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II.1 Consider a continuous blending process where the water is mixed with slurry to give
slurry the desired consistency (Figure II.1). The streams are mixed in a constant volume
(V) blending tank, and the mass fraction of the solids in the inlet slurry stream is given as
xs, with a volumetric flow rate of qs. Since xs and qs vary, the water make-up mass flow
rate w is adjusted to compensate for these variations. Develop a model for this blender
that can be used to predict the dynamic behavior of the mass fraction of solids in the exit
stream xe for changes in xs, qs, or w. What is the number of degrees of freedom for this
process?
Figure II.1: Schematic of the blending process
Solution:
Let us assume that we have perfect mixing and no volume changes due to mixing. Water
stream is considered to be pure water and t is the density of solid. The mass flow rates of
each stream are designated by w and the volumetric rates are by q. Then, by definition,
we have the following,
ws = sqs
s =
ws 1
=
 xsws
+
(1−xs )ws   xs (1−xs )

 t
  + 
   t  
e =
1
 xe
+
(1−xe )
 
 t  
The total mass balance yields the following equation,

d( V)
e
= w
dt s + w − we
And since the volume is constant, we have,
d( )
V
e
= w
dt s + w − we
A component balance on the solids will give,
d(x  V)
e e
= w x − w x
Or,
dt s s e e
d(x ) d( )
V
e
= −V
e
+ w x − w x = −x (w + w − w ) + w x − w x
dt dt s s e e e s e s s e e
This equation along with the definitions of the densities, forms the model of this process
to help predict the variations in the mass fraction of solids in the exit slurry as a function
of other process variables.
For a degree of freedom analysis, we have,
• Constants: V, , t
• Number of Equations: 4 (one mass balance + one component balance + two
algebraic relations)
• Number of variables: s, e, w, we, ws, xe, xs
The number of degrees of freedom is 3. Note that one usually needs to specify the
upstream solids content (density or solids fraction) and the flow rate as well as the water
flow rate to fully define the system.
II.2. A binary mixture at its saturation point is fed to a single-stage flash unit (Figure
II.2), where the mixture is heated at an unknown rate (Q). The feed flow rate and feed mole
fractions are known and may vary with time. Assume that x represents the mole fraction of
the more volatile component (e.g., xf is the mole fraction of the more volatile component in
the feed stream) and the molar heat of vaporization is the same for both components. Flow
rate is given in moles per unit time. H represents the molar liquid holdup.

Figure II.2: Schematic of a flash unit.
1. Derive the modeling equations for this system. State your assumptions clearly
and explicitly.
2. Derive the transfer function between the overhead mole fraction of the more
volatile component and its feed mole fraction. (Hint: Assume constant molar
holdup.)
Solution:
The control volume is the flash tank. We make the following assumptions:
• Negligible vapor holdup in the unit
• Constant stage temperature and pressure
• No heat loss to surroundings
• Negligible heat transfer resistance for transfer of Q.
The equilibrium relationship is given by:
xD = K(T,P)xB
where is K the equilibrium constant
For the energy balance, the quantity of interest is:
Total Energy = U + K + P

Here, U, K, P represent the internal, kinetic and potential energies of the system,
respectively. Assuming thermal equilibrium between the vapor and the liquid streams, we
can also neglect the energy balance on the vapor phase.
Since the liquid in the tank can be considered stationary

dK
=
dP
= 0 and
dE
=
dU
dt dt dt dt
For liquid systems, one can assume that
dU
dt

dH
dt
H denotes the total enthalpy of the liquid in the tank (vapor holdup neglected).
Furthermore,
Where:
H = Hcp,B,av (T −Tref )
cp,B,av : average molar heat capacity of the liquid in the tank
Tref : reference temperature where the specific enthalpy of the liquid is assumed to
be zero.
The average molar heat capacities of the liquid streams can be expressed as:
cp,F,av
cp,B,av
= xF cp,A
= xB cp,A
+ (1− xF ) cp,C
+ (1− xB ) cp,C
Total energy balance can be formulated as:
Accumulation of total energy Input of total energy Output of total energy
time
= −
time time
Or
dHcp,B,av (T −Tref )
Energy supplied by steam
+
time
= Fcp,F ,av (Tin −Tref ) − Bcp,B,av (T −Tref ) − D[cp,D,av (T −Tref ) + ]+ Q
dt
where  is the molar heat of vaporization, and Tin = T . At steady-state, this reduces to,
Q = D
Overall material balance yields,
d(HM )
= M
dt
F − M D − M B

where HM is the mass holdup of the unit and Mi are the mass flow rates. We can express
the mass flow rate as, for example:
M F = MWA FxF + MWC F(1− xF ) = FMWC + xF (MWA − MWC )
This results in the following material balance (molar balance) expression:
d(HMWC +xB (MWA −MWC ))
= FMW + x (MW − MW )
dt
C F A C
− BMWC + xB (MWA − MWC )
− DMWC + xD (MWA − MWC )
The component balance for component A yields,
d(HxB MWA )
= Fx
dt
F
MWA − BxB MWA − DxD MWA
d(HxB )
= Fx
dt
F
− BxB − DxD
II.3. An oil stream is heated as it passes through two well-mixed tanks in series (Exercise
I.11). Assuming constant physical properties, develop the nonlinear state-space model for
this process to predict the time evolution of the temperatures in both tanks. State your
assumptions clearly and explicitly.
Solution:
In this problem the state variables are
input and the oil flow rate.
T1,T2 . Possible time-varying inputs are the heat
Since the volumes are assumed constant we only need to perform an energy balance
around each tank
Total energy balance can be formulated as:
Accumulation of
time
total energy Input of
=
total energy Output of
−
time
total energy
Energy supplied by the coil
+
time
time
E = U + KE + PE ,
where U is the internal energy, KE is the kinetic energy and PE is the potential energy.
Since the tank is not moving,

ref
= T
1
F
dKE
=
dPE
= 0 .
Thus
dt dt
dE
=
dU
,
dt dt
and for liquid systems,
dU
=
dHT
dt dt
where, HT is the total enthalpy of material in the tank. H may be written as,
AhCp (T −T )
where Tref : is the reference temperature. The energy balance for Tank 1 may be written
as:
d(V1Cp (T1 −Tref )) = FCp
dt
(Tin −Tref )− FCp (T1 −Tref )+ Q
Assuming Tref = 0, we will have:
V1
d(T1 )
dt
= FT − FT +
Q
in 1
C
p
dT1
dt
= (Tin −T1 )+
V1
Q
CpV1
Similarly for Tank 2 we have
dT2
=
F
dt V2
(T1 −T2 )
Thus the set of Equation representing the dynamic of the temperatures in the tanks is
given by
d(T1 ) F
in − FT1 +
Q
dt V1 CpV1
d(T2 )
=
F
(T −T2 )
dt V2

