1. Revised for: NDF Workshop III Trondheim, Norway, Aug. 22, 1991 Additional Literature: Dynamics: p. 75 in NDF Notes 1990 Control: p.57 in NDF Notes 1990 (AIChE, May 1990) Dynamics and Control of Distillation Columns 1. Distillation Dynamics (45%)……….p.2 2. Distillation Control (30%)……….....p.23 3. Some basic control theory (25%)…..p.37 SIGURD SKOGESTAD NDF Workshop I Lyngby, Denmark, May 24, 1989

2. 1. Distillation Dynamics 1.1 Introduction3 1.2 Degrees of freedom/steady state4 1.3 Dynamic Equations5 1.4 composition Dynamics 1.4 composition Dynamics7 1.5 Flow dynamics 19 1.6 Overall Dynamics 21 1.7 Nonlinearity 21A 1.8 Linearization 22 1.4.1 Dominating time constant 8 1.1.4 Internal flows 17 2

3. 1.1 Introduction Typical column: 1 feed, 2 products, no intermediate cooling 3

4. 3a Manipulated Variables (Valves): (μ) V (indirectly), L, D, B, VT (indirectly)(5) Disturbances: (d) F, zF, qF (= fraction liquid in feed, feed enthalpy), rain shower, + all u’s above Controlled Variables (y) M O, M B, P,Y P, X B (5) InventorySeparation (composition, temperature )

5. ( 1.2) STEADY-STATE OPERATION Must keep holdups (MD,MB,MV) constant This uses 3 degrees of freedom (u’s); only two left. The two degrees of freedom may be used to specify two product specifications 4 e.g.,Y D and x B T top and T atm D (product rate) and Y D etc. Recall specifications for steady-state simulations (PROCESS)

6. 4a Steady-state behavior (design) Infinite reflux, exact (Fenske): Finite reflux, good approximation: holds for columns with feed at optimal location Example: 3-Stage Column Example: 3-Stage Column Constant p Constant p Constant holdup liquid and level Constant holdup liquid and level Negligible vapor h. Negligible vapor h. Constant molar flows Constant molar flows Constant α=10 Constant α=10 2 Vicor stages + total condenser 2 Vicor stages + total condenser Mi = 1 kmol (i = 1,2,3) Mi = 1 kmol (i = 1,2,3) (slight modification of Jafarey, Douglas, McAvoy, 1979) Overall separation, binary

7. 4b Get: Total reflux: Approximation: Actual column: Stage Condenser Feedstage Reboiler i123i123 L i 3.05 4.05 V i 3.55 33.55 X i 0.9000 0.4737 0.1000 y i 0.9000 0.5263 D = 0.5 Y D =0.9 M2M2 M2M2 V=3.5 M2M2 C = 1k mol / min Z F =0.5 B = 0.5 X B = 0.1 L=3

8. 4c i+1 L X L+1 VYiVYi B X B (x0) <1 <1 x1 = 0

9. 5 Additional Assumptions (not always) A4. Neglect vapor holdup (M vi ≈ 0) A5. Constant pressure (vapor holdup constant) A6. Flow dynamics immediate (M vi constant) A8. Constant molar flow A9. Linear tray hydraulics Assumptions (always used) A1. Perfect mixing on all stages A2. Equilibrium between vapor on liquid on each stage (adjust total no. of stages to match actual column) A3. Neglect heat loss from column, neglect heat capacity of wall and trays 1.3 Dynamics of Distillation Columns Balance equations Accumulated = in – out =D/DT (inventory) in out

10. 6 stage i+1 i i-1 M i+1 ViVi L i+1 MiMi V i-1 LiLi Balance equation for stage without feed/side draw INDEP. DIFF.Component balance (index for component not Shown EQ. FLASH (Strongly Nonlimear)

11. 6a Total balance (sum of component balances) Energy balance Tray Hydraulics (Algebraic) A9.Simplified (linearized): τ L : time constant for change in liquid holdup (≈2-10sek.) λ: effect of increase in vapor rate on L L i0,M i0,V i0 : steady-state values (t=0). (ALGEBRAIC) Pressure drop: Δp i = f(M i,V i,…)

