‫وزارة التعليم العالي والبحث العلمي‬
Ministry of Higher Education and
Scientific Research
Al-Furat Al-Awsat Technical
University
Al-Mussaib Technical College
Mechanical Power Engineering
Department
‫جامعة الفرات االوسط التقنية‬
‫ المسيب‬/ ‫الكلية التقنية‬
‫قسم هندسة تقنيات ميكانيك القدرة‬
Measurement and Control Processes
Fourth Stage Students
By:
Dr. Fadhil Alrubaiy
E-mail: fadhil.alrubaiy@atu.edu.iq
2023-2024
1
Chapter One
Introduction to Control Systems
Process control refers to the methods that are used to control process variables
when manufacturing a product. For example, factors such as the proportion of
one to another, the temperature of the materials, how well the ingredients are
mixed, and the pressure under which the materials are held can greatly affect
the quality of an end product.
Control System Objectives
 Economic Incentive
 Safety
 Equipment Protection
 Reduce variability
 Increase efficiency
 Ensure the stability of a process
 Elimination of routine
1.1 Definitions and Terminologies




System: it is a combination of components that act together and
perform a certain objective.
Process: is defined as the changing or refining of raw materials that
pass through or remain in a liquid, gaseous, or slurry state to create
end products.
Control: in process industries refers to the regulation of all aspects of
the process. Precise control of level, pH, temperature, pressure and flow
is important in many process applications.
Sensor: a measuring instrument, the most common measurements are
of flow (F), temperature (T), pressure (P), level (L), pH and
composition (A, for analyzer).The sensor will detect the value of the
measured variable as a function of time.
2
 Set point: The value at which the controlled parameter is to be
maintained.

Controller: A device which receives a measurement of the process
variable, compares with a set point representing the desired control
point, and adjusts its output to minimize the error between the
measurement and the set point.
 Error Signal: The signal resulting from the difference between the set
point reference signal and the process variable feedback signal in a
controller.
 Feedback Control: A type of control whereby the controller receives
a feedback signal representing the condition of the controlled process
variable, compares it to the set point, and adjusts the controller output
accordingly.
 Steady-State: The condition when all process properties are constant
with time, transient responses having died out.
 Transmitter: A device that converts a process measurement (pressure,
flow, level, temperature, etc.) into an electrical or pneumatic signal
suitable for use by an indicating or control system.

Controlled variable: process output which is to be maintained at a
desired value by adjustment of a process input.

Manipulated variable: process input which is adjusted to maintain
the controlled output at set point.

Disturbance: a process input (other than the manipulated parameter)
which affects the controlled parameter.

Process Time Constant (τ): Amount of time counted from the
moment the variable starts to respond that it takes the process variable
to reach 63.2% of its total change.
3