V
1
= s 1
T 2
V
2
F
2
V T
 V
The equations are ‘slightly’ nonlinear due to the multiplication between the flow rate and
the temperatures. Rearranging and taking the Taylor series expansion,
dT1
=
F
dt V1
(Tin − T1) +
Q
=
cpV1
f1(F,T1,Q)
 F Q 
 
V
(Tin − T1) +
c V
 + a1(F − Fs ) + a2 (T1 − T1s ) + a3 (Q − Qs )
1
dT F
p 1 ss
2
=
dt
(T1 − T2 ) =
2
 F 
f2 (F,T1,T2 )
  (T1 − T2 )
 2 ss
+ b1(F − Fs ) + b2 (T1 − T1s ) + b3 (T2 − T2s )
We can see that the constant coefficients are given as:
f
a1 =
F ss
f2
=
(Tin −T1s )
;a
V1
(T1s −T2s )
f1
T1 ss
f2
−F
= ;a3
V1
Fs
f
=
Q ss
f2
=
1
cpV1
−Fs
b1 = =
ss V2
;b2 =
1 ss
= ;b3 = =
ss 2
By subtracting the steady-state equation and defining deviation variables (like
F = F − Fs ), we obtain the following equations:
dT1
dt
dT2
dt
= a1F + a2T1 + a3Q
= b1F + b2T1 + b3T2
II.4. Consider the stirred-tank heater shown in Figure II.3. The steam is injected directly
in the liquid. A1 is the cross sectional area of the tank. Assume that the effluent flow rate
is proportional to the liquid static pressure that causes its flow.
1. Identify the state variables of the system.
2. Determine what balances you should perform.
3. Develop the state model that describes the dynamic behavior of the system.

Figure II.3: Stirred tank heater
Solution:
a) State Variables: h , T2
b) Total mass and energy balance.
Total mass balance
accumulation
=
input
−
output
time
d(Ah)
time time
At constant density:
= F1 − F2 + Q
dt
A
dh
= F − F +
Q
Equation 1
dt
1 2

Total energy balance
accumulation
=
input
−
output
time time time
E = U + KE + PE ,
where U is the internal energy, KE is the kinetic energy and PE is the potential energy.
Since the tank is not moving,

ref
dKE
=
dPE
= 0 .
Thus
dt dt
dE
=
dU
,
dt dt
and for liquid systems,
dU
=
dHT
dt dt
where, HT is the total enthalpy of material in the tank. Total mass in the tank is
V = Ah.
H may be written as,
AhCp (T −T )
where Tref : is the reference temperature. The input of total energy into the tank is:
F1H1 + H
where, H is the heat supplied by 40 psi steam per unit volume. The output of total
energy from the tank is: F2 H2 . The energy balance may be written as:
d(VC (T −T ))
p 2 ref
= F C (T −T )− F C (T −T )+ H
dt
1 p 1 ref 2 p 2 ref
Substituting for V = Ah, we get
d(AhC (T −T ))
p 2 ref
= F C (T −T )− F C (T −T )+ H
dt
1 p 1 ref 2 p 2 ref
Assuming Tref = 0, we will have:
A
d(hT2 )
dt
= F1T1 − F2T2 +
H
Cp
Using the product rule:
A
d(hT2 )
dt
= Ah
dT2
dt
dh
+ AT2
dt

2 2
1 2 2
Substituting this into the above equation, we get:
Ah
dT2
= F T − F T +
H
− AT
dh
dt
1 1
Cp
2
dt
From Equation 1, we have the term A
dh
dt
in the above equation. Therefore, the energy
balance results in the following equation:
dT2 H  Q 
Ah
dt
= F1T1 − F2T2 +
C
−T2  F1 − F2 +


p  
Simplifying results in the following equation:
Ah
dT2
= F T − F T +
H
−T
Q
Equation 2
dt
1 1
Cp 
II.5. Most separation processes in the chemical industry consist of a sequence of stages.
For example, sulfur dioxide present in combustion gas may be removed by the use of a
liquid absorbent (such as dimethylalanine) in a multistage absorber. Consider the three-
stage absorber displayed in Figure II.4.

Figure II.4: Schematic of a three-stage absorber.
This process is modeled through the following equations2
:

dx1
dt
dx2
= K(y f − b) − (1+ S)x1 + x2

dt

dx3
dt
= Sx1 − (1+ S)x2 + x3
= Sx2 − (1+ S)x3 + x f
H is the liquid holdup in each stage and assumed to be constant, and x and y represent
liquid and vapor compositions, respectively. Also,  = H / L is the liquid residence time,
S = aG / L
constants.
is the stripping factor and K = G / L is the gas-to-liquid ratio. A and b are
a. How many variables are there? How many equations (relationships)? What is
the degree of freedom?
b. Is this system underdetermined or overdetermined? Why?
2
Seborg, D.E, T.F. Edgar. D.A. Mellichamp, Process Dynamics and Control, Wiley

