1. The document describes the components of a closed loop control system including the process, measuring element, comparator, controller, and control valve. Block diagrams and transfer functions are developed for each component. 2. Transfer functions relating the output Co(s) to the input Ci(s) and setpoint Csp(s) are derived for the example of a mixing process. 3. For a step change in input Ci of 2 units, the final output Co is calculated to be 1.515 units.

1. Process Control
Course II
Lecture
Closed Loop system

2. Development of Block Diagram for closed system
Consider the system shown below . We try to construct the block diagram for the CSTH and to find the
transfer function ( Regulator and Servo)
The main components of the loop are :
- Process
- Measurement
- Controller
- Control Valve
Now we will discuss each components in details
m
cp
Ti
Steam in
m
cp
To
M
Q
Tm
Tsp
E
Comparator
Process
Final control element
Temperature
measuring
element
Controller
Control Valve
P

3. 1. Process
Consider the heating tank which shown in Figure below. The tank represents the process in the closed loop system.
Now we will try to find the transfer function for the process which relate the output variable 𝑇𝑜 𝑠 with input variables
𝑇𝑖 𝑠 and the rate of input heat 𝑄 𝑠 . Here we will assume that the mass flowrate input to the system (m) is constant. So
𝑇𝑜 𝑠 = 𝑓 𝑇𝑖 𝑠 , 𝑄 𝑠 .
Now we try to find the transfer function of the process
which relate the output variable to the inlet variables.
m
Ti
cp
m
To
cp
M
Q
Process
Heat balance
In = out + Accumulation
𝑚 . 𝑐𝑝 . 𝑇𝑖 + 𝑄 = 𝑚. 𝑐𝑝. 𝑇𝑜 + 𝑀 𝑐𝑝
𝑑𝑇𝑜
𝑑𝑡

4. 𝑀
𝑚
𝑑𝑇𝑜
𝑑𝑡
+ 𝑇𝑜 = 𝑇𝑖 +
1
𝑚 . 𝑐𝑝
𝑄
𝜏
𝑑𝑇𝑜
𝑑𝑡
+ 𝑇𝑜 = 𝑇𝑖 + 𝑘 𝑄 𝜏 =
𝑀
𝑚
𝑘 =
1
𝑚 . 𝑐𝑝
𝜏𝑠 + 1 𝑇𝑜 𝑠 = 𝑇𝑖 𝑠 + 𝑘 𝑄 𝑠
𝑇𝑜 𝑠 =
1
𝜏𝑠 + 1
𝑇𝑖 +
𝑘
𝜏𝑠 + 1
𝑄 𝑠
∑
+
+
𝑇𝑖(s)
𝑇𝑜 (s)
1
𝜏𝑠 + 1
𝑘
𝜏𝑠 + 1
𝑄(s)

5. 2 Measuring Element
The temperature-measuring element, which senses the bath temperature 𝑇𝑜 and transmits a signal 𝑇𝑚 to the
comparator.
𝑇𝑜(s)
𝐺𝑚
𝑇𝑚(s)
In general the measuring element can be represented by one of the
following forms:
𝐺𝑚 𝑠 =
𝑇𝑚(𝑠)
𝑇𝑜(𝑠)
=
𝑘𝑚
𝜏𝑚𝑠 + 1
… … . . 1
Where 𝑘𝑚 is the steady – state gain of measuring element
𝜏𝑚 is time constant or lag of measuring lag.
Sometimes, the time constant of the measuring device 𝜏𝑚 is so small that could be neglected, and then the transfer
function becomes:
𝐺𝑚 𝑠 =
𝑇𝑚(𝑠)
𝑇𝑜(𝑠)
= 𝑘𝑚 … … . . 2

6. 3- Comparator
The action of the comparator is to compare between the two signals; the output signal from the measuring
element 𝑇𝑚 and the set point 𝑇𝑠𝑝 which represent the desired value for the output variable. The output of the
comparator represents the error signal 𝐸.
+ E(s)
𝟏
Tm(s)
-
Tsp(s)
𝐸(𝑠) = 𝑇𝑚(𝑠) − 𝑇𝑠𝑝(𝑠)

