Supervisor Dr Hayley Wyatt
Laboratory H
Location S0.14 (CUSP Laboratory)
HI
Control Laboratory --- Liquid Level Control
1. INTRODUCTION
The liquid level control laboratory aims to introduce second year students to control theory and more importantly the use of control in industrial applications. Specific aims however, are to give students an understanding of:
· The importance of sensor characteristics in control applications
. The benefit of using closed loop control when compared with open loop (feed forward) control systems
· The effects of proportional and integral feedback on control systems
2. EXPERIMENTAL EQUIPMENT
Prior to starting the experimental programme it is advised you familiarise yourself with the experimental equipment used in the experiment additional handouts will be provided by lab support.
2.1 The CE120 Controller
The controller is an instrument comprising a wide selection of amplifiers, signal conditioning circuits and power supplies which can be connected together in a variety of ways to provide control for any system.
2.2 The Coupled Tanks Apparatus
This apparatus relates specifically to fluid transport and liquid level problems as they would occur in process control industries. It comprises two separate vertical tanks which are connected by a flow channel. In this laboratory only the left hand tank (tank A) is considered.
Figure 1 shows the front panel of the apparatus, which provides a schematic functional detail of the unit, and easy access, via 2 mm terminals, to the flow-rate transducers, liquid level transducers and pump control circuits.
3. CONTROL THEORY
To complete this laboratory you will need to know some basic control theory. This will be taught to you during the control and instrumentation module however as this section is not lectured until later in the term the basic theory is outlined below.
3.1 System Modelling - 1st Order System
In order to control a system we must understand the dynamics and express them mathematically. It can be shown that the transfer function of the system is given by:
yı(s)_G
Vi(s) TS+1
1
TANK A
TANK B
H
Water Level
Calibration
Output from
Output from
Scales
Right-Hand Level
Left-Hand Level Sensor
Sensor
Output from
Output from Left-Hand Flow
Right-Hand Flow Sensor (Optional)
Sensor
Vale A
0 to +10V
0 to +10V
Variable D.C.
1.1
Variable D.C.
Motor Control
Motor Control Voltage
Voltage
(Optional
1
1
Common Earth)
Valve B
Valve C
Terminal
Prismatic Level
Indicators
Figure 1 Detail of the Coupled Tanks Apparatus
(where y (s) = level sensor output, vi (s) = set point voltage or input) and that G = K1KsKb and T = AKb
(where K1 = pump input gain, Ks = level sensor gain, Kb = output characteristics for a particular operating level and A = cross sectional area of tank.)
The step response method of determining G and t is simple and accurate. The gain G is determined by applying a step input of U volts, the steady state change in output is then U x G. The time constant t is defined as the time required for the system to reach 0.632 of its final value.
Y(1)
UXG
0.632 × UXG
T
Time
Figure 2 Step response for first order system
2
Common Earth)
Terminal
3. 2 Open Loop Control
Consider the system shown in Figure 3; a pump fills the tank at a constant rate. An outlet, controlled by a valve at a set position, allows water to leave the tank such that eventually the liquid will achieve a constant level. At this point, PUMP FLOWRATE = TANK DISCHARGE RATE or Water in = Water out and the system is in equilibrium or steady state.
This is an example of OPEN LOOP CONTROL, i.e. no information concerning the liquid level is fed back to the pump circuit to compensate for changes in liquid level. If this system were used, it would require an operator to vary the pump voltage to account for any changes in the system.
Water Level
Tank
Pump
Water
Water
Out
In
Discharge
Valve
Pump Motor
Amplifier
Figure 3 Single tank system
A more acceptable control system would use a transducer to produce a signal proportional to liquid level. Electronic circuits would generate an error signal equal to the difference between the measured signal and the reference signal (or set point) which is chosen to achieve the desired liquid level or OUTPUT. Figure 4 shows a typical arrangement for a closed loop system in which the use of a signal measured at output is used to adjust the input.
Error Signal
Differencing
Reference Signal.
