1

Chapter 4

Examples of Mathematical Models of Chemical Engineering Systems

(1) I ntroduction
This chapter is devoted to more complete examples. We will start with simple systems
and progress to more realistic and complex processes.

It would be impossible to include in this chapter mathematical models for all types of
chemical engineering systems. The main idea here is to learn good strategies and
procedures so that you can apply them to your specific problem.

The examples cover a number of very commonly encountered equipment tanks,
reactors of several types, and distillation columns.

Remember just go back to basics when faced with a new situation. Use the dynamic
mass and energy balances that apply to your system.

In each case set up all the equations required to describe the system. The purpose at
this stage is to translate the important phenomena occurring in the physical process
into quantitative mathematical equations.

2

(2) Series of I sothermal, Constant-Holdup CSTRs
The system is sketched in the following Figure, product B is produced and reactant A
is consumed in each of the three perfectly mixed reactors by a first-order reaction
occurring in the liquid.

Both temperatures and the liquid volumes are assumed to be constant (isothermal and
constant holdup). Density is assumed constant throughout the system, which is a binary
mixture of A and B.


Figure 1: Series of CSTRs.

With these assumptions in mind, start formulating the model. If the volume and density
of each tank are constant, the total mass in each tank is constant. Thus the total
continuity equation for the first reactor is:

d(V
1
)/dt = F
o
F
1
= 0 (1)
3

or simply, F
1
= F
o


Likewise total mass balances on tanks 2 and 3 give:

F
3
= F
2
= F
1
= F
o
= F (2)

Where F is defined as the throughput (m
3
/min).

In order to keep track of the amounts of reactant A and product B in each tank,
component continuity equations are needed.

However, since the system is binary a total mass of material in each tank and only one
component continuity equation are required. Either B or A can be used. If A is
arbitrarily chosen, the equations describing the dynamic changes in the amounts of
reactant A in each tank are (with units of k-mol of A/min):

V
1
dC
A1
/dt = F(C
Ao
- C
Al
) - V
1
k
1
C
A1



V
2
dC
A2
/dt = F(C
Al
- C
A2
) - V
2
k
2
C
A2


V
3
dC
A3
/dt = F(C
A2
- C
A3
) - V
3
k
3
C
A3
(3)

4

The specific reaction rates, k
n
, are given by the Arrhenius equation as:

k
n
= k
no
e
-E/RT
for n = 1, 2, 3 (4)

If the temperatures in the reactors are different, the ks are different. The n refers to the
stage number.

The volumes V
n
can be taken out of the time derivatives because they are constant. The
flows are all equal to F but can vary with time.

An energy equation is not required because isothermal operation is assumed. Any heat
addition or heat removal required to keep the reactors at constant temperature could be
calculated from a steady-state balance (zero-time derivatives of temperature).

The three first-order nonlinear ordinary differential equations given in Eqs. (3) are the
mathematical model of the system. The parameters that must be known are V
1
, V
2
, V
3
,
k
1
, k
2
, and k
3
.

The variables that must be specified before these equations can be solved are F and
C
Ao
. The initial conditions of the three concentrations (their values at time equal zero)
must also be known.
5

Check degrees of freedom of the system: There are three equations and, with the
parameters and forcing functions specified, there are only three unknowns or
dependent variables: C
Al
, C
A2
, and C
A3
. Consequently a solution should be possible, as
will be demonstrated later.


(3) CSTRs with Variable Holdups

If the previous example is modified slightly to permit the volumes in each reactor to
vary with time, both total and component continuity equations are required for each
reactor. To show the effects of higher-order kinetics, assume the reaction is now n
th
-
order in reactant A.


