Unformatted text preview: 1234567898
5.1 a) xDP(t) = hS(t) – 2hS(ttw) + hS(t2tw)
h
xDP (s) = (1 − 2etws + e2tws)
s b) Response of a firstorder process, K h
t s
2t s
Y (s) = (1 − 2e w + e w ) τs + 1 s
α α
or
Y(s) = (1 − 2etw s + e2tw s) 1 + 2 s τs + 1
Kh
Kh
α1 =
= Kh
α2 =
τs + 1 s =0
s s =− 1
τ = − Khτ Kh Khτ Y(s) = (1 − 2etw s + e2tw s) −
τs + 1 s Kh(1−et/τ)
y(t) = , 0 < t < tw Kh(–1 – et/τ + 2e(ttw)/τ) , tw < t < 2tw Kh(–et/τ + 2e(ttw)/τ − e(t2tw)/τ ) , 2tw < t Response of an integrating element,
Y (s) = K h
(1 − 2etw s + e2tw s)
s s Kht c) 0 < t < tw Kh(t + 2 tw) , tw < t < 2tw 0 y(t) = , , 2tw < t This input gives a response, for an integrating element, which is zero after
a finite time. Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp. 51 5.2 a) For a step change in input of magnitude M
y(t) = KM (1 et/τ) + y(0)
We note that KM = y(∞) – y(0) = 280 – 80 = 200°C
Then K = 200 1 C
= 400 °C/Kw
0.5Kw At time t = 4, 230 − 80
= 1 − e − 4 / τ or τ = 2.89 min
280 − 80 Thus T ′( s )
400
=
[°C/Kw]
P ′( s ) 2.89 s + 1 ∴ For an input ramp change with slope a = 0.5 Kw/min
Ka = 400 × 0.5 = 200 °C/min
This maximum rate of change will occur as soon as the transient has died
out, i.e., after
5 × 2.89 min ≈ 15 min have elapsed.
1500 1000 T' a) y(4) = 230 °C 500 0 0 1 2 3 4 5 6 7 8 9 10 time(min) Fig S5.2. Temperature response for a ramp input of magnitude 0.5 Kw/min. 52 5.3 The contaminant concentration c increases according to this expression:
c(t) = 5 + 0.2t
Using deviation variables and Laplace transforming,
c′(t ) = 0.2t C ′( s ) = or 0 .2
s2 Hence
′
C m ( s) = 1
0 .2
⋅ 2
10 s + 1 s and applying Eq. 521
′
cm (t ) = 2(e− t /10 − 1) + 0.2t ′
As soon as cm (t ) ≥ 2 ppm the alarm sounds. Therefore,
∆t = 18.4 s (starting from the beginning of the ramp input) The time at which the actual concentration exceeds the limit (t = 10 s) is
subtracted from the previous result to obtain the requested ∆t .
∆t = 18.4 − 10.0 = 8.4 s
2.5 2 c'm 1.5 1 0.5 0 0 2 4 6 8 10 12 14 16 18 20 time
Fig S5.3. Concentration response for a ramp input of magnitude 0.2 Kw/min. 53 5.4 a) Using deviation variables, the rectangular pulse is
0
2
0 c′ =
F t<0
0≤t<2
2≤t≤∞ Laplace transforming this input yields
c′ (s) =
F ( 2
1 − e −2s
s ) The input is then given by c ′( s ) = 8
8e −2 s
−
s (2s + 1) s (2s + 1) and from Table 3.1 the time domain function is
c ′(t ) = 8(1 − e −t / 2 ) − 8(1 − e − (t −2 ) / 2 ) S (t − 2) (1) 6 5 C' 4 3 2 1 0 0 2 4 6 8 10 12 14 16 18 20 time Fig S5.4. Exit concentration response for a rectangular input. b) By inspection of Eq. 1, the time at which this function will reach its
maximum value is 2, so maximum value of the output is given by
54 c ′(2) = 8(1 − e −1 ) − 8(1 − e −0 / 2 ) S (0) (2) and since the second term is zero, c ′(2) = 5.057
c) By inspection, the steady state value of c ′(t ) will be zero, since this is a
firstorder system with no integrating poles and the input returns to zero.
To obtain c ′(∞) , simplify the function derived in a) for all time greater
than 2, yielding
c ′(t ) = 8(e − (t −2) / 2 − e −t / 2 ) (3) which will obviously converge to zero.