c. What additional relationships, if necessary, can you suggest to reduce the
degrees of freedom to zero?
Solution:
All relevant symbols are given below:
a,b,H (Constants)
x1, x2, x3, x f , y1, y2 , y3, y f ,S,K,G,L, (13 variables)
Here we also included the gas phase compositions (of SO2) although they do not appear
explicitly in the modeling equations. We have three equations that result from the
application of the component balances in each stage and three defining equations for
three variables (given in the problem statement). One can also write the following
equilibrium relationships that must be satisfied at each stage:
xi = fi (yi ) i =1,2,3
With these, we have a total of nine equations. The degree of freedom analysis yields:
F =13− 9 = 4
This is an underdetermined system. To fully define the system and have a feasible
control problem, we need to remove four degrees of freedom. We can do that by the
following specifications:
1. The SO2 content of the liquid feed should be zero (there is no reason why
dimethylalanine should contain any SO2). x f = 0
2. The feed gas composition y f can be considered as a disturbance as it would be
defined by the operation of upstream units.
3. Similarly, the flow rate of the gas stream may be a considered as a disturbance
because the operation of upstream units (furnaces) may vary.
4. A control problem can be defined. One can suggest a feedback control mechanism
that would measure the SO2 composition in the gas phase, y3, and according to the
specified target, y3,t arg et , manipulate the flow rate of the liquid, L. That establishes a
relationship through the feedback mechanism as follows:
L = f (y3)
Hence, we now have one specification, two disturbances, and a feedback mechanism,
resulting in four new relationships, thereby reducing the degrees of freedom to zero.
II.6. Consider a liquid chromatography for the separation of a mixture containing N
components. Assuming that the process is isothermal, and there are no radial

1
2
q 
concentration gradients, the following governing equations for solute j in the mobile
phase and on the adsorbent can be obtained:
c j
u0 + t
c j
+ (1− )
q j
= DL
 c j
z t t
  N 
z2
q j
= ka, j c  −
j m, j 
qi  −

kd, j q j
t  i=1 qm,i 
In this model, c is the concentration of solute in the mobile phase, and q is the adsorbate
concentration. Also, u0 is the superficial velocity,  and t are column void fraction and
total void fraction respectively, DL is the axial dispersion coefficient, qm is the maximum
adsorbate concentration, and ka, j and kd, j are the adsorption and desorption rate
constants for solute j respectively.
1. How would you classify this system of equations? Why?
2. How many variables are there? How many equations (relationships)? What is the
number of degrees of freedom?
3. Is this system underdetermined or overdetermined? Why?
4. What additional relationships, if necessary, can you suggest to reduce the degrees
of freedom to zero?
Solution:
a. This model should be classified as a nonlinear, distributed model. Distributed
models provide relationships for state variables as functions of both space and time,
whereas a non-distributed (lumped) model will only depend on time. It is also
nonlinear as one can see the terms involving multiplication of state variables.
a. For N components, we have cj and q j as the state variables. One can also consider
the velocity u0 to be a variable as the throughput for the chromatography column
may change. Then, we have the following parameters:
ka, j ,kd, j ,qm, j ,,t ,DL
This yields 5N+4 variables. We have 2N equations. The degrees of freedom at this
point are:
F = (5N + 4) − 2N = 3N + 4
Can we come up with more relationships? Following assumptions are appropriate:
• Void fractions (,t ) are constant.

• Maximum adsorbate concentration qm, j is a constant.
This yields N + 2 additional relationships. The adsorption and desorption rate constants
can vary with time during the chromatographic process. They can also be related to the

intrinsic adsorption/desorption rate constants (Lin et al., Ind. & Eng. Chem. Research,
1998). We will assume that they can be expressed as:
kd, j =
ka, j =
f (kd, j ,qm, j ,c0,i ,.....)
f (ka, j ,qm, j ,c0,i ,.....)
This yields 2N more relationships. Finally, the dispersion coefficient can be expressed
as:
In summary, we have
d pu0
= 0.2 + 0.011Re0.48
DL
F = (3N + 4) − (N + 2) − 2N −1= 1
Thus, the degree of freedom is one.
b. The system is underdetermined because F = 1  0 .
c. What we, as process control engineers, would do is to use a controller to affect one
variable by manipulating another variable, thus providing one additional relationship
and reducing F to 0. For example, it might be advantageous to control the exit
concentration of one of the species by manipulating the velocity (or the flow through)
u0 . The feedback yields one additional relationship between two variables, thus
reducing the degrees of freedom to zero.
II.7. Consider a distillation process (Figure II.5) with the following assumptions: binary
mixture, constant pressure, constant relative volatility, constant molar flows, no vapor
holdup, equilibrium on all stages, and a total condenser. The modeling equations are
given as follows:
Figure II.5: Schematic of the distillation column

Total material balance on stage i:
dmi
= L − L +V −V
dt
i+1 i i−1 i
Material balance for light component on stage i:
d(mi xi )
= L
dt i+1
xi+1 − Li xi + Vi−1 yi−1 −Vi yi
The above equations apply to all stages except the top (condenser), the feed and the
bottom (reboiler) stages.
From the assumption of constant molar flows and no vapor dynamics, we arrive at the
following expression for the vapor flows:
Vi−1 = Vi = V
The liquid flows depend on the liquid holdup on the stage above. We may use Francis'
Weir formula:
Li = f (mi )
The vapor composition yi is related to the liquid composition xi on the same stage
through the vapor-liquid equilibrium relationship:
x
y = i
i
1+ ( −1)xi
Feed Stage: i=nF
dmi
dt
= Li+1 − Li +Vi−1 −Vi + F
d(mi xi )
= L
dt i+1
xi+1 − Li xi + Vi−1 yi−1 −Vi yi + FzF
Total Condenser: i=nT
dmi
dt
= −Li +Vi−1 − D = Vi−1 − R − D
d(mi xi )
= −L x +V y − Dx = V y − (R + D)x
dt
i i i−1 i−1 D i−1 i−1 D

Reboiler: i=1
dmi
dt
= Li+1 −Vi − B = Li+1 − B −V
d(mi xi )
= L
dt i+1
xi+1 −Vi yB − BxB = Li+1
xi+1 − (B +V)xB
1. How many variables are there in this model? How many equations
(relationships)? What is the degree of freedom?
2. Is this system underdetermined or overdetermined? Why?
3. What additional relationships, if necessary, can one suggest to reduce the degrees
of freedom to zero?
Solution:
Variables:
mi ; Li ; xi ; yi ;V;B,D;F; zF ;R
Thus, we have 4N+6 variables and  is a parameter to be specified.
Equations:
2N differential equations and 2N algebraic equations →4N Equations
Degrees of freedom DOF=6
System is underdetermined since DOF>0
We need to specify some variables and/or define possible control loops to reduce the
DOF to zero.
Feed conditions F and zF are specified from conditions elsewhere in the plant
(disturbances) this reduces the degrees of freedom to 4.
We can define the following control loops which will add additional relationships among
the variables:
• Distillate flow rate (D) can be adjusted to control the level of the condenser drum
• Bottom flow rate (B) can be adjusted to control the level of the reboiler
• Reboiler heat duty can be adjusted to control the amount of vapor in the system
• Reflux flow rate can be adjusted to control the composition on the top of the
column
This will reduce the degrees of freedom (DOF) to zero