12. 6b Numerical Solution (Integration) More details: p.82 in NDF Notes 1990 Moles of component on tray i Solution (at given time) Given value of state variable Perform constant nU-flash (given internal energy and phase split (M Li, M vi ), compositions (x i, y i ), temperature, pressure (ρ i ), spec energy (hi, Vi) Compute Li and Vi from tray hydraulics and pressure drop relations 2. Simplified Approach, neglect vapor holdup (M vi = 0) State Variables : (N L on each tray) Mole fraction total holdup → X i → M i (note: n i = M i x i )

13. 6a.1 Solution : 1. Given value of state variables, guess pressure, P i 2. Perform bubble point flash (given x i,p i ) → y i,T i, enthalpies 3. Compute Vi from energy balance (gives “index problem”: LHS (derivative) is known) Common simplification: Use dV i 1dt = dh i l /dt from previous step. (Do not set d/dt (M i V i ) ≈ 0 constant molar flows 4. Compute L i and Pi from tray hydraulics and pressure drop.

14. 7 (1.4) Composition Dynamics SUM: (N C independent differential equations) xN No. of componentsNo. of trays Might expect very high-order complicated behavior. Surprise: The dominant composition dynamics is approximately 1 order! Fig. Response in Y D to step change τ 1 : approximately independent of what we step (reflux, feed rate, boilup,…) and what and where we measure (Y D, T top, T btm, etc,…) 100% τ1τ1 time

15. 7a EXAMPLE: 3-stage column (see p.4B)  Neglect flow dynamics (M u = 1 = constant)  1 state on each tray  Constant molar flows (V 2 = V 3 ) Fig. Response for 3-stage column to feed composition change. Note: Composition change inside column much larger than at column ends. This is the main reason for the “slow” composition response ∆X 2 (t) Feed tray ∆X 3 (t) reboiler ∆X 1 (t) condenser 63% τ 1 =4.5 min. Time (min) VLE Step in Z F from 0.50 50 0.51 Dist.m: (matlab subroutine)

16. Composition Response of an Individual Tray Component material balance, tray i Assume the column is at steady-state, and consider the effect of an increase in x i f i to x i+1 +∆x i+1. Assume flows constant, and neglect interactions between the trays (y i-1 : constant). In terms of deviation variables Where the linearized VLE-constant is 7b i+1 i i-1 MixiMixi x i+1

17. Overall response time from top to bottom of column (neglecting “vapor” interactions) total HOLDUP Inside Column Example: 3-stage Column Conclusion: Do not yet correct overall response time (4.5 min) by simply adding together individual trays. 7c Collecting ∆K i terms we get a 1 st order response with time constant

18. 8 Response to step change in reflux )This does not imply that flow dynamics are not important for composition control!) In particular, assume liquid holdup (Mi) constant time Assumptions : Use A1-A6 A4. M vi is negligible (OK when pressure is low; at 10 bar M vi will be about 10% of liquid holdup) A6. Flow dynamics much faster than composition dynamics. L LB xBxB LBLB θLθL xBxB Objectives:  Understand why overall response 1 st order  Develop formula for τ 1  When does τ 1 not apply? τ1τ1 1.4.1 τ 1 Dominant Time Constant

19. F1zFF1zF Bf x B f Assumption 6: D(t) = D f B(t) = B f Component balance whole column; Column at Steady-State at t ≤ 0 Something Happens at t = 0 (not steady-state) New Steady-State t= (subscript f=final) F t,z Ft B,x B B(t) x B (t) D(t) Y D (t) Df YDf D,Y D t = 0t > 0t =

20. Assumption A7. “All trays have some dynamic response”, that is, (4) Justification: Large interaction between trays because of liquid and vapor streams. (Reasonable if Substitute (4) into (3): 10 5

21. 10a

22. 11 4. (5) applies to any given component 5. τ1 may be extremely large if both products pure (Reason: Numerator>>Denominator because compositions inside column change a lot, while product compositions change very little). 6. Limitation: τ1 Does not apply to changes in INTERNAL FLOWS ONLY, that is, L and V increase with ∆D=0 and ∆B=0. Reason: Denominator (τ)=0, (will find τ2<τ1!)q 3. Only steady-state data needed! (+holdups) Need steady- state before (t=0) and after (t=∞) upset. Comments on τ 1 -Formula (5)

