control of multivariable processes · step change to reflux with constant reboiler ... control of...

CONTROL OF MULTIVARIABLE PROCESSES

Process plants ( or complex experiments) have many variables that must be controlled. The engineer must

1. Provide the needed sensors2. Provide adequate manipulated variables3. Decide how the CVs and MVs are paired

(linked via the control design)

Fortunately, most of what we learned about single-loop systems applies, but we need to learn more!

v1

Hot Oil

v2

v3

L1

v7

v5 v6

Hot Oil

F1 T1 T3

T2

F2

T4T5

F3 T6

T8

F4

L1

v8

T7

P1F5

F6T9


Two control approaches are possible

L

F

T

A

Multiloop: Many independent PID controllers

Centralized: Method covered in Chapter 23

L

FT

A

Centralized

Controller


Two control approaches are possible

We will concentrate on MULTILOOP

A

Multiloop: Many independent PID controllers

IdtCVdTdttE

TtEKtMV

tCVtSPtEt

dI

c +

−+=

−=

∫0

')'(1)()(

)()()(IdtCVdTdttE

TtEKtMV

tCVtSPtEt

dI

c +

−+=

−=

∫0

')'(1)()(

)()()(

IdtCVdTdttE

TtEKtMV

tCVtSPtEt

dI

c +

−+=

−=

∫0

')'(1)()(

)()()(

FC

LCTC

v1

v3

v2


v1

Hot Oil

v2

v3

L1

v7

v5 v6

Hot Oil

F1 T1 T3

T2

F2

T4T5

F3 T6

T8

F4

L1

v8

T7

P1F5

F6T9

Let’s assume (for now) that we have the sensors and valves.

How do we ask the right questions in the best order to have a systematic method for pairing multiloop control?


Some key questions whose answers help us design a multiloop control system.

1. IS INTERACTION PRESENT?- If no interaction ⇒ All single-loop problems

2. IS CONTROL POSSIBLE?- Can we control the specified CVs with the

available MVs?

3. WHAT IS S-S AND DYNAMIC BEHAVIOR?- Over what range can control keep CVs near the set points?

Let’s starthere to build

understanding


T

A

What is different when we have multiple MVs and CVs?

INTERACTION!!

Definition: A multivariable process has interaction when input (manipulated) variables affect more than one output (controlled) variable.

Does this process have interaction?


How can we determine how much interaction exists?

One way - Fundamental modelling

FA, xA

FS, xAS = 0FM, xAM

AMMASSAA

MSA

xFxFxFFFF

=+

=+'S

ssAs

A'A

ssAs

SAM

SAM

F)FF(

FF)FF(

F'x

'F'F'F

+

−+

+=

+=

22

Fundamental (n-l) Fundamental linearized

v1

v2




time

v1, %

ope

n

FA, xA

FS, xAS = 0FM, xAM

v1

v2

time

v2, %

ope

ntime

Tota

l flo

w, F

M

timeC

once

nt,

xA

M

Sketch the responses of

the dependent variables




time

v1, %

ope

n

FA, xA

FS, xAS = 0FM, xAM

v1

v2

time

v2, %

ope

ntime

Tota

l flo

w, F

M

timeC

once

n.,

x AM

Sketch the responses of

the dependent variables



Second way - Empirical modelling (process reaction curve)

0 20 40 60 80 100 1200.98

0.982

0.984

0.986

0.988

0.99

Time (min)

XD

(mol

frac)

0 20 40 60 80 100 1200.02

0.025

0.03

0.035

0.04

Time (min)

XB

(mol

frac)

0 20 40 60 80 100 1201.12

1.125

1.13

1.135x 10

4

Time (min)

R (m

ol/m

in)

0 20 40 60 80 100 1201.5613

1.5613

1.5614

1.5614

1.5615

1.5615x 10

4

Time (min)

V (m

ol/m

in)