Block diagram: it is relationship between the input and the output of
the system. It is easier to visualize the control system in terms of a block
diagram.
Transfer Function: it is the ratio of the Laplace transform of output
(response function) to the Laplace transform of the input (driving force)
under assumption that all initial conditions are zero unless that given
another value. e.g. the transfer function of the above block diagram is
G (s) = Y(s)/X(s)
1.2 The Control Loop
Control loops in the process control industry work in the same way.
It requires three tasks to occur:
 Measurement
 Comparison
 Adjustment
1.2.1 Manual and Automatic Control
4
1.3 Closed and Open Control loops
Closed-loop control system there is a feedback control system which the
output signals has a direct effect upon the control action.
Advantage: more accurate than the open-loop control system.
Disadvantages: (1) Complex and expensive, (2) The stability is the major
problem in closed-loop control system.
5
Open-loop control system: it is a control system in which the output has no
effect upon the control action. (The output is neither measured nor fed back
for comparison with the input).
Advantages:
(1) Simple construction and ease of maintenance.
(2) Less expensive than closed-loop control system.
(3) There is no stability problem.
Disadvantages:
(1) Disturbance and change in calibration cause errors; and output may be
different from what is desired.
(2) To maintain the required quality in the output, recalibration is necessary
from time to time.
6
1.4 Laplace Transforms
In this part, we introduce a mathematical tool, the Laplace transform, which
can significantly reduce the effort required to solve and analyze linear
differential equation models.
1.4.1 The definition of the Laplace Transforms
The Laplace transform of a function f(t) is defined as :
………………. (1.1)
Where:
F(s) is the symbol for the Laplace transform,
(s) is a complex independent variable,
f(t) is some function of time to be transformed, and
ℒ is an operator of Laplace transform, defined by the integral.
Example 1.1:
7
1.4.2 Laplace Transform of Some Common Functions
Note that in the same way as in Ex.1.1, we can determine the Laplace
transform of some elementary functions:
Example 1.2:
8
1.5 The First Shifting Theorem:
9
1.6 The Second Shifting Theorem:
10
1.7 Laplace of Integral:
Example 1.7
1.8 Laplace Transform of the Derivatives
……………….1.10
……………….1.11
……………….1.12
…….1.13
11
Example 1.8: Find the Laplace Transform of the functionfollowing
differential equation and initial condition:
12
Problems:
Q1: Find the Laplace Transform of 𝑓(𝑡) = 𝑡𝑒 −2𝑡
Q2: Find the Laplace Transform of 𝑓(𝑡) = sin⁡(3𝑡)𝑒 5𝑡
Q3: Find the Laplace Transform of following equation:
1.9 Inverse of Laplace Transform
……………………. (1.14)
1.9.1 Laplace Inverse of Some Common Functions
Below table shows inverse Laplace transform of some common functions:
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Example 1.9: Find the Laplace inverse of the following functions:
Solution:
1.9.2 Laplace Inverse by partial fraction:
Sometimes the function whose inverse is required is not recognisable as a
standard type, such as those listed in Table 1.1. In such cases it may be
possible, by using partial fractions, to resolve the function into simpler
fractions which may be inverted on sight. For example:
14
For problems in control system analysis, the Laplace transforms, in general,
have the form:
f ( s) 
B( s)
A( s)
Where A(s) and B(s) are polynomials in s. In expansion F(s) = B(s)/A(s)
In the form of a partial fraction, it is important that the highest force of s in
A(s) is greater than the highest power of s in B(x).
Let a function in the s domain be given by:
f ( s) 
B( s )
B( s )

A( s) ( s  a)(s  b)(s  c)
Poles:
The values a, b, c,…that makes the denominator zero, and hence f (s) infinite,
are called the system poles of f (s). If there are no repeated factors, the poles
are simple poles. If there are repeated factors, the poles are multiple poles.
Zeros:
Values of s that make the numerator B(s) zero, and hence f (s) zero, are
called the system zeros of f (s).
For example:
s4
has simple poles at s=−1, and s=+2, and a zero
( s  1)( s  2)
at s=4.
There are three cases for the partial fraction depending on types of the poles,
as shown below:
Case (1): F(s) has simple poles:
15
Example 1.10: Find the Laplace inverse of the following function:
F ( s) 
s3
( s  1)( s  2)
Solution:
16
Case (2): F(s) has multiple repeated poles:
In this case the transfer function can be written in below form:
f ( s) 
f ( s) 
B( s )
B( s )

A( s) ( s  si )(s  si ) 2 ..............(s  si ) r
A1
A2
B( s )
Ar


 ........
2
A( s) ( s  si ) ( s  si ) .
( s  si ) r
Where (r) number of repeated poles.
The coefficient of multiple repeated poles ( A1, A2,……Ar) are described as
follows:
Ar  [(s  si )
r
F (s)]
s  si
Ar  1 
d
r
[(
s

s
)
i F (s)] s  s
ds
i
Ar  2 
1 d2
r
2 [(s  si ) F (s)]
21 ds
s  si

A1 
1
d r 1
r
2 [(s  si ) F (s)]
(r  1)1 r 1
s  si
 ds
Partial fracture expansion when F(s) involves multiple poles. Instead of
discussing the general case, we will use an example to show how to obtain
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the expansion of a fraction Partial of F(s). (See also Problem A-2-16.)
Consider the following statements:
Also, differentiation of both sides of Equation (1) with respect to s yields:
If we let s = -1 in equation (2) , then
By differentiation both sides of Equation (2) with respect to s, the result is:
18
From the preceding analysis it can be seen that the values of coefficients are
found systematically as follows:
19
Case (3): F(s) has complex-conjugated poles:
If the function F(s) involves a pair of complex-conjugate poles, it is
convenient not to expand F(s) into the usual partial fractions but expand to the
sum of the damped sine and damped cosine functions, with referring to below
Laplace inverse relationships:
Example 1.11: Find the Laplace inverse of the following function:
F ( s) 
2s  12
s  2s  5
2
Solution:
Notice that the denominator polynomial can be factored as shown:
s 2  2s  5  s 2  2s  1  4  (s  1  2 j )(s  1  2 j )
F ( s) 
2s  12
10  2s  2
10  2( s  1)
 2

