1234567898
1234567898
5
2.1
a)
Overall mass balance:
d (ρV )
= w1 + w2 − w3
dt
(1)
Energy balance:
C
d 1V 2T3 − Tref 3 
dt
= w1C (T1 − Tref ) + w2C (T2 − Tref )
− w3C (T3 − Tref )
(2)
Because ρ = constant and V = V = constant, Eq. 1 becomes:
w3 = w1 + w2
b)
(3)
From Eq. 2, substituting Eq. 3
1CV
d (T3 − Tref )
dt
= 1CV
dT3
= w1C 2T1 − Tref 3 + w2C 2T2 − Tref 3
dt
− ( w1 + w2 ) C (T3 − Tref )
(4)
Constants C and Tref can be cancelled:
ρV
dT3
= w1T1 + w2T2 − ( w1 + w2 )T3
dt
The simplified model now consists only of Eq. 5.
Degrees of freedom for the simplified model:
Parameters : ρ, V
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.
2-1
(5)
Variables : w1, w2, T1, T2, T3
NE = 1
NV = 5
Thus, NF = 5 – 1 = 4
Because w1, w2, T1 and T2 are determined by upstream units, we assume
they are known functions of time:
w1 = w1(t)
w2 = w2 (t)
T1 = T1(t)
T2 = T2(t)
Thus, NF is reduced to 0.
2.2
Energy balance:
Cp
d 1V 2T − Tref 3 
dt
= wC p (Ti − Tref ) − wC p (T − Tref ) − UAs (T − Ta ) + Q
Simplifying
1VC p
dT
= wC p Ti − wC p T − UAs 2T − Ta 3 + Q
dt
dT
1VC p
= wC p 2Ti − T 3 − UAs 2T − Ta 3 + Q
dt
b)
T increases if Ti increases and vice versa.
T decreases if w increases and vice versa if (Ti – T) < 0. In other words, if
Q > UAs(T-Ta), the contents are heated, and T >Ti.
2.3
a)
Mass Balances:
2-2
ρA1
dh1
= w1 − w2 − w3
dt
(1)
dh2
= w2
dt
(2)
ρA2
Flow relations:
Let P1 be the pressure at the bottom of tank 1.
Let P2 be the pressure at the bottom of tank 2.
Let Pa be the ambient pressure.
w2 =
Then
P1 − P2
ρg
=
(h1 − h2 )
R2
g c R2
(3)
P1 − Pa
ρg
=
h1
R3
g c R3
(4)
w3 =
b)
Seven parameters: ρ, A1, A2, g, gc, R2, R3
Five variables : h1, h2, w1, w2, w3
Four equations
Thus NF = 5 – 4 = 1
1 input = w1 (specified function of time)
4 outputs = h1, h2, w2, w3
2.4
Assume constant liquid density, ρ . The mass balance for the tank is
d (ρAh + m g )
dt
= ρ( q i − q )
Because ρ, A, and mg are constant, this equation becomes
2-3
A
dh
= qi − q
dt
(1)
The square-root relationship for flow through the control valve is


ρgh
q = C v  Pg +
− Pa 
gc


1/ 2
(2)
From the ideal gas law,
Pg =
(m g / M ) RT
(3)
A( H − h)
where T is the absolute temperature of the gas.
Equation 1 gives the unsteady-state model upon substitution of q from Eq. 2 and
of Pg from Eq. 3:
1/ 2
 (mg / M ) RT ρ gh

dh
+
− Pa 
A = qi − Cv 
dt
gc
 A( H − h)

