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Chapter 2 Traditional Advanced Control Approaches – Feedforward, Cascade and Selected Control.

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Presentation on theme: "Chapter 2 Traditional Advanced Control Approaches – Feedforward, Cascade and Selected Control."— Presentation transcript:

1 Chapter 2 Traditional Advanced Control Approaches – Feedforward, Cascade and Selected Control

2 2-1 Feed Forward Control (FFC)
Block Diagram Design of FFC controllers Examples Applications

3 Why Feedforward ? Advantages of Feedback Control Disadvantages
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 the settling time of the process. Then process may never settle down.

4 Feedforward Control Feedforward Controller Disturbance Output Process
Manipulated Variable

5 Feedforward Control Advantages Disadvantages
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

6 EXAMPLES LI FB Boiler Feed control steam FI LI FFC steam FI
Feedback control Feedforward control LI steam FI FFC FB Σ Combined feedforward-feedback control

7 Design Procedures (Block diagram Method)
FF Controller GL(s) Load transfer function GF(s) Load Manipulated Variable Gp(s) Process X2 C Output L M

8 Derivation

9 Examples Example 1 Example 2 Let Gp(s)=Kp/τps+1, GL(s)=KL/τLs+1
Then, GF(s)=-(KL/Kp)(τps+1)/(τLs+1) Therefore, feedforward controller is a “lead-lag” unit. Example 2 Let Gp(s)=Kpe-Dps/τps+1, GL(s)=KLe-DLs/τLs+1 Then, GF(s)=-(KL/Kp)(τps+1)/(τLs+1)e(-DL+DP)s If -DL+DP is positive, then this controller is unrealizable. However, an approximation would be to neglect the delay terms, and readjusting the time constants. In this case, perfect FF compensation is impossible.

10 Tuning feedforward controllers
Let This has three adjustable constants, K, τ1, τ2 Tuning K, K is selected so that for a persistent disturbance, there is no steady state error in output. Adjustingτ1, τ2 can be obtained from transfer functions. Fine tune τ1, τ2 such that for a step disturbance, the response is somewhat symmetrical about the set point.

11 Example: A simulated disturbed plant
Disturbed flow rate DV Waste water treatment Chemicals MV BOD (CV)

12 Simulated Block Disgram
Disturbed flow rate + Chemicals

13 Feedforward v.s. Feedback Control

14 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, increase distillate rate.

15 Design of Feedforward Control Using Material and Energy Balances
Consider the hear exchanger Energy Balance yields Q=WC(T2-T1)=Wsλ Where λ=hear of vaporization Ws=WC(T2-T1)/λ This equation tells us the current stream demand based on (1) current flow rate, W, (2) current inlet temperature, T1, (3) desired value of outlet temperature T2. Ws Steam w, T1 T2 Condensate

16 Control Law and Design Implementation:
Note no dynamics are incorporated Σ K X measured Tset Gain Ws w T1 - +

17 When to use Feedforward ?
Feedback control is unsatisfactory Disturbance can be measured and compensated for Frequency of disturbance variations are comparable to frequency of oscillation of the system Output variable cannot be measured. There are large time delays in the system

18 2-2 Cascade Control Block Diagram Design Considerations Applications

19 Illustrative Example : Steam Jacket

20 Illustrative Example: Steam Jacket - Continued
Energy Balance of the Tank: Energy Balance of Jacket: Material Balance of the Jacket

21 Illustrative Example: Steam Jacket - Continued
Assume: Where X=valve position

22 Jacket Pressure Controller
Block Diagram Feed back Controller Steam Valve Stirred Tank Tset Steam supply pressure Valve position Jacket steam pressure primary Tank Temp. secondary Primary Controller Jacket Pressure Controller Steam Valve Stirred Tank Tset secondary Jacket pressure supply pressure Tank Temp. Secondary loop Primary loop

23 Principal Advantages and Disadvantages
Disturbances in the secondary loop are corrected by secondary controllers Response of the secondary loop is improved, thus increasing the speed of response of the primary loop Gain variations in secondary loop are compensated by secondary loop Disadvantages Increased cost of instrumentation Need to tune two loops instead of one Secondary variable must be measured

24 Design Considerations
Secondary loop must be fast responding otherwise system will not settle Time constant in the secondary loop must be smaller than primary loop Since secondary loop is fast, proportional action alone is sufficient, offset is not a problem in secondary loop Only disturbances within the secondary loop are compensated by the secondary loop. Hence, cascading improves the response to these disturbances

25 Applications: 1. Valve Position Control
Valve Motor Desired position Secondary loop Valve position Air Pressure to Valve Motor Valve motion is affected by friction and pressure drop in the line. Friction causes dead band. High pressure drop also causes hysteresis in the valve response Useful in most loops except flow and pressure

26 Application 2. Cascade Flow Loop
Output From Primary Controller “ no cascade “ Output From Primary Controller FC FT Fset DP “ cascade “

27 GC2 GC1 GP1 GP2 Σ c m2 e2 mset Secondary loop Primary loop cset GC2
controller Secondary process primary c m2 e2 mset Secondary loop Primary loop cset GC2 GCL GP2 Σ mset m2 c cset

28 θ Gc 12 Σ + - Primary GC2 Secondary G2(S) G3(S) For a cascade system (open-loop) Without cascade control θc

29 Illustrative Example: Steam Jacket – Continued – Cascade Case
Wu = 0.53 Mag = 20*log10(AR) = -30 (dB)  AR =

30 Illustrative Example: Steam Jacket – Continued – No Cascade Case
Wu = 0.25 Mag = 20*log10(AR) = 0 (dB)  AR = 1

31 Illustrative Example: Steam Jacket – Continued – No Cascade Case
Ku = 1;wu = 0.25;Pu = 2*Pi / wu =  Kc = Ku/1.7 =  Taui = Pu / 2 =  Taud = Pu /8 =

32 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

33 2-3 Selective Control Systems
Override Control Auctioneering Control Ratio Control Change from one controlled (CV) or manipulated variables (MV) to another

34 1. Override Control – Example Boiler Control
LT LC LSS PC Normal loop water steam LSS: Low Selective Switch – Output a lower of two inputs Prevents: 1. Level from going too low, 2. Pressure from exceeding limit (lower)

35 Example: Compressor Surge Control
motor SC HSS PC FC Gas out Gas in Normal loop

36 Example: Steam Distribution System
High Pressure Line Low Pressure Line PC HSS

37 2. Auctioneering Control Systems
Length of reactor Temperate T1 T2 Hot spot Temperature profiles in a tubular reactor

38 Auctioneering Control Systems
TT HSS Cooling flow TC

39 Temperature Control Split Range Control: More than one manipulated variable is adjusted by the controller TC Bypass Exchanger T2 Steam

40 Example: Steam Header: Pressure Control
Boiler 2 Steam Header PC

41 3. Ratio Control – Type of feedforward control
Wild stream FA FT Disadvantage: Ratio may go To erratic Desired Ratio ε Gc Driver FT FB Controlled Stream B However, one stream in proportion to another. Use if the ratio must be measured and displayed

42 Another implementation of Ratio Control
Wild stream FA FT Multiplier Desired Ratio + FC ε - FT FB Controlled stream

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