Faculty of Engineering
Helwan
Prepared by
Dr. Mohiy Bahgat
Faculty of Engineering
Helwan University
Course Outlines
1.
Introduction and Characteristics of
Industrial Processes :
a. Process control : necessity ,
possibility and implementation.
b. Discrete processes .
c. Batch processes .
d. Continuous processes .
e. Hybrid processes .
2.
Mathematical Modeling of
Industrial Processes :
a. Modeling procedure.
b. Linearization & numerical
solutions.
c. Laplace transform & T.F.
d. Block diagrams & response
e. Examples.
3. Measurement of Control
System Parameters :
a. Temperature Sensors.
b. Position Sensors.
c. Pressure Sensors.
d. Force Sensors.
e. Fluid Sensors.
4. Industrial Controllers :
a. On/Off Controller .
b. P , I , D , PI , PD and PID
controllers.
c. Temperature Control .
d. Pressure Control .
e. Flow Rate Control .
f. Level Control .
5. Introduction to process
automation :
a. Introduction.
b. PLC Operation Scan.
c. PLC Addressing.
d. Relay Ladder Logics (RLL).
6. Application to process
automation using PLC:
a. Signal Lamp Process.
b. Machine Safety Process.
c. Central Heating Process.
d. Automatic Mixing Process.
e. Automatic Packing Process.
Chapter 1
Introduction and
Characteristics of
Industrial Processes
Chapter (1) Outlines
1. Process control : necessity ,
possibility and implementation.
2. Discrete processes .
3. Batch processes .
4. Continuous processes .
5. Hybrid processes .
To make a good introduction to the
process control, we should answer
some questions such as :
1. What is a process?
2. What does a control system do?
3. Why is control necessary?
4. Why is it possible?
5. How it can be done?
6. Where it can be implemented?
7. What are the control engineers
interests?
8. How can the process control be
documented?
9. What are control strategies?
1.What is a process control?
Process control is an engineering discipline
that deals with architectures, mechanisms,
and algorithms for controlling the output of
a specific process.
This
can
be
simple
as
making
the
temperature in a room kept constant or as
complex as manufacturing an integrated
circuit.
The
industrial
processes
are
different in behavior, architecture
and characteristics as follows :
a. Discrete processes.
b. Batch processes.
c. Continuous processes.
d. Hybrid processes.
a. Discrete processes :
It can be found in many manufacturing,
motion and packaging applications.
Robotic assembly, such as that found in
automotive
characterized
production,
as
discrete
can
be
process
control.
Most discrete manufacturing involves
the production of discrete pieces of
product, such as metal stamping.
Robot arm as a part of a discrete process
b. Batch processes :
It can be found in some applications
require
specific quantities of raw
materials to be combined in specific
ways for particular durations to
produce an intermediate or end result.
One
example
is
the
production
of
adhesives and glues, which normally
require the mixing of raw materials in a
heated vessel for a period of time to
form a quantity of end product.
Other important examples are the
production of food, beverages and
medicine.
Batch processes are generally used
to produce a relatively low to
intermediate quantity of product per
year.
Mixing or batching processes
c.Continuous processes :
often,
a
physical
system
is
represented through variables that
are smooth and uninterrupted in
time.
The
control
of
the
water
temperature in a heating jacket, for
example, is a form of continuous
process control.
Some
important
continuous
processes are the production of
fuels, chemicals and plastics.
Continuous
processes,
manufacturing, are used
produce very large quantities
product per year, millions
billions of pounds.
in
to
of
to
Continuous reject process
d. Hybrid processes :
Applications having elements of
discrete, batch and continuous
process control are often called
hybrid applications.
Product line as a hybrid process
2. What does a control system do?
A control system normally perform
three main steps :
a. Measurement process for the
variable to be controlled, or
collecting
data
from
the
controlled plant.
This is done by sensors or data
acquisition cards.
b. Comparison
between
the
measured
variable
and
a
reference value , doing some
calculations to get the change
in
the
variable,
or
data
processing for the collected
data.
This is done by comparators, or
through running of an algorithm
or program.
c. Making a final decision in order
to maintain the sensed variable
within
a
desired
range,
or
sending some control signals to
the controlled plant.
This is done via the system
actuators
elements.
or
final
control
Manual level control steps
Automatic level control system
Manual heating control steps
Automatic heating control system
So,
the final goal of the control is to
maintain or adapt desired conditions in
a physical system by adjusting selected
variables in that system.
This
can be done by making a use of
an output signal of a system to
influence an input signal of the same
system, which called feedback.
3. Why is control necessary?
The industrial processes needs some degree
of control for two main reasons :
1. To maintain the controlled conditions
in a physical system at the desired
values when disturbances occur.
2. To respond to changes in the desired
values by adjusting selected variables
in the process.
Finally, the process control will assure
the following aspects :
a. Safety.
b. Environmental protection.
c. Equipment protection.
d. Smooth plant operation.
e. Product quality.
f. Profit optimization.
g. Monitoring and diagnosis.
4. Why is control possible?
When designing an industrial process or a plant,
several considerations must be accounted such
as :
a. Expected changes in the plant variables.
b. Providing adequate equipment.
c. Adding a percentage extra capacity for
the equipment sizing.
If the previous considerations are not correct,
or the plant design is not accurate, the control
may not be possible.
5. How can control be done ?
 In simple process control, it can be
done using the human feedback.
 In complex processes the feedback
actions are automated by sensing,
calculating, manipulating the controlled variables by communicated
parts of the control system.
So,
one can say that, the process
control is done automatically using
instrumentation
and
computation
that perform all the features of
feedback control without requiring,
but allowing the human intervention.
6. Where can control be implemented?
In order to operate an industrial
process on a minute-to-minute basis,
a lot of information from much of the
process has to be available at a
central location which known as the
control room or control center.
Such control scheme is generally
known as SCADA system.
 Sensors
and control elements are
located in the process.
 Signals
which are mostly electronic
or communications with the control
center to be viewed to the operator.
 Distances
between the process and
the control center ranges from few
hundred feet to a mile or more.
 In
some processes, small control
panels
are
used
nearby
the
equipment to allow access to them.
Local and centralized control equipment
7. What are the control engineers
interests?
The main interests
of
the process
control engineers are :
a. Process design :
where the process
must be designed such that being with
rapid response and minimal disturbances.
b. Measurements :
where the sensors
has to be selected with rapid response
and high accuracy.
c. Final elements :
where the final control
elements must be provided and handled so that the
manipulated variables can be adjusted by the
control calculation.
d. Control structure :
where the basic issues in
designing the controller must be considered such
as which control element should be manipulated to
control which measurement.
e. Control calculations :
where equations are
used to handle the measurements and the desired
values in calculating the manipulated variables.
8. How can the process control be
documented?
The process control can be documented in many
forms :
a. Equipment specifications and sizing.
b. Operating manuals.
c. Technical
experiments
and
control
equations.
d. Engineering drawings.
e. ROMs for storing the control algorithms.
f. Additional EPROMs.
Stirred-tank with
composite control
Flow controller
Tank level controller
Mixing process with
composite control
The process drawings include some
symbols such as :
•A
•F
•L
•P
•T
analyzer.
flow rate.
level of liquid or solids in a vessel.
pressure.
temperature.
and so on … … …
9. What are the control strategies?
Control strategies
Classical
control
Industrial
controllers
+
PLCs
Modern
control
Adaptive
control
Optimal
control
Robust
control
Computer control
A.I
control
Components of Industrial Process
The industrial processes comprises
several types of components :
a. Process.
b. Measuring elements (sensors).
c. Error detectors (comparators).
d. Controllers (industrial or computer).
e. Final control elements(actuators).
Chapter 2
Mathematical
Modeling of
Industrial Processes
Chapter (2) Outlines
1. Modeling procedure.
2. Linearization & numerical
solutions.
3. Laplace transform & T.F.
4. Block diagrams & response
5. Examples.
Modeling Procedure
The general steps for building a
mathematical model of a process
can be summarized as follows :
1. Define goals :
a. Specific design decisions.
b. Numerical values.
c. Functional relationship.
d. Required accuracy.
2. Prepare information :
a. Sketch process and identify
the system.
b. Identify variables of interest.
c. State assumptions and data.
3. Formulate the model:
a. Formulate the conservation
balances
Eqns.)
(Energy
balance
b. Formulate
the
matrial
constitutive equations.
c. Combine
equations
collect terms.
and
d. Check degrees of freedom.
e. Convert to the dimensionless
form.
4. Determine the solution:
a. Analytical.
b. Numerical.
5. Analyze the results :
a. Check results for correctness:
• Limiting and approximate answers.
• Accuracy of numerical methods.
b. Interpret the results :
• Plot solutions.
• Characteristic
behavior like
oscillations or extrema .
• Relate
results to data and
assumptions.
• Evaluate sensitivity.
• Answer “ what if ” questions.
6. Validate the model :
a. Select key values for validation.
b. Compare
with
experimental
results.
c. Compare with results from more
complex model.
The previous procedure can be
divided into two main sections :
a. Model
development
steps
(steps 1 to 3).
b. Model solution and simulation
(steps 4 to 6).
Model
Development
Steps
1. Define Goals
• The goal should be specific and clear.
• Sometimes the goal is represented
by numerical values.
•
In some cases the system’s behavior
is the goal.
•
The model accuracy should
included in the goal definition.
also
2. Prepare Information
The information needed to be prepared
are :
1. Identifying
the process, the key
variables and the system boundaries.
2. The assumptions on which the model
will be built on.
3. The