7. 4. Controller
Actually, there are many types of controllers depending on their actions.
(Proportional, Integral, Derivative, Proportional- derivative, Proportional-Integral and Proportional-Derivative –Integral)
The input to the controller is the error signal E(s)
The output of the controller is hydraulic, pneumatic or electric signal.
The controller will manipulate the error signal in such a way to reduce or eliminate the error signal.
In general, The output of the controller is hydraulic, pneumatic or electric signal that transfers to the final control
element. Here, we shall take the type of the controller as proportional and the output signal is pneumatic p(s), so the transfer
function of the controller 𝐺𝑐(𝑠) will be:
𝐺𝑐(𝑠) =
𝑃(𝑠)
𝐸(𝑠)
= 𝐾𝑐
𝐺𝑐
E (s) P (s)
𝐾𝑐
E (s) P (s)

8. Final control Element (Control valve)
Usually, the final control element may be
- control valve which operates pneumatically or
- solenoid valve which operates electrically.
The input variable to the valve is pressure signal
The output will affect the manipulating variable ( in our example is Q)
Mostly, the transfer function of the final control element can be represented by either
- first - order lag 𝐺𝑣 𝑠 =
𝑄 𝑠
𝑝 𝑠
=
𝐾𝑣
𝜏𝑣𝑠+1
or
- constant. 𝐺𝑣 𝑠 =
𝑄 𝑠
𝑝 𝑠
= 𝐾𝑣
Gv (s)
P (s) 𝑄 (s)
Control Valve Solenoid Valve

9. The Block diagram of the closed loop system can be given as in Figure below
𝑘𝑐
∑ ∑
+
+
+
_
𝑇i (s)
𝑇𝑜 (s)
𝑇sp (s)
1
𝜏𝑠 + 1
𝑘
𝜏𝑠 + 1
𝑘𝑣
𝜏𝑣𝑠 + 1
𝑘𝑚
𝜏𝑚𝑠 + 1
𝑇m (s)
𝐸(s) 𝑃(s) 𝑄(s)

10. Representation of closed loop system by symbols and signal transmitters
The elements of the closed loop system can be represented by symbols and signals transmitters.
Figure below shows the signal transmitter and symbols between measuring element and control valve
Variable X
Temperature T
Flow F
Pressure P
Level L
Concentration C
Measuring
element
Controller
Transmitter
Control valve
X T X C
set point

11. Measuring
element Temperature
Controller
Temperature
Transmitter
Control valve
T T T C
set point
Measuring
element Flow
Controller
Flow
Transmitter
Control valve
F T F C
set point

12. Measuring
element Level
Controller
Level
Transmitter
Control valve
L T L C
set point
Measuring
element concentration
Controller
concentration
Transmitter
Control valve
C T C C
set point

13. The type of Signal transmitter can be represented by lines as follows:
Signal type symbol
Electric signal
Pneumatic signal
Hydraulics signal
Measuring
element Flow
Controller
Flow
Transmitter
Control valve
F T F C
set point

14. m
cp
Ti
Steam
in
m
cp
To
M
Q
Tm
Tsp
E
Comparator
Process
Final control element
Temperature
measuring
element
Controlle
r
Control Valve
P
m
cp
Ti
Steam
in
m
cp
To
M
Q
Process
T T
Tc
Set point

15. Example 1 Consider the mixing system shown in Fig below
Given the following data:
𝑞𝑖 = 7 𝑙𝑖𝑡/𝑠𝑒𝑐 , 𝑉 = 14 𝑙𝑖𝑡 𝐺𝑣 = 1.5 , 𝐺𝑚 = 2, 𝐺𝑐 = 0.75
Find the followings:
a. The transfer functions:
𝐶𝑜 𝑠
𝐶𝑖 𝑠
𝑎𝑛𝑑
𝐶𝑜 𝑠
𝐶𝑠𝑝 𝑠
and sketch the
signal flow diagram
b. The final value of 𝐶𝑜 for step change in 𝐶𝑖 of value 2 units.
c. The final value of 𝐶𝑜 for step change in 𝐶𝑠𝑝 of value 3 units.
co
qi
Ci
q
V
d=2 cm
d=3 cm
Pure A
6 m
9 m
𝐶𝑜
′
CC
CT
𝐶𝑖
′
m (mole/s)
Set point

