CONTROL OF MULTIVARIABLE PROCESSES

Transcription

1 Process plants ( or complex experiments) have many variables that must be controlled. The engineer must. Provide the needed sensors 2. Provide adequate manipulated variables 3. Decide how the CVs and MVs are paired (linked via the control design) Fortunately, most of what we learned about single-loop systems applies, but we need to learn more! v8 F2 F T T3 T4 v3 T5 F3 T6 F4 F5 P v v2 L v5 v7 v6 T7 L T2 T8 Hot Oil T9 Hot Oil F6

2 Two control approaches are possible Multiloop: Many independent PID controllers Centralized: Method covered in Chapter 23 F T L F T Centralized Controller L

3 Two control approaches are possible We will concentrate on MULTILOOP Multiloop: Many independent PID controllers E( t) = SP( t) CV ( t) t d CV MV ( t) = Kc E( t) + E t dt Td + I T ( ') ' I dt v FC E( t) = SP( t) CV ( t) t d CV MV ( t) = Kc E( t) + E t dt Td + I T ( ') ' I dt E( t) = SP( t) CV ( t) t d CV MV ( t) = Kc E( t) + E t dt Td + I T ( ') ' I dt TC LC v3 v2

4 Let s assume (for now) that we have the sensors and valves. How do we ask the right questions in the best order to have a systematic method for pairing multiloop control? v8 F2 F T T3 T4 v3 T5 F3 T6 F4 F5 P v v2 L v5 v7 v6 T7 L T2 T8 Hot Oil T9 Hot Oil F6

5 Some key questions whose answers help us design a multiloop control system.. IS INTERCTION PRESENT? - If no interaction ll single-loop problems Let s start here to build understanding 2. IS CONTROL POSSIBLE? - Can we control the specified CVs with the available MVs? 3. WHT IS S-S ND DYNMIC BEHVIOR? - Over what range can control keep CVs near the set points?

6 What is different when we have multiple MVs and CVs? INTERCTION!! Definition: multivariable process has interaction when input (manipulated) variables affect more than one output (controlled) variable. T Does this process have interaction?

7 How can we determine how much interaction exists? One way - Fundamental modelling F, x F S, x S = v v2 F M, x M F F + x Fundamental (n-l) F S + = F S F x M S = F M x M F' x' M M = F' = (F Fundamental linearized + F' s F S S + F ) 2 ss F ' F + (Fs + F ) 2 ss F ' S

8 How can we determine how much interaction exists? One way - Fundamental modelling F, x Total flow, F M F S, x S = v v2 Concent, x M F M, x M Sketch the responses of the dependent variables time time v, % open v2, % open time time

9 How can we determine how much interaction exists? One way - Fundamental modelling F, x Total flow, F M F S, x S = v v2 Concen., x M F M, x M Sketch the responses of the dependent variables time time v, % open v2, % open time time

10 How can we determine how much interaction exists? Second way - Empirical modelling (process reaction curve) Step change to reflux with constant reboiler.99 Sketch the responses of the dependent variables.4 R x D XD (mol frac) XB (mol frac) V Time (min).35 x Time (min).565 x R (mol/min).3.25 V (mol/min) x B Time (min) Time (min)

11 How can we determine how much interaction exists? Second way - Empirical modelling (process reaction curve) Step change to reflux with constant reboiler.99.4 R x D XD (mol frac) XB (mol frac) V.35 x 4 Time (min).565 x 4 Time (min).565 R (mol/min).3.25 V (mol/min) x B Time (min) Time (min)

12 How can we determine how much interaction exists? Second way - Empirical modelling (process reaction curve) ) (.. ) (.. ) ( ) (. ) (. ) (. s F s e s F s e s x s F s e s F s e s x V s R s B V s R s D = + + = F R F V x B x D Which models can be determine from the experiment shown in the previous slide?

13 How can we determine how much interaction exists? Use the model; if model can be arranged so that it has a diagonal form, no interaction exists. CV... CV n = K K 22 K nn MV MV MV 2 3 If any off-diagonal are non-zero, interaction exists.

