Post on 14-Apr-2018
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Chemical Process
Dynamics and Control
Cheng-Liang ChenPSELABORATORY
Department of Chemical EngineeringNational TAIWAN University
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Chen CL 1
Some Notes
- Instructor:
Cheng-Liang Chen (CCL@ntu.edu.tw)
- Teaching Assistant:
Ying-Jyuan Ciou (D93524013@ntu.edu.tw)
-
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%)
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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
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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
Cheng-Liang ChenPSELABORATORY
Department of Chemical EngineeringNational TAIWAN University
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
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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
A Process Heat ExchangerFour Basic Elements
- 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
A Process Heat ExchangerFour Basic Elements
- 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
A Process Heat ExchangerFour Basic Elements
- 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
A Process Heat ExchangerFour Basic Elements
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 ?
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Design Aspects of A Process Control System
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
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Design Aspects of A Process Control System
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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Why Laplace Transform
-
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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Why Laplace Transform
-
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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Why Laplace Transform
- 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)
Cheng-Liang ChenPSELABORATORY
Department of Chemical EngineeringNational TAIWAN University
Chen CL 1
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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
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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
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Rise Time
Chen CL 12
Ti C
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Time Constant
Chen CL 13
S li Ti
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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
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Force TransducerBounded Resistance Strain Gage
Stress: =force
area=
F
A
Strain: = deformationlength
= LL
=
Chen CL 20
Fo ce T a sd ce
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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:
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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:
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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
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Pressure TransducerDiaphragm with Strain Gage
Chen CL 25
Pressure Transducer
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Pressure TransducerDiaphragm with Strain Gage
Chen CL 26
Pressure Transducer
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Pressure TransducerCapacitive Absolute-Pressure Transducer
Chen CL 27
Pressure Transducer
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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
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Pressure TransducerPotentiometer Sensing Pressure Transducer
Chen CL 30
Flow Transducer
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Flow TransducerDifferential Pressure Obstruction Flow Sensors
Orifice plate, Venturi, Pitot tube
f = CoAo
po
1
dD
4
Chen CL 31
Flow Transducer
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Flow TransducerDeflection-type Flow Transducers
Cantilever beam, Variable-area rotameter
Chen CL 32
Flow Transducer
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Flow TransducerSpin-type Flow Transducers
Paddle wheel, Flow turbine
Chen CL 33
Flow Transducer
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Flow TransducerElectromagnetic Flow Meter
Chen CL 34
Flow Transducer
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Flow TransducerUltrasonic Flow Transducer
Chen CL 35
Flow Transducer: Characteristics
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Flow Transducer: Characteristics
Chen CL 36
Level Transducer
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Level TransducerDiscrete-type Level Transducer
Float switch, Photoelectric
Chen CL 37
Level Transducer
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Level TransducerBy Sensing Pressure with
Offset transducer, Sealed tank
Chen CL 38
Level Transducer
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Level TransducerBy Sensing differential Pressure
Chen CL 39
Level Transducer
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Level Transducerwith Float-driven Control System
Chen CL 40
Level Transducer
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Level TransducerCapacitance Level Transducer
Chen CL 41
Level Transducer
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Level TransducerTop-mounted Ultrasonic Level Sensor
Chen CL 42
Common Temperature Transducers
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Common Temperature Transducers
Chen CL 43
Common Temperature Transducers
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p
Chen CL 44
Common Temperature Transducers
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p
Chen CL 45
Thermocouple
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pSeebeck Effect
- Current in a closed circuit
- Voltage across an open circuit
Chen CL 46
Thermocouple
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pEquivalent Circuits
Chen CL 47
Thermocouple
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pwith External Reference
Chen CL 48
Thermocouple
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pExternal Reference Junction with No Ice Bath
Chen CL 49
Thermocouple
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pScanning Many Thermocouples
Chen CL 50
Thermocouple
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pElectronic Ice-point Compensation
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Chen CL 1
Why Focus on Valve ?
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- 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
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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
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Chen CL 6
Efect of Vaporization on Flow Rate
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Chen CL 7
Energy and Pressure Grades Across A Valve
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Chen CL 8
Valves Cost Energy
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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
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Chen CL 10
Variable Speed Drive Saves Energy
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Chen CL 11
Affinity Laws
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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
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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
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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
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Globe Valve
A Single-Ported Two Way Valve
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Chen CL 17
Types of Control ValvesGl b V l
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Globe Valve
An Angle Valve and A Y-Pattern Valve
Chen CL 18
Types of Control ValvesGl b V l
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Globe Valve
Valves with Three-Way Bodies
Chen CL 19
Types of Control ValvesGl b V l
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Globe Valve
Anti-Cavitation Valve
Chen CL 20
Types of Control ValvesB ll V l
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Ball Valve
Full Ball Valve
Chen CL 21
Types of Control ValvesB ll V l
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Ball Valve
Three-Way Ball Valve
Chen CL 22
Types of Control ValvesBall Val e
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Ball Valve
Segmented Ball Valve
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Chen CL 24
Types of Control ValvesDi h V l
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Diaphragm Valve
Chen CL 25
Types of Control ValvesDi i l V l
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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
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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
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- 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...)