Substituting c ′(t ) = 0.05 in the previous equation and solving for t gives
t = 9.233 5.5 a) Energy balance for the thermocouple,
mC dT
= hA(Ts − T )
dt (1) where m is mass of thermocouple
C is heat capacity of thermocouple
h is heat transfer coefficient
A is surface area of thermocouple
t is time in sec
Substituting numerical values in (1) and noting that
Ts = T and 15 dT dT ′
=
,
dt
dt dT ′
= Ts′ − T ′
dt Taking Laplace transform, T ′( s )
1
=
Ts′( s ) 15s + 1 55 Ts(t) = 23 + (80 − 23) S(t)
Ts = T = 23 From t = 0 to t = 20,
Ts′(t ) = 57 S(t)
T ′( s ) = , Ts′( s ) = 57
s 1
57
Ts′( s ) =
15s + 1
s (15s + 1) Applying inverse Laplace Transform,
T ′(t ) = 57(1 − e −t / 15 )
Then
T (t ) = T ′(t ) + T = 23 + 57(1 − e −t / 15 )
Since T(t) increases monotonically with time, maximum T = T(20).
Maximum T(t) = T(20) = 23 + 57 (1e20/15) = 64.97 °C 50 45
41.97 º
40 35 30 25
T' b) 20 15 10 5 0 0 5 10 15 20 25 30 35 40 time Fig S5.5. Thermocouple output for part b) 56 45 50 5.6 a) The overall gain of G is G s =0
= b) K1
K2
⋅
= K1 K 2
τ1 × 0 + 1 τ 2 × 0 + 1 If the equivalent time constant is equal to τ1 + τ2 = 5 + 3 = 8, then
y(t = 8)/KM = 0.632 for a firstorder process. 5e −8 / 5 − 3e −8 / 3
y(t = 8)/KM = 1 −
= 0.599 ≠ 0.632
5−3
Therefore, the equivalent time constant is not equal to τ1 + τ2
c) The roots of the denominator of G are
1/τ1 and 1/τ2
which are negative real numbers. Therefore the process transfer function
G cannot exhibit oscillations when the input is a step function. 5.7 Assume that at steady state the temperature indicated by the sensor Tm is
equal to the actual temperature at the measurement point T. Then,
′
Tm ( s )
K
1
=
=
T ′( s ) τs + 1 1.5s + 1
Tm = T = 3501 C ′
Tm (t ) = 15sin ωt
where ω =2π × 0.1 rad/min = 0.628 rad/min
At large times when t/τ >>1, Eq. 526 shows that the amplitude of the
sensor signal is 57 Am = A
ω2 τ2 + 1 where A is the amplitude of the actual temperature at the measurement
point.
Therefore A = 15 (0.628)2 (1.5) 2 + 1 = 20.6°C Maximum T = T + A =350 + 20.6 = 370.6
Maximum Tcenter = 3 (max T) – 2 Twall
= (3 × 370.6)−(2 × 200) = 711.8°C
Therefore, the catalyst will not sinter instantaneously, but will sinter if
operated for several hours. 5.8 a) Assume that q is constant. Material balance over the tank,
A dh
= q1 + q 2 − q
dt Writing in deviation variables and taking Laplace transform ′
′
AsH ′( s ) = Q1 ( s ) + Q2 ( s )
H ′( s )
1
=
Q1′ ( s ) As b) ′
q1 (t ) = 5 S(t) – 5S(t12)
5 5 −12 s
− e
s s
1
5/ A 5/ A
H ′( s ) =
Q1′ ( s ) = 2 − 2 e −12 s
As
s
s Q1′ ( s ) = 58 5
5
t S(t) − (t − 12) S(t12)
A
A h ′(t ) = 4+ 5
t = 4 + 0.177t
A 0 ≤ t ≤ 12 h(t) =
5 4 + × 12 = 6.122
A 12 < t 2.5 2 h'(t) 1.5 1 0.5 0 0 5 10 15 20 25 30 35 40 45 50 time Fig S5.8a. Liquid level response for part b) c) h = 6.122 ft at the new steady state t ≥ 12 d) ′
q1 (t ) = 5 S(t) – 10S(t12) + 5S(t24) ; tw = 12 ( ) 5
1 − 2e −12 s + e − 24 s
s
5 / A 10 / A
5/ A
H ′( s ) = 2 − 2 e −12 s + 2 e − 24 s
s
s
s Q1′ ( s ) = h(t) = 4 + 0.177tS(t) − 0.354(t12)S(t12) + 0.177(t24)S(t24)
For t ≥ 24
h = 4 + 0.177t − 0.354(t − 12) + 0.177(t − 24) = 4 ft at t ≥ 24
59 2.5 2 h'(t) 1.5 1 0.5 0 0.5 0 5 10 15 20 25 30 35 40 time Fig S5.8b. Liquid level response for part d) 5.9 a) Material balance over tank 1.
