dynamic control2

7/27/2019 Dynamic Control2

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Chemical Process

Dynamics and Control

Cheng-Liang ChenPSELABORATORY

Department of Chemical EngineeringNational TAIWAN University


2/921

Chen CL 1

Some Notes

- Instructor:

Cheng-Liang Chen ([email protected])

- Teaching Assistant:

Ying-Jyuan Ciou ([email protected])

-

Textbook:Smith, C.A., and A. Corripio (2006).

Principles and Practice of Automatic Process Control (3rd Ed.)

- Lecture Time: Mon 9 : 10 10 : 00; Wed 10 : 20 12 : 10

- Examination: 2 Mid-terms and 1 Final (20 2 + 30 = 70%)

- Homework: 10 (20%)

- Computer Exercise: Simulink, (PControLab3) (10%)


3/921

Chen CL 2

Learning Objectives

- Understand how the basic components of control systemswork (Sensor, Valve, Controller, Process)

- Develop dynamic mathematical process models that will help

in the analysis, design, and operation of control systems

- Design and tune feedback controllers

- Apply a variety of techniques that enhance feedback control,

including cascade control, selective control, override control,

ratio control, and feedforward control- Master the fundamentals of dynamic simulation of process

control systems using MATLAB and Simulink


4/921

Chen CL 3

Course Outline

- Process Control: Essentials Chapter 1

- Basic Control Elements: Sensor Chapter 5- Basic Control Elements: Valve Chapter 5

- Basic Control Elements: Controller Chapter 5

- Basic Control Elements: Process Chapters 3,4

- Analysis of Feedback Control Loops Chapters 6,2

- Adjusting Controller Parameters Chapter 7

- Frequency Response Techniques Chapter 8

- Enhanced PID Control: Cascade Control Chapter 9

- Enhanced PID Control: Selective Control Chapter 10

- Enhanced PID Control: Feedforward Control Chapter 11

-

Control Systems and Dynamic Simulation Chapter 13


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Basic Concept of

Process Control



C C 1


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Chen CL 1

A Process Heat ExchangerThe Problem

- Control Objectives: to keep T at TD (and q at qD) T: controlled variable TD: set point

- Environment: varying qs, Ps, Ta, Ti, q , E (efficiency)

Ps, Ta, Ti, E: hard to handle disturbances

qs, q: easy for adjusting manipulated variable

Ch CL 2


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Chen CL 2

A Process Heat ExchangerThe Tools

- We need one sensor to know current status of T

- We need one valve to adjust qs

- We need one method to make decision

- We have to check if CV = SP from time to time

Ch CL 3


8/921

Chen CL 3

A Process Heat ExchangerManual Method to Achieve Control Objective

- To know: reading T (influenced by Ps, Ta, Ti, E)

- To decide: comparing T with TD and decide adjusting action

- To do: implementing new qs manually by a field operator

- Repeat above actions for every second

Ch CL 4


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Chen CL 4

A Process Heat ExchangerAutomatic Method to Achieve Control Objective

- To know: reading T (influenced by Ps, Ta, Ti, E)

- To decide: comparing T with TD and decide adjusting action

- To do: implementing new qs automatically by a controller

- Repeat above actions for every second

Ch CL 5


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Chen CL 5

A Process Heat ExchangerFour Basic Elements

- Primary/Secondary Element (sensor/transmitter)

to know current status for CV

(fast, accurate, standard)

T (oC) TE= T (mV)TT

= y (4 20 mA; 1 5V; 0% 100%)

Example: desired zero = 50o

C, span = 100o

C

50oC 4 mA 1 V 0%

TE/TT or or

150o

C 20 mA 5 V 100%

Chen CL 6


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Chen CL 6


- Decision-making Element (operator or controller)

to calculate the trial corrective action

(simple to use, acceptable performance, robust, reliable)

inputs: set-point ysp (mA or %), (from TDo

C)measured PV y(t) (mA or %), (from T oC)

output: control action u(t) (mA or %)

Chen CL 7


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Chen CL 7


- Final Control Element (I/P transducer + valve)

to realize operators or controllers decision

u(t) (4 20 mA)I/P

= u(t) (3 15 psi)valve

= 0% 100% valve opening

= qs(t) (kg/sec) ( MV)

Chen CL 8


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Chen CL 8


- Process

to wait for new value of CV

inputs: qs (kg/sec) (MV)

Ps, Ti, Ta, (Disturbances) output: T (oC) (PV, CV)

Chen CL 9


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Chen CL 9


Summary

- Primary/Secondary Element (sensor/transmitter)

- Decision-making Element (controller)

- Final Control Element (I/P transducer/valve)

- Process (the heat exchanger)

Chen CL 10


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Chen CL 10

Chen CL 11


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Chen CL 11

Basic Concept of Process ControlSummary

- Process Control:

adjusting a Manipulated Variable ( MV)

to maintain the Controlled Variable ( CV)

at desired operating value ( Set Point) ( SP)

in the presence of output Disturbances ( Ds)

Chen CL 12


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Chen CL 12

Control StrategiesFeedback Control

- Adjusting MV if CV is not equal to SP

- Advantage: simple, can compensate all disturbances

-

Disadvantage: CV is not equal to SP in most time

Chen CL 13


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Chen CL 13

Control StrategiesFeed-forward Control

- Adjusting MV to compensate influence of multiple Ds on CV

- Advantage:

simultaneously consider influence of multiple Ds and MV on CV

detecting Ds adjusting MV BEFORE CV deviates from SP

- Disadv.s: modeling error ?; not considering ALL disturbances ?

Chen CL 14


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Chen CL 14

Control StrategiesFeed-forward Control with Feedback Trim

Chen CL 15


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Chen CL 15

Incentives for Chemical Process Control

-

Safety:temperature, pressure, concentration of chemicals

should be within allowable limits

- Production Specifications:

a plant should produce desired amounts and quality of final products

- Environmental Regulations:

various laws specify concentrations of chemicals of effluent from a

plant be within certain limits

Chen CL 16


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Chen CL 6

Incentives for Chemical Process Control

-

Operational Constraints:various types of equipment have constraints inherent to their

operation

- Economics:

operating conditions are controlled at given optimum levels ofminimum operating cost and maximum profit

Chen CL 17


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Design Aspects of A Process Control System

Define Control Objectives:

Q 1: What are the operational objectives that a control system is called

upon to achieve ?

Ensuring stability of the process, or

Suppressing the influence of external disturbances, or Optimizing the economic performance of a plant, or

A combination of the above

Select Measurements:

Q 2: What variables should we measure to monitor the operational

performance of a plant ?

Chen CL 18


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Select Manipulated Variables:

Q 3: What are the manipulated variables to be used to control a

chemical process ?

Select Control Configuration: (control structure)

Q 4: What is the best control configuration for a given chemical

process control situation ?

Feedback control cascade ? override ? Feedforward control feedback trim ? Inferential control

Chen CL 19


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Design the Controller: (control law)

Q 5: How is the information, taken from the measurements, used to

adjust the values of the manipulated variables ?

Control law (controller structure, P - PI - PID ?)

Controller tuning (Kc, I, D ?)

Chen CL 20


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Why Laplace Transform

-

Ex: PID Controller 4 signals 2 signals

PID Controller: u(t) = Kc

e(t) + 1

TI

t0

e()d + TDde(t)

dt

+ ub

Steady States: u = Kc

e + 1

TI

t0

e d + TDde

dt

+ ub [u = ub; e = 0]

Deviation Variables: U(t) = Kc

E(t) + 1

TI

t0

E()d + TDdE(t)

dt

U(t) u(t) ub; E(t) = e(t) 0)

Chen CL 21


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-

PID Controller: (cont)

Laplace Transform: U(s) = Kc

E(s) + 1

TI

E(s)

s+ TDsE(s)

= Kc 1 + 1TI

1s

+ TDsE(s)

Transfer Function:U(s)

E(s)= Kc

1 + 1

TI

1s

+ TDs

Gc(s)

Chen CL 22


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-

Ex: Simple Process 3 signals 2 signals

First-Order Model: Tdy(t)dt

+ y(t) = Ku(t d)

Steady States: Tdydt

+ y = Ku

Deviation Variables: Td[y(t)y]dt

+ [y(t) y] = K[u(t d) u]

TdY(t)

dt+ Y(t) = KU(t d)

Laplace Transform: T sY(s) + Y(s) = KU(s)e

ds

Transfer Function:Y(s)

U(s)=

Keds

T s + 1 Gp(s)

Chen CL 23


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- Dynamic relation (ysp to y): time-domain vs. s-domain

Time domain: simultaneous dynamic equations

S-domain:y(s)

ysp(s)=

GcGp1 + GcGp


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Sensors and Transmitters

(Transducer)



Chen CL 1


30/921

Transducer

- Transducer:

to convert a physical quantity into an electrical signal

The sensor produces a phenomenonmechanical,

electrical, or the likerelated to the process variable

it measures The transmitter converts this phenomenon into a

signal that can be transmitted

- Measured Quantities:

position, force, velocity, acceleration,

pressure, level, flow, temperature,

- Output Signals:

current, voltage, resistance, capacitance, or frequency

Chen CL 2


31/921

Transducer Specifications

-

Static Specifications: Accuracy

Resolution

Repeatability

Hysteresis

Linearity

- Dynamic Specifications

Dead time Rise time

Time constant

Frequency response

Chen CL 3


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AccuracyPercent of Full Scale Output (%FPO)

