Modelling of a simple continuous neutralisation reactor

Transcription

1 Modelling of a simple continuous neutralisation reactor Serge Zihlmann ThirdWay February 23, 205

2 Foreword When I was browsing the web for a decent and intuitive model of a neutralisation reactor, I could not nd the appropriate theorethical descripton nor model les for download. At that time I started to develop an own model including its implementation in a Simulation software. Quickly, this ended up in a rather extensive project. The theory developed and the models used are described in this little script in two dierent ways: One more intuitively and the other more academic. The models can of course be downloaded read to use. S. Zihlmann February 23, 205 Page of 8

3 Introduction and Modelling Figure. illustrates a CSTR used for neutralisation of waste water. In the given example, the waste water enters the reactor with a feedrate of V F = l/h at F =3. Acid and base are dosed by a PI-controller, which varies the ows of VA and V B. A probe measures the current -value and causes feedback to the controller with a lag time of 50 s. The reactor volume is V = l and the maximum feed of the acid (25 % Wastewater feed F, V F Acid feed V A C A [H + ],C A [OH - ] M Base feed V B C B [H + ],C B [OH - ] Outlet V F +V A +V B -probe Figure.: Continuously stirred tank reactor (CSTR) with acid and base dosing for neutralisation. value is equivalent to concentration of H + and OH ions. HCl) is 30 l/h. As the setpoint (goal value) is 0, only acid addition will be used in the following.. Intuitive modelling of the reactor All parameters are subject to time change, except the volume V. Especially V A, VB are used for control of the neutralisation. A simplication is the assumption of linear mixing volume (no miscibility gap) of acid and waste water. S. Zihlmann February 23, 205 Page 2 of 8

4 In order to model the dynamic behavior of the CSTR, usually a balance for each compound is set up. The accumulation term is then given by sum of inputs and outputs of the respective compound. Accumulation = Input Output Let us assume for simplication that we only have an acid pump available to neutralise the alkaline waste water, thus V B = 0. For the both species involved, the balances read: V d dt C[OH ] = V F C F [OH ] ( V F + V A ) C[OH ] + V A C A [OH ] V d dt C[H+ ] = V F C F [H + ] ( V F + V A ) C[H + ] + V A C A [H + ] The diculty of a neutralisation reactor is that a reaction occurs, i.e. the both equations are coupled by the autoprotolysis of water. H + + OH H 2 O 4 mol2 This reaction has the approximate equilibrium constant of 0 l 2 water concentration is constant. C[H + ] C[OH ] = 0 4 assuming that the To keep things simple let us assume that this reactions does not take place and the H + and OH ions remain in the solution, because for the matter of neutralisation we only need to balance the both ions to the same level. Still, if we want to calculate the, we need to nd the equilibrium of the autoprotolysis. Say, if we start from the concentrations as calculated by the dynamic models above, then the ion prodcut is not yet fullled. The above reaction consumes or produces a certain amount H of OH and H + ions, until the equation is satised and we have an equilibrium level of C[H + ] and C[OH ]. (C[H + ] + H) (C[OH ] + H) = 0 4 We can eliminate the irrelevant H by renaming C[H + ] + H = C[H + ]. It follows: C[H + ] (C[OH ] + C[H + ] C[H + ]) = 0 4 This equations must be solved for C[H + ] using a root solver to nd the equilibrium concentration. Luckily, the equation is only quadratic, thus a straight forward solution exists. The equation in standard shape is: with: 0 = C[H + ] 2 + (C[H + ] C[OH ]) C[H + ] a = b = C[H + ] C[OH ] c = 0 4 S. Zihlmann February 23, 205 Page 3 of 8

5 The solution is: C[H + ],2 = b ± b 2 4ac 2a While only one of the two solutions retrieves a positive concentration. It can further be simplied to depend only on the input concentration of H + and OH ions: And the value will read: C[H + ] = 2 (b + b ) = log 0 ( C[H + ]) We see that the dierence b of the ion concentration plays a central role to the solution of the problem. Let's try to combine the two dierential equations by subtracting the rst from the second. V d dt (C[H+ ] C[OH ]) = V F (C F [H + ] C F [OH ]) ( V F + V A ) (C[H + ] C[OH ]) + V A (C A [H + ] C A [OH ]) We can identify some terms by b, b F and b A : V d dt b = V F b F ( V F + V A ) b + V A b A The neutralisation problem can now be solved by one simple ODE. Note: For the modelling of base addition, the term V B b B must be added to the right side... Dynamic behavior The model was implemented in Simulink (see Appendix) and the dynamic behavior in terms of step response and PI-controlled CSTR are shown in gure.2. The step response when suddenly acid is fed to the reactor with otherwise constant inow, can be interpreted as titration curve. The controlled plant is quite sensitive to oscillation due to the lag time of the -electrode (here 50 s) and the high nonlinearity, for which the PID's are not well designed..2 Second approach One might ask whether there is a more sophisticate approach to the modelling part in contrast to the previous rather intuitive model. Of course there is a more strict, mathemathical way but the intuitive has its advantage that the user understands the process behind well. To proceed with the model in the academic way, I propose a more simplied model structure (gure.3) which contstitutes of one lumped system S. Zihlmann February 23, 205 Page 4 of 8

