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Distillation Modeling and Dynamics Distillation Part 2.

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Presentation on theme: "Distillation Modeling and Dynamics Distillation Part 2."— Presentation transcript:

1 Distillation Modeling and Dynamics Distillation Part 2



4 Dynamics of Distillation Columns Additional Assumptions (not always) A4. Neglect vapor holdup (M vi ≈ 0) A5. Constant pressure (vapor holdup constant) A6. Flow dynamics immediate (M li constant) A8. Constant molar flow (simplified energy balance) 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 Balance equations Accumulated = in – out =d/dt (inventory) in out NOT GOOD FOR CONTROL!


6 τ L : time constant for change in liquid holdup (≈2-10sek.) λ: effect of increase in vapor rate on L (usually close to 0) L i0,M i0,V i0 : steady-state values (t=0). 1 Pressure drop (algebraic): Δp i = f(M i,V i,…) 1 Tray Hydraulics (Algebraic). Trays: Francis weir formula L i = k M i 2/3. A9 Simplified (linearized) (for both tray and packings):

7 Numerical solution (“integration”)

8 Composition Dynamics

9 Component balance whole column;

10 Assumption A7. “ 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)


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

13 4.  1 (C) applies to any given component 5. τ 1 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: τ 1 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 τ 2 <τ 1 ! See later)

14 Example: Three-stage column Estimated Dominant Time Constant using (C): Excellent agreement with observed 4.5 minute Stage Condenser Feed Stage Reboiler i321i321 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 M i =1 mol on all stages

15 All flows kept constant Example 2. Propane-propylene (C3-splitter) Simulation 1 (t=0) Simulation 2 (t=∞ ) z F y D x B 0.650.9950.1000.714 0.600.9580.0300.495 110 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

16 Simulated response to step change in z F : 0.65 to 0.60 x feed stage 63% Observed   close to 480 min

17 Variation in τ 1 with operating point (at least for small changes)

18 Detailed simulations for propane-propylene (C3) splitter (”column D”)

19 Propane-propylene. Simulated composition response with detailed model. Change in feed composition (z F from 0.65 to 0.60). About 7 min “delay” because composition change has to go through 39 stages (by chance this happens to be similar to the liquid flow dynamics in this case) 20 min Top: 20 min 2”delay” + second order 30h

20 Simulated response. Mole fraction of light component on all 110 stages Change in z F from 0.65 to 0.60. min 30h

21 Same simulations using Aspen.. Almost the same xB: initial xB xF xD

22 Aspen… very similar

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

24 Propane-propylene. Simulated composition response with detailed model. Increase both L and V by same amount (  V =  L = +0.05) Very small effect feed stage

25 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 0.9950 0.9951 XDXD XBXB X feed stage 0.100 0.0998 0.0992 Both products get purer. Response for x B a bit “strange” (with very large overshoot) because it takes some time for L to reach the bottom

26 Aspen… very similar again

27 External and Internal Flows Large effect on composition (large “gain”) One products get purer – the other less pure Effect on composition obtained by assuming separation factor S constant Small effect on composition (small “gain”) Both products get purer or both less pure Effect on composition obtained by considering change in S Steady–state composition profiles (column A) COMP 0 0 stage External flows Internal flows  L =  B = -  D = (-0.1, -0.01, 0, 0.01, 0.1)  V=  L = (1, 0, -1).  B=  D=0 MAIN EFFECT ON COMPOSITION BY ADJUSTING D/F; “FINE TUNE” WITH INTERNAL FLOWS

28 Dynamics External Flows Internal Flows with no flow dynamics Step ΔL = Δ B = -ΔD Step ΔL = ΔV Conclusion: Large S.S. effect Slow (τ 1 ) Small S.S. Effect Faster (τ 2 ). Can show: time 63% 0 Δx B Δy D τ1τ1 τ2τ2 0 Δx B Δy D But: Derived when flow dynamics neglected (doubtful since τ 2 is relatively small) See simulations

29 Internal flows dynamics (unrealistic)

30 SUMMARY External flows (change D/F) Internal flows (D/F constant) LV-configuration DV-configuration Increase V (and L) with D constant

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

32 Individual Tray Hydraulics A9.: Assume simplified linear tray hydraulics = hydraulic time constant (because M i varies with L) = effect of change in V i on L i (λ>0 when vapor pushes liquid off tray. For λ>0.5: inverse response)

33 Flow dynamics for Column section Consider Deviations from Initial Steady-State (ΔL i =L i -L io,…) 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 ΔL B ΔL N “Each tray one tank” Column: Combine all trays

34 Approximate N small lags  L in series as a time delay 

35 Linearization of “full” model Need linear models for analysis and controller design Obtain experimentally or 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 pages) 2) Linearize non-linear model (not as difficult as people think)

36 = + + Equations for dM i /dt=……  Can derive transfer matrix G(s) A Δx + B ΔL ΔV State matrix (eigenvalues determine speed of response) Input matrix “states” (tray compositions) inputs

37 Are dynamic really so simple? 1970’s and 1980’s: Mathematical proofs that dynamics are always stable  Based on analyzing dynamic model with L and V [mol/s] as independent variables In reality, independent variables are  L w [kg/s] = L [mol/s] ¢ M [kg/mol]  Q B [J/s] = V [mol/s] ¢  H vap [J/mol] Does it make a difference? YES, in some cases!

38 More about mass reflux and instability t=0: z F is decreased from 0.5 to 0.495. L w [kg/s]= L[mol/s]/M where M [kg/mol] is the molecular weight, Data: M L =35, M H =40. What is happening? Mole wt. depends on composition: more heavy ! M up ! L down ! even more heavy...) Can even get instability! With M H =40, instability occurs for M L <28 (Jacobsen and Skogestad, 1991)

39 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

40 I IIIII I IV II Reflux back again.... but not composition !? Reflux Top composition

41 Multiple steady state solutions IIIIV I II V


43 V IV I

44 Actually not much of a problem with control! 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) V IV I Conclusion: “Simple” dynamics OK for our purposes

45 Nonlinearity The dynamic response of distillation column is strongly nonlinear. However, simple logarithmic transformations counteract most of the nonlinearity. Derivation: ln x i : Logarithmic composition CONCLUSION: Response nearly linear (constant gain) with log. comp.

46 Ref. S. Skogestad. “Dynamics and control of distillation columns: A tutorial introduction”, Trans. IChemE, Vol. 75, 1997, p. 553 X i+1 XiXi

47 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:

48 Initial Response to 10% ∆L: (V constant) (Column A with Flow Dynamics) nonline ar Linear model nonline ar Linear model Extremely non linear ∆xB∆xB ∆yD∆yD Log: Counteracts Nonlinearity -∆ln(1- y D ) ∆ ln x B


50 Conclusion dynamics 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

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