an external signal applied to the summing point of the control system to cause the plant to produce a specified action. This signal represents the desired value of a controlled variable and is also called the “setpoint.”

Actuating Signal
-The actuating signal represents the control action of the control loop and is equal to the algebraic sum of the reference input signal and feedback signal. This is also called the “error signal.”

Control Element
-The control elements are components needed to generate the appropriate control signal applied to the plant. These elements are also called the “controller.”

Manipulated Variable
-The manipulated variable is the variable of the process acted upon to maintain the plant output (controlled variable) at the desired value.

Disturbance
-The disturbance is an undesirable input signal that upsets the value of the controlled output of the plant.

Plant
-The plant is the system or process through which a particular quantity or condition is controlled. This is also called the controlled system.

Controlled variable (output)
-The controlled output is the quantity or condition of the plant which is controlled. This signal represents the controlled variable.

Feedback Element
-The feedback elements are components needed to identify the functional relationship between the feedback signal and the controlled output.

Feedback Signal
-The feedback signal is a function of the output signal. It is sent to the summing point and algebraically added to the reference input signal to obtain the actuating signal.


  • Draw a conclusion base on the system?
A block diagram of a system is a pictorial representation of the functions performed by each component and of the flow of signals. In a block diagram all system variables are linked to each other through functional blocks. The functional block or simply block is a symbol for the mathematical operation on the input signal to the block that produces the output.
Figure 1.1 shows an element of the block diagram. Such arrows are referred to as signals.


Figure 1.1      Element of a block diagram
(Source : Katsuhiko Ogata (1990), Modern Control Engineering)
The advantages of the block diagram representation of a system lies in the fact that it is easy to form the overall block diagram for the entire system by merely connecting the blocks of the components according to the signal flow and that it is possible to evaluate the contribution of each component to the overall performance of the system.






Tuesday, 1 October 2013

EXPERIMENT (LAB 2)

FLOW CONTROL DEMONSTRATION PANEL

3)   Safety Instructions

3.1 Danger to humans

  • DANGER! Exercise caution when handling electrical system components. There is a danger of electric shock. Disconnect the mains plug before accessing any electrical components. Such work should only be performed by qualities personnel.

  • DANGER! Never operate the device without a correctly installed ground conductor. A failure to observe this instruction might result in harm to humans and equipment.

3.2 Danger to Equipment and Functionality
  • Caution! Do not fill the water tank with more than 15 litres of water. Excess water might overflow into the device and damage it.
  • Caution! Never operate the pump without water. Dry running can damage the pump.
  • Caution! Do not change the sensor's basic setting. Altered signals can result in a loss of process control.
  • Caution! Drain the water tank prior to shutdown periods of more than 3 weeks.
  • Caution! Store the model under frost-free conditions. Frost can damage individual components.

4)   Experiments

4.1 Response of the control system
  • This experiment is intended to ascertain the properties and response of the control system underlying the model.
  • For this purpose, steps in control value are applied successively to the system in the non-regulated mode and the system's response is observed.
Experiment procedure: 
  • Turn on the demonstration model via its main switch.
  • Fully open the adjustment cock (6). Setting: 0 degree.
  • Set the controller to manual operation and the manipulated variable y to 10% =(2.5 litre/min).
  • Turn on the pump. After a certain time, the flow rate assumes a constant value. Read and note this value.
  • Increment the manipulated variable successively by 10%, waiting briefly each time until the flow rate has attained a constant value. Read and note these values too.
Manipulated variable y in %
10
20
30
40
50
60
70
80
90
100
Flow rate in (litre/min)
0.25
0.27
0.51
0.78
1.47
2.0
3.5
5.8
7.1
7.65




Result: 
  • The system evidently responds very quickly to change in the valve setting, much faster than temperature, filling-level and pressure control systems.
  • The system is of a compensatory nature, resulting in constant final values each time.
  • This characteristics was expected, because every pipe system possesses an intrinsic resistance to flow, thus preventing flow rates from rising indefinitely.
  • The characteristic of the control variable x clearly indicates the equal-percentage response of the control valve.