A dh
= C (qi − 8.33h)
dt where A = π × (4)2/4 = 12.6 ft2
C = 0.1337 ft 3 /min
USGPM AsH ′( s ) = CQi′ ( s ) − (C × 8.33) H ′( s )
H ′( s )
0.12
=
Qi′ ( s ) 11.28s + 1 For tank 2,
A dh
= C (qi − q)
dt 510 45 50 AsH ′( s ) = CQi′( s )
b) H ′( s ) 0.011
=
Qi′ ( s )
s , Qi′( s ) = 20 / s
For tank 1, H ′( s ) = 2.4
2.4
27.1
=
−
s (11.28s + 1)
s 11.28s + 1 h(t) = 6 + 2.4(1 – et/11.28)
For tank 2, H ′( s ) = 0.22 / s 2
h(t) = 6 + 0.22t 9 8 7 6 h'(t) 5 4 3 2 1
Tank 1
Tank2
0 0 5 10 15 20 25 30 35 40 time Fig S5.9. Transient response in tanks 1 and 2 for a step input. c) h(∞) = 6 +2.4 – 0 = 8.4 ft For tank 2,
d) For tank 1, h(∞) = 6 + (0.22 × ∞) = ∞ ft For tank 1, 8 = 6 + 2.4(1 – et/11.28)
h = 8 ft at t = 20.1 min
8 = 6 + 0.22t
h = 8 ft at t = 9.4 min For tank 2, Tank 2 overflows first, at 9.4 min. 511 5.10 a) The dynamic behavior of the liquid level is given by
d 2 h′
dh ′
+A
+ Bh ′ = C p ′(t )
dt
dt
where
A= 6µ
R 2 B= 3g
2L and C = 3
4ρL Taking the Laplace Transform and assuming initial values = 0
s 2 H ′( s ) + AsH ′( s ) + BH ′( s ) = C P ′( s )
or H ′( s ) = C/B
P ′( s )
1 2 A
s + s +1
B
B We want the previous equation to have the form
H ′( s ) = K
P ′( s )
τ s + 2ζτs + 1
2 Hence K = C/B = 2 1
2ρg 1
τ =
B 2L then τ = 1 / B = 3g A
2ζτ =
B 3µ 2 L then ζ = 2 R ρ 3g 2 b) 1/ 2 1/ 2 The manometer response oscillates as long as 0 < ζ < 1 or
1/ 2 0 <
b) 3µ 2 L R 2ρ 3 g < 1 If ρ is larger , then ζ is smaller and the response would be more
oscillatory.
If µ is larger, then ζ is larger and the response would be less oscillatory. 512 5.11 Y(s) = K
K2
KM
= 21 +
s (τs + 1)
s (τs + 1) s
2 K1τs + K1 + K2s = KM
K1 = KM
K2 = −K1τ = − KMτ
Hence
Y(s) =
or KM
KMτ
−
2
s (τs + 1)
s y(t) = KMt − KMτ (1et/τ) After a long enough time, we can simplify to
y(t) ≈ KMt  KMτ (linear) slope = KM
intercept = −KMτ
That way we can get K and τ y(t)
Slope = KM −ΚΜτ Figure S5.11. Time domain response and parameter evaluation 513 5.12 a) 11 + Ky + 4 y = x
1
y
1
Assuming y(0) = y (0) = 0 Y (s)
1
0.25
= 2
=
2
X ( s ) s + Ks + 4 0.25s + 0.25 Ks + 1
b) Characteristic equation is
s2 + Ks + 4 = 0
The roots are s = − K ± K 2 − 16
2 10 ≤ K < 4 Roots : positive real, distinct
Response : A + B e t / τ1 + C et / τ 2
K = 4 Roots : positive real, repeated
Response : A + Bet/τ + C et/τ 4 < K < 0 Roots: complex with positive real part.
t
t
Response: A + eζt/τ (B cos 1 − ζ 2
+ C sin 1 − ζ 2 )
τ
τ K=0 Roots: imaginary, zero real part.