(Set Level)
Amplifier
Liquid
Level
Level Transducer
Signal
Liquid
Level
-Tank
Pump
Water
Water
.. In
Discharge
Out
Valve
Pump Motor
Feedback
Amplifler
Controller
Figure 4 Diagram of a closed loop control system
The schematic diagram shown in Figure 5 represents this system.
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3
Error
Actuation Signal
Signal
Differencing
Amplifier
Output
Reference
+
Controller
System
(Process
Signal
Varlable
(Set Point)
Feedback Loop
Figure 5 Block diagram of control system
If the system is in equilibrium and the output valve is opened, the liquid level will fall and therefore the error signal will increase. This will increase the supply to the pump motor and hence the flow rate. As the liquid returns to the reference or set level, the error signal reduces causing the energy supplied to the motor to be reduced. In theory, the motor supply would become so small that the pump would stop. In practice the pump rate reduces until a new equilibrium is produced where pump rate = discharge rate and the error achieves a new value. The difference between the Actual level and the Set level is the Steady State Error of the system (Figure 6).
If the gain of the controller is increased (termed proportional control) the steady state error can be reduced, i.e. the system level will get closer to the required level. The time constant t would also be reduced. If too much gain were introduced the system may become unstable (i.e. the level would oscillate and not achieve equilibrium). Proportional control is so called since the output from the controller is proportional to the input, so in order to maintain a non-zero input to the pump drive there must be a non-zero error signal at the controller.
In order to operate with zero error we must introduce integral control. The integrator is designed such that its output is proportional to the integral of its input. In practice proportional and integral control are used together (termed PI control) to achieve zero error as shown below.
With a low level of integral control, the system will be slow to respond to change and is termed over damped. With a high level of integral control, the system will be under damped, will overshoot when responding to change and may even become unstable
Controlled
Varlable
Set-Point
Steady State
Error - P Only
PI
P Only
Time
Figure 6 PI Control
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4 EXPERIMENTATION
In this laboratory the given units are sometimes not expressed as standard SI units - this is for convenience.
4.1 Circuit Calibration.
For an effective control system all aspects within the system need to be calibrated. Calibration allows systems to be 'controlled' in terms of voltages. It is evident that a control system can not utilise a level of water in mm or a flow rate in cm3/min but require voltages, in proportion to the level or flow rate.
Tables 1 and 2 present the calibration data for the pump and the water level respectively. The pump calibration constant K1 has been found to be approximately 500 cm3.min-1. V-1 whilst the level sensor constant Ks has been found to be approximately 0.04V.mm-1.
Plot the calibration curves using the data provided (flow rate against supply voltage and tank height against voltage output)
Table 1: Calibration data for the pump system
Flow Rate [cm3/min]
Voltage
[V]
0
1.95
0
0
1000
3.15
50
1.93
2000
4.95
100
3.94
3000
6.95
150
5.88
4000
8.92
3000
6.87
200
7.87
2000
4.90
1000
3.07
Table 2: Calibration data for the water level
Tank Level
Voltage
[mm]
[V]
Using the plotted curves to:
· Demonstrate that the constants described in section 4.1 are correct.
· Write an expression for Q (flowrate) in terms of the Gain constant of the pump K1, the pump supply voltage Vs, and Vmin. (Hint y=mx + c).
. Can you identify why the flow rate vs. supply voltage plot does not pass through zero?
4.2 Open and closed loop systems
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4.2.1 Open-loop control
The single tank is an example of a first order system, whose transfer function can be written in the general form
G
1+TS
G is the gain constant of the system and t is the time constant of the system. G and t are to be determined by examining the system's open loop response to a step input.
Connect the equipment as shown in Figure 7 (supplied by lab support). The connections to the supply potentiometer allow us to supply the system with a constant voltage and then increase this instantaneously, i.e. a step input. By monitoring the fluid level sensor output, we are monitoring the system response as illustrated in Figure 8.
INPUT
OUTPUT
Vo
Vi
Vi
G
Vo
1 + TS
t
t
Figure 8 Response of first order system to step input
Valve settings: Valve A - 0, Valve B - 3, Valve C - 5.