16. Solution
a. Process
𝑞𝑖𝐶𝑖
′
+ 𝑚 = 𝑞𝑖𝐶𝑜
′
+ 𝑉
𝑑𝐶𝑜
′
𝑑𝑡
𝑉
𝑞𝑖
𝑑𝐶𝑜
′
𝑑𝑡
+ 𝐶𝑜
′
= 𝐶𝑖
′
+
1
𝑞𝑖
𝑚
𝜏𝑠 + 1 𝐶𝑜
′ 𝑠 = 𝐶𝑖
′
𝑠 +
1
𝑞𝑖
𝑚 𝑠 … … . . (1)
But
𝐶𝑜 𝑠
𝐶𝑜
′ 𝑠
= 𝑒−𝜏𝐷2𝑠
𝐶𝑜
′ 𝑠 =
𝐶𝑜 𝑠
𝑒−𝜏𝐷2𝑠 … … … . . (2) co
qi
Ci
q
V
d=2 cm
d=3 cm
Pure A
6 m
9 m
𝐶𝑜
′
𝐶𝑖
′
m (mole/s)
𝐶𝑖
′
𝑠
𝐶𝑖 𝑠
= 𝑒−𝜏𝐷1𝑠
𝐶𝑖
′
𝑠 = 𝐶𝑖 𝑠 . 𝑒−𝜏𝐷1𝑠 … … . . (3)
Subs Eq. (3) and (2) in (1)

17. 𝜏𝑠 + 1 𝐶𝑜 𝑠 . 𝑒𝜏𝐷2𝑠 = 𝐶𝑖 𝑠 . 𝑒−𝜏𝐷1𝑠 +
1
𝑞𝑖
𝑚 𝑠
(𝜏𝑠 + 1)𝐶𝑜 𝑠 . = 𝐶𝑖 𝑠 . 𝑒−(𝜏𝐷1+𝜏𝐷2)𝑠
+
𝑒−𝜏𝐷2𝑠
𝑞𝑖
𝑚(𝑠)
𝐶𝑜 𝑠 =
𝑒−(𝜏𝐷1+𝜏𝐷2)𝑠
𝜏𝑠+1
𝐶𝑖 𝑠 +
𝑒−𝜏𝐷2𝑠
𝑞𝑖 𝜏𝑠+1
𝑚 𝑠
𝜏 =
𝑉
𝑞𝑖
=
14
7
= 2,
𝜏𝐷1 =
𝑉𝑖
𝑞𝑖
=
𝜋
4 𝑑𝑖
2
𝐿1
𝑞𝑖
=
𝜋
4 0.02 2 6
7
1000
= 0.27
𝜏𝐷2 =
𝑉
𝑜
𝑞𝑜
=
𝜋
4
𝑑2
2
𝐿2
𝑞𝑜
=
𝜋
4
0.03 2(9)
7/1000
= 0.9
𝐶𝑜 𝑠 =
𝑒−(0.9+0.27)𝑠
2𝑠+1
𝐶𝑖 𝑠 +
𝑒−0.9𝑠
7(2𝑠+1)
𝑚 𝑠
co
qi
Ci
q
V
d=2 cm
d=3 cm
Pure A
6 m
9 m
𝐶𝑜
′
CC
CT
𝐶𝑖
′
m (mole/s)
Set point
𝐶𝑜 𝑠 =
𝑒−1.17𝑠
2𝑠+1
𝐶𝑖 𝑠 +
𝑒−0.9𝑠
7(2𝑠+1)
𝑚 𝑠

18. 𝐺𝑣 = 1.5 , 𝐺𝑚 = 2, 𝐺𝑐 = 0.75
0.75
∑
∑
+
+
+
_
𝐶i (s)
𝐶𝑜 (s)
𝐶sp (s)
𝑒−1.17𝑠
2𝑠 + 1
𝑒−0.9𝑠
7(2𝑠 + 1)
1.5
2
𝐶m (s)
𝐸(s) 𝑃(s) 𝑚(s)
𝐶𝑜 𝑠 =
𝑒−1.17𝑠
2𝑠+1
𝐶𝑖 𝑠 +
𝑒−0.9𝑠
7(2𝑠+1)
𝑚 𝑠
a. Transfer function
𝐶𝑜 𝑠
𝐶𝑖 𝑠
=
𝑒−1.17𝑠
2𝑠+1
1+
0.75∗1.5∗𝑒−0.9𝑠∗2
7(2𝑠+1)
=
𝑒−1.17𝑠
2𝑠+1
1+
0.32𝑒−0.9𝑠
2𝑠+1
𝐶𝑜 𝑠
𝐶𝑠𝑝 𝑠
=
(0.75)(1.5)
𝑒−1.9𝑠
)
7(2𝑠 + 1
1 + 2(0.75)(1.5)
𝑒−0.9𝑠
)
7(2𝑠 + 1
=
0.16
𝑒−1.9𝑠
2𝑠 + 1
1 +
0.3214𝑒−1.9𝑠
2𝑠 + 1

