1 Problem From table 7.1 in the textbook, the Laplace transform of y(t)=te-t is: L te t = y(s) = 1 (s + 1) 2 (S1.1)
Moreover, the Laplace transform of the unit-impulse (t) is:
L [ (t) ] = u(s) = 1
(S1.2)
Since u(t)=0 all the time but at t=to, assuming that y(t) is in deviation form (i.e. ys=0), the transfer function of the system in exam can be written as follows: G(s) = y(s) 1 = u(s) (s + 1) 2 (S1.3)
Problem 2 Assuming constant density and reactor volume, the overall material balance can be written as follows:
A
dh(t) = Fi (t) F = Fi (t) 8h(t)1 2 dt
(S2.1)
As initial condition for equation (S2.1), we chose the steady state value of the hydrostatic pressure hs given by:
2 h(0) = h s = Fis 64
(S2.2)
Linearizing the right hand side of equation (S2.1) about the steady state value (S2.2), one obtains:
Fi (t) 8h(t)1 2 [ Fi (t) Fs ] 4h s 1 2 [ h(t) h s ]
(S2.3)
Defining the deviation variables H(t)=h(t)-hs and Q(t)=Fi(t)-Fs, and considering (S2.3), equation (S2.2) becomes Ah1 2 dH(t) h1 2 s + H(t) = s Q(t) 4 dt 4 subject to the following initial condition (S2.4)
H(0) = 0
(S2.5)
Therefore, by comparison to the standard form of 1st order systems, the time constant and the gain k are the following:
Ah1 2 s = 4 Hence:
a) h s = 3ft b) h s = 9ft
h1 2 k= s 4
(S2.6)
=3 3 4 =9 4
(S2.7)
Problem 3
The material balance on the component A is the following:
V dc A (t) = F [ c Ai (t) c A (t) ] kc A (t)V dt (S3.1)
subject to the initial condition cA(0)=cAss=F/(F+kV)cAi,ss. As the differential equation describing the dynamics of the system is linear, it is already in deviation form when one chooses the deviation variables x(t)=cA(t)-cASS and y(t)=cAi(t)-cAi,SS. Thus, (S3.1) can be rewritten as follows: V dx(t) F + x(t) = y(t) F + kV dt F + kV (S3.2)
subject to the initial condition x(0)=0. Therefore, by comparison to standard the form of 1st order systems, the time constant and the gain k of the system are given by: = V F + kV k= F F + kV (S3.3)
and the transfer function G(s) can be expressed as follows: x(s) = G(s) = y(s) F F + kV = k V s +1 s +1 F + kV
(S3.4)
Unit-step change y(s) = 1 s . Therefore
x(s) = 1 k 1 = k [s + 1 ] s s s + 1
(S3.5)
which yields: x(t) = k(1 e t ) Thus, the sketch to a unit-step change is shown in figure S3.a (S3.6)
Unit-impulse change y(s) = 1 . Therefore
x(s) =
k s +1
(S3.7)
which yields: k x(t) = e t Thus, the sketch to a unit-impulse change is shown in figure S3.b
Sinusoidal input y(s) = A (s 2 + 2 ) . Therefore
(S3.8)
x(s) =
k A s + 1 s 2 + 2
(S3.9)
Repeating what we did in class (see also page 318 of the textbook), one gets: x(t) = kA +1
2 2
sin(t + )
= tan 1 (t)
(S3.10)
Thus, the sketch to a sinusoidal input is shown in figure S3.c
Figure S3
Problem 4
Assuming constant density and reactor volume, the overall material balance can be written as follows: AR dh(t) + h(t) = RFi (t) dt (S4.1)
As initial condition for equation (S4.1), we chose the steady state value of the hydrostatic pressure hs given by:
h(0) = h s = RFis
(S4.2)
Since (S4.1) is linear, defining the deviation variables H(t)=h(t)-hs and Q(t)=Fi(t)-Fis (S4.1) can be immediately rewritten in deviation form as follows: AR dH(t) + H(t) = RQ(t) = Ra sin(t) dt (S4.3)
subject to the initial condition H(0) = 0 (S4.4)
The first order system given by (S4.3)-(S4.4) has time constant and gain respectively given by:
= AR k = aR
(S4.5)
We know from what we did in class that the output of a first order system subject to a sinusoidal input is a sinusoidal wave with the same frequency and with an amplitude ratio given by:
ARatio = b = k 22 + 1 = aR A 2 R 22 + 1 (S4.6)
Solving (S4.6) for A gives: A= 1 (aR) 2 b 2 R b (S4.7)