Chapter 2

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 Controller Disturbance Output Process Manipulated Variable Feedforward Control

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

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

8. Derivation

9. Examples • Example 1 • 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 FB FF

14. 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, increase distillate rate.

15. Design of Feedforward Control Using Material and Energy Balances Ws • 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. Steam T2 w, T1 Condensate

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

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

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. Air Pressure to Valve Motor Valve Motor Valve position Desired position Control Secondary loop 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

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

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

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

29. Illustrative Example: Steam Jacket – Continued – Cascade Case • Wu = 0.53 • Mag = 20*log10(AR) = -30 (dB)  AR = 0.0316 

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 = 25.1327 •  Kc = Ku/1.7 = 0.5882 •  Taui = Pu / 2 = 12.5664 •  Taud = Pu /8 = 3.1416

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. steam PC LC LSS LT water Normal loop 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. Normal loop HSS SC PC FC Gas in motor Gas out Example: Compressor Surge Control

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

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

38. TT TT TT TT HSS Cooling flow Auctioneering Control Systems TC

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

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

41. 3. Ratio Control – Type of feedforward control A Wild stream FA Disadvantage: Ratio may go To erratic FT 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 A Wild stream FA FT Multiplier Desired Ratio + FC ε - FT FB Controlled stream