6 Time / h Time / h (a) Step response when acid is fed continuously at full power to a batch tank with initially 3. The behavior is the same as for a titration curve. (b) Controlled neutralisation reactor to reach 0.5 with an initial and input of 3. Feedback has a lag time of 50 s and the PID-controller parameters are P =.0, I = 0.5 s Figure.2: Intuitive model: Step response and controlled CSTR. which represents the CSTR (in the lumped system all properties are averaged, thus well stirred). The inows origin from the acid, base and the waste water. The reactor overows to an outlet. For each arrow there is a volumetric owrate and the ow composition vector c. WW A VA R CA R VWW R CWW VB R CB B VR O CR O Figure.3: Model of the neutralisation reactor as one lumped system (R=reactor) and four reservoirs (A=acid, WW=waste water, B=base, O=outow). First of all we have three species involved and we dene a species vector as follows: [H + OH H 2 O] T From the general conservation Accumulation = Flows + Reaction S. Zihlmann February 23, 205 Page 5 of 8

7 There follows directly the equation in matrix notation: ṅ R = F ˆn + ˆn r Where ṅ R is the total change in moles of each compound in the reactor, ˆn is the molar ow of each compound and in each stream, F represents the direction matrix of the ows (in or out) and is only a selector matrix, ˆn r represents the production/consumption by reaction (= index r) of each compound within the reactor. More specic, the ow vector consists of (omitting the base ow) ˆn = ˆn A R ˆn W W R ˆn R O And the ow selector matrix: F = = [ ] I The notation is very ecient using the kronecker product. The molar ow in each stream is simply ˆn x y = V x y c x for an exemplary ow from tank x to y. The input ows V A R and V W W R are well dened as well as their composition c A, c W W. The outlet ow must be obtained from an assumption. Here it is, that the volume during mixing of the two streams remains constant and the reactor has constant volume, thus V R O = V A R + V W W R. The composition of the outlet c R ow changes with time and must be obtained from the dynamic model c R = V n R where V is the volume of the reactor..2. Including the reaction The reaction term ˆn r mentioned in the last section is represented by the reaction stochiometry matrix N and some extent of reaction ñ. For the autodissociation of the water H 2 O the reaction stochiometri matrix reads ˆn r = N T ñ kf kb H + + OH N = [ ] forward backward S. Zihlmann February 23, 205 Page 6 of 8

8 for the forward and backward reaction. The reaction extent is formulated by the reaction rate law ˆn r = N T V K r g(c r ) where V is the reactor volume, K r contains the reaction rate constants and g(c r ) is the formulation of the rate law. Joined in one equation: ṅ R = F ˆn + N T V K r g(c r ) The central assumption for the further procedure is that the autoprotolysis reaction of the water is very fast. In other words, the reaction reaches always the equilibrium. Therefore we will eliminate the reaction term by nding the left nullspace Ω of N T and it is [ ] 0 Ω = 0 Of course there exists an innite amount of possible nullspaces, it is selected in that way that the equations used during root solving become nice. By multiplying the whole equation with the left null space, we eliminate the reaction terms. Ω ṅ R = Ω F ˆn + Ω N T V K r g(c r ) Ω ṅ R = Ω F ˆn After the nullspace transformation we can identify a new state z R = Ω n R ż R = Ω F ˆn Now the dynamic model is nished and simply the ODE for the state z R has to be solved. ˆn is dened by the input variables (acid ow, waste water ow) and the reactor composition (feedback). That requires the reconstruction of the reactor composition c R from the state z R. That involves solving the roots of the equation 0 = z R Ω n R = z R Ω V c R In this case we need to solve for three dimensions in the space of c R but the above expression returns only two equations. Thus we need another condition and it originates from the reaction term, more precisely, the equilbrium relation: 0 = N T V K r g(c r ) (.) [ ] [ ] 0 = kf 0 c V H2 O (.2) 0 k b c H + c OH k f c H2 O k b c H + c OH 0 = V k f c H2 O k b c H + c OH (.3) k f c H2 O + k b c H + c OH (.4) S. Zihlmann February 23, 205 Page 7 of 8