23. Estimated Dominant Time Constant Excellent agreement with observed 4.5 minute Check of Assumption (7): “All trays have same dynamic response”(because of tray interactions) Example (Continued): Three-stage column Stage Condenser Feed Stage Reboiler i123i123 Li 3.05 4.05 Vi 3.55 Flows Compositions with Z F =0.50with Z F =0.51 X i Y i 0.9000 0.4737 0.900 0.1000 0.5263 0.9091 0.5001 0.9091 0.1109 0.5549 12

24. Reasonable if Overall response time (incl. Tray interactions) Response time neglecting tray interactions

25. 13 Example 2. Propane-propylene splitter  110 theoretical stages   = 1.12 (relative volatility)  Assume constant molar flows  L/D = 19, D/F = 0.614   Find τ 1 when Z F decreases from 0.65 to 0.60 All flows kept constant Simulation 1 (t=0) Simulation 2 (t=∞ ) z F y D x B 0.650.9950.1000.714 0.600.9580.0300.495

26. Response: 0.05.63 Δy D 0 01h 8h10h time

27. 14 τ 1 : Shortcut Formula Make some simplifying assumptions which hold best for columns with large reflux pure products Small changes Binary separation (or use pseudobinary) (ln is typically from 4 to 15) Columns with pure products I S is small Varies with oper. Cond!

28. 15 For small changes assume (*) applies. Have Reasonable agreement: NOTE: Contribution small from condenser because purity is high so absolute changes in compositions are small

29. 16 Variation in τ 1 with operating point Typical plot. May be derived from shortcut formula. Conclusion: Time constant depends on operating conditions – mainly on purity of least pure product (I S ). -Get “asymmetric” behavior: a.) Least pure product gets purer: Time constant gets longer (slow response) b.) Least pure product gets less pure: Time constant gets smaller (fast response) τ1τ1 1 Peak is large if both products are pure Example Pure “Fast” “slow” time getting pore ΔZ F

30. 17 Large effect on composition (large “gain”) Effect on composition obtained by assuming separation factor constant Small effect on composition (small “gain”) Effect on composition obtained by considering change in S: (“separation unchanged split changed”) (“separation changed split unchanged”) MAIN EFFECT ON COMPOSITION BY ADJUSTING D/F; “FINE TUNE” WITH INTERNAL FLOWS (1.4.2) τ 2 External and Internal Flows Steady -state COMP tray COMP100 AB=ΔDAB=ΔD=0

31. Dynamics External FlowsInternal Flows Step ΔB = -ΔDStep ΔL = -ΔV Conclusion: Large S.S. effect Slow (τ 1 ) Small S.S. Effect Faster (τ 2 ) Initial Response: Not as Different (see more accurate formulas in Skogestas & Morari, 1988 I E E C Res, 27, 1848-1862) But: Derived when flow dynamics neglected (doubtful since τ 2 is relatively small) For columns with pure products Recall: time 63% 0 Δx B Δy D τ1τ1 τ2τ2 0 Δx B

32. 19 FLOW DYNAMICS (variations in liquid holdup neglected so far) A8. Constant molar flows A4. Neglect vapor holdup Total material balance becomes i+1 i Mi Li Li+1

33. Tray Hydraulics A9.: Assume simplified linear tray hydraulics = hydraulic time constant = effect on change in Vi on Li (vapor may push liquid off tray, λ>0.5: inverse response) 19a

34. 20 Consider Deviations from Initial Steady-State (Δli=Li-Lio,…) Then Consider response in L B to change in L: N tanks in series, each time constant τ L 0.5ΔL ΔL ΔL B t θ L =N·τ L (“almost” a dead time) ΔV V Response in LB to change in V: “Vapor pushes liquid off each tray” ΔL B ΔL N ΔV θ L =N·τ L t ΔL B 0 λ·ΔV

35. 21 1.6Overall Dynamics Composition Dynamics Typical value External flows, τ 1 250 min Internal flows, τ 2 20 min + Liquid Flow Dynamics “Dead time” from top to bottom, θ L 3 minute + Pressure dynamics + Top level – “ – + BTM level – “ – + Valve dynamics, + Heat transfer dynamics (V and V t indirectly controlled) + Measurements Dynamics Depend on tuning of pressure and level controls. Typical time constant: 0.5 minute 0.2 minute 0.1 minute Essential for control ! Simplified Model Describe each effect independently “Add” together to give overall dynamics (Alternative: Linearize all equations) NOTE: Exact value of τ 1 not important for control!

