1 Problem a) For noninteracting capacities with linear resistances subject to a unit-step change in the input of the first tank, the material balance can be written as follows:
A1R1
A2R 2
dy1 + y1 = R1u(t) dt
dy 2 R + y 2 = 2 y1 dt R1
(S1.1)
(S1.2)
subject to the initial conditions y1 (0) = y1s = R1u s y 2 (0) = y 2s = R 2 R1 y1s = R 2 u s (S1.3)
Defining the deviation variables Y1=y1-y1s, Y2=y2-y2s and Q=u(t)-us, equations (S1.1)(S1.2) become:
A1R1
dY1 + Y1 = R1Q dt
(S1.4)
A2R 2
dY2 R + Y2 = 2 Y1 dt R1
(S1.5)
subject to the initial conditions Y1 (0) = 0 Y2 (0) = 0 (S1.6)
Hence, the transfer functions for equations (S1.4)-(S1.5) are:
G1 (s) = Y1 (s) R1 = Q(s) (A1R1 )s + 1
(S1.7)
G 2 (s) =
Y2 (s) R 2 R1 = Y1 (s) (A 2 R 2 )s + 1
(S1.8)
Since the two noninteracting tanks are placed in series, the overall transfer function is the following: G(s) = G1 (s)G 2 (s) = Y2 (s) R2 = Q(s) [ (A1R1 )s + 1][ (A 2 R 2 )s + 1] (S1.9)
Rewriting (S1.9) in the standard form, gives:
G(s) = G1 (s)G 2 (s) = Y2 (s) R2 = 2 Q(s) (A1R1 )(A 2 R 2 )s + 2(A1R1 + A 2 R 2 ) + 1
(S1.10)
For critically damped systems, =1. Therefore, comparing equation (S1.10) with the standard form yields:
2 = 12 = (A1R1 )(A 2 R 2 ) = 1 + 2 = A1R1 + A 2 R 2 Equations (S1.12)-(S1.11) are simultaneously solved if: 1 = 2 A1R1 = A 2 R 2
(S1.11) (S1.12)
(S1.13)
In other words, the system is critically damped if the roots of the denominator in (S1.9) are in reality one root with multiplicity equal to 2. This immediately implies (S1.13). Thus: R1 A 2 1 = = R 2 A1 2 (S1.14)
b) From the textbook, solution to the critically damped second order system subject to a unit-step change is:
t Y2 (t) t = 1 e 1 + R2
(S1.15)
Since it takes 1 min for the change in level of the second tank to reach 50 percent of the total change, one obtains:
1 1 0.5 = 1 e 1 +
= 0.59 min
(S1.16)
c) From the textbook, solution to a first order system (1st tank) subject to a unit-step change is:
t Y1 (t) = 1 e R1
(S1.17)
Therefore, in order for the level of the first tank to reach 90 percent of the total change it takes:
0.9 = 1 e
t 0.59
t = 0.59 ln(10) min = 1.36 min
(S1.18)
Problem 2
Bringining the transfer function in standard form, yields: G(s) = 20 4s + 0.6s + 1
2
(S2.1)
Therefore the natural period of oscillation and the damping factor are the following: 2 = 4 = 2 (time)
= 0.6 (2) = 0.15 (D-less)
(S2.2) (S2.3)
2 = 0.6 (time)
Hence, from the page 191 of textbook one obtains:
= 0.62 (D-less) OS = exp 1 2
(S2.4)
=
1 2 cicles = = 0.079 2 2 time
(S2.5)
Problem 3
Let 1/Rt=1/Ra+1/R1=3/2. Then, the material balance on the first tank yields: A1 dh1 h = q (q A + q1 ) = q 1 dt Rt (S3.1)
subject to the initial h1(0)=h1s=qsRt. condition Equation (3.1) can be rewritten as follows: A1R t dh1 + h1 = R t q dt dh 2 R + h 2 = 2 h1 dt R1 (S3.2)
Figure 3.1
The material balance on the second tank yields: A2R 2
(S3.3)
subject to the initial condition h2(0)=R2/R1h1s. Defining the deviation variables H1=h1-h1s, H2=h2-h2s and Q=q-qs, equations (S3.2)-(S3.3) become: A1R t A2R 2 dH1 + H1 = R t Q dt dH 2 R + H 2 = 2 H1 dt R1 (S3.4)
(S3.5)
subject to the initial conditions
H1 (0) = 0
H 2 (0) = 0
(S3.6)
Hence, the transfer functions for equations (S3.4)-(S3.5) are:
G1 (s) = G 2 (s) = Rt H1 (s) 23 = = Q(s) (A1R t )s + 1 4 3s + 1 H 2 (s) R 2 R1 1 = = H1 (s) (A 2 R 2 )s + 1 s + 1
(S3.7)
(S3.8)
Thus, the overall transfer function is the following: G(s) = G1 (s)G 2 (s) = H 2 (s) 23 = Q(s) ( 4 3s + 1)( s + 1) (S3.9)
Problem 4
a) Linearizing cAcR about cAscRs yields:
CA CR = CAs CRs + (CA CR ) (CA CR ) (CA CAs ) + (CR CRs ) + HOT = CA CA =CAs CR CA =CAs (S4.1) C =C C =C
R Rs R Rs
CAs CRs + CRs (CA CAs ) + CAs (C R CRs )
Substituting (S4.1) into (4.1a) and (4.2b) gives: dCA = CAi CA kCAs CRs kCRs (CA CAs ) kCAs (CR CRs ) dt dCR = CRi CR + kCAs CRs + kCRs (CA CAs ) + kCAs (CR CRs ) dt (S4.2) (S4.3)
At steady state, one obtains: CAis CAs = kCAs CRs CRis + CRs = kCAs CRs (S4.4) (S4.5)
Defining the deviation variables A=CA-CAs, R=CR-CRs, QA=CAi-CAis and QR=CRi-CRis equations (S4.2)-(S4.3) become: dA = QA A kC Rs A kCAs R dt dR = QR R + kCRs A + kCAs R dt (S4.6) (S4.7)
Subject to:
A(0) = 0
R(0) = 0
(S4.8)
b) Taking the Laplace transform of both sides of (S4.6)-(S4.7) and applying the initial conditions (S4.8) yields:
sA(s) = Q A (s) A(s) kCRs A(s) kCAs R(s) sR(s) = Q R (s) R(s) + kCRs A(s) + kCAs R(s) (S4.9) (S4.10)
Notice that in equation (S4.9) QR(s)=0 since CRi is constant. Thus, solving (S4.10) for A gives: A(s) = s + 1 kCAs R(s) kCRs (S4.11)
which substituted into (S4.9) provides the transfer function between R(s) and QA(s):
G(s) = R(s) = Q A (s) kCRs 2 k(CRs CAs ) 1 + k(CRs CAs ) s 2 + 1 + s + 2 2
(S4.12)
c) Notice that the denominator of G(s) can be written as:
as 2 + bs + c
(S4.13)
where a=1. The condition (CAs-CRs)<1/k ensures that the 1st order and 0th order coefficient of the polynomial denominator of G(s) are positive. Hence, since s1s2=c>0, then the two roots have the same sign (either positive or negative). Moreover, as (s1+s2)=b, then the two roots are both negative. Therefore, the condition (CAs-CRs)<1/k guarantees stability for the system. d) From (S4.12) one obtains: = 1+ k(CRs CAs ) 2 (S4.14)
Since the term CRs-CAs is positive, so is the term k(CRs-CAs)/2. Therefore, the damping factor is bigger than 1 and the system in overdamped.