Distillation Part 2 Distillation Modeling and Dynamics April

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Distillation Part 2 Distillation Modeling and Dynamics April 4 -8, 2004 KFUPM-Distillation Control Course

Distillation Part 2 Distillation Modeling and Dynamics April 4 -8, 2004 KFUPM-Distillation Control Course

April 4 -8, 2004 KFUPM-Distillation Control Course 2

April 4 -8, 2004 KFUPM-Distillation Control Course 2

April 4 -8, 2004 KFUPM-Distillation Control Course 3

April 4 -8, 2004 KFUPM-Distillation Control Course 3

Dynamics of Distillation Columns Balance equations in out Accumulated = in – out =d/dt

Dynamics of Distillation Columns Balance equations in out Accumulated = in – out =d/dt (inventory) Assumptions (always used) A 1. Perfect mixing on all stages A 2. Equilibrium between vapor on liquid on each stage (adjust total no. of stages to match actual column) A 3. Neglect heat loss from column, neglect heat capacity of wall and trays Additional Assumptions (not always) A 4. Neglect vapor holdup (Mvi ≈ 0) A 5. Constant pressure (vapor holdup constant) NOT GOOD FOR CONTROL! A 6. Flow dynamics immediate (Mli constant) A 8. Constant molar flow (simplified energy balance) A 9. Linear tray hydraulics April 4 -8, 2004 KFUPM-Distillation Control Course 4

April 4 -8, 2004 KFUPM-Distillation Control Course 5

April 4 -8, 2004 KFUPM-Distillation Control Course 5

1 Tray Hydraulics (Algebraic). Trays: Francis weir formula Li = k Mi 2/3 .

1 Tray Hydraulics (Algebraic). Trays: Francis weir formula Li = k Mi 2/3 . A 9 Simplified (linearized) (for both tray and packings): τL: time constant for change in liquid holdup (≈2 -10 sek. ) λ: effect of increase in vapor rate on L (usually close to 0) Li 0, Mi 0, Vi 0: steady-state values (t=0). 1 Pressure drop (algebraic): Δpi = f(Mi, Vi, …) April 4 -8, 2004 KFUPM-Distillation Control Course 6

Numerical solution “)integration(” April 4 -8, 2004 KFUPM-Distillation Control Course 7

Numerical solution “)integration(” April 4 -8, 2004 KFUPM-Distillation Control Course 7

Composition Dynamics April 4 -8, 2004 KFUPM-Distillation Control Course 8

Composition Dynamics April 4 -8, 2004 KFUPM-Distillation Control Course 8

Component balance whole column; April 4 -8, 2004 KFUPM-Distillation Control Course 9

Component balance whole column; April 4 -8, 2004 KFUPM-Distillation Control Course 9

Assumption A 7. “All trays have some dynamic response”, that is, (B) Justification: Large

Assumption A 7. “All trays have some dynamic response”, that is, (B) Justification: Large interaction between trays because of liquid and vapor streams. (Reasonable if 1 À x) Substitute (B) into (A): (C) April 4 -8, 2004 KFUPM-Distillation Control Course 10

April 4 -8, 2004 KFUPM-Distillation Control Course 11

April 4 -8, 2004 KFUPM-Distillation Control Course 11

Comments on τ1 -Formula (C) 1. No linearization (change may be large)! 2. More

Comments on τ1 -Formula (C) 1. No linearization (change may be large)! 2. More convenient formula for denominator: Example: Change in feed composition (with F=0, D=0, B=0): Denominator = F z. F 3. Only steady-state data needed! (+holdups). Need steady -state before (t=0) and after (t=∞) upset. April 4 -8, 2004 KFUPM-Distillation Control Course 12

4. 1 (C) applies to any given component 5. τ1 may be extremely large

4. 1 (C) 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, for increase in L and V with ∆D=0 and ∆B=0. Reason: Denominator =0, (will find τ2<τ1! See later) April 4 -8, 2004 KFUPM-Distillation Control Course 13