(4)
Because the model contains Pa, operation of the system is not independent of Pa.
For an open system Pg = Pa and Eq. 2 shows that the system is independent of Pa.
2.5
a)
For linear valve flow characteristics,
Pd − P1
P − P2
, wb = 1
,
Ra
Rb
Mass balances for the surge tanks
wa =
dm1
= wa − wb ,
dt
wc =
P2 − Pf
Rc
dm2
= wb − wc
dt
where m1 and m2 are the masses of gas in surge tanks 1 and 2,
respectively.
If the ideal gas law holds, then
2-4
(1)
(2)
P1V1 =
m1
RT1 ,
M
P2V2 =
m2
RT2
M
(3)
where M is the molecular weight of the gas
T1 and T2 are the temperatures in the surge tanks.
Substituting for m1 and m2 from Eq. 3 into Eq. 2, and noticing that V1, T1,
V2, and T2 are constant,
V1M dP1
V2 M dP2
= wa − wb and
= wb − wc
RT1 dt
RT2 dt
(4)
The dynamic model consists of Eqs. 1 and 4.
b)
For adiabatic operation, Eq. 3 is replaced by
γ
V 
V
P1  1  = P2  2
 m1 
 m2
or
 PV γ
m1 =  1 1
 C




γ

 = C , a constant

1/ γ
and
 PV γ
m2 =  2 2
 C




(5)
1/ γ
(6)
Substituting Eq. 6 into Eq. 2 gives,
1
γ
 V1 γ

 C





1/ γ
1
γ
 V2 γ

 C





1/ γ
P1
(1− γ ) / γ
dP1
= wa − wb
dt
(1− γ ) / γ
dP2
= wb − wc
dt
P2
as the new dynamic model. If the ideal gas law were not valid, one would
use an appropriate equation of state instead of Eq. 3.
2.6
a)
Assumptions:
1. Each compartment is perfectly mixed.
2. ρ and C are constant.
3. No heat losses to ambient.
Compartment 1:
2-5
Overall balance (No accumulation of mass):
0 = ρq − ρq1
thus
q1 = q
(1)
Energy balance (No change in volume):
V11C
dT1
= 1qC 2Ti − T1 3 − UA2T1 − T2 3
dt
(2)
Compartment 2:
Overall balance:
0 = ρq1 − ρq2
thus
q2 = q1= q
(3)
Energy balance:
V21C
b)
dT2
= 1qC 2T1 − T2 3 + UA2T1 − T2 3 − U c Ac 2T2 − Tc 3
dt
(4)
Eight parameters: ρ, V1, V2, C, U, A, Uc, Ac
Five variables: Ti, T1, T2, q, Tc
Two equations: (2) and (4)
Thus NF = 5 – 2 = 3
2 outputs = T1, T2
3 inputs = Ti, Tc, q (specify as functions of t)
c)
Three new variables: ci, c1, c2 (concentration of species A).
Two new equations: Component material balances on each compartment.
c1 and c2 are new outputs. ci must be a known function of time.
2.7
Let the volume of the top tank be γV, and assume that γ is constant.
Then, an overall mass balance for either of the two tanks indicates that the
flow rate of the stream from the top tank to the bottom tank is equal to
q +qR. Because the two tanks are perfectly stirred, cT2 = cT.
2-6
Component balance for chemical tracer over top tank:
4V
dcT 1
= qcTi + qR cT − 2 q + qR 3cT 1
dt
(1)
Component balance on bottom tank:
(1 − 43V
dcT 2
= 2q + qR 3cT 1 − qR cT − qcT
dt
or
(1 − 43V
dcT
= 2 q + qR 32cT 1 − cT 3
dt
(2)
Eqs. 1 and 2 constitute the model relating the outflow concentration, cT, to
inflow concentration, cTi. Describing the full-scale reactor in the form of
two separate tanks has introduced two new parameters into the analysis, qR
and γ. Hence, these parameters will have to be obtained from physical
experiments.
2.8
Additional assumptions:
(i) Density of the liquid, ρ, and density of the coolant, ρJ, are constant.
(ii) Specific heat of the liquid, C, and of the coolant, CJ, are constant.
Because V is constant, the mass balance for the tank is:
ρ
dV
= q F − q = 0 ; thus q = qF
dt
Energy balance for tank:
ρVC
dT
0.8
= q F ρC (TF − T ) − Kq J A(T − TJ )
dt
(1)
Energy balance for the jacket:
ρ J VJ C J
dTJ
dt
= q J ρ J C J (Ti − TJ ) + Kq J
2-7
0.8
A(T − TJ )
(2)
where A is the heat transfer area (in ft2) between the process liquid and the
coolant.
Eqs.1 and 2 comprise the dynamic model for the system.
2.9
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. Ts is a function of Ps , Ts = f(Ps)
Mass balance for the tank:
dV
= qF − q
dt
Energy balance for the tank:
1C
d V (T − Tref ) 
dt
= qF 1C 2TF − Tref 3 − q1C 2T − Tref 3
(1)
(2)
+UA(Ts − T )
where: Tref is a constant reference temperature
A is the heat transfer area
Eq. 2 is simplified by substituting for (dV/dt) from Eq. 1, and replacing Ts
by f(Ps), to give
ρVC
dT
= q F ρC (TF − T ) + UA[ f ( Ps ) − T ]
dt
Then, Eqs. 1 and 3 constitute the dynamic model for the system.
2-8
(3)
2.10
Assume that the feed contains only A and B, and no C. Component
balances for A, B, C over the reactor give.
dc A
= qi c Ai − qc A − Vk1e− E1 / RT c A
dt
(1)
dcB
= qi cBi − qcB + V (k1e − E1 / RT c A − k2e− E2 / RT cB )
dt
(2)
dcC
= − qcC + Vk2e − E2 / RT cB
dt
(3)
V
V
V
An overall mass balance over the jacket indicates that qc = qci because the