Example:

A load cell is a transducer used to measure weight (100 kg 20

mV). A calibration record is given in table. Plot calibration curve

(skip). Determine the accuracy of the transducer. Express answer inboth %FSO (percent of full scale output) and % of reading. Assume

linear relationship between output.

vtrue =vfull scale

loadfull scale load

=20 mV

100 kg load = 0.2

mV

kg

load

Chen CL 4


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True output Actual output Error Accuracy

(mV) (mV) (mV) %FSO %reading

0 0 0.08 0.08 0.40 -

5 1.00 0.45 0.55 2.75 55.00

10 2.00 1.02 0.98 4.90 49.00

15 3.00 1.71 1.29 6.45 43.00

20 4.00 2.55 1.45 7.25 36.25

25 5.00 3.43 1.57 7.85 31.40

30 6.00 4.48 1.52 7.60 25.33

35 7.00 5.50 1.50 7.50 21.43

40 8.00 6.53 1.47 7.35 27.01

45 9.00 7.64 1.36 6.80 15.11

50 10.00 8.70 1.30 6.50 13.00

55 11.00 9.85 1.15 5.75 10.45

60 12.00 11.01 0.99 4.95 8.25

65 13.00 12.40 0.60 3.00 2.77

70 14.00 13.32 0.68 3.40 7.14

75 15.00 14.35 0.65 3.25 4.33

80 16.00 15.40 0.60 3.00 3.75

85 17.00 16.48 0.52 2.60 3.06

90 18.00 17.66 0.34 1.70 1.89

95 19.00 18.90 0.10 0.50 0.53

100 20.00 19.93 0.07 0.35 0.35

Chen CL 5


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Resolution

This optical encoder has a 90o resolution

Chen CL 6


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Resolution: Example

A 2.5-m-long vane is rotated slowly in a circle. The motor and gears

attach to the vane at its center. It is necessary to know the position

of the vane within 2 cm. What must be the resolution of the optical

encoder attached to the shaft that positions the vane ?

Solution:

c = d = (2.5 m) = 7.854 m = 785.4 cm

arc

360o=

2 cm

785.4 cm

arc =(360o)(2 cm)

785.4 cm= 0.917o

360o

0.917o= 392.6 ( 400 500) pulses per revolution

Chen CL 7


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Repeatability

accurate repeatable both accurate

not repeatable not accurate and repeatable

Chen CL 8


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Hysteresis ( Increasing= Decreasing)


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Chen CL 10

D d i


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Dead Time

Chen CL 11

Ri Ti


40/921

Rise Time

Chen CL 12

Ti C


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Time Constant

Chen CL 13

S li Ti


42/921

Settling Time

Chen CL 14

F R


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Frequency Response

Chen CL 15

Bl k Di f A T d


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Block Diagram of A Transducer

H(s) =C(s)

PV(s)

=K

T

Ts + 1K

T=

(20 4) mA

(200 0) pisg= 0.08 mA/pisg

KT

=(100 0) %TO

(200 0) pisg= 0.5 %TO/pisg

Chen CL 16

P iti T d


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Position TransducerLinear and Angular Potentiometers

Chen CL 17

Li d A l P t ti t E l


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Linear and Angular Potentiometers: Example

It is necessary to measure the position of a panel. It moves 0.8 m.

Its position must be known within 0.1 cm. Part of the mechanismwhich moves the panel is a shaft that rotates 250o when the panel is

moved from one extreme to the other. A control potentiometer has

been found which is rated at 300o full-scale movement. It has 1000

turns of wire. Can this be used ?

Solution:

300o

1000

= 0.300o (resolution of the potentiometer)

250o

0.8 m= 312.5o/m or 3.125o/cm (conversion of shaft)

0.1 cm 3.125o/cm = 0.3125o (panel required resolution)

> 0.300o

the potentiometer will work

Chen CL 18

P iti T d


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Position TransducerLinear Variable Differential Transformer (LVDT)

One primary coil,

two secondary coils,one free-moving rod-shaped

magnetic core inside coil

assembly

Chen CL 19

F T d


48/921

Force TransducerBounded Resistance Strain Gage

Stress: =force

area=

F

A

Strain: = deformationlength

= LL

=

Chen CL 20

Fo ce T a sd ce


49/921

Force TransducerBounded Resistance Strain Gage

Stress Strain Electrical Resistance

Gage Factor, GF = fractional change in resistancefractional change in length =R/R

L/L

Chen CL 21

Example:


50/921

Example:

A strain gage is bounded to a steel beam which is 10.00 cm long and

has a cross-sectional area of 4.00 cm2. Youngs modulus (E = /)

of elasticity for steel is 20.7 1010 N/m2. The strain gage has a

nominal (unstrained) resistance of 240 and a gage factor of 2.20.

When a load is applied, the gages resistance changes by 0.013.

Calculate the change in length of the steel beam and the amount of

force applied to the beam.

Chen CL 22

Solution:


51/921

Solution:

L =1

GF L

R

R

= 12.20

0.1(m) 0.013240

= 2.46 106 m

F

A = = E = EL

L

= E1

GF

R

R

= 20.7 1010(N/m2)1

2.20

0.013

240 F = 2.037 103 N

1 b

4.482N

= 454 b

Chen CL 23

Force Transducer


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Force TransducerTwo Strain Gages

Chen CL 24

Pressure Transducer


53/921

Pressure TransducerDiaphragm with Strain Gage

Chen CL 25

Pressure Transducer


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Pressure TransducerDiaphragm with Strain Gage

Chen CL 26

Pressure Transducer


55/921

Pressure TransducerCapacitive Absolute-Pressure Transducer

Chen CL 27

Pressure Transducer


56/921

Pressure TransducerPressure-to-Displacement Transducer

Capsule, Bellow, Bourdon tube, Spiral, Helix

Chen CL 28

Pressure Transducer


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Pressure TransducerLVDT Sensing Pressure Transducer

Chen CL 29

Pressure Transducer


58/921

Pressure TransducerPotentiometer Sensing Pressure Transducer

Chen CL 30

Flow Transducer


59/921

Flow TransducerDifferential Pressure Obstruction Flow Sensors

Orifice plate, Venturi, Pitot tube

f = CoAo

po

1

dD

4

Chen CL 31

Flow Transducer


60/921

Flow TransducerDeflection-type Flow Transducers

Cantilever beam, Variable-area rotameter

Chen CL 32

Flow Transducer


61/921

Flow TransducerSpin-type Flow Transducers

Paddle wheel, Flow turbine

Chen CL 33

Flow Transducer


62/921

Flow TransducerElectromagnetic Flow Meter

Chen CL 34

Flow Transducer


63/921

Flow TransducerUltrasonic Flow Transducer

Chen CL 35

Flow Transducer: Characteristics


64/921

Flow Transducer: Characteristics

Chen CL 36

Level Transducer


65/921

Level TransducerDiscrete-type Level Transducer

Float switch, Photoelectric

Chen CL 37

Level Transducer


66/921

Level TransducerBy Sensing Pressure with

Offset transducer, Sealed tank

Chen CL 38

Level Transducer


67/921

Level TransducerBy Sensing differential Pressure

Chen CL 39

Level Transducer


68/921

Level Transducerwith Float-driven Control System

Chen CL 40

Level Transducer


69/921

Level TransducerCapacitance Level Transducer

Chen CL 41

Level Transducer


70/921

Level TransducerTop-mounted Ultrasonic Level Sensor

Chen CL 42

Common Temperature Transducers


71/921


Chen CL 43



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p

Chen CL 44



73/921

p

Chen CL 45

Thermocouple


74/921

pSeebeck Effect

- Current in a closed circuit

- Voltage across an open circuit

Chen CL 46

Thermocouple


75/921

pEquivalent Circuits

Chen CL 47

Thermocouple


76/921

pwith External Reference

Chen CL 48

Thermocouple


77/921

pExternal Reference Junction with No Ice Bath

Chen CL 49

Thermocouple


78/921

pScanning Many Thermocouples

Chen CL 50

Thermocouple


79/921

pElectronic Ice-point Compensation


80/921

Chen CL 1

Why Focus on Valve ?


