4_tanksinseries

# 4_tanksinseries - Spring 2006 Process Dynamics Operations...

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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, C Ai F, C A1 volume V 1 volume V 2 F, C A2 Our component A mass balance is written over each tank. 2 A 1 A 2 A 2 1 A Ai 1 A 1 FC FC C V dt d FC FC C V dt d = = (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 1 A 2 A 2 A 2 Ai 1 A 1 A 1 C C dt dC C C dt dC = + τ = + τ (4.1-2) revised 2006 Mar 6 1

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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, C A1 . If we view the tanks as separate systems, we see that C A1 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, C A1 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 , 2 A r , 1 A r , Ai r , A C C C C = = = We subtract the reference condition from (4.1-2) and thus express the variables in deviation form. ' 1 A ' 2 A ' 2 A 2 ' Ai ' 1 A ' 1 A 1 C C dt dC C C dt dC = + 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 dt C e e 1 C t 0 ' Ai t t 1 ' 1 A 1 1 τ τ τ = (4.2-1) ∫∫ τ τ = τ τ τ τ t 0 t 0 ' Ai t t 1 t t 2 ' 2 A dt dt C e e 1 e e 1 C 1 1 2 2 (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

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## This note was uploaded on 11/27/2011 for the course CHEMICAL E 20.410j taught by Professor Rogerd.kamm during the Spring '03 term at MIT.

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4_tanksinseries - Spring 2006 Process Dynamics Operations...

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