  D
II.8. For the single-stage flash unit introduced earlier in Exercise II.2, derive the transfer
function between the overhead mole fraction of the more volatile component and its feed
mole fraction.
Solution:
We assumed constant molar holdup, hence, we have the following component balance:
H
d(xB )
= Fx
dt
F
− BxB − DxD
Using the equilibrium relationship (and also the fact that T and P are constant), we have:
This results in,
xD = KxB
H d(xD )
= Fx
K dt
F −
B
x
K
D
 B
− DxD

= FxF − 
K
+ DxD
 
In standard form
H d(xD )
K dt
= FxF −
 B
 K
+ D

x

 B + KD 
= FxF − 
 K
xD

H d(xD )
=
KF
x − x
B + KD dt B + KD
F D

d(xD )
+ xD
dt
= kxF
where 
H
B + KD
and k =
KF
.
B + KD
This is a linear equation (as all flows are constant now). Defining deviation variables,
xD = xD − xD,s
xF = xF − xF,s
And taking Laplace transform and rearranging, we have the following transfer function:

0
sxD (s) + xD (s) = kxF (s)
k
xD (s) =
s +1
xF (s)
g(s) =
xD (s)
=
k
xF (s) s +1
where we have

H
B + KD
and k =
KF
B + KD
II.9. A liquid-phase isothermal reaction takes place in a continuous stirred-tank reactor.
The reaction is first-order,
A → B r = kCA
We assume that the vessel has a constant volume, operates isothermally (constant
temperature) and is well mixed.
For this system:
1. Derive the process transfer function between the outlet (tank) concentration and
the feed concentration of component A.
2. Obtain the time evolution of the concentration as function of the feed
concentration and the process parameters. Hint: use partial fraction expansion.
3. For the design and operating parameters,
3
F = 0.1mol/m3
,
3
V = 2 m3
,
CA0 = 0.1mol/m , k = 0.050 1/min and CA0 =1 mol/m , calculate the outlet
concentration when t = V /(F +Vk) and when t = 40 min .
Solution:
From Example 4.5 in the book, the state equation for our reactor that provides the time
evolution of the reactant composition is given as
dC A
=
F
C −
F
C − kC
dt V
A0
V
A A
Rewriting

dCA
+ 
F
+ k C =
F
C

dt V


A
V
A0
 V  dCA
C
F
C
  +
F +Vk dt
A = A0
F +Vk
 

dCA
+ C
dt
A = kCA0
Note that k in the last equation is the steady-state gain. Defining deviation variables and
taking Laplace transform to both sides of the equation
sCA (s) + CA(s) = kCA0 (s)
(s + +1)CA (s) = kCA0 (s)
Finally,
CA(s) = CA − CAs and CA0 = CA0 - CA0s
CA(s)
= g(s) =
k
CA0 (s) (s + +1)
where:  =
V
F0 +V
and k =
F0
.
F0 +V
To obtain the time domain solution, we use partial fraction technique
C s
M CA0 k M
A0 ( ) = =
s
CA (s) =
s (s + +1) s
1
=
A
+
B
(s +1)s s (s +1
s=0
s
=
As
+
Bs
= A
1
= A
(s +1)s s (s +1 (1)
s=-1/
(s +1)
=
A(s +1)
+
B(s +1)
= B
1
= − = B
(s +1)s s (s +1 −1/
 A B  1  

CA (s) = kM +  = kM − 
 s (s +1)   s (s +1) 
Inverting (using Table of Laplace functions)

A0
A0
A
A
 −t /

C (t) = kM1−
e
 = kM (1− e−t /
)
Substituting
A
  
C (t) = C +
F0
C (1− e−t /
)
A As
F0 +V
For the conditions
F = 0.1m3
/min V = 2m3
CA0 = 0.8mol/m3
 = 0.0501/min C = 1mol/m3
First we need to find the steady-state value for the concentration CAs.
At steady-state

dCA
+ C
dt
A = kCA0
CAs = kCA0s =
0.1
0.1+ 20.05
0.8 = 0.4
Substituting for this value
C (t) = 0.4 +
0.1
1(1− e−t /
)
A
0.1+ 2 0.05
C (t) = 0.4 + 0.5(1− e−t /
)
For t= CA (t) = 0.4 + 0.5 (1- 0.3679) = 0.7161
 =
V
=
2
= 10
F +Vk 0.1+ 2 0.05
For t=40 min
C (t) = 0.4 + 0.5(1− e−t /10
)= 0.4 + 0.5(1− e−40 /10
)
= 0.4 + 0.5(1− 0.0183)
= 0.4 + 0.498 = 0.8908
Figure II.S1 illustrates a plot of the concentration as function of time.