Step change to reflux with constant reboiler

Sketch the responses of the dependent variables

xD

xB

V

R




0 20 40 60 80 100 1200.98

0.982

0.984

0.986

0.988

0.99

Time (min)

XD

(mol

frac)

0 20 40 60 80 100 1200.02

0.025

0.03

0.035

0.04

Time (min)

XB

(mol

frac)

0 20 40 60 80 100 1201.12

1.125

1.13

1.135x 10

4

Time (min)

R (m

ol/m

in)

0 20 40 60 80 100 1201.5613

1.5613

1.5614

1.5614

1.5615

1.5615x 10

4

Time (min)

V (m

ol/m

in)

Step change to reflux with constant reboiler

xD

xB

V

R




)(.

.)(.

.)(

)(.)(.)(

.

sFsesF

sesx

sFsesF

sesx

V

s

R

s

B

V

s

R

s

D

121012530

171111730

11506670

11207470

233

23

+−

+=

+−

+=

−−

−−FR

FV

xB

xD

Which models can be determine from the experiment shown in

the previous slide?



Use the model; if model can be arranged so that it has a “diagonal” form, no interaction exists.

=

3

2

1

22

111

000000

MVMVMV

KK

K

CV

CV

nnn

...

If any off-diagonal are non-zero, interaction exists.



Let’s use qualitative understanding to answer the question.

T

A

Does this CSTR reactor process haveinteraction?v1

v2

Feed has CA0

cooling

Information:

1. Exothermic reaction

2. -rA = k0 e-E/RT CA

LC



Let’s use qualitative understanding to answer the question.

T

A

v1

v2

Feed has CA0

cooling

LC

v1 → feed flow rate → residence time → conversion → “heat generated” → T

→ Q/(FρCP) → → Tv2 → coolant flow → heat transfer → conversion → Analyzer

From the cause-effect analysis, v1 affects both sensors; interaction exists!

What about v2?

+ +

+ +

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)D(s)

CV1(s)

CV2(s)MV2(s)

MV1(s)


We will use a block diagram to represent the dynamics of a 2x2 process, which involves multivariable control.

The blocks (G) cannot be found in the process. They are the models that represent the behavior between one input and one output.

FD

XD

FRB

FB

XB

FXF

q

FR

FV

+ +

+ +

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)XF(s)

XD(s)

XB(s)FRB(s)

FR(s)

Let’s relate the block diagram to a typical physical process. What are the MVs, CVs, and a disturbance, D?


+ +

+ +

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)


T

A

Reactor


v1

v2

LC

Better control level!

+ +

+ +

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)


T

A

Reactor


v1

v2

v1

v2

A

T

Disturbances = feed composition, feed temperature, coolant temperature, …

LC

Better control level!




2. IS CONTROL POSSIBLE?- How many degrees of freedom exist?- Can we control CVs with MVs?


Let’s learn how to answer thiskey question


DEGREES OF FREEDOM

How do we determine the maximum # variables that be controlled in a process?

v1

Hot Oil

v2

v3

L1

v7

v5 v6

Hot Oil

F1 T1 T3

T2

F2

T4T5

F3 T6

T8

F4

L1

v8

T7

P1F5

F6T9


DEGREES OF FREEDOM

A requirement for a successful design is:The number of valves ≥ number of CV to be controlled

v1

Hot Oil

v2

v3

L1

v7

v5 v6

Hot Oil

F1 T1 T3

T2

F2

T4T5

F3 T6

T8

F4

L1

v8

T7

P1F5

F6T9

OK, but this does not ensure that we cancontrol the CVs that we want to control!


CONTROLLABILITY

A system is controllable if its CVs can be maintained at their set points, in the steady-state, in spite of disturbances

entering the system.

A system is controllable when the matrix of process gains can be inverted, i.e., when the determinant of K ≠ 0.

DKK

MVMV

KKKK

CVCV

d

d

+

=

=

2

1

2

1

2221

1211

2

1

00Model

for 2x2 system

I am in desperate need of examples


For the autos in the figure

• Are they independently controllable?

