1

Example regulatory control: Distillation

5 dynamic DOFs (L,V,D,B,VT)

Overall objective: Control compositions (xD and xB)

“Obvious” stabilizing loops:1. Condenser level (M1)2. Reboiler level (M2)3. Pressure

E.A. Wolff and S. Skogestad, ``Temperature cascade control of distillation columns'', Ind.Eng.Chem.Res., 35, 475-484, 1996.

2

LV-configuration used for levels (most common)

“LV-configuration”: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)

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Temperature control: Ts is DOF for layer above

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Partial control

y1 = P1 u1 + Pr1 (y2s-n2) + Pd1 dP1 = G11 – G12 G22

-1 G21

Pd1 = Gd1 – G12 G22-1 Gd2 - WANT SMALL

Pr1 = G12 G22-1

Supervisory control:

Primary controlled variables y1 = c = (xD xB)T

Regulatory control:

Control of y2=Tusing u2 = L (original DOF)

Setpoint y2s = Ts : new DOF for supervisory control

u1 = V

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Disturbance sensitivity with partial control (Pd1)

10

11

RGA with temperature loop closed

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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 layerDifference smaller if we use multivariable control (MPC)

16

No need to close two inner temperature loops

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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 distillation1. Condenser level

2. Reboiler level

3. Pressure (sometimes left “floating” for optimality)

4. 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)

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Conclusion stabilizing control:Remaining supervisory control problem

TCTs

Ls

+ may adjust setpoints for p, M1 and M2 (MPC)

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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 examplecolumnn A. u=L. Which temperature to control?

1. Nominal simulation (gives x0)

2. 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)

3. 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

1. F (from 1 to 1.2; has effect!) yoptf

2. zF from 0.5 to 0.6 yoptz

3. qF from 1 to 1.2 yoptq

4. xB from 0.01 to 0.02yoptxb=0 (no effect)

4. Control (implementation) error. Assume=0.05 on all stages

5. 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

Assume here that both xD + xB are important>> res=[x0' gainl' gainf' gainz' gainq']

res =

0.0100 1.0853 0.5862 1.1195 1.0930 0.0143 1.5396 0.8311 1.5893 1.5506 0.0197 2.1134 1.1397 2.1842 2.1285 0.0267 2.8310 1.5247 2.9303 2.8513 0.0355 3.7170 1.9990 3.8539 3.7437 0.0467 4.7929 2.5734 4.9789 4.8273 0.0605 6.0718 3.2542 6.3205 6.1155 0.0776 7.5510 4.0392 7.8779 7.6055 0.0982 9.2028 4.9126 9.6244 9.2694 0.1229 10.9654 5.8404 11.4975 11.0449 0.1515 12.7374 6.7676 13.3931 12.8301 0.1841 14.3808 7.6202 15.1675 14.4857 0.2202 15.7358 8.3133 16.6530 15.8509 0.2587 16.6486 8.7659 17.6861 16.7709 0.2986 17.0057 8.9194 18.1436 17.1311 0.3385 16.7628 8.7524 17.9737 16.8870 0.3770 15.9568 8.2872 17.2098 16.0758 0.4130 14.6955 7.5832 15.9597 14.8058 0.4455 13.1286 6.7215 14.3776 13.2280 0.4742 11.4148 5.7872 12.6286 11.5021 0.4987 9.6930 4.8542 10.8586 9.7680 0.5265 11.0449 5.4538 12.1996 11.1051 0.5578 12.2975 6.0007 13.4225 12.3410 0.5922 13.3469 6.4473 14.4211 13.3727 0.6290 14.0910 6.7478 15.0929 14.0989 0.6675 14.4487 6.8669 15.3588 14.4397 0.7065 14.3766 6.7871 15.1800 14.3528 0.7449 13.8797 6.5135 14.5676 13.8438 0.7816 13.0098 6.0724 13.5807 12.9654 0.8158 11.8545 5.5060 12.3136 11.8051 0.8469 10.5187 4.8635 10.8764 10.4677 0.8744 9.1066 4.1929 9.3765 9.0566 0.8983 7.7071 3.5346 7.9044 7.6602 0.9187 6.3866 2.9184 6.5261 6.3443 0.9358 5.1876 2.3624 5.2829 5.1505 0.9501 4.1314 1.8756 4.1942 4.1000 0.9617 3.2233 1.4592 3.2631 3.1975 0.9712 2.4574 1.1098 2.4816 2.4368 0.9789 1.8211 0.8208 1.8355 1.8053 0.9851 1.2990 0.5848 1.3077 1.2875 0.9900 0.8747 0.3938 0.8805 0.8670

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>> 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!