1
0.9
0.8
0.7
0.6
0.5
0.4
0 10 20 30 40 50 60
Figure II.S1: Concentration response as function of time.
II.10. Consider the same liquid-phase, isothermal, continuous stirred-tank reactor as in
Exercise II.9 where the component balance can also be expressed in terms of the product
concentration.
1. Derive the process transfer function between the outlet (tank) concentration
for component B (product) and the feed concentration of component A.
2. Obtain the time evolution of the concentration as a function of the feed
concentration and the process parameters and compare your results with those
of Exercise II.9.
3. Assuming the same design and operating conditions as before what is the
value of the concentration when t = V /(F +Vk) and t = 40 min ?
Solution:
Balance on CB
dCB
= −
F
C + kC
dt V
B A
dCB
+ 
F
C = kC
 
dt V 
B A
V dC
 
B
+ C =
Fk
C
 
 F  dt
B
V
A

dCB
+ C
1
dt
B = k1CA

1
1
1

B
2
C
2
A0
where k1 is the new steady-state gain.
1sCB + CB = k1CA
(1s +1)CB = k1CA
CB
=
k1
CA (1s +1)
 =
V
1
F
k =
Fk
1
V
Thus,
CA (s)
CA0 (s)
=
k
(s + +1)
CB CA (s)
CA CA0 (s)
=
k k1
(s + +1) (1s +1)
CB
=
kk1
CA0 (s) (s + +1)(1s +1)
CA0 (s) =
M
s
=
CA0
s
C (s) =
kk1 M
B
(s + +1)(1s +1) s
Inverting using the Laplace Table,
   
C (t) = kk M1+ 1
e−t /1
− 2
e−t /2

 2 −1
 1
2 −1 

= kk1M 1+
 2 −1
( e−t /1
− e−t /2
)

F
CB (t) =
F
A0

1+
1
( e−t /1
− e−t /2
)
(F + Vk) Vk  2 −1 
F = 0.1m3
/min V = 2m3
CA0 = 0.8mol/m3
 = 0.0501/min C = 1mol/m3

 =
V
=
2
= 10
F +Vk 0.1+ 2 0.05
 =
V
1
F
=
2
= 20
0.1

A
B
B
C (t) =
0.1 0.1
1+
1
(10e−t /10
− 20e−t / 20
)

B
(0.1+ 2 0.05) (2 0.05) 

20 −10 
= 0.5(1+ 0.1(10e−t /10
− 20e−t / 20
))
Steady-state value
CBs = kCAs
CAs=0.4 then
C =
F
C =
0.1
0.4 = 0.4
Bs
Vk
As
2 0.05
For t=
For t=40
C (t) = 0.4 + 0.5(1+ 0.1(10e−10 /10
− 20e−10 / 20
))= 0.4774
C (t) = 0.4 + 0.5(1+ 0.1(10e−40 /10
− 20e−40 / 20
))= 0.7738
1
0.9 CA
0.8
CB
0.7
0.6
0.5
0.4
0 20 40 60 80 100 120
Figure II.S2: Plot of the concentrations as function of time.
II.11. Consider the same liquid-phase isothermal continuous stirred-tank reactor as in
Exercise II.9 but now the reaction is second-order,
A → B r = kC 2
1. Obtain a linear state-space model for this system.
2. Derive the process transfer function between the outlet (tank) concentration and
the feed concentration of component A.
3. Compare the characteristic parameters with those of Exercise II.9 and discuss.

As
A0s As As
As
As
As
Solution:
From Example 4.5 in the book, the state equation for our reactor that provides the time
evolution of the reactant composition is given as
dC A
=
F
C −
F
C − kC 2
dt V
A0
V
A A
We have to linearize,
VkC2
= (VkC2
) + (2VkC )(C − C )
Substituting
A As As A As
V
dC A
= F(C
dt
A0 − CA ) − (VkC 2
) + (2VkC )(CA − CAs )
At steady-state,
Substracting
0 = F(C − C ) −VkC2
V
dC A
= F
dt
(CA0 − CA )− (CA0s − CAs )+(2VkCAs )(CA − CAs )
V
dC A
= F(C
dt
A0 − CA
)+ (2VkC )CA
V
dCA
= FC
dt
A0 − FCA + (2VkCAs
)CA
= FCA0 − F + (2VkCAs )CA
V dC A
+ C =
F
C
F + 2VkCAs dt
A
F + 2VkC
A0
where:

dCA
+ C
dt
A = kCA0
 =
V
F + 2VkCAs
and k =
F
F + 2VkCAs
and the definition of the steady-state gain should be clear. Taking Laplace transform of
both sides of the equation

sCA (s) + CA (s) = kCA0 (s)
(s + +1)CA (s) = kCA0 (s)

A0

A0
Finally,
CA (s)
CA0 (s)
= g(s) =
k
(s + +1)
To obtain the time domain solution, we use partial fraction technique
C s
M CA0 k M
A0 ( ) = =
s
CA (s) =
s (s + +1) s
1
=
A
+
B
(s +1)s s (s +1
s=0
s
=
As
+
Bs
= A
1
= A
(s +1)s s (s +1 (1)
s=-1/
(s +1)
=
A(s +1)
+
B(s +1)
= B
1
= − = B
(s +1)s s (s +1 −1/
 A B  1  
CA (s) = kM +  = kM − 
 s (s +1)   s (s +1) 
Inverting (using Table of Laplace functions),
 −t /

C (t) = kM1−
e
 = kM (1− e−t /
)
Substituting
A  
 
For the conditions
CA (t) = CAs +
F
F + 2VkCAs
C (1− e−t /
)
F = 0.1m3
/min V = 2m3
CA0 = 0.8mol/m3
 = 0.0501/min C =1mol/m3
First we need to find the steady-state value for the concentration CAs. At steady-state,
0 =
F
C
V
A0
0 =
FC −
F
C
V
A

A
As
− FC
− kC 2
− kVC
2
A0s As As
0 = 0.08 − 0.1CAs − 0.1C 2

A
2
Roots are -1.5247 and 0.5247, Thus CAs=0.5247 since the other root is negative.
Substituting this value
C (t) = 0.5247 +
0.1
(1− e−t /
)
A
0.1+ 2 2 0.05 0.5247
C (t) = 0.5247 + 0.4879(1− e−t /
)
II.12. Consider the same liquid-phase isothermal, continuous, stirred-tank reactor as in
Exercise II.11 and now allow for the possibility that the vessel volume may also change.
The reaction is still second-order and the outlet flow rate depends linearly on the liquid
volume in the tank.
1. What are the state variable(s), input variable(s) and output variable(s)? Obtain a
linear state-space model for this system.
2. Derive the process transfer function between the outlet (tank) concentration of
component A and the feed flow rate.
3. What are the poles and zeros of this transfer function?
Solution:
The model equations are given as:
dC
=
F0
(C − C) − kC2
dt V
dV
= F0
dt
0
− F = F0 − V
Here C is the tank concentration, V is the tank volume, F is the flow rate, and the
subscript 0 refers to inlet conditions. k and  are constants.
The state variables are composition C and volume V. Input variables would be inlet
concentration and inlet flow rate. The output variables would depend on control objectives.
We would typically be interested in maintaining a constant yield in the reactor (hence
constant outlet composition) and constant level (or volume) to ensure constant residence
time. Outputs can be the outlet composition and the volume (the state variables).
The first equation (component balance) can be classified as nonlinear, hence requiring the
application of Taylor expansion. The second equation (total mass balance) is already in
linear form.
dC
=
dt
dV
F0
V
(C0 − C) − kC = f1 (F0 ,C0 ,V,C)
= F0 − F = F0 − V =
dt
f2 (F0 ,C0 ,V,C)