s  2s  5 s  2s  1  4 ( s  1) 2  4
F ( s)  5
2
2
s 1
2
2
2
( s  1)  2
( s  1) 2  4
Then:
20
1.10 Initial Value Theorem (IVT)
The initial value of a function means the value of that function at time equal
to zero, f (0). The initial value theorem can provide the initial value from its
Laplace transforms without going through the inverse Laplace transform
procedure.
Example 1.12: Find the Find the initial value of the function using initial
value theorem:
f (t )  2(1  e 2t )
Answer:
1.11 Final Value Theorem (FVT)
The final value of a function means the value at time (∞) ,This value equal
steady-state value of a time-domain. Final value theorem is applicable if F(s)
satisfy the following conditions:
a. No poles on imaginary axis (except simple pole at origin).
b. No poles having positive real part (i.e., the value of s should not be a
positive)
21
1.12 Solved problems
Answer:
22
23
24
25
Chapter Two
Transfer Function and Response of first order systems
2.1 Transfer Function
The transfer function is an algebraic relation in the s-domain that represents
the dynamic of the system. It is the ratio of the output variable to the input
variable and denoted by T.F or G(s). Figure 2.1 shows a block diagram that
represents the transfer function G(s) that relates the output variable y(s) and
the input variable x(s).
The mathematical formula for the transfer function is:
2.1.1 Conditions of the Transfer Function T.F
Any system has a transfer function if;
a. It can be represented by a linear differential equation in the time- domain.
b. The initial values of the dependent variable and its derivatives are all zero.
26
Example 2.1:
27
28
2.2 Definitions and Terminologies
2.2.1 Open System
Open system (or Open-loop): Any system which is not controlled i.e., the
system has no controllers, so the output variables are subjected to changes
against any change in the inlet variables. In other words, in an open-loop
control system, the output is neither measured nor “fed back” for comparison
with the input. Let’s take the continuous stirred tank heating system (CSTH)
as an example as shown in Fig.2.4.
In the continuous stirred tank heater (CSTH), a liquid of heat capacity (c p,
kJ/kg oC) enters the vessel with flowrate (m, kg/s) and inlet temperature (𝑇𝑖,
o
C ). Amount of heat (Q, kW) inlet to the system by a steam coil. The outlet
temperature of the liquid is (𝑇𝑜 , oC). The quantity of liquid inside the tank
(holdup) is (M, kg). In the steady-state operation, all the inlet and outlet
variables values (i.e., Ti, m, Q, To) are constant with time.
Note that any change in the value of 𝑇𝑖 or m or Q will affect the value of To.
That is 𝑇𝑜= f( 𝑇𝑖 ,𝑄,𝑚). For example, if the inlet temperature is increased
suddenly by 5 oC and keeping all other variables (m and 𝑇𝑖) constants, this
will cause the outlet temperature (𝑇𝑜) to increase to a new steady-state value.
The outlet temperature will also change if any change occurs in the inlet
flowrate (m) or inlet rate of heat (Q). So, we can say that the outlet temperature
(𝑇𝑜) is a function of the inlet variables (𝑇𝑖, m, Q). Since the outlet temperature
28
(𝑇𝑜) will vary against any variation in the inlet variables (𝑇𝑖, m, Q), therefore;
this system is called an uncontrolled or open-loop system.
2.2.2 Closed System
In the continuous stirred tank heater (CSTH), if the system was built to control
the value of 𝑇𝑜 at a certain value by a control process which permits us to
maintain the outlet temperature 𝑇𝑜 constant at a certain value against any
change that occurs in any of the input variables (𝑇𝑖 .𝑄.𝑚). In this case the
system is called closed system. See Fig 2.5 which shows how the outlet
temperature is controlled.
29
