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UNIVERSITI TUN HUSSEIN ONN MALAYSIA Faculty of Mechanical and
Manufacturing Engineering
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DEPARTMENT OF ENGINEERING MECHANICS
CONTROL LABORATORY
LAPORAN MAKMAL/LABORATORY REPORT
Kod M/Pelajaran/ Subject Code
ENGINEERING LABORATORY VI BDA 37101
Kod & Tajuk Ujikaji/ Code & Title of Experiment
Kod Kursus/ Course Code
Seksyen /Section
Kumpulan/Group No. K.P / I.C No. Nama Pelajar/Name of
Student
No. Matrik
Lecturer/Instructor/Tutors Name
1. 2.
Nama Ahli Kumpulan/ Group Members
No. Matrik Penilaian / Assesment
1. Teori / Theory 10 %
2. Keputusan / Results 15 %
3. Pemerhatian /Observation 20 %
4. Pengiraan / Calculation 10 %
5. Perbincangan / Discussions 25 %
Tarikh Ujikaji / Date of Experiment
Kesimpulan / Conclusion 15 %
Tarikh Hantar / Date of Submission
Rujukan / References 5 %
JUMLAH / TOTAL 100%
COP DITERIMA/APPROVED STAMP
ULASAN PEMERIKSA/COMMENTS
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UNIVERSITI TUN HUSSEIN ONN MALAYSIA Faculty of Mechanical and
Manufacturing Engineering
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BDA37101-Edition III/2011
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COURSE INFORMATION
COURSE TITLE: ENGINEERING LABORATORY VI (BDA 37101) TOPIC 5:
PROPORTIONAL CONTROL (LIQUID LEVEL AND FLOW SYSTEM) 1. OBJECTIVES
In this activity you will accomplish the following:
i. Describe the action of the proportional control algorithm ii.
Define offset iii. Describe the system using Laplace
transformations iv. Run the system with the proportional
controller
2. SKILLS In this activity you will develop the following
skills:
i. Technology: apply technology. ii. System: understand systems.
iii. Resources: manage time, manage materials/facilities. iv.
Interpersonal: work as a member of a team. v. Information: use
computers to process information. vi. Personal: responsibility,
self-esteem, self-management.
3. MATERIALS In this activity you will need the following
materials:
i. Process Line panel for Level Control ii. PC computer with
Process Motion software.
4. THEORY 4.1 PROPORTIONAL CONTROL ALGORITHM The proportional
control algorithm output m is defined as: m = Kc*e + mo Eq. 1-1
Where:
m proportional control algorithm output, which is expressed in
terms of percent input to the pump.
Kc proportional control algorithm constant (gain) e error mo
controller bias. This is a fixed value that will provide a
constant
value to the control algorithm.
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BDA37101-Edition III/2011
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When the error is 0, m will be equal to mo. When the error is
positive, m will be greater than mo, and when the error is negative
m will be less than mo. Using the automobile cruise control
example, when the cruise control is set to 55 miles per hour and
the car is travelling at 55 miles per hour, the error is zero.
However, the controller needs to send a constant amount of gas to
the engine to keep the car going 55 mph. This constant amount of
gas is the m for the cruise controller. Suppose that the car starts
climbing a hill. Its speed will decrease and an error will develop.
The cruise controller will add an amount of gas equal to Kc*e to
the constant amount mo. As the car speeds up and approaches 55 mph,
the error will decrease, and the controller will decrease the
amount of gas sent to the engine back to the mo value. This example
illustrates an important principle of the proportional controller.
When a load is presented to a system, or when the reference value
changes, an error is created. The proportional control algorithm
output m changes proportionally to the error e. The ratio between
the changes in the control algorithm output m and the error e is
determined by the controller gain Kc, as expressed in Kc = m e The
block diagram that describes the level control system with a
proportional control algorithm is as shown in Fig1-1:
Figure 1-1: Level control system block diagram. With the
proportional control algorithm is used the output m is proportional
to the error (e) thus for a small error the signal will be
relatively small and for a larger error it will be bigger. 4.2
LAPLACE TRANSFORMATIONS To examine the effect of changes in Kc over
the system, you will use Laplace transforms. The Laplace transform
is a technique for translating certain complicated differential
equations into simple algebraic equations. By substituting
- +
Controller Algorithm Tank
error
Level Sensor
Actual Level q in
Final Control Element (pump)
m Reference level
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BDA37101-Edition III/2011
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a complex number S for t, we translate the equation from the
real time plane into an imaginary plane called the Laplace space.
