1234567898
6.1
a)
By using MATLAB, the poles and zeros are: Zeros: (-1 +1i) , (-1 -1i) Poles: -4.3446 (-1.0834 +0.5853i) (-1.0834 –0.5853i) (+0.7557 +0.5830i) (+0.7557 −0.5830i) These results are shown in Fig E6.1
Figure S6.1. Poles and zeros of G(s) plotted in the complex s plane. Solution Manual for Process Dynamics and Control, 2nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.
6-1
b) c)
Process output will be unbounded because some poles lie in the right half plane. By using Simulink-MATLAB 8
2
x 10
0
-2
-4
Output
-6
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-10
-12
-14
-16
0
5
10
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30
Time
Figure E6.1b. Response of the output of this process to a unit step input.
As shown in Fig. S6.1b, the right half plane pole pair makes the process unstable.
6.2 a)
Standard form =
K (τ a s + 1) (τ1 s + 1)(τ 2 s + 1)
b)
0.5(2s + 1)e −5 s Hence G ( s ) = (0.5s + 1)(2s + 1) Applying zero-pole cancellation:
0.5e −5 s G (s) = (0.5s + 1) c)
Gain = 0.5 Pole = −2 Zeros = No zeros due to the zero-pole cancellation. 6-2
d)
1/1 Pade approximation: e −5 s =
(1 − 5 / 2 s ) (1 + 5 / 2 s )
The transfer function is now G (s) =
0.5 (1 − 5 / 2 s ) × 0.5s + 1 (1 + 5 / 2 s )
Gain = 0.5 Poles = −2, −2/5 Zeros = + 2/5
6.3
Y ( s ) K (τ a s + 1) = X ( s) (τ1 s + 1)
X (s) =
,
M s
From Eq. 6-13
a)
b)
τ τ − τ1 −t / τ1 y(t) = KM 1 − 1 − a e −t / τ1 = KM 1 + a e τ1 τ1 τ − τ1 τ a y (0 + ) = KM 1 + a KM = τ1 τ1
Overshoot → y(t) > KM
τ − τ1 −t / τ1 KM 1 + a e > KM τ1 or τa − τ1 > 0 , that is, τa > τ1 y1 = − KM c)
( τ a − τ1 ) τ1
2
e −t / τ1 < 0
for
KM > 0
Inverse response → y(t) < 0
τ − τ1 −t / τ1 KM 1 + a e <0 τ1 τ a − τ1 τa < −e +t / τ1 or < 1 − e +t / τ1 < 0 τ1 τ1 Therefore τa < 0. 6-3
at t = 0.
6.4
K (τ a s + 1) Y (s) = , τ1>τ2, X ( s ) (τ1 s + 1)(τ 2 s + 1)
X(s) = M/s
From Eq. 6-15
τ − τ1 −t / τ1 τ a − τ 2 −t / τ2 y (t ) = KM 1 + a e e − τ − τ τ − τ 1 2 1 2 a)
Extremum → y1 (t ) = 0 1 τ − τ1 −t / τ1 1 τ a − τ 2 e KM 0 − a + τ τ − τ τ 2 τ1 − τ 2 1 1 2 1
1
−t − 1 − τa / τ2 τ τ = e 1 2 ≥1 1 − τ a / τ1
b)
−t / τ 2 e =0
since
τ1>τ2
Overshoot → y (t ) > KM
τ − τ1 −t / τ1 τ a − τ 2 −t / τ2 KM 1 + a e − e > KM τ1 − τ 2 τ1 − τ 2 1
1
−t − τ a − τ1 τ τ > e 2 1 > 0 , therefore τa>τ1 τa − τ2
c)
Inverse response → y1 (t ) < 0 at t = 0+ 1 τ − τ1 −t / τ1 1 τ a − τ 2 −t / τ2 + e e KM 0 − a + < 0 at t = 0 τ1 τ1 − τ 2 τ 2 τ1 − τ 2 1 τ − τ1 1 τ a − τ 2 + <0 − a τ1 τ1 − τ 2 τ 2 τ1 − τ 2
1 1 τ a − τ 2 τ1 < 0 τ1 −τ 2
6-4
Since τ1 > τ2, τa < 0. d)
If an extremum in y exists, then from (a) 1 1 − t − τ1 τ 2
1 − τa τ2 = 1 − τ a τ1 1 − τa τ2 ττ t = 1 2 ln τ1 − τ 2 1 − τ a τ1 e
6.5
Substituting the numerical values into Eq. 6-15 Case (i) : y(t) = 1 (1 + 1.25e-t/10 − 2.25e-t/2) Case (ii(a)) : y(t) = 1 (1 − 0.75e-t/10 − 0.25e-t/2) Case (ii(b)) : y(t) = 1 (1 − 1.125e-t/10 + 0.125e-t/2) Case (iii) : y(t) = 1 (1 − 1.5e-t/10 + 0.5e-t/2)
1.6 case(i) case(ii)a case(ii)b case(iii)
1.4
1.2
1
y(t)
0.8
0.6
0.4
0.2
0
-0.2
0
5
10
15
20
25
30
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40
45
50
Time
Figure S6.5. Step response of a second-order system with a single zero.
6-5
Conclusions: τa > τ1 gives overshoot. 0 < τa < τ1 gives response similar to ordinary first-order process response. τa < 0 gives inverse response.
