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 ofDistillation Columns

1. Distillation Dynamics (45%)……….p.2

2. Distillation Control (30%)……….....p.23

3. Some basic control theory (25%)…..p.37

SIGURD SKOGESTADNDF Workshop I

Lyngby, Denmark, May 24, 1989

1.1. Distillation DynamicsDistillation Dynamics

1.1 Introduction1.1 Introduction 33

1.2 Degrees of freedom/steady state1.2 Degrees of freedom/steady state 44

1.3 Dynamic Equations1.3 Dynamic Equations 55

1.4 composition Dynamics1.4 composition Dynamics 7

1.5 Flow dynamics1.5 Flow dynamics 1919

1.6 Overall Dynamics1.6 Overall Dynamics 21 21

1.7 Nonlinearity1.7 Nonlinearity 21A21A

1.8 Linearization1.8 Linearization 2222

1.4.1 Dominating time constant 8 1.1.4 Internal flows

17 1τ

2

L

( m α , m 3 )

P

N

N - 1

3 2

1 M B ( m o l )

M D

O v e r h e a d V a p o r V T

C o n d e n s e r

C o n d e n s e r H o l d u p

R e f l u x D i s t i l l a t e

D , y D

1s

mol

F e e d

B o i l u p

R e b o i l e r

R e b o i l e r h o l d u p

B o t t o m P r o d u c t

F , z F

1s

mol

M o l e f r a c t i o n l i g h t c o m p o n e n t

B , x B ? 0 ( m o l / s )

Q B ( k N )

V = Q D / λ

1.1 Introduction1.1 IntroductionTypical column: 1 feed, 2 products, no intermediate coolingTypical column: 1 feed, 2 products, no intermediate cooling

3

3a

Manipulated Variables (Valves): (Manipulated Variables (Valves): (μμ))

V (indirectly), L, D, B, VT (indirectly)V (indirectly), L, D, B, VT (indirectly) (5)(5)

Disturbances: (d)Disturbances: (d)

F, zF, qF (= fraction liquid in feed, feed enthalpy), rain F, zF, qF (= fraction liquid in feed, feed enthalpy), rain shower, + all u’s aboveshower, + all u’s above

Controlled Variables (y)Controlled Variables (y)

MO, MB, P, YP, XB (5)

Inventory Separation

(composition, temperature)

((1.2) STEADY-STATE OPERATION1.2) STEADY-STATE OPERATION

Must keep holdups (MD,MB,MV) constantMust keep holdups (MD,MB,MV) constant

This uses 3 degrees of freedom (u’s); only two left.This uses 3 degrees of freedom (u’s); only two left.

The two degrees of freedom may be used to specify two The two degrees of freedom may be used to specify two product specificationsproduct specifications

4

D B

V V

V

V

Inventories

M ,M :holdup liquid in condenser and reboiler

RTp=M -pressure;M :mol vapor insidecolumn

VV :volume vapor insidecolumn.

That is, pressureis a measure for holdup vapor (M )

e.g.,e.g., YYDD and x and xBB

TTtoptop and T and Tatmatm

D (product rate) and YD (product rate) and YDD etc. etc.

Recall specifications for Recall specifications for steady-state simulations steady-state simulations (PROCESS)(PROCESS)

D D

B B

y /(1-y )S=

x /(1-x )

T

B

N

T TNN

B B

L /VS=α

L V

4a

Steady-state behavior (design)Steady-state behavior (design)

NS=αInfinite reflux, exact (Fenske):Infinite reflux, exact (Fenske):

Finite reflux, good approximation:Finite reflux, good approximation:

holds for columns with feed at optimal location

Example: 3-Stage ColumnExample: 3-Stage Column Constant pConstant p Constant holdup liquid and levelConstant holdup liquid and level Negligible vapor h.Negligible vapor h. Constant molar flowsConstant molar flows Constant Constant αα=10=10 2 Vicor stages + total condenser2 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)(slight modification of Jafarey, Douglas, McAvoy, 1979)

Overall separation, binaryOverall separation, binary

4b

Get:

Total reflux:

Approximation:

T

B

N 1

T TNN 1

B B

L /V 3.05/3.55S=α =100 =75

L V 4.05/3.55

N 2S=α 10 =100

D D

B B

1-Y /Y 0.9/0.1S= = =81

x /(1x ) 0.1/0.9Actual column:

Stage

Condenser Feedstage Reboiler

i

1 2 3

Li

3.05 4.05

Vi

3.55 33.55

Xi

0.9000 0.4737 0.1000

yi

0.9000 0.5263

D = 0.5

YD=0.9 M2

M2

V=3.5

M2

C = 1kmol/min

ZF=0.5

B = 0.5XB = 0.1

L=3

B

T TNN

B B

L VDerivation of S=α

L V

Bottom Section

i+1 Yi BLx =V +Bx (Mass Balance)

Nii i

Ni

YY =α x (VLE)(Constant Relative Volatiliy)

X

4c

i+1

L

XL+1

V

Yi

B

XB(x0)

Ni Ni Nii+1 B

i Ni+1 Ni+1 i Ni+1

Combine:

X Y XX Vα B X= +

X X L X L X X

<1 <1 x1

= 0i i+1

iNi i

x S VαS =

x S L

stripping factor bottom

5

Additional Assumptions (not always)

A4. Neglect vapor holdup (Mvi ≈ 0)

A5. Constant pressure (vapor holdup constant)

A6. Flow dynamics immediate (Mvi 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

6

stage

i+1

i

i-1

Mi+1

Vi

Li+1

Mi

Vi-1 Li

"state"

Balance equations:

d IN OUTinventory = -

dt sec sec

Balance equation for stage without feed/side drawINDEP.DIFF. Component balance (index for component notShown EQ.

i i vi i i i+1 i-1 i-1 i i i i+1Î

dN -1 M x +M Y =L x +V Y -L x -V Y

dt

i i1 i, i

i i2 i i

L VL i i3 i i

(Allgebraic) VLE:

Y = K x p

T =K x ,p

h ,H =K x p

FLASH

(Strongly Nonlimear)

6a

Total balance (sum of component balances)

i vi i+1 i-1 i i

d1 M +M = L +V -L -V

dt

Energy balance

L v L Vi i vi ri i+1 i+1 i-1 i-1 i i i i

d0or1 M V +M U =L h +V H -L h -V H

dt

Assume Mvi*0 (negl. vapor holdup)

Tray Hydraulics(Algebraic) A9.Simplified (linearized):

i ioi io i9 i-1,0

L

M -ML =L + +λ V -V

τ

τL: time constant for change in liquid holdup (≈2-10sek.)

λ: effect of increase in vapor rate on L

Li0,Mi0,Vi0: steady-state values (t=0).

(ALGEBRAIC) Pressure drop: Δpi = f(Mi,Vi,…)

6b

Numerical Solution (Integration) More details: p.82 in NDF Notes 1990

i i i i i C

i i vi

toti i i vi vi

1.Rigorous Approach

State variables:

n =μ x +μv y N on each tray

(note:Sum of n is giveμ +μ )

U =μ U +μ U (1 on each tray)

total internal energy on tray i

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 (MLi, Mvi), compositions (xi, yi), temperature, pressure (ρi), spec energy (hi, Vi)

Compute Li and Vi from tray hydraulics and pressure drop relations

2. Simplified Approach, neglect vapor holdup (Mvi = 0)

State Variables: (NL on each tray)

Mole fraction total holdup

→ Xi → Mi

(note: ni = Mixi)

6a.1

SolutionSolution::

1. Given value of state variables, guess pressure, Pi

2. Perform bubble point flash (given xi,pi) → yi,Ti, enthalpies

3. Compute Vi from energy balance (gives “index problem”: LHS (derivative) is known) Common simplification: Use dVi 1dt = dhi

l /dt from previous step. (Do not set d/dt (MiVi) ≈ 0 constant molar flows

4. Compute Li and Pi from tray hydraulics and pressure drop.

7

(1.4) Composition Dynamics(1.4) Composition Dynamics

SUM: (NC independent differential equations) xNv1(when M 0)

L1 (when M const.no flow dynamic)

No. of components No. of trays

Might expect very high-order complicated behavior.

