1 dynamics and control of distillation columns 1. distillation dynamics (45%)……….p.2 2....
TRANSCRIPT
1
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
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)
2
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 Dominant time constant 8 1.1.4 Internal flows 17
1τ
3
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
4
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)
5
((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 The two degrees of freedom may be used to specify two product specifications two product specifications
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)
6
D D
B B
y /(1-y )S=
x /(1-x )
T
B
N
T TNN
B B
L /VS=α
L V
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
7
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
8
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:
9
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
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
10
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
11
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 Nonlinear)
12
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,…)
13
Numerical Solution (Integration) More details: p.82 in NDF Notes 1990
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
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
14
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.
2. Simplified Approach, neglect vapor holdup (Mvi = 0)
State Variables: (NL on each tray)
Mole fraction total holdup
→ Xi → Mi
(note: ni = Mixi)
15
(1.4) Composition Dynamics(1.4) Composition Dynamics
SUM: (NC independent differential equations) xN
v1(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!
1-t/to oΔy t = 1-l Δy
oΔy
DΔy
(or temp.,etc.)
100%
-11-l =63% Dy t
τ1
time
16
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,…)
EXAMPLE: 3-stage column (see p.4B)
Neglect flow dynamics (Mu = 1 = constant) 1 state on each tray Constant molar flows (V2 = V3)
∆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)
17
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
18
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
i+1
i
i-1
Mixi
xi+1
19
Overall response time from top to bottom of column (neglecting “vapor” interactions) total HOLDUP Inside Column
Ix x I
Mθ Nτ 0.5 M /L
L V
Example: 3-stage Column
31 2x
1 2 2 2 3 3
condenser reboiler
MM Mθ = + + =0.282+0.171+0.097=0.55 min
L +D L +V V B+k V
Conclusion: Do not yet correct overall response time (4.5 min) by simply adding together individual trays.
Collecting ∆Ki terms we get a 1st order response with time constant
ix
i i i
Mτ =
L +K V
20
)This does not imply that flow dynamics are not important for composition control!)
In particular, assume liquid holdup (Mi) constant
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.
Objectives: Understand why overall response 1st order Develop formula for τ1 When does τ1 not apply?
Response to step change in reflux
time
L
LB
xB
LB
θL
xB
τ1
1.4.1 τ1 Dominant Time Constant
21
Assumption 6:
D(t) = Df
B(t) = Bf
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 =
F1zF
Bf
xBf
22
Component balance whole column;
N
i i f Ff f B f Di=1
imbalance in supplyof componentchange in components
holdup in column
dt 0 : M x (t) F Z -B x t -D y t
dt
i i f f Di
D D D
dSubtract: M Δx t =-B t -D Δy t
dt
t>0:
where Δy t =y t -y f
Bx
23
Assumption
A7. “All trays have some dynamic response”, that is,
i i
D D
B B
Δx t =Δx 0 k t ; k 0 =1
k(t) is the same for all trays, for example:
Δy t =Δy 0 k t
Δx t =Δx 0 k t
(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
i i
i1
f B f D
where
M Δx 0τ
B Δx 0 +D Δy 0
(5)
24
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)
25
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 //
3. Only steady-state data needed! (+holdups) Need steady-state before (t=0) and after (t=∞) upset.
Comments on τ1-Formula (5)
26
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
27
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
28
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
29
All flows kept constant
Example 2. Propane-propylene splitter
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
tot i
MM
= =111min have neglected holdup in reboiler and condenserF F
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 i tot1
f D f B F
M Δx 111min 0.714 0.495Mτ = = =
D Δy +B Δx FΔz 0.05
480min 8hours
30
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
31
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
CI CD CBτ is large (may neglect τ , τ )
(ln is typically from 4 to 15)
Columns with pure products
IS is small
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
Varies with oper. Cond!