F ,C ,V ,C
F
F V
The Taylor expansion of the first equation yields:
dC

dt
f1 (F0,s ,C0,s ,Vs ,Cs ) +
f1
C0 F0,s ,C0,s ,Vs ,Cs
(C0 − C0,s ) +
f1
Fin F0,s ,C0,s ,Vs ,Cs
(F0 − F0,s )
f1
C 0,s 0,s s s
(C − Cs ) +
f1
V F0,s ,C0,s ,Vs ,Cs
(V −Vs )
The derivatives can be calculated as follows:
f1
F0,s 
C0 F0,s ,C0,s ,Vs ,Cs
= 
 Vs
 = a

f1
C0,s −Cs 
F0
f1

F0,s ,C0,s ,Vs ,Cs
= 


= −
Vs
F0,s
 = b


− 2kCs  = c
C 0,s
,C0,s ,Vs ,Cs  Vs 
f1
 F0,s (C0,s −Cs )
V 0,s ,C0,s ,Vs ,Cs
= − 2
 s
 = d

The first equation becomes
dC

dt
f1(F0,s ,C0,s ,Vs ,Cs ) + a(C0 − C0,s ) + b(F0 − F0,s ) + c(C − Cs ) + d(V −Vs )
By defining deviation variables like C = C − Cs , and recognizing that
0 = f1(F0,s ,C0,s ,Vs ,Cs ), we have,
dC
dt
= aC0 + bF0 + cC + dV
The second equation can also be manipulated by subtracting the steady-state equation,
dV
= F − V − (F − V )= (F − F )− (V −V )
dt
0 0,s s 0 0,s s
And by defining deviation variables,
dV
= F − V

To define the state-space model, define
This leads to
x1 = C ,x2 = V ,u1 = C0 ,u2 = F0 , y1 = x1, y2 = x2 .
x 1 
=
c d x1 
+
a bu1 
       
x 2  0 −  x2  0 1u2 
y =
 y1 
=
1 0x1 
    
y2  0 1x2 
Taking the Laplace transform of both linear equations, we get,
sC (s) = aC0 (s) + bF0 (s) + cC (s) + dV (s)
sV (s) = F0 (s) − V (s)
We need to find the transfer function,
From the second equation,
C (s)
= g(s)
F0 (s)
V (s) =
1
F (s)
Substitute in first equation,
s +  0
sC (s) = aC (s) + bF (s) + cC (s) + d
1
F (s)
0 0
Collecting terms,
s +  0
C (s) =
a
C (s) +
b
F (s) +
d 1
F (s)
s − c
0
s − c
0

s − c s +  0

C (s) =
a
C (s) +
1
b +
d
F (s)
s − c
0
s − c  s +  
0
Hence,
g(s) =
1
b +
+ +
 =
 d  bs b d
s − c  s +   (s − c)(s + )

The poles and zeros come from the roots of the following polynomials:
bs + b + d = 0 and (s − c)(s + ) = 0
II.13. A bioreactor is represented by the following model that uses the Monod kinetics:

11
2
dx1
= ( − D)x
dt
dx2
= (4− x
dt
2
1
)D − 2.5x1
Here x1 is the biomass concentration, x2 is the substrate concentration, and D is the
dilution rate. The specific growth rate  depends on the substrate concentration as
follows:
 =
0.53x2
0.12 + x2
In this system, we are interested in controlling the biomass concentration using the
dilution rate, at the steady-state defined by, x1s =1.4523,x2s = 0.3692,Ds = 0.4.
1. Obtain a linear state-space model for this system.
2. Derive the process transfer function for this system.
3. What is the order of this process?
Solution:
We start with the following expansion,
dx1
f1
= ( − D)x1 = f1 (x1,x2 ,D) 
dt
f1 (x1s ,x2s ,Ds ) +
x1 ss
(x1 − x1s ) +
+
f1
x2 ss
(x2 − x2s
) + +
f1
D ss
(D − Ds )
dx2
dt
= (4 − x2 )D − 2.5x1 = f2 (x1,x2 ,D)  f2 (x1s , x2s ,Ds ) +
f2
x1 ss
(x1 − x1s ) +
+
f2
x2 ss
The derivative terms can be calculated as follows,
(x2 − x2s
) + +
f2
D ss
(D − Ds )
a =
f1  0.53x 
= 2
− D 0
x1 ss 
0.12 + x2
 =
ss
a =
f1 0.53x (0.12 +x ) −0.53x x 
= 1 2 1 2 0.3866
12
x 
f
1

2 ss  (0.12 + x2 )

 =
ss
b1 =
D ss
= − x1 ss
= −1.4523

x
21
2
2
1 1
22
a =
f2
 2.5(0.53x )
= − 2
1

x1 ss 
0.12 + x2
 = −
ss
a =
f2  2.5x (0.53(0.12 +x ) −0.53x )
= − 0.4+ 1 2 2 1.3649
22
x
f2

2 ss  (0.12+ x2 )
 = −
ss
b2 =
D ss
= 4− x2 ss
= 3.6308
This yields the following state space form of the model, after defining the deviation
variables:
x1 = x1 − x1s ,x2 = x2 − x2s ,u = D − Ds
x 1
 
0
0.3866x1  −1.4523
x =   =    +  u = Ax + bu
x 2  −1 −1.36 x2   3.6308 
y = 1 0x1 
= cx
 
 2 
To obtain the transfer function model, we need to take Laplace transform of the linear
model,
sx1(s) = a11x1(s) + a12 x2 (s) + b1u(s) = a12 x2 (s) + b1u(s)
sx2 (s) = a21x1(s) + a22 x2 (s) + b2u(s)
Solve the second equation for x2 (s)and replace in the second,
x (s) =
a21x1(s)
+
b2u(s)
s − a22 s − a22
sx (s) =
a12a21x1(s)
+
a12b2u(s)
+ b u(s)
s − a22 s − a22
Now, we have to collect the terms for the transfer function, and recognize that
y(s) = x1(s),
y(s)
= g(s) =
b1s +(b2a12 −b1a22 )
u(s) This is a second-order system.

