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Chapter 4. Multi-Variable Control. Topics. Synthesis of Configurations for Multiple-Input, Multiple-Output Processes Interaction Analysis and Decoupling Methods Optimal Control Approaches. Examples of Multivariable Control - (1) Distillation Column. Available MV ’ s Reflux Flow

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chapter 4

Chapter 4

Multi-Variable Control

topics
Topics
  • Synthesis of Configurations for Multiple-Input, Multiple-Output Processes
  • Interaction Analysis and Decoupling Methods
  • Optimal Control Approaches
slide3

Examples of Multivariable Control

  • - (1) Distillation Column
  • Available MV’s
  • Reflux Flow
  • Distillate Flow
  • Steam: Flow to Reboiler
  • (Heat Duty)
  • 4. Bottom: Flow
  • 5. Condenser Duty
  • Available CV’s
  • Level in Accumulator
  • Level in Reboiler
  • Overhead Composition
  • Bottom Composition
  • Pressure in the Column

Problem: How to ‘pair’ variables?

slide4

Stream A

FT

CT

  • Examples of Multivariable Control
  • (2) Blending or two Streams

Stream C

Stream B

MV’s: Flow of Stream A;

Flow of Stream B.

CV’s: Flow of Stream C

Composition of Stream C

slide5

Hot

Cold

LT

TT

Examples of Multivariable Control:

(3) Control of a Mixing Tank

MV’s: Flow of Hot Stream CV’s: Level in the tank

Flow of Cold Stream Temperature in the tank

2 issue of degree of freedom dof
2. Issue of Degree of Freedom (DOF)
  • Question: How many variables can be measured and controlled?
  • Answer: The number of variables to be controlled is related to the degree of freedom of a system.
        • F=V-E
  • If F>0, then F variables must be either (1) externally defined or (2)throught a control system
  • To control n CV’s, we need n MV’s.
2 issue of degree of freedom dof continued the level system

Fi

h

F0

2. Issue of Degree of Freedom (DOF)-Continued (The Level System)

h

Equations

Variables: h, Fi, Fo, x

F=4-2=2

2 issue of degree of freedom dof continued the level system1
2. Issue of Degree of Freedom (DOF)-Continued (The Level System)
  • Externally Specified: Fi
  • Controlled Variables: 2-1=1
  • Hence, we can control one variable: (1) control flow; (2) control level
  • The manipulated variable is valve position or we may choose to have no control; externally specified x
2 issue of degree of freedom dof continued the flash drum

vapor

Fv

y i

Feed

P,T

Z f ,Pf,T f,Ff

h

Steam Ts

FL

Xi

Liquid

2. Issue of Degree of Freedom (DOF)-Continued – The flash drum

Equations:

2 issue of degree of freedom dof continued the flash drum1
2. Issue of Degree of Freedom (DOF)-Continued – The flash drum
  • More Equations: yi=Kixi (Phase Equilibrium)
  • More Equations: Σyi=1; Σxi=1
  • Total number of Equations=2N+3
  • Variables: zi (N), Tf, ρf, Ff, P, T,y(N-1),FL,xL (N-1),Ts: Total 3N+8
  • External Specified: zi (N), Tf, ρf, Ff, Total: N+1
2 issue of degree of freedom dof continued the flash drum2
2. Issue of Degree of Freedom (DOF)-Continued – The flash drum
  • DOF=3N+8-(N+1)-(2N+1)=4
  • There are four variables that must be fixed by controllers or other means.
  • Typically: Specify or control T, P, h, Ff
  • Ff determines the production rate
  • Ff may also be externally specified
2 issue of degree of freedom dof continued the flash drum3

PC

TC

.

2. Issue of Degree of Freedom (DOF)-Continued – The flash drum
  • What are the manipulated variables? Look for possible valve positions:

LC

Paring : 4!=24 possibilities;

If Ff is specified 3!=6 possibilities

practical tips
Practical Tips
  • Choose the manipulated variable (MV) that has a fast directed effect on the controlled variable (CV) (Calculate or estimate steady state gain, check for controllability)
  • Avoid long dead times in the loop
  • Reduce or avoid interaction (use relative gain array)
3 interaction problem
3. Interaction Problem
  • Consider a process with two inputs and two outputs.
  • Two control loops can be established.
  • Question: Will actions in one loop affect the other loop?
  • Answer: Most often, yes. This is called interaction
  • Interaction is usually decremented to control loop performance
approaches
Approaches
  • Design single loop controllers and detune them such that one loop does not degrade another loop performance
          • not always possible
          • selection of pairing very important
  • Modify controlled or manipulated variables such that interaction is reduced
          • Steady state interaction
          • Dynamic interaction
  • Determine interacting element’s transfer functions and try to compensate for them
  • Solve the multivariable control problem (such as model predictive control –MPC)
block diagram analysis