• Does interaction exist?

Let’s do the toyautos first; then,

do some processes





Connected by spring





Connected by beam


FA, xA

FS, xAS = 0FM, xAM

For process Example #1: the blending process

• Are the CVs independently controllable?


'S

ssAs

A'A

ssAs

SAM

SAM

F)FF(

FF)FF(

F'x

'F'F'F

+

−+

+=

+=

22


FA, xA

FS, xAS = 0FM, xAM

For process Example #1: the blending process



'S

ssAs

A'A

ssAs

SAM

SAM

F)FF(

FF)FF(

F'x

'F'F'F

+

−+

+=

+=

22

02

2

2

2≠

+−

+

−=

)FF(F

)FF(F)K(Det

SA

S

SA

A

Yes, this system is controllable!


For process Example #2: the distillation tower



FR

FV

xB

xD

)(.

.)(.

.)(

)(.)(.)(

.

sFsesF

sesx

sFsesF

sesx

V

s

R

s

B

V

s

R

s

D

121012530

171111730

11506670

11207470

233

23

+−

+=

+−

+=

−−

−−


For process Example #2: the distillation tower

FR

FV

xB

xD

)(.

.)(.

.)(

)(.)(.)(

.

sFsesF

sesx

sFsesF

sesx

V

s

R

s

B

V

s

R

s

D

121012530

171111730

11506670

11207470

233

23

+−

+=

+−

+=

−−

−−

Det (K) = 1.54 x 10-3 ≠ 0

Small but not zero (each gain is small)

The system is controllable!


For process Example #3: the non-isothermal CSTR



+ +

+ +

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)

A → B-rA = k0 e -E/RT CA

T

ACB

v1

v2





T

ACB


v1

v2

The interaction can be strong

In general, the temperature and conversion (extent of reaction) can be influenced.

The system is controllable.

(See Appendix 3 for examples)


For process Example #4: the mixing tank



T

ACA

+ +

+ +

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)v1

v2


For process Example #4: the mixing tank



T

ACA

+ +

+ +

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)v1

v2 0

0

Nothing affects composition at S-S; the system is NOT controllable.

CONTROL OF MULTIVARIABLE PROCESSESFor process Example #5: the non-isothermal CSTR




T

ACB

+ +

+ +

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)v1 v2

CONTROL OF MULTIVARIABLE PROCESSESFor process Example #5: the non-isothermal CSTR




T

ACB

v1 v2

Solution continued on

next slide

=

=

2

1

2121

1111

00

MVMV

KKKK

TCB

µµ

Both valves have the same effects on both variables; the only difference is the magnitude of the flow change (µ = constant).

Det (K) = 0; not controllable!


In this case, both MVs affect ONE common variable, and this common variable affects both CVs. We can change both CVs, but we cannot move the CVs to independentvalues!

+ +

+ +

G11(s)

G22(s)

Gd2(s)

Gd1(s)D(s)

CV1(s)

CV2(s)MV2(s)

MV1(s)

G(s)

G(s)

“InputContraction”


next slide

SP2(s)

+-

++

+ +-+

Gc1(s)

Gc2(s)

G11(s)

G22(s)

Gd2(s)

Gd1(s)

D(s)

CV1(s)

CV2(s)MV2(s)

MV1(s)SP1(s)

Does not

work!!


For input contraction, multivariable feedback control is not possible; the system is not controllable! We can change both CVs, but we cannot move the CVs to independent values!

Solutioncomplete




A → B + 2C-rA = k0 e -E/RT CA

A

ACB

CC


+ +

+ +

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)v1

v2





A

ACB

CC


v1

v2

Using the symbol Ni for the number of moles of component “i” that reacts, we have the following.

ACAB NNNN 2−=−=

Because of the stoichiometry,

NC = 2 NB

and the system is not controllable!


next slide





A

ACB

CC


v1

v2


next slide

=

=

2

1

2121

1111

2200

MVMV

KKKK

CC

C

B

Det (K) = 0; not controllable!

