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Chapter 4

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

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  1. 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. 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

  5. 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

  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. 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

  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. 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:

  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. 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

  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. + + + + Process Block Diagram Analysis 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 P22have a common denominator: Roots of Q(s) determine stability

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

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

  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. 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

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

  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. 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)

  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. 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

  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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