+

+

+

+

Process

Block Diagram Analysis

Loop 1

-

+

+

+

+

+

+

-

Loop 2

block diagram analysis continued both loops closed
Block Diagram Analysis-Continued (Both Loops Closed)

P11,P12,P21 and P22have a common denominator:

Roots of Q(s) determine stability

slide20

process

m1

y1

y2

m2

process

m1

y1

Controller

Set

point

y2

m2

Measuring Interaction

measuring interactions brisol s relative gain continued
Measuring Interactions – Brisol’s Relative Gain - Continued
  • If λ11 is unity the other loops do not affect loop 1 and hence there is no interaction
  • If λ11 is zero then m1 cannot use to control c1
  • if λ11 →∞then other loops will make c1 uncontrollable with m1
slide22

Hot

Cold

LT

TT

Examples of Multivariable Control:

(3) Control of a Mixing Tank

MV’s: Flow of Hot Stream CV’s: Level in the tank

Flow of Cold Stream Temperature in the tank

relative gain array
Relative Gain Array
  • All rows add up to 1. All columns add up to 1.
  • λij is dimensionless. Λ can be used to find suitable
  • pairings of input and output
  • (3) If Λ is diagonal, there is no interaction between loops
  • (4) Elements of Λ significantly different from 1 indicate
  • problems with interaction
derivation of
Derivation of Λ

Choose control loop pairs such that λij is

Positive and as close to unity as possible.

slide25

XA

F

CT

FT

XA

FA

F

FB

Example: Blending Problem

example blending problem continued1
Example – Blending Problem - Continued

If FB/F≒1, FA/F≒0 then pair FA→xA; FB→F

If FA/F≒1, FB/F≒0 then pair FB→xA; FA→F

steady state decoupling
Steady State Decoupling
  • In the Blending Problem:
  • Hence to control F, without changing x is possible by changing ratio constant of FA and FB.
  • Change ratio of FA and FB while keeping FA + FB constant
  • Let m1= FA + FB, m2=FB/ FA, then FA =m1/ (1 + m2)

FB =m1 m2 / (1 + m2)

FA

COMPUTER

FB

F

X

dynamic decoupling

c1(s)

m1(s)

g11

g12

g21

m2(s)

c2(s)

g22

Dynamic Decoupling
  • Let C(s)=G(s)m(s)
  • Or c1(s)=g11(s)m1(s)+g12(s)m2(s)

c2(2)=g21(s)m1(s)+g22(s)m2(s)

  • One way to decouple the system is to define new input variables m(s)=D(s)u(s)
dynamic decoupling continued
Dynamic Decoupling-Continued
  • If D(s) is a matrix, then c(s)=G(s)D(s)u(s)
  • If D(s) is chosen such that G(s)D(s)=diagonal, then
  • u1(s) affects only c1, u2 only affects c2.
  • The system is called decoupled.
slide31

y1s

-

+

y1

+

m1

u1

plant

Gc1

+

D2

D1

y2

+

+

Gc2

u2

m2

+

-

y2s

dynamic decoupling continued1
Dynamic Decoupling-Continued
  • For simplicity, we use
  • Then
  • We want g11D1+g12=0;g21+g22D2=0
  • Hence,D1=-g12/g11;D2=-g21/g22
dynamic decoupling continued2

m1(s)

c1(s)

g11

g12

g21

m2(s)

c2(s)

g22

Dynamic Decoupling-Continued
  • This leads to m1=u1-g12/g11u2; m2=u2-g22/g21u2
  • c1=(g11+D2g12)u1;c2=(g21D1+g22)u2
conclusion
Conclusion
  • Multivariable control is essential in the industrial applications
  • Steady state decoupling is very useful in case of no dynamic system such as mixing
  • Dynamic decoupling is very important for high valued-added system such as distillation systems
homework 4 multivariable systems
Homework #4 Multivariable Systems
  • Textbook p698
  • 21.3, 21.4, 21.7, 21.8