6.6 Y (s) =
K1 K2 K2 K U (s) + U ( s) = 1 + U ( s ) s τs + 1 s τs + 1
Y ( s ) K 1 τs + K 1 + K 2 s ( K 1τ + K 2 ) s + K 1 = = U ( s) s (τs + 1) s (τs + 1)
Put in standard K/τ form for analysis:
K K 1 τ + 2 s + 1 K1 Y (s) G( s) = = U ( s) s (τs + 1) a)
Order of G(s) is 2 (maximum exponent on s in denominator is 2)
b)
Gain of G(s) is K1. Gain is negative if K1 < 0.
c)
Poles of G(s) are: s1 = 0 and s2 = –1/τ s1 is on imaginary axis; s2 is in LHP.
d)
Zero of G(s) is:
sa =
If
− K1 −1 = K K1τ + K 2 τ + 2 K1
K1 < 0 , the zero is in RHP. K1τ + K 2
6-6
Two possibilities: 1. K1<0 and K1τ + K2 >0 2. K1 > 0 and K1τ + K2 < 0 e)
Gain is negative if K1 < 0 Then zero is RHP if K1τ + K2 > 0 This is the only possibility.
f)
Constant term and e-t/τ term.
g)
If input is M/s, the output will contain a t term, that is, it is not bounded.
6.7
a)
2 s −3 −3 2 Q ′( s ) = P ′( s ) = 20 s + 1 20 s + 1 s p ′(t ) = (4 − 2) S (t )
, P ′( s ) =
Q ′(t ) = −6(1 − e −t / 20 )
b)
R ′( s ) + Q ′( s ) = Pm′ ( s ) r ′(t ) + q ′(t ) = p ′m (t ) = p m (t ) − p m (0) r ′(t ) = p m (t ) − 12 + 6(1 − e − t / 20 )
K=
r ′(t = ∞) 18 − 12 + 6(1 − 0) = =6 p (t = ∞) − p (t = 0) 4−2
Overshoot,
OS =
r ′(t = 15) − r ′(t = ∞) 27 − 12 + 6(1 − e −15 / 20 ) − 12 = = 0.514 r ′(t = ∞) 12
6-7
− πζ OS = exp 1− ζ2
= 0.514
,
ζ = 0 .2
Period, T, for r ′(t ) is equal to the period for pm(t) since e-t/20 decreases monotonically.
c)
Thus,
T = 50 − 15 = 35
and
τ=
T 1 − ζ 2 = 5.46 2π
Pm′ ( s ) K K′ = 2 2 + P ′( s ) τ s + 2ζτs + 1 τ′s + 1
(K ′τ )s = 2
d)
+ ( Kτ′ + 2 K ′ζτ) s + ( K + K ′) (τ s 2 + 2ζτs + 1)(τ′s + 1) 2
2
Overall process gain =
Pm′ ( s ) P ′( s )
= K + K′ = 6 −3 = 3 s =0
% psi
6.8
a)
Transfer Function for blending tank:
Gbt ( s ) =
K bt τ bt s + 1
where K bt = τ bt =
qin ≠1 ∑ qi
2m 3 = 2 min 1m 3 / min
Transfer Function for transfer line Gtl ( s ) =
K tl
(τ tl s + 1)5
where K tl = 1 τ tl =
6-8
0.1m 3 = 0.02 min 5 × 1m 3 / min
′ (s) C out K bt = C in′ ( s ) (2s + 1)(0.02s + 1) 5
∴
a 6th-order transfer function. b)
Since τbt >> τtl [ 2 >> 0.02] we can approximate
1 by e-θ s 5 (0.02 s + 1)
5
where θ = ∑ (0.02) = 0.1 i =1
′ ( s ) K bt e −0.1s C out ≈ C in′ ( s ) 2s + 1
∴ c)
Since τbt ≈ 100 τtl , we can imagine that this approximate TF will yield results very close to those from the original TF (part (a)). We also note that this approximate TF is exactly the same as would have been obtained using a plug flow assumption for the transfer line. Thus we conclude that investing a lot of effort into obtaining an accurate dynamic model for the transfer line is not worthwhile in this case. [ Note that, if τbt ≈ τtl , this conclusion would not be valid] By using Simulink-MATLAB, 1.2
1
0.8
Output/Kbt
d)
0.6
0.4
0.2
0 Exact model Approximate model -0.2
0
5
10
15 Time
20
25
30
Fig S6.8. Unit step responses for exact and approximate model.
6-9
6.9
a), b) Represent processes that are (approximately) critically damped. A step response or frequency response in each case can be fit graphically or numerically. c)
θ = 2, τ = 10
d)
Exhibits strong overshoot. Can’t approximate it well.
e)
θ = 0.5, τ = 10
f)
θ = 1, τ = 10
g)
Underdamped (oscillatory). Can’t approximate it well.
