Chapter 5 :
Control Systems and Their
Components (Instrumentation)
Professor Shi-Shang Jang
Department of Chemical Engineering
National Tsing-Hua University
Hsinchu, Taiwan
March, 2013
Overview

Need of Instrumentation.

Signals and Signal Levels.

Sensing Element.

What are Final Control Elements?

Characteristics of some transducers and transmitters.

Transmission Signals and representation.

Accuracy of Instrumentation.

What are Transducers and Transmitters?
Need of Instrumentation

As already discussed in earlier chapter, we need to measure
some parameters to control the process.
 Instrumentation is a methodology through which we obtain
value of a desired parameter.
 Type of instrument to be used is very vital issue and is
decided by Type of the process.
 Instrument’s dead-time & velocity lag.
 Accuracy in measurement.
 Sampling rate (for digital systems).
 Measuring range.
 Safety and hazards associated with it.
An Example- Blending Process
(Instrumentation)
Transducer
Controller
Final Control Element
Sensor/transmitter
Gas Process Control
fi (t )  0.16mi (t )
f o (t )  0.00506mo (t ) p(t )  p(t )  p1 (t ) 
fi  f o  8scfm; p  40 psia; p1  1atm; mi  mo  50%
Temperature Control of Non-isothermal CSTR
3-1 Signals and Signal Levels

Any Data or instruction carrying entity is called a signal.

Signals could be characterized by the nature of information it
transport and the medium of Transport.

On the basis of nature of information, the signal could be a
continuous signal or a Discrete/Digital Signal.

On the basis of medium of transport, the signal could be an
Electric Signal, Light/Laser Signal, Sound Signal, Radio Signal,
Pneumatic Signal etc.
3-1 Signals and Signal Levels cont..

Signal level is a physical range within which an information is
transmitted as a signal.

If the signal is continuous, the signal level are generally continuous.

If the signal is digital, the signal level is a set of discrete values.

Signal levels are an industry standard and may change time to time.

Signal range/level used in industry are• 1~5 mA
• 4~20 mA
• 10~50 mA
• 0~5 V DC
• ± 10 V DC
• 3-15 psig
3-1 Sensors/Sensing Element
Sensing Elements may be in physical contact with the system and
are responsible for determining the parameter. For example•Temperature-Thermocouples, Filled-bulbs, Resistance (RTD).
•Pressure-Bourdon tubes, Strain gauges.
•Level－ Float position, Difference Pressure, Ultra-sonic level meters.
•Flow rate－ Orifice plates, Venturi meters, Rotameter.
•Composition－ Gas chromograph (GC), pH meter, Conductivity, IR
absorption, UV absorption.
3-1 Sensors/Sensing Element
H s 
KT
T s 1
Sensors/Sensing Element
Characteristics of a linear temperature-current sensing element.
KT 
 20  4 
 200  0 
mA
 0.08 mA
psig
psig
Example: A Typical Experiment
Time
(second)
Y(temperature,oC,70100oC)
Y(temperature,mA,4-20mA)
Y (temperature, %)
0.
ln(1-Y)
0
70
4
1
71.74
4.928
0.058
-0.0598
2
76.51
7.472
0.217
-0.2446
3
80.8
9.76
0.360
-0.4463
4
84.64
11.808
0.488
-0.6694
5
88
13.6
0.600
-0.9163
6
90.76
15.072
0.692
-1.1777
7
93.16
16.352
0.772
-1.4784
8
94.99
17.328
0.833
-1.7898
9
96.64
18.208
0.888
-2.1893
10
97.75
18.8
0.925
-2.5903
0
100
Temp.C
95
90
85
80
75
70
0
1
2
3
4
5
6
7
8
9
Time, sec.
10
3-2 Final-control Element

After the data for the control variable (CV) and other parameters is
processed by the designed controller, signals are sent to Final-control
element which manipulates other variables like flow-rate etc. for the
system.

Generally, control is done by changing flow rates for the inlet/outlet of
material and energy to achieve control and control valves are widely used
for it.

Control Valves are of different kinds like Pneumatic.
 Ball Valve.
 Electro-mechanical.
 Butterfly Valve.
 Manual.
3-2 Control Valve- Final Control Element
There are generally two type of valve On/Off Control Valve
 Proportional
Control Valve Action
3-2 Control Valve- Final Control Element Cont..

Air to Close（AC）
- Valve action to close as the pressure increases.

Air to Open（AO）
- Valve action to open as the pressure increases as the previous figure.

The selection of AC or AO control valve is based on the
consideration of the emergent need, for instance, the emergent
shut down.

