chapter 5 : control systems and their components (instrumentation) professor shi-shang jang...

Chapter 5 : Chapter 5 : Control Systems Control Systems and Their and Their Components Components (Instrumentation)(Instrumentation)

Professor Shi-Shang JangDepartment of Chemical EngineeringNational Tsing-Hua UniversityHsinchu, TaiwanMarch, 2013

OverviewOverview

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 InstrumentationNeed 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 An Example- Blending Process Process (Instrumentation)(Instrumentation)

Sensor/transmitter

Transducer

Controller

Final Control Element

Gas Process ControlGas Process Control

1

1

( ) 0.16 ( )

( ) 0.00506 ( ) ( ) ( ) ( )

8 ; 40 ; 1 ; 50%

i i

o o

i o i o

f t m t

f t m t p t p t p t

f f scfm p psia p atm m m

Temperature Control of Temperature Control of Non-isothermal CSTRNon-isothermal CSTR

3-1 Signals and Signal Levels3-1 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 3-1 Signals and Signal Levels cont..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-

• 0~5 V DC• ± 10 V DC• 3-15 psig

• 1~5 mA• 4~20 mA• 10~50 mA

3-1 Sensors/Sensing 3-1 Sensors/Sensing ElementElement

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 3-1 Sensors/Sensing ElementElement

1

T

T

KH s

s

Sensors/Sensing Sensors/Sensing ElementElement

20 40.08

200 0T

mA mAK psigpsig

Characteristics of a linear temperature-current sensing element.

Example: A Typical Example: A Typical ExperimentExperiment

Time (second)

Y(temperature,oC,70-100oC)

Y(temperature,mA,4-20mA) Y (temperature, %) ln(1-Y)

0 70 4 0. 0

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 1 2 3 4 5 6 7 8 9 1070

75

80

85

90

95

100

Temp.C

Time, sec.

3-2 Final-control Element3-2 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- Ball Valve. Butterfly Valve.

Pneumatic. Electro-mechanical. Manual.

3-2 Control Valve- Final 3-2 Control Valve- Final Control ElementControl Element

There are generally two type of valve- On/Off Control Valve Proportional

Control Valve ActionControl Valve Action

3-2 Control Valve- Final Control 3-2 Control Valve- Final Control Element Element Cont..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 3-2 Control Valve- Final Control Element Element Cont..Cont..

Control Valve- Final Control Valve- Final Control Element Control Element Cont.. Cont.. SizingSizing

For all valves, the liquid flow rate going through is following theequation below:

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=0.60.7 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.

( ) Vv

f

PF C vp G (5-1)

Control Valve- Final Control Control Valve- Final Control Element Element Cont.. Cont.. Valve Valve CharacteristicsCharacteristics

1

,max

Linear :

Quick opening :

Equal percentatge :

ln

vp

vp

vpvp

vp

vp

v v vp

A vp

A vp

A

dAdx

A

C C A

The graph shows the flow-rate as the function of opening of the valve.

Flow at 95% valve positionRangeability

Flow at 5% valve position

Control Valve- Final Control Control Valve- Final Control Element Element Cont..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)


22

2

20 2

0,max2

,max 0max2 2

,max

2,max

2max ,max

;

1costant

; ( )1

;1

1

1

v f L L fv

L v f Lv

vv v vp

fL v

vLL

f fL v

L vv

v L v

fp G p k G f

C

p p p G f kC

C pf C C A vp

Gk C

C ppk f

G f Gk C

k CCf

f C k C

Control Valve- Final Control Valve- Final Control Element Control Element Cont.. Cont.. ExampleExample

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 Control Valve- Final Control Element Element Cont.. ExampleCont.. Example

1/2 1/2,max ,max

62

0

max 26

,max

2,max

0.94700 ; 2 607 / ( ) Masoneilan valve 640 / ( )

56

13.0 100.94 700

5 6 11

640 11870

0.941 13.0 10 640

Case 1(linear trim) :

1

v v v v

L

vp P

v P

L v P

C C C gpm psi C gpm psi

k

p psi

f gpm

A v

C vf

k C v

0

2 26 2

1

1 1,max 0

2 2 2 2 16,max

0.95

0.05

640 11

0.941 13.0 10 640

Case 2(equal percentage trim) :

640 11

0.941 1 13.0 10 640

839Rangeability 15.8

53

In case of linear valve

P

P P

P

P

fP

vvp

v vv

vfL v P

p v

G v

A

C pf

Gk C v

f

f

:

863Rangeability= 8

109

Control Valve- Final Control Control Valve- Final Control Element Element Cont.. ExampleCont.. Example

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

100

200

300

400

500

600

700

800

900

linear

Equal percentage

vpMatlab code:>> alpha=50;vp=linspace(0,1);for i=1:100f(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 Control Valve- Final Control Element Cont.. Transfer FunctionElement Cont.. Transfer Function

1,max

,max

1 .