14 How can we determine how much interaction exists? Let s use qualitative understanding to answer the question. Feed has C v Does this CSTR reactor process have interaction? T LC Information:. Exothermic reaction v2 2. -r = k e -E/RT C cooling

15 How can we determine how much interaction exists? Let s use qualitative understanding to answer the question. Feed has C v T LC From the cause-effect analysis, v affects both sensors; interaction exists! cooling v2 What about v2? v feed flow rate residence time conversion heat generated T v2 coolant flow heat transfer Q/(FρC P ) T conversion nalyzer

16 We will use a block diagram to represent the dynamics of a 2x2 process, which involves multivariable control. MV G + + CV The blocks (G) cannot be found in the process. They are the models that represent the behavior between one input and one output. G 2 G 2 G d G d2 D MV 2 G CV 2

17 Let s relate the block diagram to a typical physical process. What are the MVs, CVs, and a disturbance, D? F R F R G + + XD F XF q XD F D G 2 G 2 G d G d2 XF F V FRB F RB G XB XB FB

18 Let s relate the block diagram to a typical physical process. What are the MVs, CVs, and a disturbance, D? Reactor v Better control level! G + + T LC G 2 G d G 2 G d2 v2 G

19 Let s relate the block diagram to a typical physical process. What are the MVs, CVs, and a disturbance, D? Reactor v Better control level! v G + + T LC G 2 G d G 2 G d2 v2 v2 G T Disturbances = feed composition, feed temperature, coolant temperature,

20 Some key questions whose answers help us design a multiloop control system.. IS INTERCTION PRESENT? - If no interaction ll single-loop problems 2. IS CONTROL POSSIBLE? - How many degrees of freedom exist? - Can we control CVs with MVs? Let s learn how to answer this key question 3. WHT IS S-S ND DYNMIC BEHVIOR? - Over what range can control keep CVs near the set points?

21 DEGREES OF FREEDOM How do we determine the maximum # variables that be controlled in a process? v8 F2 F T T3 T4 v3 T5 F3 T6 F4 F5 P v v2 L v5 v7 v6 T7 L T2 T8 Hot Oil T9 Hot Oil F6

22 DEGREES OF FREEDOM requirement for a successful design is: The number of valves number of CV to be controlled v8 F2 F T T3 T4 v3 T5 F3 T6 F4 F5 P v v2 L v5 v7 v6 T7 L T2 T8 Hot Oil T9 Hot Oil F6 OK, but this does not ensure that we can control the CVs that we want to control!

23 CONTROLLBILITY system is controllable if its CVs can be maintained at their set points, in the steady-state, in spite of disturbances entering the system. Model for 2x2 system CV CV 2 = = K K 2 K K 2 22 MV MV 2 + K K d d 2 D system is controllable when the matrix of process gains can be inverted, i.e., when the determinant of K. I am in desperate need of examples

24 Let s do the toy autos first; then, do some processes For the autos in the figure re they independently controllable? Does interaction exist?

25 For the autos in the figure re they independently controllable? Does interaction exist? Connected by spring

26 For the autos in the figure re they independently controllable? Does interaction exist? Connected by beam

27 For process Example #: the blending process re the CVs independently controllable? Does interaction exist? F, x F S, x S = F M, x M F' x' M M = F' = (F + F' s F S S + F ) 2 ss F ' + F (Fs + F ) 2 ss F ' S

28 For process Example #: the blending process re the CVs independently controllable? Does interaction exist? F, x F S, x S = F M, x M F' x' M M = F' = (F + F' s F S S + F ) 2 ss F ' + F (Fs + F ) 2 ss F ' S Det(K) = (F F 2 + F S ) 2 (F F 2 S + F S ) 2 Yes, this system is controllable!