9 The three equations are equal and result in k b k f = c H2 O c H + c OH The concentration of the water is assumed constant and the equilibrium constant equal to 0 4 as generally for the water dissoziation. k = c H + c OH = 0 4 Thus the third required equation to reconstruct the composition c R from the state z R is 0 = k c H + c OH Finally we have to fulll all three equations: [ ] zr Ω V c 0 = R k c H + c OH This can be solved for c R by a numerical root solver, but in this simple case explicit expressions are available. In detailed formulation this reads: z V (c H + c OH ) 0 = z 2 V (c H + + c H2 O) k c H + c OH After solving this system one obtains: c H + = 2 c OH = ( z V + k c H + (z ) ) V k (.5) (.6) c H2 O = z 2 V c H + (.7) Now the modelling is complete and the structure can be implemented into a numerical software. The concentration vector for each input ow must be obtained from the or acid/base concentration. For the water always a constant concentration of 55.6 mol/l is assumed. One central dierence of this model to the intuitive one is that also the water is balanced..2.2 Dynamic behavior In gure.4 the results of the dynamic Simulink model for the academic approach are shown. The results is equal to the intuitive approach. S. Zihlmann February 23, 205 Page 8 of 8

10 Time Series Plot: 3 Time Series Plot: Time / h Time / h (a) Step response when acid is fed continuously at full power to a batch tank with initially 3. The behavior is the same as for a titration curve. (b) Controlled neutralisation reactor to reach 0.5 with an initial and input of 3. Feedback has a lag time of 50 s and the PID-controller parameters are P =.0, I = 0.5 s Figure.4: Academic model: Step response and controlled CSTR..3 Linearizing the plant The aim is to abtain a linearized model of the plant in standard shape as depicted for the system Σ in gure.5 r(t) u(t) Σ ẋ = Ax + bu y = dx + cu x y(t) Figure.5: The model block for a linear system Σ with the input u, state x and output y. r is a disturbance eect, e.g. noise. We shall work here with the dierential equation from the intuitive approach, as obtained before: V d dt b = V F b F ( V F + V A ) b + V A b A Identifying b as state and call it x. Also identify the acid ow V A as the input u(t). This retrieves the model in standard shape: ẋ(t) = V F b F V V F V x(t) V x(t) u(t) + b A V u(t) S. Zihlmann February 23, 205 Page 9 of 8

11 The output of the plant is not b but the value. Therefore we needed the root solver: C[H + ] = 2 (b + b ) and thus = log 0 ( 2 (b + b ) ) In standard form, where is the output y(t) and b again the state x(t): ( y(t) = log 0 (x(t) + ) ) x 2 2 (t) Thus we have the complete non linear model in standard shape: ẋ(t) = V F b F V F V V x(t) x(t) u(t) + b A V V u(t) = f (x(t), u(t)) ( y(t) = log 0 (x(t) + )) x 2 2 (t) = g (x(t), u(t)) The linearization will bring the model into the shape: [ ẋ(t) = A x(t) + b u(t) y(t) = c x(t) + d u(t) While A, b, c and d are matrices in general, but here simple constants. The linearization is done around the equilibrium point with x = xe and u = ue. In this equilibrium point we have: 0 =! ẋ(t) = V F b F V F V V xe V xe ue + b A V ue Usually for this equilibrium point the plants steady state point without interactions is chosen (this would be in our case that the reactor has the same composition as the inlet ow). Because the model is highly non linear, this would not be a good choice since we want to control the in detail around 0. Thus we select 0 as linearization point and have to search for the appropriate ue. b is readily calculated from the : ] be = C2 [H + ] 0 4 C[H + ] = = and thus xe = be = = 0 4 (0 6 ) 0 4 Now we have to nd the acid feed ue to reach steady 0 (xe). Reformulation of the equation above: ue = V F xe b F b A xe S. Zihlmann February 23, 205 Page 0 of 8

12 The constants are then calculated: A = f u=ue = V F x x=xe V u(t) u=ue V x=xe b = f u=ue = x(t) + b A u=ue u x=xe V V c = g u=ue = x x=xe ln(0) x 2 (t) x=xe = u=ue x=xe = V F +ue V b A xe V = ln(0) xe d = g u=ue = 0 u x=xe The factor d appears to be zero, because the acid feed has no instantaneous eect onto the value..3. Dynamic behavior and linearization error As the model is linearized around, it is not a good idea to start the control at 3 due to the high non-linearity. For this reason, two scenarios are compared here: ˆ Control from 2 to (strong disturbance) ˆ Control from. to (slight disturbance) Each scenario is compared for the controlled linearized and full model (as from intuitive approach). Th results are depicted in gure.6. What is obvious is that the linearized model starts at the wrong due to linearization error. This error becomes much less when the starting point is.. For the big step in control variable, the dierence between the control characteristic of the full/linearrized model is quite large, whilst at a small step the error is negligible (less linearization error). As a consequence, the modeling with this linear model is only appropriate if the does not vary much from the setpoint (e.g. when keeping the level constant under disturbance).s.4 Transfer function The transfer function represents the dynamic input/output behavior. The transfer function in the laplace domain can not readily be calculated for the full model, as x and y occur in a multiplication (which leads to a convolution in the laplace domain). Anyhow, it can be obtained quickly from the linearized model: [ ] ẋ(t) = A x(t) + b u(t) y(t) = c x(t) + d u(t) S. Zihlmann February 23, 205 Page of 8