After the Laplace transform, time t is removed from the system and
is replaced with S. The system is now said to be in the Laplace
space, also known as the Laplace plane or Laplace domain. 4.3 LEVEL
CONTROL AND ITS LAPLACE TRANSFORM The following is a block diagram
in Fig 1-2 describes the level control system and specifically for
this case it is a first order system in the Laplace domain.
Figure 1-2 : Control system in Laplace space - block diagram 4.4
DERIVING THE SYSTEM GAIN The four blocks of the system shown in fig
1-2 and their functions are:
1. Controller block.The function in the controller block in time
domain is m = Kc *e mo. The controller multiplies the error e by
the controller gain Kc and then adds the constant mo to determine
the proportional algorithm output m. In the Laplace domain the
controller function is m = Kc.
e 2. The Final Control Element (controlled pump gain).The
controlled pump
gain is Kp. The dynamic properties of the pump will be ignored.
In the Laplace domain the controller function is qin = K p .
m 3. The tank is a first order system whose gain is K and whose
time constant
is . The gain K and time constant need to be determined first.
The
Laplace transform of the tank is qin
kh tan = 1S
K
4. The level sensor gain is Km. The level sensor gain in this
system can be regarded as 1 and it will not affect the overall gain
in the tank. However, the sensors inability to measure itself leads
to a very significant difference between the actual system output
and the reference output. This difference
Controller Algorithm Tank
error
Level Sensor
q in
Final Control Element (pump)
m Ref s Kp
Km
ms=F(e)s
+
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UNIVERSITI TUN HUSSEIN ONN MALAYSIA Faculty of Mechanical and
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__________________________________________________________________
BDA37101-Edition III/2011
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is called the offset error. The relation between the level in
the tank and the reference value is ;-
You should note here, that the reference value is always going
to be slightly larger than the tank pressure. This will cause a
condition called an offset error. In actuality, this means that a
proportional controller can never really direct the system all the
way up to the true reference value, and the system output will
always be slightly lower than the reference. You will observe the
offset when you run the system.
Simplifying equation 1-2 using algebra yields equation 1-3:
Where;-
4.5 CONCLUSIONS From the analysis of the proportional control
system, it can be concluded as follows;-
1. When a proportional control algorithm is used to control a
first order system the entire system acts also like a first order
system (as demonstrated in equation 1-3).
2. The entire system gain is
3. The system gain is always smaller than 1. Thus, for a step
input, the system will always display an offset error where the
actual output is slightly less than the reference value.
4. Increasing Kc (the controller gain) increases the system gain
and decrease
the offset error.
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5. The system time constant is . The value of the systems time
constant system is smaller then the tank time constant.
6. Increasing the controller gain, Kc decreases the systems time
constant and the system response time.
4.6 ADDITIONAL THEORY
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BDA37101-Edition III/2011
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5. TASK In this activity you will:
1. Run the system using the proportional control algorithm. 2.
Introducing a step input in the reference value and monitor the
resulting
system response (time constant and error). 3. Vary the
controller gain to test the resultant system response (in both
dynamic and static systems). 5.1 TASK 1 ; INVENTORY AND SAFETY
CHECKS1
1. Check whether all materials required for this activity are
available at your lab station.
2. Check whether your lab station conforms to the Safety
Guidelines. 3. Complete the Inventory and Safety Check List on the
Worksheet for this
activity. 5.2 TASK 2 ; SETTING UP
1. Turn on the computer. 2. Run the ProcessMotion Software. 3.
Activate the PressureLine panel, making sure that the panel is
connected to
a COM port on your computer and that the Communications key is
turned to PC.
4. Drain the tank so that it is at its minimum level. Then close
valve A and keep valve B open. Valve C should remain closed
throughout.
5. Click On-Line (at modes). 6. Click the Level Control / Proses
button. 7. Select the PID Controller from the dialog box and enter
the following
parameters: Setpoint : 110 Kc : 5 Ti : 100000
Td : 0 Mo : 0
In setting these parameters, you have created a system where the
Ti, Td, and Mo parameters will not contribute to the final m. m
will be equal to Kc * e. This is a strict proportional control
algorithm with no bias.
8. Click Run. 9. Run the system until it reaches steady
state.
Question: Is there an offset? 10. Record the system output H in
column 1 of the chart on your worksheet.
This column is marked Initial reference value. Record this value
in every row of this column.
11. Record the controller output m in column 5 of the chart on
your worksheet.