19. b.
𝐶𝑜 𝑠 = 𝐶𝑖 𝑠
𝑒−1.17𝑠
2𝑠 + 1
1 +
0.32𝑒−0.9𝑠
2𝑠 + 1
=
2
𝑆
[
𝑒−1.17𝑠
2𝑠 + 1
1 +
0.32𝑒−0.9𝑠
2𝑠 + 1
]
𝐶𝑜 ∞ = lim
𝑠→0
𝑠 ∗ 𝐶𝑜(𝑠) = 1.515
𝐶𝑜 𝑠
𝐶𝑖 𝑠
=
𝑒−1.17𝑠
2𝑠 + 1
1 +
0.32𝑒−0.9𝑠
2𝑠 + 1
𝐶𝑖 𝑠 =
2
𝑠
𝑠𝑡𝑒𝑝 𝑐ℎ𝑎𝑛𝑔𝑒
𝐶𝑜 𝑠 = 𝐶𝑠𝑝 𝑠 ∗
0.16
𝑒−1.9𝑠
2𝑠 + 1
1 +
0.3214𝑒−1.9𝑠
2𝑠 + 1
=
3
𝑠
∗
0.16
𝑒−1.9𝑠
2𝑠 + 1
1 +
0.3214𝑒−1.9𝑠
2𝑠 + 1
C.
𝐶𝑜 ∞ = lim
𝑠→0
𝑠 ∗ 𝐶𝑜(𝑠) = 0.363

20. Example 2
For the liquid- level system shown in the Figure below, find
a. The transfer function of the process.
b. Sketch the signal flow block diagram.
c. The T.F
𝐻 𝑠
𝐻𝑠𝑝 𝑠
d. The response of H if a unit step change occurs in set point
and sketch it.
Given the following data: 𝐴 = 1 𝑚2
,𝑅1 = 2, 𝑞 = 10 𝑙𝑖𝑡/𝑚𝑖𝑛,
𝑉 = 5 𝑙𝑖𝑡, 𝐺𝑣 = 1.5, 𝐺𝑚 = 1, 𝐺𝑐 = 1
𝑞𝑜
𝑞𝑖
𝐻
V 𝑅1
LC
LT

21. Material balance
𝑞𝑖 = 𝑞𝑜 + 𝐴
𝑑𝐻
𝑑𝑡
𝑞𝑖 =
𝐻𝑖
𝑅1
+ 𝐴
𝑑𝐻
𝑑𝑡
𝑅1𝐴
𝑑𝐻
𝑑𝑡
+ 𝐻 = 𝑅1𝑞𝑖
𝜏
𝑑𝐻
𝑑𝑡
+ 𝐻 = 𝑅1𝑞𝑖
𝜏𝑠 + 1 𝐻 𝑠 = 𝑅1𝑞𝑖(𝑠)
𝐻 𝑠 =
𝑅1
𝜏𝑠+1
𝑞𝑖 𝑠 , 𝑅 = 2, 𝜏 = 𝑅. 𝐴 = 2
𝐻 𝑠 =
2
2𝑠 + 1
𝑞𝑖 𝑠
a.
Process.
𝑞𝑜
𝑞𝑖
𝐻
V 𝑅1
C.V
𝐺𝑝 𝑠 =
𝐻 𝑠
𝑞𝑖 𝑠
=
2
2𝑠 + 1
There is only
Process Gp, no load
GL exist
𝑞𝑜
𝑞𝑖
𝐻
V 𝑅1
LC
LT
Set point

22. c.
1.5
2
2𝑠 + 1
𝐻(s)
+ E(s)
-
𝐻𝑠𝑝(s)
Note that Gc=1 and
Gm=1 , and no need to
put them in the block
diagram.
b.
𝐻 𝑠
𝐻𝑠𝑝 𝑠
=
1.5 ∗ 2
2𝑠 + 1
1 +
1.5 ∗ 2
2𝑠 + 1
=
3
2𝑠 + 4
𝐺 𝑠 =
𝐻 𝑠
𝐻𝑠𝑝 𝑠
=
3/4
0.5𝑠 + 1
=
0.75
0.5𝑠 + 1
d.
Response for unit step change in servo
0
0.75
𝐻(𝑡)
𝑡
𝐻 𝑠 = ℒ−1
1
𝑠
.
0.75
0.5𝑠 + 1
= 0.75(1 − 𝑒−2𝑡)
𝐻 𝑠 = ℒ−1
𝐻𝑠𝑝 𝑠 . 𝐺 𝑠