Reactor 1:
dV
1
/dt = F
o
- F
1

d(V
1
C
A1
)/dt = F
o
C
Ao
F
1
C
Al
- V
1
k
1
(C
A1
)
n
(3.6)

Reactor 2:
dV
2
/dt = F
1
F
2

d(V
2
C
A2
)/dt = F
1
C
Al
F
2
C
A2
V
2
k
2
(C
A2
)
n
(3.7)


Reactor 3:
dV
3
/dt = F
2
F
3

d(V
3
C
A3
)/dt = F
2
C
A2
F
3
C
A3
V
3
k
3
(C
A3
)
n
(3.8)
6



Figure 1: Series of CSTRs.


Comments:
Our mathematical model now contains six first-order nonlinear ordinary
differential equations. Parameters that must be known are k
1
, k
2
, k
3
, and n.

Initial conditions for all the dependent variables that are to be integrated must be
given: C
Al
, C
A2
, C
A3
, V
1
, V
2
, and V
3
. The forcing functions C
Ao
(t) and F
o
(t) must
also be given.

Check degrees of freedom of this system (i.e. closed system of equations):

7

There are 6 equations. But there are 9 unknowns: C
Al
, C
A2
, C
A3
, V
1
, V
2
, V
3
, F
1
, F
2
,
and F
3
. Clearly, this system is not sufficiently specified and a solution could not
be obtained.

What is missing in our modeling?

We have not specified how the flows out of the tanks are to be set. Physically there
would probably be control valves in the outlet lines to regulate the flows. How are
these control valves to be set?

A common configuration is to have the level in the tank controlled by the outflow,
i.e., a level controller opens the control valve on the exit line to increase the outflow
if the level in the tank increases. Thus there must be a relationship between tank
holdup and flow.



The f functions will describe the level controller and the control valve. These three
equations reduce the degrees of freedom to zero.

F
1
= f(V
1
), F
2
= f(V
2
), F
3
= f(V
3
) (3.9)
8

The reactors shown in this example would operate at atmospheric pressure if they
were open to the atmosphere.

If the reactors are not vented, they would run at the bubble point pressure for the
specified temperature and varying composition. Therefore the pressures could be
different in each reactor, and they would vary with time, even though temperatures
are assumed constant, as the C
A
s change.


(4) Two Heated Tanks

Let us consider a process in which two energy balances are needed to model the
system. We have the following situation:

F: the flow rate F of oil passing through two perfectly mixed tanks in series is
constant (90 ft
3
/min).
: the oil density is constant (40 lb/ft
3
).
C
p
: the oil heat capacity C
P
is constant (0.6 Btu/lb
o
F)
V
1
: the volume of the first tank is constant (450 ft
3
).
V
2
: the volume of the second tank is constant (90 ft
3
).
T
o
: the temperature of the oil entering the first tank at the initial steady state (150
o
F).
T
1
and T
2
: the temperatures in the two tanks (250
o
F).
9

A heating coil in the first tank uses steam to heat the oil. Let Q
1
be the heat addition
rate in the first tank.

Modeling:
There is one energy balance for each tank, and each will be similar to Eq. (2.26)
except there is no reaction involved in this process.



Since the flow rate is constant F
o
= F
1
= F
2
= F. Since volumes, densities, and heat
capacities are all constant, Eqs. (3.10) and (3.11) can be simplified as:


Energy balance for tank 1:

d( V
1
C
P
T
1
)/dt = C
P
(F
o
T
o
- F
1
T
1
) + Q
1
(3.10)

Energy balance for tank 2:

d(V
2
C
P
T
2
)/dt = C
P
(F
1
T
1
F
2
T
2
) (3.11)
V
1
C
P
dT
1
/dt = FC
P
(T
o
- T
1
) + Q
1
(3.12)
V
2
C
P
dT
2
/dt = FC
P
(T
1
T
2
) (3.13)
10

Check the degrees of freedom of this system: The parameter values that are known
are , C
P
, V
1
, V
2
, and F. The heat input to the first tank Q
1
would be set by the
position of the control valve in the steam line.

Thus we are left with two dependent variables, T
1
and T
2
, and we have two
equations, so the system is correctly specified.