h)
θ = 2, τ = 0 By using Simulink-MATLAB, models for parts c), e), f) and h) are compared: (Suppose K = 1) Part c)
1
0.9
0.8
0.7
Output
0.6
0.5
0.4
0.3
0.2
0.1 Exact model Approximate model 0
0
5
10
15
20
25 time
30
35
40
45
50
Figure S6.9a. Unit step responses for exact and approximate model in part c)
6-10
Part e)
1
0.9
0.8
0.7
Output
0.6
0.5
0.4
0.3
0.2
0.1 Exact model Approximate model 0
0
5
10
15
20
25 Time
30
35
40
45
50
Figure S6.9b. Unit step responses for exact and approximate model in part e)
Part f)
1
0.9
0.8
0.7
Output
0.6
0.5
0.4
0.3
0.2
0.1 Exact model Approximate model 0
0
5
10
15
20
25 Time
30
35
40
45
50
Figure S6.9c. Unit step responses for exact and approximate model in part f)
6-11
Part h) 1.5
1
Output
0.5
0
-0.5
Exact model Approximate model -1
0
5
10
15
20
25
30
35
40
45
50
Time
Figure S6.9d. Unit step responses for exact and approximate model in part h)
6.10
a)
The transfer function for each tank is
C i′( s ) 1 = C i′−1 ( s ) V s + 1 q
i = 1, 2, …, 5
,
where i represents the ith tank. co is the inlet concentration to tank 1. V is the volume of each tank. q is the volumetric flow rate. 5 C ′(s) 1 C 5′ ( s ) = ∏ i , = C 0′ ( s ) i =1 C i′−1 ( s ) 6s + 1 5
Then, by partial fraction expansion,
6-12
t 1 t 2 1 t 3 1 t 4 −t / 6 c5 (t ) = 0.60 − 0.15 1 − e 1 + + + + 6 2! 6 3! 6 4! 6 Using Simulink,
b)
0.6 c5 c4
0.58
c3 c2
Concentration
0.56
c1
0.54 0.52 0.5 0.48 0.46 0.44
0
5
10
15
20
25 time
30
35
40
45
50
Figure S6.10. Concentration step responses of the stirred tank.
The value of the expression for c5(t) verifies the simulation results above: 52 53 54 c5 (30) = 0.60 − 0.15 1 − e −5 1 + 5 + + + = 0.5161 2! 3! 4!
6.11
a)
Y (s) =
− τa s + 1 E A B C = + 2 + 2 τ1 s + 1 s s s τ1 s + 1
We only need to calculate the coefficients A and B because Ce −t / τ1 → 0 for t >> τ1. However, there is a repeated pole at zero.
6-13
E (− τ a s + 1) B = lim =E s →0 τ1 s + 1 Now look at E (− τ a s + 1) = As (τ1 s + 1) + B (τ1 s + 1) + Cs 2 − Eτ a s + E = Aτ1 s 2 + As + Bτ1 s + B + Cs 2
Equate coefficients on s: − Eτ a = A + Bτ1 A = − E ( τ a + τ1 ) Then the long-time solution is y (t ) ≈ Et − E (τ a + τ1 ) Plotting
(τa+τ1)
y
y(t)=Et
=Et-E(τa+τ1)
actual response -E(τa+τ1)
time
b)
For a LHP zero, the apparent lag would be τ1 − τa
c)
For no zero, the apparent lag would be τ1
6-14
6.12
a)
Using Skogestad’s method
G ( s ) approx =
b)
5e − ( 0.5+0.2) s 5e −0.7 s = (10s + 1)((4 + 0.5) s + 1) (10s + 1)(4.5s + 1)
By using Simulink-MATLAB 5
4
Output
3
2
1
0
Exact model Approximate model -1
0
5
10
15
20
25 Time
30
35
40
45
50
Figure S6.12a. Unit step responses for exact and approximate model.
c)
Using MATLAB and saving output data on vectors, the maximum error is Maximum error = 0.0521 at = 5.07 s This maximum error is graphically shown in Fig. S6.12b
6-15
Exact model Approximate model
0.9 0.8 0.7
Output
0.6 0.5 0.4 0.3 0.2 0.1 0 -0.1
0
1
2
3
4 Time
5
6
7
8
Figure S6.12b. Maximum error between responses for exact and approximate model.
6.13
From the solution to Problem 2-5 (a) , the dynamic model for isothermal operation is
V1 M dP1 Pd − P1 P1 − P2 = − RT1 dt Ra Rb
(1)
V2 M dP2 P1 − P2 P2 − Pf = − RT2 dt Rb Rc
(2)
Taking Laplace transforms, and noting that Pf′ ( s ) = 0 since Pf is constant,
K b Pd′ ( s ) + K a P2′ ( s ) τ1 s + 1 K P ′( s ) P2′ ( s ) = c 1 τ2 s + 1 P1′( s ) =
where
6-16
(3) (4)
K a = Ra /( Ra + Rb ) K b = Rb /( Ra + Rb ) K c = Rc /( Rb + Rc )
τ1 =
V1 M Ra Rb RT1 ( Ra + Rb )
τ2 =
V2 M Rb Rc RT2 ( Rb + Rc )
Substituting for P1′( s ) from Eq. 3 into 4,
Kb Kc 1 − K K Kb Kc P2′( s ) a c = = Pd′ ( s ) (τ1 s + 1)(τ 2 s + 1) − K a K c τ1 τ 2 2 τ1 + τ 2 s + 1 − K K a c 1− Ka Kc
(5)
s + 1
Substituting for P2′ ( s ) from Eq. 5 into 4,
Kb (τ2 s + 1) 1 − Ka Kc P1′( s ) = Pd′ ( s ) τ1τ2 2 τ1 + τ2 s + 1 − Ka Kc 1 − Ka Kc
s +1
(6)
To determine whether the system is over- or underdamped, consider the denominator of transfer functions in Eqs. 5 and 6.
τ1 τ 2 τ 2 = 1 − Ka Kc
τ + τ2 , 2ζτ = 1 1− Ka Kc
Therefore, ζ=
τ 1 (τ1 + τ 2 ) (1 − K a K c ) 1 τ1 1 = + 2 2 (1 − K a K c ) 2 τ2 τ1 (1 − K a K c ) τ1 τ 2
Since x + 1/x ≥ 2 for all positive x,
6-17
ζ≥
1 (1 − K a K c )
Since KaKc ≥ 0, ζ ≥1
Hence the system is overdamped.