volume of coolant in jacket and the density of coolant are constant.
Energy balance for the reactor:
d (Vc A M A S A + VcB M B S B + VcC M C SC ) T 
dt
= ( qi c Ai M A S A + qi cBi M B S B ) (Ti − T )
−UA(T − Tc ) + (−∆H1 )Vk1e − E1 / RT c A + (−∆H 2 )Vk2 e− E2 / RT cB
(4)
where MA, MB, MC are molecular weights of A, B, and C, respectively
SA, SB, SC are specific heats of A, B, and C.
U is the overall heat transfer coefficient
A is the surface area of heat transfer
Energy balance for the jacket:
1 j S jV j
dTc
= 1 j S j qci 2Tci − Tc 3 + UA2T − Tc 3
dt
where:
ρj, Sj are density and specific heat of the coolant.
Vj
is the volume of coolant in the jacket.
Eqs. 1 - 5 represent the dynamic model for the system.
2-9
(5)
2.11
Model (i) :
Overall mass balance (w=constant= w ):
d ( ρV )
dh
= Aρ
= w1 + w2 − w
dt
dt
(1)
A component balance:
d (ρVx)
= w1 − wx
dt
or
Aρ
d (hx)
= w1 − wx
dt
(2)
Note that for Stream 2, x = 0 (pure B).
Model (ii) :
Mass balance:
d (1V 3
dh
= Aρ
= w1 + w2 − w
dt
dt
(3)
Component balance on component A:
d (ρVx)
= w1 − wx
dt
or
Aρ
d (hx)
= w1 − wx
dt
(4)
2-10
2.12
a)
Note that the only conservation equation required to find h is an overall
mass balance:
dm d (ρAh)
dh
=
= ρA = w1 + w2 − w
dt
dt
dt
Valve equation: w = C v′
ρg
h = Cv h
gc
where C v = C v′
ρg
gc
(1)
(2)
(3)
Substituting the valve equation into the mass balance,
dh
1
=
( w1 + w2 − C v h )
dt ρA
(4)
Steady-state model :
0 = w1 + w2 − C v h
b)
c)
Cv =
w1 + w2
h
=
2.0 + 1.2 3.2
kg/s
=
= 2.13 1/2
1.5
2.25
m
Feedforward control
2-11
(5)
Rearrange Eq. 5 to get the feedforward (FF) controller relation,
w2 = C v hR − w1
where hR = 2.25 m
w2 = (2.13)(1.5) − w1 = 3.2 − w1
(6)
Note that Eq. 6, for a value of w1 = 2.0, gives
w2 = 3.2 –1.2 = 2.0 kg/s
which is the desired value.
If the actual FF controller follows the relation, w2 = 3.2 − 1.1w1 (flow
transmitter 10% higher), w2 will change as soon as the FF controller is
turned on,
w2 = 3.2 –1.1 (2.0) = 3.2 – 2.2 = 1.0 kg/s
(instead of the correct value, 1.2 kg/s)
Then C v h = 2.13 h = 2.0 + 1.0
or
h=
3
= 1.408 and h = 1.983 m (instead of 2.25 m)
2.13
Error in desired level =
2.25 − 1.983
×100% = 11.9%
2.25
The sensitivity does not look too bad in the sense that a 10% error in flow
measurement gives ~12% error in desired level. Before making this
2-12
conclusion, however, one should check how well the operating FF
controller works for a change in w1 (e.g., ∆w1 = 0.4 kg/s).
2.13
a)
Model of tank (normal operation):
dh
= w1 + w2 − w3
dt
π (2) 2
A=
= π = 3.14 m 2
4
ρA
(800)(3.14)
(Below the leak point)
dh
= 120 + 100 − 200 = 20
dt
20
dh
=
= 0.007962 m/min
dt (800)(3.14)
Time to reach leak point (h = 1 m) = 125.6 min.
b)
Model of tank with leak and w1 , w2 , w3 constant:
1A
dh
= 56 − δ q4 = 56 − 12676583 h − 9 = 20 − 20 h − 1 , h ≥ 1
dt
To check for overflow, one can simply find the level hm at which dh/dt =
0. That is the maximum value of level when no overflow occurs.
0 = 20 − 20
hm − 1 or
hm = 2 m
Thus, overflow does not occur for a leak occurring because hm < 2.25 m.
2.14
Model of process
Overall material balance:
2-13
ρAT
dh
= w1 + w2 − w3 = w1 + w2 − C v h
dt
(1)
Component:
ρAT
d (hx3 )
= w1 x1 + w2 x2 − w3 x3
dt
ρAT h
dx3
dh
+ ρAT x3
= w1 x1 + w2 x2 − w3 x3
dt
dt
Substituting for dh/dt (Eq. 1)
ρAT h
dx3
+ x3 ( w1 + w2 − w3 ) = w1 x1 + w2 x2 − w3 x3
dt
ρAT h
dx3
= w1 ( x1 − x3 ) + w2 ( x 2 − x3 )
dt
dx3
1
=
[w1 ( x1 − x3 ) + w2 ( x2 − x3 )]
dt ρAT h
or
a)
(2)
(3)
At initial steady state ,
w3 = w1 + w2 = 120 + 100 = 220 Kg/min
220
Cv =
= 166.3
1.75
b)
If x1 is suddenly changed from 0.5 to 0.6 without changing flowrates, then
level remains constant and Eq.3 can be solved analytically or numerically
to find the time to achieve 99% of the x3 response. From the material
balance, the final value of x3 = 0.555. Then,
dx3
1
=
[120(0.6 − x3 ) + 100(0.5 − x3 )]
dt (800)(1.75)π
=
1
[ (72 + 50) − 220 x3 )]
(800)(1.75)π
= 0.027738 − 0.050020x3
Integrating,
2-14
x3 f
∫
x3 o
t
dx3
= dt
0.027738 − 0.050020 x3 ∫0
where x3o=0.5 and
x3f =0.555 – (0.555)(0.01) = 0.549
Solving,
t = 47.42 min
c)
If w1 is changed to 100 kg/min without changing any other input variables,
then x3 will not change and Eq. 1 can be solved to find the time to achieve
99% of the h response. From the material balance, the final value of the
tank level is h =1.446 m.
800π
dh
= 100 + 100 − Cv h
dt
dh
1 
=
200 − 166.3 h
dt 800π 