100 %

ln ln equal percentage trim

linear trim

1

vv

v

vpvv v

vv

vv

v

dCdf dvp dfK

dm dm dvp dC

dvp frn vp

dm CO

dCC C

dvp

dCor C

dvp

KG s

s

What are Transducer and What are Transducer and Transmitter?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 What are Transducer and Transmitter? Transmitter? Cont..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 An Example- Blending Process (Instrumentation)Process (Instrumentation)

Sensor/transmitter

Transducer

Controller

Final Control Element

Block DiagramBlock Diagram

Assume m is small

3-3. Conventional Feedback 3-3. Conventional Feedback Controllers - Proportional Controllers - Proportional ControllersControllers

In the s-domain

( )

c

c

m t m K e t

M s K E s

R(t) =error=R(t)-C(t)

Kc is called controller gain

3-3 Conventional Feedback 3-3 Conventional Feedback Controllers -Proportional Controllers -Proportional Controllers Controllers Cont..Cont..

Proportional Band=PB=50/100=50%

0

100

50

Kc=100/50=2

PB=100/Kc

control valve saturation

control valve saturation

Kc<0: Direct Acting Control p increases with y increasesKc>0: Reverse Acting Control p decreases with y increases

Conventional Feedback Conventional Feedback ControllersControllers

0

1( ) ( ) * *

In the s domain

11

t

cI

cI

m t m K e t e t dt

M s K E ss

Proportional-Integral 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.


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.


0

1* *

11

Modification for physical realizability

11

1

t

c DI

c DI

Dc

I D

de tm t m K e t e t dt

dt

M sK s

E s s

M s sK

E s s s

Proportional-Integral-Derivative Controllers

Conventional Feedback Conventional Feedback ControllersControllersProportional-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 3-3 Typical Responses of Feedback Control SystemsControl Systems

3-3 Typical Responses of Feedback 3-3 Typical Responses of Feedback Control Systems - ContinuedControl Systems - Continued

time

No control

P-control

PI-controlPID-control

3-3 Typical Responses of Feedback 3-3 Typical Responses of Feedback Control Systems - ContinuedControl Systems - Continued

time

No control

Kc=1

Kc=2

Kc=5

3-4 Process Control 3-4 Process Control ApplicationsApplications

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 3-4 Process Control Applications- continuedApplications- continued

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 I10-20 seconds


Liquid level control:

Applications: reactor volume control, buffer tank level control, reboiler

level control accumulator level control, steam generator level control,

…,etc.

Implementation:


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 P- controller should be tuned to very low.


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.


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.


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 Summary3-5 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

HomeworkHomework

Text p1925-3, 5-5, 5-10, 5-16, 5-18

Supplemental Supplemental MaterialMaterial


Constant 8 3

40pisg

T he yP P P ( - )

Pressure drop vs flow Pressure drop vs flow raterate

Matlab code:k=30/(200)^2 f=linspace(0,300); for i=1:100dp(i)=k*f(i)^2;endplot(f,dp)dp_valve=100-dp;plot(f,dp_valve)

0 50 100 150 200 250 3000

10

20

30

40

50

60

70

flow rate (gal/min)

pre

ss

ure

dro

p (

ps

ig)

0 50 100 150 200 250 30030

40

50

60

70

80

90

100

flow rate (gal/min)

pre

ss

ure

dro

p a

cc

ros

s t

he

va

le (

ps

i)

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 positionArea vs valve position

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

100

valve position

are

a(%

)

x=linspace(0,1);R=50;for i=1:100fl(i)=R^(x(i)-1)*100;endplot(x,fl)

Solution

2

2

30 200

30200

hl

s hl

P q

qP P

2

40 40 30200v hl

qP P

Given any flow rate q, pressure drop across the heat exchanger

Pressure drop across the valve:

(a) Calculate rated Cv

200126.5

0.5 10vC

2

1

200*1.1 220

30 1.1 36.3 40 36.3 3.7

220114.4

3.7

1 log / log

s v

v

l

v v

v v

q gpm

P psi P psi

C

qR

C P

ql R

C P

(b) Calculate the rated Cv at 110% of qd

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

Solution

Example: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?

20 4(1) 0.16

200 100100

(2) 425

25(3) 4 20 min1

15 9Nominal signal of the valve is 9psig, 0.5

15 3

. . Nominal flow rate through the valve is 20 0.5 10 ##minA sudden change of temper

mAKm F

Kc

galF x x

x

gali e F

ature 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 3The signal to the valve is changed by 9 3.2

20 4

6.6 ##

15 6.60.7

15 3

The flow rate will change to 20 0.7 14 ##min

psig

x

galF

Solution

chapter 5 : control systems and their components (instrumentation) professor shi-shang jang...

Documents

signal levels

continuous signal

sound signal

radio signal

lightlaser signal

electric signal

pneumatic signal

discretedigital signal