6.14
a)
For
Y (s) =
X (s) =
M s
KM A B C = + + s (1 − s )(τs + 1) s 1 − s τs + 1 A = lim =
KM = KM (1 − s )(τs + 1)
B = lim =
KM KM = s (τs + 1) τ + 1
s →0
s →1
KM − KM τ2 KM C = lim = = s →−1/ τ s (1 − s ) τ +1 − 1 1 + 1 τ τ Then,
et τ −t / τ y1 (t ) = KM 1 − − e τ +1 τ +1 For M =2 , K = 3, and τ = 3, the Simulink response is shown:
6-18
8
0
x 10
-2
Output
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-6
-8
-10
-12
0
2
4
6
8
10
12
14
16
18
Time
Figure S6.14a. Unit step response for part a).
b)
If
G2 ( s) =
Ke −2 s (1 − s )(τs + 1)
then,
e t −2 τ −( t − 2 ) / τ y 2 (t ) = KM 1 − − e S (t − 2) τ +1 τ +1 Note presence of positive exponential term. c)
Approximating G2(s) using a Padé function G2 ( s) =
K (1 − s ) K = ( s + 1)(τs + 1)(1 − s ) ( s + 1)(τs + 1)
Note that the two remaining poles are in the LHP. d)
For
X (s) =
Y (s) =
M s
KM s ( s + 1)(τs + 1)
Using Table 3.1 τ1 = 1
,
τ2 = τ
6-19
20
1 y3 (t ) = KM 1 + (e −t − τe −t / τ ) τ −1
Note that no positive exponential term is present. e)
Instability may be hidden by a pole-zero cancellation.
f)
By using Simulink-MATLAB, unit step responses for parts b) and c) are shown below: (M = 2 , K = 3 , τ = 3) 7
0
x 10
-1
-2
-3
Output
-4
-5
-6
-7
-8
-9
-10
0
2
4
6
8
10
12
14
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18
20
Time
Figure S6.14b. Unit step response for part b). 6
5
Output
4
3
2
1
0
0
2
4
6
8
10 Time
12
14
16
18
Figure S6.14c. Unit step response for part c) .
6-20
20
6.15
From Eq. 6-71 and 6-72, ζ=
Since x +
R2 A2 + R1 A1 + R2 A1 2 R1 R2 A1 A2
=
1 R1 A1 + 2 R2 A2
R2 A2 R1 A1
1 + 2
R2 A1 R1 A2
1 ≥ 2 for all positive x and since R1, R2, A1, A2 are positive x
ζ≥
1 (2) + 1 R2 A1 ≥ 1 2 2 R1 A2
6.16
a)
If w1 = 0 and ρ = constant ρA2
dh2 = w0 − w2 dt
w2 =
1 h2 R2
[ Note: could also define R2 by q 2 =
1 ρ h2 → w2 = ρq 2 = h2 ] R2 R2
Substituting, ρA2
dh2 1 = w0 − h2 dt R2
or
ρA2 R2
dh2 = R2 w0 − h2 dt
Taking deviation variables and Laplace transforming ρA2 R2 sH 2′ ( s ) + H 2′ ( s ) = R2W0′( s )
6-21
H 2′ ( s ) R2 = W0′( s ) ρA2 R2 s + 1 Since W2′ ( s ) =
1 H 2′ ( s ) R2
W2′ ( s ) R2 1 1 = = W0′( s ) R2 ρA2 R2 s + 1 ρA2 R2 s + 1 Let τ2 = ρA2R2
W2′ ( s ) 1 = W0′( s ) τ 2 s + 1
b)
ρ = constant dh ρA1 1 = − w1 dt w1 =
c)
1 (h1 − h2 ) R1
ρA2
dh2 = w0 + w1 − w2 dt
w2 =
1 h2 R2
Since this clearly is an interacting system, there will be a single zero. Also, we know the gain must be equal to one. ∴
τ s +1 W1′( s ) = 2 2 a W0′( s ) τ s + 2ζτs + 1
W2′ ( s ) 1 = 2 2 W0′( s ) τ s + 2ζτs + 1
or
τa s + 1 W1′( s ) = W0′( s ) (τ1′ s + 1)(τ′2 s + 1)
W2′ ( s ) 1 = W0′( s ) (τ1′ s + 1)(τ′2 s + 1)
where τ1′ and τ′2 are functions of the resistances and areas and can only be obtained by factoring. f)
Case b will be slower since the interacting system is 2nd-order, "including" the 1st-order system of Case a as a component.
6-22
6.17 The input is Ti′(t ) = 12 sin ωt 2π radians where ω= = 0.262 hr −1 24 hours The Laplace transform of the input is from Table 3.1, Ti′( s ) =
12ω s + ω2 2
Multiplying the transfer function by the input transform yields Ti′( s ) =
(−72 + 36 s )ω (10 s + 1)(5s + 1)( s 2 + ω 2 )
To invert, either (1) make a partial fraction expansion manually, or (2) use the Matlab residue function. The first method requires solution of a system of algebraic equations to obtain the coefficients of the four partial fractions. The second method requires that the numerator and denominator be defined as coefficients of descending powers of s prior to calling the Matlab residue function: Matlab Commands >> b = [ 36*0.262 −72*0.262] b= 9.4320 −18.8640 >> a = conv([10 1], conv([5 1], [1 0 0.262^2])) b= 50.0000
15.0000
4.4322
>> [r,p,k] = residue(b,a) r= 6.0865 − 4.9668i 6.0865 + 4.9668i 38.1989 −50.3718 6-23
1.0297
0.0686
p= −0.0000 − 0.2620i −0.0000 + 0.2620i −0.2000 −0.1000 k= [] Note: the residue function recomputes all the poles (listed under p). These are, in reverse order: p1 = 0.1( τ1 = 10) , p2 = 0.2( τ 2 = 5) , and the two purely imaginary poles corresponding to the sine and cosine functions. The residues (listed under r) are exactly the coefficients of the corresponding poles, in other words, the coefficients that would have been obtained via a manual partial fraction expansion. In this case, we are not interested in the real poles since both of them yield exponential functions that go to 0 as t→ ∞. The complex poles are interpreted as the sine/cosine terms using Eqs. 3-69 and 3-74. From (3-69) we have: α1= 6.0865, β1 = 4.9668, b = 0, and ω=0.262. Eq. 3-74 provides the coefficients of the periodic terms: y (t ) = 2α1e − bt cos ωt + 2β1e −bt sin ωt + ... Substituting coefficients (because b= 0, the exponential terms = 1) y (t ) = 2(6.068) cos ωt + 2(4.9668) sin ωt + ...