data regarding the physical
process
components
and
the
external inputs to the process.
3. Formulate the Model
When formulating a model of an industrial
process, the first step is to select the
variables whose behavior is predicted and
then deriving the equations based on the
conservation balance in mass and energy
in addition to the accumulation as follows:
Accumulation = In – Out + Generation
Hence :
Material Balance :
Accumulation of mass = Mass in – Mass out
Energy Balance :
Accumulation of energy =
(H + PE +KE)in – (H + PE + KE)out + Q – Ws
Where :
H : enthalpy = E + pv
E : internal energy
pv : flow work
PE : potential energy
KE : kinetic energy
Q : heat transferred to the process
from the surroundings.
Q = h . A . ∆T
Ws : work done by the process on the
surroundings
Model Solution
and Simulation
Steps
4. Determine Solution
• Determining
the
process
solution is very important.
model
• The
analytical solution under some
approximations is usually sought first.
• If
such solution results in unacceptable errors, numerical solutions are
then sought. Despite they are not
exact but errors can be made less.
The analytical solution steps are :
a. Calculate
the
required
specific
numerical values.
b. Determine the important functional
relationships
model,
among
the
process
variables
and
system
sensitivity
study
of
behavior.
c. Make
a
results
changes.
associated
with
the
data
5. Analyze Results
1. The first step of result analysis is to evaluate
whether the solution is correct or not. This can
be done by ensuring the following :
a. The results satisfy initial and final conditions
b. Obey the process bounds.
c. Contains
negligible
errors
associated
with
numerical solutions.
d. Obey the process semi-quantitative expectations
such as output change sign.
2. The second step of result analysis is to analyze
the process behavior. This can be done by :
a. Determining
the
numerical
results
quantitavelly to help in making decisions
regarding the equipment operation and
sizing.
b. Plotting the results.
c. Observing the process oscillations in case
of max or min oscillations.
d. Studying the process behavior associated
with change in data or important variables.
6. Validate Model
The model validation involves determining
whether
the
results
obtained
in
the
previous steps are truly represent the
physical process.
This
can
be
done
by
comparing
the
obtained results with some experimental
results taken from the process at different
operating points to assure the model
validity in representing the process.
Example (1)
For the mixing tank shown in figure :
1. The goal is to determine the dynamic
response due to a step change in the
inlet concentration. In other words,
determining the time needed for the
outlet reaches 90% of change in
concentration after the step change in
the inlet.
2. The information :
a. The process is the tank with its fluid in it, its
design and shape and the speed of making the
fluid uniform.
b. Assumptions
:
well-mixed
vessel,
density is the same for A and solvent S
in addition the flow in is constant.
c. Data : F0 = 0.085 m3/min , CAinit = 0.925
mole/m3 , ∆CA0 = 0.925 mole/m3 and
CA0 = 1.85 mole/m3 after the step.
The system is initially at steady state.
3. The model formulation :
problem
involves
since the
concentration, hence
using the material balance equation we can
get :
a. Accumulation of mass = Mass in – Mass out
(ρV)(t+∆t) – (ρV)(t) = Fo ρ.∆t – F1 ρ.∆t
Dividing by ∆t and taking the limit as ∆t
d(ρV )
dt
0
dV
ρ
= ρ F0 - ρ F1
dt
Assuming that the level in the tank is almost
constant, which means that the flows in and
out are equal, i.e :
Fo = F1 = F
dV/dt = Fo – F1 = 0 ,
i.e :
or :
V = constant
(1)
b. applying the same material balance for
component A :
Accumulation of comp A = Comp Ain – Comp Aout
(M WA V CA)(t+∆t) – (M WA V CA)(t) = ( M WA F CAo
– M WA F CA )∆t
Dividing by ∆t and taking the limit as ∆t
dC A
M WA V
= M WA F (CA0 - C A )
dt
applying
the
component S :
same
material
balance
0
(2)
for
dC s
M Ws V
= M Ws F (Cs0 - Cs )
dt
Accordingly :
• The process variables are :
CA
and
F0
and
F1
• The external variables are :
•
CA0
The process model is represented by
equations (1) & (2) .
4. Determine the solution :
as it can
been seen from eqn.(2) the process model is
a
linear
1st
order
ordinary
differential
equation that can be transformed to the
separable form using an integral factor as
follows :
dC A
M WA V
= M WA F (CA0 - C A )
dt
dC A
V
= F (CA0 - C A )
dt
dC A 1
1
+ CA =
C A0
dt
τ
τ
, with V/F = τ = time constant
Use the integrating factor I.F = exp(∫ (1/τ)dt = et/ τ
C A0 t/τ
dC A 1
e (
+ CA ) =
e
dt
τ
τ
t/τ
e
t/τ
C A0 t/τ
dC A
d(e C A )
de
+ CA
=
=
e
dt
dt
dt
τ
t/τ
t/τ
t/τ
C
e
C A0
A0
t/τ
∫d(CA e ) = ∫ τ dt = τ
∫e
t/τ
dt
CAe
t/τ
C A0 τ t/τ
=
e +K
τ
C A = C A0 + K e
-t/ τ
Using the initial conditions, we get :
K = CAinit – CA0
C A = C A0 + (CAinit - C A0 ) . e
C A - C Ainit = ΔC A0 (1 - e
- t/τ
-t/ τ
)
Substituting with the given numerical values :
CA - 0.925 = 0.925 (1 - e
-t/24.7
)
(3)
Two aspects of the process dynamic response
have to be considered :
a. The
speed
of
response
which
characterized by the time constant
τ.
b. The steady state gain which is :
Δ CA
Δ output
Kp =
=
=1
Δ input
Δ C A0
is
5. Result analysis :
the solution of the
process model described in eqn. (3) is an
exponential curve as displayed in the Fig.
The process response from the change
beginning to the end is affected by the time
constant (
τ ), where the large time constant
the slow process response and vice versa.
According to the goal, it is needed to know
the time taken to get 90% of the change in
outlet concentration.
This time can be calculated from eqn. (3).
Dynamic response of the process
6.Validation :
By performing an experiment on a stirred
tank as
described in the controlled process and taking samples
of the outlet material, analyzing the obtained samples
and drawing the data points, one can get the shown Fig.
By visual evaluation, one can say the model is valid in
representing the process.
Example (2)
On-off room heating process :
1. The goal :
is to determine the dynamic
response of the room temperature.
Also, ensure that the furnace does not
switches on or off more than once per 3
minutes
2. The information :
a. The process is the air inside the room.
The important variables are the room
temperature
status.
and
the
furnace
on-off
b. Assumptions :
1. The air in the room is well mixed.
2. No transfer of material to or from the room.
3. The heat transferred depends only on the
temp. difference between the room and the
outside environment.
4. No heat is transferred from the floor to the
ceiling.
5. Effects of kinetic and potential energies are
negligible.
c. Data :
1. The heat capacity of the air CV = 0.17 cal/g Cº.
2. The overall heat transfer coefficient UA = 45 x
103 cal/Cº hr.
3. The size of the room is 5 m by 5 m by 3 m
high.
4. The furnace heating capacity Qh is 0 (of) or
1.5 x 103 cal/hr (on).
5. The furnace switches inst. At 17 Cº (on) and at
23 Cº (off).
6. The initial room temp. is 20 Cº.
7. The outside temp. is 10 Cº.
3. The model formulation :
since the
process is defined as the air inside the
room, hence using the energy balance
equation one can get :
dE/dt = KE + PE + Q – Ws
KE = PE = Ws = 0
dE/dt = Q
from assumptions
…………
but dE/dt = ρ V CV dT/dt
and Q = - UA (T – Ta) + Qh
and Qh is represented by :
(1)
0
Qh =
when T > 23 Cº
1.5 x 106
when T < 17 Cº
unchanged
when T < 23 Cº
Finally, the process model is :
dT
ρ V CV
= - UA ( T - Ta ) + Qh
dt
Finally, the process model is :
dT
ρ V CV
= - UA ( T - Ta ) + Qh
dt
• The process variables are :
T
and
Qh
• The external variable is :
•
Ta
The process parameters are :
UA , CV , V and ρ
(2)
4. Determine the solution :
rearranging
eqn. (2) gives the following linear O.D.E :
UA Ta + Qh
dT 1
+ T=
dt
τ
V ρ Cv
, with
V ρ Cv
τ=
UA
Use the integrating factor IF = exp(∫ (1/τ)dt = et/ τ
dT 1
t/τ UA Ta + Q h
e (
+ T)= e .
dt
τ
V ρ Cv
t/τ
t/τ
t/τ
dT
de
d(e
. T)
t/τ
t/τ UA Ta + Q h
e .
+ T.
=
= e .
dt
dt
dt
V ρ Cv
UA Ta + Qh
∫d(e . T) = ∫e . V ρ C v dt
t/τ
t/τ
UA Ta + Qh
∫d(e . T) = V ρ C v
t/τ
∫e
t/τ
dt
UA Ta + Qh t/τ
e . T = τ.
e +K
V ρ Cv
t/τ
UA Ta + Qh
- t/τ
T = τ.
+K.e
V ρ Cv
Using the initial conditions, we get :
T - Tinit = ( Tfinal - Tinit ) ( 1 - e
-t/ τ
)
Where : t = time from step in Qh
τ = time constant = 0.34 hr
Tfinal = final value of T as t
∞ = Ta+Qh/UA
= 10 Cº when Qh = 0
= 43.3 Cº when Qh = 1.5 x 106
Tinit = the value of T when a step in Qh
occurs.
Process response
5. Result analysis :
from the previous
figure, it can noticed that the room temp
decreases until it reaches
17 Cº, the
furnace will start heating and the temp.
increases until it reaches
23 Cº. This
process will be repeated with the heater On
and Off periodically.
•
Linearization
If
the
developed
process
model
is
linear,
analytical
solutions
can
be
obtained easily.
•
Most of the physical system models are
nonlinear.
•
The
analytical
solutions
of
the
nonlinear models are not available, thus
the numerical simulations are sought.
•
•
Instead
of
obtaining
nonunderstandable solutions for the nonlinear
models by numerical simulations,
approximate linearized solutions can
be used for representing realistic
processes.
A model is to be linear if it satisfies
the
properties
of
additivity,
proportionality and superposition as
follows :
•
A system satisfies the property of
additivity, if a sum of inputs results
in a sum of outputs. If there is an
input of :
x3(t) = x1(t) + x2(t)
it should result in an output of :
y3(t) = y1(t) + y2(t)
•
A system satisfies the property
of superposition, if a sum of
scaled inputs results in a sum
of scaled outputs. i.e :
f(Ax + By) = f(Ax) + f(By)
= A.f(x) + B.f(y)
•
If the system has the following
performance equation :
f(x) = k . X½
f(Ax1 + Bx2) = k.(Ax1 + Bx2)½
≠ k.(Ax1)½ + k.(Bx2)½
Thus the above system in not linear,
it is nonlinear system. And so on …
•
If
the
dynamic
process
following
is
behavior
of
considered,
example
a
the
can
be
illustrated.
Consider the following
stirred
tank heat exchanger when being
subjected to a change in the feed
temperature and cooling fluid flow
rate.
Stirred tank heat exchanger
Process response
due to a change in
cooling fluid flow
rate
Process response
due to a change in
feed temperature
The total process response due to a change in
both feed temperature and cooling fluid flow rate
•
According to the dynamic behavior
of the process, one can say that
this process is linear because it
obeys the superposition principle.
•
In general a nonlinear process
model
can
approximated
be
linearized
and
by a linear model
using Taylor series expansion as
follows :
•
A process nonlinear model with one
variable can be linearized around its
S.S point as :
dF
1 dF
2
F ( x )  F(xs ) 
( x  xs ) 
( x  xs )  R
dx xs
2 dx xs
•
A process nonlinear model with two
variables can be linearized around
its S.S point as follows :
F
F(x 1 , x 2 )  F( x 1s , x 2 s ) 
x 1
F