Surprise: The dominant composition dynamics is approximately 1 order!

Fig. Response in YD to step change

τ1: approximately independent of what we step (reflux, feed rate, boilup,…) and what and where we measure (YD, Ttop, Tbtm, etc,…)

1-t/to oΔy t = 1-l Δy

oΔy

DΔy

(or temp.,etc.)

100%

-11-l =63% Dy t

τ1

time

7a

EXAMPLE: 3-stage column (see p.4B)

Neglect flow dynamics (Mu = 1 = constant) 1 state on each tray Constant molar flows (V2 = V3)

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

∆X2(t)

Feed tray

∆X3(t)reboiler

∆X1(t)condenser

63%

τ1=4.5 min. Time (min)

VLE

Step in ZF from 0.50 50 0.51

Dist.m: (matlab subroutine)

Composition Response of an Individual Tray

Component material balance, tray i

i i i+1 i+1 i-1 i-1 i i i i

dM x =L x +V y -L x -V y

dt

Assume the column is at steady-state, and consider the effect of an increase in xifi to xi+1+∆xi+1. Assume flows constant, and neglect interactions between the trays (yi-1: constant). In terms of deviation variables

i

ii i+1 i+1 i i i i i

dΔxM =L Δx +0-L Δx -V K Δx

dtΔy

Where the linearized VLE-constant is

i i2

i

αK average K 1

1 α 1 x

7b

i+1

i

i-1

Mixi

xi+1

Overall response time from top to bottom of column (neglecting “vapor” interactions) total HOLDUP Inside Column

/LM0.5VL

MNτθ I

Ixx

Example: 3-stage Column

min55.0097.0171.0282.0VkB

M

VVL

M

DL

M

reboiler

33

3

222

2

condenser

1

1x

Conclusion: Do not yet correct overall response time (4.5 min) by simply adding together individual trays.

7c

Collecting ∆Ki terms we get a 1st order response with time constant

ix

i i i

Mτ =

L +K V

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. Mvi is negligible (OK when pressure is low; at 10 bar Mvi will be about 10% of liquid holdup)

A6. Flow dynamics much faster than composition dynamics.

L

LB

xB

LB

θL

xB

Objectives: Understand why overall response 1st order Develop formula for τ1 When does τ1 not apply?

τ1

1.4.1 τ1 Dominant Time Constant

F1zF

Bf

xBf

Assumption 6:

D(t) = Df

B(t) = Bf

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)

Ft ,zFt Ft ,zFt

B,xBB(t)

xB(t)

D(t)

YD(t)

Df

YDfD,YD

t = 0 t > 0 t =

componentofsupplyin imbalance

DfBfFff

columninholdupcomponentsinchange

N

1iii tyDtxBZF)t(xM

dt

d:0t

fytytywhere

:0t

tyDtxBtxMdt

d:Subtract

DDD

DfBfii

i

Assumption

A7. “All trays have some dynamic response”, that is,

tk0xtx

tk0yty

:examplefor,traysallforsametheis)t(k

10k;tk0xtx

BB

DD

ii

)4(

Justification: Large interaction between trays because of liquid and vapor streams. (Reasonable if

Substitute (4) into (3):

1 1

i i f B f Di

-t τ -t τ

1 1

dk tM Δx 0 B Δx 0 D Δy 0 k t

dt

1st order differential equation for k(t).

dk t 1 1Solution: k t l =- l =- k t

dt τ τ

10

0yD0xB

0xMτ

where

DfBf

ii

i

1

5

1

Physically

"change in holdup of component (kmol)"t =

"Imbalance in supply of component (kmol/min)"

(5) was first "derived" by Davidson (1956). Rederived Moczek

(1963), Wahl and Harriot (1970), Waller (19 SMALL D's only.