Example: Three-stage column Compositions Flows Stage i with z. F=0. 50 Li Vi Xi

Example: Three-stage column Compositions Flows Stage i with z. F=0. 50 Li Vi Xi Yi with z. F=0. 51 Xi Yi Condenser 3 3. 05 0. 9000 Feed Stage 2 4. 05 3. 55 0. 4737 0. 900 0. 5001 0. 9091 Reboiler 0. 1000 0. 5263 0. 1109 0. 5549 1 3. 55 0. 9091 Mi=1 mol on all stages Estimated Dominant Time Constant using (C): Excellent agreement with observed 4. 5 minute April 4 -8, 2004 KFUPM-Distillation Control Course 14

Example 2. Propane-propylene (C 3 -splitter) 110 theoretical stages = 1. 12 (relative volatility)

Example 2. Propane-propylene (C 3 -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 April 4 -8, 2004 z. F y. D x. B Simulation (t=0) 1 0. 65 0. 995 0. 100 0. 714 Simulation (t=∞ ) 2 0. 60 0. 958 0. 030 0. 495 KFUPM-Distillation Control Course 15

Simulated response to step change in z. F: 0. 65 to 0. 60 xfeed

Simulated response to step change in z. F: 0. 65 to 0. 60 xfeed stage Observed 1 close to 480 min 63% April 4 -8, 2004 KFUPM-Distillation Control Course 16

Variation in τ1 with operating point (at least for small changes) April 4 -8,

Variation in τ1 with operating point (at least for small changes) April 4 -8, 2004 KFUPM-Distillation Control Course 17

Detailed simulations for propane-propylene (C 3) splitter (”column D”) April 4 -8, 2004 KFUPM-Distillation

Detailed simulations for propane-propylene (C 3) splitter (”column D”) April 4 -8, 2004 KFUPM-Distillation Control Course 18

Propane-propylene. Simulated composition response with detailed model. Change in feed composition (z. F from

Propane-propylene. Simulated composition response with detailed model. Change in feed composition (z. F from 0. 65 to 0. 60). 20 min Top: 20 min 2”delay” + second order 30 h April 4 -8, 2004 About 7 min “delay” because composition change has to go through 39 stages (by chance this happens to be similar to the 19 KFUPM-Distillation Control liquid flow. Course dynamics in this case)

Simulated response. Mole fraction of light component on all 110 stages Change in z.

Simulated response. Mole fraction of light component on all 110 stages Change in z. F from 0. 65 to 0. 60. min 30 h April 4 -8, 2004 KFUPM-Distillation Control Course 20

Same simulations using Aspen. . Almost the same x. D x. F x. B:

Same simulations using Aspen. . Almost the same x. D x. F x. B: initial x. B April 4 -8, 2004 KFUPM-Distillation Control Course 21

n What happens if we increase both L and V at the same time?

n What happens if we increase both L and V at the same time? April 4 -8, 2004 KFUPM-Distillation Control Course 23

Propane-propylene. Simulated composition response with detailed model. Increase both L and V by same

Propane-propylene. Simulated composition response with detailed model. Increase both L and V by same amount ( V = L = +0. 05) feed stage Very small effect April 4 -8, 2004 KFUPM-Distillation Control Course 24

Propane-propylene. Simulated composition response with detailed model. Increase both L and V by same

Propane-propylene. Simulated composition response with detailed model. Increase both L and V by same amount ( V = L = +0. 05) ( D= B=0) Very small effect! Note axis: Have “blown up” to see details 0. 9951 Xfeed stage XD 0. 9950 0. 100 0. 0998 Both products get purer. XB 0. 0992 April 4 -8, 2004 Response for x. B a bit “strange” (with very large overshoot) because it takes some time for L to reach the bottom KFUPM-Distillation Control Course 25

External and Internal Flows Steady–state composition profiles (column A) External flows Internal flows stage

External and Internal Flows Steady–state composition profiles (column A) External flows Internal flows stage 0 0 COMP L = B = - D = (-0. 1, -0. 01, 0, 0. 01, 0. 1) V= L = (1, 0, -1). COMP B= D=0 Large effect on composition (large “gain”) Small effect on composition (small “gain”) One products get purer – the other less pure Both products get purer or both less pure Effect on composition obtained by assuming separation factor S constant Effect on composition obtained by considering change in S MAIN EFFECT ON COMPOSITION BY ADJUSTING D/F; “FINE TUNE” WITH INTERNAL FLOWS April 4 -8, 2004 KFUPM-Distillation Control Course 27