x 2
x1 s , x 2 s
x1 s , x 2 s
1  2F
(x 2  x 2 s ) 
2
2 x 1
2
1F

2
2 x 2
(x 1  x 1s )
(x 1  x 1s )
x1 s , x 2 s
2
1 F
(x 2  x 2 s ) 
2 x 1 x 2
2
x1 s , x2 s
2
(x 1  x 1s )( x 2  x 2 s )  R
x1 s , x2 s
Comparison between linear and
exact nonlinear models
Function examples :
1. F(x) = x½
F(x) ≈ xs½ + ½ xs-½ (x – xs)
2.
x
F(x) =
1 a x
xs
1
F(x) 

(x  x s )
2
1  a x s 2 (1  a x s )
Example (3)
Tank draining process :
1. The goal :
is to determine the model of this
tank process.
Also,
evaluate
the
accuracies
of
the
linearized model at small (10 m3/hr) and
large (60 m3/hr) step changes in the inlet
flow rate.
2. The information :
a. The process is the liquid in the tank.
The important variables are the level
and the flow out.
b. Assumptions :
1. The density is constant.
2. The cross sectional area of the tank A does not
change with height.
3. The system is at quasi-steady state because the
pipe dynamics is fast with respect to that of the
tank level.
4. The pressure is constant at inlet and outlet.
c. Data :
1. The initial steady state conditions are :
i. Flows F0 = F1 = 100 m3/hr
ii. Level L = 7 m.
2. The cross sectional area A = 7 m2
3. The model formulation :
since the
process is defined as the liquid in the tank
and the level depends on the total amount
of liquid, thus using the material balance
equation one can get :
dL
A
  Fo -  F1
dt
Another
................... (1)
eqn. is required, one can relate
the outlet flow to the head as follows :
F1  kF1(Pa - L - Pa )
0.5
 kF1 L
0.5
.... (2)
Finally, combining the two eqns., the process
model will be :
dL
0.5
A
 Fo - k F1L
dt
................... (3)
This model has a nonlinear term which can be
linearized as :
L
0.5
 L s  0.5 L s (L - Ls ) ........ (4)
0.5
-0.5
Replacing the nonlinear term in Eqn.(3) and
Subtracting the S.S conditions and putting the
input as a constant step : F’o = ∆Fo , one can get :
Finally, the process model is :
dL'
0.5
A
 Fo - (0.5kF1L s ) L' ...... (5)
dt
• The process variable is :
L’
• The external variable is :
•
∆Fo
The process parameters are :
A
and kF1
4. Model solution :
rearranging eqn. (5)
gives the following linear O.D.E :
dL' 1
1
 L' 
Fo
dt

A
A
, with  
0.5 k F1 L-s0.5
Use the integrating factor IF = exp(∫ (1/τ)dt = et/ τ
dL' 1
1
t/
e (
 L' )  e . Fo
dt

A
t/
e
t/
dL'
de
d(e L' )
t/ 1
 L'

 e
Fo
dt
dt
dt
A
t/
t/
 d(e
t/
 d(e
t/
e
t/
L' ) 
e
t/
1
Fo dt
A
Fo
t/
L' ) 
e dt

A
 Fo t/
L' 
e K
A
 Fo
L' 
 K . e - t/
A
  Fo
K
A
Using the initial conditions, we get :
 Fo
- t/
L' 
( 1- e )
A
L'  Fo K p ( 1 - e
-t/ 
)

1
with K p 

A
0.5k F1Ls0.5
Where : t = time from step in Fo
τ = time constant = 0.98 hr
kp = 0.14 hr/m2
Process response to a small change in inlet flow rate
Process response to a large change in inlet flow rate
5. Result analysis :
from the previous
figure, it can be noticed that the solution of
the linearized model is quite accurate with
the small change in the inlet flow rate.
On the other hand, it is inaccurate with the
large change in the inlet flow rate and it
gives impossible negative level at S.S.
The general trend is the linearized model be
more accurate with small changes than with
the large ones.
Example (4)
Stirred tank heat exchanger process :
1. The goal :
is to determine the dynamic
response of the tank temperature due to a
step change in the coolant flow rate.
2. The information :
a. The process is the liquid in the tank.
The
important
temperature.
variables
are
tank
b. Assumptions :
1. The tank is well insulated, i.e, no heat
transfer to the surroundings.
2. The energy accumulation in the tank
walls
and
the
cooling
coil
are
negligible to that of the liquid.
3. The tank is well mixed.
4. Physical properties are constant.
5. The process is initially at steady state.
c. Data :
F = 0.085 m3/min.
,
V = 2.1 m3
Ts = 85.4 Cº
,
ρ = 106 g/m3
Cp = 1 cal/g.Cº
,
To = 150 Cº ,
Tcin = 25 Cº
,
Fcs= 0.5 m3/min
Cpc = 1 cal/g.Cº
,
ρc = 106 g/m3
3. The model formulation :
since the
process is defined as the liquid in the tank
and the tank temperature depends on the
amount of liquid.
Thus using the material balance equation
one can get :
Fo = F1 = F
assuming constant level.
Thus using the energy balance equation
one can get :
dE
 {Ho } - {H1 }  Q - Ws
dt
.... (2)
Since the tank is well insulated, i.e, Ws = 0
the following thermodynamic relations can
be written :
dE
dT
dT
  V Cv
  V Cp
dt
dt
dt
Hi   Cp Fi (Ti - Tref )
..... (3)
.......... (4)
Which means that heat capacity at
constant volume is approximately as the
heat capacity at constant pressure.
Subs. From (3) and (4) into (2) gives :
dT
VC p
 CpF [(To - Tref ) - (T1 - Tref )]  Q .... (5)
dt
To complete the process model, the heat
transferred Q should be related to tank
temperature by applying the energy
balance on the cooling coil as :
Tcout  Tcin
Q
cCpcFc
.......... (6)
Where : c denotes the coolant fluid.
Furthermore, the heat transferred can be
expressed as :
 (T - Tcin )  (T - Tcout ) 
Q  UA
 .... (7)
2


Also, the heat coefficient UA depends on
film coefficients and wall resistance as :
UA  a F
b
c
.......... (8)
Substituting from (6) and (8) in (7), the
heat transferred Q can be expressed as :
aFcb 1
Q
(T - Tcin ) .... (9)
b
aFc
Fc 
2 cCpc
Substituting the eqn. (9) into eqn. (5), one
can get the process final model as :
dT
VC p
 CpF (To - T) dt
aFcb 1
(T - Tcin ) .... (10)
b
aFc
Fc 
2 cCpc
The process model described in Eq. (10) is
nonlinear model due to the variable Fc raised to
the power b or b+1 and due to the product of
Fc and T with the following variables :
• The process variable is :
T
• The external variables are :
•
Tcin , To , F and Fc
The process parameters are :
V , ρ , Cp , a , b , ρc and Cpc
Linearizing the Eq. of the heat transferred Q as :
Q = Qs + KT(T – Ts) + KFC(Fc – Fcs) …. (11)
where :




b 1
 - aFc (T - Tcin ) 
Qs  
b

aFc
 Fc 

2 cCpc  s

and


- aFcb 1

KT  
aFcb
 Fc 
2 cCpc






s
also :
K Fc
b




aF
 - aFb  F 

c

(
T

T
)
c c
cin 


2 b cCpc 



2
b




aF
c
 Fc 







2

C
c
pc



s
Finally, the linear model of the process in terms
of the variable deviations is :
dT
VC p
 CpF (-T' )  K T T'  KFC Fc' .... (12)
dt
4. Model solution :
rearranging eqn. (12)
gives the following linear 1st order D.E :
K Fc
dT' 1
+ T' =
Fc'
dt
τ
V ρ Cp
F
with τ =
V
KT
V ρ Cp
1
Use the integrating factor IF = exp(∫ (1/τ)dt = et/ τ
KFc
dT' 1
t/
e (
 T' )  e
Fc
dt

V  Cp
t/
e
t/
KFc
dT'
de t/ d(et/ T' )
t/
 T'

e
Fc
dt
dt
dt
V  Cp
 d(e
 d(e
e
t/
t/
t/
T' ) 
e
t/
KFc
Fc dt
V  Cp
KFc Fc
T' ) 
V  Cp
t/
e
 dt
KFc Fc  t/
T' 
e K
V  Cp
K Fc Fc 
T' 
 K e - t/
V  Cp
Using the initial conditions, we get :
 K Fc Fc 
K 
V  Cp
K Fc Fc 
T' 
(1 - e - t/ )
V  Cp
T'  Fc K p ( 1 - e
-t/ 
)
K Fc 
C
with K p 
 33.9 3
V  Cp
m / min

