1

ChE / MET 43311 Apr 12

Process Linearity, Integral Windup, PID Controllers

Linearity, Windup, & PID

2

ChE / MET 433 Quiz Solutions

Process LinearityTest the Heat Exchanger process linearity by:• Starting Loop Pro trainer• Set %CO to 80%• Make steps down (say 10% down) to the %CO• Measure the response • Calculate the process gain

3

SCK

4

K = -0.15

K = -1.09

K = -0.69

K = -0.26

K = 0.-45K = -0.33

Adaptive Control ?

Integral (Reset) Windup

5

• “Windup” can occur if integral action present• Most modern controllers have anti-windup protection• If doesn’t have windup protection, set to manual when reach point

of saturation, then switch back to auto, when drops below sat. level

• IE: LoopPro Trainer, select Heat Exchanger• Set %CO to 90%; SP to 126; Kc to 1 %/deg C; Tau I to 1.0 min• Set Integral with Anti-Reset Windup ON• Change Set Point to 120 deg. C. (~10 min); then change back to

126 deg. C• Repeat with controller at ON: (Integral with Windup)

Integral (Reset) Windup

6

In-Class PID Controller Exercise

Tune the Heat Exchanger for a PID Controller:• Use the built in IMC, and choose Moderately Aggressive• Start Loop Pro trainer• Tune at the initial %CO and exit temperature• Compare PI with PID• Compare PID with PID with filter

7

8

ChE / MET 433

11 Apr 12Cascade Control: Ch

9

Advanced control schemes

Improve Feedback Control

Feedback control:• Disturbance must be measured before action taken• ~ 80% of control strategies are simple FB control• Reacts to disturbances that were not expected

9

We’ll look at:• Cascade Control (Master – Slave)• Ratio Control• Feed Forward

Cascade Control• Control w/ multiple loops• Used to better reject specific disturbances

10

Take slow process:

PGcG-

sE+ sR sC)(sM

??PG

Split into 2 “processes” that can measure intermediate variable?

2PGcG-

sE+ sR sC

A-

+1PG

2TK

2CG

Gp2 must be quicker responding than GP1. • Inner (2nd-dary) loop faster

than primary loop• Outer loop is primary loop

11

Material Dryer Example

PGcG-

sE+ sR sCmoisture%

Heat Exchger

T

airblower

MC

spMT

steam

% moisture

VG TK

12

Separate Gp into 2 blocks

Heat Exchger

T

airblower

MC

spMT

steam

% moisture

sp

TTTC

TPG1cG

-

sE+ sR sC

A-

+MPG

TTK

2CG VGMTK

primaryset point Secondary

Controller

secondary process variable

primary process variable

Final ControlElement

SecondaryProcess

PrimaryController

Primary Process

secondaryset point

DisturbanceProcess I

–+ ++–+

DisturbanceProcess II

secondaryprocessvariable

++

primaryprocessvariable

disturbancevariable I

disturbancevariable II

cascade control can improve rejection of this disturbance

but can not help rejection of this disturbance

13

14

Problem Solving Exercise: Heat

Exchanger

Heat Exchger

T

Hot water

TC

sp

TTsteam

Single feedback loop.Suppose known there will be steam

pressure fluctuations…

Design cascade system that measures (uses) the steam pressure in the HX shell.

Heat Exchger

T

Hot water

TTsteamPT

Temperature Control of a Well-Mixed Reactor (CSTR)

Responds quicker to Tichanges than coolant temperature changes.

Ti

15

Temperature Control of a Well-Mixed Reactor (CSTR)

Ti

If Tout (jacket) changes it is sensed and controlled before “seen” by primary T sensor.

Use Cascade Control to improve control.

Secondary Loop• Measures Tout (jacket)• Faster loop• SP by output primary loop

Primary Loop:• Measures controlled var.• SP by operator

16

Cascade Control

• Disturbances in secondary loop corrected by 2ndary loop controller• Flowrate loops are frequently cascaded with another control loop• Improves regulatory control, but doesn’t affect set point tracking • Can address different disturbances, as long as they impact the

secondary loop before it significantly impacts the primary (outer loop).

Benefits:

• Secondary loop must be faster than primary loop• Bit more complex to tune• Requires additional sensor and controller

Challenges:

17

18

Cascade Control

Examples

Objective:

Regulate temperature (composition)

at top and bottom of

column

Distillation Columns

19

Objective:

Keep T2 outat the

set point

T2 out

Objective:

Keep TP

outat the

set point

TP out

Heat Exchanger

Furnace

20

In-Class Exercise: Cascade Control System Design

What affects flowrate?• Valve position• Height of liquid• P (delta P across valve)

Design a cascade system to control level (note overhead P can’t be controlled)

21

In-Class Exercise: Cascade Control System Design

Does this design reject P changes in the overhead vapor space?