A transducer is needed to convert the electronic signal to the
pneumatic signal and thus finally controlling the flow rate.
3-2 Control Valve- Final Control Element Cont..
Control Valve- Final Control Element
Cont.. Sizing
For all valves, the liquid flow rate going through is following the
equation below:
F  Cv (vp )
PV
Gf
(5-1)
where, F＝flow rate；gal/min.
P＝Pressure drop；psi.
Gf ＝specific gravity of the fluid.
Cv ＝Size, choose the valve size such that at normal
operation, the valve is nearly half opened, i.e.
vp=0.60.7 vp is the fractional area of the valve that
allows the fluid going through, if vp=1, then all area
is available (valve fully open), if vp=0, the valve is
fully closed.
Control Valve- Final Control Element Cont.. Valve
Characteristics

The graph shows the flow-rate as the function of
opening of the valve.
Linear : Avp  vp
Quick opening : Avp  vp
Equal percentatge : Avp   vp 1
dAvp
dx  ln 
Avp
Cv  Cv ,max Avp
Rangeability 
Flow at 95% valve position
Flow at 5% valve position
Control Valve- Final Control Element Cont..

Valve Characteristics
◦ In most practical cases, equal percentages valves are selected to make sure
that the flow rate through the control valve is proportional to the signal vp by
choosing correct values of .
◦ This is due to the friction of the fluid through the pipe lines, and in most cases
this is non-negligible.
pL＝kLGf f2
(5-2)
Control Valve- Final Control Element Cont..
f2
pv  G f 2 ; pL  k LG f f 2
Cv
 1

p0  pL  pv  G f f 2  2  k L   costant
 Cv

f 
kL 
f
f max
Cv
1  k L Cv2
p0
; Cv  Cv ,max Avp (vp )
Gf
Cv ,max
pL
; f max 
2
Gf f
1  k L Cv2,max
1  k L Cv2,max
Cv

Cv ,max
1  k LCv2
p0
Gf
Control Valve- Final Control Element
Cont.. Example

Figure 5-2.3 shows a process for transferring an oil from a strage tank to a
separation tower. Nominal oil flow is 700 gpm, friction pressure drop is 6 psi,
available pressure drop for the control valve is 5 psi.
Control Valve- Final Control Element Cont.. Example
0.94
; Cv ,max  2  Cv  607 gpm / ( psi )1/2  Masoneilan valve Cv ,max  640 gpm / ( psi )1/2
5
Cv  700 
kL 
6
 0.94  700 
2
 13.0 106
p0  5  6  11 psi
f max 
640
1  13.0 106   640 
2
11
 870 gpm
0.94
Case 1(linear trim) : Avp  vP
f 
Cv ,max vP
1  kLC
2
2
v ,max P
v
p0
640vP

2
Gf
1  13.0 106   640  vP2
11
0.94
Case 2(equal percentage trim) : Avp   1vP
f 
Cv ,max 1vP
1  kLC
2
2
v ,max P
Rangeability 
v
p0
640 1vP

2
2 1 v
Gf
1  13.0 106   640    P 
f 0.95 839

 15.8
f 0.05 53
In case of linear valve:
863
Rangeability=
8
109
11
0.94
Control Valve- Final Control Element Cont.. Example
900
800
Flow, f
700
600
linear
500
400
300
200
Equal percentage
100
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
vp
Matlab code:
>> alpha=50;
vp=linspace(0,1);
for i=1:100
f(i)=640*alpha^(vp(i)-1)/(sqrt(1+13e-6*(640*alpha^(vp(i)-1))^2))*sqrt(11/0.94);
f1(i)=640*vp(i)/(sqrt(1+13e-6*(640*vp(i))^2))*sqrt(11/0.94);
end
>> plot(vp,f1,vp,f)
Control Valve- Final Control Element Cont.. Transfer
Function
Kv 
dvp dCv df
df

dm dm dvp dCv
1 frn.vp
dvp

dm 100 %CO
dCv
  ln   Cv ,max vp 1   ln   Cv
dvp
dCv
 Cv ,max
dvp
Kv
Gv  s  
vs 1
or
 equal percentage trim 
 linear trim 
What are Transducer and Transmitter?

Transducer is a device which convert one type of signals into another. In
other worlds, it may convert one form of energy to another form.

Eg 1: A Digital thermometer’s Transducer convert thermal energy into
equivalent electrical signals.

A typical Transducer consist of a sensing element combined with a driving
element (transmitter).

Transducers for process control measurements convert the magnitude of a
process variable (e.g., flow rate, pressure, temperature, level, or concentration)
into a signal that can be sent directly to the controller.
What are Transducer and Transmitter? Cont..