69)

10a

11

o F O BO O OO0

f Ff f Bf f Df

F f B OO OO

denominator in

1. No linearization (change may be large)!

2. More convenient formula for denominator.

Use: t 0 : F z =B x +D y

t= : F z =B x +D y

Subtract: Δ Fz = B Δx 0 +ΔDy -ΔBx

F OO OO

F

F

that is Denominator in (5) Fz ΔDy Bx (6)

//example : change in feed composition Δz ¹0, but ΔF=0, ΔD=0,ΔB=0

Denominator F z //

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 τ1-Formula (5)

Estimated Dominant Time Constant

N+1

i ii=1

1F D B

=0(flowsconstant)10.01

M Δx 0.091 0.0264 0.0109τ =

Δ FZ - y ΔD-x ΔB

0.04644.46min

0.01

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

i

1

2

3

Li

3.05

4.05

Vi

3.55

3.55

Flows

Compositions

with ZF=0.50 with ZF=0.51

Xi Yi Xi Yi

0.9000

0.4737 0.900

0.1000 0.5263

0.9091

0.5001 0.9091

0.1109 0.5549

12

xIn our case θ 0.55min 4.5min OK!

x

1

i large reflux (θ small)1 ii both products pure large

In generalθx τ for

Reasonable if

x x 1Nτ θ τ Overall response time (incl. Tray interactions)

Response time neglecting tray interactions

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 ZF decreases from 0.65 to 0.60

i

tot i

MM

= =111min have neglected holdup in reboiler and condenserF F

All flows kept constant

Simulation 1(t=0)

Simulation 2(t=∞ )

zF yD xB

0.65 0.995 0.100 0.714

0.60 0.958 0.030 0.495

i itot

tot

M xx =

M

i i tot1

f D f B F

M Δx 111min 0.714 0.495Mτ = = =

D Δy +B Δx FΔz 0.05

480min 8hours

Response:

0.05

.63

ΔyD

00 1h 8h 10h

time

Ix

OK!

M 111Check of Assumptions 7:θ 4.6min 480min

L+v 11.8 12.3

14

τ1: Shortcut Formula

Make some simplifying assumptions which hold best for columns with

large reflux pure products Small changes Binary separation (or use pseudobinary)

SC SCI SCD SCB

Derive

τ τ τ τ

ISCI

s

DSCD D D

S

BSCB B B

S

S B B D D

D

M 1Fτ = contribution from holdup inside column

I lnS

M /Fτ = 1y y contribution from holdup in condenser

I

M /Fτ = 1-x x contribution from holdup in reboiler

I

B DI = x 1-x + y 1y "impurity sum"

F Fy /1

S=

D

B B

-y"separation factor"

x /1-K

CI CD CBτ is large (may neglect τ , τ )

(ln is typically from 4 to 15)

Columns with pure products

IS is small

Varies with oper. Cond!

15

I

D B

Example :Propane - Propylene

MAsssume : = 111min insideholdupcolumn

FM M

= 60min, = 20minF F

D

B

B=0.386

Fy =0.995

x =0.100

For small changes assume (*) applies.

Have

S

SCI

SCD

SCB

I =0.386×0.1×0.9+0.614×0.005×0.995=0.035+0.003=0.038

0.995/0.005S= =1791ÞlnS=7.49

0.1/0.9Then

111t = min=390min

0.038×7.4960

t = ×0.005×0.995min=8min0.63820

t = ×0.1×0.9min=47min0.038

Sum:t=445min

Reasonable agreement:

NOTE: Contribution small from condenser because purity is high so absolute changes in compositions are small

16

Variation in τ1 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 (IS).