F

K
T


and  

V V C 
p 

1
 11.9 min
Process response to a step change in coolant flow rate
5. Result analysis :
from the previous
figure, it can be noticed that the solution of
the linearized model is accurate along with
the nonlinear model of the process for the
considered small change in the coolant flow
rate.
Numerical Solutions of O.D.E
• As
seen before, most of the practical
modeling of process and process
control would result in nonlinear
models.
• The nonlinear algebraic and differential
models can not be solved analytically.
• Such models are solved using different
methods of numerical solutions.
• The
numerical solutions do not give
expressions as before, but they give
points close to the exact solutions of
the process models.
• The concept of
the numerical methods
is to use initial values and an
approximation of the derivatives over a
step of integration, and hence calculate
the variables after that step as follows.
• The
most numerical methods for solving
differential equations consider the Taylor
series
expansion
and
make
approximations by choosing a specified
terms of the series.
• The most commonly used methods are :
a.Euler’s
method which considers the
first two terms of the Taylor series.
yi+1 = yi + f(ti , yi)*∆t
b.Heun’s
method which considers the
first three terms of the Taylor series.
k1 = f(ti , yi)
k2 = f(ti+∆t , yi+k1*∆t)
yi+1 = yi + ( k1/2 + k2/2 )*∆t
c. Runge-Kutta
fourth order method
which considers the first four
terms of the Taylor series.
yi+1= yi+(∆t/6)* (k1+2k2+2k3+k4)
where :
k1 = f (ti , y i )
k2 =
Δt
Δt
f ( ti +
, yi +
k1)
2
2
k3 =
Δt
Δt
f ( ti +
, yi +
k2 )
2
2
k4 = f(ti + Δt , yi + k 3 * Δt)
• The selection of the step size ∆t is very
important in reaching the approximate
solution of the process model using the
numerical solutions.
• In
Euler’s
method
the
error
is
proportional to the step size ∆t,
however, in Runge-Kutta method the error
is proportional to (∆t)4.
• In
most engineering applications, the
appropriate step size is ∆t = 0.01 sec.
Model Analysis of Processes
• The process model which comprises a
linear differential equation can be
solved by analytical solution.
• The process model which comprises a
set of linear differential equations
with constant coefficients can be
solved by Laplace transform method.
•A
control system involves several
simultaneous processes and control
calculations can be modeled using
input and output variables with the
aid of block diagrams and transfer
functions.
• The process behavior to sine inputs
can be carried easily using the
frequency response method to
illustrate the influence of input
frequency.
• When
a process is subjected to a
step disturbance , it is required to
determine whether its behavior is
stable or not.
Laplace Transform
• The
Laplace transform is a very
powerful method for engineers to
analyze the process control and
control systems.
• It
converts the constant coefficient
differential equations to algebraic
equations which can be solved easily.
• It
replaces the time domain by a
frequency domain or complex domain.
• The Laplace transform is defined as :

L( f (t ))  F(s)   f(t).e
•
The Laplace
operator, i.e :
0
transform
- st
is
dt
linear
L[af1(t) + bf2(t)] = aL[f1(t)] + bL[f2(t)]
• Tables
of Laplace transform and its
inverse transform are available for most
commonly used functions as follows :
• The inverse Laplace :
L-1[F(s)] = f(t)
for t ≥0
• Laplace transform for a constant :

C
e

L(C) =
- st
0
C -st
dt  - e
s
∞
0
C

s
• Laplace transform for an exponential :

e

L(eat) =
0
at
e
- st
1
-(s- a)t
dt 
e
a-s
∞
0
1

s-a
• Laplace transform for a step function :
x(t)
 A; t  0
x( t )  
0 t  0

X(s)   x(t).e
t
- st
0
A - st
X(s )  - e
s

dt   A.e
0

0
A

s
- st
dt
Function f(t)
δ
unit impulse
U(t) unit step or
constant
 A;
u( t )  
0
t 0
t 0
 At; t  0
r(t )  
t 0
0
tn
Laplace F(s)
1
1/s
A/s
A / s2
n! / sn+1
Function f(t)
Laplace F(s)
at
1 / (s-a)
e
1 t / 
e

1 / (τs+1)
sin (ωt)
ω / (s2 + ω2)
cos (ωt)
s / (s2 + ω2)
f (t  a) t  a
f ( t)  
t a
0
e
 as
F(s)
Function f(t)
Laplace F(s)
df ( t)
dt
s F(s)
dn f ( t)
dtn
sn F(s)
t
 f (t) dt
0
F(s) / s
Laplace Transform Properties
Linearity
Time Change
L[ af (t) + bg(t) ] = aF(s) + bG(s)
1
s
L [ f (at)] 
F( )
a
a
Time – axis
displacement
L( f (t − T )) = e−sT F(s)
S – axis
displacement
L(eat f (t)) = F(s − a)
Initial Value
Theorem
Limt0 f(t)  Lims s F(s)
Final Value
Theorem
Limt f(t)  Lims0 s F(s)
Partial Fractions
C1
C2
N(s)
Y(s) 


 .....
D(s) H1(s) H2 (s)
Taking the inverse Laplace for both sides,
one can get :
1
1
-1
Y(t)  C1 L [
]  C2 L [
]  ....
H1 (s)
H2 (s)
-1
where Ci are constants and Hi(s) are the
factors of the characteristic polynomial
D(s) = 0.
If H(s) is a 1st order term, then :
N(s)
A
B
Y(s) 


 .....
D(s) H1(s) H2 (s)
If H(s) is a 2nd order term, then :
N(s) A s  B
C
Y(s) 


 .....
D(s)
H1(s)
H2 (s)
If there are repeated factors :
C1
C2
M(s)
Y(s) 


 .....
n
2
(s  a)
(s  a) (s  a)
Example (1)
For the stirred-tank
deviation variables is :
V
dC'A
dt
 F (C
'
A0
mixing
model
-C )
'
A
using Laplace transform :
V s C’A(s) = F [ C’AO(s) – C’A(s) ]
τ s C’A(s) + C’A(s) = C’AO(s)
C’A(s) ( τ s + 1) = C’AO(s)
in
Considering a step change in the inlet
concentration, i.e :
C’A0(s) = ∆CA0/s
C (s)  C A0
1
s (s  1)
C (s)  C A0
1

( 
)
s
s  1
'
A
'
A
Using the inverse Laplace, one can get :
C A (t)  C A0 (1 - e
'
-t/ 
)
Example (2)
Consider an industrial process having a
model in deviation variables as :
1

y" (t)  2 2 y' (t)  y(t)  G.x(t)
2
0
0
using Laplace transform :
1 2

s Y(s)  2 2 sY(s)  Y(s)  G.x(s)
2
0
0
G . 02
Y(s)  2
. x(s)
2
s  2 s  0
Take the input x(t) as a step function,
then its Laplace transform will be :
A
X(s) 
s
and the output final equation is :
GA . 
Y(s) 
2
2
s.[s  2 s  0 ]
2
0
by
then,
we
have
described as follows :
four
conditions
1. if α > ωo
the process Eqn. will have
two real distinct roots as :
r1,2  -  
 -
2
2
2
GA . 0
Y(s) 
s (s  r1 ) (s  r2 )
k1
k2
GA



s
(s  r1 )
(s  r2 )
Finally, the process output y(t) will be :
y(t)  G.A  k1e
r1t
 k 2e
r2 t
This condition is called the over damping
condition where the output response does
not have any oscillations as shown :
y(t)
yss = G.A
α > ωo
y(0) = 0
y’(0) = 0
time
2. if α = ωo
the process Eqn. will have
two real equal roots as :
r1,2  r  -
2
GA . 0
Y(s) 
s (s  r )2
k1
k2
GA



s
(s  r )
(s  r )2
Finally, the process output y(t) will be :
y(t)  G.A  (k1  k 2 ) e
t
This condition is called the critically
damped where the output also does not
have oscillations as shown :
y(t)
yss = G.A
α = ωo
y(0) = 0
y’(0) = 0
time
3. if α < ωo
the process Eqn. will have
two complex conjugate roots as :
r1,2  -   j   
2
2
2
GA . 0
Y(s) 
s (s  r1 ) (s  r2 )
k1
k2
GA



s
(s  r1 )
(s  r2 )
Finally, the process output y(t) will be :
y(t)  G.A  k e
t
sin(d t  )
This condition is called the under damped
where the output has decayed oscillations
as shown :
y(t)
yss = G.A
α < ωo
y(0) = 0
y’(0) = 0
time
4. if α = 0
the process Eqn. will have
two complex conjugate roots as :
r1,2  r   j
2
GA . 0
Y(s) 
s (s  j) (s  j)
k1
k2
GA



s
(s  j)
(s  j)
Finally, the process output y(t) will be :
y( t )  G.A  k cos( 0 t)
This condition is called the oscillatory
condition where the output will have
continuous oscillations as shown :
y(t)
α=0
y(0) = 0
y’(0) = 0
time
Transfer Function
The process transfer function is defined
as the Laplace transform of the output
Y(s) divided by the Laplace transform of
the input X(s).
Transfer Function =
T.F
Y ( s)
= G(s) =
X( s )
Basic Definitions
 Order :
the system order is the highest
power of s in the denominator of the T.F.
 Pole :
it is a root of the denominator of
the T.F or a root of the system
characteristic equation.
 Zero : it is a root of the numerator of
the T.F.
Steady state gain :
it is the ratio
∆Y/∆X at steady state or yss/xin and
usually denoted by K.
Example
For a system or a process whose T.F is :
Y ( s)
6 s - 45.83
 2
X(s) s  1.789s  35.8
• The system is 2 order
• The poles are : s = - 8.95 ± j 5.92
• The zero is : s = 7.64
• The S.S gain is : K = 45.83/35.8 = 1.28
nd
Block Diagram
The block diagram method is a powerful
graphical representation for the system or
process individual components based on
their T.F. It has some advantages :
a. It retains individual systems and allows
b.
c.
model simplification and changes.
It provides a visual representation of the
relationships
between
the
system
components.
It gives insight into the effect of
components
on
the
overall
system
performance.
Block Diagram for a General
Control System
Physical process control
Block diagram of the process control
On–Off control of a heating or cooling process
Analog control of the heating process
Digital control of the heating process
PLC control of the heating process
Block Diagram Notations
G(s)
X(s)
X1(s)
X1(s)
+
X2(s)
Y(s)
X3(s)
X3(s)
X2(s)
Block Diagram Algebra
1.Series or Cascaded Blocks :
X1(s)
X1(s)
G1(s)
Y1(s)
X2(s)
Y2(s)
G2(s)
G(s) = G1(s).G2(s)
Y2(s)
G(s) = G1(s).G2(s)
2. Parallel Blocks :
X(s)
G1(s)
Y1(s)
Yt(s)
+
G2(s)
Y2(s)
Yt(s)
X(s)
G(s) = G1(s) +G2(s)
G(s) = G1(s) + G2(s)
3.Feedback System Blocks :
+
R(s)
-
G (s)
C(s)
H(s)
X(s)
Gt(s)
Yt(s)
C(s)
G(s)
G(s) 