(5) Gas-phase, Pressurized CSTR

Suppose a mixture of gases is fed into the reactor sketched in Figure 2. The reactor
is filled with reacting gases, which are perfectly mixed. A reversible reaction occurs:



The forward reaction is 1.5
th
order in A; the reverse reaction is first-order in B. Note
that the stoichiometric coefficient for A and the order of the reaction are not the
same. The mole fraction of reactant A in the reactor is y. The pressure inside the
k
1

2A B
k
2

11

vessel is P (absolute). Both P and y can vary with time. The volume of the reactor V
is constant.



Figure 2: Gas-phase CSTR

Assume an isothermal system, so the temperature T is constant. Perfect gases are
also assumed. The feed stream has a density
o
and a mole fraction y
o
of reactant A.
Its volumetric flow rate is F
o
.

The flow out of the reactor passes through a restriction (control valve) into another
vessel, which is held at a constant pressure P
D
(absolute). The outflow will vary with
the pressure and the composition of the reactor.

12

Flows through control valves can be simply given using the following formula (Eq.
14):
(14)
Where C
V
is the valve-sizing coefficient.

Density varies with pressure and composition (Eq. 15):

(15)

Where M = average molecular weight,
M
A
= molecular weight of reactant A, and
M
B
= molecular weight of product B.

The concentration of reactant in the reactor (Eq. 16) is:

(16)
with units of moles of A per unit volume.

The overall reaction rates (forward and reverse reactions) are:

F = C
V
{(P-P
D
)/}
0.5
(3.14)
= MP/RT = [y M
A
+ (1 - y) M
B
]P/RT (3.15)
C
A
= Py/RT
13


Now we can write the total and component continuity equations (Equations 17 &
18) as:

Total continuity: V d/dt =
o
F
o
F

Component A continuity: V dC
A
/dt =F
o
C
Ao
FC
A
2Vk
1
C
A
1.5
+Vk
2
C
B



There are five equations [Eqs. (14) - (18)] that make up the mathematical model of
this system.

The parameters that must be known are V, C
V
, k
1
, k
2
, R, M
A
, and M
B
.

The forcing functions (or inputs) could be P
D
,
o
, F
o
, and C
Ao
. This leaves five
unknowns (dependent variables): C
A
, , P, F, and y.






R
F
= k
1
C
A
1.5
R
R
= k
2
C
B

14

(6) Non-I sothermal CSTR
In the reactors studied so far, energy equations were not needed because we assumed
isothermal operations. Let us now consider a system in which temperature can change
with time.

An irreversible, exothermic reaction is carried out in a single perfectly mixed CSTR as
shown in Figure 3 A B

The reaction is n
th
-order in reactant A, has a rate constant k, and a heat of reaction
(Btu/lbmol of A reacted). Negligible heat losses and constant densities are assumed.



Figure 3: Non-isothermal CSTR.


15

To remove the heat of reaction, a cooling jacket surrounds the reactor. Cooling water is
added to the jacket at a volumetric flow rate F
J
and with an inlet temperature of T
Jo
.
The volume of water in the jacket V
J
is constant.

Assume a perfectly-mixed jacket, i.e., the temperature everywhere in the jacket is
constant T
J
. The heat transfer between the process at temperature T and the cooling
water at temperature T
J
is described by an overall heat transfer coefficient.

Q = UA
H
(T-T
J
) (19)

Where Q: heat transfer rate,
U : overall heat transfer coefficient,
A
H
: heat transfer area.

In general the heat transfer area could vary with the holdup in the reactor if some area
was not completely covered with reaction mass liquid at all times.






16

Equations describing the system are:



Where
J
: density of cooling water,
h: enthalpy of process liquid,
h
J
= enthalpy of cooling water.