Response: A + B cos t/τ + C sin t/τ 0<K<4 Roots: complex with negative real part.
t
t
Response: A + eζt/τ (B cos 1 − ζ 2
+ C sin 1 − ζ 2 )
τ
τ K=4 Roots: negative real, repeated.
Response: A + Bet/τ + C t et/τ 4 < K ≤ 10 Roots: negative real, distinct
Response: A + B e −t / τ1 + C e −t / τ 2 Response will converge in region 0 < K ≤ 10, and will not converge in
region –10 ≤ K ≤ 0
514 5.13 a) The solution of a criticallydamped secondorder process to a step change
of magnitude M is given by Eq. 550 in text. t y(t) = KM 1 − 1 + e −t / τ τ Rearranging
y t
= 1 − 1 + e − t / τ
KM τ 1 + t −t / τ
y
= 1−
e
τ
KM When y/KM = 0.95, the response is 0.05 KM below the steadystate value. KM
0.95KM
y 0 ts t s −t / τ
= 1 − 0.95 = 0.05
1 + e
τ t t
ln1 + s − s = ln(0.05) = −3.00
τ τ t t
Let E = ln1 + s − s + 3
τ τ 515 time and find value of ts
that makes E ≈ 0 by trialanderror.
τ
E
0.6094
0.2082
0.2047
0.0008 ts/τ
4
5
4.5
4.75 ∴
b) a value of t = 4.75τ is ts, the settling time. Y(s) = a
a a2
a4
Ka
= 1+ 2 + 3 +
2
s s
τs + 1 (τs + 1) 2
s (τs + 1)
2 We know that the a3 and a4 terms are exponentials that go to zero for large
values of time, leaving a linear response.
a2 = lim
s →0 Define Q(s) = Ka
= Ka
(τs + 1) 2 Ka
(τs + 1) 2 dQ − 2 Kaτ
=
ds (τs + 1) 3 Then a1 = − 2 Kaτ 1
lim 1! s →0 (τs + 1) 3 (from Eq. 362)
a1 = − 2 Kaτ
∴ the longtime response (after transients have died out) is
y 2 (t ) = Kat − 2 Kaτ = Ka (t − 2τ)
= a (t − 2τ)
for K = 1
and we see that the output lags the input by a time equal to 2τ. 516 2τ
y x=at 0 yl =a(t2τ) actual response time 5.14 a) 11.2mm − 8mm
= 0.20mm / psi
31psi − 15psi Gain = 12.7 mm − 11.2mm
= 0.47
11.2mm − 8mm − πζ = 0.47
Overshoot = exp , 1− ζ2 2πτ = 2.3 sec
Period = 1− ζ2 Overshoot = τ = 2.3 sec × ζ = 0.234 1 − 0.234 2
= 0.356 sec
2π R ′( s )
0.2
=
2
P ′( s ) 0.127 s + 0.167 s + 1
b) (1) From Eq. 1, taking the inverse Laplace transform,
11
1
0.127 R ′ + 0.167 R ′ + R ′ = 0.2 P ′
11 11
R′ = R 1
1
R′ = R R ′ = R8 11
1
0.127 R + 0.167 R + R = 0.2 P + 5
11
1
R + 1.31 R + 7.88 R = 1.57 P + 39.5 517 P ′ = P15 5.15
P ′( s )
3
=
2 2
T ′( s ) (3) s + 2(0.7)(3) s + 1 [ºC/kW] Note that the input change p ′(t ) = 26 − 20 = 6 kw
Since K is 3 °C/kW, the output change in going to the new steady state
will be T ′ = (31 C / kW )(6 kW ) = 18 1 C t →∞ a) Therefore the expression for T(t) is Eq. 551
0 .7 t 1 − ( 0 .7 ) 2
− 3 T (t ) = 70 + 18 1 − e
cos 3 1 1 1 − (0.7 ) 2
0. 7
+
t
sin 2 τ
1 − ( 0 .7 ) 25 20 T'(t) 15 10 5 0 0 5 10 15 20 25 30 35 40 45 50 time Fig S5.15. Process temperature response for a step input b) The overshoot can obtained from Eq. 552 or Fig. 5.11. From Figure 5.11
we see that OS ≈ 0.05 for ζ=0.7. This means that maximum temperature is
Tmax ≈ 70° + (18)(1.05) = 70 + 18.9 = 88.9°
From Fig S5.15 we obtain a more accurate value.