-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)

τ1

1 B

D

x

1-y

Peak is large if both products are pure

Example

Pure

“Fast”

“slow”

time

getting pore

ΔZF

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:

D D

B B

y /1-yS=

x /1-x

(“separation unchanged split changed”)

(“separation changed split unchanged”)

D D

B B

y /1-yS=

x /1-x

MAIN EFFECT ON COMPOSITION BY ADJUSTING D/F; “FINE TUNE” WITH INTERNAL FLOWS

(1.4.2) τ2 External and Internal Flows

Steady -state

Steady -state Steady -state

COMP

tray

COMP10 0

AB=ΔD AB=ΔD=0

Dynamics

External Flows Internal Flows

Step ΔB = -ΔD Step ΔL = -ΔV

Conclusion:

Large S.S. effect

Slow (τ1) Small S.S. Effect

Faster (τ2)

Initial Response: Not as Different

I1

S

M /Fτ

I lnS

2 Iτ M /F

(see more accurate formulas in Skogestas & Morari, 1988 I E E C Res, 27, 1848-1862)

But: Derived when flow dynamics neglected (doubtful since τ2 is relatively small)

For columns with pure products

IM /F

Recall:

time time

63%

63%

0

ΔxB

ΔyD

τ1 τ2

0ΔyD

ΔxB

19

FLOW DYNAMICS

(variations in liquid holdup neglected so far)

A8. Constant molar flows

A4. Neglect vapor holdup i-1 i i+1V =V =V

Total material balance becomes

ii+1 i-1 i i

dM=L +V +L -V (1)

dt

i+1

iMi

Li

Li+1

2/3L oi i

oi iL

L

Estimates

τ :Francis weir formula M =k×L

M2 1 Mτ = (assume Moi Mi/2)

3 L 3 LMi

(Packed:τ =0.6 )L

λ:Difficult. trays, λ>0usually

packed, λ 0

Tray Hydraulics

A9.: Assume simplified linear tray hydraulics

i io i io i ioL

1L =L + M -M +λ V -V (2)

τ

iL

V

i

i Mi

ΔMτ =

ΔL

Lλ=

V

Lio,Mio,Vio: Steady state values

= hydraulic time constant

= effect on change in Vi on Li (vapor may push liquid off tray, λ>0.5: inverse response)

19a

20

Consider Deviations from Initial Steady-State (Δli=Li-Lio,…)

Theni

i+1 i

dM=ΔL -ΔLeach tray one tank combine all t

dtrays

Consider response in LB to change in L:

N tanks in series, each time constant τL

0.5ΔL

ΔL

ΔLB

tθL=N·τL

(“almost” a dead time)

ΔVV

B N

L

Laplace

1ΔL S = ΔL s

1+ τ S

Response in LB to change in V:

“Vapor pushes liquid off each tray”

1

B N

L

Laplace

1ΔL S = x ΔV s

1+ τ S

ΔLB

ΔL

N

ΔV

θL=N·τL

t

ΔLB

0

λ·ΔV

21

1.6 Overall Dynamics

Composition Dynamics

Typical value

External flows , τ1 250 min

Internal flows , τ2 20 min

+ Liquid Flow Dynamics

“Dead time” from top to bottom, θL 3 minute

+ Pressure dynamics

+ Top level – “ –

+ BTM level – “ –

+ Valve dynamics,

+ Heat transfer dynamics

(V and Vt indirectly controlled)

+ Measurements Dynamics

Depend on tuning of pressure and level controls. Typical time constant: 0.5 minute

0.2 minute

0.1 minute

Ess

enti

al f

or c

ontr

ol !

Simplified Model

Describe each effect independently

“Add” together to give overall dynamics

(Alternative: Linearize all equations)

NOTE: Exact value of τ1 not important for control!

1.7 Nonlinearity

The dynamic response of distillation column is strongly nonlinear. However, simple logarithmic transformations counteract most of the nonlinearity.

xii i i i-1 i+1

xii i+1 i

depends strongly on operating point

iKi +1i

i

dDerivation.component balance: M =L x - x +V y - y

dtinitial response to charge in L :

dM = x - x

dt

"Trick":DividebyK :

xdlnM = -1 DL

dt x

More details: p.132-133 in NDF Notes 1991

Li

NI

vα (bottom)-nearly indep.of operating point

LIn general:

XUse: Xi=ln

X

Light key component on tray i

Heavy component

May also be used for temperatures!