or
y (t ) = 12.136 cos ωt + 9.9336 sin ωt + ...
The amplitude of the composite output sinusoidal signal, for large values, of t is given by A = (12.136) 2 + (9.9336) 2 = 15.7
Thus the amplitude of the output is 15.7° for the specified 12° amplitude input.
6-24
6.18
a)
Taking the Laplace transform of the dynamic model in (2-7)
[( γVs + (q + q R )]CT′ 1 ( s) = qCTi′ ( s) + q R CT′ (s)
(1)
[(1 − γ)Vs + (q + q R )]CT′ ( s) = (q + q R )CT′ 1 ( s)
(2)
Substituting for CT′ (s ) from (2) into (1),
CT′ 1 ( s ) q[(1 − γ )Vs + (q + q R )] = CTi′ ( s ) [γVs + (q + q R )][(1 − γ )Vs + (q + q R )] − q R (q + q R ) (1 − γ )V s + 1 (q + q R ) = γ (1 − γ )V 2 2 V s + s + 1 q q(q + q R )
(3)
Substituting for CT′ 1 ( s ) from (3) into (2),
CT′ ( s ) 1 == CTi′ ( s ) γ (1 − γ )V 2 2 V s + s + 1 q q(q + q R ) b)
Case (i), γ → 0 V s + 1 q + qR CT′ 1 ( s ) == CTi′ ( s ) V q s + 1
CT′ ( s ) 1 == CTi′ ( s ) V q s + 1 Case (ii), γ → 1
CT′ ( s ) C ′ ( s) 1 = = T1 CTi′ ( s ) V CTi′ ( s ) q s + 1
6-25
(4)
Case (iii), q R → 0
CT′ ( s ) 1 == , CTi′ ( s ) γ (1 − γ )V 2 2 V s + s + 1 q2 q
CT′ 1 ( s ) 1 == CTi′ ( s ) γV q s + 1
Case (iv), q R → ∞
CT′ ( s ) C ′ ( s) 1 = = T1 CTi′ ( s ) V CTi′ ( s ) q s + 1 c)
Case (i), γ → 0 This corresponds to the physical situation with no top tank. Thus the dynamics for CT are the same as for a single tank, and CT′ 1 ≈ CTi′ for small qR. Case (ii), γ → 1 Physical situation with no bottom tank. Thus the dynamics for CT 1 are the same as for a single tank, and CT = CT 1 at all times. Case (iii), q R → 0 Physical situation with two separate non-interacting tanks. Thus, top tank dynamics, CT 1 , are first order, and bottom tank, CT , is second order. Case (iv), q R → ∞ Physical situation of a single perfectly mixed tank. Thus, CT = CT 1 , and both exhibit dynamics that are the same as for a single tank.
d)
In Eq.(3),
(1 − γ )V ≥0 ( q + q ) R Hence the system cannot exhibit an inverse response. From the denominator of the transfer functions in Eq.(3) and (4),
6-26
1V ζ= 2q
γ (1 − γ )V 2 q(q + q R )
−
1 2
1
(q + q R ) 2 = 4qγ (1 − γ )
Since γ (1 − γ ) ≤ (0.5)(1 − 0.5) for 0 ≤ γ ≤ 1 , 1
(q + q R ) 2 ζ= ≥1 q
Hence, the system is overdamped and cannot exhibit overshoot. e)
Since ζ ≥ 1 , the denominator of transfer function in Eq.(3) and (4) can be written as (τ1 s + 1)(τ 2 s + 1) where, using Eq. 5-45 and 5-46, 1
γ (1 − γ )V 2 2 q( q + q R )
τ1 =
1 2
(q + q R ) (q + q R ) 4qγ (1 − γ ) − 4qγ (1 − γ ) − 1
1 2
1
γ (1 − γ )V 2 2 q(q + q R )
τ2 =
1
1
(q + q R ) 2 (q + q R ) 2 + − 1 4qγ (1 − γ ) 4qγ (1 − γ ) It is given that CTi′ ( s ) =
[
]
h h h 1 − e −t w s = − e − t w s s s s
Then using Eq. 5-48 and (4)
τ e− t / τ1 − τ2 e− t / τ2 cT (t ) = S (t )h 1 − 1 τ1 − τ2 τ1e − (t −t w ) / τ1 − τ 2 e − (t −tw ) / τ 2 − S (t − t w )h 1 − τ1 − τ 2
6-27
And using Eq. 6-15 and (3)
τ − τ1 −t / τ1 τ a − τ 2 −t / τ2 CT 1 (t ) = S (t )h 1 + a e + e τ 2 − τ1 τ1 − τ 2 τ − τ1 −(t −tw ) / τ1 τ a − τ 2 −(t −t w ) / τ 2 e e − S (t − t w )h 1 + a + τ 2 − τ1 τ1 − τ 2 where
(1 − γ )V τa = (q + q R ) The pulse response can be approximated reasonably well by the impulse response in the limit as t w → 0 , keeping htw constant.
