ChE 441 Washington State University Process Control School of ...

ChE 441
Process Control
Fall, 2007
Washington State University
School of Chemical Engineering and Bioengineering
Richard L. Zollars
Problem 3-24, Smith and Corripio
Assume that the following equation describes a certain process
Y( s )
X( s )
−0.
5s
3e
=
5s
+ 0.
2
(a) Obtain the steady-state gain, time constant, and dead time of this process.
(b) The initial condition of the variable y is y(0) = 2. For a forcing function as shown in
Fig. P3-15, what is the final value of y(t)?
SOLUTION
(a) The standard form of the transfer function for a first order system is
Y( s )
X( s )
−t D
s
Ke
=
τ s + 1
To get the equation above into this form multiply both the numerator and denominator by
5 (or divide by 0.2) to get
Y( s )
X( s )
−0.
5s
15e
=
25s
+ 1
Thus the steady-state gain (K) is 15, the time constant (τ) is 15, and the dead time (t D ) is
0.5.
(b) The forcing function is a step function of magnitude A. In the LaPlace space this has
the form
Thus
The final value is then determined by
Y( s )
X ( s ) =
A
s
−0.
5s
15e
=
25s
+ 1
A
s

⎡
−0.
5s
15e
y( ∞ ) = lim [ sY( s )] = lim⎢s
s→0
s→0
⎣ 25s
+ 1
A⎤
⎥ = 15A
s ⎦
This should be a deviation variable, however. To find the real value we need to add the
prior steady-state. Thus the true final value of y(t) is 15A + 2.