518 t The time at which this maximum occurs can be calculated by taking
derivative of Eq. 551 or by inspection of Fig. 5.8. From the figure we see
that t / τ = 3.8 at the point where an (interpolated) ζ=0.7 line would be.
∴ tmax ≈ 3.8 (3 min) = 11.4 minutes 5.16 For underdamped responses, 1 − ζ2 − ζt / τ
cos
y (t ) = KM 1 − e τ a) 1 − ζ2
ζ
+
sin t
2 τ
1− ζ t At the response peaks,
2 dy ζ −ζ t / τ 1 − ζ
cos = KM e
dt τ
τ −e −ζt / τ 1− ζ2
ζ
t+
sin τ
1 − ζ2 1− ζ2 1− ζ2
−
sin τ
τ ζ 1 − ζ2
t + cos τ τ t t = 0 Since KM ≠ 0 and e − ζt / τ ≠ 0
2 ζ ζ 1− ζ
0 = − cos
τ τ τ ζ2
1 − ζ2
t + + τ 1 − ζ2
τ 1− ζ2 πτ
0 = sin
t = sin nπ , t = n τ 1 − ζ2 where n is the number of the peak.
Time to the first peak,
b) Overshoot, OS = tp = πτ
1 − ζ2 y (t p ) − KM
KM 519 1 − ζ2 sin τ t (551) ζ − ζt OS = − exp
sin(π) cos(π) + τ 1 − ζ2 − ζτπ − πζ = exp = exp 2
2
τ 1− ζ 1− ζ c) Decay ratio, DR =
where y (t 3 p ) = KM e DR = d) y (t p ) − KM 3πτ
1− ζ2 − ζt 3 p / τ KM e y (t 3 p ) − KM − ζt p / τ is the time to the third peak. ζ 2πτ ζ = exp − (t 3 p − t p ) = exp − 1 − ζ 2 τ τ −2πζ = exp = (OS) 2
2 1− ζ Consider the trigonometric identity
sin (A+B) = sin A cos B + cos A sin B 1− ζ2 Let B = t , sin A = 1 − ζ 2 ,
cos A = ζ τ 1 y (t ) = KM 1 − e −ζt / τ
1 − ζ 2 cos B + ζ sin B 1− ζ2 − ζt / τ e = KM 1 −
sin( A + B)
2 1− ζ [ Hence for t ≥ t s , the settling time,
e − ζt / τ
1− ζ 2 ≤ 0.05 , or Therefore, ts ≥ ( t ≥ − ln 0.05 1 − ζ 2 τ 20
ln
ζ 1 − ζ2 520 τ
)ζ 5.17 a) Assume underdamped secondorder model (exhibits overshoot)
K= 1756456
10 − 6 ft
ft
=
= 0 .2
123456 140 − 120 gal/min
gal/min Fraction overshoot = 11 − 10 1
= = 0.25
10 − 6 4 From Fig 5.11, this corresponds (approx) to ζ = 0.4
From Fig. 5.8 , ζ = 0.4 , we note that tp/τ ≈ 3.5
Since tp = 4 minutes (from problem statement), τ = 1.14 min ∴ G p(s) = 0.2
0 .2
=
2
(1.14) s + 2(0.4)(1.14) s + 1 1.31s + 0.91s + 1
2 2 b) In Chapter 6 we see that a 2ndorder overdamped process model with a
numerator term can exhibit overshoot. But if the process is underdamped,
it is unique. a) Assuming constant volume and density, 5.18 Overall material balances yield: q2 = q1 = q (1) Component material balances:
dc1
= q (ci − c1 )
dt
dc
V2 2 = q (c1 − c 2 )
dt (2) V1 b) (3) Degrees of freedom analysis
3 Parameters : V1, V2, q
521 3 Variables : ci, c1, c2
2 Equations: (2) and (3)
NF = NV − NE = 3 − 2 = 1
Hence one input must be a specified function of time.