36. 1.7 Nonlinearity The dynamic response of distillation column is strongly nonlinear. However, simple logarithmic transformations counteract most of the nonlinearity. More details: p.132-133 in NDF Notes 1991 Light key component on tray i Heavy component May also be used for temperatures! Temp. BTM. Of Column Temp. on Tray i Temp. Top of Column 21A X i+1 XiXi tray

37. Initial Response to 10% ∆L: (V constant) (Column A with Flow Dynamics) nonlinear Linear model nonlinear Linear model Extremely non linear ∆xB∆xB ∆yD∆yD Log: Counteracts Nonlinearity -∆(n(1-y D ) ∆ ln x B 21b

38. 22 1.8Linearization Need linear models for controller design Obtain by Given Tray Linearize, introduce deviation variables, simplification here: assume: i) const., ii) const. molar flows   L i =  L i+1 =  L  V i-1 =  V i =  V 1)Put together simple models of individual effects (previous page) 2) Linearize non-linear model

39.  Can derive transfer matrix G(12) + Equations for dMi/dt=…… A Δx + B ΔL ΔV State matrix (eigen values determine speed of response) Output matrix “states” (tray compositions) inputs 22a = +

40. 23 (2) Distillation Control Multivariable vs. Single-Loop Control…25 Choice of Control Configurations…27 LV DV, L DB 1) Disturbance Sensitivity………….31 2) Feedback Control Properties…….34 One-Point Two-Point 1) Implementation, Level Control…..35

41. 24 Distillation Control Manipulated inputs, μ 1 s: L,V,D,B,UT Controlled outputs, Y 1 s: Mo, Mb,,Y D, x B COLUMN LVDBUTLVDBUT MDMBMVYDXBMDMBMVYDXB 5X5

42. 25 Controller Design “Full” 5x5 multivariable controller? NO! Instead use simpler scheme (decentralized control) i)3 single loops (PI, PID) for Md, MB, MV (levels and pressure) ii)There are now left two degrees of freedom for quality controller (keep Y D and x B at desired values). Design as two single loops or 2x2 controller (e.g. decoubler) PROBLEM 1: Choice of Control Configuration (Structure). Which two degrees of freedom should be left for equality control? (Same as choice inputs for level control ) PROBLEM 2: Design of Controllers i) Level controllers ii) Composition controllers remaining 2x2 system

43. 26 Options Composition Control a)“No control” that is, manual operation (e.g. reflux L and boilup V are set manually by the operator b) “One-point” control: One composition controlled automatically. (e.g. y D controlled with L using PI-controller, the other input is n manual, x B “floats”). Most common in industry, often necessary because of constraints. c) “Two-point” control: Both compositions unclear feedback control. $$! Potential for large economic savings. (larger throughput, more products, less energy) Possible Controller (2x2) (1)Single Loops Problem: Interactions (performance) Advantage: Robustness (2) M i variable, for example, decoupler Problem: Often not robust (Sensitive to errors)

44. 27 Problem 1. 1)LV-configuration (“conventional”, “energy balance”, “indirect material-balance”) Reflux L and boilup V used for composition control Level control: M D  D M B  B M V (P)  VT (cooling) 2) DV-configuration (“(direct) material balance”) because D is used Level control: M D  L M B  B 3) LB-configuration (“(direct) material balance”) Level control: M D  D M B  V 4)L/D V/B-configuration (“double ratio”) Level control: MD  both L and D (with L/D constant) MB  both V and B (with V/B constant)

45. 28 Top of Column LV-configuration LB-configuration LC LSLS VTVT L+DD L DV-configuration DB-configuration LC L L+D VTVT D DSDS

46. Comment: Will usually use cascade control on inputs for composition control L LSLS No cascade (manipulated valve position) L P LmLm LSLS With cascade using flow measurements (remove nonlinear valve characteristics 28a C L L+D VTVT D LC DIV (40)m (L/D) s Set manually or from composition controller

