MIT 10 450 - Process Dynamics, Operations, and Control (37 pages)

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Process Dynamics, Operations, and Control



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Process Dynamics, Operations, and Control

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37
School:
Massachusetts Institute of Technology
Course:
10 450 - Process Dynamics, Operations, and Control
Process Dynamics, Operations, and Control Documents
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Spring 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 CAi F CA1 F CA2 volume V1 volume V2 Our component A mass balance is written over each tank d V1C A1 FC Ai FC A1 dt d V2 C A 2 FC A1 FC A 2 dt 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 dC A1 C A1 C Ai dt dC A 2 2 C A 2 C A1 dt 1 revised 2006 Mar 6 4 1 2 1 Spring 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 C A r C Ai r C A1 r C A 2 r 4 1 3 We subtract the reference condition from 4 1 2 and thus express the variables in deviation form dC A1 1 C A1 C Ai dt dC A 2 2 C A 2 C A1 dt 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 t C A1 1 t 1 t 1 e e C Ai dt 1 0 4 2 1 C A 2 t t 1 t 2 t 2 1 t 1 t 1 e e e e C Ai dt dt 2 0 0 1 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 2 Spring 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 1 2 d 2 C A 2 dC A 2 C A 2 C Ai 1 2 dt 2 dt 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 C A 2 A1e t 1 A 2e t 2 C A 2 part C Ai 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 C occurs in the inlet concentration at time td Either 4 2 2 or 4 2 4 yields t t d t t d 1 2 1 2 C A2 U t t d C 1 e e 1 2 1 2 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 secondorder system with 1 1 and 1 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 3 Spring 2006 Process Dynamics Operations and Control 10 450 Lesson 4 Two Tanks in Series C Ai C 1 0 5 0 1 0 1 2 3 4 5 6 0 1 2 3 4 5 6 C A2 C 0 8 0 6 0 4 0 2 0 t 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 makes the math easier so we will use it in derivations some of the terminology in linear systems and process control is based on formulating the equations with Laplace transforms Definition the Laplace transform turns a function of time y t into a function of the complex variable s Variable s has dimensions of reciprocal time All the information contained in the time domain function is preserved in the Laplace domain y s L y t y t e st dt 4 4 1 0 In these notes we use the notation y s merely to indicate that y t has been transformed we do not mean that y s has the same functional dependence on s that it does on t Functional transforms textbooks for example Marlin Sec 4 2 usually include tables of transform pairs so these derivations from definition 4 4 1 are primarily to demonstrate how the tables came to be revised 2006 Mar 6 …


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