Universiti
P Malaysia
PAHANG
EngInoing . Tohnoogy . CrotMty
FACULTY OF ELECTRICAL & ELECTRONICS ENGINEERING
FINAL EXAMINATION
COURSE
:
PROCESS CONTROL
COURSE CODE
:
BEE4343
LECTURER
:
YASMIN BINTI ABDUL WAHAB
DATE
:
6 JANUARY 2012
DURATION
:
3 HOURS
SESSION/SEMESTER :
SESSION 2011/2012 SEMESTER I
PROGRAMME CODE :
BEE/BEP
INSTRUCTIONS TO CANDIDATES
1. This question paper consists of FIVE (5) questions. Answer ALL the questions.
2. All answers to a new question should start on new page.
3. All the calculations and assumptions must be clearly stated.
EXAMINATION REQUIREMENTS
1. Appendix I — Table of Formula
2. Appendix II - Table of Laplace Transform.
3. Appendix III - Open loop response of the process plant due to unit-step input for
answering Question 4 (a).
4. Appendix IV - Controlled variable response for several PB values for answering
Question 4 (c).
5. Appendix V — Expected responses for CV and MV for answering Question 5 (b)(ii).
DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO
This examination paper consists of SIXTEEN (16) printed pages including front page
BEEIBEP 11 1211BEE4343
CONFIDENTIAL
QUESTION 1
(a) All seven categories of control objectives must be achieved simultaneously; failure to do
so leads to unprofitable or, worse, dangerous plant operation. Briefly describe any FIVE
(5) major categories of control objectives.
[5 Marks]
[COl, P01, C21
(b) Consider the isothermal Continuous Stirring Tank Reactor given in Figure 1. The
chemical reaction rate is given by rAY = —
kf.
From Figure 1:
(i) Derive the non-linear first order differential model for the system.
(ii) Derive the non-linear term in the differential model in (i) as a Taylor series
expansion.
(iii)Derive the linearized first order differential model for the system in deviation
variables, C,, i = y where-Ai'' =
'-A/
- C,
(iv)Determine the transfer function of the linearized model for the system.
[15 Marks]
[CO2, P03, C41
C:
CAy
Figure 1: Isothermal CSTR
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BEE/BEP 11 121/BEE4343
QUESTION 2
Consider the two continuous stirred-tank reactors (CSTRs) in Figure 2. The linearized
individual transfer functions are given by G1 (s) = C 1 (s)/C 0 (s) and G2 (s) = C 2 (s)/C 1 (s),
respectively. For CSTR 1, the chemical reaction rate is first-order with rAl =
meanwhile the chemical reaction rate in CSTR 2 is a second-order given by rA2 = —kC2.
(a)
From the first principle, derive the individual transfer function of each reactor, GI(s)
and G2(s).
(b) Find the time constant, t and steady state gain, K for each tank if
k = 0.5[(mole/m 3 )min] -', F= 0.075m3/min, V1 2.5 m3 and V2 3.0 m3 . Let CA2s
0.22 mole/m3.
[22 Marks]
[CO2, P03, C4]
F
CAO
CSTR2
Figure 2: Two CSTRs in series
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CONFIDENTIAL
BEEIBEP 11 1211BEE4343
QUESTION 3
(a) Consider the CSTR in Figure 3.1. No product is present in the feed stream, a single
chemical reaction occurs in the reactor, and the heat of reaction is zero.
(i) Explain whether each of the following single loop control designs for that CSTR is
possible. [Hint: Consider each question separately and does a causal process
relationship exist?]
a. Control the temperature in the reactor by adjusting the valve in the coolant flow
pipe.
b. Control the temperature in the reactor by adjusting the valve in the solvent pipe.
[4 Marks]
[CO3, P03, C41
Figure 3.1: CSTR process
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BEE/BEP 11 12UBEE4343
(ii) If the CSTR in Figure 3.1 is represented in block diagram as shown in Figure 3.2,
derive the overall transfer function CV(s)/SP(s) and determine stability of the
system.