i top

BTM i

T -TXi ln

T -T

Temp. BTM. Of Column

Temp. on Tray i

Temp. Top of Column

21A

Xi+1

Xi

tray

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

nonlinear

Linear model

nonlinear

Linear model

Extremely non linear

∆xB

∆yD

Log: Counteracts Nonlinearity

-∆(n(1-yD)

∆ ln xB

21b

22

1.8 Linearization

Need linear models for controller design Obtain by

Given Tray

i

ixi i+1 i+1 i-1 i-1 i i i i ii

d α xM =L x +V y L x -V y ; y =

dt 1+ α-1 x

Linearize, introduce deviation variables, simplification here: assume: i) const., ii) const. molar flows

Li = Li+1 = L Vi-1= Vi = V

ii i1 i i i i i-1 i-1 i+1 i i i-1

Δxμ =L L +k V Δx +V k + x -x ΔL- y -y ΔV

dt

i

i 2ii

α dyK = =

dK1+ α=1 x

1) Put together simple models of individual effects (previous page)

2) Linearize non-linear model

Can derive transfer matrix G(12)

+ Equations for dMi/dt=……

d ΔxMi

dt A Δx + B ΔL

ΔV

State matrix (eigen values determine speed of response)

Output matrix “states” (tray compositions)

inputs

22a

dt

xdμi

dt

xd1μi

i

1i

1i1iiii

iii1i1ii

VKVKLL

VKVKLL

1iii1i

i1i1i2i

yyxx

yyxx

L

L=

i

1i

x

x +

23

(2) Distillation Control

Multivariable vs. Single-Loop Control…25

Choice of Control Configurations…27

LV

DV, L

L υ

D B

DB

1) Disturbance Sensitivity………….31

2) Feedback Control Properties…….34

One-Point

Two-Point

1) Implementation, Level Control…..35

24

Distillation Control

Manipulated inputs, μ1s: L,V,D,B,UT

Controlled outputs, Y1s: Mo, Mb, ,YD, xB(P)Mv

COLUMN

L

V

D

B

UT

MD

MB

MV

YD

XB

5X5

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 YD and xB 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

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. yD controlled with L using PI-controller, the other input is n manual, xB “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) Mi variable, for example, decoupler

Problem: Often not robust (Sensitive to errors)

D

B

e.g.L y

V x

27

Problem 1.

1) LV-configuration (“conventional”, “energy balance”, “indirect material-balance”)

Reflux L and boilup V used for composition control Level control:

MD D

MB B

MV (P) VT (cooling)2) DV-configuration (“(direct) material balance”)

because D is usedLevel control:

MD L

MB B3) LB-configuration (“(direct) material balance”)

Level control:

MD D

MB V4) 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)

28

Top of Column

LV-configurationLB-configuration

LCLS

VT

L+D D

L

DV-configurationDB-configuration

LC

L

L+D

VT

D

DS

Comment: Will usually use cascade control on inputs for composition control

L

LS

No cascade (manipulated valve position)

L

PLm

LS

With cascade using flow measurements (remove nonlinear valve characteristics

28a

L V- configuration

D BC

L

L+D

VT

D

LC

DIV

(40)m

(L/D)s

Set manually or from composition controller

29

Open-loop response to flow disturbance:

VD F + V- L

VD O F +O- V

30

BAD

V)

V

V

ΔD = 0×ΔF + 0×ΔV

(but = L = F

Previously 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)

V

D 1D = ΔF + ΔV

L VF 1+ +D B

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 ZF

High-purity columns: Composition extremely sensitive to small changes in D/F:

O B

FO B XB O

relative change in composition small for high-purity columns

dy dx 1 Dx × dZ -d

1-y x B +0 1-y F

Consequently: Disturbances which change D/F are “bad”

The effect of a given disturbance on D/F depends on the configuration

32

Configuration Disturbance

ΔFV

(increase in flow rate of vapor in feed)

ΔV

(increase in boilup)

Optimal V

DDD DF

FDD O

LV

DV

DB

L V

D B

V

V

DD=DF

DD=0

DD=0

DDD= DF

F

ΔD = ΔV

ΔD = 0

ΔD = 0

1ΔD = ΔV

L V1 + +

D B

(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

33

1. Summary Disturbance Sensitivity

Feed enthalpy

Disturbance in

V, L, qF

boilup

reflux

Disturbance in

F

Feed forward from Fm

LV

DV

LB

L V

D BDB

Good

Poor Poor

Good Poor

Good Poor

Good

Initially:Good Poor

t :Good Terrible L,V

1

Difficult

Easy D F

Easy B F

Not needed

D BEasy

F F

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).