= 0.079577 − 0.066169 h
where ho=1.75 and
hf =1.446 + (1.446)(0.01) = 1.460
By using the MATLAB command ode45 ,
t = 122.79 min
Numerical solution of the ode is shown in Fig. S2.14
1.8
1.7
h(m)
1.6
1.5
1.4
0
50
100
150
200
time (min)
250
300
Figure S2.14. Numerical solution of the ode for part c)
2-15
d)
In this case, both h and x3 will be changing functions of time. Therefore,
both Eqs. 1 and 3 will have to be solved simultaneously. Since
concentration does not appear in Eq. 1, we would anticipate no effect on
the h response.
a)
The dynamic model for the chemostat is given by:
2.15
dX
= Vrg − FX
dt
or
dX
F
= rg −   X
dt
V 
(1)
Product: V
dP
= Vrp − FP
dt
or
dP
F
= rp −   P
dt
V 
(2)
Substrate:
V
Cells:
V
dS
1
1
= F (S f − S ) −
Vrg −
VrP
dt
YX / S
YP / S
or
1
1
dS  F 
rg −
rP
=  ( S f − S ) −
YX / S
YP / S
dt  V 
b)
At steady state,
then,
dX
=0
dt
∴
rg = DX
µX = DX
∴
µ=D
A simple feedback strategy can be implemented where the growth rate
is controlled by manipulating the mass flow rate, F.
c)
Washout occurs if dX/dt = 0 is negative for an extended period of time;
that is,
rg − DX < 0
or
µ<D
Thus, if µ < D the cells will be washed out.
d)
(3)
At steady state, the dynamic model given by Eqs. 1, 2 and 3 becomes:
2-16
(4)
0 = rg − DX
(5)
0 = rp − DP
(6)
0 = D(S f − S ) −
1
YX / S
rg −
1
YP / S
rP
(7)
From Eq. 5,
DX = rg
(8)
From Eq. 7
rg = Y X / S ( S f − S ) D +
YX / S
rP
YP / S
(9)
Substituting Eq. 9 into Eq. 8,
YX / S
rp
YP / S
From Eq. 6 and the definition of YP/S in (2-92),
DX = Y X / S ( S f − S ) D +
rp = DP = DYP / S ( S f − S )
From Eq. 4
S=
DK S
µ max − D
Substituting these two equations into Eq. 10,