81/921

- Valves are often the least understood component of the

control loop

-

Valves are often the most neglected component of thecontrol loop

-

Valves are often the biggest contributor to poor controlloop performance

Chen CL 2

A Typical Globe Valve


82/921

Chen CL 3

Energy and Pressure Grades


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Ideal/Real Flows Across An Ideal Restriction

Chen CL 4

Energy and Pressure Grades


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A Valve or Orifice Is Not Ideal

Chen CL 5

Inception of Cavitation in An Orifice


85/921

Chen CL 6

Efect of Vaporization on Flow Rate


86/921

Chen CL 7

Energy and Pressure Grades Across A Valve


87/921

Chen CL 8

Valves Cost Energy


88/921

A control valve is simply an orifice with a variable area of flow

all control valves and regulators cost energy

Chen CL 9

A Full Speed Pump and Throttling Valve Cost Energy


89/921

Chen CL 10

Variable Speed Drive Saves Energy


90/921

Chen CL 11

Affinity Laws


91/921

Const Impeller Diameter:Q1Q2 =

N1N2

H1H2

=

N1N2

2bhp1bhp2 =

N1N23

Const Pump Speed:Q1Q2

=

D1D2H1

H2=

D1D2

2bhp1bhp2

=

D1D2

3

Chen CL 12

- Efficiency remains virtually constant for changes in speed

d ll i ll di h


92/921

and small impeller diameter changes

-

A 10% speed reduction (90% of nominal speed)

capacity = 90% of original operating conditions

head = 81% of original required head

bhp = 73% of nominal brake horsepower

Chen CL 13

Comparative Performances of Valve and


93/921

Variable Speed DriveItem

Control

Valve

Variable Speed

Drive

Ease of Installation - -

Equipment Efficiency - Better

Motor Efficiency - Better

Power Factor - Better

Operating Costs - Better

Flexibility of Location - Better

Exposure to Process - Better

Specification - Better

Ability to Control - Better

Potential for Leaks - Better

Installed Cost: small - Better

large Better -

Shutoff Capability Better -

Maintenance: expertise Better -

valve/drive - Better

equipment - Better

spare parts - -

Why Control Valve ?

Chen CL 14

Important Issues in Control Valves


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- Types

- Actions: air-failure-open ( FO), air-failure-close ( FC)

- Sizing

- Characteristics: inherent (manufactured) and installed

- Valves in common loops: flow, temperature, level,

pressure

- Software characterizer

- Diagnosis

Chen CL 15

Types of Control ValvesGl b V l


95/921

Globe Valve

A Single-Ported Two Way Valve


96/921

Chen CL 17



97/921

Globe Valve

An Angle Valve and A Y-Pattern Valve

Chen CL 18



98/921

Globe Valve

Valves with Three-Way Bodies

Chen CL 19



99/921

Globe Valve

Anti-Cavitation Valve

Chen CL 20

Types of Control ValvesB ll V l


100/921

Ball Valve

Full Ball Valve

Chen CL 21

Types of Control ValvesB ll V l


101/921

Ball Valve

Three-Way Ball Valve

Chen CL 22

Types of Control ValvesBall Val e


102/921

Ball Valve

Segmented Ball Valve


103/921

Chen CL 24

Types of Control ValvesDi h V l


104/921

Diaphragm Valve

Chen CL 25

Types of Control ValvesDi i l V l


105/921

Digital Valve

- A digital valve is made up of a group of on-off valve elements

installed in a common body

- Each valve element has a different capacity sequence of sizes form a binary series

- These valves have the capability to change from one capacity to

another instantaneously

Chen CL 26

- Example:

6 valve elements with capacity ratios 1 2 4 8 16 32


106/921

6 valve elements with capacity ratios 1, 2, 4, 8, 16, 32

Each incremental step will vary by1

63(1 + 2 + + 32 = 63)

Ratio of maximum to minimum capacity will be 63 : 1

8 elements 1, 2, . . . , 128 255 : 1

- Limited to use with clean fluids and moderate temperatures

Chen CL 27

Digital Valve Operation


107/921

- A collection of individual on-off valves arranged in parallel

- All elements have different flowrates

- All elements are controlled by pneumatic or electrical means upon

command from a computer/controller

- By adjusting the combination of open and closed valves

= desired flowrate: quick, exact, high resolution

- Digital Valve: adds or subtracts fixed area orifices responds quickly

- Analog Valve: seeks a new orifice (opening) size

responds slowly with overshoot

Chen CL 28

- IF increasing number of individual on/off valves, and

valve openings are sized in a binary progression (1-2-4...)


108/921

p g y p g ( )

(doubling area openings doubling flowrates)

THEN flow range in increased substantially

- EX: 6-bit = range - 63 : 1; resolution - 163EX: 20-bit =

range - 1, 048, 575 : 1; resolution - 1

1,048,575

Chen CL 29

Advantages of Digital Valves- Fast Res o se


109/921

- Fast Response

Analog Valve: 3

8 sec.

Digital Valve: 50 100 msec. (milli-sec) Instantaneous response from zero flow to full flow Same speed from 1 to 2, and 0 to full

- Wide Rangeability

Analog valve: far less than 100 to 1

Digital valve: 7-bit

exceeds 100 to 1

Chen CL 30

- Precise Repeatability

no position error absolutely repeatable


110/921

- High Resolution

Analog valve: 1% of maximum flow

Digital valve:

6-bit: r = 1 part in 63 (1.59%)

8-bit: r = 1 part in 255 (0.39%)

10-bit: r = 1 part in 1,023 (0.09%) 20-bit: r = 1 part in 1,048,575 (0.0001%)

- Flow Measurement Capability

- Computer Compatibility

Chen CL 31

A Globe Valve again


111/921

Chen CL 32

Overview


112/921

- Some Descriptive Terms:

single-seated, air-actuated, spring-opposed,fail-closed with a plug-and-seat trim

Air pressure under the diaphragm causes the stem to rise

As stem rises, area of opening between valve plug and seat

increases Area of opening is maximum at maximum seat position

Pressure drop across valve and valve opening determine flow rate

through valve

Chen CL 33

- Valve Coefficient: Cv (gal/psig) (Cvmax ?)

flow rate of water through the valve


113/921

flow rate of water through the valve

in US gallons per minute at 70oF

when p across the valve is 1 psig(function of valve opening)

- Flow of Liquid Through Valve:

F = Cv

p

= Cvmaxf(m)

p

F: flow rate, US gpm (or, q)

p: differential pressure, psig

: specific gravity of fluid, relative to water at 70oF

m%: 0 100%; or vp: valve position, 0 1

Chen CL 34

- Note: instrumentation schematic for a control valve


114/921

Chen CL 35

Action of Control ValvesSafety Consideration


115/921

Safety Consideration

- SAFETY is the only consideration in selecting the

action of the control valve

- Control valve action directly affects the action of the

feedback controller

Air-Failure-Close (AFC) Air-Failure-Open (AFO)

orAir-to-Open (ATO) Air-to-Close (ATC)

vp =m

100 vp = 1 m

100

Chen CL 36

Example:


116/921

Chen CL 37

Sizing of Control ValveA Basic Trade off Problem


117/921

A Basic Trade-off Problem

Process Engineer: bigger valve lower pv smaller pump lower operating cost

lower operability !!Control Engineer: smaller valve larger pv

larger pump

higher operating cost

higher operability

- Design Logic:

to design (select) the valve and the pump on having

a process that can obtain specified qmin, qmax

Chen CL 38

Sizing of Control ValveSelect A Larger Valve


118/921

Select A Larger Valve

- Larger Valve

q = 100 gpm, ph = 40 psi, f(x) = 0.5

p1 = pressure after pump

p2 = 150 psi, pressure at system output

pt = p1 p2 = f(q) (constant)let q = q

minat f = 0.1

Chen CL 39

- Select one valve such that pv = 20 psi


119/921

p1 = 150 + 40 + 20 = 210 psi pumpq = C

vmaxf(x)pv/G

Cvmax =100

0.5

201

= 44.72 gpm/

psi valve

assume ph = 40 q100

2

pv = p1 p2 ph= 210 150 40 q1002

q(x) = 44.72 f(x)210 150 40q(x)1002

Chen CL 40

f ( ) 1 44 72 1

210 150 40qmax2


120/921

f(x) = 1 qmax = 44.72 1

210 150 40 qmax100 qmax = 115 gpm

f(x) = .1 qmin

44.72 .1

210 150 40 qmin100 2q

min= 33.3 gpm

qmaxq

min

=115

33.3= 3.46 (turndown ratio)

Chen CL 41

Sizing of Control ValveSelect A Small Valve


121/921

- Smaller Valve

q = 100 gpm, ph = 40 psi, f(x) = 0.5

p1 = pressure after pump

p2 = 150 psi, pressure at system output

pt = p1 p2 = f(q) (constant)let q = q

minat f = 0.1

Chen CL 42

- Select one valve such that pv = 80 psi


122/921

p1 = 150 + 40 + 80 = 270 psi

pump

q = Cvmaxf(x)

pv/G

Cvmax =100

0.5

801

= 22.36 gpm/

psi valve

assume ph = 40 q1002 pv = p1 p2 ph

= 270 150 40 q1002 q(x) = 22.36 f(x)270 150 40q(x)1002

Chen CL 43

f(x) = 1 qmax = 22.36 1

270 150 40 qmax100 2q = 141 gpm


123/921

qmax = 141 gpm

f(x) = .1 qmin

22.36 .1270 150 40 qmin1002q

min = 24.2 gpm

qmaxqmin

= 14124.2 = 5.83 > 3.46

Chen CL 44

Control Valve Capacity


124/921

- Cv coefficient: flow in US gpm of water that flowsthrough a valve at a pressure drop of 1 psi across valve

- Liquid:

q = CvpvGf

= Cvmaxf(vp)pvGf

(liquid flow, US gpm)

pv : pressure drop across valve, psi Gf : specific gravity

w = qgal

min 60min

h 8.33Gflb

gal = 500CvGfpv lb/h

Chen CL 45

- Compressible Flow: (Masoneilan Inst. Inc.)