(2) Feedback Control Properties

(2a) One-Point Control(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(2a) Two-Point Control

* Use RGA as a function of frequency as a tool, want λRGA

(wB)≈1. See Facohsen et.al. (AIChEF, 1990 May)

Configuration

LV ←strong interaction

DV

LB

L/D V/B

DB

Single loops (yD, xB) (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

Not worth it

2x2 Controller (“Problem 2”)

34

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 measure Good L,D,V,B!)

Easy OK, but level btm: V may give inverse response

Almost impossible to control level with small flow (“no power”)

L L VHave not mentioned V ("Ryskump") : Usually between and LV

D D B

LC

D (small)LS (large)

VT (large)

36

Effect of Level Control tuning on Composition Response

LV-configuration: Response time for level control has almost no effect on composition response

B

V

D

LDB,DV, 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

DS

Level controller sets L

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.

B

V

D

L

Changing DS directly gives opposite change in L (since Ltd is constant)

LC

L

DS

FC

Level controller sets L

L D

L+D

3. Some Basic Control Theory3. 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

gc1

1S

38Transfer Functions/Matrices

1. SISO (Single Input Single Output)

Steady-state

“Increasing L from 1.0 to 1.1 changes yD from 0.95 to 0.97” (for example, run Process)

Dy =g o L

0.97- 0.95stead- stategain= =0.2

1.1-1.0Dynamics

“The response has a dead time of 2 minutes and then rises with a time constant of 50 minutes”

Laplace: DΔy s = g s ΔL s

-θs0.2

g s =1+τs

=2min

=50min

Transfer function:

time

yD

10

0.97

63%

0.96

0.95

Θ=2 50

τ=50

39

Alternative Method of Obtaining Transfer Function:

Take laplace of linearized modeld

sdt

i i i+1 i

i i i+1 i

i

i+1 ii

Materialbalance

d ln outinventory -

dt sek sekd

M x =Lx -Lxdt(Linearize) + Deviation variables

M s x s L L x s

Solve for x

x M capacityx , τ =

1+τ s L flow

x

Note: Gain is obtained by using s = o: “gain” = g(o)

i+1 i

1Effect of x on x :g s

1 τ s

“gain”=1

Time constant τ=Mi/L

Example L (constant)

Ki+1

Mi=constant

L

xi

39

Alternative Method of Obtaining Transfer Function:

Take laplace of linearized modeld

sdt

i i i+1 i

i i i+1 i

i

i+1 ii

Materialbalance

d ln outinventory -

dt sek sekd

M x =Lx -Lxdt(Linearize) + Deviation variables

M s x s L L x s

Solve for x

x M capacityx , τ =

1+τ s L flow

x

Note: Gain is obtained by using s = o: “gain” = g(o)

i+1 i

1Effect of x on x :g s

1 τ s

“gain”=1

Time constant τ=Mi/L

Example L (constant)

Ki+1

Mi=constant

L

xi

Response:

Xi+1

Xi

timeτ

Direct Generalization:

“Increasing L from 1.0 to 1.1 changes yD from 0.95 to 0.97, and xB from 0.02 to 0.03”

“Increasing V from 1.5 to 1.6 changes yD from 0.95 to 0.94, and xB from 0.02 to 0.01”

Steady-State Gain Matrix

2. MIMO (multivariable case)

B

2221

1211

B

D

Δx2outputonΔL1inputofEffect

0.10.1

0.10.2

1.51.6

0.020.01

1.01.1

0.020.031.51.6

0.950.94

1.01.1

0.950.97

0gg

0gg0G

ΔV

ΔL0G

Δx

ΔY

40s1

0.1

40s1

0.150s1

0.1

50s1

20.

