= 1 τ (C Af C A ) kc A. τ = V q. k = k(t )=k. T0 e E 1. γ = H ρc p. β = (ρc p) c Vρc p. Ua (ρc p ) c. α =

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1 TTK485 Robust Control Department of Engineering Cybernetics Spring 27 Solution - Assignment Exercise : Unstable chemical reactor a) A CSTR-reactor with heat exchange is shown in figure. Mass- and energy balances for this ideal reactor with a. order reaction A B are given as: where dc A dt = τ (C Af C A ) kc A dt dt = τ (T f T ) γkc A βq c ( e α qc )(T T c,inn ) τ = V q k = k(t )=k. T e E R T T γ = H ρc p β = (ρc p) c Vρc p α = Ua (ρc p ) c The following constants and steady-state operating conditions are given: E = 6352 cal/mole U =. cal/cm 2 sk V =3 cm 3 a = cm 2 k T =. s T =368K H = 6465 cal/mole ρc p =. cal/cm 3 K (ρc p ) c =2. cal/cm 3 K R =.98 cal K mole C Af,s =. mole/cm 3 C A,s =.25 mole/cm 3 T f,s =3K T c,inn,s =273K q s =cm 3 /s q c,s = 298 cm 3 /s T s =368K

2 C, T, q Af f f V T T c,inn q c C B Figure : CSTR-reactor with heat exchange. Atthesteady-stateoperatingpointwehave k(t s )=k T. e E R Ts T = k T τ = V q s We introduce deviation variables and linearizes the mass- and energy balances using the givensteady-stateoperatingpointasthereferencepoint: d C A dt = C Af,s C A,s q + V τ ( C Af C A ) k(t s ) C A k(t s ) E C RTs 2 A,s T d T dt = T f,s T s q + V τ ( T f T ) γk(t s ) C A γk(t s ) E C RTs 2 A,s T β(t s T c,inn,s ) q c + β(t s T c,inn,s )e α α qc,s ( + qc,s ) q qc,s 2 c βq c,s ( e α qc,s )( T Tc,inn ) 2

3 Write the model on the state space form:. x = Ax + Bu + Cd " τ = k(t s) γk(t s ) τ γk(t s) E RT 2 s k(t s ) E RT 2 s C A,s C A,s βq c,s ( e α qc,s ) # x " + β(t s T c,inn,s )[ e α qc,s ( + α q c,s )] # u + " τ βq c,s ( e α τ qc,s ) C Af,s C A,s V T f,s T s V # d where x = CA T u = q c C Af d = T f T c,inn q b) Stability: The poles of the system is the eigenvalues of the A-matrix. If the system is stable, all eigenvalues must have a negative real part, that is Re(λ) < (all the poles must lie in the left half plane). We calculate eigenvalues in matlab λ A = eig(a) = The system is unstable. The step resonses for C A and T after a unit step change in q c are given in figure 2. Figure 3 shows the step responses for the two output variables after a unit step change in the disturbance C Af. c) Controller design: It is difficult to achieve good control of this system. As can be seen from figures 2 and 3, a step change in the input has little effect on the outputs compared toanequivalentchange in the disturbance C Af. The flow rate in the heat exchanger can not be inncreased (or decreased) without bounds, so in practice it would be impossible to keep the temperature 3

4 x 4 step respons, qc.8 Amplitude To: Out() To: Out(2) Time (sec) Figure 2: Step Response, step change in the manipulated variable qc, open loop system.5 step response, disturbance Caf Amplitude To: Out() x 4.5 To: Out(2) Time (sec) Figure 3: Step response, unit step change in Caf, open loop system 4

5 within acceptable limits for large changes in C Af. Due to safty, large temperature changes (here several degrees) is unwanted so tight control of temperature is probably most important. With only one degree of freedom available tight control of both temperature and concentration is difficult However, it is possible to stabilize the system using feedback. That is, we must design a controller K so that the eigenvalues of the closed loop system, here represented by the matrix A BK, have a negative real part. We are not asked to design an optimal controller, so a simple P - controller would do. For example use u = Kx where K = k k 2.Thegain matrix K can be found in matlab using the lqr-function. (See matlab code, matlab.zip). you can also use the matlab function place and determine K so that the closed loop system gets the desired eigenvalues in the left half plane (pole placement). An example of the response after a unit step change in C Af is given in figure 4. (We assume here that there is no fysical restrictions on q c ). K is determined primarily to prevent excessive temperature variations. The eigenvalues for the closed loop system is then and the system is stable. λ A = eig(a reg ) However, the response given in figure 4 requires a change in the manipulated variable of several (cm 3 /s), which of cours can not be implemented in practice. What should have been done, was scaling the system based on the expected variation i C Af (and the other variables as well). Given the steady-state value of C Af it is not likely that any change in C Af would be as large as (mol/cm 3 ). If maximum expected variation in C Af is. (= %) for example, this should be the basis for the controller design. That is, it makes no sense designing a controller for responding to a step change in C Af when no such disturbance is expected. 5