125/921

qs = 836CvCfp1GT

(y 0.148y3)

w = 2.8CvCfp1

G520T (y 0.148y3) gas or vapor1.83CvCf

p1(1+0.0007T

SH)(y 0.148y3) steam flow

y =1.63

Cf pvp1

qs : gas flow, scfh (ft3/h at standard conditions of 14.7 psia, 60oF)

G : gas specific gravity w.r.t. air, = M W/29

T : temperature at valve inlet, oR =o F + 460

Cf : critical flow factor (0.6 0.95)p1 : pressure at valve inlet w : gas flow, lb/h

TSH

: degree of superheat

p = p1

p2 pressure drop across valve

Chen CL 46


126/921

Chen CL 47

- Compressible Flow: (Fisher Controls)


127/921

qs = Cg502GTp1 sin

59.64C1

pvp1 rad

- Note: the above sine function is basically the same

function of y

Chen CL 48


128/921

Chen CL 49


129/921

Chen CL 50

Control Valve Capacity: Example


130/921

From Fig. C-10-1a, a 3-in. Masoneilan valve with full trim has a

capacity factor of 110 gpm/(psi)

1/2

when fully opened. The pressuredrop across the valve is 10 psi.

(a) Calculate the flow of a liquid solution with density 0.8 g/cm3

(1.0 g/cm3 for water).

q = 110

100.8 = 389 gpm

w = 500(110)

(0.8)(10) = 155, 600 lb/h

Chen CL 51

(b) Calculate the flow of gas with average MW of 35 when valve inlet

conditions are 100 psig and 100oF.


131/921

G = 35/29 = 1.207 p1 = 100 + 14.7 = 114.7 psia

T = 100 + 460 = 560oR Cf = 0.9

y = 1.630.9

10

114.7 = 0.535

qs = 836(110)(0.9)114.7

(1.207)(560)[0.535 0.148(0.535)3]

0.512= 187, 000 scfh

w = 2.8(110)(0.9)(114.7)

1.207520560(0.512) = 17, 240 lb/h

Chen CL 52

(c) Calculate the flow of gas from part (b) when the inlet pressure is

5 psig. Calculate the flow both in volumetric and mass rate units,

and compare the results for a 3 in Fisher Control valve


132/921

and compare the results for a 3-in. Fisher Control valve.

p1 = 5 + 14.7 = 19.7 psia

y = 1.290 [y 0.148y3] = 0.972qs = 836(110)(0.9)

19.7(1.207)(560)

(0.972) = 61, 000 scfh

= 5, 620 lb/h

Fisher: qs = 4280

520(1.207)(560)(114.7)sin

59.6435.7

10

114.7

= 204, 000 scfh (9% higher)

qs = 4280

520(1.207)(560)(19.7) sin

59.6435.7

1019.7

= 68, 700 scfh (13% higher)

Chen CL 53

Sizing of Control Valves: Example

A t l l i t l t th fl f t i t di till ti


133/921

A control valve is to regulate the flow of steam into a distillation

column reboiler with a design heat transfer rate of 15 million Btu/h.

The supply steam is saturated at 20 psig. Size the control valve fora pressure drop of 5 psi and 100% over-capacity.

H = 930 Btu/lb latent heat of cond., from steam table

qs = 15, 000, 000/930 = 16, 130 lb/hp1 = 20 + 14.7 = 3 4.7 (valve inlet pressure)

Cf = 0.8

y = 1.63

0.8534.7 = 0.773 y 0.148y3 = 0.705Cv = f

Gf

pv= 16,130(1.83)(0.8)(34.7)(.705) = 450

gpmpsi

Cvmax = 2.0Cv = 900gpm

psi

Cv = 1000 (select a valve with this Cv)

Chen CL 54

Fisher: G = 18/2 9 = 0.621 C1 35

q (16 130)(380)/18 341 000 scfh (380 scf/lbmole)


134/921

qs = (16, 130)(380)/18 = 341, 000 scfh (380 scf/lbmole)

sin 59.6435 534.7 = sin(0.647) = 0.603Cg =

341,000520

(0.621)(710)(34.7)(0.603)

= 15, 000

Cg = 30, 000 Cv = Cg/C1 = 30, 000/35 = 856

gpmpsi

Chen CL 55

Sizing of Control Valves: ExampleThe following figure shows a process for transferring an oil from a


135/921

storage tank to a separation tower. The tank is at atmospheric

pressure, and the tower works at 25.9 in.Hg absolute (12, 7 psia).Nominal oil flow is 700 gpm, its specific gravity is 0.94, and its

vapor pressure at the following temperature of 90oF is 13.85 psia.

The pipe is 8-in. Schedule 40 commercial steel pipe, and the

efficiency of the pump is 75%. Size a valve to control the flow of oil.

From liquid flow correlations, the frictional pressure drop in the line

is found to be 6 psi.

Chen CL 56

Note: the liquid may flash if we place valve at entrance of tower

(12.7 psia < 13.85 psia)

Place valve at pump discharge: hydrostatic pressure is


136/921

Place valve at pump discharge: hydrostatic pressure is

12.7 + (62.3 lb/ft2

)(0.94)(60 ft)/(144 in2/ ft2) 24.4 psia

= 37.1 psia

Select pv = 5 psi

annual oper cost =700 gal1 min

1 ft3

7.48 gal5 lbf1 in2

144 lbf1 ft2

1 kW-min44250 ft-lbf

8200 h1 yr

$0.031 kW-h

10.75

= $500/yr

Cvmax = 2(700)0.945 = 607 gpmpsi select an 8-in. Masoneilan valve with Cv = 640

Chen CL 57

Valve Inherent Characteristics

( )


137/921

- Inherent Characteristics: q(x)

qmaxpv=c

q(x) = Cvmax f(x) Cv

pvG x : 0 1 (= vp)qmax = Cvmax 1

pv

G(valve full open)

q(x)

qmax pv=c = f(x) (ratio of flow area)=

x linear

Rx1 equal percentage

quick opening

Chen CL 58



138/921

Chen CL 59



139/921

- Note: equal percentage ?

df(x)

dx= kf(x)

df(x)f

= kdx

a + ln[f(x)] = kx

a = k (x = 0 f = 0; x = 1 f = 1)

ln[f(x)] = k(x 1)f(x) = ek(x1) = Rx1

Chen CL 60

Valve Installed Characteristics


140/921

- Installed Characteristics: q(x)

qmax pt=cq(x) = Cvmax f(x)

pv

Gx : 0 1

= Cvmax f(x)pt ph

G

= Cvmax

f(x)

pt pmaxh q(x)qmax

2

G

qmax = Cvmax 1

pt pmaxh 1G

Chen CL 61



141/921

q(x)

qmax

pt=c

= f(x)

pt pmaxh q(x)qmax2pt pmaxh

q(x)qmax pt=c = f(x)pt

pt [1 f2(x)] pmaxh > f(x)

= f(x)1

1 [1 f2(x)]

pmaxh

pt = f(x)

11 [1 f2(x)](1 )

=f(x)

+ (1 )f2(x)

Chen CL 62



142/921

where = pvptq(x)max =

pvpt

qmax

=

= pvptqnote:

q(x)

qmax

pt=c

=f(x)

1 [1 f2(x)] 1

qmaxq

Chen CL 63

Valve Installed Characteristics again


143/921

pL = kLGfq2 kL =

pLGfq

2

pv = Gfq2

C2v(Cv = Cvmaxf(x))

po = pv + pL =

1C2v

+ kL

Gfq2

q = Cv1+kLC

2v

p0Gf

qmax =

Cvmax

1+kLC2vmaxp0Gfq

qmax

po=c

=Cv

Cvmax

f(x)

1+kLC2vmax

1+kLC2v

Chen CL 64

Valve Installed Characteristics: Example

F l l fi d h i fl h h h l h


144/921

For last example, find the maximum flow through the valve, the

installed flow characteristics, and the rangeability of the valve.

Assume both linear and equal percentage characteristics with

rangeability parameter of R = 50. Analyze the effect of varying the

pressure drop across the valve at nominal flow.

Chen CL 65

kL =6 psi

(0.94)(700 gpm)2= 13.0 106 psi

(gpm)2


145/921

( )( g ) (g )

po = pv + pL = 5 + 6 = 11 psi (constant)

qmax = 6401+(13.0106)(640)2

110.94 = 870 gpm (< 2 700)

Linear:

q.95 =(640)(0.95)

1+(13.0

106)(640)2

110.94 = 862 gpm

q.05 = (640)(0.05)1+(13.0106)(640)2

110.94 = 109 gpm

rangeability = 862109 = 7.9 (inherent range =.95.05 = 19)

Equal %:

q.95 = (640)(500.95

1

)1+(13.0106)(640)2

110.94 = 839 gpm

q.05 =(640)(500.051)

1+(13.0106)(640)2

110.94 = 53.2 gpm

rangeability = 83953.2 = 15.8 (inherent range =500.951

500.051 = 34.8)

Chen CL 66

valve pressure drop, psi

2 5 10


146/921

total pressure drop 8 11 16

calculated Cvmax 960 607 429

required valve size 10-in. 8-in. 8-in.

actual Cvmax 1000 640 640

maximum flow, gpm 779 870 1049linear rangeability 5.4 7.9 7.9

Equal % rangeability 10.8 15.8 15.8

Chen CL 67

Installed flow characteristics:

(a) linear inherent characteristics.

