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ChE / MET 433

11 Apr 12Process Linearity, Integral Windup, PID Controllers

Linearity, Windup, & PID

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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

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SCK

4

K = -0.15

K = -1.09

K = -0.69

K = -0.26

K = 0.-45

K = -0.33

Adaptive Control ?

Integral (Reset) Windup

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• “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

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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

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ChE / MET 433

11 Apr 12Cascade Control: Ch

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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

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

Cascade Control

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

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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

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Material Dryer Example

PGcG-

sE+ sR sCmoisture%

Heat Exchger

T

airblower

MC

sp

MT

steam

% moisture

VG TK

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Separate Gp into 2 blocks

Heat Exchger

T

airblower

MC

sp

MT

steam

% moisture

sp

TT

TC

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

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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

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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

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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:

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Cascade Control

Examples

Objective:

Regulate temperature (composition)

at top and bottom of

column

Distillation Columns

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Objective:

Keep T2 out

at the set point

T2 out

Objective:

Keep TP

out

at the set point

TP out

Heat Exchanger

Furnace

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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)

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In-Class Exercise: Cascade Control System Design

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

Tuning a Cascade System

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• 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

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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

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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?

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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

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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

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FT

FC

sp

FY

FT

FC

sp

FY

Ratio ControlPossible cascade control systems:

“wild” stream

A

B

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FT

FT

FY FC

sp

A

B

AB

Desired Ratio

A

B

FT

FT

FY

FC

Bsp

A

B

AB

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

BA

“wild” stream

AB

Ratio Control Uses:

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• 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

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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.

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ChE / MET 433

16 Apr 12Feed Forward Control: Ch 11

Advanced control schemes

Feed Forward Control

Suppose qi is primary disturbance

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Heat Exchanger

TC

TT)(tqi

)(tTi

? What is a drawback to this feedback control loop?

? Is there a potentially better way?

Heat Exchanger

TTFT

FF

)(tTi

)(tqi

?What if Ti changes?

FF must be done with FB control!

steam

steam

Feed Forward and Feedback Control

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Heat Exchanger

TTFT

TY

)(tTi)(tqi

steamTC

FF

?

TY

PI

)(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!

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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.

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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

GK

GFFC

D

TO%

TO%

Feed Forward Control: FFC Identification

Set by traditional means:

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DTKMT

D

GK

GFFC

D

Model fit to FOPDT equation: MD GG &

1

s

eKG

D

st

DD

Do

1

s

eKG

M

st

MM

Mo

gpm

TO%

CO

TO

%

%

gpm

TOD%

stt

D

M

MT

D oMDo

D

es

s

KK

KFFC

1

1

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

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ChE / MET 433

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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