II.14. A process model is given:
s2
− a s − a12 a21

  j
 j  j
2
3

3

3

1
2
3
2 

d 2
x
dt2
+ a
dx
dt
+ x = 1
with the initial conditions, x(0) = x'(0) = 0 .
1. Using Laplace transformation, find the solutions of this model when a = 1 and a
= 3.
2. Plot the solution
the solution.
Solution:
x(t) on one graph and discuss the effect of the parameter a on
Taking the Laplace transform of this expression yields
s2
X (s) + asX (s) + X (s) =
1
s
And,
When
X (s)(s2
+ as +1) =
1
s
X (s) =
1
s(s2
+ as +1)
a = 1, using the quadratic formula for the second order polynomial, we have the
complex roots,
s = −
1

3
j
2 2
Thus, the Partial Fraction Expansion will look like:
X (s) =
1
=
A
+
B
+
C
 1
ss + +
 2
3  1 
js + − 
2  2 2 
s  1 
s + + 
 2 2 
 1 
s + − 
 2 2 
Multiplying both sides by s and evaluating the expression at s = 0 ,
A =
1
= 1
 1 3
0 + +
 1 3 
j0 + − j
 
 
 
 
2 2 
Multiplying both sides by
have,

s + −
 2
j

, and
evaluating
the
expression
at this
root, we


 
2 2
 2 2 
1

 
2

B = −
1
−
3
j
2 6
And similarly,
C = −
1
+
3
j
2 6
Thus, we have,
−
1
−
3
j −
1
+
3
j −
1 3
j
X (s) =
1
+ 2 6 + 2 6 =
1
+ 2 − 6 + ....
s  1 3 
s + + j
 1 3 
s + − j
s  1 3 
s + + j
 1 3 
s + + j
  
   2 2 
  
   2 2 
From the Laplace Transform tables, we observe that,
L−1  p 
= pe−r
Thus, we have
s + r
 
 −   1 3 
− + j 
  1  2 2 
L−1

2
1 3  = −
2
e
 s + +


 2 2
j


We also note the identity,
e(C1+C2 j)t
= eC1t
(cosC t + jsinC2t)
By completing the inverse transform, this results in,

x(t) = 1+ 2e−t / 2
− 0.5cos

3
t −
2
3
sin
3
t

6 2 
The case with a = 3 can be done in a similar manner. The result will be:

x(t) = 1+ 0.17e−2.62t
−1.17e−0.38t

Figure II.S3 shows the behavior of x(t) as a function of time for both cases. Note the
oscillatory response when a = 1 (blue line).
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0 2 4 6 8 10 12 14 16 18 20
Figure II.S3: Plot of x(t) for the cases in Exercise II.14.
II.15. For the process discussed earlier in Exercises I.11 and II.3, where an oil stream is
heated as it passes through two well-mixed tanks in series and assuming constant
physical properties, develop the transfer functions between the second tank temperature
(output), T2 , and the heat input (manipulated variable), Q and the flow rate (disturbance),
F . T0 can be assumed as constant (what if it is not?).
Solution:
The equations are ‘slightly’ nonlinear due to the multiplication between the flow rate and
the temperatures. Rearranging and taking the Taylor series expansion,

V
3
Q
V
2
V
3
1
1 2
1
dT1
dt
=
F
(T −T ) +
Q
=
V
0 1
c V f1(F,T1,Q)
1 p 1
F Q 
 
V
(T0 −T1) +
c V
 + a1(F − Fs ) + a2 (T1 −T1s ) + a3 (Q − Qs )
1
dT2 F
p 1 ss
=
dt V2
(T1 −T2 ) = f2 (F,T1,T2 )

 F

V

(T1 −T2 ) + b1(F − Fs ) + b2 (T1 − T1s ) + b3 (T2 −T2s )
 2 ss
We can see that the constant coefficients are given as:
a =
f1
F ss
=
(T0 −T1s )
;a
V1
=
f1
T1 ss
=
−Fs
;a
1
=
f1
ss
=
1
cpV1
b =
f2
F ss
=
(T1s −T2s )
;b
2
=
f2
T1 ss
=
Fs
;b
2
=
f2
T2 ss
=
−Fs
V2
By subtracting the steady-state equation and defining deviation variables (like
F = F − Fs ), we obtain the following equations:
dT1
= a F + a T + a Q
dt
1 2 1 3
dT2
= b F + b T + b T
dt
1 2 1 3 2
Taking the Laplace transformation,
and rearranging,
sT1(s) = a1F(s) + a2T1(s) + a3Q (s)
sT2 (s) = b1F(s) + b2T1(s) + b3T2 (s)
T (s) =
a1
F(s) +
a3
Q(s)
T2 (s) =
s − a2
b1
s − b3
F(s) +
s − a2
b2
s − b3
T1(s)
Replacing the first equation into the second equation and rearranging again,

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“I never felt such gratitude to God, as I did this day, for bringing
me from the idolatry of the Romish church. My heart was grieved in
reading some of their horrid doctrines¹ about saints, and images. O
God, thou hast done this for me; and thou hast done many thousand
things beside for me; and now I beseech thee do this for me, give
me an humble, thankful and penitent heart.”
¹ He was about this time, employed partly in reading Bishop
Usher’s famous disputation with the Jesuits in Ireland. And
of this book he says, ‘I cannot think that a papist who has
learning and the fear of God, can, after reading it, remain
in the errors of Popery.’
“This was a feast, and a fast day to my soul. All the ordinances of
God are exceeding precious and profitable to me.”
“I was all day deeply engaged with God: wept much, and prayed
earnestly, yet I had not much joy. I had a full and firm confidence
that he would fulfil his word of promise to my soul. My weakness
can do nothing without thy power. I lay hold on thy strength, and
offer myself to thy holy will. O let me glorify thee, as well by
suffering as by doing.”
“This morning I met with a woman where I breakfasted, who
was exceeding happy in God. A few weeks ago I met her in the
same place, but she was then utterly dead and careless. I then
spoke plainly to her, and at parting, after prayer, said ‘I pray God you
may never rest till you rest in Christ.’ The words were applied to her
heart, and her burden increased every day, so that she was brought
almost to black despair, when God revealed his love in her heart.
She could now scarce tell it thro’ weeping. O what a God is the God
of the Christians!”