R(s) 1  G(s) H(s)
Example on B.D Reduction
X(s)
+
+
C1s
1
C1s
C2 s
R 2C 2 s  1
C1s
R1
R1
C2s
1
C2s
Y(s)
Using the series blocks rule, one can get :
X(s)
+
+
Y(s)
1
R 2C 2 s  1
1
R1 C1 s
R1 C2 s
Using the feedback rule, one can get :
X(s)
+
Y(s)
1
R1C1s  1
1
R 2C 2 s  1
R1 C2 s
X(s)
1
R1C1R 2C2s2  (R1C1  R 2C2  R1C2 ) s  1
Y(s)
Frequency Response
 The
frequency
response
is
very
important when studying the system
dynamic
behavior
associated
with
sinusoidal input at different frequencies.
 For the linear systems, the output Y’(t)
will be a sine with the same frequency
as the input X’(t).
 The relationship between input and
output can be characterized by :
Outputmagn itude
Amplituder atio 
Inputmagnitude

Y ' ( t) max
X' ( t) max
 G(j)
 Re(G(j))2  Im(G(j))2
Pahseangle    G(j)
 Im(G(j)) 

 tan 
 Re(G(j)) 
where the frequency  is in rad/sec.
-1
Example
For the mixing process shown in Fig :
The model of the stirred tank was written
before as :
dC A
V
 F (CA0 - C A )
dt
Since the inlet will be used as a sinusoidal
input, i.e : CA0 = A sin(ωt) , one can rewrite
the system model as :
dC A
V
 F. A sin( t) - F . C A
dt
Using
τ =
V/F and taking the Laplace
transform, the system model becomes as :

 s C A (s)  A . 2
- C A (s)
2
s 

C A (s) (1   s)  A . 2
2
s 
A . / 
C A (s) 
1
(s  ) (s 2  2 )

k3
k1
k2



1
(s  ) (s  j) (s  j)

Using the partial fraction method, the
system model becomes as :
k3
k2
C A (s) 


1
(s  ) (s  j) (s  j)

k1
Where :
A 
k1 
1  2 2
A
1
k2 
e j
j2 1   2 2
k3
A

j2
1
1   2 2
  tan 1(  )
e j
Finally, the system model becomes as :
CA (t)  k1 e
-t/ 
 k2 e
j( t  )
 k3 e
-j( t  )
Which can be rewritten in the form :
A
A
- t/
C A (t) 
e 
sin( t  )
2 2
2 2
1  
1  
Note that the input was : CA0(t) = A sin(ωt)
Chapter 3
Measurement of
Control System
Parameters
(Sensors
Chapter (3) Outlines
1. Temperature Sensors.
2. Position Sensors.
3. Pressure Sensors.
4. Force Sensors.
5. Fluid Sensors.
6. Signal Conversion.
Temperature Sensors
• Temperature
process
sensors are used in the
control
that
concerns
with
temperature regulation.
• It depends on the electrical methods of
measuring temperature.
• The basic types of
are :
temperature sensors
a. Bimetal temperature sensor.
b. Resistance–Temperature
Detectors
(RTD).
c. Thermistors,
which represent the
semiconductor usage in measuring
temperature.
d. Thermocouples.
e. Solid state temperature sensors.
Bimetal Temperature Sensors
• The
bimetallic temperature sensors have
some advantages :
• simplicity and low cost.
• Disadvantages are :
• Existence of hysteresis.
• Inaccuracy.
• Slow time response.
• They are used in ON/OFF
applications.
cyclic
• The
sensor basic operation is built on the
thermal linear expansion which is the
change in dimensions of a material due to
temperature changes.
L = Lo ( 1 +
where :
γ . Δt )
L = the final length.
Lo = the initial length.
Δt = T – To = temp. difference.
γ
= the linear thermal
expansion coefficient.
• The
bimetallic
sensor
consists
of
two
materials with grossly different thermal
expansion coefficients bounded together.
• When
the sensor is being subjected to
heating, the different expansion rates of the
two
materials
will
cause
the
sensor
assembly to be curved as shown in Figure.
• This
effect can be used to close switch
contacts
mechanism
or
to
when
actuate
the
an
ON/OFF
temperature
increases to some appropriate set-point.
γ1
γ2 < γ1
at
To
Resistance Temperature Detectors
• One
of
the
most
important
electrical measurement of
methods
for
temperature is
based on the electrical resistance change of
a conducting material.
• So,
the principle of measuring or sensing
temperature is to place a conducting material
with sensitive change of
resistance with
respect to temperature in contact with the
environment whose temperature is to be
measured or sensed.
• The
resistance of a conductor
according to the following factors :
varies
1. The resistance is directly proportional
to the conductor length :
Rαl
2. The resistance is inverse proportional
to the conductor cross section area :
R α 1/a
3. The resistance depends on the type of
the conductor material :
R=ρ.l/a
4. The resistance is affected by
surrounding resistance such that :
the
RT = RTo ( 1 + α . Δt )
where :
RT = the conductor resistance at a
temperature T.
RTo = the conductor resistance at a
temperature To.
Δt = T–To = temperature difference.
α
= linear change coefficient in
resistance w.r.t temperature.
• The
resistance-temperature detector (RTD)
is a temperature sensor whose operation is
based on the resistance variation of a metal
conductor with temperature.
• Metal
used in such a sensor vary from
platinum which is quit sensitive and
expensive to nickel which is more sensitive
and less expensive.
• The
sensitivity of the RTD sensor depends
on the value of the linear change coefficient
in resistance with respect to temperature
(α).
• Typical
values
of
such
coefficient
for
different materials are :
• In
α = 0.004 /Cº
for
platinum.
α = 0.005 /Cº
for
nickel.
general the RTD has a time response
ranges between 0.5 to 5 seconds or more.
• The RTD sensor construction is basically in
the form of a wire wound as a coil to
achieve
small
size,
improved
thermal
conductivity and decreased time response.
The effective range of RTD sensors depends on
the type of the effective element wire :
Platinum RTD has the range of : - 100 to 650 Cº
Nickel RTD has the range of : - 180 to 300 Cº
Thermistor Sensors
• The thermistor represents another class of
temperature
sensor
that
measures
temperature through changes of material
resistance.
• The
characteristics of such devices are
very different from those of the RTDs and
depend on the behavior of semiconductor
resistance with temperature.
• The resistance of a thermistor is a function
of the ambient temperature.
• The
change
in
resistance
ΔR
of
the
thermistor is proportional to the change in
temperature ΔT.
• The
thermistor
construction
may
take
several forms including discs, beads and
rods varying in size from a bead 1 mm in
diameter to a disc of several cm in diameter
and thick.
• The effective range of thermistor sensors is
: - 80 to 300 Cº.
Thermocouple Sensors
• The thermocouple is a device that converts
thermal energy into electrical energy.
•A
thermocouple
is
constructed
of
two
dissimilar metal wires joined at one end.
• The most important factor to be considered
when selecting a pair of materials is the
thermoelectric difference between the two
materials.
•
Other materials may be used, for example:
Chrome-Constantan
is
excellent
for
temperatures up to 2000 F° and TungstenRhenium is used for temperatures up to 5000 F°.
•
When a thermocouple is subjected to changes
in temperature, it will cause an electric current
to flow in the attached circuit.
•
The amount of produced current depends on
the
temperature
difference
between
measurement and reference junction.
the
Simple thermocouple circuit
Thermocouple circuit with temperature
control and signal conditioning
Position Sensors
• The measurement of displacement, position,
or location is important in the process
industries. For examples :
a) Location
and
position
on
conveyor
systems.
b) Orientation of steel plates in a rolling
mill.
c) Liquid or solid level monitoring, … etc
Potentiometers :
• The
simplest type of displacement sensor
involves the action of moving the wiper of a
potentiometer.
• This
device converts the linear or angular
motion into a changing in resistance that may
converted
directly
into
a
voltage
and/or
current signals.
• The
output voltage of the sensor can be
calculated from the following formula :
Motion
Wiper
r
R
Vout
Vin
Vout
r

