Presentation is loading. Please wait.

Presentation is loading. Please wait.

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

Similar presentations


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 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 Process Output Manipulated Variable

5 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

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

7 Design Procedures (Block diagram Method) G L (s) Load transfer function G F (s) Load Manipulated Variable G p (s) Process X2X2 C Outp ut L M FF Controller

8 Derivation

9 Examples Example 1 Let G p (s)=K p /τ p s+1, G L (s)=K L /τ L s+1 Then, G F (s)=-(K L /K p )(τ p s+1)/(τ L s+1) Therefore, feedforward controller is a lead-lag unit. Example 2 Let G p (s)=K p e -Dps /τ p s+1, G L (s)=K L e -DLs /τ L s+1 Then, G F (s)=-(K L /K p )(τ p s+1)/(τ L s+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, τ 1, τ 2 Tuning K, K is selected so that for a persistent disturbance, there is no steady state error in output. Adjustingτ 1, τ 2 can be obtained from transfer functions. Fine tune τ 1, τ 2 such that for a step disturbance, the response is somewhat symmetrical about the set point.

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

12 Simulated Block Disgram Disturbed flow rate + Chemicals

13 Feedforward v.s. Feedback Control FB FF

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(T 2 -T 1 )=W s λ Where λ=hear of vaporization W s =WC(T 2 -T 1 )/λ This equation tells us the current stream demand based on (1) current flow rate, W, (2) current inlet temperature, T 1, (3) desired value of outlet temperature T 2. Steam T2T2 w, T 1 Condensate WsWs

16 Control Law and Design Implementation: Note no dynamics are incorporated Σ K X measured Tset Gain measured 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 PCPT TC

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

23 Principal Advantages and Disadvantages Advantages 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 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 Control Valve Motor Desired position Secondary loop Valve position Air Pressure to Valve Motor

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

27 G C2 G C1 G P1 G P2 Σ Primary controller Secondary controller Secondary process primary process c m2m2 e2e2 m set Secondary loop Primary loop c set G C2 G CL G P2 Σ m set m2m2 c c set

28 Gc12 Σ Σ + - Primary + - G C2 Secondary G 2 (S)G 3 (S) For a cascade system (open-loop) Without cascade control θcθc θ

29 Illustrative Example: Steam Jacket – Continued – Cascade Case W u = 0.53 Mag = 20*log 10 (AR) = -30 (dB) AR = 0.0316

30 Illustrative Example: Steam Jacket – Continued – No Cascade Case W u = 0.25 Mag = 20*log 10 (AR) = 0 (dB) AR = 1

31 Illustrative Example: Steam Jacket – Continued – No Cascade Case K u = 1;w u = 0.25;P u = 2*Pi / w u = 25.1327 K c = K u /1.7 = 0.5882 Tau i = P u / 2 = 12.5664 Tau d = P u /8 = 3.1416

32 Illustrative Example: Steam Jacket – Continued – Cascade Case K u = 20;w u = 0.53;P u = 2*Pi / w u = 12 K c = K u /1.7 = 11.8 Tau i = P u / 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 LT LCLSS PC Normal loop water steam 1. Override Control – Example Boiler Control LSS: Low Selective Switch – Output a lower of two inputs Prevents: 1. Level from going too low, 2. Pressure from exceeding limit (lower)

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

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

37 Length of reactor Temperate T1T1 T2T2 Hot spot 2. Auctioneering Control Systems Temperature profiles in a tubular reactor

38 TT HSS Cooling flow Auctioneering Control Systems TC

39 Temperature Control Steam T2T2 Bypass Exchanger TC Split Range Control: More than one manipulated variable is adjusted by the controller

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

41 FT Driver FT A Wild stream Controlled Stream B FBFB ε GcGc Desired Ratio FAFA 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 go To erratic

42 A Wild stream FT Desired Ratio Multiplier ε FC FT Controlled stream FBFB FAFA Another implementation of Ratio Control + -


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

Similar presentations


Ads by Google