Control settings: Upper potentiometer 5V, lower potentiometer 1V.
Apply a constant voltage from the upper potentiometer only, and wait for system to reach equilibrium (steady state) (15 minutes). Attach the output from appropriate points on the apparatus to the input channels of the data logger as follows (as shown in Figure 7),
i) The pump input voltage
ii) The level sensor output
iii) The flow-meter output (n.b. gain = 2.16x10-3 V. cm-3.min-1)
Start recording and after a few seconds apply a step increase of 1 V by making the link as shown as a dotted line in Figure 7. Keep the test running until the steady state is again reached, and print the responses of above measured signals. Name graph as the control scheme used and mark each line in the graph. Using the plotted graph:
· Determine G (gain of system) and t (time constant).
· Can you identify what factors affect G and T ?
. Can you think of another way to calculate G and t ?
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4.2.2 Closed-loop proportional control
This section of the laboratory will demonstrate the clear benefits of closed loop proportional control. Connect the equipment as shown in Figure 9 as supplied by the laboratory support.
Valve settings: Valve A - 0, Valve B - 3, Valve C - 5. Controller settings: Proportional gain Kp = 1. (Note: Integral and differential switches off)
Build a circuit for calculating the control error, the difference of the set point and the measured liquid level, and connect the error signal to the inlet of the PID controller. Connecting the outputs from appropriate points on the apparatus to the input channels of the data logger as follows,
i) The set point (desired output of the liquid level sensor)
ii) The level sensor output
iii) The flow-meter output
iv) The PID controller output
Set the set point to 5 volts and wait until the liquid level is in steady state. Start recording and after a few seconds apply a step increase of 1 V. Keep the test running until the steady state is reached, and print the responses of above measured signals. Name graph as the control scheme used and mark each line in the graph.
Repeat for values of Kp of 5 and 10.
Using the plotted graphs can you:
· Comment on the effectiveness of proportional control?
· Discuss how increasing the proportional gain Kp influences system parameters such as the time constant t and the steady-state error?
· Determine what would happen if too large proportional gain was used?
. Can you identify an industry that could use closed loop control to monitor volumes of liquid in tanks?
· Can you think why it would not be possible to have zero steady state error using only proportional gain?
· Verify that the graphs plotted are in accordance with your expectations and briefly describe the key features of graphs, and how they change as the proportional gain in increased.
From the completed work can you:
· Draw a block diagram to represent the system, inserting the values calculated from the graph recorded at Kp = 5.
4.2.3 Closed-loop proportional and integral control
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Experiment 4.2.2 demonstrated steady state error under proportional control. Proportional control will not remove the error in a system, no matter how high the gain Kp.
In order to obtain zero error we must also include an integrator in the feedback loop, which will respond to changes in the error signal such that it will attempt to reduce the error to zero. Connect the equipment as for the previous, and switch on the integral controller.
Conduct test on Kp = 5 & Ki = 1; and repeat with Kp & 5, Ki = 5
Using the plotted graphs can you on:
· Discuss the effect of integral action on steady state error.
· Comment on How increasing the integral gain Ki affects the system?
· What would happen if too much integral gain was used?
· Verify that the graphs plotted by the data logger are in accordance with your expectations and briefly describe the key features of the graphs.
5. Control Laboratory Short Report Hints
Make good laboratory notes, but do not repeat anything already detailed in this handout, you will not get a chance to immediately write a long report, and hence you may need to refer back to your notes at a later date.
Ensure all figures are labelled correctly (titles, units, headings etc) remembering titles for figures go below the plot/image but above for tables.
Don't forget to leave space at the start of your report for an abstract. An abstract should include information on the work you have completed and your main conclusions. A good abstract would allow a reader to know if the report is applicable to their work and worth further reading. Your discussion can be in the form of bullet points in your lab notebook but should be detailed enough to allow a fuller discussion if you are requested to submit a long report on this laboratory.
Please add a conclusion section in your lab notebook, again in bullet points if desired.
If you need help ask!
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