>> [yoptf yoptq yoptz span]

ans =

0.0077 0.0014 0.0011 0.0601 0.0108 0.0019 0.0016 0.0644 0.0146 0.0027 0.0025 0.0698 0.0192 0.0036 0.0038 0.0765 0.0246 0.0047 0.0056 0.0849 0.0309 0.0061 0.0082 0.0951 0.0379 0.0077 0.0117 0.1073 0.0456 0.0096 0.0163 0.1215 0.0536 0.0118 0.0222 0.1375 0.0612 0.0141 0.0294 0.1546 0.0678 0.0164 0.0379 0.1720 0.0724 0.0186 0.0474 0.1884 0.0742 0.0204 0.0575 0.2022 0.0726 0.0217 0.0676 0.2119 0.0673 0.0222 0.0769 0.2164 0.0584 0.0220 0.0847 0.2151 0.0467 0.0211 0.0906 0.2085 0.0332 0.0196 0.0945 0.1973 0.0190 0.0177 0.0964 0.1831 0.0052 0.0155 0.0966 0.1673 -0.0076 0.0134 0.0955 0.1665 -0.0242 0.0102 0.0915 0.1758 -0.0412 0.0066 0.0858 0.1837 -0.0578 0.0029 0.0784 0.1892 -0.0728 -0.0008 0.0696 0.1932 -0.0851 -0.0042 0.0596 0.1990 -0.0938 -0.0072 0.0491 0.2001 -0.0984 -0.0095 0.0387 0.1965 -0.0988 -0.0111 0.0288 0.1887 -0.0954 -0.0119 0.0202 0.1775 -0.0891 -0.0120 0.0129 0.1640 -0.0807 -0.0115 0.0072 0.1494 -0.0711 -0.0107 0.0030 0.1347 -0.0610 -0.0095 0.0001 0.1206 -0.0512 -0.0083 -0.0017 0.1112 -0.0419 -0.0070 -0.0027 0.1016 -0.0335 -0.0057 -0.0030 0.0923 -0.0261 -0.0045 -0.0029 0.0835 -0.0197 -0.0035 -0.0025 0.0756 -0.0142 -0.0025 -0.0020 0.0686 -0.0095 -0.0017 -0.0013 0.0626

u=L

Assume here that both xD + xB are important

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>> span = abs(yoptf)+abs(yoptz)+abs(yoptq)+0.05;>> plot(gainl’./span)

u=L: max. on stage 15 (below feed stage)In practice (because of dynamics): Will use tray in top (peak at 27)

5 10 15 20 25 30 35 4010

20

30

40

50

60

70

80

Assume here that both xD + xB are important

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Gwv =

-1.0975 -0.8625

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)

u=V >> [yoptf yoptq yoptz span]

ans =

0.0065 -0.0008 -0.0000 0.0573 0.0091 -0.0011 0.0001 0.0603 0.0123 -0.0014 0.0004 0.0641 0.0162 -0.0018 0.0010 0.0690 0.0208 -0.0022 0.0021 0.0750 0.0260 -0.0025 0.0038 0.0823 0.0320 -0.0029 0.0062 0.0911 0.0384 -0.0032 0.0097 0.1013 0.0450 -0.0033 0.0144 0.1128 0.0513 -0.0033 0.0205 0.1251 0.0567 -0.0030 0.0280 0.1377 0.0604 -0.0023 0.0367 0.1495 0.0617 -0.0013 0.0464 0.1594 0.0601 0.0001 0.0565 0.1667 0.0553 0.0019 0.0664 0.1736 0.0476 0.0040 0.0754 0.1770 0.0376 0.0063 0.0830 0.1769 0.0262 0.0086 0.0888 0.1736 0.0143 0.0109 0.0929 0.1681 0.0028 0.0130 0.0952 0.1609 -0.0078 0.0148 0.0962 0.1688 -0.0221 0.0165 0.0946 0.1832 -0.0367 0.0180 0.0915 0.1962 -0.0510 0.0191 0.0866 0.2067 -0.0639 0.0197 0.0800 0.2136 -0.0744 0.0198 0.0718 0.2160 -0.0819 0.0192 0.0625 0.2137 -0.0858 0.0182 0.0527 0.2067 -0.0860 0.0167 0.0430 0.1957 -0.0831 0.0149 0.0338 0.1818 -0.0775 0.0130 0.0256 0.1662 -0.0702 0.0111 0.0187 0.1500 -0.0618 0.0093 0.0131 0.1341 -0.0530 0.0076 0.0088 0.1194 -0.0444 0.0061 0.0056 0.1061 -0.0364 0.0048 0.0033 0.0945 -0.0291 0.0037 0.0018 0.0846 -0.0226 0.0028 0.0008 0.0763 -0.0170 0.0021 0.0003 0.0694 -0.0123 0.0015 0.0001 0.0638 -0.0083 0.0010 0.0000 0.0593

Assume here that both xD + xB are important

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>> span = abs(yoptf)+abs(yoptz)+abs(yoptq)+0.05;>> plot(gain’./span)

u=V: sharp max. on stage 14 (below feed stage)

5 10 15 20 25 30 35 4010

20

30

40

50

60

70

80

90

100

Assume here that both xD + xB are important

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COMMENT!

NEED to be careful about nonlinearity!Effect of step size in V:

delta=0.005Gw = -0.7600 -1.1956

delta=0.001Gw = -1.0276 -0.9326

delta=0.0002Gw = -1.0843 -0.8758

delta=0.00005Gw = -1.0948 -0.8653

delta=0.00001Gw = -1.0975 -0.8625