. Vin
R
Capacitive sensor :
• The
basic operation of a capacitive sensor
can be derived from the capacitance equation
of the parallel plate capacitor :
 o r A
C
d
where :
εo
is the air permittivity = 8.85 pF/m
εr
is the dielectric constant.
A
is the plate common area.
d
is the plate separation.
• The
capacitance of the capacitor can be
changed by varying the distance between the
plates (d), or by varying the shared area of
the plates (A) .
• An
A.C bridge or other active electronic
circuit is employed to convert the capacity
change to a current or voltage signal.
d
Capacity C
A
Capacity C
Linear Variable Differential Transformer
(LVDT) sensor
Using the LVDT sensor to produce a bipolar D.C
voltage that varies with core displacement
Speed Sensors
1. Tachometer as a speed sensor :
• The tachometer is a permanent magnet D.C
generator,
when
driven
mechanically;
it
generates an output voltage proportional to
shaft speed.
Since :
EαΦ.N
For permanent magnet Tacho : Φ= constant
Hence :
EαN
• Therefore,
the tachometer will generate an
output voltage proportional to shaft speed.
• The
other
main
requirements
for
a
tachometer are :
1. The output voltage should be smooth over the
operating range.
2. The
output
should
temperature variations.
be
stabilized
against
2. Optical Encoder sensor :
• Absolute encoders.
• Incremental encoders.
Incremental
Encoders
Absolute encoder s
Pressure Sensors
1. Bellow type sensor :
The metallic bellows pressure sensors are
used when it is needed to sensing low
pressures and providing power for activating
recording and indicating mechanisms. Such
sensors are most accurate when measuring
pressures from 0.5 to 75 psi.
2. Bourdon Tube Pressure sensor :
•
The bourdon tube pressure instrument is
one
of
the
oldest
pressure
sensing
instruments in use today.
• The
tube
bourdon tube consists of a thin-walled
that
is
flattened
diametrically
on
opposite sides to produce a cross-sectional
area elliptical in shape, having two long flat
sides and two short round sides.
• The tube is bent lengthwise into an arc of a
circle from 270 to 300 degrees.
Force Sensors (Strain Gauge)
• One of
the most important force sensors or
transducers is the strain gauge.
• The
simple
strain
gauge
is
used
for
measuring the external force or pressure
applied to a fine wire.
• The fine wire is usually arranged in the form
of a grid or a folded wire.
Fluid Sensors
Liquid level measuring devices are classified
into:
a) Direct method.
• An
example of
b) Inferred method.
the direct method is the
dipstick in the car which measures the height
of the oil in the oil pan.
• On the other hand, an example of the inferred
method is a pressure gauge at the bottom of a
tank which measures the hydrostatic head
pressure from the height of the liquid.
The level sensors can be classified as follows :
1. Mechanical sensors :
• Float methods
• Buoyancy method
• Vibrating level systems
2. Hydrostatic pressure methods :
• Differential pressure level detectors
• Bubbler systems
3.Electrical methods :
• Conductivity probes
• Capacitance probes
• Optical level switches
• Ultrasonic level detectors
• Microwave level systems
• Nuclear level systems
The float level sensor
The buoyancy level sensor
π r2 (Δ h - Δ L ) ρ g = k . Δ L
Ultrasonic level measurement :
The measuring equipment consists of the
following elements:
•A
transmitter : which periodically sends an
ultrasonic pulse to the surface of the liquid
• A receiver : which receives and amplifies the
returning pulse.
•A
time interval counter : which measures
the time elapsing between the transmission
of
a
pulse
and
reception
corresponding pulse echo.
of
the
The travelling distance can be calculated as :
L=c.t/2
And consequently, the head can be as :
h = Lmax – L = Lmax – c . t / 2
where :
c = sonar pulse velocity (m/sec).
t = time in sec.
L = travelling distance (m).
a) Solid or liquid above
surface measurement
b) Liquid material below
surface measurement
Sensor Placement
A number of factors must be considered before
a specific means of measuring the process
variable (PV) can be selected for a particular
loop :
• The
normal range over which the PV may
vary, and if there are any extremes to this
range.
• The
accuracy,
precision
and
sensitivity
required for the measurement.
• The required dynamics of the sensor.
• The required reliability.
• The costs involved, including
installation
and operating costs as well as purchase
costs.
• The installation requirements and problems
such as :
 Size and shape restraints.
 Remote transmission.
 Corrosive fluids.
 Explosive mixtures, etc ... … …
• The rules to be applied in sensors placement
are :
 As
the importance of the data being
watched increases, the importance of
the sensor increases where the most
important data should be monitored first
and then the less important comes later
and so on.
 As
the importance of the data being
watched increases, the sensitivity of the
sensor increases which means that the
sensor near the important data needs to
be the most sensitive whilst the other
outside becomes less sensitive and so
on.
 Tuning the sensors is an ongoing process
where
each
sensor
will
need
to
be
adjusted to selectively watch and ignore
network traffic and should be different
because of the data it is protecting and
what kind of traffic it is watching.
Chapter 4
Industrial
Controllers
Chapter (4) Outlines
1. On/Off Controller .
2. P , I , D , PI , PD and PID
controllers.
3. Temperature Control .
4. Pressure Control .
5. Flow Rate Control .
6. Level Control .
Control Objectives
The main objectives of the control system are :
1. Stability : the controlled variables do not grow
without limits.
2. Accuracy : the controlled variables reach the
desired values with minimal error.
3. Speed of response : the controlled variables
reach the desired values within a suitable time.
4. Cost : the cost of the control process should
not be high.
Control Loops
There are two main types of control
loops :
1. Open loop control systems
2. Closed loop control systems
Open Loop Control Systems
The general form of such system is :
inpu
t
controller
Control
signal
Final
control
element
Controlled
Plant
outpu
t
Properties :
 The control signal comes from a separate
system and do not affected by the
controlled variable.
 The controller is designed according to the
system history.
 The controller input is the system reference
input.
 All the timer based systems are open loop
control systems such as :
 A/C machines without thermostat
 Automatic washing machines
 Toasters.
 Such systems are simple and cheep.
Closed Loop Control Systems
The general form of such system is :
Control
signal
p(t)
e(t)
inpu
r(t)
t
+
_
controller
Feedback
signal
Final
control
element
b(t)
Sensor
Controlled
Plant
outpu
t
c(t
)
Types of controllers
1. Two Position Controller
2. Proportional Controller
3. Integral Controller
4. Differential Controller
5. PI – Controller
6. PD – Controller
7. PID-Controller
ON / Off or Two Position Control
• The
oldest strategy for control is to use a
switch giving simple ON / Off control which is
a discontinuous form of control action and
also known as two position control.
• The
technique
is
primitive,
cheap
and
effective method of control if a fairly large
fluctuation
acceptable.
of
the
process
variable
is
•
A perfect ON / Off controller is ON when the
measurement is below the setpoint and the
manipulated variable is at its maximum value,
however, above the setpoint, the controller is Off
and the MV is a minimum.
•
ON/Off control is widely used in both industrial
and domestic applications and most people are
familiar with the technique as it is commonly
used in home heating systems and domestic
water heaters.
•
There is usually a dead zone due to mechanical
delays in the process, this is often introduced to
reduce the frequency of operation and wear on
the components.
● The time equation is :
p( t) =
P;
t ≥ 0
0
t < 0
● Characteristic
curve
p(t)
e(t)
● The controller block diagram :
e(t)
P(t)
● Although it is simple with low cost, it is
subjected to chattering which might
destroy the devices. It can be used with
feedback systems.
p(t)
e(t)
-E
+E
et)
P(t)
Proportional Control
•
This principal of control is employed where the
automatic controller needs to correct the
controller
output
(CO)
with
an
action
proportional to ERR.
•
The correction starts from a CO value at the
beginning of the automatic control action.
•
Although this indicates that the setpoint (SP)
can be time variable, in most process control it
is kept constant for long periods of time.
•
For a proportional controller
proportional to the error signal.
the
output
is
● It is commonly used in industries with a
time equation of :
p(t) = Kp e(t)
● Characteristic
curve
where
p(t)
Kp is the slope
of the line.
e(t)
● The block diagram of the proportional
controller is :
e(t)
Kp
P(t)
● The T.F of the controller is
P ( s )  K p . E(s)
P(s)
Gc ( s ) 
 Kp
E(s)
Kp = - R2/R1
● Its advantages are : simplicity and fast
●
response
Drawback is : it does not eliminate ess
Integral Action
•
Integral control describes a controller in which
the output rate of change is dependent on the
magnitude of the input.
•
Specifically, a smaller amplitude input causes a
slower rate of change of the output.
•
This controller is called an integral controller
because
it
approximates
the
mathematical
function of integration.
•
The integral control method is also known as
reset control.
● The integral controller is commonly used
in industries to eliminate the steady
state error.
● Time equation of the integral controller
is :

p( t )  K I
e(t)
dt

0
where KI is the integral constant.
● The controller block diagram is :
e(t)
KI
1/s
P(t)
● The controller T.F is :
E(s)
P( s )  K I .
s
P(s)
KI
Gc ( s ) 

E(s)
s
●K = - 1/RC
I
The electronic circuit is
C
● :
e(t)
p(t)
● The important advantage of the integral
controller is that it eliminates the steady
state error, on the other hand it is slow and
chattering.
•
The major advantage of integral controllers is
that they have the unique ability to return the
controlled variable back to the exact setpoint
following a disturbance (i.e, eliminating
•
ess).
Disadvantages of the integral control mode are
that it responds relatively slowly to an error
signal and that it can initially allow a large
deviation at the instant the error is produced
which might lead to system instability and cyclic
operation.
•
For this reason, the integral control mode is not
normally used alone, but is combined with
another control mode.
Derivative Controller
•
The only purpose of derivative control is to add
stability to a closed loop control system.
•
The magnitude of derivative control D-control is
proportional to the rate of change or speed of
the PV.
•
Since the rate of change of noise can be large,
using D-control as a means of enhancing the
stability of a control loop is done at the expense
of amplifying noise.
•
It is always used in combination with P-control
or PI-control which result in a PD-control or PIDcontrol.
● It can not be used individually
in
industries, but it is usually used with
other controllers.
● Time equation is :
d e(t)
p( t )  K D
dt
where :
KD is the differentiation constant.
● The controller block diagram is :
e(t)
KD
s
P(t)
● The controller T.F is :
P ( s )  K D . s . E(s)
P(s)
Gc ( s ) 
 KD .s
E(s)
●
KD = - RC
The electronic circuit is :
● The important advantage of the derivative controller
is that it has a fast response, on the other hand it
does not eliminate the steady state error in addition
to amplifying the disturbances.
Derivative action in practice :
•
In practice the output will be changed to +8
times the value of the change of the ERR value
and the output will decrease at a rate of 63.2%
in every derivative time unit, as displayed below:
Proportional-Integral controller
● It is commonly used in industries to
eliminate the steady state error.
● Time equation is :