DK S 
 D
DX = 2Y X / S  S f −
µ
−
D
max


2-17
(10)
1
DX (g/L.h)
0.8
0.6
MAXIMUM
PRODUCTION
0.4
0.2
WASHOUT
0
0
0.05
0.1
0.15
0.2
0.25
D (1/h)
Figure S2.15. Steady-state cell production rate DX as a function of dilution rate D.
From Figure S2.15, washout occurs at D = 0.18 h-1 while the maximum
production occurs at D = 0.14 h-1. Notice that maximum and washout points
are dangerously close to each other, so special care must be taken when
increasing cell productivity by increasing the dilution rate.
2.16
a)
We can assume that ρ and h are approximately constant. The dynamic
model is given by:
rd = −
dM
= kAc s
dt
(1)
Notice that:
M = ρV
∴
dM
dV
=ρ
dt
dt
(2)
V = πr 2 h
∴
dV
dr
dr
= (2πrh)
=A
dt
dt
dt
(3)
2-18
Substituting (3) into (2) and then into (1),
− ρA
dr
= kAc s
dt
−ρ
∴
dr
= kc s
dt
Integrating,
r
∫ r dr = −
o
kcs t
dt
1 ∫0
∴
r (t ) = ro −
kc s
t
ρ
(4)
Finally,
M = ρV = ρπhr 2
then
kc

M (t ) = ρπh ro − s
ρ

b)

t 

2
The time required for the pill radius r to be reduced by 90% is given by
Eq. 4:
0.1ro = ro −
kc s
t
ρ
∴
t=
0.9ro ρ (0.9)(0.4)(1.2)
=
= 54 min
kc s
(0.016)(0.5)
Therefore, t = 54 min .
2.17
For V = constant and F = 0, the simplified dynamic model is:
dX
S
= rg = µ max
X
dt
Ks + S
dP
S
= rp = YP / X µ max
X
dt
Ks + S
dS
1
1
rg −
rP
=−
dt
YX / S
YP / X
Substituting numerical values:
dX
SX
= 0 .2
dt
1+ S
2-19
dP
SX
= (0.2)(0.2)
dt
1+ S
dS
SX
= 0 .2
dt
1+ S
 1 0 .2 
 − 0 .5 − 0 .1 


By using MATLAB, this system of differential equations can be solved.
The time to achieve a 90% conversion of S is t = 22.15 h.
Figure S2.17. Fed-batch bioreactor dynamic behavior.
2-20