(b) equal percentage characteristics with = 59


147/921

( ) q p g

Chen CL 68

Application I: Fluid TransferSelf-Regulated Processes


148/921

Chen CL 69

Application I: Fluid TransferLiquid Transferred by Pressure Difference


149/921

Chen CL 70

Application I: Fluid TransferLiquid Transferred by Pressure Difference


150/921

Chen CL 71

Application I: Fluid TransferA Flow-Controlled Pump


151/921

Chen CL 72

Application I: Fluid TransferSteam Ejector


152/921

Chen CL 73

Application II: Heat TransferSteam Heaters and Dryers


153/921

Chen CL 74

Application II: Heat TransferHeat Exchangers


154/921

Chen CL 75

Application II: Heat TransferAntifreeze Applications


155/921

Chen CL 76

Application III: Chemical ReactionsA pH Control System


156/921

Chen CL 77

Application III: Chemical ReactionsCombustion Processes


157/921


158/921

Proportional-Integral-Derivative


159/921

PID Controllers

Cheng-Liang Chen

PSELABORATORYDepartment of Chemical Engineering

National TAIWAN University

Chen CL 1

Outline

- Proportional Integral Derivative Controller:


160/921

- Proportional-Integral-Derivative Controller:

PID Control based on CurrentFuture Error with ConstantReset Bias



PIDControl based on CurrentFuture Error with ConstantReset Bias




Chen CL 2

- Steady Offset of PID when using Constant bias


161/921

-Series PID Parallel PID

- Response of PID Controllers to Typical Inputs

-

Operational Aspects of PID Controllers problems of D action: sensitivity to noise

problems of I action: moving PB and reset windup

manual control requirement

bump-less transfer

Chen CL 3

PID Controller: A Survey

More Than 95% of Controllers Are of PID Type


162/921

- Bialkowski (1993): paper mills in Canada

A typical mill has more than 2000 control loops

97% of loops use PI control

Only 20% of control loops were found to work welland decrease process variability

Reasons for poor performance were Poor tuning (30%)

Valve problems (30%)

Others (20%): sensor, bad sampling rates

Chen CL 4

PID Controller: A Survey

More Than 95% of Controllers Are of PID Type


163/921

- Ender (1993)

30% of installed process controllers operate in manual

20% of loops use default parameters (factory tuning)

30% of loops function poorly because of equipmentproblems (valves, sensors )

Chen CL 5

Why PID Controllers Are So Popular ?

- St t i l


164/921

- Structure: simple

Easy for understanding

A few adjustable parameters, easy for tuning

- Performance: good or acceptable level

- Robustness: strong to uncertain operating conditions

- Applicability: wide to different processes/industries

Chen CL 6

A Heat Exchanger with Feedback ControlValve: AFC (ATO); Controller: Reverse Action


165/921

Chen CL 7

Simplest Controller: On-Off Control


166/921

u(t) =

ub + u if y(t) < ysp d

ub u if y(t) > ysp + d

- Problem: oscillatory response !On-off cares about control direction only;

On-off gives same control action for different error magnitudes

- Solution: control action considering error magnitude

Chen CL 8

Current-Error-Based P Controlwith Constant Bias (SS value)


167/921

current control corrective action, u(t) current error magnitude, e(t) ysp y(t)

u(t) e(t) ysp y(t)

u(t) = Kc e(t) p(t) (Kc: adjustable proportionality) u(t) = Kc e(t)

u(t)

+ ub

SS value

for e(t) = 0

Chen CL 9

Problem of Current-Error-Based P Control

u(t) = Kc e(t) + ub


168/921

( ) c ( ) u(t)+ bSS valuefor e(t) = 0

- Too late to correct current error

- It takes time to know effect of control on output

conservative action to guarantee stability limited achievable control performance

Chen CL 10

Solution: Future-Error-Based P ControlBack to the Future


169/921

Control action based on future condition (error) take control action in advance better control performance

u(t) = Kc e(t+Td) u(t)

+ ubSS value

for e(t) = 0

u(t) = Kc e(t+Td) u(t)

+ ubSS value

for e(t) = 0

Chen CL 11

Future-Error-Based P ControlGives Better Performance

HE: pulse decrease in input temperature


170/921


- At t1: e(t1 + Td) > e(t1) P control based on e(t1 + Td) has larger action more steam smaller undershoot in T

( Pre-Act P) Kc e(t1 + Td) + ub > Kc e(t1) + ub

( Normal P)

Chen CL 12

Future-Error-Based P ControlGives Better Performance



171/921


- At t2: e(t2 + Td) < e(t2) P control based on e(t2 + Td) has smaller action less steam smaller overshoot in T

( Pre-Act P) Kc e(t2 + Td) + ub < Kc e(t2) + ub

( Normal P)

Chen CL 13

Implement Future-Error-Based P ControlAs Current-Error-Based PD Control


172/921

Original P: u(t) = Kc e(t) + ub

PreAct P: u(t) = Kc e(t + Td) + ub

Kc

e(t) + Td

de(t)

dt

e(t)e(t+Td)

+ ub

Chen CL 14


173/921

u(t) = Kc e(t + Td) + ub

Kc e(t) + Tdde(t)

dt e(t)e(t+Td)

+ ub

Chen CL 15

Sub-SummaryP control based on current error with constant bias


174/921

e(t) e(t + Td)

P control based on future error with constant bias

u(t) = Kce(t + Td) + ub

Kc

e(t) + Td

de(t)

dt

e(t)

+ub

PD control based on current error with constant bias

Chen CL 16

Steady Offset for P (PD) Control

- Non zero Steady State Offset (I): Setpoint Change


175/921

- Non-zero Steady State Offset (I): Setpoint Change

Initial steady state: (all variables [0, 100%])

y = ysp = y e = 0 u = ub = u =

Chen CL 17

Case I: Setpoint change ysp : y y + New steady state: y =?, u =?, e =?

(change in y) = K {change in process input}


176/921

(change in y) = Kp {change in process input}

= Kp {(change in CO) + (change in load)}

y y = Kp

u u

change in CO

+

change in load=0

= Kp

Kc

e (y + ) new sp

y

+ub u

u + =0

y = y + KcKp1 + KcKp

new steady state

= y + desired sp

Chen CL 18

- Non-zero Steady State Offset (II): Load Change


177/921

Initial steady state: (all variables [0, 100%])

y = ysp = y e = 0 u = ub = u =

Chen CL 19

Case II: Load change : + New steady state: y =?, u =?, e =?

(change in y) K {change in process input}


178/921

(change in y) = Kp {change in process input}

= Kp {(change in CO) + (change in load)}

y y = Kp

u u

change in CO+

change in load=

= Kp

Kc

e (y)

sp

y

+ub

uu +

y = y + Kp1 + KcKp

new steady state

= ydesired sp

Chen CL 20

Current-Error-Based P Controlwith Reset Bias


179/921

- Question:using P or PD control with constant bias,

how to guarantee zero steady-state offset ?

(at steady state: e = 0, y = ysp)

- Answer: u = ub at SS

desired: e = 0 (y = ysp) at SS

u = ub at SS


180/921

Chen CL 22

Reset Bias for Zero SS OffsetFeasible Method


181/921

-

Feasible method to guarantee u = ub at SS:desired: ub = u at SS

simplest method: ub(t) equals u(t) t

feasible method: ub(t) tracks u(t) t

- ub(t) tracks u(t) with first-order dynamics:

(use Ti to adjust tracking velocity)

Tidub(t)

dt+ ub(t) = u(t)

equal

tracking

Chen CL 23

P control based on current error with reset biasvia first-order dynamics


182/921

Chen CL 24

Reset Bias Integral Action


183/921

(1) u(t) = Kce(t) + ub(t) (P with reset bias)

(2) u(t) = Tidub(t)

dt + ub(t) (ub(t) tracks u(t))

Tidub(t)

dt= Kce(t) = p(t)

ub(t) ub =1

Ti

t0

p(t)dt

change in bias

=KcTi

t0

e(t)dt

=ub(t)

Chen CL 25

(1) u(t) = Kce(t)p(t)

+

ub(t) 1

Ti

t0

p(t)(t)dt +ub (PI)


184/921

p(t) ub(t)= Kce(t) +

KcTi

t0

e(t)dt + ub

= Kc

e(t) + 1Ti

t0

e(t)dt

+ ub

- Reset bias: interpreted as integral action

current-error-based P control with reset bias

(reseting bias according to integral of error)

= current-error-based PI control with constant bias

Chen CL 26

Reset Bias Integral Action


185/921

Chen CL 27


186/921

Chen CL 28

Future-Error-Based P Control with Reset Bias

- P control based on future error with reset bias


187/921

PD PI Interpreted As PD +PI (series PID ) Controller

e(t + Td) e(t) + Tdde(t)

dt

e(t)

Chen CL 29

ub(t) =1

Ti

Kce

(t)dt + ub

ub(t)


188/921

u(t) = Kce(t) p(t)

+

b( ) 1TiKce(t)

p(t)

dt +ub

ub(t)

Chen CL 30

Series PID Parallel PID


189/921

Kpc = Kc

1 + TdTi

Kc = Kpc

0.5 +

0.25

Tpd

Tpi

T

p

i = Ti 1 + TdTi Ti = Tpi 0.5 + 0.25 Tpd

Tpi Tpd = Td/

1 + TdTi

Td = T

pd /

0.5 +

0.25

Tpd

Tpi

Chen CL 31

Series PID Parallel PID


190/921

p(t) = Kc

e(t) + Td

de(t)dt

u(t) = p(t) + ub(t)

