1. Example regulatory control: Distillation
5 dynamic DOFs (L,V,D,B,VT)
Overall objective:
Control compositions (xD and xB)
“Obvious” stabilizing loops:
Condenser level (M1)
Reboiler level (M2)
Pressure
E.A. Wolff and S. Skogestad, ``Temperature cascade control of distillation columns'', Ind.Eng.Chem.Res., 35, , 1996.

2. LV-configuration used for levels (most common)
L and V remain as degrees of freedom
after level loops are closed
Other possibilities:
DB, L/D V/B, etc….

3. BUT: To avoid strong sensitivity to disturbances:
Temperature profile must also be “stabilized”
Temperature
Ttop

4. Data “column A” Binary ideal mixture with relative volatility 1.5
Equimolar feed (zF=0.5)
Both products 99% purity (xD, xB)
41 stages, feed at stage 20
Liquid lag: 1.5 min (from top to bottom)
Level control: Tight
Temperature control: Kc=1.84 (1 min closed-loop time constant)
Assume top product (xD) most important → Close loop at top (L)
Composition control (supervisory layer): Decentralized PID
Measurement delay for composition: 6 min

5. Temperature control: Which stage x?

6. Singular value rule: Tray 27 is most sensitive (6 above feed)

7. Temperature control: Ts is DOF for layer above

8. Partial control Primary controlled variables y1 = c = (xD xB)T u1 = V
Supervisory control:
Primary controlled
variables y1 = c = (xD xB)T
Regulatory control:
Control of y2=T
using u2 = L (original DOF)
Setpoint y2s = Ts : new DOF
for supervisory control
u1 = V

9. Disturbance sensitivity with partial control (Pd1)

11. RGA with temperature loop closed

12. Disturbance sensitivity with decentralized control: CLDG

13. Closed-loop response with decentralized PID-composition control
Interactions much smaller with “stabilizing” temperature loop closed
… and also disturbance sensitivity is expected smaller

14. Integral action in temperature loop has little effect

15. May get improvement with L/D V/B - configuration for level loops
Simulations are with decentralized PID-control in supervisory layer
Difference smaller if we use multivariable control (MPC)

16. No need to close two inner temperature loops

17. Conclusion: Stabilizing control distillation
Control problem as seen from layer above becomes much simpler if we control a sensitive temperature inside the column (y2 = T)
Stabilizing control distillation
Condenser level
Reboiler level
Pressure (sometimes left “floating” for optimality)
Column temperature
Most common pairing:
“LV”-configuration for levels
Cooling for pressure
V for T-control (alternatively if top composition most important: L for T-control)

18. Conclusion stabilizing control: Remaining supervisory control problemTC Ts Ls
+ may adjust setpoints for p, M1 and M2 (MPC)

19. Why control temperature at stage 27? Rule: Maximize the scaled gain
Scalar case. Minimum singular value = gain |G|
Maximize scaled gain: |Gs| = |G| / span
|G|: gain from independent variable (u) to candidate controlled variable (c)
span (of c) = variation (of c) = optimal variation in c + control error for c

20. Distillation example Assume here that both xD + xB are important
columnn A. u=L. Which temperature to control?
Nominal simulation (gives x0)
One simulation: Gain with constant inputs (u)
Make small change in input (L)
with the other inputs (V) constant
Find gainl =  xi/ L
- have also given gains for disturbances (F,z,q)
From process gains: Obtain “optimal” change for candidate measurements to disturbances,
  (with input L adjusted to keep both xD and xB close to setpoint, i.e. min ||xD-0.99; xb-0.01|| , with V=constant).
w=[xB,xD]. Then (- Gy pinv(Gw) Gwd + Gyd) d
(Note: Gy=gainl, d=F: Gyd=gainf etc.)
Find xi,opt (yopt) for the following disturbances
F (from 1 to 1.2; has effect!) yoptf
zF from 0.5 to 0.6 yoptz
qF from 1 to 1.2 yoptq
xB from 0.01 to 0.02 yoptxb=0 (no effect)
Control (implementation) error. Assume=0.05 on all stages
Find
scaled-gain = gain/span
where span = abs(yoptf)+abs(yoptz)+abs(yoptq)+0.05
“Maximize gain rule”: Prefer stage where scaled-gain is large
>> res=[x0' gainl' gainf' gainz' gainq']
res =

21. Assume here that both xD + xB are important
u=L
>> [yoptf yoptq yoptz span]
ans =
>> Gw
Gw =
1.0853
0.8747
Gwf =
0.5862
0.3938
>> yoptf = (-gain'*pinv(Gw)*Gwf + gainf')*0.2
>> yoptq = (-gain'*pinv(Gw)*Gwq + gainq')*0.2
>> yoptz = (-gain'*pinv(Gw)*Gwz + gainz')*0.1
have checked with nonlinear simulation. OK!

22. Assume here that both xD + xB are important
u=L: max. on stage 15 (below feed stage)
In practice (because of dynamics): Will use tray in top (peak at 27)
>> span = abs(yoptf)+abs(yoptz)+abs(yoptq)+0.05;
>> plot(gainl’./span)

23. Assume here that both xD + xB are important
>> [yoptf yoptq yoptz span]
ans =
u=V
Gwv =
Gwf =
0.5862
0.3938
>> yoptf = (-gainv'*pinv(Gwv)*Gwf + gainf')*0.2
>> yoptq = (-gainv'*pinv(Gwv)*Gwq + gainq')*0.2
>> yoptz = (-gainv'*pinv(Gwv)*Gwz + gainz')*0.1
>> span = abs(yoptf)+abs(yoptz)+abs(yoptq)+0.05;
>> plot(gainv'./span)

24. Assume here that both xD + xB are important
u=V: sharp max. on stage 14 (below feed stage)
>> span = abs(yoptf)+abs(yoptz)+abs(yoptq)+0.05;
>> plot(gain’./span)

25. COMMENT!
NEED to be careful about nonlinearity!
Effect of step size in V:
delta=0.005
Gw =
delta=0.001
delta=0.0002
delta=
delta=