Chapter 2 Traditional Advanced Control Approaches – Feedforward, Cascade and Selected Control 2-1 Feed Forward Control (FFC) Block Diagram Design of FFC controllers Examples Applications 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. Feedforward Control Feedforward Controller Disturbance Output Process Manipulated Variable 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 EXAMPLES steam FI steam LI Boiler Feed control FI LI FB Feedback control FFC steam Feedforward control FI LI FB FFC Σ Combined feedforward-feedback control Design Procedures (Block diagram Method) Load transfer function Load GL(s) L FF Controller GF(s) Process M Manipulated Variable Gp(s) X2 ∑ C Outp ut Derivation C ( s ) GL ( s ) L ( s ) GP ( s ) M ( s ) GL ( s ) L( s ) GP ( s )GF ( s ) L( s ) GL s GP s GF ( s )L( s ) We want C ( s ) 0 for all L( s ). Hence GL s GP s GF ( s ) 0 or G F s GL s GP s need : (1)GL ( s ), load transfer function (2)GP s , process transfer function 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. Tuning feedforward controllers 1s 1 GF ( s ) K 2s 1 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. Example: A simulated disturbed plant Disturbed flow rate DV Chemicals Waste water treatment MV BOD (CV) Simulated Block Disgram Disturbed flow rate 1 s 1 s 2 s 3 + Chemicals 1 s 1 3 Feedforward v.s. Feedback Control 0.15 FB 0.1 0.05 0 -0.05 -0.1 -0.15 -0.2 FF 0 5 10 15 20 25 30 Example: Distillation Column Example: Distillation Column Mass Balance: F=D+B Fz=Dy+Bx D=F(z-x)/(y-x) In practice D F ( z xset ) yset xset For example: If light key increase in feed, increase distillate rate. Design of Feedforward Control Using Material and Energy Balances Consider the hear exchanger Ws Steam T2 w, T1 Condensate 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. Control Law and Design Implementation: Tset measured T1 Σ + - K measured Gain X w Note no dynamics are incorporated Ws 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 2-2 Cascade Control Block Diagram Design Considerations Applications Illustrative Example : Steam Jacket TC PT PC Illustrative Example: Steam Jacket Continued Energy Balance of the Tank: dT V hATJ T Heat Loss dt Energy Balance of Jacket: dTJ d PJVJ dt dt ns R VJ d PJ / ns dt R Material Balance of the Jacket dns nin condensate dt Illustrative Example: Steam Jacket Continued Assume: T s 1 PJ ( s ) 30 s 13s 1 PJ s 1 X s 10 s 1s 12 Where X=valve position Block Diagram Steam supply pressure Tset Feed back Controller Valve position Steam Valve secondary Tset Primary Controller secondary Jacket Pressure Controller Jacket steam pressure supply pressure Jacket Steam Valve Secondary loop Primary loop pressure Stirred Tank Tank Temp. primary Stirred Tank Tank Temp. 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 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 Applications: 1. Valve Position Control Air Pressure to Valve Motor Desired position Control Valve Motor Valve position Secondary loop 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 Application 2. Cascade Flow Loop Output From Primary Controller “ no cascade “ Output From Fset Primary Controller FC DP FT “ cascade “ Primary controller cset Σ mset GC2 Secondary controller GC1 Secondary process e2 GP1 primary process m2 c GP2 Secondary loop Primary loop GC1GP1 m2 GC 2 mset 1 GC1GP1 cset Σ GC2 mset GCL GC 2GCL GP 2 c cset 1 GC 2GCL GP 2 m2 GP2 c θc + - Σ Gc Primary + Σ - GC2 G2(S) G3(S) 12 1 ( S 1) (10S 1) 1 (30 S 1)(3S 1) 2 Secondary For a cascade system (open-loop) 12 G2G3 c 1 12 G2 Without cascade control G2G3 c θ Illustrative Example: Steam Jacket – Continued – Cascade Case Wu = 0.53 Mag = 20*log10(AR) = -30 (dB) AR = 0.0316 Ku 1 31.6228 AR Illustrative Example: Steam Jacket – Continued – No Cascade Case Wu = 0.25 Mag = 20*log10(AR) = 0 (dB) AR = 1 Ku 1 1 AR 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 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 2-3 Selective Control Systems Override Control Auctioneering Control Ratio Control Change from one controlled (CV) or manipulated variables (MV) to another 1. Override Control – Example Boiler Control steam PC water LT LC LSS Normal loop LSS: Low Selective Switch – Output a lower of two inputs Prevents: 1. Level from going too low, 2. Pressure from exceeding limit (lower) Example: Compressor Surge Control Normal loop HSS SC PC FC Gas in Gas out motor Example: Steam Distribution System High Pressure Line PC HSS PC Low Pressure Line 2. Auctioneering Control Systems Hot spot Temperate T1 T2 Length of reactor Temperature profiles in a tubular reactor Auctioneering Control Systems TT TT TT HSS TC Cooling flow TT Temperature Control Split Range Control: More than one manipulated variable is adjusted by the controller TC Bypass Exchanger T2 Steam Example: Steam Header: Pressure Control Boiler 2 PC Boiler 2 Boiler 2 Steam Header 3. Ratio Control – Type of feedforward control A Wild stream FA Disadvantage: Ratio may go To erratic FT Desired Ratio ε Driver Gc FT B FB Controlled Stream However, one stream in proportion to another. Use if the ratio must be measured and displayed Another implementation of Ratio Control A Wild stream FA FT Multiplier Desired Ratio + - ε FC FT FB Controlled stream