Tidub(t)dt + ub(t) = u(t)

P(t) = Kc

E(t) + Td

dE(t)dt

U(t) = P(t) + Ub(t)

TidUb(t)dt + Ub(t) = U(t)

P(s) = Kc (E(s) + TdsE(s)) = Kc(1 + Tds)E(s)

U(s) = P(s) + Ub(s)

TisUb(s) + Ub(s) = U(s) Ub(s) =1

Tis+1U(s)

Chen CL 32

U(s) = P(s) + Ub(s) = P(s) +1

Tis+1U(s)

U(s) = Tis+1Tis P(s)

Tis+1K (1 + T )E( ) ( i PID)


191/921

= Tis+1TisKc(1 + Tds)E(s) (series PID)

= KcTis

TiTds2 + (Ti + Td)s + 1

E(s)

= KcTi+TdTi

1 + 1(Ti+Td)s

+ TiTdTi+Tds

E(s)

= Kc 1 + TdTi1 + 1Ti1+TdTi s + Td 1 + TdTi sE(s)= Kpc

1 + 1

Tpi s

+ Tpd s

E(s) (parallel PID)

Chen CL 33

P Control Based on {Current, Future} Errorwith {Constant, Reset} Bias

const. bias reset bias


192/921

current error P PI

future error PD PID

Chen CL 34

P Action to Step Error


193/921

Chen CL 35

PD Action to Step Error


194/921

Chen CL 36

PI (P-Reset) Action to Step Error


195/921

Chen CL 37

PID Action to Step Error


196/921

Chen CL 38

PD Action to Ramp Error


197/921

Chen CL 39

PID Action to Ramp Error


198/921

Chen CL 40

Problem of D Action (I):Sensitive to High-Frequency Noise


199/921

- Derivative action without filter:

- Problem: sinusoidal input output mag. frequency

e(t) = A sin(t) D(t) = Tdde(t)dt = TdA cos(t)

Chen CL 41

Solution: (first-order) low-pass filter

TdN

dD(t)

dt+ D(t) = Td

de(t)

dt


200/921

D action equal

follow

Chen CL 42

- Problem: (again)derivative + filtering large inner signal (e

d)

Solution: filtering derivative


201/921

Solution: filtering derivative

Chen CL 43

Problem of D Action (II):Bump Response to Step Input

- Solution 1: D action on measurement


202/921

Derivative on measurement:y(t + Td) y(t) + Td

dy(t)

dte(t + Td) = ysp y(t + Td)

= ysp y(t) + Td

dy(t)

dt e(t) Tddy(t)

dt For noise: filtering + derivative action

TdN

dyf(t)

dt+ yf(t) = y(t) (yf(t) follows y(t))

D(t) = KcTddyf(t)

dt (derivative on yf(t))

Chen CL 44

- Solution 2: filtering on setpoint signal(use r(t) as the practical setpoint)


203/921

- Solution 3: setpoint weighting (skip)

Chen CL 45

Problem of Reset (I):Moving Proportional Band


204/921

- (Reverse) P with reset bias: ([umin, umax] = [0%, 100%])

u(t) = Kc

ysp y(t)

+ ub(t) (Kc > 0)

y(t) = ymax = ? u(t) = Kc

ysp ymax

+ ub(t) = umin = 0%

y(t) = ymin

= ? u(t) = Kc ysp ymin

+ ub(t) = umax = 100%

y(t) > ymax u(t) < umin (saturation !)

y(t) < ymin u(t) > umax (saturation !)

y(t) {ymin, ymax} u(t) {umax, umin} (normal operation)

Chen CL 46

- Proportional Band: (for normal operation)

PB = [ymin, ymax] =

ysp +ub(t)umin

Kc, ysp +

ub(t)umaxKc

ub(t)0 ub(t)100


205/921

= ysp + ub( ) 0Kc , ysp + ub( ) 00Kc - If y(t) PB, Then controller output is normal

width of PB: affected by Kc

ymax ymin =umax umin

Kc=

100

Kc%

position of PB: affected by Kc, ysp, ub(t)

Chen CL 47

- Example:ysp = 50%, Kc = 2, ub(t) = 50% t,

umin = 0%, umax = 100%


206/921

PB =

50 + ub(t) 100Kc

% , 50 + ub(t) 0Kc

%

= [ 25% , 75% ]

Chen CL 48

Moving PB due to Resetting Bias: Example

- Process: G(s) = 0.0163s+150

30s+11

10s+1


207/921

- ZN-PI tuning: Kc = 3.40, Ti = 37 ( PB =

100%Kc%/%

= 29.4%)

Chen CL 49


- Process: G(s) =

0.0163s+1

50

30s+1

1

10s+1


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- Faster reset: Kc = 3.40, Ti = 18.5 ( PB = 100%Kc%/% = 29.4%)

Chen CL 50


- Process: G(s) = 0.0163s+150

30s+11

10s+1


209/921

- Slower reset: Kc = 3.40, Ti = 74 ( PB =

100%Kc%/%

= 29.4%)

Chen CL 51

Problem ofReset (II)


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Reset Windup

HE Example

Chen CL 52

Anti Reset-Windup:Use A Saturation Model to Limit Control Signal


211/921

u(t) =

100% for v(t) 100%

v(t) for 0% v(t) 100%

0% for v(t) 100%

Chen CL 53

Anti Reset-Windup for Parallel PID


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If v > u = umax

Then es = u v < 0

1Tt

esdt < 0 v until v = umax

Chen CL 54

Modification of PID (I):High-Frequency Noise


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- Problem of D Action: sensitive to high frequency noiseSolution: low-pass filter on D (P & D)

Chen CL 55

Modification of PID (II):Anti Reset-Windup


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- Problem of Reset Action: reset windup (saturation)Solution: limiter

Chen CL 56

Modification of PID (III):Manual Control

- Manual control: with an integrator


215/921

Chen CL 57

Modification of PID (IV):Bumpless A/M Transfer


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- When in automatic control: m(t) (manual) tracks u(t) (auto) m(t): always ready to control

Chen CL 58

Modification of PID (V):Bumpless M/A Transfer

- Method 1: resetting bias during M/A (keep original ysp)


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let ub(t) = m(t) Kce(t) (when M/A transfer)

thus u(t) = ub(t) + Kce(t) (auto control action)

= m(t) (final manual action)

Chen CL 59

- Method 2: set-point tracking (ysp(t) tracks y(t) when manual control)

Tsdysp(t)

dt+ ysp(t) = y(t) (when manual control)


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Disadvantage: ysp moved to undesired value ?

Chen CL 60

A Complete PID Controller (I) Automatic control Anti-reset windup Bias tracking: bias tracks final output u(t) Manual tracking: m(t) tracks final output u(t)


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Manual tracking: m(t) tracks final output u(t) Manual setpoint adjustment

Chen CL 61

A Complete PID Controller (II) Manual control Anti-reset windup Bias tracking: bias tracks final output u(t) Manual tracking: m(t) tracks final output u(t)


220/921

a ua t ac g m(t) t ac s a output u(t) (Manual setpoint adjustment)

Chen CL 62

A Complete PID Controller (III) External setpoint (used in Cascade Control) Anti-reset windup Bias tracking: bias tracks final output u(t) Manual tracking: m(t) tracks final output u(t)


221/921

g ( ) p ( ) Setpoint tracking:

Chen CL 63

A Complete PID Controller (IV) Automatic control (used in Override Control) Anti-reset windup Bias tracking: bias tracks external signal z(t) Manual tracking: m(t) tracks final output u(t)


222/921

g ( ) p ( ) Setpoint tracking:

Chen CL 64


223/921

Thank You for Your Attention

Modeling Dynamic and StaticBehavior of Chemical Processes


224/921

Cheng-Liang Chen

PSELABORATORYDepartment of Chemical Engineering

National TAIWAN University

Chen CL 1

State Variables and State Equations

- State Variables:


225/921

A set of fundamental dependent quantities whose valueswill describe the natural state of a given system

(temperature, pressure, flow rate, concentration )

- State Equations:A set of equations in the state variables above which

will describe how the natural state of a given system

changes with time

Chen CL 2

Principle of Conservation of A Quantity S


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S =

total mass

mass of individual componentstotal energy

momentum

Chen CL 3

accumulation of S

within a system

time period=

flow of S

in the system

time period


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flow of S

out the system

time period

+

amount of S generated

within the system

time period

amount of S consumed

within the system

time period

Chen CL 4

- Total Mass Balance:

d(V)

dt=i:inlet

iFi

j:outlet

jFj


228/921

- Mass Balance on Component A:

dnA

dt=

d(cA

V)

dt=

i:inlet

cAi

Fi

j:outlet

cAj

Fj rV

- Total Energy Balance:

dEdt

=d(U + K+ P)

dt=i:inlet

iFihi

j:outlet

jFjhjQWs

Chen CL 5

Mathematical ModelA Stirred Tank Heater

- Mathematical model of a process


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= state equations with associated state variables

Chen CL 6

- Total mass in tank: V = Ah

- Total energy of liquid in tank:


230/921

E = U + K+ PdU

dt

dH

dt;

dK

dt=

dP

dt= 0

H = Ahcp

T Tref

- State variables: h, T

- Total mass balance:

d(Ah)

dt= Fi F

=c A

dh

dt= Fi F

Chen CL 7

- Total energy balance:


231/921

d

Ahcp

T Tref

dt

= Ficp

Ti Tref F cp

T T

ref

+ Q

Tref=0 A

d(hT)

dt

= FiTi F T +Q

cp

Ad(hT)

dt= Ah

dT

dt+ T A

dh

dt

=FiF= FiTi F T + Q

cp

AhdT

dt= Fi (Ti T) +

Q

cp

Chen CL 8

- Summary: State equations

Adh

dt= Fi F


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AhdTdt = Fi (Ti T) + Qcp

- Summary: variables

state variables: h, T

output variables: h, T

disturbances: Ti, Fi

manipulated variables: Q, Fparameters: A,,cp

Chen CL 9

Mathematical ModelA Stirred Tank Heater (cont)


233/921

- Assumed initial steady states:

0 = Adh

dt= Fi,s Fs

0 = AhdTdt

= Fi,s (Ti,s Ts) +Qscp

Chen CL 10

- Temperature response to a step decrease in inlet temperature:


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- Dynamic response to a step decrease in inlet flow rate:

Chen CL 11

Additional Element:Transport Rate Equations


235/921

- Transport Rate Equations:To describe rate of mass, energy, and momentum transfer between

a system and its surroundings

- Example: a stirred tank heater

heat supplied by steam:

Q = U At (Tst T)

Chen CL 12

Additional Element:Kinetic Rate Equations

-


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- Kinetic Rate Equations:To describe rates of chemical reactions taking place in a system

- Example: a 1st-order reaction in a CSTR

reaction rate equation:

r = k0eE/RTc

A

Chen CL 13

Additional Element:Reaction and Phase Equilibrium Relationships

- Reaction and Phase Equilibrium Relationships:


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To describe equilibrium situations reached during a chemicalreaction or by two or more phases

- Example: a flash drum temperature of liquid phase

= temperature of vapor phase

pressure of liquid phase

= pressure of vapor phase

chemical potential of component i

in liquid phase =

chemical potential of component i

in vapor phase

Chen CL 14

Additional Element:Equations of States

- Equations of States:

To describe the relationship

i i i bl

Ideal gas law for vapor phase:


238/921

among intensive variablesdescribing the

thermodynamic state

of a system

- Example: a flash drum

pVvapor

= (moles of A + moles of B)RT

=mass of A + mass of B

average MW

RT

=mass of A + mass of B

yA

MA

+ yB

MB

RT

vapor =mass of A + mass of B

Vvapor

= [yA

MA

+ yB

MB

] pRT

liquid

= (T, xA

)

Chen CL 15

Dead Time

- Dead Time:

Whenever an input variable of a system changes


239/921

there is a time interval (short or long) during which

no effect is obsrved on outputs of the system

- dead time, transportation lag, pure delay,

distance-velocity lag

Chen CL 16

- Example: liquid through a pipe


240/921

A: temperature of inlet changes

B: temperature of outlet response

dead time: d

d =volume of pipe

volumetric flow rate=

A L

A Uav=

L

UavTout(t) = Tin(t d)

Chen CL 17

Modeling Difficulties

- Poorly understood processes


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- Imprecisely known parameters

- Size and complexity of a model

Chen CL 18

Additional Examples of Mathematical ModelingContinuous Stirred Tank Reactor (CSTR)


242/921

Exothermic Rx: A B

Chen CL 19

- Total Mass Balance:

d(V)

dt= iFi F 0

=c=

dV

dt

= Fi F


243/921

dt

- Mass Balance on Component A:

(r: rate of reaction per unit volume)

dnAdt

= d(cAV)dt

= cAi

Fi cAF rV

Vdc

A

dt+ c

A

dV

dt

=FiF= c

AiFi cAF k0e

E/RTcA

V

dcAdt

= FiV

cAi c

A

k0e

E/RTcA

Chen CL 20

- Total Energy Balance:total energy E = U + K+ P = U H(T, nA

, nB

) (enthalpy)


244/921

dE

dt=

dU

dt

dH

dt= iFihi(Ti) F h(T)Q (1)

also dHdt

= HTV cp

dTdt

+ Hn

AHA(T)

dnAdt

+ Hn

BHA(T)

dnBdt

notedn

A

dt=

d(cA

V)

dt= c

AiFi cAF rV

dnB

dt =d(c

B

V)

dt = cBiFi =0

cBF + rV

dH

dt= V cp

dT

dt+ H

A

cAi

Fi cAF rV

+ HB

c

BF rV

(2)


245/921

Chen CL 22

- Summaries:

state var.s: V, cA

, T

state eqn.s:dV

dt= Fi F

d F


246/921

dcAdt

=FiV

cAi c

A

k0e

E/RTcA

dT

dt= FiV (Ti T) + Jk0e

E/RTcA QcpV

output var.s: V, cA, Tinput var.s: c

Ai, Fi, Ti, Q , F

manip. var.s: Q, F

disturbances: cAi

, Fi, Ti

const. par.s: , cp, (Hr), k0, E , R

Chen CL 23

Additional Examples of Mathematical ModelingAn Ideal Binary Distillation Column


247/921

Chen CL 24

-Assumptions: constant vapor holdup:

equal molar heats of vaporization for A and B

negligible heat loss

constant relative volativility


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constant relative volativility 100% tray efficiency

V = V1 = = VN

yi =

xi

1 + ( 1)xi

neglect dynamics of condenser and reboiler

neglect momentum balance for each tray

leaving liquid = Li = f(Mi), i = 1, , Nliquid holdup = Mi

Chen CL 25

- State Equations (1): feed tray (i = f)


249/921

total mass:dMf

dt= Ff + Lf+1 + Vf1 Lf Vf

= Ff + Lf+1 Lf

comp A: d(Mfxf)dt

= Ffcf + Lf+1xf+1 + Vf1yf1 Lfxf Vfyf

Chen CL 26

- State Equations (2): top tray (i = N)


250/921

total mass:dM

N

dt= F

R+ V

N1 L

N V

N

= FR L

N

comp A:

d(MN

xN

)

dt = FRxD + VN1yN1 LNxN VNyN

Chen CL 27

- State Equations (3): bottom tray (i = 1)


251/921

total mass:dM1

dt= L2 L1 + V V1

= L2 L1

comp A: d(M1x1)dt

= L2x2 + V yB L1x1 V1y1

Chen CL 28

- State Equations (4): ith tray (i = 2, , N 1; i = f)


252/921

total mass:dMi

dt= Li+1 Li + Vi1 Vi

= Li+1 Li

comp A:d(Mixi)

dt= Li+1xi+1 Lixi + Vi1yi1 Viyi

Chen CL 29

- State Equations (5): reflux drum


253/921

total mass:dM

RD

dt= V

N F

R F

D

comp A:d(M

RDxD

)

dt= V

N

yN

(FR

+ FD

)xD

Chen CL 30

- State Equations (6): column base


254/921

total mass:dM

B

dt= L1 V FB

comp A:

d(MB

xB

)

dt = L1x1 V yB FBxB

Chen CL 31

-Relationships: equilibrium relationships:

yi =xi

1 + ( 1)xii = 1, , f, , N; B


255/921

hydraulic relationships: (Francis weir formula)

Li = f(Mi) i = 1, , f, , N

- State Variables:

liquid holdups:

M1, M2, , Mf, , MN; MRD, MB

liquid concentrations:

x1, x2, , xf, , xN; xD, xB

Chen CL 32

-Summaries: 2N + 4 nonlinear differential equations (state eqn.s)

2N + 1 algebraic equations (equilibrium and hydraulic)

example: N = 20 trays


256/921

example: N = 20 trays

2N + 4 = 2(20) + 4 = 44 nonlinear diff. eqn.s

2N + 1 = 2(20) + 1 = 41 algebraic equations

Chen CL 33

Modeling Considerationsfor Control Purposes

- State variables model


257/921

- State-variables model input-output model (convenient for control)

- Degrees of freedom ( df) inherent in the process

extent of control problem to be solved

Chen CL 34

- Input-Output Model:


258/921

output = f(input variables)

yi = f(m1, , mk; d1, , dt) i = 1, , m

Chen CL 35

- Example: Input-Output Model for CSTR


259/921

Assumptions: Fi = F dV/dt = 0

Chen CL 36

Total Energy Balance:

VdT

dt= Fi(Ti T) +

Q

cpQ = U At(Tst T)

dTdt +FiV + U AtV c T = FiV Ti + U AtV c Tst


260/921

dTdt +FV + U AV cp

a1/+K

T FV1/

T + U AV cp K

Tst

dT

dt+ aT = 1Ti + KTst

SS: 0 + aTs =1Ti,s + KTst,s

d(T Ts)

dt+ a (T Ts)

T

= 1 (Ti Ti,s) T

i,s

+K(Tst Tst,s) T

st

dT

dt+ aT

= 1T

i + KT

st

T

(t) = c1eat + t

0

1T

i + KT

st dt

initial: T

(t = 0) = 0 c1 = 0

T

(t) =

t0

1

T

i + KT

st

dt

Chen CL 37

Block Diagram: inputs (T

i(t), T

st(t)) output (T

(t))