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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
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- 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
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- 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
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Chen CL 32
Overview
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- 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
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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
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Chen CL 35
Action of Control ValvesSafety Consideration
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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:
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Chen CL 37
Sizing of Control ValveA Basic Trade off Problem
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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
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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
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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
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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
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- 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
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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
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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
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- 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.)
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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
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Chen CL 47
- Compressible Flow: (Fisher Controls)
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qs = Cg502GTp1 sin
59.64C1
pvp1 rad
- Note: the above sine function is basically the same
function of y
Chen CL 48
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Chen CL 49
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Chen CL 50
Control Valve Capacity: Example
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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.
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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
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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
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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)
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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
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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
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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
( )
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- 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
Valve Inherent Characteristics
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Chen CL 59
Valve Inherent Characteristics
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- 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
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- 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
Valve Installed Characteristics
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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
Valve Installed Characteristics
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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
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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
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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
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( )( 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
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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
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( ) q p g
Chen CL 68
Application I: Fluid TransferSelf-Regulated Processes
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Chen CL 69
Application I: Fluid TransferLiquid Transferred by Pressure Difference
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Chen CL 70
Application I: Fluid TransferLiquid Transferred by Pressure Difference
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Chen CL 71
Application I: Fluid TransferA Flow-Controlled Pump
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Chen CL 72
Application I: Fluid TransferSteam Ejector
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Chen CL 73
Application II: Heat TransferSteam Heaters and Dryers
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Chen CL 74
Application II: Heat TransferHeat Exchangers
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Chen CL 75
Application II: Heat TransferAntifreeze Applications
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Chen CL 76
Application III: Chemical ReactionsA pH Control System
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Chen CL 77
Application III: Chemical ReactionsCombustion Processes
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Proportional-Integral-Derivative
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PID Controllers
Cheng-Liang Chen
PSELABORATORYDepartment of Chemical Engineering
National TAIWAN University
Chen CL 1
Outline
- Proportional Integral Derivative Controller:
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- Proportional-Integral-Derivative Controller:
PID Control based on CurrentFuture Error with ConstantReset Bias
PID Control based on CurrentFuture Error with ConstantReset Bias
PID Control based on CurrentFuture Error with ConstantReset Bias
PIDControl based on CurrentFuture Error with ConstantReset Bias
PID Control based on CurrentFuture Error with ConstantReset Bias
PID Control based on CurrentFuture Error with ConstantReset Bias
PID Control based on CurrentFuture Error with ConstantReset Bias
Chen CL 2
- Steady Offset of PID when using Constant bias
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-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
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- 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
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- 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
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- 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
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Chen CL 7
Simplest Controller: On-Off Control
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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)
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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
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( ) 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
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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
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HE: pulse decrease in input temperature
- 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
HE: pulse decrease in input temperature
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HE: pulse decrease in input temperature
- 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
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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
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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
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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
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- 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}
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(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
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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}
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(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
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- 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
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Chen CL 22
Reset Bias for Zero SS OffsetFeasible Method
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-
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
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Chen CL 24
Reset Bias Integral Action
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(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)
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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
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Chen CL 27
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Chen CL 28
Future-Error-Based P Control with Reset Bias
- P control based on future error with reset bias
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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)
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u(t) = Kce(t) p(t)
+
b( ) 1TiKce(t)
p(t)
dt +ub
ub(t)
Chen CL 30
Series PID Parallel PID
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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
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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)
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= 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
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current error P PI
future error PD PID
Chen CL 34
P Action to Step Error
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Chen CL 35
PD Action to Step Error
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Chen CL 36
PI (P-Reset) Action to Step Error