[10 Marks]
[CO3, P03, C41
D(s)
CV(s)
Figure 3.2: Block Diagram of CSTR process
Where
G, (s) = K (i
+ T1s1
\
Ga(s)
K
(Ts+1)(Ts+1)
=
G, (s) =
1
with
K = 0.1
mole/rn3
Ta(s) = 1.0min
,
r = 0.5 mm, K = 15 (%open)/(mole/m3),
CONFIDENTIAL
BEE/BEP 11 12UBEE4343
(b)
Pure A
Figure 3.3: Mixing of two liquids
Figure 3.3 illustrates the mixing of two liquids which is commonly found in the process
industries: The two liquids contain different concentrations of A:
. Stream B=1 %A
Stream A100% A
The liquids mix to form
XAO.
The resulting mix can be represented by a mathematical
equation as:
XAO FB + XAO FA = FB XAB + FA XAA
where
XAO
=concentration of mixing two liquids.
XAB
=concentration of A in stream B = 2%A.
XAA
=concentration of A in stream A = 100%A.
FB
=flow rate of stream B = 6.5 m3/min.
FA
=flow rate of stream A 0.185 m3/min.
Defining x'A O = XAO- xA oand F A = FA - FAR , examine the transfer function that relates
x'Ao
and PA.
[6 Marks]
[CO3, P03, C41
CONFIDENTIAL
BEE/BEP 11 1211BEE4343
QUESTION 4
Figure 4.1 shows the closed-loop block diagram of a process plant with Proportional
controller. Meanwhile, Figure 4.2 (Appendix III) shows the open loop response of the
process plant due to unit-step input which was obtained experimentally in manual mode. The
controlled variable responses for several proportional band (PB) values in the automatic
mode are provided in Figure 4.3(a)-(f) of Appendix IV. The control system is to be tuned
with these conditions:
Case (1) without a filter, r f =0 mm.
Case (2) with a first-order filter where rf =0.5 mm.
Case (3) with a first-order filter where r3.O mm.
Find the following:
(a) Estimate the process transfer function, G(s) for Figure 4.2 in Appendix III.
[4 Marks]
(b) Based on the given information, estimate the P1 tuning constants for all the three
cases using Cohen-Coon tuning rule.
[9 Marks]
(c) By using information in Appendix IV, determine the PID tuning constant using
Method II of Ziegler Nichols tuning rule.
[4 Marks]
(d) Given a temperature transmitter having range of 50 0C to 2000C whereby the input
current signal is in the range of 4-2OmA. Calculate the temperature measurement if
14.7 mA current is output from the transmitter.
[3 Marks]
[CO3, P03, C51
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BEE/BEP 111211BEE4343
Controller
Process
SP(s
Figure 4.1: Closed Loop system with P controller
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BEE/BEP 11 1211BEE4343
QUESTION 5
(a) Figure 5.1 shows the block diagram of a cascade control structure. The disturbance
occurs in the secondary loop where this loop is encircled by the dotted line.
(i) Describe TWO (2) conditions when the cascade control system can be
implemented.
(ii) Derive the transfer function of set-point response
CV, (S)
SPj(s) ,
and disturbance
C V, (s)
response
D(s)
[6 Marks]
[CO3, P03, C41
D(s)
Gd(s)
+(
FG,
GC
le
SP(s)
Figure 5.1: Block diagram of a cascade control structure
BEE/BEP 11 1211BEE4343
CONFIDENTIAL
(b) Feedforward control uses a measured input signal to determine an adjustment to an
input manipulated variable. Figure 5.2 shows a block diagram of feedforward control
system with feedback. From the block diagram:
(i)
Design a feedforward controller Gif (s) that suitable with the system in Figure
5.2.
(ii)
Sketch the expected responses for controlled variable (CV) and manipulated
variable (MV) if the disturbance response is as given in Figure 5.3 (Appendix
V).
(iii) Describe the complementary of feedforward and feedback controller in terms of
advantages and disadvantages.