2 Outputs = c1, c2
1 Input = ci
c) If a recycle stream is used Overall material balances:
q1 = (1+r)q (4) q2 = q1 = (1+r)q (5) q3 = q2 – rq = (1+r)q − rq = q (6) Component material balances:
V1 dc1
= qci + rqc 2 − (1 + r )qc1
dt (7) V2 dc 2
= (1 + r )qc1 − (1 + r )qc 2
dt (8) Degrees of freedom analysis is the same except now we have
4 parameters : V1, V2, q, r 522 d) If r → ∞ , there will a large amount of mixing between the two tanks as a
result of the very high internal circulation.
Thus the process acts like ci
q
c2
q Total Volume = V1 + V2
Model :
dc 2
= q (c i − c 2 )
dt
c1 = c2 (complete internal mixing) (V1 +V2) Degrees of freedom analysis is same as part b) 5.19 a) For the original system, dh1
h
= Cqi − 1
dt
R1
dh
h
h
A2 2 = 1 − 2
dt
R1 R2
A1 where A1 = A2 = π(3)2/4 = 7.07 ft2 ft 3 /min
gpm
h
2.5
ft
= 0.187 3
R1 = R2 = 1 =
Cqi 0.1337 × 100
ft /min
C = 0.1337 523 (9)
(10) Using deviation variables and taking Laplace transforms,
H 1′ ( s )
=
Qi′ ( s ) C = CR1
0.025
=
A1 R1 s + 1 1.32 s + 1 1
R1
′
H 2 ( s)
1 / R1
R2 / R1
1
=
=
=
1
′
H 1 (s)
A2 R2 s + 1 1.32 s + 1
A2 s +
R2
′
H 2 ( s)
0.025
=
Qi′( s ) (1.32s + 1) 2
A1 s + For step change in qi of magnitude M,
h1′max = 0.025M
′
h2 max = 0.025M since the secondorder transfer function
0.025
is critically damped (ζ=1), not underdamped
(1.32 s + 1) 2
2.5 ft
Hence Mmax =
= 100 gpm
0.025 ft/gpm For the modified system,
A dh
h
= Cq i −
dt
R A = π(4) 2 / 4 = 12.6 ft 2
V = V1 + V2 = 2 × 7.07ft 2 × 5ft = 70.7ft3
hmax = V/A = 5.62 ft
R= h
Cq i H ′( s )
=
Qi′ ( s ) = 0.5 × 5.62
ft
= 0.21 3
0.1337 × 100
ft /min
C As + 1
R = CR
0.0281
=
ARs + 1 2.64 s + 1 ′
hmax = 0.0281M
2.81 ft
Mmax =
= 100 gpm
0.0281 ft/gpm 524 Hence, both systems can handle the same maximum step disturbance in qi.
b) For step change of magnitude M, Qi′( s ) = M
s For original system,
′
Q2 ( s ) = 1
1
0.025 M
′
H 2 ( s) =
R2
0.187 (1.32 s + 1) 2 s 1 1.32
1.32
= 0.134M −
−
2 s (1.32s + 1) (1.32s + 1) t −t / 1.32 q ′ (t ) = 0.134 M 1 − 1 +
e
2 1.32 For modified system,
Q ′( s ) = 1
1
0.0281 M
2.64 1
H ′( s ) =
= 0.134 M − R
0.21 (2.64 s + 1) s s 2.64 s + 1 [ q ′(t ) = 0.134 M 1 − e −t / 2.64 ′
Original system provides better damping since q 2 (t ) < q ′(t ) for t < 3.4. 5.20 a) Caustic balance for the tank,
ρV dC
= w1c1 + w2 c 2 − wc
dt Since V is constant, w = w1 + w2 = 10 lb/min
For constant flows, ′
ρVsC ′( s ) = w1C1′ ( s ) + w2 C 2 ( s ) − wC ′( s )
w1
C ′( s )
5
0.5
=
=
=
′
C1 ( s ) ρVs + w (70)(7) s + 10 49s + 1 525 ′
C m (s)
K
,
=
C ′( s ) τs + 1 K = (30)/3 = 1 τ ≈ 6 sec = 0.1 min , (from the graph) ′
C m ( s)
1
0.5
0.5
=
=
C1′ ( s ) (0.1s + 1) (49s + 1) (0.1s + 1)(49s + 1)
b) 3
s C1′ ( s ) = 1.5
s (0.1s + 1)(49 s + 1) 1
c ′ (t ) = 1.51 +
(0.1e −t / 0.1 − 49e −t / 49 )
m (49 − 0.1) ′
C m ( s) = c) ′
C m ( s) = 0.5 3
1.5
=
(49 s + 1) s s (49 s + 1) ( c ′ (t ) = 1.5 1 − e − t / 49
m The responses in b) and c) are nearly the same. Hence the dynamics of the
conductivity cell are negligible.