For output contraction, both MVs affect both CVs, but the CVs are related through the physics and chemistry. We can change both CVS, but we cannot move the CVs to independent values!

+

+

G11(s)

G22(s)

Gd2(s)

Gd1(s)D(s)

CV1(s)

CV2(s)MV2(s)

MV1(s)

G12(s)

+

G(s)

“output contraction”


next slide

+-

++

+ +-

+

Gc1(s)

Gc2(s)

G11(s)

G12(s) Gd2(s)

Gd1(s)

D(s)

CV1(s)

CV2(s)

MV2(s)

MV1(s)

SP1(s)

SP2(s)

Does not

work!!

In this case, multivariable feedback control is not possible; the system is uncontrollable!

Solutioncomplete


CONTROLLABILITY

Conclusions about determining controllability

1. One CV cannot be affected by any valve

3. Lack of independent effects.Look for “contractions”

This is generallyeasy to determine.

This requires care andprocess insight or modelling to

determine.

Lack of controllability when




2. IS CONTROL POSSIBLE?- How many degrees of freedom exist?- Can we control CVs with MVs?


Let’s see howgood the

performancecan be


How does interaction affect the steady-state behavior of the blending process?

FA, xA

FS, xAS = 0 FM, xAM

0

10

20

30

40

50

60

70

80

90

100

0.00 0.10 0.20 0.33 0.40 0.50 0.60 0.70 0.80 0.90 1.00

xAM, Composition (fraction A)

Tota

l flo

w, F

3

Please sketch the achievable

values

FA maximum = 30

Fs maximum = 60


How does interaction affect the steady-state behavior of the blending process?

FA, xA

FS, xAS = 0 FM, xAM

0

10

20

30

40

50

60

70

80

90

100

0.00 0.10 0.20 0.33 0.40 0.50 0.60 0.70 0.80 0.90 1.00

Composition (fraction A)

Tota

l flo

w, F

3

infeasible

feasible

Please explainthis shape

Note:

This shows a range of set points that can be achieved (without disturbances).

FA maximum = 30

Fs maximum = 60


STEADY-STATE OPERATING WINDOW

Conclusions about the steady-state behavior of a multivariable process

1. Shape of the Window

2. What influences the size of the Window?

3. How does Window relate to Controllability?

Summarize your conclusions here.

The operating window is not generally a simple shape, like a rectangle.

The operating window is influenced by process chemistry, equipment capacity, disturbances …

Process cannot be “moved” (controlled) outside of the window. If operating window is empty, the system is not controllable.


1. How many experiments are needed to tune controllers?

2. Which controller should be implemented first?

+ +

+ +

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)D(s)

CV1(s)

CV2(s)MV2(s)

MV1(s))()(

1

1

sMVsCV

)()(

2

2

sMVsCV

To tune each controller, we need

NOW, LET’S LOOK AT THE DYNAMIC BEHAVIOR


3. We have implemented one controller. What do we do now?

+ +

+ +-+ Gc2(s)

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)

D(s)

CV1(s)

CV2(s)

MV2(s)

MV1(s)

SP2(s)


4. What if we perform another experiment to learn the dynamics between MV1 and CV1, with controller #2 in automatic?

Is the dynamic behavior different from without second controller (GC2)? What elements does the behavior depend upon?

+ +G11(s)

G21(s)

G12(s) Gd2(s)

Gd1(s)

D(s)

CV1(s)

MV1(s)


4. Is the dynamic behavior different from without second controller (GC2)? What elements does the behavior depend upon?

)s(G)s(G)s(G)s(G)s(G

)s(G)s(MV)s(CV

c

c

222

2211211

1

1

1+−=

The process “seen” by the controller

changes!

+

+ +-+ Gc2(s)

G11(s)

G21(s)

G12(s)

G22(s)

CV1(s)

CV2(s)

MV2(s)

MV1(s)

SP2(s)

“direct path”

“interaction path”


In general, the behavior of one loop depends on the interaction and the tuning of the other loop(s).

