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Multi-Variable Control

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Presentation on theme: "Multi-Variable Control"— Presentation transcript:

1 Multi-Variable Control
Chapter 4 Multi-Variable Control

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

3 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?

4 Examples of Multivariable Control (2) Blending or two Streams
FT CT Stream A Stream C Stream B MV’s: Flow of Stream A; Flow of Stream B. CV’s: Flow of Stream C Composition of Stream C

5 Examples of Multivariable Control: (3) Control of a Mixing Tank
Hot Cold LT TT MV’s: Flow of Hot Stream CV’s: Level in the tank Flow of Cold Stream Temperature in the tank

6 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.

7 2. Issue of Degree of Freedom (DOF)-Continued (The Level System)
Fi F0 h Equations Variables: h, Fi, Fo, x F=4-2=2

8 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

9 2. Issue of Degree of Freedom (DOF)-Continued – The flash drum
Feed Z f ,Pf,T f,Ff P,T h vapor Fv y i Liquid FL Xi Steam Ts Equations:

10 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

11 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

12 2. Issue of Degree of Freedom (DOF)-Continued – The flash drum
What are the manipulated variables? Look for possible valve positions: PC . TC LC Paring : 4!=24 possibilities; If Ff is specified 3!=6 possibilities

13 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)

14 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

15 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)

16 Block Diagram Analysis
+ Process Loop 1 - + + + + + + - Loop 2

17 Block Diagram Analysis-Continued (One Loop Closed)
Changing y1sp will affects y2

18 Block Diagram Analysis-Continued (Both Loops Closed)
P11,P12,P21 and P22 have a common denominator: Roots of Q(s) determine stability

19 Measuring Interaction – Brisol’s Relative Gain
Process FBC C1 C2 m1 m2

20 Measuring Interaction
process m1 m2 y1 y2 process m1 m2 y1 y2 Controller Set point

21 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

22 Examples of Multivariable Control: (3) Control of a Mixing Tank
Hot Cold LT TT MV’s: Flow of Hot Stream CV’s: Level in the tank Flow of Cold Stream Temperature in the tank

23 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

24 Derivation of Λ Choose control loop pairs such that λij is
Positive and as close to unity as possible.

25 Example: Blending Problem
CT FT XA F FA FB XA F

26 Example – Blending Problem - Continued

27 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

28 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

29 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) g11 g21 g12 g22 m1(s) m2(s) c2(s) c1(s)

30 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.

31 y1s - + y1 + m1 u1 plant Gc1 + D2 D1 y2 + + Gc2 u2 m2 + - y2s

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

33 Dynamic Decoupling-Continued
This leads to m1=u1-g12/g11u2; m2=u2-g22/g21u2 c1=(g11+D2g12)u1;c2=(g21D1+g22)u2 g11 g21 g12 g22 m1(s) m2(s) c2(s) c1(s)

34 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

35 Homework #4 Multivariable Systems
Textbook p698 21.3, 21.4, 21.7, 21.8


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