1234567898 5.1
a)
xDP(t) = hS(t) – 2hS(t-tw) + hS(t-2tw) h xDP (s) = (1 − 2e-tws + e-2tws) s
b)
Response of a first-order process, K h -t s -2t s Y (s) = (1 − 2e w + e w ) τs + 1 s α α or Y(s) = (1 − 2e-tw s + e-2tw 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 − 2e-tw s + e-2tw s) − τs + 1 s
y(t) =
Kh(1−e-t/τ)
,
0 < t < tw
Kh(–1 – e-t/τ + 2e-(t-tw)/τ)
,
tw < t < 2tw
Kh(–e-t/τ + 2e-(t-tw)/τ − e-(t-2tw)/τ ) ,
2tw < t
Response of an integrating element, Y (s) =
y(t) =
c)
K h (1 − 2e-tw s + e-2tw s) s s
Kht
,
0 < t < tw
Kh(-t + 2 tw)
,
tw < t < 2tw
0
,
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.
5-1
5.2
a)
For a step change in input of magnitude M y(t) = KM (1- e-t/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.
5-2
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. 5-21 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.
5-3
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 ′F ( s ) =
(
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 5-4
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 first-order 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
5-5
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 (1-e-20/15) = 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)
5-6
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 first-order 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. 5-26 shows that the amplitude of the sensor signal is
5-7
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(t-12) 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 ) =
5-8
5 5 t S(t) − (t − 12) S(t-12) 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(t-12) + 5S(t-24) ; 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(t-12)S(t-12) + 0.177(t-24)S(t-24) For t ≥ 24 h = 4 + 0.177t − 0.354(t − 12) + 0.177(t − 24) = 4 ft at t ≥ 24 5-9
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
5-10
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 – e-t/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)
d)
For tank 1,
h(∞) = 6 +2.4 – 0 = 8.4 ft
For tank 2,
h(∞) = 6 + (0.22 × ∞) = ∞ ft
For tank 1,
8 = 6 + 2.4(1 – e-t/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.
5-11
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 2ρ
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 =
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)
2
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.
5-12
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 (1-e-t/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
5-13
5.12
a)
1y1 + Ky1 + 4 y = x
Assuming y(0) = y1 (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
Roots: complex with negative real part. t t Response: A + e-ζt/τ (B cos 1 − ζ 2 + C sin 1 − ζ 2 ) τ τ
K=4
Roots: negative real, repeated. Response: A + Be-t/τ + C t e-t/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 5-14
5.13
a)
The solution of a critically-damped second-order process to a step change of magnitude M is given by Eq. 5-50 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 steady-state 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
5-15
time
and find value of
ts that makes E ≈ 0 by trial-and-error. 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 a a4 Ka = 1 + 22 + 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. 3-62) a1 = − 2 Kaτ ∴ the long-time 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.
5-16
2τ y
x=at
0
yl =a(t-2τ)
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′ + 0.167 R1 ′ + R ′ = 0.2 P ′ 0.127 R 11′ = R 11 R
R1 ′ = R1
R ′ = R-8
11 + 0.167 R1 + R = 0.2 P + 5 0.127 R 11 + 1.31 R1 + 7.88 R = 1.57 P + 39.5 R
5-17
P ′ = P-15
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. 5-51 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. 5-52 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. 5-18
t
The time at which this maximum occurs can be calculated by taking derivative of Eq. 5-51 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 = KM e cos 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
5-19
1 − ζ2 sin τ
t
(5-51)
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 − 2 τ 1 − ζ 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 e − ζt / τ = KM 1 − sin( A + B) 1 − ζ2
[
]
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 − ζ 2
5-20
)ζτ
5.17
a)
Assume underdamped second-order 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 2nd-order 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 5-21
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
5-22
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
5-23
(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 second-order 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
′ = 0.0281M hmax 2.81 ft Mmax = = 100 gpm 0.0281 ft/gpm
5-24
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 ′2 (t ) = 0.134 M 1 − 1 + e 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
5-25
C m′ ( s ) K , = C ′( s ) τs + 1
K = (3-0)/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 ′m (t ) = 1.51 + (0.1e −t / 0.1 − 49e −t / 49 ) (49 − 0.1)
C m′ ( s ) =
c)
C m′ ( s ) =
0.5 3 1.5 = (49 s + 1) s s (49 s + 1)
(
c ′m (t ) = 1.5 1 − e − t / 49
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)
5-26
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 ′A 20000 = qc ′A i − (q + Vk (T ))c ′A − Vc A k (T ) T′ dt T2
ρVC
(3)
dT ′ 20000 = − ρqC + ∆H RVc A k (T ) T ′ − ∆H RVk (T )c ′A (4) 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 ′A (s ) from Eq. 5 into Eq. 6 and rearranging, T ′( s ) = C A′i ( s )
(−∆H R )Vk (T )q 20000 20000 Vs + q + Vk (T ) ρVCs + ρqC − (−∆H R )Vc A k (T ) + (−∆H R )V 2 c A k 2 (T ) 2 T T2
(7)
c A is obtained from Eq. 1 at steady state,
5-27
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 ′A i ( s ) (0.0722s + 1)(50s + 1) For step response, C ′Ai ( s ) = 1 / s
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 first-order approximation of the transfer function is
T ′( s ) 11.38 = C ′A i ( s ) 50s + 1 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 first-order 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.
5-28
]
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)
5-29
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)
Second-order 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/(1-Kb), 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 τ a τ b
1 1 − Kb
(11)
Investigating Eq. 11 we see that the quantity in brackets is the same as ζ for an overdamped 2nd-order system (ζOD) [ from Eq. 5-43 in text]. ζ=
ζ OD 1 − Kb
where ζ OD =
1 τa + τ b 2 τa + τb
5-30
(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 critically-damped 2nd-order 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 critically-damped 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 K 1′ which has the form = ( s indicates presence of integrator) s (τ′s + 1) =
5-31
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.
5-32