Note that in the closed system, we want to control the outlet temperature To,
so this variable is called the controlled variable. The control valve is put on
the inlet steam line in order to control the quantity of heat input to the system,
so the variable (Q) is called manipulating variable. The other inlet variables
(Ti and m) are called load variables since they may cause a disturbance to the
system.
2.3 Terminologies
2.3.1 Disturbance
Disturbance means the change that occurs in the input variables (open system)
or in the load variables (closed system). This disturbance has different forms,
some are regular and others are irregular. Below are some common forms of
the disturbances which will be mentioned in detail in the subsequent section.
1. Step change
2. Ramp change
3. Pulse change
4. Impulse change
5. Sine change
6. Cosine change
7. Other irregular changes
2.3.2 Response
Response means the variation with time of the output variable (open-loop) or
the controlled variable (closed-loop). This variation occurs as a result of any
disturbance or change that occurs in the input or load variables.
2.4 Special Functions of Disturbance
30
31
32
33
2.6 Development of T.F for first order system:
2.6.1 Mercury Thermometer
It is a measuring device used to measure the temperature of a stream.
Consider a mercury in glass thermometer to be
located in a flowing stream of fluid for which the
temperature x varies with time.
The object is to calculate the time variation of the
thermometer reading y for a particular change of
x.
The following assumptions will be used in this analysis:1. All the resistance to heat transfer resides in the film surrounding the bulb
(i.e. the resistance offered by the glass and mercury is neglected).
2. All the thermal capacity is in the mercury. Furthermore, at any instant the
mercury assumes a uniform temperature throughout.
3. The glass wall containing the mercury does not expand or contract during
the transient response.
It is assumed that the thermometer is initially at steady state. This means
that, before time zero, there is no change in temperature with time. At time
zero the thermometer will be subjected to some change in the surrounding
temperature x(t).
(i.e at t<0 x(t)= y(t) =constant there is no change in temperature with time).
At t=0 there is a change in the surrounding temperature x(t).
Unsteady state energy balance:
This equation can be briefly written as:
In = Out+ Accumulation
34
35
Example (2.3):
A thermometer having a time constant of 0.1 min is at a steady state
temperature of 90 o F. At time t = 0, the thermometer is placed in a temperature
bath maintained at 100°F. Determine the time needed for the thermometer to
read 98 o F.
36
2.6.2 Liquid-Level Tank
Consider a system of the liquid-level tank with a uniform cross-sectional area
A (m2) shown in Figure below. The liquid enters the tank at a rate 𝑞𝑖 (L/min)
and exits at a rate 𝑞𝑜 (L/min). The level of the liquid inside the tank is h (m).
The flow through the valve in the exit stream is assumed to be linear so, the
outlet flow rate (𝑞𝑜) is related to the liquid level (h) inside the tank and the
resistance of the valve (R) through a linear equation:
𝑞 ∝ ℎ⁡
𝑞=
ℎ
𝑅
……..(1) for linear valve
Now we will make a mass balance to obtain the differential equation that
relates the liquid level inside the tank (h) to the inlet variables and time. The
control volume in the balance equation is the tank.
Mass Balance:
Let’s define the parameters related to the mass balance process.
37
ρ: Density of liquid (kg/L)
𝑑ℎ
𝑞𝑖 𝜌 = 𝑞𝑂 𝜌 + 𝐴𝜌 … … … … … … … … … . (1) dividing by (ρ), use equation
𝑑𝑡
(1), and, then re-arrange to get:
ℎ
𝑑ℎ
𝑞𝑖 − = 𝐴 … … … … … … … … … . (2)
𝑅
𝑑𝑡
At steady state
ℎ𝑠
𝑑ℎ𝑠
𝑞𝑖𝑠 − = 𝐴
… … … … … … … … … . (3)
𝑅