1.5 1 Cm'(t) d) ) 0.5 Part b)
Part c)
0 0 20 40 60 80 100
time 120 140 160 Fig S5.20. Step responses for parts b) and c) 526 180 200 5.21 Assumptions: a) 1) Perfectly mixed reactor
2) Constant fluid properties and heat of reaction Component balance for A,
V dc A
= q (c A i − c A ) − Vk (T )c A
dt (1) Energy balance for the tank,
ρVC dT
= ρqC (Ti − T ) + (− ∆H R )Vk (T )c A
dt (2) Since a transfer function with respect to cAi is desired, assume the other
inputs, namely q and Ti, are constant. Linearize (1) and (2) and note that
dc A dc ′A
dT dT ′
=
,
=
,
dt
dt
dt
dt
V dc ′
20000
A
= qc ′ i − (q + Vk (T ))c ′ − Vc A k (T )
T′
A
A
dt
T2 ρVC (3) dT ′
20000 = − ρqC + ∆H RVc A k (T )
T ′ − ∆H RVk (T )c ′ (4)
A
dt
T2 Taking Laplace transforms and rearranging [Vs + q + Vk (T )]C ′ (s) = qC ′ (s) − Vc
A Ai A k (T ) 20000
T ′( s )
T2 (5) 20000 ′
ρVCs + ρqC − (−∆H R )Vc A k (T ) T 2 T ′( s ) = (−∆H R )Vk (T )C A ( s ) (6) Substituting C ′ (s ) from Eq. 5 into Eq. 6 and rearranging,
A
T ′( s )
=
′
C A( s )
i (−∆H R )Vk (T )q
20000 20000 2
2
Vs + q + Vk (T ) ρVCs + ρqC − (−∆H R )Vc A k (T )
2 + (−∆H R )V c A k (T ) T 2
T (7) c A is obtained from Eq. 1 at steady state, 527 cA = qc Ai
= 0.001155 lb mol/cu.ft.
q + Vk (T ) Substituting the numerical values of T , ρ, C, –∆HR, q, V, c A into Eq. 7
and simplifying, T ′( s )
11.38
=
C ′ i ( s ) (0.0722s + 1)(50s + 1)
A
For step response, C ′ ( s ) = 1 / s
Ai T ′( s ) = 11.38
s (0.0722s + 1)(50s + 1) 1
T ′(t ) = 11.381 +
(0.0722e −t / 0.0722 − 50e −t / 50 ) (50 − 0.0722) A firstorder approximation of the transfer function is T ′( s )
11.38
=
C ′ i ( s ) 50s + 1
A
For step response, T ′( s ) = [ 11.38
or T ′(t ) = 11.38 1 − e −t / 50
s (50s + 1) The two step responses are very close to each other hence the
approximation is valid.
12 10 T'(t) 8 6 4 2 Using transfer function
Using firstorder approximation
0 0 20 40 60 80 100
time 120 140 160 180 200 Fig S5.21. Step responses for the 2nd order t.f and 1st order approx. 528 5.22 (τas+1)Y1(s) = K1U1(s) + Kb Y2(s)
(τbs+1)Y2(s) = K2U2(s) + Y1(s)
a) (1)
(2) Since the only transfer functions requested involve U1(s), we can let U2(s)
be zero. Then, substituting for Y1(s) from (2)
Y1(s) = (τbs+1)Y2(s) (3) (τas+1)(τbs+1)Y2(s) =K1U1(s) + KbY2(s) (4) Rearranging (4)
[(τas+1)(τbs+1) − Kb]Y2(s) =K1U1(s) Y2 ( s )
K1
=
U 1 ( s ) (τ a s + 1)(τ b s + 1) − K b
Also, since ∴ (5) Y1 ( s )
= τb s + 1
Y2 ( s ) (6) From (5) and (6) K 1 (τ b s + 1)
Y1 ( s ) Y2 ( s ) Y1 ( s )
=
×
=
U 1 ( s ) U 1 ( s ) Y2 ( s ) (τ a s + 1)(τ b s + 1) − K b
b) (7) The gain is the change in y1(or y2) for a unit step change in u1. Using the
FVT with U1(s) = 1/s. K1
y 2 (t → ∞) = lim s
s →0 (τ a s + 1)(τ b s + 1) − K b K1
1
=
s 1 − Kb This is the gain of TF Y2(s)/U1(s).
