Systems Engineering/Process Control L10

Transcription

1 1 / 29 Systems Engineering/Process Control L10 Controller structures Cascade control Mid-range control Ratio control Feedforward Delay compensation Reading: Systems Engineering and Process Control:

2 2 / 29 Cascade control Cascade control can be used for systems that can be split: u G p2 y 2 G p1 y 1 where bothy 2 andy 1 can be measured G p2 is (or can be made) at least 10 times faster thang p1 Example:G p1 = K 1 1+T 1 s andg p2= K 2 1+T 2 s witht 2<0.1T 1

3 3 / 29 Cascade control block diagram r G c1 u 1 G c2 u 2 y 2 G p2 G p1 y 1 Secondary controllerg c2 controlsy 2 Inner loop is fast compared to outer loop Often P-controller with high gain For outer loop we havey2 u 1 Primary controllerg c1 controlsy 1 Often PI or PID controller

4 4 / 29 Example: Heat exchanger TIC TT Steam Water Control may work poorly if, e.g.,: valve in nonlinear steam pressure varies (load disturbance)

5 5 / 29 Example: Heat exchanger with cascade control Börvärde TIC FIC FT TT Steam Water The inner loop controls the steam flow Setpoint to flow controller given by temperature controller

6 Example: Heat exchanger simulation With cascade control (solid) and without (dashed); disturbance att=5: 100 Temperature Flow Valve position Time 6 / 29

7 7 / 29 Mid ranging Useful replacements for processes with two inputs and one measurement, e.g.,: u 2 G p2 u 1 G p1 Σ y u 1 high precision but little working range u 2 low precision but big working range

8 8 / 29 Mid ranging Example Flow control with two controlled valves: replacements v 1 v 2 FT Valvev 1 is small and has high accuracy big risk of saturation Valvev 2 is big but has worse accuracy How can they cooperate?

9 Mid ranging Example Mid ranging: G R2 G R1 v 1 v 2 FT Fast controllerg R1 controls flow with little valvev 1 Slow controllerg R2 adjusts big valvev 2 such thatv 1 is in the middle of its working range 9 / 29

10 10 / 29 Mid ranging simulation Big valve (dashed) keeps little valve (solid) at50% Flow Valve position Time

11 11 / 29 Mid ranging Block diagram r u1 G c2 u 2 G p2 r y G c1 u 1 G p1 Σ y G c1 andg p1 forms a fast and accurate loop Input fromg c1 is measurement forg c2 r u1 chosen to middle ofu 1 :s working range G c2 has low gain, maybe only I part Rule of thumb: at least 10 times bigger time constant than fast loop

12 12 / 29 Ratio control Example: Keep constant air/fuel ratio Suppose we wanty l /y b =a. Naive solution (control ratioa directly): r b u b Regulator Process y b a u l Regulator Process y l Nonlinear, gain in second loop varies withy b

13 13 / 29 Ratio control Better solution: r b u b Regulator Process y b a r l Regulator u l Process y l Setpoint for flow to first loop that is assumed slow Second loop is made fast and maintains desired ratio

14 14 / 29 Feedforward Example Concentration control Framkoppling QIC Återkoppling Bas Syra QT QT Blandning Feedforward can compensate for sudden changes in acid concentration

15 15 / 29 Feedforward Simulation of example With feedforward (solid) and without (dashed); disturbance att=5: 100 Concentration Valve position Time

16 16 / 29 Feedforward Block diagram l G ff r u y Σ G c Σ G p 1 How to choose compensatorg ff (s)? Depends on where disturbancelenters the system.

17 17 / 29 Feedforward Tank example Control of lower tank u l 1 1 r G c l 2 y 2

18 18 / 29 Feedforward Tank example Feedforward froml 1 : l 1 G ff r u y Σ G c Σ Σ G p1 G p2 1 ChooseG ff (s)= 1 to eliminate effect of disturbance

19 19 / 29 Feedforward Tank example Feedforward froml 2 : l 2 G ff r u y Σ G c Σ G p1 Σ G p2 1 ChooseG ff (s)= 1 G P1 to eliminate effect of disturbance

20 20 / 29 The inverse Example: Implementation of feedforward 1 can be problematic to implement G p1 (s) 1 G p1 (s)= 1+sT e sl 1 G p1 (s) =(1+sT)esL (derivation and neg. time delay) Common solutions: Introduce lowpass filter (compare D part in PID-controller) Approximate negative time delays with 0 Implement the static gain only

21 21 / 29 Dead time compensation Example of dead time process: G p (s)= K p 1+sT e sl Hard to control if L > T (dead time dominated) Frequency analysis: G p (s)=g p0 (s)e sl G p (iω c ) = G p0 (iω c ) argg p (iω c )=argg p0 (iω c ) ω c L The largerl, the smaller the phase margin

22 Example: Control of paper machine G p (s)= 2 1+2s e 4s Simulation with cautious PI controller (K=0.2,T i =2.6); disturbance at t = 25: Input Output Time 22 / 29

23 23 / 29 Example: Control of paper machine Simulation with more aggressive PI controller (K=1,T i =1): Input Output Time

24 24 / 29 Dead time compensation with Smith predictor r Regulator u Process y Modell Modell utan dödtid y 1 + Σ y 2 + Controller designed after model without delay. Model must be: asymptotically stable accurate enough

25 25 / 29 Analysis of Smith predictor Smith-predictor r e u y Σ Σ G c G p Ĝ p Ĝp0 1 G p =G p0 e sl real process Ĝ p =Ĝ p0 e sˆl model of process Ĝ p0 model of process without dead time G c controller designed forĝp0

26 Control signal: Closed loop system: Analysis of Smith predictor Y= U= G c 1 G c (Ĝp Ĝp0) E G p G c 1 G c (Ĝp Ĝp0)+G p G c R Suppose G p =Ĝp (perfect model): Y= G p0 e sl G c 1 G c (G p0 e sl G p0 )+G p0 e sl G c R = G p0g c 1+G p0 G c e sl R Like control of process without delay, but with delayed response 26 / 29

27 Example: Control of paper machine Model without delay:g p0 (s)= 2 1+2s Simulation with aggressive PI controller (K=1,T i =1) and Smith predictor with perfect process model: Output Input Time 27 / 29

28 28 / 29 Example: Control of paper machine Simulation with aggressive PI controller and Smith predictor with not perfect process model (ˆL=0.9L, ˆT=0.9T): Output Input Time

29 29 / 29 The Smith predictor conclussions Works only for asymptotically stable systems Works only if process model is accurate Controller should be designed such that closed-loop time constant larger than process dead time (Better variations for dead time compensation exist, but all rely on prediction using a process model)