6.19
Let
VR = volume of each tank A1 = ρ1Cp1VR A2 = ρ2Cp2VR B1 = w1Cp1 B2 = w2Cp2 K = UA
Then energy balances over the six tanks give dT8 = B2 (T6 − T8 ) + K (T3 − T8 ) dt dT A2 6 = B2 (T4 − T6 ) + K (T5 − T6 ) dt dT4 A2 = B2 (T2 − T4 ) + K (T7 − T4 ) dt A2
(1) (2) (3)
A1
dT7 = B1 (T5 − T7 ) + K (T4 − T7 ) dt
(4)
A1
dT5 = B1 (T3 − T5 ) + K (T6 − T5 ) dt
(5)
6-28
A1
dT3 = B1 (T1 − T3 ) + K (T8 − T3 ) dt
(6)
Define vectors
T ′( s ) = [T8′( s ), T7′ ( s ), T6′ ( s ), T5′( s ), T4′ ( s ), T3′( s )]
T
T ′ ( s ) * T (s) = 2 T1′( s ) Using deviation variables, and taking the Laplace transform of Eqs.1 to 6, we obtain an equation set that can be represented in matrix notation as s I T ′( s ) = A T ′( s ) + B T ( s ) *
(7)
where I is the 6 × 6 identity matrix
− K − B2 A 2 0 0 A= 0 0 K A1
0 0 0 0 B= 0 0 0 0
B2 A2
0
0
0
− K − B2 A2 K A1
B1 A1 K A2 − K − B1 A1
K A1 B2 A2
K A2
0
0
− K − B2 A2
0
0
0
0
− K − B1 A1 0 0
0
0 B1 A1
B2 A2 0
From Eq. 7, T ′( s ) = ( s I − A) −1 B T ( s ) *
6-29
0
0 0 B1 A1 0 − K − B1 A1 K A2
Then
T8′( s ) T ′ ( s ) = 7
1 0 0 0 0 0 * −1 0 1 0 0 0 0 ( s I − A) B T ( s )
6.20
The dynamic model for the process is given by Eqs. 2-45 and 2-46, which can be written as dh 1 = ( wi − w) dt ρA
(1)
w dT Q = i (Ti − T ) + dt ρAh ρAhC
(2)
where h is the liquid-level A is the constant cross-sectional area System outputs: h , T System inputs : w, Q Hence assume that wi and Ti are constant. In Eq. 2, note that the nonlinear dT term h can be linearized as dt
or
h
dT ′ dT + h′ dt dt
h
dT ′ since dt
dT =0 dt
Then the linearized deviation variable form of (1) and (2) is dh′ 1 =− w′ dt ρA dT ′ − wi 1 = T′+ Q′ dt ρAh ρAh C
6-30
Taking Laplace transforms and rearranging, H ′( s ) =0 , Q ′( s )
H ′( s ) K 1 = W ′( s ) s
,
where K 1 = −
1 ; ρA
and K 2 =
Unit-step change in Q: h(t ) = h
K2 T ′( s ) = Q ′( s ) τ 2 s + 1
T ′( s ) =0 , W ′( s ) 1 wi C
,
, τ2 =
ρAh wi
T (t ) = T + K 2 (1 − e − t / τ 2 )
Unit step change in w: h(t ) = h + K 1t , T (t ) = T 6.21 Additional assumptions: (i) The density, ρ, and the specific heat, C, of the process liquid are constant. (ii) The temperature of steam, Ts, is uniform over the entire heat transfer area. (iii) The feed temperature TF is constant (not needed in the solution). Mass balance for the tank is dV = qF − q dt
(1)
Energy balance for the tank is ρC
d [V (T − Tref )] dt
= q F ρC (TF − Tref ) − qρC (T − Tref ) + UA(Ts − T ) (2)
where Tref is a constant reference temperature A is the heat transfer area dV from Eq. 1. Also, replace dt V by AT h (where AT is the tank area) and replace A by pT h (where pT is the perimeter of the tank). Then,
Eq. 2 is simplified by substituting for
6-31
AT
dh = qF − q dt
ρCAT h
(3)
dT = qF ρC (TF − T ) + UpT h(Ts − T ) dt
(4)
Then, Eqs. 3 and 4 constitute the dynamic model for the system. a)
Making Taylor series expansion of nonlinear terms in (4) and introducing deviation variables, Eqs. 3 and 4 become: AT
dh ′ = q ′F − q ′ dt
ρCAT h
(5)
dT ′ = ρC (TF − T )qF′ − (ρCqF + UpT h )T ′ dt + UpT hTs′ + UpT (Ts − T )h′
(6)
Taking Laplace transforms, H ′( s ) =
1 1 QF′ ( s ) − Q′( s ) AT s AT s
ρCAT h ρCqF + UpT h
(7)
ρC (TF − T ) s + 1 T ′( s ) = QF′ ( s ) ρCqF + UpT h
UpT (Ts − T ) UpT h + Ts′( s ) + H ′( s ) ρ Cq + Up h ρ Cq + Up h F T F T
(8)
Substituting for H ′( s ) from (7) into (8) and rearranging gives
ρCAT h ρCqF + UpT h
[ AT s ]
ρC (TF − T ) AT s QF′ ( s ) s + 1 T ′( s ) = ρCqF + UpT h
UpT hAT s UpT (Ts − T ) + Ts′( s ) + [QF′ ( s ) − Q′( s )] ρ Cq + Up h ρ Cq + Up h F T F T Let τ =
ρCAT h ρCqF + UpT h
Then from Eq. 7
6-32
(9)
H ′( s ) 1 = QF′ ( s ) AT s
H ′( s ) 1 =− Q′( s ) AT s
,
,
H ′( s ) =0 Ts′ ( s )
And from Eq. 9 UpT (Ts − T ) ρC (TF − T ) AT T ′( s ) ρCqF + UpT h UpT (Ts − T ) = QF′ ( s ) ( AT s )( τs + 1)
s + 1
UpT (Ts − T ) − ρCqF + UpT h T ′( s ) = Q′( s ) ( AT s )( τs + 1) UpT h T ′( s ) ρCqF + UpT h = Ts′( s ) τs + 1
Note:
τ2 =
ρC (TF − T ) AT UpT (Ts − T )
is the time constant in the numerator.