“January 1758, Sunday 1. We met at four, and after prayer I
preached on Psalms xc. 12. We had the good Mr. ―― at the chapel,
whose humility and fervour, more than compensated for the
irregularity of his sermon. I have had much more happiness on other
days, than on this sabbath, tho’ not more sincerity and resignation. I
feel my weakness and confess my ignorance, and implore the
wisdom and power of God.”
“After being some hours in my room, the fire from heaven went
through me, and I could praise the Lord continually, for his goodness
to me. I find such an impression of his power and love, as cannot be
expressed in words.”
*“This whole sabbath was both a delight and honourable to me.
Such revelations of God’s goodness; such manifestations of his
Spirit, and such operations of his love, I never felt. My very outward
man was affected and refreshed. It cannot be declared what I then
felt. Oh there is much in these words, Ye shall be baptised with the
Holy Ghost and with fire. Whatsoever I did, the Lord made it to
prosper. O holy Father, let all the hosts of heaven praise and adore
thy name!”
“God is love. This is the foundation of all my hopes. I feel much
shame, because of my infirmities; but I have also sweet
consolations.”
“I was seized with a violent pain in my stomach, and was
exceeding ill; however by the mercy and power of God, I went
through the duties of the day with delight, and could thank God for
pain, so as I never could before.”
“As I read my Greek testament this morning, my soul magnified
the Lord for the description, and progress of his work, contained in
the acts of the apostles. And while I am now writing, my soul is so
cheared with the fire of love, as I cannot describe, unless to such as
experience the same.”

“Lord, I have not publickly preached for thee this day, but I have
had many blessings from thee, and mine heart has been in thy work.
I beseech thee bless the labours of thy more faithful servants, whom
I have heard.”
*“I have great cause to praise God, that I am free from worldly
care. Surely I was appointed to this work in which I am engaged. O
that I may obtain mercy of the Lord, to be found faithful! O Jesus,
plead thou my cause in the heavens, and fill me with thy grace here
upon earth. All my hope of heaven stands in thee! O shew me, if
there be ought in me which thou abhorrest? And let me hear thee
say, ‘Thou art all fair, my love, there is no spot in thee.’”
“O that I could love and obey, as fast as I learn. Truth appears to
me every day with new lustre: new springs are opened, and the best
wine kept until last.”
“Sunday 19. After asking help from God, I preached my farewel
sermon (farewel indeed; It was the last he preached in London, and
the last day of his being there) at the Foundery, from Acts xx. 32.
And now, brethren I commend you to God, and to the word of his
grace, which is able to build you up, and to give you an inheritance
among all them which are sanctified. And in the evening, I bid them
farewel at the chapel in West-street, from Colossians ii. 6. As ye
have therefore received Christ Jesus the Lord, so walk ye in him. In
all the duties of the day, public and private, God was exceeding
gracious to me. I believe I never felt such strength of love. I was in
truth sick of love. I could not sufficiently praise him. All words come
far short of what I felt. Lord, thou hast given me much favour in the
eyes of this people. They shew it by words and deeds; yea prayers
and tears! Reward them a thousand fold. Bring me safe to Bristol,
that there I may shew forth the praises of the Lord, and declare thy
righteousness, and thy salvation. Amen, Lord Jesus.”

“Monday 20. After prayer with our family, I set out in the
machine. I read my Hebrew psalter, and the Christian pattern. I
found great tranquility of mind, and my spirit was refreshed with the
goodness of God. I conversed with three gentlemen, my companions
in the coach, on divine subjects. I prayed earnestly to God before I
set out, that my fellow travellers might not swear or curse, and the
Lord heard me; for so it was; they rather approved of scripture
subjects and studies. O the joy of a good conscience! And the rest
which the soul finds in the love of God. The Lord supplies the
absence of friends, and all things that are dear to us. His presence
makes our paradise. It is not where but what we are, which is the
great matter.”
“Thursday 23. At Bristol, I met Mr. W. T. under whose preaching
(as has been related) God gave me the clear witness of his forgiving
love. Our meeting was for the better. As iron sharpneth iron, so doth
the countenance of a man his friend. We remembered the years of
the right hand of the most high; and how the Lord filled our mouths
with laughter, when he brought back our captivity. Lord bless this
man, and make him faithful in all things! And now that I am come to
this city to preach the gospel of the kingdom, and spend my life and
strength in thy service; assist me O Lord, and make thy goodness
known to me. Give me wisdom and strength, O help me Lord Jesus,
to glorify thy name. Amen.”
“I read through to day, the epistle of St. James. And I do not
wonder that the proud, the sensual, and the lovers of the world, yea
all the ungodly of the earth, should find fault with it. In prayer with
the family, the spirit was poured out from on high upon us, and
great grace rested upon us all.”

“After prayer this morning, I began and read through, in Greek
and Latin the 2d. Epistle to Timothy, and found much instruction,
and reproof for my soul. O what a man ought a minister to be! How
holy, and how wise! What courage, zeal, patience and temperance,
are necessary for him in an especial manner in order to give account
of himself and others to God with joy.”
“Preaching on 1 John iv. 18. My mind was more clearly
enlightened than ever, to see that perfect love is Christian perfection.
By simple, but powerful faith, I desire to attain it; and to live and
grow in this love, till my spirit returns to God.”
With such desires, and in such meditations as these did he spend
his days and nights, longing and sighing for the sight of God
continually; and in his prayers, the violence of his affections, did not
a little increase the weakness of his body.
The End of the Eleventh Volume.

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