+-+

+

+ +-

+

Gc1(s)

Gc2(s)

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)

D(s)

CV1(s)

CV2(s)

MV2(s)

MV1(s)

SP1(s)

SP2(s)

• Tuning that is stable for each loop might not be stable when both are in operation!

• We need to tune loops iteratively, until we obtain good performance for all loops!

I think that I need an example again!



Let’s look at a simple example with interaction,

++

++=

−−

−−

)s(MV)s(MV

se.

se.

se.

se.

)s(CV)s(CV

s.s.

s.s.

2

10101

0101

2

1

2101

21750

21750

2101

We will pair the loops on the “strongest” gains,

MV1(s) ⇒ CV1(s) and MV 2(s) ⇒ CV2(s)

Results with only one controller in automatic (KC1 = 2.0, TI1 = 3)

This system is stable and perhaps a bit too aggressive.

0 20 40 60 80 1000

0.5

1

1.5IAE = 2.5003 ISE = 1.5448

CV

1

0 20 40 60 80 1000

0.2

0.4

0.6

0.8

1IAE = 66.45 ISE = 49.8627

CV

2

0 20 40 60 80 100 1200

0.5

1

1.5

2

2.5

3SAM = 7.1543 SSM = 87.4545

Time

MV

1

0 20 40 60 80 100 120-1

-0.5

0

0.5

1SAM = 0 SSM = 0

Time

MV

2

Feedback control for SP change

Why did this change?

Results with both controllers in automatic (Kc = 2.0, TI = 3)

This system is unstable!! Each was stable by itself!!

0 20 40 60 80 100-15

-10

-5

0

5

10

15IAE = 251.0259 ISE = 1426.773

CV

1

0 20 40 60 80 100-15

-10

-5

0

5

10

15IAE = 250.3684 ISE = 1425.127

CV

2

0 20 40 60 80 100 120-20

-10

0

10

20SAM = 854.3713 SSM = 16670.7031

Time

MV

1

0 20 40 60 80 100 120-20

-10

0

10

20SAM = 852.3648 SSM = 16618.2934

Time

MV

2



)]s(G)s(G)s(G)s(G)[s(G)s(G)s(G)s(G)s(G)s(G)s(numerator

)s(SP)s(CV

cccc 21122211212221111

1

1 −+++=

+-+

+

+ +-

+

Gc1(s)

Gc2(s)

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)

D(s)

CV1(s)

CV2(s)

MV2(s)

MV1(s)

SP1(s)

SP2(s)

I recall that elementsin the denominator

affect stability.



0

1

2

3

4

Kc2

, LO

OP

2 C

ON

TRO

LLER

GA

IN

0 0.5 1 1.5 2 2.5 3 3.5 Kc1, LOOP 1 CONTROLLER GAIN

Stable for 2x2

Unstable for 2x2

Notes:1. TI = 3 for both

controllers (reasonable = t63%)

2. KC < 3.75 stable for single-loop feedback

3. KC = 2.0 stable for one loop

4. KC = 2.0 unstable for two loops!!

5. These numerical results are for the example only; concepts are general

Single-loop inside the dashed lines would be stable.



0

1

2

3

4

Kc2

, LO

OP

2 C

ON

TRO

LLER

GA

IN

0 0.5 1 1.5 2 2.5 3 3.5 Kc1, LOOP 1 CONTROLLER GAIN

Stable for 2x2

Unstablefor 2x2

We can have any values in the stable region.

How do we choose the “best”.