Tuning a Cascade System

22

• Both controllers in manual• Secondary controller set as P-only (could be PI, but this might slow

sys)• Tune secondary controller for set point tracking• Check secondary loop for satisfactory set point tracking

performance• Leave secondary controller in Auto• Tune primary controller for disturbance rejection (PI or PID)• Both controllers in Auto now• Verify acceptable performance

23

In-Class Exercise: Tuning Cascade Controllers

• Select Jacketed Reactor• Set T cooling inlet at 46 oC (normal operation temperature; sometimes it drops to 40 oC)• Set output of controller at 50%.• Desired Tout set point is 86 oC (this is steady state temperature)

• Tune the single loop PI control• Criteria: IMC aggressive tuning• Use doublet test with +/- 5 %CO• Test your tuning with disturbance from 46 oC to 40 oC

24

In-Class Exercise: Tuning Cascade Controllers• Select Cascade Jacketed Reactor• Set T cooling inlet at 46 oC (again)• Set output of controller (secondary) at 50%.• Desired Tout set point is 86 oC (as before)

• Note the secondary outlet temperature (69 oC) is the SP of the secondary controller

• Tune the secondary loop; use 5 %CO doublet open loop• Criteria: ITAE for set point tracking (P only)• Use doublet test with +/- 5 %CO• Test your tuning with 3 oC setpoint changes• Tune the primary loop for PI control; make 3 oC set point changes (2nd-dary controller)• Note: MV = sp signal; and PV = T out of reactor• Criteria: IAE for aggressive tuning (PI)• Implement and with both controllers in Auto… change disturbance from 46 to 40 oC.• How does response compare to single PI feedback loop?

25

ChE / MET 433

13 Apr 12Ratio Control: Ch 10

Advanced control schemes

Ratio Control• Special type of feed forward control

•Blending/Reaction/Flocculation

•A and B must be in certain ratio to each other

A B

26

Ratio ControlPossible control system:

•What if one stream could not be controlled?

• i.e., suppose stream A was “wild”; or it came from an upstream process and couldn’t be controlled.

A B

27

FT

FC

sp

FY

FT

FC

sp

FY

Ratio ControlPossible cascade control systems:

“wild” stream

A

B

28

FT

FT

FY FC

sp

A

B

AB

Desired Ratio

A

BFT

FT

FY

FCBsp

A

B

AB

Desired RatioThis unit multiplies A by the desired ratio; so output = A

BA

“wild” stream

AB

Ratio Control Uses:

29

• Constant ratio between feed flowrate and steam in reboiler of distillation column

• Constant reflux ratio

• Ratio of reactants entering reactor

• Ratio for blending two streams

• Flocculent addition dependent on feed stream

• Purge stream ratio

• Fuel/air ratio in burner

• Neutralization/pH

30

In-Class Exercise: Furnace Air/Fuel Ratio• Furnace Air/Fuel Ratio model• disturbance: liquid flowrate• “wild” stream: air flowrate• ratioed stream: fuel flowrate

• Minimum Air/Fuel Ratio 10/1• Fuel-rich undesired (enviro, econ, safety)• If air fails; fuel is shut down

Independent MV

PV

Ratio set point

Dependent MV

Disturbance var.

TC

TC output

Desired 2 – 5% excess O2

Check TC tuning to disturbance & SP changes.

31

ChE / MET 433

16 Apr 12Feed Forward Control: Ch 11

Advanced control schemes

Feed Forward ControlSuppose qi is primary disturbance

32

Heat Exchanger

TC

TT)(tqi

)(tTi

? What is a drawback to this feedback control loop?? Is there a potentially better way?

Heat ExchangerTTFT

FF

)(tTi

)(tqi

? What if Ti changes?

FF must be done with FB control!

steam

steam

Feed Forward and Feedback Control

33

Heat ExchangerTTFT

TY

)(tTi)(tqi

steamTC

FF?

TYP

I)(tM FF )(tM

)(tM

FFFF MtMtMtM )()()(

Block diagram:

TPGCG

sE sT++

FFG

TTK

VG

DTKLG

sQi

++

M

FFM

M-

+ sR

FFCGFF

Feed Forward Control

No change; perfect compensation!

34

PGCG-

sE+ sR sT++

FFG

TTK

VG

DTKLG

sQi

++

M

FFM

M

t0

DT

PT

tT

PT

MFF

DT

tqi

Response to MFF

Feed Forward Control

Examine FFC T.F.

35

MGCG-

sE+ sR sC++

FFC

DTKDG

sQi

++FFM

M

MG sC

FFC

DTKDG

sQi

++

FFM

gpm

TO%

DTO%

FFCO%

)()( sQKFFCGsQGsC iTMiD D

For “perfect” FF control: 0sC

)()(0 sQKFFCGsQG iTMiD D

MT

D

GKGFFCD

TO%

TO%

Feed Forward Control: FFC Identification

Set by traditional means:

36

DTKMT

D

GKGFFCD

Model fit to FOPDT equation: MD GG &

1

seKG

D

stD

D

Do

1

seKG

M

stM

M

Mo

gpmTO%

COTO

%%

gpmTOD%

stt

D

M

MT

D oMDo

D

ess

KKKFFC

11

FF Gain

Lead/lag unit

Dead time compensator

{ FFC ss }steady state FF control

{ FFC dyn }dynamic FF control

Accounts for time differences in 2 legs

Often ignored; if set term to 1

oMo ttD

37

ChE / MET 433

38

Problem Solving Exercise: Heat Exchanger

Draw the block diagram: what is the primary and what is the secondary loop?

Heat Exchger

T

Hot water

TC

sp

TTsteamPT

PC

PPGTcG

-

sE+ sR sT

-

+TPG

PTK

PCG VG

TTK

sP