Transducer is a device which convert one type of signals into another.
In other worlds, it may convert one form of energy to another form.
A transmitter is usually required to convert sensor output compatible
with the controller input and to drive the transmission lines
connecting the two.
Pneumatic (air pressure) signals were used extensively up till 1960s
but currently Digital instrumentation is widely used.
An Example- Blending Process
(Instrumentation)
Transducer
Controller
Final Control Element
Sensor/transmitter
Block Diagram
Assume m is small
3-3. Conventional Feedback Controllers Proportional Controllers
R(t) =error=R(t)-C(t)
m t   m  Kce t 
In the s-domain
M ( s)  K c E  s 
Kc is called controller gain
3-3 Conventional Feedback Controllers -Proportional
Controllers Cont..
control valve saturation
100
Kc=100/50=2
control valve saturation
Proportional Band=
PB=50/100=50%
0
PB=100/Kc
50
Kc<0: Direct Acting Control p increases with y increases
Kc>0: Reverse Acting Control p decreases with y increases
Conventional Feedback Controllers
Proportional-Integral Controllers

1
m(t )  m  K c  e(t ) 
I

In the s  domain

e
t
*
dt
*



0

t

1 
M  s   K c 1 
 E s

s

I 
Conventional Feedback Controllers
Proportional-Integral Controllers Cont..
• The integral actions contribute positive or negative as the signs
appeared shown above. However further build up of integral term
becomes quite large and the controller is saturated is referred to as
reset windup.
Conventional Feedback Controllers
Proportional-Integral Controllers Cont..
• Reset windup occurs when a PI or PID controller encounters a
sustained error. In this situation, a physical limitation prevents the
controller from reducing the error signal to zero.
• It is undesirable to have integral term continue to build up after the
controller output saturates.
• Commercial controllers available provides anti-reset windup.
Conventional Feedback Controllers
Proportional-Integral-Derivative Controllers
t

de  t  
1
m  t   m  K c e  t    e  t * dt *   D

I 0
dt 

M s


1
 K c 1 
 Ds
E s
 Is

Modification for physical realizability
M s

 Ds 
1
 K c 1 


E s

s

s

1

I
D

Conventional Feedback Controllers
Proportional-Integral-Derivative Controllers Cont..
• The direct implementation of the derivative term of a PID controller
is basically undesirable due to:
 Noisy signal is normally received. The derivatives of these signals
are meaningless.
 The derivative element is physically unrealizable.
3-3 Typical Responses of Feedback Control Systems
3-3 Typical Responses of Feedback Control Systems Continued
Response
No control
P-control
PID-control
PI-control
time
3-3 Typical Responses of Feedback Control Systems Continued
Response
No control
Kc=1
Kc=2
Kc=5
time
3-4 Process Control Applications

Applications of process control are mostly in the areas of:





Flow rate control
Level control
Air pressure control
Temperature control
Composition control
3-4 Process Control Applicationscontinued

Flow rate control
 Applications: inlet flow control, outlet flow control of processes, reflux flow
rate, pipeline flow rate,…,etc.
 Implementation
 Remarks:
 (i) Due to the effect of turbulence and pressure fluctuation, the measurement is noisy.
 (ii) no offset is allowed in flow rate control – integral action is necessary.
 (iii) flow rate process is fast – no need for derivative actions.
 Conclusion: PI control is needed with low gain and I10-20 seconds
3-4 Process Control Applications- continued

Liquid level control:
 Applications: reactor volume control, buffer tank level control, reboiler
level control accumulator level control, steam generator level control, …,etc.
 Implementation:
3-4 Process Control Applications- continued

Remarks:
◦ (i) the level process is basically noisy – fluctuation of the liquid level – low controller gain
◦ (ii) in case of important levels such as reboiler level, accumulator, integral action is needed,
◦ (iii) the level processes are basically a first order system.

Conclusions: The tank level control is basically loose, for instance, to maintain the tank is not
completely empty at low inlet flow rate and to maintain tank is not full at high inlet flow rate.
Thus, a low gain P controller is frequently implemented. However, in case of important level
system such as reboiler level, accumulator, PI controllers should be used.

Other tips: If the outlet of the tank is very important for example, the flow rate to the reactor.
Then, the level controller should not influence the flow rate. The controller gain of the Pcontroller should be tuned to very low.
3-4 Process Control Applications- continued

Air pressure control
• Applications: Gas storage tank, air phase reactors.
• Remarks:
 (i) vapor pressure control is not this case, it should be considered as temperature
control in the next slide,

(ii) Gas pressure system is fast – no derivative action is needed,
 (iii) the measurement is not noisy.
• Conclusions: Use high gain P-only controllers.
3-4 Process Control Applications- continued