Dynamics Internal Flows External Flows with no flow dynamics Δx. B 63% Δy. D

Dynamics Internal Flows External Flows with no flow dynamics Δx. B 63% Δy. D 0 63% 0 Δy. D Δx. B τ1 time Step ΔL = ΔB = -ΔD τ2 time Step ΔL = ΔV Conclusion: Large S. S. effect Small S. S. Effect Slow (τ1) Faster (τ2). Can show: But: Derived when flow dynamics neglected (doubtful since τ2 is relatively small) See simulations April 4 -8, 2004 KFUPM-Distillation Control Course 28

Internal flows dynamics (unrealistic) April 4 -8, 2004 KFUPM-Distillation Control Course 29

Internal flows dynamics (unrealistic) April 4 -8, 2004 KFUPM-Distillation Control Course 29

SUMMARY External flows (change D/F) Internal flows (D/F constant) Increase V (and L) with

SUMMARY External flows (change D/F) Internal flows (D/F constant) Increase V (and L) with D constant LV-configuration April 4 -8, 2004 DV-configuration KFUPM-Distillation Control Course 30

FLOW DYNAMICS (variations in liquid holdup) A 8. Constant molar flows A 4. Neglect

FLOW DYNAMICS (variations in liquid holdup) A 8. Constant molar flows A 4. Neglect vapor holdup Total material balance becomes i+1 Li+1 i Mi Li April 4 -8, 2004 KFUPM-Distillation Control Course 31

Individual Tray Hydraulics A 9. : Assume simplified linear tray hydraulics = hydraulic time

Individual Tray Hydraulics A 9. : Assume simplified linear tray hydraulics = hydraulic time constant (because Mi varies with L) = effect of change in Vi on Li (λ>0 when vapor pushes liquid off tray. For λ>0. 5: inverse response) April 4 -8, 2004 KFUPM-Distillation Control Course 32

Flow dynamics for Column section Consider Deviations from Initial Steady-State (ΔLi=Li-Lio, …) “Each tray

Flow dynamics for Column section Consider Deviations from Initial Steady-State (ΔLi=Li-Lio, …) “Each tray one tank” ΔL N Column: Combine all trays Consider response in LB to change in L: N tanks in series, each time constant τL ΔL 0. 5ΔL V ΔV ΔLB t θL=N·τL (“almost” a dead time) April 4 -8, 2004 KFUPM-Distillation Control Course 33

Approximate N small lags L in series as a time delay April 4 -8,

Approximate N small lags L in series as a time delay April 4 -8, 2004 KFUPM-Distillation Control Course 34

Linearization of “full” model Need linear models for analysis and controller design Obtain experimentally

Linearization of “full” model Need linear models for analysis and controller design Obtain experimentally or by 1) 2) Put together simple models of individual effects (previous pages) Linearize non-linear model (not as difficult as people think) Given Tray Linearize, introduce deviation variables, simplification here: assume: i) const. , ii) const. molar flows Li = Li+1 = L Vi-1= Vi = V April 4 -8, 2004 KFUPM-Distillation Control Course 35

= + A Δx + State matrix (eigenvalues determine speed of response) B ΔL

= + A Δx + State matrix (eigenvalues determine speed of response) B ΔL ΔV inputs Input matrix “states” (tray compositions) + Equations for d. Mi/dt=…… ÞCan derive transfer matrix G(s) April 4 -8, 2004 KFUPM-Distillation Control Course 36

Are dynamic really so simple? n 1970’s and 1980’s: Mathematical proofs that dynamics are

Are dynamic really so simple? n 1970’s and 1980’s: Mathematical proofs that dynamics are always stable q n In reality, independent variables are q q n Based on analyzing dynamic model with L and V [mol/s] as independent variables Lw [kg/s] = L [mol/s] ¢ M [kg/mol] QB [J/s] = V [mol/s] ¢ Hvap [J/mol] Does it make a difference? YES, in some cases! April 4 -8, 2004 KFUPM-Distillation Control Course 37

More about mass reflux and t=0: z is decreased from 0. 5 to 0.