p( t )  K p .e( t )  K I  e(t) dt
0
● Where Kp is the proportional constant
and KI is the integral constant.
● The PI-controller block diagram is :
e(t)
p(t)
● The PI-controller transfer function
is :
E(s)
P ( s )  K p .E(s)  K I .
s
P(s)
KI
Gc ( s ) 
 Kp 
E(s)
s
e(t
)
p(t)
Proportional-Derivative controller
● It is commonly used in industries .
● Time equation is :
d e(t)
p( t )  K p .e( t )  K D
dt
● Where Kp is the proportional constant
and KD is the derivative constant.
● The PD-controller block diagram :
e(t)
p(t)
● The PD-controller T.F is :
P ( s )  K p .E(s)  K D . s E(s)
P(s)
Gc ( s ) 
 K p  KD .s
E(s)
PID controller
● It is commonly used in industries and can be
considered as the most powerful controller.
● Time equation is :
d e(t)
p( t )  K p e( t )  K I  e( t )dt  K D
dt
● Where Kp is the proportional constant, KI is
the
integral
constant
derivative constant.
and
KD
is
the
● The PID controller block diagram is :
Kp
p(t)
e(t)
KI
1/S
KD
S
+
● The
PID controller
function is :
transfer
E(s)
P(s) = K pE(s) + K I
+ K D sE(s)
s
KI
P(s)
Gc ( s ) =
= Kp +
+ KD s
E(s)
s
● The PID electronic circuit is :
Ideal PID Controller
O / P = K c * (Tder
1
+
+ 1)
Tint
O/P = Gain * (D-control + I-control + P-control
Real PID Controller
Tder + 1
1 + Tint
O / P = Kc *
+
α Tder + 1
Tint
O/P = Gain * Lead * PI-control
Control Algorithms
•A
control
algorithm
is
a
mathematical
expression of a control function to be used
with a computerized controlled process or
among a modern discrete controlled process.
• Control
the
algorithms can be used to calculate
requirements
of
much
more
complex
control loops.
• In such more complex control loops, questions
such as :
 How
far should the valve be opened or
closed in response to a given change in
setpoint?
 How
long should the valve be held in the
new position after the process variable
moves back toward setpoint?
All
and
such
questions
need
to
be
answered.
• The
following chart indicates the modern
control strategies and algorithms to be used in
an industrial process control.
Chapter 5
Introduction to
Process Automation
Chapter (5) Outlines
1. Introduction.
2. PLC Operation Scan.
3. PLC Addressing.
4. Relay Ladder Logics (RLL).
Introduction
• the
purpose
of
automation
has
shifted
from increasing productivity and reducing
costs, to broader issues, such as increasing
quality and flexibility in the manufacturing
process.
• Automation is now often applied to increase
quality in the manufacturing process, where
it can increase quality substantially.
• For
example, automobile and truck pistons
used to be installed into engines manually.
This
is
rapidly
being
transitioned
to automated machine installation, because
the error rate for manual installment was
around 1-1.5%, but has been reduced to
0.00001% with automation.
• Hazardous
operations such as oil refining,
manufacturing of industrial chemicals, and
all forms of metal working were always
early contenders for automation.
Control System Task
• The main task of
a control system is to
control a sequence of events or maintain
some variable constant or follow some
prescribed change.
• The
inputs to such control systems
might come from switches or sensors,
however the outputs of the controller
might go to run a motor in order to move
an object, or to turn a valve, or perhaps
some heater on or off.
• In
the traditional form of control systems,
the governing rules and the control actions
depend on the wiring of the control circuit.
• When changing the rules used for giving the
control
actions,
the
wiring
has
to
be
changed too. This leads to expensive cost
of replacing the controllers.
• Instead
of hardwiring each control circuit
for each control rule or action, the basic
system for all situations can be used with a
microprocessor based controller.
• So, by changing the program instructions, the
same control circuit may be used with a wide
variety of control rules or actions, which
saves the cost.
• This
was the main idea behind inventing the
programmable logic controllers (PLC).
• The
PLC was invented in response to the
needs
of
the
automotive
manufacturing
industry where software revision replaced the
re-wiring of hard-wired control panels when
production models changed.
The basic internal construction of a PLC is
includes the main components of a PLC as :
1. Rack or mounting part.
2. Processor or central processing unit
(CPU).
3. Input assembly.
4. Output assembly.
5. Power supply.
6. Programming unit.
A.C main supply
Power Supply
Module
Inputs
Interfacing
and
Multiplexing
D.C main supply
to other modules
Central
Processing
Unit
Memory devices
ROM – RAM – EEPROM
Communication link to personal
computer or programmer
Outputs
Interfacing
and
Multiplexing
Empty Rack
Mounting a module
Rack with modules
Large scale modular type PLC system
Large scale modular type PLC system
Bricks or shoebox PLCs
Input/Output Unit
• The input/output unit provides the interface
between the PLC system and the outside
world allowing the connections to be made
through
input/output
channels
to
input
devices such as sensors or output devices
such as motors and solenoids.
• The
input/output
signal conditioning.
provides
isolation
and
Typical input devices used with PLCs
include :
1. Mechanical switches for position
detection.
2. Proximity switches.
3. Photoelectric switches.
4. Encoders.
5. Temperature & pressure switches.
6. Potentiometers.
7. Linear variable differential Tr.
8. Strain gauges.
9. Thermistors.
10. Thermotransistors.
11. Thermocouples.
On the other hand, typical output
devices used with PLCs include :
1. Relays.
2. Contactors.
3. Solenoid valves.
4. Motors.
PLC Operation Scan
Scan all inputs
Repeat
sequence
Running the
program
Updating all
outputs
PLC Addressing
1. Mitsubishi PLC :
Inputs
: X400 , X401 , X402 , … … etc
Outputs : Y430 , Y431 , Y432 , … … etc
2. Toshiba PLC :
Inputs
: X000 , X001 , X002 , … … etc
Outputs : Y000 , Y001 , Y002 , … … etc
3. Allen Bradley :
I = input
O=
output
Module
number
x : xxx / xx
Rack
number
Terminal
number
4. Siemens SISMATIC S5 :
I = input
Q=
output
X
xx . x
Byte
number
Bit
number
Programming Rules
• Programs
for microprocessor-based
controllers usually being loaded in
machine code as binary numbers and
representing the instructions.
• Assembly
language can be used in
the form of mnemonics to indicate
the operations, e.g : LD , OUT , OR , …
… etc.
PLC Programming Methods
1. IL
(Instruction List Programming) :
This
is
effectively
mnemonic
programming.
2. ST (Structured Text) - A BASIC like
programming language.
3. LD
(Ladder Diagram) - Relay logic
diagram based programming.
4. FBD
(Function Block Diagram) - A
graphical
method
5. SFC
dataflow
programming
(Sequential Function Charts) -
A graphical method for structuring
programs
Relay Ladder Logics (RLL)
• Ladder
logic is a drawing of
electrical logic schematics which
results from the usage of relays.
• It is now a graphical language very
popular for programming PLCs,
where sequential control of a
process or manufacturing operation
is simulated.
L2
L1
2
1
M
Holding switch
Motor stop – start circuit
Ladder Programming Symbols
Several symbols are used to enter a ladder
program either using a the keypad of a
programming device with symbols or using a PC
software. The following are samples of such
symbols :
input
output
Start of a
junction
Horizontal circuit link
End of a
junction
Instruction Lists
• The instruction list programming for a
PLC differs according to the type of
the used PLC.
• The
following
different
types
table
of
shows
PLCs
and
the
the
corresponding instructions to be used
with them.
PLCs Types
Command
Start a rung with
a NOC
Start a rung with
a NCC
Series element
with a NOC
Series element
with a NCC
Parallel element
with a NOC
Parallel element
with a NCC
An Output
IEC11 Mitsub Omro Sieme Telemec Sphere+
31-3
ishi
n
ns
anique
Schuh
LD
LD
LD
A
L
STR
LDN
LDI
LD
NOT
AN
LN
STR
NOT
AND
AND
AND
A
A
AND
ANDN
ANI
AND
NOT
AN
AN
AND
NOT
O
OR
OR
O
O
OR
ORN
ORI
OR
NOT
ON
ON
OR NOT
ST
OUT
OUT
=
=
OUT
Normally Open Contact (NOC)
This can be used to represent any input to
the control logic such as : a switch or
sensor, a contact from an output, or an
internal
output.
When
solved,
the
referenced input is examined for an ON
(logical 1) condition :
• If
it is ON, the contact will close
and allow power (logic) to flow
from left to right.
• If the status is OFF (logical 0), the
contact is Open, power (logic)
will NOT flow from left to right.
Normally Closed Contact (NCC)
When solved, the referenced input is
examined for an OFF condition :
• If the status is OFF (logical 0) power
(logic) will flow from left to right.
• If the status is ON, power will not
flow.
LD coils & their types
Also, the coils (output of a rung) may
represent
a
physical
output
which
operates some device connected to the
programmable controller such as solenoid
valves, lights, motor starters and servo
motors, or may represent an internal
storage bit for use elsewhere in the
program.
Normally Open Coil