6 .5 step response, disturbance Caf To: Out().5 Amplitude -5 To: Out(2) Time (sec) Figure 4: Step Response, Unit step change in Caf, closed loop system Exercise 2: Fluid Catalytic Cracking a) Allowed variation in F s and F a is given in the interval and 2 3 respectively. Assume that the variables vary about a nominal point in the middle of the interval so allowed variation is /2 and /2 respectively This gives the scaling matrix D u = An error e i = y i r i =should correspond to a change in the riser inlet temperature, T ri =3K, a change in the regenerator cyclon temperature, T cy =2K and a change in the temperature in the top of the riser, T rg =3K in the original model. This gives the scaling matrix D y = Equivalently, a scaled disturbance change d =should correspond to a change in the temperature of the air entering the regenerator, T a =5K and a change in feed oil composition given by k c =2.5%. This gives the scaling matrix 5 D d =.25 6

7 We then have y = D y y s u = D u u s d = D d d s Substitute for y, u and d in the state space model:. x = Ax + BD u u s + ED d d s = Ax + Bu b s + Êd s y s = D y = b Cx + b Du s Cx + D DD u u s y We can also describe the system with the transfer functions: b) Need for control? G(s) =D y C(sI A) BD u + D DD u G d (s) =Dy C(sI A) ED d Step responses after a unit step change in the manipulated variables F a and F s and the disturbances T a and k c is given in figure 5. y 7

8 5 Sprangrespons, sprang i Fs 35 Sprangrespons, sprang i Fa Tri Tcy Trg Sprangrespons, sprang i Ta Sprangrespons, sprang i kc Figure 5: Step response The steady state gain in the system is G() = G d () = d (T a ) has little effect on the output response compared to d 2 (k c ). d 2 has a considerable effect on all the three outputs so we definitely need to ontrol the system. Choice of variables for control We now want to design a controller for this system based on two of the three measurements T ri, T cy and T rg the manipulated variables F s and F a (input-output pairing). i) Effect of disturbances and changes in the manipulated variables The step responses given in figure 6 imply that the disturbances have the most effect on T ri and T rg. However, T cy is also considerably affected by changes in k c. Manipulated variable, F s, has relatively little effect on output 3, T rg. 8

9 ii) Right half plane zeros Zeros in the right half plane limit achievable bandwidth for the system, no matter what kind controller we choose. Existence and placement of right half plane zeros therfore are a measure of controllability. Here, the combination of T rg and T cy willhaveazerointheright half plane. This combination therefore seems to be a bad choice for input-output pairing. The other two combinations T ri /T cy and T ri /T rg have no right half plane zeros. iii) Process knowledge In the cracker process it is important to stop the regnerator cyclon temperature T cy from getting too high because this may lead to material problems in the regenerator and the downstream piping. Also, good control of production capacity is important to ensure that maximal production limits are not exceeded The amount gas produced is strongly dependent on the outlet temperature in the riser, T ri. Based on the response after the step changes, it may seem best to choose to control T ri /T rg. Based on process knowledge one could argue that choosing T ri /T cy would be better. However, based on the data given in the problem statment, it is not expected that the students should have a thorough understanding of the process so choosing T ri /T rg is OK. iii) Paring of variables Based on process knowledge (pair close!) pairing F s T ri and F a T rg (alternatively F a T cy ) seems to be a rational choice. See also the step responses in the figure above. A closer examination of the RGA-values in the bandwidth area for the different alternative control structures confirms that there is only minor interactions in the system. iv) Controller design We are not asked to design an optimal controller. Try for example two single loop PIcontrollers or a feedback loop with a lqr controller. See the matlab code for examples. 9