Page 156 - Practical Control Engineering a Guide for Engineers, Managers, and Practitioners

Matrices and Higher-Order Process Models 131

Before the reader gets too impressed, I did not do this algebra by
hand. Rather, I used the Matlab Symbolic toolbox-the algebra is just
too tedious and the opportunity to make an algebraic bookkeeping
mistake is too large. Note that all of the coefficients c; in the denomi-
nator's polynomial are positive.
As an aside, the Matlab script that I used to develop Eq. (5-13) is
% interconnect.m
clear
syms Al A2 A3 R12 R23 R3 s I2 U rho
syms X y z Zl Z2 R33
syms Rl2Al Rl2A2 R23A2 R23A3 R3A3
syms lam w
syms Tl T2 T3 % declare these variables as symbolic

% tanks with back flow
S=solve( rho*Al*s*x+(x-y)/Rl2-U, rho*A2*s*y-((x-y)/
R12-(y-z)/R23) , rho*A3*s*z-((y-z)/R23-z/R3));
Zl=S.z;
Zl=collect(Zl,s);
pretty(Zl)
Although I do not expect you to be adept at creating Matlab
scripts, I do think you can browse the above code and get a feel for
how simple it is to have the computer do the algebra.
On looking at Eq. (5-13) carefully, one sees the combination
pAR occurs frequently. This combination has units of seconds
and could be considered a time constant of sorts. However, find-
ing the poles of the transfer function is not as straightforward as
for Eq. (5-4).
In any case, Eq. (5-13) shows that the highest power of the Laplace
operator s is three, meaning that the equations describe a third-order
system. Figure 5-6 shows the response of the three tank levels for a
step in the input flow rate. The parameter values used were A = 0.1,
1
A = 0.1, A = 0.1, R = 10, R = 10, and~= 10. Note how the steady-
2 3 12 23
state levels are all different.
Question 5-2 Why are the steady-state levels different?
Answer At steady state all the flows must be the same. The net flow between
tanks one and two is

Since U is nonzero and positive, X must be greater than X • A similar argument
1 2
shows that X must be greater than X • Since the levels drive the flows, they must
2 3
also be different.
The frequency domain behavior can be obtained in a manner
similar to that for the three-tank process with no backflow and is