Because TF − T < 0 (heating) and Ts − T > 0 , τ2 is negative. We can show this property by using Eq. 2 at steady state: ρCqF (TF − T ) = −UpT h (Ts − T )
or ρC (TF − T ) =
−UpT h (Ts − T ) qF
Substituting τ2 = −
hAT qF
Let V = hAT so that τ2 = − For
V = − (initial residence time of tank) qF
T ′( s ) T ′( s ) and the “gain” in each transfer function is QF′ ( s ) Q′( s )
6-33
Up (T − T ) T s K = AT ( ρCqF + UpT h )
and must have the units temp/volume . (The integrator s has units of t-1). To simplify the transfer function gain we can substitute UpT (Ts − T ) = −
ρCqF (TF − T ) h
from the steady-state relation. Then K=
−ρCqFT (TF − T ) hAT ( ρCqF + UpT h )
or K =
T − TF Up h V 1 + T ρCqF
and we see that the gain is positive since T − TF > 0 . Further, it has dimensions of temp/volume. (The ratio
b)
UpT h is dimensionless). ρCqF
h − qF transfer function is an integrator with a positive gain. Liquid level accumulates any changes in qF , increasing for positive changes and viceversa. h − q transfer function is an integrator with a negative gain. h accumulates changes in q, in opposite direction, decreasing as q increases and vice versa.
h − Ts transfer function is zero. Liquid level is independent of Ts , and of the steam pressure Ps . T − q transfer function is second-order due to the interaction with liquid level; it is the product of an integrator and a first-order process.
6-34
T − qF transfer function is second-order due to the interaction with liquid level and has numerator dynamics since qF affects T directly as well if TF ≠ T .
T − Ts transfer function is simple first-order because there is no interaction with liquid level. c)
h − qF : h increases continuously at a constant rate. h − q : h decreases continuously at a constant rate.
h − Ts : h stays constant. T − qF : for TF < T , T decreases initially (inverse response) and then increases. After long times, T increases like a ramp function. T − q : T decreases, eventually at a constant rate.
T − Ts : T increases with a first-order response and attains a new steady state.
6.22 a)
The two-tank process is described by the following equations in deviation variables: dh1' 1 ' 1 ' = w1 − (h1 − h2' dt ρA1 R dh2' 1 = dt ρA2
1 ' ' R (h1 − h2
(1)
(2)
Laplace transforming ρA1 RsH1' ( s ) = RWi ' ( s ) − H1' ( s ) + H 2' ( s )
(3)
ρA2 RsH 2' ( s ) = H1' ( s ) − H 2' ( s )
(4)
6-35
From (4) (ρA2 Rs + 1) H 2' ( s ) = H1' ( s )
(5)
H 2' ( s ) 1 1 = = ' H1 ( s ) ρA2 Rs + 1 τ2 s + 1
(6)
or
where τ2 = ρA2 R Returning to (3) (ρA1 Rs + 1) H1' ( s ) − H 2' ( s ) = RWi ' ( s )
(7)
Substituting (6) with τ1 = ρA1 R
1 ' ' (τ1s + 1) − H1 ( s ) = RWi ( s ) τ s + 1 2
(8)
or (τ1τ2 ) s 2 + (τ1τ2 ) s H1' ( s ) = R (τ2 s + 1)Wi ' ( s ) H1' ( s ) R(τ2 s + 1) = ' W1 ( s ) s [ τ1τ2 s + (τ1 + τ2 )]
(9) (10)
Dividing numerator and denominator by (τ1 + τ2 ) to put into standard form H1' ( s ) [ R /(τ1 + τ2 )](τ2 s + 1) = W1' ( s ) ττ s 1 2 s + 1 τ1 + τ2
(11)
Note that K=
R R 1 1 = = = τ1 + τ2 ρA1 R + ρA2 R ρ( A1 + A2 ) ρA
since A = A1 + A2 Also, let
6-36
(12)
τs =
τ1τ2 ρ2 R 2 A1 A2 ρRA1 A2 = = τ1 + τ2 ρR( A1 + A2 ) A
(13)
so that H1' ( s ) K (τ2 s + 1) = Wi ' ( s ) s (τ3 s + 1)
(14)
and H 2' ( s ) H 2' ( s ) H1' ( s ) 1 K (τ2 s + 1) = ' = ' ' Wi ( s ) H1 ( s ) Wi ( s ) (τ2 s + 1) s (τ3 s + 1) K = s (τ3 s + 1)
(15)
Transfer functions (6), (14) and (15) define the operation of the two-tank process. The single-tank process is described by the following equation in deviation variables:
dh' 1 ' = wi dt ρA
(16)
Note that ω , which is constant, subtracts out. Laplace transforming and rearranging: H ' ( s ) 1/ ρA = Wi ' ( s ) s
(17)
Again K=
1 ρA
H ' ( s) K = Wi ' ( s ) s
(18)
which is the expected integral relationship with no zero.