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__________________________________________________________________
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5.3 TASK 3 ; USING PROPORTIONAL CONTROL
1. Click the Level Control button and change the Kc setting to
10. 2. Now perform a step input. Increase the set point, by
changing it to the
Initial reference value + 30% (e.g. if your initial reference
was 100mm, change the set point to 130mm.)Record the new set point
value in column 2 of your chart. Click Run.
3. Wait till the system reaches steady state. Record the error
in column 4 of the graph on your worksheet. Record the controller
output m in column 5 of your graph.
4. Record the total response time (Time at reaching steady state
minus Time of step input) in column 6.
5. Record the percentage of the offset error (Error/Reference)
in column7.
Question: Describe the system output response
6. Reset the setpoint back to the initial reference value and
allow the system to return to the reference value. Repeat the
experiment, for step inputs of +40%, +50%, -20%, -30%and 40%,
respectively. Record your results in the chart.
Question: When is the offset percentage highest? Lowest?
Question: What is the cause of offset? 5.4 TASK 4 ; VARY THE Kc
VALUE Change the Kc value will change the system response. You will
now present the system with a step input, and vary the Kc setting
to see how the proportional control affects the system response.
You will also add system loads to see how the system responds.
1. Reset the Setpoint to the initial reference value in Task
1-3. Click Run and allow the system to come to steady state. You
will now raise the Kc value and test the system with step inputs
and various loads.
2. Change the Kc to a value of 100. Perform a step input by
increasing the setpoint by 30%.
3. Now present the system with a load, by opening valve A and
closing valve B. Valve A drains more quickly than valve B, thus
putting an added load on the system and Click Run.
Q: Describe the general system response, the response time, and
the percentage of offset.
4. Now open both valve A and valve B.
Q: How does the system respond?
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UNIVERSITI TUN HUSSEIN ONN MALAYSIA Faculty of Mechanical and
Manufacturing Engineering
__________________________________________________________________
BDA37101-Edition III/2011
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5. Change the Kc to a value of 1. Perform a step input by
increasing the
setpoint by 30%. Click Run
Q: Describe the general system response, the response time, and
the percentage of offset.
6. Change the Kc to a value of 10. Perform a step input by
increasing the
setpoint by 30%. Click Run.
Q: Describe the general system response, the response time, and
the percentage offset.
Q: How does the Kc affect the system response time? The offset?
(Kc
100,Kc1 and Kc 10) 6. OBSERVATIONS
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UNIVERSITI TUN HUSSEIN ONN MALAYSIA Faculty of Mechanical and
Manufacturing Engineering
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BDA37101-Edition III/2011
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Manufacturing Engineering
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7. CALCULATIONS
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8. GRAPH ANALYSIS
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UNIVERSITI TUN HUSSEIN ONN MALAYSIA Faculty of Mechanical and
Manufacturing Engineering
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9. QUESTIONS
TASK 2 (REFERE TO TASK 2 PROCEDURES): Question: Is there an
offset?
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TASK 3 (REFER TO TASK 3 PROCEDURES): Question: Describe the system
output response
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Manufacturing Engineering
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BDA37101-Edition III/2011
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Question: When is the offset percentage highest? Lowest?
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Question: What is the cause of offset?
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TASK 4 (REFER TO TASK 4 PROCEDURES): Q: Describe the general
system response, the response time, and the percentage of offset.
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Q: How does the system respond?
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Q: Describe the general system response, the response time, and
the percentage of offset.
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Q: Describe the general system response, the response time, and the
percentage of offset.
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Q: How does the Kc affect the system response time? The offset?
(Kc 100,Kc1 and Kc 10)
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10. CONCLUSION Deduce conclusions from the experiment. Please
comment on your experimental work in terms of achievement, problems
faced throughout the experiment and suggest recommendation for
improvements.
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UNIVERSITI TUN HUSSEIN ONN MALAYSIA Faculty of Mechanical and
Manufacturing Engineering
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BDA37101-Edition III/2011
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UNIVERSITI TUN HUSSEIN ONN MALAYSIA Faculty of Mechanical and
Manufacturing Engineering
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BDA37101-Edition III/2011
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11. REFERENCES
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UNIVERSITI TUN HUSSEIN ONN MALAYSIA Faculty of Mechanical and
Manufacturing Engineering
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BDA37101-Edition III/2011
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RESULTS WORKSHEET
Column 1
Column 2
Column 3
Column 4
Column 5
Column 6
Column 7
Initial reference
value (e=0)
Reference value
(after step input)
Percentage of step
Error Controller output
Total response
time
Error/ reference
value
Same 0%
30%
40%
50%
-20%
-30%
-40%
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