29 F R x D For process Example #2: the distillation tower re the CVs independently controllable? Does interaction exist? F V x x D B ( s) = ( s) =. 747e 2s + 3s 3. 3s. 73e. 7s + F R F ( s) R ( s). 667e 5s + 2s 2s. 253e. 2s + F V F ( s) V ( s) x B

30 For process Example #2: the distillation tower x D x D ( s) =. 747e 2s + 3s F R ( s). 667e 5s + 2s F V ( s) F R x B ( s) = 3. 3s. 73e. 7s + F R ( s) 2s. 253e. 2s + F V ( s) F V Det (K) =.54 x -3 Small but not zero (each gain is small) The system is controllable! x B

31 For process Example #3: the non-isothermal CSTR re the CVs independently controllable? Does interaction exist? B -r = k e -E/RT C G + + v G 2 G d T G 2 G d2 C B v2 G

32 For process Example #3: the non-isothermal CSTR re the CVs independently controllable? Does interaction exist? B -r = k e -E/RT C v T The interaction can be strong In general, the temperature and conversion (extent of reaction) can be influenced. The system is controllable. C B v2 (See ppendix 3 for examples)

33 For process Example #4: the mixing tank re the CVs independently controllable? Does interaction exist? G + + v G 2 G d T G 2 G d2 C G v2

34 For process Example #4: the mixing tank re the CVs independently controllable? Does interaction exist? G + + v G 2 G d T G 2 G d2 C G v2 Nothing affects composition at S-S; the system is NOT controllable.

35 For process Example #5: the non-isothermal CSTR re the CVs independently controllable? Does interaction exist? v B -r = k e -E/RT C v2 C B T G G 2 G 2 G G d G d2 + +

36 For process Example #5: the non-isothermal CSTR re the CVs independently controllable? Does interaction exist? Solution continued on next slide B -r = k e -E/RT C Both valves have the same effects on both variables; the only difference is the magnitude of the flow change (µ = constant). v v2 T C B T = = K K 2 µ K µ K 2 MV MV 2 C B Det (K) = ; not controllable!

37 For process Example #5: the non-isothermal CSTR In this case, both MVs affect ONE common variable, and this common variable affects both CVs. We can change both CVs, but we cannot move the CVs to independent values! MV G G + + Input Contraction G d CV D Solution continued on next slide G d2 G MV 2 G CV 2

38 For process Example #5: the non-isothermal CSTR For input contraction, multivariable feedback control is not possible; the system is not controllable! We can change both CVs, but we cannot move the CVs to independent values! SP - + G c Does not work!! MV G + + G d G d2 CV D Solution complete SP G c2 MV 2 G CV 2

39 For process Example #6: the non-isothermal CSTR re the CVs independently controllable? Does interaction exist? v C B C C B + 2C -r = k e -E/RT C G G 2 G 2 G G d G d2 + + v2

40 For process Example #6: the non-isothermal CSTR re the CVs independently controllable? Does interaction exist? B + 2C -r = k e -E/RT C Solution continued on next slide Using the symbol N i for the number of moles of component i that reacts, we have the following. v C B C C v2 N = N N = 2N B Because of the stoichiometry, C N C = 2 N B and the system is not controllable!

41 For process Example #6: the non-isothermal CSTR re the CVs independently controllable? Does interaction exist? B + 2C -r = k e -E/RT C Solution continued on next slide v C C B C = = K 2K 2 K 2 K 2 MV MV 2 C B C C v2 Det (K) = ; not controllable!

42 For output contraction, both MVs affect both CVs, but the CVs are related through the physics and chemistry. We can change both CVS, but we cannot move the CVs to independent values! MV G + output contraction + CV G 2 G d D Solution continued on next slide G G d2 MV 2 G 22 + CV 2

43 In this case, multivariable feedback control is not possible; the system is uncontrollable! SP - + G c Does not work!! MV 2 MV G G G d G d2 CV D SP G c2 + + CV 2 Solution complete

44 CONTROLLBILITY Conclusions about determining controllability Lack of controllability when. One CV cannot be affected by any valve This is generally easy to determine. 3. Lack of independent effects. Look for contractions This requires care and process insight or modelling to determine.