G

:dynamicsincludealsoCan

s

B

D

x

y

)Time constant 50 min for yD(

)time constant 40 min for xB(

40

41Important Advantages With Transfer

Matrices:

1 .G(s) is independent of input μ!

For given u(s) compute output y(s) as sμsGsy

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!

jwLgwtsinAjwgtywtsinAtMLet 111111

Response (as things settle)

time

y1(t)

μ1(t)

11Lg jwΔt = -

w

AA. |g11(jw)|

g11(s)μ y1

42

Note Tells directly how much a sine of frequency w is amplified by process

jwgμ

y11

1

1

11

2Typical : g (s)

1 50 s

w (log scale)

Fast sinusoids are “filtered by process (don’t come through

BODE PLOT (magnitude only)

0.1

1

2|g11|

|g11(jw)| (log scale)|

0.01 0.02 0.1

=1/50

3.2 Effect of Feedback Control

43

C(s)ys μ

G(s)d

y

G(s): process (distillation column) C(s): controller (multivariable or single-loop PI’s)

y: output , ys: setpoint for output μ: input d: effect of disturbance on output

s

Process : y s G s μ s d s (1)

Controller : μ s C s y s y s (2)

No disturbance supression

No Control μ s 0

y s d s

Negative feedback

With feedback control Eliminate μ(s) in (1) and (2), ys=0

sdC(s)sGIsy 1

S(S)=“sensitivity function” = surpression disturbances

44

s s s

s

y d

Ideal:-Want 0

In practice:

Cannot do anything with fast changes, that is, ς(jw)= I at high frequency

S is often used as performance measure

1 11 12 1

2 21 2

y S (s) S (s) d (s)

y S (s) . . d (s)

21 jwS

One Direction

Log- scale

1

0.1

Small resonance peak (small peak = large GM and PM)

Bandwidth (Feedback no help)

Slow disturbance d1: 90% effect on y2 removed

w (log)w B

τ

44a

“Summing up of Channels”

“maximum singular value” (worst direction)

0.1

1 w (log)wB

“minimum singular value” (best direction)

w (s)

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

y1

y2

g12

g21

g11

g22μ2

μ1

sμsgsgsy

sμsgsgsy

"loopOpen"

222212

212111

Consider Effect of μ, on y1

1) “Open-loop” (C2 = 0): y1 = g11(s)·μ1

2) Closed-loop” (close loop 2, C2≠0)

Change cause by “interactions”

1222

22112111 μ

Cg1

Cggsgy

46

Limiting Case C2→∞ (perfect control of y2)

RGA11

def

2211

2112

22

211211

11

11

0L11 λ

gg

gg1

1

g

ggg

g

/μy

/μyGainRelative

122

2112111 μ

g

ggsgy

CL22

OL22

CL12

OL12

CL11

OL21

CL11

OL11

2221

2111

/μy

/μy

μy

μy

/μy

/μy

/uy

/uy

λλ

λλΛRGA

How much has “gain” from u1 to y1 changed by closing loop 2 with perfect control

The relative Gain Array (RGA) is formed by considering all the relative gains

21

12RGA

2

0.10.2

0.10.11

1λ,

0.10.1

0.10.2G

0.5

11

Example from before

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!

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

Example:

0.2 0.1G o

0.1 0.1

1 1 1

2 1 1

y = 0.2 u -0.1 u

y = 0.1u -0.1μ

1 1

2 2

1 2

2 1

2 -1RGA =

-1 2

Only acceptablepairings :

μ y

μ y

Will not work :

μ y

μ y

With integral action :

Negative RGA individual

loop unstable or overall system unstable

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

example

1 1 91 90G= RGA

0.9 0.91 90 91

Large (BAD!)

21

12 12

0.9

1Relativechange- makesmatrixsingular!

x

1ˆThen g g 1 0.91

90

11.1%

90