Temperature Control
 Applications: reactor temperature control, heat exchanger temperature
control, temperature control of pre-heaters, vapor pressure control,…,etc.
 Manipulated variables: cooling water flow rate, steam flow rate.
 Remarks:
 (i) the quality of sensor is crucial for this type of control, for instance, sensor noise and
time lag will influence the control quality,
 (ii) the system is quite slow (heat transfer mechanism), derivative action is needed,
 (iii) off set of the temperature is not allowed – integral action is needed,
 (iv) there exists an inherent upper bound of the controller gain, process stability is an
issue.
 Conclusion: A typical PID control situation.
3-4 Process Control Applications- continued

Composition control
• Applications: pH control, reactor composition control, distillation
composition control, …,etc.
• Remarks:
‾ (i) in some cases, the measurements are noisy with time lag (e.g. GC) makes the
control very difficult,
‾ (ii) the process is typically slow – derivative action,
‾ (iii) no offset is allowed – integral action.
• Conclusion: PID control situation, PI may be implemented in some
cases, controller settings are case by case.
3-5 Summary

Instrumentation is a part of the manufacturing process.

The sensors and transmitters are introduced

The sizing and characteristics of control valves are essential for
instrumentations

The conventional controllers are derived and analyzed

The applications of the controllers are introduced
Homework
Text p192
 5-3, 5-5, 5-10, 5-16, 5-18

Supplemental Material
Control Valve- Final Control Element Cont..
PT  Phe  Py  Constant
 40pisg
( 8- 3)
Pressure drop vs flow rate
70
60
pressure drop (psig)
50
40
30
20
10
0
0
50
100
50
100
150
200
flow rate (gal/min)
250
300
pressure drop accross the vale (psi)
100
90
80
70
60
50
40
30
0
150
200
flow rate (gal/min)
250
300
Matlab code:
k=30/(200)^2
f=linspace(0,300);
for i=1:100
dp(i)=k*f(i)^2;
end
plot(f,dp)
dp_valve=100-dp;
plot(f,dp_valve)
Example
A pump furnishes a constant head of 40 psig, the heat exchanger pressure
drop is 30 psig at 200 gal/min. Select a Cv of the valve and plot the installed
characteristic for:
1.
A linear valve that is half open at the design flow rate.
2.
An equal percentage valve (R=50) that is sized to be completely
open at 110% of the design flow rate.
Area vs valve position
100
90
80
x=linspace(0,1);
R=50;
for i=1:100
fl(i)=R^(x(i)-1)*100;
end
plot(x,fl)
70
area(%)
60
50
40
30
20
10
0
0
0.1
0.2
0.3
0.4
0.5
0.6
valve position
0.7
0.8
0.9
1
Solution
Given any flow rate q, pressure drop across the heat exchanger
Phl  q 


30  200 
2
 q 
Ps  Phl  30 

 200 
2
Pressure drop across the valve:
 q 
Pv  40  Phl  40  30 

 200 
(a) Calculate rated Cv
200
Cv 
 126.5
0.5 10
2
Solution
(b) Calculate the rated Cv at 110% of qd
q  200*1.1  220 gpm
Ps  30 1.1  36.3 psi  Pv  40  36.3  3.7 psi
2
Cv 
220
 114.4
3.7
q
 R l 1
Cv Pv


q
 l  1  log 
 / log R
 C P 
v 
 v
To plot for q over l, (i) set q, (ii) get Ps, (iii) get Pv, then get l
cv=115;R=50;
for i=1:11
q(i)=20*i;
dps=30*(q(i)/200)^2;
dpv=40-dps;
l(i)=1+log(q(i)/(cv*sqrt(dpv)))/log(R);
end
Example:

The temperature of a CSTR is controlled by a pneumatic feedback control
system containing
◦ (1) a 100 to 200oF temperature transmitter,
◦ (2) a PI controller with integral time set at 3 minutes/repeat and proportional band a 25, and
◦ (3) a control valve with linear trim, air to close action, and a Cv=4 through which cooling water
flows. The pressure drop across the valve is a constant 25 psi.
If the steady-state controller output pressure is 9 psig, how much cooling
water is going through the valve? If a sudden disturbance increases reactor
temperature by 5oF, what will be the immediate effect on the controller output
pressure and the water flow rate?
Solution
20  4
 0.16 mA
F
200  100
100
(2) Kc 
4
25
25
(3) F  4 x
 20 x gal
min
1
(1) Km 
Nominal signal of the valve is 9psig,  x 
15  9
 0.5
15  3
i.e. Nominal flow rate through the valve is F  20  0.5  10 gal
min
##
A sudden change of temperature of 5F will cause
a sudden change of signal by 5  0.16=0.8mA ,i.e. e=-0.8mA will cause a change
of signal to the controller by 4  ( 0.8)=-3.2mA
15  3
The signal to the valve is changed by 9  3.2 
 6.6 psig ##
20  4
15  6.6
x
 0.7
15  3
The flow rate will change to F  20  0.7  14 gal
##
min