If both CVs are of equal importance, we would detune both controllers equally. (Kc1 = Kc2 = 0.95; TI1 = TI2 = 3.0)

0 20 40 60 80 1000

0.2

0.4

0.6

0.8

1IAE = 7.2124 ISE = 2.8115

CV

1

0 20 40 60 80 1000

0.1

0.2

0.3

0.4

0.5IAE = 5.4079 ISE = 1.1407

CV

2

0 20 40 60 80 100 1200

0.5

1

1.5

2

2.5SAM = 2.8521 SSM = 18.89

Time

MV

1

0 20 40 60 80 100 120-2

-1.5

-1

-0.5

0SAM = 1.734 SSM = 0.27062

Time

MV

2

Somewhat slow to reach SP1

Interaction affects CV2

If CV1 is more important, we would make Gc1 aggressive and detune Gc2 more. (Kc1 = 1.40 and Kc2 = 0.50; TI1 = TI2 = 3.0)

0 20 40 60 80 1000

0.2

0.4

0.6

0.8

1IAE = 4.8803 ISE = 1.8496

CV

1

0 20 40 60 80 1000

0.2

0.4

0.6

0.8IAE = 10.2283 ISE = 3.1023

CV

2

0 20 40 60 80 100 1200

0.5

1

1.5

2

2.5SAM = 4.3516 SSM = 41.4664

Time

MV

1

0 20 40 60 80 100 120-2

-1.5

-1

-0.5

0SAM = 1.7163 SSM = 0.1792

Time

MV

2

Fast tracking of the set point

Increased disturbance from interaction

If CV2 is more important, we would make Gc2 aggressive and detune Gc1 more. (Kc1 = 0.50 and Kc2 = 1.40; TI1 = TI2 = 3.0)

0 20 40 60 80 1000

0.2

0.4

0.6

0.8

1IAE = 13.6475 ISE = 5.9352

CV

1

0 20 40 60 80 1000

0.1

0.2

0.3

0.4IAE = 3.653 ISE = 0.3957

CV

2

0 20 40 60 80 100 1200

0.5

1

1.5

2

2.5SAM = 2.2765 SSM = 5.2708

Time

MV

1

0 20 40 60 80 100 120-2

-1.5

-1

-0.5

0SAM = 1.7163 SSM = 0.1792

Time

MV

2

Very slow to reach the SP1

Reduced disturbance due to interaction



Some conclusions for multiloop PID tuning:

1. For multiloop, we generally have to tune the controllers in a less aggressive manner than for single-loop.

2. Textbook gives tuning approach, (Kc)ml ≈ 1/2 (Kc)sl

3. We can tune important loop tightly, if we also detune (make less aggressive) other loops.


Summary of key questions for multiloopcontrol system

1. IS INTERACTION PRESENT?- No interaction ⇒ All single-loop problems

- We use models to determine interaction

Fundamental or empirical

2. IS CONTROL POSSIBLE?

3. WHAT IS S-S AND DYNAMIC BEHAVIOR?



1. IS INTERACTION PRESENT?

2. IS CONTROL POSSIBLE?- DOF: The # of valves ≥ # of controlled variables

- Controllable: Independently affect every CV

Check if det [K] ≠ 0

Look for “contractions”

3. WHAT IS S-S AND DYNAMIC BEHAVIOR?





3. WHAT IS S-S AND DYNAMIC BEHAVIOR?- The operating window is:

strongly affected by interaction

strongly affected by equipment capacities

not a rectangle or symmetric





3. WHAT IS S-S AND DYNAMIC BEHAVIOR?- Interaction affects loop stability

- We have to detune to retain stability margin

- We can improve some loops by tight tuning, but we have to detune and degrade other loops


Workshop in Interaction in Multivariable Control Systems


Workshop Problem # 1

You have been asked to evaluate the control design in the figure.

Discuss good and poor aspects and decide whether you would recommend the design.



We need to control the mixing tank effluent temperature and concentration.

You have been asked to evaluate the design in the figure.

Discuss good and poor aspects and decide whether you would recommend the design.

F1T1CA1

F2T2CA2

T

A



Summarize the underlying principles that can lead to a “contraction” and to a loss of controllability. These principles will be applicable to many process examples, although the specific variables could be different.

Hint: Review the examples in the lecture and identify the root cause for each.

control of multivariable processes · step change to reflux with constant reboiler ... control of...

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