More about mass reflux and t=0: z is decreased from 0. 5 to 0. 495. instability F Lw[kg/s]= L[mol/s]/M where M [kg/mol] is the molecular weight, Data: ML=35, MH=40. What is happening? Mole wt. depends on composition: more heavy ! M up ! L down ! even more heavy. . . ) Can even get instability! With MH=40, instability occurs for ML<28 (Jacobsen and Skogestad, 1991) April 4 -8, 2004 KFUPM-Distillation Control Course 38

Instabiliy for “ideal” columns: Many people didn’t believe us when we first presented it

Instabiliy for “ideal” columns: Many people didn’t believe us when we first presented it in 1991! Likely to happen if the mole weights are sufficiently different April 4 -8, 2004 KFUPM-Distillation Control Course 39

II Reflux III I IV IV Reflux back again. . but not composition !?

II Reflux III I IV IV Reflux back again. . but not composition !? Top composition II I April 4 -8, 2004 KFUPM-Distillation Control Course 40

Multiple steady state solutions IV III V II I April 4 -8, 2004 KFUPM-Distillation

Multiple steady state solutions IV III V II I April 4 -8, 2004 KFUPM-Distillation Control Course 41

IV III V II I April 4 -8, 2004 KFUPM-Distillation Control Course 42

IV III V II I April 4 -8, 2004 KFUPM-Distillation Control Course 42

V IV I April 4 -8, 2004 KFUPM-Distillation Control Course 43

V IV I April 4 -8, 2004 KFUPM-Distillation Control Course 43

Actually not much of a problem with control! IV V I This is why

Actually not much of a problem with control! IV V I This is why you are not likely to notice it in practice. . . unless you look carefully at the reflux. . will observe inverse response in an unstable operating point (V) Conclusion: “Simple” dynamics OK for our purposes April 4 -8, 2004 KFUPM-Distillation Control Course 44

Nonlinearity The dynamic response of distillation column is strongly nonlinear. However, simple logarithmic transformations

Nonlinearity The dynamic response of distillation column is strongly nonlinear. However, simple logarithmic transformations counteract most of the nonlinearity. Derivation: ln xi: Logarithmic composition CONCLUSION: Response nearly linear (constant gain) with log. comp. April 4 -8, 2004 KFUPM-Distillation Control Course 45

Xi+1 Xi Ref. S. Skogestad. “Dynamics and control of distillation columns: A tutorial introduction”,

Xi+1 Xi Ref. S. Skogestad. “Dynamics and control of distillation columns: A tutorial introduction”, Trans. IChem. E, Vol. 75, 1997, p. 553 April 4 -8, 2004 KFUPM-Distillation Control Course 46

In general: Use logarithmic compositions Mole fraction of Light key component on stage i

In general: Use logarithmic compositions Mole fraction of Light key component on stage i Mole fraction of Heavy key component on stage i May also be used for temperatures! Temp. bottom of column (or boiling point heavy) Temp. on stage i Temp. top of column (or boiling point light) Derivation: April 4 -8, 2004 KFUPM-Distillation Control Course 47

Initial Response to 10% ∆L: (V constant) (Column A with Flow Dynamics) nonline ar

Initial Response to 10% ∆L: (V constant) (Column A with Flow Dynamics) nonline ar Linear Extremely non linear model ∆x. B ∆y. D nonline ar Linear model Log: Counteracts Nonlinearity -∆ln(1 y. D) ∆ ln x. B April 4 -8, 2004 KFUPM-Distillation Control Course 48

April 4 -8, 2004 KFUPM-Distillation Control Course 49

April 4 -8, 2004 KFUPM-Distillation Control Course 49

Conclusion dynamics n n n Dominant first order response – often close to integrating

Conclusion dynamics n n n Dominant first order response – often close to integrating from a control point of view Liquid dynamics decouples the top and bottom on a short time scale, and make control easier Logarithmic transformations linearize the response April 4 -8, 2004 KFUPM-Distillation Control Course 50