45 Some key questions whose answers help us design a multiloop control system.. IS INTERCTION PRESENT? - If no interaction ll single-loop problems 2. IS CONTROL POSSIBLE? - How many degrees of freedom exist? - Can we control CVs with MVs? Let s see how good the performance can be 3. WHT IS S-S ND DYNMIC BEHVIOR? - Over what range can control keep CVs near the set points?

46 How does interaction affect the steady-state behavior of the blending process? Please sketch the achievable values Total flow, F F, x F S, x S = x M, Composition (fraction ) F maximum = 3 F s maximum = 6 F M, x M

47 How does interaction affect the steady-state behavior of the blending process? Please explain this shape Total flow, F feasible F, x F S, x S = infeasible F maximum = 3 F s maximum = 6 Note: F M, x M This shows a range of set points that can be achieved (without disturbances) Composition (fraction )

48 STEDY-STTE OPERTING WINDOW Conclusions about the steady-state behavior of a multivariable process. Shape of the Window The operating window is not generally a simple shape, like a rectangle. 2. What influences the size of the Window? The operating window is influenced by process chemistry, equipment capacity, disturbances 3. How does Window relate to Controllability? Process cannot be moved (controlled) outside of the window. If operating window is empty, the system is not controllable. Summarize your conclusions here.

49 NOW, LET S LOOK T THE DYNMIC BEHVIOR. How many experiments are needed to tune controllers? 2. Which controller should be implemented first? MV G + + CV CV ( s) MV ( s) G 2 G 2 G d G d2 D To tune each controller, we need MV 2 G CV 2 CV2( s) MV ( s) 2

50 3. We have implemented one controller. What do we do now? G + + CV MV G 2 G d D MV 2 G 2 G d2 SP 2 + G c2 - G CV 2

51 4. What if we perform another experiment to learn the dynamics between MV and CV, with controller #2 in automatic? Is the dynamic behavior different from without second controller (G C2 )? What elements does the behavior depend upon? G + + CV MV G 2 G 2 G d G d2 D

52 4. Is the dynamic behavior different from without second controller (G C2 )? What elements does the behavior depend upon? G + direct path interaction path CV MV G 2 CV MV = G G 2 G + G c2 2 G G 22 c2 MV 2 G 2 The process seen by the controller changes! SP 2 + G c2 - G CV 2

53 In general, the behavior of one loop depends on the interaction and the tuning of the other loop. SP - + G c MV G G G d CV Tuning that is stable for each loop might not be stable when both are in operation! SP MV 2 G c2 G 2 G 22 G d2 + + D CV 2 We need to tune loops iteratively, until we obtain good performance for all loops! I think that I need an example again!

54 In general, the behavior of one loop depends on the interaction and the tuning of the other loop. Let s look at a simple example with interaction, = MV MV s e. s e. s e. s e. CV CV s. s. s. s We will pair the loops on the strongest gains, MV CV and MV 2 CV 2

55 Results with only one controller in automatic (K C = 2., T I = 3) This system is stable and perhaps a bit too aggressive..5 IE = 2.53 ISE =.5448 IE = ISE = CV.5 Feedback control for SP change CV Why did this change? SM = SSM = SM = SSM = MV.5 MV Time Time

56 Results with both controllers in automatic (Kc = 2., TI = 3) This system is unstable!! Each was stable by itself!! 5 IE = ISE = IE = ISE = CV CV SM = SSM = SM = SSM = MV MV Time Time

57 In general, the behavior of one loop depends on the interaction and the tuning of the other loop. SP - + G c MV G G G d CV I recall that elements in the denominator affect stability. D MV 2 G 2 G d2 SP G c2 G CV 2 CV SP = numerator + G G + G G + G G [G G G G c c2 22 c c ]