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Chen CL 37
PID Action to Step Error
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Chen CL 38
PD Action to Ramp Error
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Chen CL 39
PID Action to Ramp Error
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Chen CL 40
Problem of D Action (I):Sensitive to High-Frequency Noise
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- 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
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D action equal
follow
Chen CL 42
- Problem: (again)derivative + filtering large inner signal (e
d)
Solution: filtering derivative
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Solution: filtering derivative
Chen CL 43
Problem of D Action (II):Bump Response to Step Input
- Solution 1: D action on measurement
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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)
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- Solution 3: setpoint weighting (skip)
Chen CL 45
Problem of Reset (I):Moving Proportional Band
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- (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
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= 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%
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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
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- ZN-PI tuning: Kc = 3.40, Ti = 37 ( PB =
100%Kc%/%
= 29.4%)
Chen CL 49
Moving PB due to Resetting Bias: Example
- 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
Moving PB due to Resetting Bias: Example
- Process: G(s) = 0.0163s+150
30s+11
10s+1
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- 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
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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
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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)
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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)
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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)
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g ( ) p ( ) Setpoint tracking:
Chen CL 64
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Thank You for Your Attention
Modeling Dynamic and StaticBehavior of Chemical Processes
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Cheng-Liang Chen
PSELABORATORYDepartment of Chemical Engineering
National TAIWAN University
Chen CL 1
State Variables and State Equations
- State Variables:
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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
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- 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:
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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:
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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)
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- 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
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- 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:
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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
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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
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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)
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Exothermic Rx: A B
Chen CL 19
- Total Mass Balance:
d(V)
dt= iFi F 0
=c=
dV
dt
= Fi F
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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)
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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)
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Chen CL 22
- Summaries:
state var.s: V, cA
, T
state eqn.s:dV
dt= Fi F
d F
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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
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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)
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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)
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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)
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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)
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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
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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
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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
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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
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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
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- 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:
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output = f(input variables)
yi = f(m1, , mk; d1, , dt) i = 1, , m
Chen CL 35
- Example: Input-Output Model for CSTR
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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
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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))
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This example: output variables = state variables
Chen CL 38
Distillation: output variables = state variables! State variables:
liquid holdups:
M1, M2, , Mf, , MN; MRD, MB
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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
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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
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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
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Chen CL 42
Degrees of Freedom of A Process
- f = DOF = V E = (# Var.s) (# Indep. Eq.s)
- Case 1: DOF = 0
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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
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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
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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
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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
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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
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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)
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( ) ( )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
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Chen CL 3
Simulation of Control InstrumentationSimulation of PI Controller
% P control with reset bias
plot(tout,p,m,linewidth,2)
hold onplot(tout ub b linewidth 2)
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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
Simulation of Control InstrumentationSimulation of PI Controller
% P control with reset bias
subplot(3,1,1)
plot(tout,p,m,linewidth,2)ylabel(\bf p(t) fontsize 14);
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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,
set(gca,linewidth,3);
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)
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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
Simulation of Control InstrumentationSimulation of PD Controller
D Action on Error e with A Low-pass Filter
TDdef(t)
dt=
1
(e(t) ef(t)) ; ef(t) ef(0) =
t0
1
TD
TD
def()
d
d
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Chen CL 7
Simulation of Control InstrumentationSimulation of PD Controller
D Action on Error e with A Low-pass Filter
% PD control
plot(tout,e,Color,[0,.5,0],linewhold on
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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);
xlabel(\bf t (min),fontsize,14);
set(gca,linewidth,3);
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
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% PD control
subplot(5,1,1)
plot(tout,e,Color,[0,.5,0],...
linewidth,2)
ylabel(\bf e(t),fontsize,14);
set(gca,linewidth,3);
ylabel(\bf e(t+T_d),fontsize,
set(gca,linewidth,3);
subplot(5,1,3)
plot(tout,p,r,linewidth,2)
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);
set(gca,linewidth,3);
subplot(5,1,5)plot(tout,PDout,m,linewidth,
ylabel(\bf u(t),fontsize,14);
xlabel(\bf t (min),fontsize,
set(gca,linewidth,3);
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)
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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
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Chen CL 11
Simulation of Control InstrumentationSimulation of PD ControllerD Action on Error y with A Low-pass Filter
% PD control
plot(tout,y,Color,[0,.5,0],linewhold on
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plot(tout,yfuture,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);
xlabel(\bf t (min),fontsize,14);
set(gca,linewidth,3);
legend(y(t),y(t+T_d),p(t),-D
hold off
Chen CL 12
Simulation of Control InstrumentationSimulation of PD ControllerD Action on Error y with A Low-pass Filter
subplot(5,1,2)
plot(tout,yfuture,Color,[.5,0,.
linewidth,2)ylabel(\bf y(t+T d) fontsize
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% PD control
subplot(5,1,1)
plot(tout,y,Color,[0,.5,0],...linewidth,2)
ylabel(\bf y(t),fontsize,14);
set(gca,linewidth,3);
ylabel(\bf y(t+T_d),fontsize,
set(gca,linewidth,3);
subplot(5,1,3)
plot(tout,p,r,linewidth,2)
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)
set(g