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    Chapter 2

    Traditional Advanced Control

    Approaches Feedforward, Cascadeand Selected Control

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    2-1 Feed Forward Control (FFC)

    Block Diagram

    Design of FFC controllers

    Examples

    Applications

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    Why Feedforward ?

    Advantages of Feedback Control Corrective action is independent of sources of

    disturbances

    No knowledge of process (process model) is required

    Versatile and robust

    Disadvantages No corrective action until disturbance has affected the

    output. Perfect control is impossible.

    Nothing can be done about known process disturbance

    If disturbances occur at a frequency comparable to thesettling time of the process. Then process may neversettle down.

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    Feedforward Control

    FeedforwardController Disturbance

    Process

    Output

    ManipulatedVariable

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    Feedforward Control

    Advantages Corrective action is taken as soon as disturbances

    arrives.

    Controlled variable need not be measured.

    Does not affect the stability of the processes

    Disadvantages Load variable must be measured

    A process model is required

    Errors in modeling can result in poor control

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    LI

    FFC

    steam

    FI

    LI

    steam

    FI

    FFCFB

    Feedback control

    Feedforward control

    Combined feedforward-feedback control

    LI

    FB

    BoilerFeed control

    steam

    FI

    EXAMPLES

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    Design Procedures (Block diagram

    Method)

    GL(s)

    Load transfer function

    GF(s)

    Load

    ManipulatedVariable

    Gp(s)

    Process

    X2

    C

    Output

    L

    M

    FF

    Controller

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    Derivation

    functiontransferprocess,(2)

    functiontransferload),()1(:need

    or

    0)(Hence).(allfor0)(wantWe

    )()(

    )()()()()(

    )()()()()(

    sG

    sG

    sG

    sGsG

    sGsGsG

    sLsC

    sLsGsGsG

    sLsGsGsLsG

    sMsGsLsGsC

    P

    L

    P

    L

    F

    FPL

    FPL

    FPL

    PL

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    Examples

    Example 1 Let Gp(s)=Kp/ps+1, GL(s)=KL/Ls+1

    Then, GF(s)=-(KL/Kp)(ps+1)/(Ls+1)

    Therefore,feedforward controller is a lead-lag unit.

    Example 2 Let Gp(s)=Kpe

    -Dps/ps+1, GL(s)=KLe-DLs/Ls+1

    Then, GF(s)=-(KL/Kp)(ps+1)/(Ls+1)e(-DL+DP)s

    If -DL+DPis positive, then this controller is unrealizable.However, an approximation would be to neglect thedelay terms, and readjusting the time constants. In thiscase, perfect FF compensation is impossible.

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    Tuning feedforward controllers

    Let

    This has three adjustable constants, K, 1, 2

    Tuning K, K is selected so that for a persistent

    disturbance, there is no steady state error inoutput.

    Adjusting1, 2 can be obtained from transferfunctions. Fine tune 1, 2 such that for a step

    disturbance, the response is somewhatsymmetrical about the set point.

    11)(

    2

    1

    ssKsGF

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    Example: A simulated disturbed plant

    Waste watertreatment

    Disturbedflow rate

    BOD

    Chemicals

    (CV)

    DV

    MV

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    Simulated Block Disgram

    1

    1 2 3s s s Disturbed flowrate

    +

    3

    1

    1s

    Chemicals

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    Feedforward v.s. Feedback Control

    0 5 10 15 20 25 30-0.2

    -0.15

    -0.1

    -0.05

    0

    0.05

    0.1

    0.15

    FB

    FF

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    Example: Distillation Column

    Example: Distillation Column Mass Balance: F=D+B

    Fz=Dy+Bx

    D=F(z-x)/(y-x)

    In practice

    For example: If light key increase in feed, increasedistillate rate.

    setset

    set

    xy

    xzFD

    )(

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    Design of Feedforward Control Using

    Material and Energy Balances Consider the hear exchanger

    Energy Balance yields Q=WC(T2-T1)=Ws

    Where =hear of vaporization Ws=WC(T2-T1)/

    This equation tells us the current stream demand based on (1)current flow rate, W, (2) current inlet temperature, T1, (3) desiredvalue of outlet temperature T2.

    Steam

    T2w, T1

    Condensate

    Ws

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    Control Law and Design

    Implementation:

    Note no dynamics are incorporated

    K

    X

    measured Tset

    Gain

    measured Ws

    w

    T1- +

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    When to use Feedforward ?

    Feedback control is unsatisfactory

    Disturbance can be measured andcompensated for

    Frequency of disturbance variations arecomparable to frequency of oscillation of thesystem

    Output variable cannot be measured. There are large time delays in the system

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    2-2 Cascade Control

    Block Diagram

    Design Considerations

    Applications

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    Illustrative Example : Steam Jacket

    PCPT

    TC

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    Illustrative Example: Steam Jacket -

    Continued

    Energy Balance of the Tank:

    Energy Balance of Jacket:

    Material Balance of the Jacket

    LossHeat TThAdt

    dTV J

    dt

    nPd

    R

    V

    Rn

    VP

    dt

    d

    dt

    dT sJJ

    s

    JJJ /

    condensatendt

    dnin

    s

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    Illustrative Example: Steam Jacket -

    Continued

    Assume:

    Where X=valve position

    21110

    113130

    1

    )(

    sssX

    sPsssP

    sT

    J

    J

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    Block Diagram

    Feed backController

    SteamValve

    StirredTank

    Tset

    Steamsupply

    pressure

    Valve

    position

    Jacket

    steampressure primary

    Tank

    Temp.

    secondary

    PrimaryController

    JacketPressureController

    SteamValve

    StirredTank

    Tset secondary

    Jacketpressure

    supplypressure

    TankTemp.