261/921

This example: output variables = state variables

Chen CL 38

Distillation: output variables = state variables! State variables:

liquid holdups:

M1, M2, , Mf, , MN; MRD, MB


262/921

liquid concentrations:

x1, x2, , xf, , xN; xD, xB

Output variables:

distillate rate and composition: FD

, xD

bottom rate and composition: FB

, xB

Chen CL 39

DOF: Degree of Freedom

- Degrees of Freedom (DOF):

# of independent variables that must be specified in order to define

a process completely


263/921

a process completely

DOF = (# Var.s) (# Indep. Eq.s)

Chen CL 40

-

Example: stirred tank heater mathematical model: # of eq.s = 2

Adh

dt= Fi F

dT( )

Q


264/921

AhdT

dt= Fi (Ti T) +

Q

cp

# of variables = 6 (h, Ti, T , F , F i, Q)

DOF = 6 - 2 = 4 specify Ti, Fi, F , Q h(t), T(t)

in order to specify a process completely

the # of DoF should be zero

Chen CL 41

-

Example: binary distillation column

DOF = (4N + 11) (4N + 5) = 6


265/921

Chen CL 42

Degrees of Freedom of A Process

- f = DOF = V E = (# Var.s) (# Indep. Eq.s)

- Case 1: DOF = 0


266/921

Case 1: DOF 0unique values of the V variables

the process is exactly specified

-Case 2: DOF > 0

multiple solutions result from the E equations

can specify arbitrarily f of the V variables

the process is underspecified by f equations

- Case 3: DOF < 0no solution to the E equations

the process is overspecified by f equations

Chen CL 43

DOF and Process Controllers

- An under-specified process with DOF = f > 0

- Q: how to reduce DOF to zero


267/921

Q Oto specify system completely with unique behavior ?

from external world: disturbances

to add control loops

- Control loop:

additional equation between MV and CV

additional variable: set-point

same: DOFdifference: specify MV specify set-point

Chen CL 44

-

Example: stirred tank heater with two control loops

DOF = 4 DOF = 0 if we specify

Ti, Fi from external world ( disturbances)

set-points of the two controllers


268/921

p

Chen CL 45

-Example: binary distillation column ( DOF = 6)

specification of disturbances (external world):

feed rate (Ff) and feed composition (cf)

DOF = 6 DOF = 4


269/921

specification of control objectives ( set-points):

(I) for products:

xD: distillate composition x

B: bottom stream composition

(II) for operational feasibility:

MRD

: liquid holdup in reflux drum

MB: liquid holdup at base of column four control loops

DOF = 6 DOF = 4 DOF = 0

Chen CL 46


270/921

Note: other alternative control objectives

(1) keep at desired FD, xD, MRD, MB(2) keep at desired F

B, x

B, M

RD, M

B

Controller Simulationusing Simulink


271/921

Cheng-Liang Chen

PSELABORATORYDepartment of Chemical EngineeringNational TAIWAN University

Chen CL 1

Simulation of Control InstrumentationP Control with Reset Bias (PI)

u(t) = Kce(t) + ub(t) u(0) = ub(0) = ub

u(t) = TIdub(t)dt

+ ub(t)


272/921

( ) ( )dt

( )

dub(t)

dt=

1

TI[u(t) ub(t)]

ub(t) =

t

0

1

TI [u

(

) u

b(

)]d

+ub(0)

=

t0

1

TI[(u() u(0)) (ub() ub(0))] d+ ub(0)

Ub(t) = t

0

1

TI

[U() Ub()] d

Ub(s) =1

s

1

TI[U(s) Ub(s)]

Chen CL 2

Simulation of Control InstrumentationSimulation of PI Controller

u(t) = Kce(t) + ub(t); ub(t) ub(0) =

t0

1

TI[u() ub()] d


273/921

Chen CL 3


% P control with reset bias

plot(tout,p,m,linewidth,2)

hold onplot(tout ub b linewidth 2)


274/921

plot(tout,ub, b , linewidth ,2)

plot(tout,u,r,linewidth,2)

ylabel(\bf PI Output Signals,...

fontsize,14);

xlabel(\bf t (min),fontsize,14);

set(gca,linewidth,3);legend(p(t),u_b(t),u(t));

hold off

Chen CL 4


% P control with reset bias

subplot(3,1,1)

plot(tout,p,m,linewidth,2)ylabel(\bf p(t) fontsize 14);


275/921

ylabel( \bf p(t) , fontsize ,14);

set(gca,linewidth,3);

subplot(3,1,2)

plot(tout,ub,b,linewidth,2)

ylabel(\bf u_b(t),fontsize,1

set(gca,linewidth,3);subplot(3,1,3)

plot(tout,u,r,linewidth,2)

ylabel(\bf u(t),fontsize,14);

xlabel(\bf t (min),fontsize,


Chen CL 5

Simulation of Control InstrumentationSimulation of PD Controller

D Action on Error e with A Low-pass Filter

u(t) = Kce(t + TD) + ub Kc e(t) + TDde(t)

dt + ubdef(t)


276/921

u(t) = Kc

e(t) + TD

def(t)

dt

+ ub

e(t) = TDdef(t)

dt+ ef(t) TD

def(t)

dt=

1

[e(t) ef(t)]

ef(t) =t0

1TD

TDde

f()d

d+ ef(0)

=0

=

t0

1

TD

1

[e() ef()]

d+ ef(0)

=0

or Ef(t) =t0

1TD

1

[E()Ef()]d (deviation var.s)

Ef(s) =1

s

1

TD

1

[E(s)Ef(s)] =

1

s

1

TDL

TD

dEf(t)

dt

Chen CL 6



TDdef(t)

dt=

1

(e(t) ef(t)) ; ef(t) ef(0) =

t0

1

TD

TD

def()

d

d


277/921

Chen CL 7



% PD control

plot(tout,e,Color,[0,.5,0],linewhold on


278/921

hold on

plot(tout,efuture,Color,[.5,0,.5],

plot(tout,p,r,linewidth,2)

plot(tout,Daction,b,linewidth,2)

plot(tout,PDout,m,linewidth,2)

ylabel(\bf PD Output Signals,...fontsize,14);



legend(e(t),e(t+T_d),p(t),D(

hold off

Chen CL 8

Simulation of Control InstrumentationSimulation of PD ControllerD Action on Error e with A Low-pass Filter

subplot(5,1,2)

plot(tout,efuture,Color,[.5,0,.

linewidth,2)l b l(\bf ( T d) f i


279/921

% PD control

subplot(5,1,1)

plot(tout,e,Color,[0,.5,0],...

linewidth,2)

ylabel(\bf e(t),fontsize,14);


ylabel(\bf e(t+T_d),fontsize,


subplot(5,1,3)


ylabel(\bf p(t),fontsize,14);set(gca,linewidth,3);

subplot(5,1,4)

plot(tout,Daction,b,linewidth

ylabel(\bf D(t),fontsize,14);


subplot(5,1,5)plot(tout,PDout,m,linewidth,

ylabel(\bf u(t),fontsize,14);

xlabel(\bf t (min),fontsize,


Chen CL 9

Simulation of Control InstrumentationSimulation of PD ControllerD Action on Measurement y with A Low-pass Filter

u(t) = Kce(t + T

D) + u

b K

c ysp y(t) + TDdy(t)

dt + ub dy (t)


280/921

u(t) = Kc

ysp

y(t) + TD

dyf(t)

dt

+ ub

y(t) = TDdyf(t)

dt+ yf(t) TD

dyf(t)

dt=

1

[y(t) yf(t)]

yf(t) =t0

1TD

TD

dyf()d

d+ yf(0)

=

t0

1

TD

1

[y() yf()]

d+ yf(0)

or Yf(t) =t0

1TD

1

[Y() Yf()]d (deviation var.s)

Yf(s) =1

s

1

TD

1

[Y(s) Yf(s)] =

1

s

1

TDL

TD

dYf(t)

dt

Chen CL 10

Simulation of Control InstrumentationSimulation of PD ControllerD Action on Error y with A Low-pass Filter

TDdyf(t)

dt=

1

(y(t) yf(t)) ; yf(t) yf(0) =

t0

1

TD

TD

dyf()

d

d


281/921

Chen CL 11


% PD control

plot(tout,y,Color,[0,.5,0],linewhold on


282/921

plot(tout,yfuture,Color,[.5,0,.5],


plot(tout,Daction,b,linewidth,2)

plot(tout,PDout,m,linewidth,2)

ylabel(\bf PD Output Signals,...fontsize,14);



legend(y(t),y(t+T_d),p(t),-D

hold off

Chen CL 12


subplot(5,1,2)

plot(tout,yfuture,Color,[.5,0,.

linewidth,2)ylabel(\bf y(t+T d) fontsize


283/921

% PD control

subplot(5,1,1)

plot(tout,y,Color,[0,.5,0],...linewidth,2)

ylabel(\bf y(t),fontsize,14);


ylabel(\bf y(t+T_d),fontsize,


subplot(5,1,3)


ylabel(\bf p(t),fontsize,14);


subplot(5,1,4)

plot(tout,Daction,b,linewidth

ylabel(\bf -D(t),fontsize,14)

set(g

dynamic control2

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