6-37
b)
For A1 = A2 = A / 2
τ2 = ρAR / 2 τ3 = ρAR / 4
(19)
Thus τ2 = 2τ3 We have two sets of transfer functions: One-Tank Process
Two-Tank Process
H ' ( s) K = Wi ' ( s ) s
H i' ( s ) K (2τ3 s + 1) = Wi ' ( s ) s (τ3 s + 1) H 2' ( s ) K = ' Wi ( s ) s (τ3 s + 1)
Remarks: -
The gain ( K = 1/ ρA) is the same for all TF’s. Also, each TF contains an integrating element. However, the two-tank TF’s contain a pole (τ3 s + 1) that will “filter out” changes in level caused by changing wi(t). On the other hand, for this special case we see that the zero in the first tank transfer function ( H i' ( s ) / Wi ' ( s )) is larger than the pole 2 τ3 > τ 3 and we should make sure that amplification of changes in h1(t) caused by the zero do not more than cancel the beneficial filtering of the pole so as to cause the first compartment to overflow easily. Now look at more general situations of the two-tank case: H1' ( s ) K (ρA2 Rs + 1) K (τ2 s + 1) = = ' Wi ( s ) ρRA1 A2 s (τ3 s + 1) s s + 1 A ' H 2 (s) K = ' Wi ( s ) s (τ3 s + 1) For either A1 → 0 or A2 → 0 , τ3 =
ρRA1 A2 →0 A
6-38
(20)
(21)
Thus the beneficial effect of the pole is lost as the process tends to “look” more like the first-order process. c)
The optimum filtering can be found by maximizing τ3 with respect to A1 (or A2) ρRA1 A2 ρRA1 ( A − A1 ) = A A ∂τ ρR Find max τ3 : 3 = [ ( A − A1 ) + A1 (−1)] ∂A1 A τ3 =
Set to 0:
A − A1 − A1 = 0 2 A1 = A A1 = A / 2
Thus the maximum filtering action is obtained when A1 = A2 = A / 2. The ratio of τ2 / τ3 determines the “amplification effect” of the zero on h1 (t ). τ2 ρA2 R A = = ρ RA A A1 τ3 1 2 A τ As A1 goes to 0, 2 → ∞ τ3
Therefore the influence of changes in wi (t ) on h1 (t ) will be very large, leading to the possibility of overflow in the first tank. Summing up: The process designer would like to have A1 = A2 = A / 2 in order to obtain
a
the maximum filtering of h1 (t ) and h2 (t ). However, the process response should be checked for typical changes in wi (t ) to make sure that h1 does not overflow. If it does, the area A1 needs to be increased until that is not problem. Note that τ2 = τ3 when A1 = A , thus someone must make a careful study (simulations) before designing the partitioned tank. Otherwise, leave well enough alone and use the non-partitioned tank.
6-39
6.23
The process transfer function is Y ( s) K = G( s) = 2 U ( s) (0.1s + 1) (4 s 2 + 2 s + 1)
where K = K1K2 We note that the quadratic term describes an underdamped 2nd-order system since τ2 = 4 → τ=2
a)
2ζτ = 2 → ζ = 0 .5 For the second-order process element with τ2 = 2 and this degree of underdamping (ζ = 0.5) , the small time constant, critically damped 2ndorder process element (τ1 = 0.1 ) will have little effect.
In fact, since 0.1 << τ2 (= 2) we can approximate the critically damped element as e −2τ1 so that G (s) ≈ b)
Ke −0.2 s 4s 2 + 2s + 1
From Fig. 5.11 for ζ = 0.5 , OS ≈ 0.15 or from Eq. 5-53
− πζ Overshoot = exp 1− ζ2
= 0.163
Hence ymax = 0.163 KM + KM = 0.163 (1) (3) + 3 = 3.5
c)
From Fig. 5.4, ymax occurs at t/τ = 3K or tmax = 6.8 for underdamped 2ndorder process with ζ = 0.5 . Adding in effect of time delay t ′ = 6.8 + 0.2 = 7.0
d)
By using Simulink-MATLAB
6-40
τ1 = 0.1 3.5
3
2.5
Output
2
1.5
1
0.5
0 Exact model Approximate model -0.5
0
5
10
15 Time
20
25
30
Fig S6.23a. Step response for exact and approximate model ; τ1 = 0.1
τ1 = 1 3.5
3
2.5
Output
2
1.5
1
0.5
0 Exact model Approximate model -0.5
0
5
10
15
20
25
30
Time
Fig S6.23b. Step response for exact and approximate model ; τ1 = 1
6-41
τ1 = 5 3.5
3
2.5
Output
2
1.5
1
0.5
0 Exact model Approximate model -0.5
0
5
10
15
20
25
30
Time
Fig S6.23c. Step response for exact and approximate model ; τ1 = 5
As noted in plots above, the smaller τ1 is, the better the quality of the approximation. For large values of τ1 (on the order of the underdamped element's time scale), the approximate model fails.
6.24 0
-0.2
-0.4
Output
-0.6
-0.8
-1
-1.2
-1.4
0
50
100
150
200
250
300
350
400
Time
Figure S6.24. Unit step response in blood pressure.
6-42
The Simulink-MATLAB block diagram is shown below
-1 40s+1 Step
Transfer Fcn
Transport Delay1= 30 s Scope
-0.4 40s+1 Step1
Transfer Fcn1
Transport Delay = 75 s
It appears to respond approx. as a first-order or overdamped second-order process with time delay.
6-43