    Secondary loop

    Primary loop

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    Principal Advantages and Disadvantages

    Advantages Disturbances in the secondary loop are corrected by

    secondary controllers

    Response of the secondary loop is improved, thusincreasing the speed of response of the primary loop

    Gain variations in secondary loop are compensated bysecondary loop

    Disadvantages Increased cost of instrumentation

    Need to tune two loops instead of one Secondary variable must be measured

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    Design Considerations

    Secondary loop must be fast responding otherwisesystem will not settle

    Time constant in the secondary loop must be smaller than primaryloop

    Since secondary loop is fast, proportional actionalone is sufficient, offset is not a problem insecondary loop

    Only disturbances within the secondary loop are

    compensated by the secondary loop. Hence,cascading improves the response to thesedisturbances

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    Applications: 1. Valve Position Control

    Valve motion is affected by friction and pressuredrop in the line. Friction causes dead band. High

    pressure drop also causes hysteresis in the valveresponse

    Useful in most loops except flow and pressure

    ControlValveMotor

    Desired

    position

    Secondaryloop

    Valveposition

    Air Pressure to

    Valve Motor

    A li i 2 C d Fl L

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    Application 2. Cascade Flow LoopOutput From

    Primary Controller no cascade

    Output From

    Primary ControllerFC

    FT

    Fset

    DP

    cascade

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    GC2 GC1 GP1 GP2

    Primarycontroller

    Secondarycontroller

    Secondaryprocess

    primaryprocess

    cm2e2mset

    Secondary loop

    Primary loop

    cset

    2

    11

    112

    1C

    PC

    PC

    set

    GGG

    GGmm

    GC2 GCL GP2 mset m2

    ccset

    22

    22

    1 PCLC

    PCLC

    set GGG

    GGG

    c

    c

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    Gc 12)110()1(

    12 SS )13)(130(

    1

    SS

    +

    -

    Primary

    +

    -

    GC2

    Secondary

    G2(S) G3(S)

    2

    32

    121

    12

    G

    GG

    c

    For a cascade system

    (open-loop)

    Without cascade control

    32GG

    c

    c

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    Illustrative Example: Steam Jacket

    Continued Cascade Case

    Wu = 0.53

    Mag = 20*log10(AR) = -30 (dB)

    AR = 0.0316

    6228.311

    AR

    Ku

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    Illustrative Example: Steam Jacket

    Continued No Cascade Case

    Wu = 0.25

    Mag = 20*log10(AR) = 0 (dB)

    AR = 1

    11

    AR

    Ku

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    Illustrative Example: Steam Jacket

    Continued No Cascade Case

    Ku = 1;wu = 0.25;Pu = 2*Pi / wu = 25.1327

    Kc = Ku/1.7 = 0.5882

    Taui = Pu / 2 = 12.5664 Taud = Pu /8 = 3.1416

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    Illustrative Example: Steam Jacket

    Continued Cascade Case

    Ku = 20;wu = 0.53;Pu = 2*Pi / wu = 12

    Kc = Ku/1.7 = 11.8

    Taui = Pu / 2 = 6

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    2-3 Selective Control Systems

    Override Control

    Auctioneering Control

    Ratio Control

    Change from one controlled (CV) or manipulated variables

    (MV) to another

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    LT LC LSS

    PC

    Normalloop

    water

    steam

    1. Override Control Example Boiler Control

    LSS: Low Selective Switch Output a lower of two inputsPrevents: 1. Level from going too low, 2. Pressure fromexceeding limit (lower)

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    motor

    SC

    HSS

    PC FC

    Gas out

    Gas in

    Normal loop

    Example: Compressor Surge Control

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    High Pressure Line

    Low Pressure Line

    PC

    PC

    HSS

    Example: Steam Distribution System

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    Length of reactor

    Tem

    perate

    T1

    T2

    Hot spot

    2. Auctioneering Control Systems

    Temperature profiles in a tubular reactor

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    TT TTTTTT

    HSS

    Cooling flow

    Auctioneering Control Systems

    TC

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    Temperature Control

    Steam

    T2

    Bypass

    Exchanger

    TC

    Split Range Control: More than one manipulated variable is

    adjusted by the controller

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    Boiler 2 Steam Header

    Boiler 2

    Boiler 2

    PC

    Example: Steam Header: Pressure Control

    3 Ratio Control Type of feedforward control

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    FT

    Driver

    FT

    A Wild stream

    Controlled StreamB

    FB

    Gc

    Desired

    Ratio

    FA

    3. Ratio Control Type of feedforward control

    However, one stream in proportion to another. Use if the ratio

    must be measured and displayed

    Disadvantage:Ratio may goTo erratic

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    A Wild stream

    FT

    Desired Ratio

    Multiplier

    FC

    FT

    Controlled streamFB

    FA

    Another implementation of Ratio Control

    +

    -