Alternatively, Y (s) K1
K1
K = lim 2 = lim =
s →0 U ( s ) 1 s →0 (τ a s + 1)(τ b s + 1) − K b 1 − K b
For Y1(s)/U1(s) 529 K 1 (τ b s + 1)
y1 (t → ∞) = lim s
s →0 (τ a s + 1)(τ b s + 1) − K b K1
1
=
s 1 − Kb In other words, the gain of each transfer function is
c) K1
1 − Kb Y2 ( s )
K1
=
U 1 ( s ) (τ a s + 1)(τ b s + 1) − K b (5) Secondorder process but the denominator is not in standard form, i.e.,
τ2s2+2ζτs+1
Put it in that form Y2 ( s )
K1
=
2
U 1 ( s) τ a τ b s + (τ a + τ b ) s + 1 − K b (8) Dividing through by 1 Kb
K 1 /(1 − K b )
Y2 ( s )
=
τ a τ b 2 (τ a + τ b )
U 1 (s)
s +
s +1
1 − Kb
1 − Kb (9) Now we see that the gain K = K1/(1Kb), as before τ2 = τa τ b
1 − Kb 2ζτ = ζ= τ= τa τb
1 − Kb (10) τa + τb
, then
1 − Kb 1 τa + τb
2 1 − Kb 1 − K b 1 τa + τb = τa τb 2 τa τb 1
1 − Kb (11) Investigating Eq. 11 we see that the quantity in brackets is the same as ζ
for an overdamped 2ndorder system (ζOD) [ from Eq. 543 in text].
ζ= ζ OD
1 − Kb where ζ OD = 1 τa + τ b
2 τa + τb 530 (12) Since ζOD>1,
ζ>1, for all 0 < Kb < 1.
In other words, since the quantity in brackets is the value of ζ for an
overdamped system (i.e. for τa ≠ τb is >1) and 1 − K b <1 for any positive
Kb, we can say that this process will be more overdamped (larger ζ) if Kb
is positive and <1.
For negative Kb we can find the value of Kb that makes ζ = 1, i.e., yields a
criticallydamped 2ndorder system.
ζ OD ζ =1= (13) 1 − K b1 ζ
or 1 = OD
1 − K b1
2 1 – Kb1 = ζOD2
Kb1 = 1 − ζOD2 (14) where
Kb1 < 0 is the value of Kb that yields a criticallydamped process.
Summarizing, the system is overdamped for 1 − ζOD2 < Kb < 1.
Regarding the integrator form, note that Y2 ( s )
K1
=
2
U 1 ( s) τ a τ b s + (τ a + τ b ) s + 1 − K b (8) For Kb = 1 Y2 ( s )
K1
K1
=
=
2
U 1 ( s ) τ a τ b s + (τ a + τ b ) s s[τ a τ b s + (τ a + τ b )]
K 1 /(τ a + τ b ) τ τ s a b s + 1 τa + τ b ′
K1
which has the form =
( s indicates presence of integrator)
s (τ′s + 1)
= 531 d) Return to Eq. 8
System A: Y2 ( s )
K1
2K
1
=
= 2 1
= 2
2
U 1 ( s ) (2)(1) s + (2 + 1) s + 1 − 0.5 4s + 6s + 1 4s + 6s + 1
τ2 = 4
2ζτ = 6 →
→ τ=2
ζ = 1.5 System B:
For system 1
1
= 2
(2 s + 1)( s + 1) 2 s + 3s + 1
τ22 = 2 → τ2 = 2ζ2τ2 = 3 → ζ2 = 2
3
2 2 = 1.5
2 ≈ 1.05 Since system A has larger τ (2 vs. 2 ) and larger ζ (1.5 vs 1.05), it will
respond slower. These results correspond to our earlier analysis. 532 ...
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 Spring '08
 Cummings
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