In Example 16.1, the ultimate gain for the primary con-

Chapter 16, Problem 16.2

(choose chapter or problem)

In Example 16.1, the ultimate gain for the primary con- troller was found to be 43.3 when Kcz = 5. (a) Derive the closed-loop transfer functions for Y1/ D1 and Y1/ Dz as a function of Kcl and Kcz (b) Examine the effect of Kc2 on the critical gain of Kc1 by varying Kcz from 1 to 20. For what values of Kcz do the benefits of cascade control seem to be less important? Is there a stability limit on Kcz? Exercises 309 Seborg, D. E., A Perspective on Advanced Strategies for Process Control (Revisited), Advances in Control, P. M. Frank (Ed.), Springer-Verlag, New York, 1999, pp. 103-134. Shinskey, F. G., Feedback Controllers for the Process Industries, McGraw-Hill, New York, 1994. Shinskey, F. G., Process Control Systems, 4th ed., McGraw-Hill, New York 1996. Singer, J. G. (Ed.), Combustion-Fossil Power Systems, 4th ed., Combustion Engineering, Windsor, CT, 1993, pp. 14-27. Smith, O.J.M., Closer Control of Loops with Dead Time, Chern. Eng. Prog., 53 (5), 217 (1957). (c) Integral action was not included in either primary or secondary loops. First set Kcz = 5, TJl = oo, and TJ2 = 5 min. Find the ultimate controller gain using the Routh array. Then repeat the stability calculation for TJl = 5 min and TJ2 = oo and compare the two results. Is offset for Y1 eliminated in both cases for step changes in D1 or Dz?