5. An open-loop unstable process can be stabilized with correct proportional control action. Consider a traditional feedback control loop (shown below) with Gm-G,=1. Y(s) The process is described as U(s) Kp -Tps+1 Demonstrate the response of the process is unstable without a controller. а. b. Find the range of Ke values that lead to a stable closed loop response. (Assume that In must be positive)
5. An open-loop unstable process can be stabilized with correct proportional control action. Consider a traditional feedback control loop (shown below) with Gm-G,=1. Y(s) The process is described as U(s) Kp -Tps+1 Demonstrate the response of the process is unstable without a controller. а. b. Find the range of Ke values that lead to a stable closed loop response. (Assume that In must be positive)
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Transcribed Image Text:5.
An open-loop unstable process can be stabilized with correct proportional
control action. Consider a traditional feedback control loop (shown below) with
Gm=G=1.
Y(s)
%D
Kp
The process is described as
U(s)
-Tps+1
а.
Demonstrate the response of the process is unstable without a controller.
b.
Find the range of Ke values that lead to a stable closed loop response. (Assume that
Tp must be positive).
The result of part b shows that KmKcKvKp>0 is a simplified statement when one of
the
c.
is unstable. An increase in setpoint in closed loop response should
processes
cause an increase in the control variable (a positive gain). This is true if the gain of
the closed loop response is positive. Show that the gain of Y(s)/Ysp(s) is positive
within the range of your answer to part a.
D(s)
Gp
Y sp(5)
Km
P(s)
Gy
Yp(s)
Y(s)
E(s)
G.
U(s)
Yuls),
Gp
Ym (s)
GM
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