4.2 Flow Control with a PI Controller
  • In this experiment, a controller with proportional and integral components is used for flow control, accompanied by variations in parameters. 
  • The controller's differential component remains inactive. 
  • The control circuit's response to changes in the reference variable w is observed.
4.2.1 Slow PI-Controller

Experiment procedure:
  • Turn on the demonstration model via its main switch.
  • Set the controller and demonstration model as shown in the following table:
Controller type
PI-Controller
Controller mode
Automatic
P-component                    >>                Pb.1
0.1
I-component                     >>                rt
4 seconds
D-component                   >>                dt
0.0 second
Controller settings                                      Start value:   
                                                                        Step value:
6 litre/min   (30%) 
12 litre/min (60%)
Adjustment cock
Half open, 45 degree

  • Observe the flow rate using the readings indicated by the controller and rotameter. After a certain time, the flow rate assumes a constant value of 6 litre/min.
  • Increment the reference variable w by setting the controller to 12 litre/min. The flow rate increases and assumes a constant value of 12 litre/min after a certain time.

Result:
  • The input signal y reveals that although the controller responds immediately to changes in the reference variable, it takes a long time to achieve a constant target value. The desired flow rate of 12 litre/min is attained very slowly ( t > 1 minute )
  • The controller's P-component achieves fast response, while the I-component eliminates persistent deviations. However, the selected integration time is still too long.

4.2.2 Fast PI-Controller
  • In this experiment,too, a controller with proportional and integral components is used for flow control.
  • The controller's differential components remains inactive.
  • Compared with experiment 4.2.1, the integration time.- i.e. the controller's I-component is set to a notably lower value.
  • The control circuit's response to changes in the reference variable w is observed.

Experiment procedure:
  • Turn on the demonstration model via its main switch.
  • Set the controller and demonstration model as shown in the following table:
Controller type
PI-Controller
Controller mode
Automatic
P-component                    >>                Pb.1
0.1
I-component                     >>                rt
0.5 seconds
D-component                   >>                dt
0.0 second
Controller settings                                  Start value:                                                              Step value:
6 litre/min   (30%)
12 litre/min (60%)
Adjustment cock
Half open, 45 degree
  • Observe the flow rate using the readings indicated by the controller and rotameter. After a certain time, the flow rate assumes a constant value of 6 litre/min.
  • Increment the reference variable w by setting the controller to 12 litre/min. The flow rate increases and assumes a constant value of 12 litre/min after a certain time.

Result:
  • The input signal y reveals that the controller quickly generates values which are notably higher than in the previous experiment. In fact, the control variable now distinctly overshoots the target value of 12 litre/min and starts to oscillate about it. 
  • The oscillations decay in 30 seconds to a permanent level of roughly +/- 5% about the target value.
  • The parameters selected here do not result in satisfactory control performance. The selected integration time is obviously too short.

4.2.3  PI-Controller with Improved Parameters
  • In this experiment, too, a controller with proportional and integral components is used for flow control.
  • The controller's differential component remains inactive.
  • The results obtained in experiments 4.2.1 and 4.2.2 are used as a basis for adapting the controller's integration time here.
  • The control circuit's response to changes in the reference variable w is observed.

Experiment procedure:
  • Turn on the demonstration model via its main switch.
  • Set the controller and demonstration model as shown in the following table:
Controller type
PI-Controller
Controller mode
Automatic
P-component                    >>                Pb.1
0.1
I-component                     >>                rt
0.75 seconds
D-component                   >>                dt
0.0 second
Controller settings                                      Start value:   
                                                                        Step value:
6 litre/min   (30%)
12 litre/min (60%)
Adjustment cock
Half open, 45 degree
  • Observe the flow rate using the readings indicated by the controller and rotameter. After a certain time, the flow rate assumes a constant value of 6 litre/min.
  • Increment the reference variable w by setting the controller to 12 litre/min. The flow rate increases and assumes a constant value of 12 litre/min after a certain time.

Result:
  • The input signal y rises immediately after the step in the reference variable to achieve a nearly constant value in just ~5seconds.
  • The control variable initially overshoots the target value by ~5% and never becomes completely stable, instead oscillating irregularly about the target value.
  • However, the control performance is acceptable for a fast system such as this one.
  • The control parameters in this operating mode are nearly ideal for responding to change in the reference variable. This configuration is a compromise between repsonse and control performance.

4.3 Note on Further Experiments
  • The experiments and associated parameters described above are a subset of the available possibilities.
  • The additiona variants of this demonstration model also easily allow a realization and evaluation of control systems with different settings and control parameters.
Possible Variants:
- Use of just a P-controller
- Use of a PID-controller
-Variations in reference variable step
- Disturbance variable control: Introduction of a disturbance variable z by means of the adjustment cock (6) and comparison of the results with those obtained from reference variable control.
- Optimization of reference and disturbance variable control parameters for various operating points.


