1. 1 Unconstrained degrees of freedom: C. Optimal measurement combination (Alstad, 2002) Basis: Want optimal value of c independent of disturbances ) –  c opt = 0 ¢  d Find optimal solution as a function of d: u opt (d), y opt (d) Linearize this relationship:  y opt = F  d F – sensitivity matrix Want: To achieve this for all values of  d: Always possible if Optimal when we disregard implementation error (n)

2. 2 Example C (measurement combination): Optimal blending of gasoline Stream 1 Stream 2 Stream 3 Stream 4 Product 1 kg/s Stream 199 octane0 % benzenep 1 = 0.1 (1 + m 1 ) $/kg Stream 2105 octane0 % benzenep 2 = 0.200 $/kg Stream 395 → 97 octane0 % benzenep 3 = 0.120 $/kg Stream 499 octane2 % benzenep 4 = 0.185 $/kg Product> 98 octane< 1 % benzene Disturbance m 1 = ? ( · 0.4) m 2 = ? m 3 = ? m 4 = ?

3. 3 Exercise 1: Optimal operation The Matlab function quadprog will be used to find optimal operation Calculate the optimal operating point for the nominal values (octane in stream 3 equal 95) Calculate the optimal operating point with disturbance (octane in stream 3 equal 97) Report value of objective function and flowrates for both cases How many degrees of freedom, active constraints and unconstrained degrees of freedom are there?

4. 4 Optimal solution Degrees of freedom u o = (m  m 2 m 3 m 4 ) T Optimization problem:Minimize J =  i p i m i = 0.1(1 + m 1 ) m 1 + 0.2 m 2 + 0.12 m 3 + 0.185 m 4 subject to m 1 + m 2 + m 3 + m 4 = 1 m 1 ¸ 0; m 2 ¸ 0; m 3 ¸ 0; m 4 ¸ 0 m 1 · 0.4 99 m 1 + 105 m 2 + 95 m 3 + 99 m 4 ¸ 98 (octane constraint) 2 m 4 · 1 (benzene constraint) Nominal optimal solution (d * = 95): u 0,opt = (0.26 0.196 0.544 0) T ) J opt =0.13724 $ Optimal solution with d=octane stream 3=97: u 0,opt = (0.20 0.075 0.725 0) T ) J opt =0.13724 $ 3 active constraints ) 1 unconstrained degree of freedom

5. 5 Exercise 1 cont.: Optimal operation There is one unconstrained degree of freedom, which can be used to optimize operation What is the loss of using this unconstrained DOF to maintain the following constant at nominal set point: –m 1 –m 2 –m 3 Find a linear combination of two variables that will give zero loss.

6. 6 Implementation of optimal solution Available ”measurements”: y = (m 1 m 2 m 3 m 4 ) T Control active constraints: –Keep m 4 = 0 –Adjust one (or more) flow such that m 1 +m 2 +m 3 +m 4 = 1 –Adjust one (or more) flow such that product octane = 98 Remaining unconstrained degree of freedom 1.c=m 1 is constant at 0.26 ) Loss = 0.00036 $ 2.c=m 2 is constant at 0.196 ) Infeasible (cannot satisfy octane = 98) 3.c=m 3 is constant at 0.544 ) Loss = 0.00582 $ Optimal combination of measurements c = h 1 m 1 + h 2 m 2 + h 3 m a From optimization:  m opt = F  d where sensitivity matrix F = (-0.03 -0.06 0.09) T Requirement: HF = 0 ) -0.03 h 1 – 0.06 h 2 + 0.09 h 3 = 0 This has infinite number of solutions (since we have 3 measurements and only ned 2): c = m 1 – 0.5 m 2 is constant at 0.162 ) Loss = 0 c = 3 m 1 + m 3 is constant at 1.32 ) Loss = 0 c = 1.5 m 2 + m 3 is constant at 0.83 ) Loss = 0 Easily implemented in control system

7. 7 Blending: Example of implementation of ”self-optimizing” constant setpoint policy Selected ”self-optimizing” variable: c = m 1 – 0.5 m 2 Changes in feed octane (stream 3) detected by octane controller (OC) Implementation is optimal provided active constraints do not change Price changes can be included as corrections on setpoint c s Comment: Example is for illustration. Use MPC in practice (changing constraints)! FC OC m tot.s = 1 kg/s m tot m3m3 m 4 = 0 kg/s Octane s = 98 Octane m2m2 Stream 2 Stream 1 Stream 3 Stream 4 c s = 0.162 0.5 m 1 = c s + 0.5 m 2 Octane varies

