5_operability - Spring 2006 Process Dynamics, Operations,...

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Spring 2006 Process Dynamics, Operations, and Control 10.450 Lesson 5: Operability of Processes 5.0 context and direction In Lesson 4, we encountered instability. We think of stability as a mathematical property of our linear system models. Now we will embed this mathematical notion within the practical context of process operability. That is, we must not forget that our system models help us operate processes. Along the way, we will encounter a special category of instability/inoperability: the non-self-regulating process. DYNAMIC SYSTEM BEHAVIOR 5.1 remember the stability criterion for linear systems In Section 4.9, we introduced a stability criterion for a linear system: non- negative poles in the transfer function (5.1-1) indicate that the system output y(t) will not remain stable in response to a system input x(t). 1 s a s a s a 1 ) s ( G ) s ( x ) s ( y 1 1 n 1 n n n ' ' + + + + = = L (5.1-1) As a simple example, consider a first-order system: () dy yK xy 00 dt ′′ τ+ = = (5.1-2) We know that the Laplace transform representation is completely equivalent. K ys xs s1 = (5.1-3) The transfer function in (5.1-3) has a single pole at - τ -1 . If the time constant τ is a positive quantity (as in our tank), the pole is negative and the response is stable (as we have seen in Lesson 3). If the time constant were a negative quantity, however, the pole would be positive. As we saw in Section 4.9, the response would be unstable because of the exponential term in the solution of (5.1-2) t y(t) e τ ± (5.1-4) This unbounded response could be in a positive or negative direction, depending on the sign of the gain K. We will address on another occasion what sort of system might have a negative time constant; for now we recognize that encountering an unstable linear system should cause us to look carefully at the process whose behavior it represents. revised 2006 Feb 1 1
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Spring 2006 Process Dynamics, Operations, and Control 10.450 Lesson 5: Operability of Processes 5.2 remember that feedback control can make stable systems go unstable Recall from Section 4.21 that we took a perfectly stable second-order process, placed it in a feedback loop with a first-order valve, and applied proportional-mode control. By increasing the controller gain too far, we could drive the system to instability. We could use our linear stability criterion to predict the onset of instability as we did in Section 5.1. That is, we compute the poles of … not the process transfer function, but the transfer function that represents the process in feedback loop with other components! 5.3 the special case of zero poles Now consider a system with a single pole whose value is zero. () K ys xs s = τ (5.3-1) This is a non-negative pole; we claim this indicates an unstable system. If we apply a step disturbance to (5.3-1), we obtain upon inversion: t yK ′ = τ (5.3-2) Certainly y’ increases without bound, so that it qualifies as unstable. You should try different bounded disturbances in (5.3-1), though, to explore whether this is always the case.
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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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5_operability - Spring 2006 Process Dynamics, Operations,...

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