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8 Rev: 12603 10.1
According to Guideline 6, the manipulated variable should have a large
effect on the controlled variable. Clearly, it is easier to control a liquid
level by manipulating a large exit stream, rather than a small stream.
Because R/D>1, the reflux flow rate R is the preferred manipulated
variable. 10.2 Exit flow rate w4 has no effect on x3 or x4 because it does not change the
relative amounts of materials that are blended. The bypass fraction f has a
dynamic effect on x4 but no steadystate effect because it also does not
change the relative amounts of materials that are blended. Thus, w2 is the
best choice. 10.3 Both the steadystate and dynamic behavior needs to be considered. From
a steadystate perspective, the reflux stream temperature TR would be a
poor choice because it is insensitive to changes in xD, due to the small
nominal value of 5 ppm. For example, even a 100% change from 5 to 10
ppm would result in a negligible change in TR. Similarly, the temperature
of the top tray would be a poor choice. An intermediate tray temperature
would be more sensitive to changes in the tray composition but may not be
representative of xD. Ideally, the tray location should be selected to be the
highest tray in the column that still has the desired degree of sensitivity to
composition changes.
The choice of an intermediate tray temperature offers the advantage of
early detection of feed disturbances and disturbances that originate in the
stripping (bottom) section of the column. However, it would be slow to
respond to disturbances originating in the condenser or in the reflux drum.
But on balance, an intermediate tray temperature is the best choice.
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp 101 10.4 For the flooded condenser in Fig. E10.4, the area available for heat
transfer changes as the liquid level changes. Consequently, pressure
control is easier when the liquid level is low and more difficult when the
level is high. By contrast, for the conventional process design in Fig. 10.5,
the liquid level has a very small effect on the pressure control loop. Thus,
the flooded condenser is more difficult to control because the level and
pressure control loops are more interacting than they are for the
conventional process design in Fig. 10.5. 10.5 (a) The larger the tank, the more effective it will be in “damping out”
disturbances in the reactor exit stream. A large tank capacity also provides
a large feed inventory for the distillation column, which is desirable for
periods where the reactor is shut down. Thus a large tank is preferred from
a process control perspective. However a large tank has a high capital cost,
so a small tank is appealing from a steadystate, design perspective. Thus,
the choice of the storage tank size involves a tradeoff of control and
design objectives. (b) After a setpoint change in reactor exit composition occurs, it would be
desirable to have the exit compositions for both the reactor and the storage
tank change to the new value as soon as possible. But the concentration in
the storage tank will change gradually due to its liquid inventory. The time
constant for the storage tank is proportional to the mass of liquid in the
tank (cf. blending system models in Chapters 2 and 4). Thus, a large
storage tank will result in sluggish responses in its exit composition, which
is not desirable when frequent setpoint changes are required. In this
situation, the storage tank size should be smaller than for case (a). 10.6 Variables : q1, q2,…. q6, h1, h2 NV = 8 Equations :
3 flowhead relations: 102 q3 = Cv1 h1 q5 = Cv 2 h2 q 4 = K (h1 − h2 )
2 mass balances:
1A1 dh1
= 12 q1 + q6 − q3 − q4 3
dt
dh
1A2 2 = 12 q2 + q4 − q5 3
dt Thus NE = 5
Degrees of freedom: NF = NV – NE = 8 − 5 = 3
Disturbance variable : q6 ND = 1 NF = NFC + ND
NFC = 3 − 1 = 2 10.7 Consider the following energy balances assuming a reference temperature
of Tref = 0 :
Heat exchanger:
Cc (1 − f ) wc (TC 0 − TC1 ) = Ch wh (Th1 − Th 2 ) (1) C c wc (TC 2 − TC1 ) = C h wh (Th1 − Th 2 ) (2) wc = (1 − f ) wc + fwc (3) Overall: Mixing point: Thus, 103 NE =3 , NV = 8 ( f , wc , wh , Tc1 , Tc 2 , Tc 0 , Th1 , Th 2 ) NF =NV − NE = 8 − 3 = 5
NFC =2 (f, wh)
also
ND = NF − NFC = 3 (wc, Tc1, Tc2)
The degrees of freedom analysis is identical for both cocurrent and
countercurrent flow because the mass and energy balances are the same
for both cases. 10.8 The dynamic model consists of the following material balances:
Mass balance on the tank:
1A dh
= 24 − f 3 w1 + w2 − w3
dt (1) Component balance on the tank:
1A d (hx3 )
= 24 − f 3 x1w1 + x2 w2 − x3 w3
dt (2) Mixing point balances:
w4 = w3 + fw1 (3) x4w4 = x3w3 + fx1w1 (4) Thus,
NE = 4 (Eqs.14) NV = 10 (h, f , w1 , w2 , w3 , w4 , x1 , x2 , x3 , x4 ) NF = NV − NE = 6
Because two variables ( w2 and f ) can be independently adjusted, it
would appear that there are two control degrees of freedom. However, the 104 fraction of bypass flow rate, f , has no steadystate effect on x4. To
confirm this assertion, consider the overall steadystate component
balance for the tank and the mixing point: x1 w1 + x 2 w2 = x 4 w4 (5) This balance does not depend on the fraction bypassed, f, either directly or
indirectly,
Conclusion : NFC = 1 (w2) 10.9 Ci C Vessel
q q Let Ci = concentration of N2 in the inlet stream = 100%
C = concentration in the vessel = exit concentration (perfect mixing)
Assumptions:
1. Perfect mixing
2. Initially, the vessel contains pure air, that is, C(0) = 79%.
N2 balance on the vessel:
V dC
= q (C i − C )
dt (1) Take Laplace transforms and let τ=V/q: 56 sC 2 s 3 − C 2t = 738 = Ci
− C 2s3
s Rearrange, 105 C ( s) = Ci
C (t = 0)
+
s (5s + 43
5s + 4 Take inverse Laplace transforms (cf. Chapter 3),
C (t ) = Ci (1 − e −t / 5 ) + C (t = 0)e−t 9 5 (2) Also,
V 20, 000 L 1 m3 5= = =
q 0.8 m3 / min 1000 L Substitute for τ, Ci and C(0) into (2) and rearrange 21%
t = (25 min) ln 100% − C (t ) (3) Let C(t) = 98% N2 (i.e., 2% O2). From (3),
t = 58.7 min 10.10 Define k as the number of sensors that are working properly. We are
interested in calculating P (k ≥ 2) , when P(E) denotes the probability that
an event, E, occurs.
Because k = 2 and k = 3 are mutually exclusive events,
P (k ≥ 2) = P (k = 2) + P (k = 3) (1) These probabilities can be calculated from the binomial distribution 1
and the given probability of a sensor functioning properly (p = 0.99): 3
1
P(k = 2) = ( 0.01) (0.99)2 = 0.0294
2 3
0
P(k = 3) = ( 0.01) (0.99)3 = 0.9703 3 106 n
where the notation, , refers to the number of combinations of n objects
r taken r at a time, when the order of the r objects is not important. Thus 3 3 = 3 and = 1 . From Eq.(1), 2 3 P (k ≥ 2) = 0.0294 + 0.9703 = 0.9997
1 See any standard probability or statistics book, e.g., Montgomery D.C and G.C. Runger,
Applied Statistics and Probability for Engineers, 3rd ed., John Wiley, NY (2003). 10.11 Assumptions:
1. Incompressible flow.
2. Chlorine concentration does not affect the air sample density.
3. T and P are approximately constant.
The time tT that is required to detect a chlorine leak in the processing area
is given by:
tT = ttube + tA
where:
ttube is the time that the air sample takes to travel through the tubing
tA is the time that the analyzer takes to respond after chlorine first
reaches it.
The volumetric flow rate q is the product of the velocity v and the crosssectional area A:
q = vA ∴ then: 107 v= q
A π D 2 3.14 ( 6.35 − 0.762 )
A=
=
= 24.5 mm 2
4
4
3
10 cm / s
v=
= 40.8 cm / s
24.5 × 10−2 cm 2
2 Thus,
ttube = 4000 cm
= 98.1 s
40.8 cm / s Finally,
tT = 98.1 + 5 = 103.1 s
Carbon monoxide (CO) is one of the most widely occurring toxic gases,
especially in confined spaces. High concentrations of carbon monoxide
can saturate a person’s blood in a matter of minutes and quickly lead to
respiratory problems or even death. Therefore, this amount of time is not
acceptable if the hazardous gas is CO. 10.12 The key safety concerns include:
1. Early detection of any leaks to the surroundings
2. Over pressurizing the flash drum
3. Maintain enough liquid level so that the pumps do not cavitate.
4. Avoid having liquid entrained in the gas. These concerns can be addressed by the following instrumentation.
1. Leak detection: sensors for hazardous gases should be located in
the vicinity of the flash drum.
2. Over pressurization: Use a high pressure switch (PSH) to shut
off the feed when a high pressure occurs.
3. Liquid inventory: Use a low level switch (LSL) to shut down
the pump if a low level occurs.
108 4. Liquid entrainment: Use a high level alarm to shut off the feed
if the liquid level becomes too high. This SIS system is shown below with conventional control loops for
pressure and liquid level. Figure S10.12. 109 10.13 The proposed alarm/SIS system is shown in Figure S10.13: The solenoidoperated valves are normally open. If the column pressure
exceeds a specified limit, the high pressure switch (PSH) shuts down both
the feed stream and the steam flow to the reboiler. Both actions tend to
reduce the pressure in the column. 1010 ...
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 Spring '08
 Cummings
 Chemical Engineering, Mass Balance, Trigraph, Mass flow rate, intermediate tray temperature

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