8. 8 Exercise 2: Singular value Assume that there is an implementation error of 0.01 kg/s in flowrates Use the singular value rule to find the approximate losses for the following cases: –m 1 constant –m 2 constant –m 3 constant –m 1 - 0.5 m 2 constant

9. 9 Singular value rule CV c span cG|G s |Loss factor 1/|G s | 2 m1m1 0.07001.00014.290.0049 m2m2 0.1310-0.4003.0530.1073 m3m3 0.1910-0.6003.1410.1013 m 1 -0.5m 2 0.01551.20077.411.668e-4

10. 10 Exercise 3: Dynamic simulations m1m1 m2m2 m3m3 m4m4 m tot

11. 11 Exercise 3: Dynamic simulations P-control of tank level: m = K c M with K c =0.001 –The tank has a variable mass M of 1000m (where m is the total flowrate) Time constant in valves are 10s Octane measurement has a delay of 60 s Each flow may vary between 0 and two times nominal value m 4 is not used so inputs are [m 1, m 2, m 3 ]' = Flowrates Output: –y 1 = Product flowrate –y 2 = Octane number in product –y 3 = c (a selected measurement, or a combination of two measurements) The linear model from m to y (for the case c = m 1 ) is: G = 1/(1000s + 1)(10s + 1) [ 1 1 1; 1e-60s 7e-60s -3e-60s; 1 0 0]

12. 12 Exercise 3: Dynamic simulations Derive the linear model for the different control policies: –m 1 constant –m 2 constant –m 3 constant –m 1 - 0.5 m 2 constant Use steady state RGA to select controller pairing for the above cases Simulate the process in Simulink with different the control structures Compare results from the different approaches and choose the best control policy

13. 13 Linear analysis Linear model with controlled variable c = m 1 - 0.5 m 2 constant: G = 1/(1000s + 1)(10s + 1) [ 1 1 1; 1e-60s 7e-60s -3e-60s; 1 -0.5 0] Steady state RGA: 00.30000.7000 0 0.3000 1.00000.0000 1.1667-0.16670.0000 -0.16671.16670.0000 1.0000 0.75000.00000.2500 0.00000.7500 0.00001.00000.0000 0.12500.25000.6250 0.04170.58330.3750 0.83330.16670.0000 c = m 1 c = m 2 c = m 3 c = m 1 - 0.5m 2

14. 14 Controller tuning: SIMC tuning rules Flow controller: G(s) = 1/((1000s+1)*(10s+1)) k = 1 τ 1 = 1000+10/2 = 1005 θ = 10/2 = 5 K = τ 1 /(2 θ k) = 100.5 τ I = min(τ 1, 8θ) = 40 Octane controller G(s) = k'*e -60s /((1000s+1)*(10s+1)) k = 7, 1, -3 depending on pairing τ 1 = 1000+10/2 = 1005 θ = 60 + 10/2 = 65 K = τ 1 /(2 θ k) = 7.73/k τ I = min(τ 1, 8 θ) = 520 K = 1.10 if m 1 controls octane K = 7.73 if m 2 controls octane K = -2.58 if m 3 controls octane

15. 15 Dynamic simulations Disturbance –Octane of stream 3 steps from 95 to 94 at t=0 –Octane of stream 3 steps from 94 to 97 at t=3000 Profit/kg = (product prize - raw material prize - TV)/amount of product –TV is a function of controller usage –Product prize = 0.2 when octane > 97.9 0.15 when octane < 97.9 Simulation time is 10 000 s

16. 16 c = m 1 : Profit/kg = 0.0683

17. 17 c = m 2 : Infeasible

18. 18 c = m 3 : Profit/kg = 0.0640

19. 19 c = m 1 - 0.5m 2 : Profit/kg = 0.0680

20. 20 Selecting control structure Loss calculations shows zero losses for linear combinations of two measurements (m 1 - 0.5 m 2 constant) Singular value rule shows a small loss factor for linear combination of two measurements (m 1 - 0.5 m 2 constant) Steady state RGA indicates interactions when using the linear combination m 1 - 0.5 m 2 constant. Are other combinations better? Dynamic simulations shows that there is better to keep m 3 constant than m 1 - 0.5m 2 constant, but is it possible to overcome this by re- tuning the controllers?