This can be used to represent any discrete
output from the control logic. When solved :
• If
the logic to the left of the coil is
TRUE, the referenced output is ON
(logical 1).
• If
the logic to the left of the coil is
FALSE, the referenced output is OFF
(logical 0).
To identify an input or an output in a
program, a numbering system is used. This
numbering system has three purposes :
• To tell contacts apart in the program.
• Serves as an address for the location
of the input module in the real world.
• Serves
as a memory address for the
contact in the processor memory.
Solving a Single Rung
Suppose a switch is wired to Input1, and a light
bulb is wired through Output1 in such a way that
the light is OFF when Output1 is OFF, and ON when
Output1 is ON.
• When
Input1 is OFF (logical 0) the contact
remains open and power cannot flow from left to
right. Therefore, Output1 remains OFF (logical 0).
• When
Input1 is ON (logical 1) then the contact
closes, power flows from left to right, and
Output1 becomes ON (the light turns ON).
Input1
Output1
controller
Load
Examples
The AND rung
The AND is a logic condition where an
output is not energized unless two NOC are
closed.
x400
x401
Y43
0
Step
0
1
2
Instruction
LD
X400
AND
X401
OUT
Y430
Step
0
1
2
Instruction
A
I0.1
A
I0.2
=
Q2.0
Mitsubishi PLC
I0.1
I0.2
Siemens PLC
Q2.0
The OR rung
The OR is a logic condition where an
output is energized when one or both of
two NOC are closed.
x400
x401
I0.1
I0.2
Y43
0
Step
0
1
2
Instruction
LD
X400
OR
X401
OUT
Y430
Step
0
1
2
Instruction
A
I0.1
O
I0.2
=
Q2.0
Mitsubishi PLC
Q2.0
Siemens PLC
The NOT rung
The NOT is a logic condition where an
output is de-energized when a NCC is
opened.
Input1
Output1
controller
Load
Simple Timers
A
timer is simply a control
built-in block
that takes an input and changes an output
based on time.
 Timers
count
fractions
of
seconds
or
seconds using internal CPU clock.
 Timers act like relays with coils which when
energized
result
in
closing
or
opening
contacts after some specified time interval.
In
PLC programming a timer is simply
being treated as an output for a rung
while
its
control
is
represented
contacts in somewhere else.
There are three basic timer types :
i.
On-Delay timer TON or T-O
ii.
Off-Delay timer
iii.
Pulse timers
TOF or O-T
TP
by
X400
T450 K5
T450
Y430
Step
0
1
2
3
4
5
Instruction
LD
X400
OUT
T450
K
5
LD
T450
OUT
Y430
END
Step
0
1
2
3
4
5
Instruction
A
I0.0
LKT
5.2
SR
T0
A
T0
=
Q2.0
END
T0
I0.0
T
0
Q2.0
KT5.2
Simple Counter
 A counter is simply a control built-in block that
takes counts the occurrence of an input signal.
 This might happen in a conveyor system, when
counting persons passing through a door,
counting cars in a parking lot or counting the
revolutions of a shaft.
 There are two basic types of counters - Up
counter and a Down counter :
• Up
Counter : as its name implies, whenever a
triggering event occurs, an up counter increments
the counter.
• Down Counter : whenever
a triggering event
occurs, a down counter decrements the
counter.
Counter Programming
The counter is used to count the events of
occurrence of an input signal and then
operates its contacts as follows :
Counter
Input
counter
CTD
Output
CV
RST
Counter
Chapter 6
Applications to
Process Automation
Chapter (6) Outlines
a. Signal Lamp Process.
b. Machine Safety Process.
c. Central Heating Process.
d. Automatic Mixing Process.
e. Automatic Packing Process.
Programming Examples
1. A signal lamp is required to be on if :
A pump is running.
And
 The pressure is satisfactory.
Or
 The test lamp is closed.
Pump Presu.
X400 X401
Test
X402
Lamp
Y430
Step
0
1
2
3
4
5
Instruction
LD
X400
AND
X401
LD
X402
ORB
OUT
Y430
END
2. A machine has 4 sensors to detect the
safety and is required to be off if :
 Any of the sensors gives input.
 when the machine is stop, an alarm is sound.
X401
X400
X403
X402
Mamchine
Y430
X400
X401
Alarm
Y431
X402
X403
Step
0
1
2
3
4
5
6
7
8
9
10
Instruction
LDI
X400
ANI
X401
ANI
X402
ANI
X403
OUT
Y430
LD
X400
OR
X401
OR
X402
OR
X403
OUT
Y431
END
Example (Central Heating)
Consider a central heating system with the
following features :
 The
boiler is
thermostatically
controlled
and
supplies the radiator system in addition to a hot
water tank.
 Pumps
are used to supply hot water to either or
both the radiator and the tank according to the
desired sensors.
 The
whole system is controlled by a clock to
operate a certain time a day.
Motorized
Pump
Radiator
System
M1
Hot water
tank
Boiler
M2
Motorized
Pump
Boiler Temp.
sensor
Hot water tank
Temp. sensor
Room
Timers
The power circuit for the central heating
system is :
Stop
Power
Run
Room sensor
Tank sensor
Outputs
Boiler sensor
Inputs
Clock
Boiler
M1
M2
 The ladder & IL program for the central heating
system using Mitsubishi PLC is :
Inputs :
X400
Clock
X401
Boiler sensor
X402
Room sensor
X403
Tank sensor
Outputs :
Y430
Boiler
Y431
Pump M1
Y432
Pump M2
Step
X402
X400
X401
Y430
X403
Y430
Y430
X402
Y431
X403
Y432
END
Instruction
0
LD
X402
1
OR
X403
2
AND
X400
3
AND
X401
4
OUT
Y430
5
LD
Y430
6
AND
X402
7
OUT
Y431
8
LD
Y430
9
AND
X403
10
OUT
Y432
11
END
 The ladder & IL program for the central heating
system using Siemens PLC is :
Inputs :
I0.0
Clock
I0.1
Boiler sensor
I0.2
Room sensor
I0.3
Tank sensor
Outputs :
Q2.0
Boiler
Q2.1
Pump M1
Q2.2
Pump M2
Step
I0.2
I0.0
I0.1
Q2.0
I0.3
Q2.0
Q2.0
I0.2
Q2.1
I0.3
Q2.2
END
Instruction
0
A
I0.2
1
O
I0.3
2
A
I0.0
3
A
I0.1
4
=
Q2.0
5
A
Q2.0
6
A
I0.2
7
=
Q2.1
8
A
Q2.0
9
A
I0.3
10
=
Q2.2
11
END
Example (Mixing Process)
 Automatic mixing processes of
liquids
and
other
compounds in the chemical and food industries are very
common.
 The mixing station goal is to mix two liquids for a
specified time and then output the final product to a
storage tank.
 The system consists of :
1. Two level sensors to monitor the flowing of the
liquids into the tank.
2. Three solenoid valves to control the flow of liquids.
3. A motor connected to an agitator to mix the liquids
into the tank.
Input Valves
VA2
VA1
MS1
Motor & Agitator
Mixing Tank
LS1
Level Sensors
LS2
Mixing
Station
VA3
Output to Storage Tank
The sequence of events for this automatic mixing process will
be as follows :
1. Open valve 1 until level 1 is reached for the first liquid .
2. Then close valve 1 .
3. Open valve 2 until level 2 is reached for the second
liquid .
4. Then close valve 2 .
5. Start the motor and agitate to mix the liquids into the
tank for a specified time .
6. Then stop the motor .
7. Open valve 3 up to a specified time to empty the mixed
product to a storage tank .
8. Then close valve 3 .
9. Repeat or end the mixing process as required .
The
automatic
mixing
station
will
components :
1. Inputs to the PLC :
 Start push button
X400
 Stop push button
X401
 Level sensor LS1
X402
 Level sensor LS2
X403
1. Outputs from the PLC :
 Valve # 1 (VA1)
Y430
 Valve # 2 (VA2)
Y431
 Motor starter (MS1)
Y432
 Valve # 3 (VA3)
Y433
require
the
following
X400
X401
M 100
M 100
M 100
X402
X403
Y432
Y433
Y430
M 100
X402
X403
Y432
Y433
Y431
M100
X403
T450 K1200
M100
Y432
T450
Y432
M100
Y433
M100
T450
M100
T451
T451 K180
T450
END
Y433
Step
Instruction
Step
Instruction
Step
Instruction
0
LD
X400
12
ANI
X403
24
ORB
1
OR
M100
13
ANI
Y432
25
ANI
T450
2
ANI
X401
14
ANI
Y433
26
OUT
Y432
3
OUT
M100
15
OUT
Y431
27
LD
M100
4
LD
M100
16
LD
M100
28
AND
T450
5
ANI
X402
17
AND
X403
29
OUT
T451
6
ANI
X403
18
OUT
T450
30
K
180
7
ANI
Y432
19
K
1200
31
LD
M100
8
ANI
Y433
20
LD
M100
32
ANI
T451
9
OUT
Y430
21
AND
Y432
33
AND
T450
10
LD
M100
22
LD
M100
34
OUT
Y433
11
AND
X402
23
AND
Y433
35
END
I0.0
I0.1
F0.1
F0.1
F0.1
I0.2
I0.3
Q2.2
Q2.3
Q2.0
F0.1
I0.2
I0.3
Q2.2
Q2.3
Q2.1
I0.3
F0.1
T0
KT1200
Q2.2
F0.1
Q2.4
Q2.4
Q2.2
F0.1
Q2.3
F0.1
Q2.4
T1
Q2.5
KT180
F0.1
Q2.5
Q2.4
END
Q2.3
Step Instruction
Step
Instruction
Step Instruction
0
A
I0.0
12
AN
I0.3
24
A
1
O
F0.1
13
AN
Q2.2
25
2
AN
I0.1
14
AN
Q2.3
3
=
F0.1
15
=
4
A
F0.1
16
5
AN
I0.2
6
AN
7
Step Instruction
36
A
T1
)
37
=
Q2.5
26
O(
38
A
F0.1
Q2.1
27
A
F0.1
39
AN
Q2.5
A
F0.1
28
A
Q2.3
40
A
Q2.4
17
A
I0.3
29
)
41
=
Q2.3
I0.3
18
LKT
1200
30
AN
Q2.4
46
END
AN
Q2.2
19
SR
T0
31
=
Q2.2
8
AN
Q2.3
20
A
T0
32
A
F0.1
9
=
Q2.0
21
=
Q2.4
33
A
Q2.4
10
A
F0.1
22
A(
34
LKT
180
11
A
I0.2
23
A
35
SR
T1
F0.1
Q2.2
Packing Process
Consider the following packing machine,
where it is required to pack 6 objects in a
box and then pack 12 objects in another
box in another path as shown :
12 in box
6 in box
X400
RESET
C461
C460
K6
X401
Out
C460
Y430
X400
RESET
C461
X401
C461
K 12
C460
Out
C461
Y431
Step
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Instruction
LD
X400
OR
C461
RST
C460
K
6
LD
X401
OUT
C460
LD
C460
OUT
Y430
LD
X400
OR
C461
RST
C461
K
12
LD
X401
AND
C460
OUT
C461
LD
C461
OUT
Y431
END
Mitsubishi PLC program
C0
I0.0
CU
C1
6
CV
Q2.0
I0.1
R
C1
I0.0
CU
C1
12
I0.1
CV
C0
Q2.1
R
Step
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Instruction
A
I0.0
O
C1
CU
C0
LCK
6
A
I0.1
R
C0
=
Q2.0
A
I0.0
O
C1
CU
C1
LCK
12
A
I0.1
R
C1
A
C0
=
Q2.1
END
Siemens PLC program