4.0 context and direction4.1 math model of continuous blending tanks4.2 solving the coupled equations - a second-order system4.3 response of system to step disturbance4.4 introducing the Laplace transform4.5 solving linear ODEs with Laplace transforms4.6 describing systems with transfer functions4.7 describing systems with block diagrams 4.8 frequency response from the transfer function4.9 stability of the two-tank system, and linear systems in general4.10 step 1 - specify a control objective for the process4.11 step 2 - assign variables in the dynamic system4.12 step 3 - proportional control 4.13 step 4 - choose set points and limits4.14 components of the feedback control loop4.15 transfer functions for loop components4.16 assembling the components into a closed loop4.17 some perspective on how we derived the closed-loop response4.18 calculating closed-loop responses4.19 calculating the response for a particular example4.20 surprise - increasing gain introduces oscillations!4.21 closed-loop stability by the Bode criterion4.22 tuning based on stability limit - gain and phase margin4.23 conclusion4.24 reference4.25 nomenclatureSpring 2006 Process Dynamics, Operations, and Control 10.450 Lesson 4: Two Tanks in Series 4.0 context and direction In Lesson 3 we performed a material balance on a mixing tank and derived a first-order system model. We used that model to predict the open-loop process behavior and its closed-loop behavior, under feedback control. In this lesson, we complicate the process, and find that some additional analysis tools will be useful. DYNAMIC SYSTEM BEHAVIOR 4.1 math model of continuous blending tanks We consider two tanks in series with single inlet and outlet streams. F, CAiF, CA1volume V1volume V2F, CA2F, CAiF, CA1volume V1volume V2F, CA2 Our component A mass balance is written over each tank. 2A1A2A21AAi1A1FCFCCVdtdFCFCCVdtd−=−= (4.1-1) As in Lesson 3, we have recognized that each tank operates in overflow: the volume is constant, so that changes in the inlet flow are quickly duplicated in the outlet flow. Hence all streams are written in terms of a single volumetric flow F. Again, we will regard the flow as constant in time. Also, each tank is well mixed. Putting (4.1-1) into standard form 1A2A2A2Ai1A1A1CCdtdCCCdtdC=+τ=+τ (4.1-2) revised 2006 Mar 6 1Spring 2006 Process Dynamics, Operations, and Control 10.450 Lesson 4: Two Tanks in Series we identify two first-order dynamic systems coupled through the composition of the intermediate stream, CA1. If we view the tanks as separate systems, we see that CA1 is the response variable of the first tank and the input to the second. If instead we view the pair of tanks as a single system, CA1 becomes an intermediate variable. The speed of response depends on two time constants, which (as before) are equal to the ratio of volume for each tank and the common volumetric flow. We write (4.1-2) at a steady reference condition to find (4.1-3) r,2Ar,1Ar,Air,ACCCC === We subtract the reference condition from (4.1-2) and thus express the variables in deviation form. '1A'2A'2A2'Ai'1A'1A1CCdtdCCCdtdC=+τ=+τ (4.1-4) 4.2 solving the coupled equations - a second-order system As usual, we will take the initial condition to be zero (response variables at their reference conditions). We may solve (4.1-4) in two ways: Because the first equation contains only C′A1, we may integrate it directly to find C′A1 as a function of the input C′Ai. This solution becomes the forcing function in the second equation, which may be integrated directly to find C′A2. That is dtCee1Ct0'Aitt1'1A11∫ττ−τ= (4.2-1) ∫∫⎥⎦⎤⎢⎣⎡ττ=ττ−ττ−t0t0'Aitt1tt2'2AdtdtCee1ee1C1122 (4.2-2) On defining a specific disturbance C′Ai we can integrate (4.2-2) to a solution. Alternatively, we may eliminate the intermediate variable C′A1 between the equations (4.1-4) and obtain a second-order equation for C′A2 as a function of C′Ai. The steps are (1) differentiate the second equation (2) solve the first equation for the derivative of C′A1 revised 2006 Mar 6 2Spring 2006 Process Dynamics, Operations, and Control 10.450 Lesson 4: Two Tanks in Series (3) solve the original second equation for C′A1 (4) substitute in the equation of the first step. The result is ()'Ai'2A'2A212'2A221CCdtdCdtCd=+τ+τ+ττ (4.2-3) Two mass storage elements led to two first-order equations, which have combined to produce a single second-order equation. A homogeneous solution to (4.2-3) can be found directly, but the particular solution depends on the nature of the disturbance: ()'Ai'part,2At2t1'2ACCeAeAC21++=τ−τ− (4.2-4) where the constants A1 and A2 are found by invoking initial conditions after the particular solution is determined. 4.3 response of system to step disturbance Suppose a step change ΔC occurs in the inlet concentration at time td. Either (4.2-2) or (4.2-4) yields ()dd1(t t ) (t t )'12A2 d12 12CUttC1 e e−− −−τ⎡⎤ττ=−Δ− +⎢τ−τ τ−τ⎣⎦2τ⎥ (4.3-1) Each tank contributes a first-order response based on its own time constant. However, these responses are weighted by factors that depend on both time constants. The result in Figure 4.3-1 looks somewhat different from the first-order responses we have seen. We have plotted the step response of a second-order system with τ1 = 1 and τ1 = 1.5 in arbitrary units. At sufficiently long time, the initial condition has no influence and the outlet concentration will become equal to the new inlet concentration; in this respect it looks like the first-order system response. However, the initial behavior differs: the outlet concentration rises gradually instead of abruptly. This S-shaped curve, often called “sigmoid”, is a feature of systems of order greater than one. Physically, we can understand this by realizing that the change in inlet concentration must spread through two tanks, and it reaches the second tank only after being diluted in the first. revised 2006 Mar 6 3Spring 2006 Process Dynamics, Operations, and Control 10.450 Lesson 4: Two Tanks in Series 00.510123456C*Ai/ C00.20.40.60.81012345tC*A2/6C00.510123456C*Ai/ C00.20.40.60.81012345tC*A2/6C Figure 4.3-1: Response to step change in inlet composition 4.4 introducing the Laplace transform We bother with the Laplace transform for two reasons: • after the initial learning pains, it actually