32
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
33
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
34
Example
Pure
“Fast”
“slow”
time
ΔZF
getting pore
35
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:
(1.4.2) τ2 External and Internal Flows
Steady -state
Steady -state Steady -state
COMP
tray
COMP10 0
AB=ΔD AB=ΔD=0
36
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
37
Dynamics
External Flows Internal Flows
Step ΔB = -ΔD Step ΔL = -ΔV
Conclusion:
Large S.S. effect
Slow (τ1) Small S.S. Effect
Faster (τ2)
time time
63%
63%
0
ΔxB
ΔyD
τ1 τ2
0ΔyD
ΔxB
38
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:
39
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
40
Tray Hydraulics
A9.: Assume simplified linear tray hydraulics
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
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)
41
Consider Deviations from Initial Steady-State (Δli=Li-Lio,…)
Theni
i+1 i
dMeach trayone tank =ΔL -ΔL combine all trays
dt
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)
B N
L
Laplace
1ΔL S = ΔL s
1+ τ S
ΔV
V
ΔLB
ΔL
N
42
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
ΔV
θL=N·τL
t
ΔLB
0
λ·ΔV
43
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 !
44
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!
45
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
iKi +1i
i
dDerivation.component balance: M =L x - x +V y - y
dtinitial response to charge in L :
dM = x - x
dtdepends strongly on operating point
"Trick":DividebyK :
xdlnM = -1 DL
dt x
Xi+1
Xi
tray
46
Li
NI
vα (bottom)-nearly indep.of operating point
LIn general:
XUse: Xi=ln
X
Light key component on tray i
Heavy component
More details: p.132-133 in NDF Notes 1991
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
47
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
48
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
1) Put together simple models of individual effects (previous page)
2) Linearize non-linear model
49
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
dt
xdμi
dt
xd1μi
i
1i
1i1iiii
iii1i1ii
VKVKLL
VKVKLL
1iii1i
i1i1i2i
yyxx
yyxx
L
L=
i
1i
x
x +
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
50
(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
3) Implementation, Level Control…..35
51
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
52
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. decoupler)
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
53
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)
54
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
55
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 used
Level control:
MD L
MB B
3) LB-configuration (“(direct) material balance”)
Level control:
MD D
MB 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)
56
Top of Column
LV-configurationLB-configuration
LCLS
VT
L+D D
L
DV-configurationDB-configuration
LC
L
L+D
VT
D
DS
57
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
L V- configuration
D BC
L
L+D
VT
D
LC
DIV
(40)m
(L/D)s
Set manually or from composition controller
58
Open-loop response to flow disturbance:
VD F + V- L
59
VD O F +O- V
60
DB(!!) (Finco & Luyben, 1989)
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)
BAD
61
V
D 1D = ΔF + ΔV
L VF 1+ +D B
62
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
63
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
64
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).
65
(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)
66
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”)
67
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
68
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)
69
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
70
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
71
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
gc1
1S
72
Transfer 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”
time
yD
10
0.97
63%
0.96
0.95
Θ=2 50
τ=50
Laplace: DΔy s = g s ΔL s
-θs0.2
g s =1+τs
=2min
=50min
Transfer function:
73
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
74
Alternative Method of Obtaining Transfer Function:
Take laplace of linearized modeld
sdt
Example L (constant)
Ki+1
Mi=constant
L
xi
Response:
Xi+1
Xi
timeτ
75
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
76
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(
77
Important 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!
1 1 11 11Let M t A sin wt y t g jw A sin wt Lg jw
g11)s(μ y1
78
Response (as things settle)
11Lg jwΔt = -
w
time
y1(t)
μ1(t)
AA. |g11(jw)|
79
Note Tells directly how much a sine of frequency w is amplified by process
111
1μ
yg jw
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
80
3.2 Effect of Feedback Control
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
81
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
82
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
83
“Summing up of Channels”
“maximum singular value” (worst direction)
0.1
1 w (log)wB
“minimum singular value” (best direction)
w (s)
84
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
1 11 12 2
2 21 22 2
" "
μ
μ
Open loop
y s g s g s s
y s g s g s s
85
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
86
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
87
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!
88
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
89
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
90
(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
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
91
Singular Matrix: Cannot take inverse, that is, decoupler hopeless.
Control difficult
92