58 In general, the behavior of one loop depends on the interaction and the tuning of the other loop. 4 Single-loop inside the dashed lines would be stable. Notes: Kc2, LOOP 2 CONTROLLER GIN 3 2 Stable for 2x2 Unstable for 2x Kc, LOOP CONTROLLER GIN. TI = 3 for both controllers (reasonable = t63%) 2. KC < 3.75 stable for single-loop feedback 3. KC = 2. stable for one loop 4. KC = 2. unstable for two loops!! 5. These numerical results are for the example only; concepts are general

59 In general, the behavior of one loop depends on the interaction and the tuning of the other loop. 4 Kc2, LOOP 2 CONTROLLER GIN 3 2 Stable for 2x2 Unstable for 2x2 We can have any values in the stable region. How do we choose the best Kc, LOOP CONTROLLER GIN

60 If both CVs are of equal importance, we would detune both controllers equally. (Kc = Kc2 =.95; TI = TI2 = 3.) IE = ISE = IE = ISE =.47 CV Somewhat slow to reach SP CV Interaction affects CV SM = SSM = 8.89 SM =.734 SSM = MV.5 MV Time Time

61 If CV is more important, we would make Gc aggressive and detune Gc2 more. (Kc =.4 and Kc2 =.5; TI = TI2 = 3.) IE = ISE = IE =.2283 ISE = 3.23 CV Fast tracking of the set point CV2.6.4 Increased disturbance from interaction SM = SSM = SM =.763 SSM = MV Time MV Time

62 If CV2 is more important, we would make Gc2 aggressive and detune Gc more. (Kc =.5 and Kc2 =.4; TI = TI2 = 3.) IE = ISE = IE = ISE =.3957 CV Very slow to reach the SP CV Reduced disturbance due to interaction SM = SSM = SM =.763 SSM = MV MV Time Time

63 In general, the behavior of one loop depends on the interaction and the tuning of the other loop. Some conclusions for multiloop PID tuning:. For multiloop, we generally have to tune the controllers in a less aggressive manner than for single-loop. 2. Textbook gives tuning approach, (Kc) ml /2 (Kc) sl 3. We can tune important loop tightly, if we also detune (make less aggressive) other loops.

64 Summary of key questions for multiloop control system. IS INTERCTION PRESENT? - No interaction ll single-loop problems - We use models to determine interaction Fundamental or empirical 2. IS CONTROL POSSIBLE? 3. WHT IS S-S ND DYNMIC BEHVIOR?

65 Summary of key questions for multiloop control system. IS INTERCTION PRESENT? 2. IS CONTROL POSSIBLE? - DOF: The # of valves # of controlled variables - Controllable: Independently affect every CV Check if det [K] Look for contractions 3. WHT IS S-S ND DYNMIC BEHVIOR?

66 Summary of key questions for multiloop control system. IS INTERCTION PRESENT? 2. IS CONTROL POSSIBLE? 3. WHT IS S-S ND DYNMIC BEHVIOR? - The operating window is: strongly affected by interaction strongly affected by equipment capacities not a rectangle or symmetric

67 Summary of key questions for multiloop control system. IS INTERCTION PRESENT? 2. IS CONTROL POSSIBLE? 3. WHT IS S-S ND DYNMIC BEHVIOR? - Interaction affects loop stability - We have to detune to retain stability margin - We can improve some loops by tight tuning, but we have to detune and degrade other loops

68 Workshop in Interaction in Multivariable Control Systems

69 Workshop Problem # You have been asked to evaluate the control design in the figure. Discuss good and poor aspects and decide whether you would recommend the design.

70 Workshop Problem # 2 We need to control the mixing tank effluent temperature and concentration. F T C F 2 T 2 C 2 You have been asked to evaluate the design in the figure. Discuss good and poor aspects and decide whether you would recommend the design. T

71 Workshop Problem # 3 Summarize the underlying principles that can lead to a contraction and to a loss of controllability. These principles will be applicable to many process examples, although the specific variables could be different. Hint: Review the examples in the lecture and identify the root cause for each.