𝑑𝑡
ℎ−ℎ𝑠
𝑑(ℎ−ℎ𝑠 )
Subtracting Eq(3) from Eq. (2): (𝑞𝑖 − 𝑞𝑖𝑠 ) −
=𝐴
𝑅
Let: 𝑄 = 𝑞𝑖 − 𝑞𝑖𝑠 , 𝐻 = ℎ − ℎ𝑠
𝐴
𝑑𝐻
𝑑𝑡
+
𝐻
𝑅
= 𝑄 ⇒ ⁡𝐴𝑅
𝑑𝐻
𝑑𝑡
𝑑𝑡
+ 𝐻 = 𝑅𝑄⁡ , taken Laplace for both sides:
(𝜏𝑠 + 1)𝐻(𝑠) = 𝑅𝑄(𝑠)
Where: 𝜏 = 𝐴𝑅 is the time constant of tank, 𝑅 is the steady-state gain of the
tank.
𝐺(𝑠) =
𝐻(𝑠)
𝑄(𝑠)
=
𝑅
𝜏𝑠+1
… … … … . (4)⁡ , Equation (4) represents transfer function
for first order system.
𝐻(𝑠) =
𝑅
𝑄 … … … … . (5)
𝜏𝑠 + 1 (𝑠)
38
39
Example 2.4:
A tank having a time constant of 1 min and a resistance of 1/9 ft/cfm is
operating at steady state with an inlet flow of 10 ft3 /min (or cfm). At time t
=0, the flow is suddenly increased to 100 ft 3/min for 0.1 min by adding an
additional 9 ft3 of water to the tank uniformly over a period of 0.1 min. (See
Figure below for this input disturbance) . Plot the response in tank level and
compare with the impulse response.
Solution:
40
41
2.6.3 Mixing Tank
Consider the mixing process shown in side
Figure. In which a stream of solution containing
dissolved salt flows at a constant volumetric flow
rate q into a tank of constant holdup volume V.
The concentration of the salt in the entering
stream x (mass of salt/volume) varies with time.
It is desired to determine the transfer function
relating the outlet concentration y to the inlet
concentration x.
If we assume the density of the solution to be constant, the flow rate in must
equal the flowrate out, since the holdup volume is fixed.
F: Volumetric flowrate
x, y : Input and output salt concentrations (mass/vol).
Unsteady state material balance
Steady state
42
Example 2.5: Find the T.F for the system shown in figure
43
2.6.4. Stirred Tank Heating System
44
2.7 Response of 1st order systems in series:
Many physical systems can be represented by several first-order processes
connected in series as shown in figure:-
1-Non Interacting System
45
46
Example 2.6:
Two non-interacting tanks are connected in series as shown in Fig. 7-1 a. The
time constants are τ2 =1, τ1 =0.5; R2 =1, and R1 =1. Find the response of the
level in tank 2 if a unit-step change is made in the inlet flow rate to tank 1.
Solution:
The transfer function for this system is found
from Equation in the previous section.
47
2- Interacting System
48
49
50
51
52
Dead time Element (Time-delay element)
Dead time element is the element whose transfer function is represented by
𝐺(𝑠)=𝑒−𝜏𝐷𝑠
𝑤ℎ𝑒𝑟𝑒 𝜏𝐷 𝑖𝑠 𝑡ℎ𝑒 𝑡𝑖𝑚𝑒 𝑜𝑓 𝑑𝑒𝑙𝑎𝑦.
In the time-delay element, the output signal is the same as the input signal in
the form and magnitude but delayed by a time equal to 𝜏𝐷. In dynamic
processes, time-delay elements are mostly represented in some parts of
52
process plant such as pipes and transport belts. So, we will take these two
elements into consideration.
Any delay in measuring, in controller action, in actuator operation, in
computer computation, and the like, is called transportation delay or dead
time, and it always reduces the stability of a system and limits the achievable
time of the system.
The Transportation Lag
The transportation lag is the delay between the time an input signal is applied
to a system and the time the system reacts to that input signal. Transportation
lags are common in industrial applications. They are often called “dead time”.
53
54
Transport Pipes
Pipes that transport liquid or gas represent a time-delay element. In Figure below fluid
enters the pipe with mass flow rate m(kg/s) and velocity u(m/s) and exits at the same values.
The transfer
Example
Pipe transporting liquid with a mass flowrate 𝑚𝑖=3𝑘𝑔/𝑠 and velocity 𝑢=2m/s.
The length of the pipe is 10 m. If a step change of value 4 occurs in the inlet
flowrate, find the response of outlet flow rate 𝑚𝑜.
Solution
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