13 2.8 Full model Linearized..08 Full model Linearized Time / h Time / h (a) Control from 2 to. (b) Control from. to. Figure.6: Response of a controlled neutralisation CSTR for dierent starting/aim values. Each plot shows the response of the full model compared to the linearized plant. Parameters of the controller are the same as before (feedback lag 50 s, P =.0, I = 0.5 s After a laplace transformation one obtains [ x(s) s = A x(s) + b u(s) y(s) = c x(s) + d u(s) because A, b, c and d are simple constants. After algebraic operations: Then And the transfer function: y(s) = x(s) = b u(s) s A G(s) = y(s) u(s) = [ ] c b s A + d u(s) c b s A + d Although derived here, this section is more for info than practical use. ] S. Zihlmann February 23, 205 Page 2 of 8

14 2 Choosing a PID controller 2. Back-linearization One might state that it is a good idea to `re-linearize' the plant, thus to take the output value and calculate it back to the concentration of H + ions. Also the setpoint shall be given as a concentration of H +. The control would then be linear, anyhow across huge magnitudes. The problem is then, that the chosen PID constants are only valid for a specic case and nding parameters for a wide range of application would be impossible. E.g. switching the setpoint from to 0 would make the control parameters useless. Figure 2. contains three examples for illustration. The basic problem with this Time / h Time / h Time / h (a) Control from 3 to 0.5. (b) Control from 3 to 0. (c) Control from 3 to. Figure 2.: Response of a controlled neutralisation CSTR with back-linearization for dierent set points of. The starting value is always 3. One set of PID parameters is not appropriate for all cases (optimized for case a) ). Feedback lag 50 s, P = 2 0 0, I = s approach is, that the overshoot (e.g. in example b) ) appears huge in the logarithmic domain but is not in the linear (controlled) one. Keeping this logarithmic behavior in mind, the reverse case, when controllong from 0, or 2 to 3 looks well for all cases. It would be good to have some anti-windup in the real plant. The integral term gets windup very quickly (especially in this logarithmic domain), thus it accumulates an output that is larger than the maximal value for the regulation variable. 3

15 3 Appendix In gures & 3.4 the models inplemented in Simulink for the intuitive model, its substructures and the linearized model are depicted. The last two pages contain the Matlab code to execute the models. All complete les can also be downloaded from (or thirdway.ch -> projects -> Neutralisation modeling). The Simulink models and code for the academic approach are not shown here, but can be downloaded from the same place. 4

16 Step(response(of(the(neutralisation Acid(pump set(point In Out Reactor( Acid(pump setpoint CSTR neutralisation reactor Controlled(neutralisation(plant Scope simout To(Workspace2 0.5 PID4s3 In Out (setpoint PI(Controller Transport Delay CSTR neutralisation reactor Scope2 simout2 To(Workspace Controlled(linearized(plant _set PID4s3 In Out (setpoint PI(Controller Linearized(CSTR Scope4 Transport Delay simout3 To(Workspace Figure 3.: Overview of the Simulink block structure for the intuitive approach. Step response, controlled plant and controlled linearized plant. S. Zihlmann February 23, 205 Page 5 of 8

17 ModeloftheneutralisationCSTR.OnebalanceforeachofH4andOH-. b_a In Setpointacidpump Acidfeedrate -C- V_A Saturation Product4 Acidpumpcapacity V_F CA_H Product3 /V Gain2 b x os integrator b In Out Out Feedrate Product2 b_init Get b_init b_f b_f Product Figure 3.2: Simulink structure of the intuitive model. CalculateGGvalue InputGconcentration H+G-GOHinGmol/l In u 2 Square 4e-4 IonGproduct FulfillGH+G+GOH-G=GH2OGequilibrium ->GRootGsolver u Sqrt /2 Gain3 H+Gconc log0 Math Function - Gain ConvertGH+GconcGtoG Gvalue Out Figure 3.3: Root solver `Get ' of the intuitive model. Linearized3plant A Acid3pump A In Saturation Setpoint acid3pump -K- u(t) Acid pump3capacity ue B B x0 init x os integrator x xe C C ye y(t) Out linearization point linearization point2 linearization point3 Figure 3.4: Structure of the linearized model. S. Zihlmann February 23, 205 Page 6 of 8

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