[12 Marks]
[CO3, P03, C41
Dm(S)
Sensor
Figure 5.2: Block diagram of feedforward control system with feedback
END OF QUESTION PAPER
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BEE/BEP 11 121/BEE4343
CONFIDENTIAL
APPENDIX I - Table of Formula
OVERALL MATERIAL BALANCE
{accumulation of mass} = { mass in} -
CONTROL SYSTEM PARAMETERS
{ mass out}
COMPONENT MATERIAL BALANCE
measured value expressed as percent of
span over range
{ accumulation of component balance =
cp
=
c—c
x1 00
C max - cmjn
{ component mass in} - { component mass
{accumulation of U+PE+KE}={H+PE+KE
error expressed as percent of span
r–b
X1 00
e =
b –b
in due to convection) -{H+PE+KE out due
controller output as percent of full scale
out} + { generation of component mass}
ENERGY BALANCE
°
max
to convection} + Q - Ws
X1 00
generation of component mass = Vr
U max -
Taylor Series for function of one variable
PID Algorithms
G (s) = K (1+ --- + Tds)
+R
T1s
21 dYj X,
^1^ X^
100
Deviation Variable
- CAIS
Final Value Theorem
f(oo) = limsf(s)
K
CPB
Initial Value Theorem
f(t)I - = limsf(s)
Non-interacting series with Dead Time
K = fl K 1 , 0 = 0i
t63%
(o, +v)
Process Reaction Curve
A
A
K =—,r=—
S
G=intercept of maximum slope with initial
value
t28% =0+, t63% =0+r,
V =
1.5(t63%
mm
t28%)
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BEE/BEP 1 1121/BEE4343
CONFIDENTIAL
PID Tuning
Ziegler-Nichols (First Method)
Type of Controller
K
P
Ti
Td
Go
0
0
0
20
0.58
0
P1
PID
T
Ziegler-Nicholes (Second Method)
P
0.5K,.
co
0
P1
0.45K
1•2Pcr
0
PID
0•6Kcr
0•5Pr
O•l25Pcr
Cohen-Coon Tuning Method
rilr rl
1-11 1+ JL 3
LKr
P1
PID
GO
e[30+3r1
L9+20r]
[J-1[o.9+z_1
12]
[KrJL
][4
Kr3
32 + 6r
1
[13+8r]
4]
where r = -0
V
12
0
0
L11+2r
CONFIDENTIAL
BEE/BEP 111211BEE4343
Appendix 11—Table of Laplace Transform
Step function, u(t)
Laplace Transform Pairs
1
S
e'
s+a
n!
tn
5n+1
f(k)
(t) -
sF(s)_sk_If(O_ ) —s" 2 f (y)
d k f(t)
di' k
F(s)
ff(t)dt
1
Impulse function 6(t)
sin wt
Co
s 2 +w2
s
2
cos cot
co
(s+a)2 +0)2
(s+a)
(s+a)2+0)2
e°' sin wt
e -at Cos Cot
2
e" sin0)y1_ç2t,
<1
s2+2cos+co 2
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CONFIDENTIAL
BEE/BEP 11 1211BEE4343
Appendix III - Open loop response of the process plant due to unit-step input for answering
Question 4 (a)
45
-
15
35
11
0
V
i/fl
25
3j
A
14-AL-1
-5
71 5
20
tfrne
30
Figure 4.2: Open loop response of the process plant due to unit-step input
Note: You must use the diagram in your answer (no need to redraw the figures) but please do not
forget to attach them to your answer booklet!
14
BEE/BEP 111211BEE4343
CONFIDENTIAL
Appendix IV - Controlled variable response for several PB values for answering Question 4 (c)
151
1
1
I
P8=24
I
P8=62
1
151
I
1
I
1
20
30
40
I
10
10
0
I.)
•0
0.
0
0
' 0
10
20
30
40
50
time
60
70
80
90
100
10
50
time
60
70
60
90
100
Figure 4.3(b)
Figure 4.3(a)
PB = 46
P6= 67
15
IS
10
10
0
0
0
0
0
10
20
30
40
50
time
60
70
60
90
100
0
10
20
30
40
50
time
60
70
80
90
100
Figure 4.3(d)
Figure 4.3(c)
P6=78
P6=95
18
IS
10
10
C-)
C-)
0
-------------- -
0
-c
0
10
20
30
40
50
time
60
70
80
90
100
time
Figure 4.3(f)
Figure 4.3(e)
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CONFIDENTIAL
BEE/BEP 11 1211BEE4343
Appendix V - Expected responses for CV and MV for answering Question 5 (b)(ii)
Figure 5.3: Expected responses for CV and MV
Note: You must use the diagram in your answer (no need to redraw the figures) but please do not
forget to attach them to your answer booklet!
16