47. 29 Open-loop response to flow disturbance:

48. 30 BADPreviously rejected from steady- state considerations (Perry, 1973; Shinskey, 1984; Skogestad & Morari, 1987; Haggblom & Waller, 1988). Works because mass may be accumulated dynamically (change in liquid level) DB(!!) (Finco & Luyben, 1989)

49. 31 DIFFERENCE BETWEEN CONFIGURATIONS 1) Disturbance sensitivity (“self regulating” properties) 2) Interactions between loops, etc. (feedback control properties) 3) Implementation level control 1. DISTURBANCE SENSITIVITY Fact: Composition are mainly dependent on D/F (external flows) should have D/F  Z F High-purity columns: Composition extremely sensitive to small changes in D/F: Consequently: Disturbances which change D/F are “bad” The effect of a given disturbance on D/F depends on the configuration

50. 32 Configuration Disturbance ΔF V (increase in flow rate of vapor in feed) ΔV (increase in boilup) Optimal (see figures p. 29-30) More detailed: See Table 3 in Skogestad, “Disturbance rejection in distillation columns” CHEMDATA’88, Goteberg, June 1988. LP. 65 Literature for 1989 NDF) or p. 40. EXAMPLE

51. 33 1. Summary Disturbance Sensitivity Feed enthalpy Disturbance in V, L, q F boilup reflux Disturbance in F Feed forward from F m Self-regulation ability Note: The above analysis was based on open-loop. A more careful analysis should take dynamics and feedback (disturbance direction into account. (See NDF Notes 1991, p. 205).

52. (2) Feedback Control Properties (2a) One-Point Control (often necessary because of constraints) Configuration Rating LV DV, LB L/D V/B DB Best Dangerous (if D or B fixed Reasonable Unworkable (material balance locked) (2a) Two-Point Control * Use RGA as a function of frequency as a tool, want λ RGA (w B )≈1. See Facohsen et.al. (AIChEF, 1990 May) Configuration LV ← strong interaction DV LB L/D V/B DB Single loops (y D, x B ) (PI or PID) Decoupler + single loops Poor if measure delay Bottom pure: OK; Top pure: POOR Bottom pure: POOR; Top pure: OK Good (except very high purity0 Good Hopeless (NO!) GOOD, (but sensitive to operating point) Not worth it 2x2 Controller (“Problem 2”) 34

53. 35 Implementation and Level Control LV DV LB L/D V/B DB Implementation Level Control (Constraints) Easy Usually good, but not for high reflux(*) Easy OK Easy Level btm: V sometimes inverse response Level top: Poor high reflux(*) Difficult (need to measureGood L,D,V,B!) EasyOK, but level btm: V may give inverse response Almost impossible to control level with small flow (“no power”) LC D (small) L S (large) V T (large)

54. 36 Effect of Level Control tuning on Composition Response LV-configuration: Response time for level control has almost no effect on composition response Configuration, etc.: (as implemented on p. 29-30): Need fast level control to avoid undesired lags in composition response. e.g. DV – or DB – configuration (top) Changing this only affects compositions indirectly through change in L LC L DSDS Level controller sets L

55. 36a Better solution if we do not want to have fast level control: Conclusion (applies to DV, DB,,-configurations, etc.) Top: Level controller always sets L+D Bottom: Level controller always sets V+B This avoids dependency on tuning of level loops. Changing D S directly gives opposite change in L (since Ltd is constant) LC L DSDS FC Level controller sets L LD L+D

56. 3. Some Basic Control Theory 3.1 Transfer Functions | Matrices…………38 1. SISO 2. MIMO 3.2 Effect of Feedback Control ……………43 The sensitivity function $ as a performance measure 3.3 Analysis of Multivariable Processes…45 Interactions Relative Gain Array (RGA) Laplace, frequency response 37

57. 38 Transfer Functions/Matrices 1.SISO (Single Input Single Output) Steady-state “Increasing L from 1.0 to 1.1 changes y D from 0.95 to 0.97” (for example, run Process) Dynamics “The response has a dead time of 2 minutes and then rises with a time constant of 50 minutes” Laplace: Transfer function: time yDyD 10 0.97 63% 0.96 0.95 Θ=2 50 τ=50