Production-related factors, fluctuations in ambient conditions and operational modofications can cause the control system's properties to change.

Wednesday, 14 August 2013

EXERCISE 1

.         Questions:

          1) Proportional Controller is used for controlling temperature in melting process. Temperature set point is 750°C and temperature measuring tool range is 0 - 1000°C .Proportional space is set  at 15%.  Pressure output range from controller is 20 –100 kN/m2 and pressure output rises when temperature increases.If pressure output is set at 60 kN/m2 for temperature set point, find
           
            i.    Temperature for pressure output at 20 kN/m2
           ii.    Temperature for pressure output at  100kN/m2    



2 ) In a process, a Proportional Controller is used to control the liquid level in the boiler. The level is set at 8 meters and the level range is 1 to 14 meters. Proportional band is at 20%.The output current has a range of  5 – 20mA. Determine:-

                      i.     The output level when current is 5 mA.
           ii.    The output current at a level of 10 meters.
          
3) An air to open valve on the inflow controls level in a tank. When the process is at the set point the valve opening is 50%. An increase in outflow results in the valve  opening increasing to a new steady state value of 80%. What is the resulting offset if the controller PB is:

                     i. 80%
                     ii. 40% 
      
                                                                                                                      

4) A temperature controller has the following characteristic curve below:


 The controller has a range of 100 0C to 400 0C. The set‐point is 250 0C. Calculate:

                                                                                                                    
                                i.    The controller Proportional Band
          ii.    The controller Proportional Gain



Answers:

1) i.  P= 20 kN/m2  , P= 0%
         
        P(t) = KpEp+P(0) , Kp = 100 , = 100
                                                PB        15

         P(t)= 100 Ep + 50
                  15    
        
           0= 100 Ep + 50
                 15     
       
        100 Ep = 0 - 50   
         15

              Ep = -50  x 15
                      100  

              Ep = -7.5 %

              Ep = SP - MV  x 100%
                         range
             
               -7.5 = 750 - MV  x 100%
                         1000 - 0

               -7.5(1000) = 750 - MV
                    100

                 -75 = 750 - MV

                 -75 - 750 = - MV

                      MV = 825°C 


ii.  P = 100 kN/m2 , P = 100%

       P(t) = KpEp + P(0) , Kp = 100  = 100
                                                 PB      15

       P(t) = 100 Ep + 50
                  15

      100 = 100 Ep + 50
                 15
      
       100 Ep = 100 - 50
        15

         Ep =  50  x 15
                  100

      Ep = 7.5%

       E = SP - MV  x 100%
               range

     7.5 = 750 - MV  x 100%
                 1000

     7.5(1000) = 750 - MV 
        100

           75 = 750 - MV 

           MV = 750 -75 
                
           MV = 675°C 


2) i.         SP - MV  x 100%
                range
       
             Ep = 8 - 10   x 100%
                     14 - 1

             Ep = -15.38 

            P(t) = KpEp + P(0)

            P = 5(-15.38) + 50 

            P = -26.9%

            -26.9 = Current - Min  x 100%
                         Max - Min 

             -26.9 = MV - 1  x 100%
                          14 - 1

                MV = -2.5

   
    ii.   Controller output = Current - Min x 100%
                                         Max - Min

                                     = 10 - 5  x 100%
                                        20 - 5

                                     =  5  x  20%
                                        15

                    Co            =  7m


3) i.  80%

        Pb = 100 
                  Kp

        80% = 100 
                    Kp  

         Kp = 100   
                   80

          Kp = 1.25

  
  ii.   40%

          Pb = 100 
                    Kp

          40 = 100 
                   Kp

          Kp = 100
                   40

          Kp = 2.5


4)  i.                Positive error band

                =        3°C            x 100%
                    (400 - 100)°C

                = 1.7%

                = 1.7 - (-1.7)
  
                = 3.4

                     Negative error band

                =        -5°C             x 100%
                   (400 - 100)°C

                = - 1.7%

   
    ii.       Pb = 100
                       Kp

             3.4 = 100 
                       Kp

               Kp = 100
                        3.4

               Kp = 29.42