58. 39 Alternative Method of Obtaining Transfer Function: Take laplace of linearized model Note: Gain is obtained by using s = o: “gain” = g(o) “gain”=1 Time constant τ=Mi/L Example L (constant) K i+1 M i =constant LxiLxi

59. 39 Alternative Method of Obtaining Transfer Function: Take laplace of linearized model Note: Gain is obtained by using s = o: “gain” = g(o) “gain”=1 Time constant τ=Mi/L Example L (constant) K i+1 M i =constant LxiLxi Response: X i+1 XiXi time τ

60. Direct Generalization: “Increasing L from 1.0 to 1.1 changes y D from 0.95 to 0.97, and x B from 0.02 to 0.03” “Increasing V from 1.5 to 1.6 changes y D from 0.95 to 0.94, and x B from 0.02 to 0.01” Steady-State Gain Matrix 2. MIMO (multivariable case) (Time constant 50 min for y D ) (time constant 40 min for x B ) 40

61. 41 Important Advantages With Transfer Matrices: 1. G(s) is independent of input μ! For given u(s) compute output y(s) as Can therefore make block diagrams G(s) μ(s) y(s) 2. Frequency Response: G(s), with s=jw (pure complex no.) gives directly steady-state response to input sinwt! Response (as things settle) time y 1 (t) μ 1 (t) A A. |g 11 (jw)| g 11 (s) μ μ  y1y1

62. 42 Note Tells directly how much a sine of frequency w is amplified by process w (log scale) Fast sinusoids are “filtered by process (don’t come through BODE PLOT (magnitude only) 0.1 1 2 |g 11 | |g 11 (jw)| (log scale)| 0.010.020.1 =1/50

63. 3.2 Effect of Feedback Control 43 C(s) ysys μ G(s) d y G(s): process (distillation column) C(s): controller (multivariable or single-loop PI’s) y: output, y s : setpoint for output μ: input d: effect of disturbance on output Negative feedback With feedback control Eliminate μ(s) in (1) and (2), y s =0 S (S) =“sensitivity function” = surpression disturbances

64. 44 In practice: Cannot do anything with fast changes, that is, ς (jw) = I at high frequency S is often used as performance measure One Direction Log- scale 1 0.1 Small resonance peak (small peak = large GM and PM) Bandwidth (Feedback no help) Slow disturbance d 1 : 90% effect on y 2 removed w (log) w B τ

65. 44a “Summing up of Channels” “maximum singular value” (worst direction) 0.1 1 w (log) wBwB “minimum singular value” (best direction) w (s)

66. 45 3.3 Analysis of Multivariable processes (Distillation Columns) What is different with MIMO processes to SISO: The concept of “ directions ” (components in u and y have different magnitude ” Interaction between loops when single-loop control is used INTERACTIONS Process Model G y1y1 y2y2 g 12 g 21 g 11 g 22 μ2μ2 μ1μ1 Consider Effect of μ, on y 1 1)“ Open-loop ” (C 2 = 0): y 1 = g 11 (s) · μ 1 2)Closed-loop ” (close loop 2, C 2 ≠0) Change cause by “ interactions ”

67. 46 Limiting Case C 2 →∞ (perfect control of y 2 ) How much has “gain” from u 1 to y 1 changed by closing loop 2 with perfect control The relative Gain Array (RGA) is formed by considering all the relative gains Example from before

68. 46a Property of RGA: Columns and rows always sum to 1 RGA independent of scaling (units) for μ and y. Note: RGA as a function of frequency tell how relative gain changes with frequency. Sigurd’s Law!

69. 47 Use of RGA: (1)Interactions From derivation: Interactions are small if relative gains are close to 1 Choose pairings corresponding to RGA elements close to 1 Traditional: Consider Steady-state Better: Consider frequency corresponding to closed- loop time constant But: Never choose pairing with negative steady-state relative gain

70. 48 (2) Sensitivity measure But RGA is not only an interaction measure: Large RGA-elements signifies a process that is very sensitive to small changes (errors) and therefore fundamentally difficult to control Singular Matrix: Cannot take inverse, that is, decoupler hopeless. Control difficult Large (BAD!)