A hydraulic relationship between reactor holdup and the flow out of the reactor is also
needed. A level controller is assumed to change the outflow as the volume in the tank
rises or falls: the higher the volume, the larger the outflow. The outflow is shut off
completely when the volume drops to a minimum value. The level controller is a
proportional-only feedback controller:
Reactor total continuity:
dV/dt = F
o
- F (3.20)

Reactor component A continuity:

d(VC
A
)/dt = F
o
C
Ao
- F C
A
- Vk(C
A
)
n
(3.21)

Reactor energy equation:

d(Vh)/dt = (F
o
h
o
F h) - Vk (C
A
)
n
- UA
H
(T - T
J
) (3.22)

Jacket energy equation:

J
V
J
dh
J
/dt =
J
F
J
(h
Jo
h
J
) + UA
H
(T - T
J
) (3.23)
17


F = K
V
(V - V
min
) (24)

Finally, we need to use the following relations for enthalpy:

h = C
P
T and h
J
= C
J
T
J
(25)
Where C
P
: heat capacity of the process liquid,
C
J
: heat capacity of the cooling water.

Also the Arrhenius relationship for k, k = k
o
e
-E/RT

The above five equations [(20) to (24)] will describe the above process.

Checking the degrees of freedom, we see that there are five equations and five
unknowns: V, F, C
A
, T, and T
J
. We must have initial conditions for these five
dependent variables. The forcing functions are T
o
, F
o
, C
Ao
, and F
J
.
The parameters that must be known are n, k
o
, E, R, , C
P
, U, A
H
,
J
, V
J
, C
J
, T
Jo
, Kv,
and V
min
.





18

(7) Single Component Vaporizer

Boiling systems represent some of the most interesting and important operations in
chemical engineering processing and are among the most difficult to model. To
describe these systems rigorously, conservation equations must be written for both the
vapor and liquid phases.

The basic problem is finding the rate of vaporization of material from the liquid phase
into the vapor phase. The equations used to describe the boiling rate should be
physically reasonable and mathematically convenient for solution.

Consider the vaporizer sketched in Figure 6. Liquefied petroleum gas (LPG) is fed into
a pressurized tank to hold the liquid level in the tank. We will assume that LPG is a
pure component of propane.

The liquid in the tank is assumed perfectly mixed. Heat is added at a rate Q to hold the
desired pressure in the tank by vaporizing the liquid at a rate W
V
(mass per time). Heat
losses and the mass of the tank are assumed negligible. Gas is drawn off the top of the
tank at a volumetric flow rate F
V
.


19



Figure 6: Single-component vaporizer.


(A) Steady-state Model
The simplest model would neglect the dynamics of both vapor and liquid phases and
relate the gas rate F
v
to the heat input by:


V
F
V
(H
V
h
o
) = Q (34)

Where H
V
: enthalpy of vapor leaving tank (Btu/lb or cal/g)
h
o
: enthalpy of liquid feed (Btu/lb or cal/g).

20

(B) Liquid-phase Dynamics Model
A more realistic model is obtained if we assume that the volume of the vapor phase is
small enough to make its dynamics negligible, i.e., a few moles of liquid have to be
vaporized. Therefore, we can consider that this pressure is always equal to the vapor
pressure of the liquid at any temperature (P = P
V
) and W
V
=
V
F
V
.

An energy equation for the liquid phase gives the temperature (as a function of time),
and the vapor-pressure relationship gives the pressure in the vaporizer at that
temperature.

A total continuity equation for the liquid phase is needed in addition to the two
controller equations relating pressure to heat input and liquid level to feed flow rate F
o

are required.

Q = f
1
(p) and F
o
= f
2
(V
L
) (35)

An equation of state for the vapor is needed to calculate density
V
from pressure or
temperature, apply the perfect gas law.

The liquid is assumed incompressible, i.e., C
v
= C
p
, and its internal energy is
(C
p
T). The enthalpy of the vapor leaving the vaporizer is (C
p
T +
v
)

21

Total Continuity: (d V
L
/dt) =
o
F
o
-
V
F
V
(36)

Energy: C
p
(d V
L
T/dt) =
o
C
p
F
o
T
o
-
V
F
V
(C
p
T +
v
) + Q (37)

Equation of State:
V
= PM/RT (38)

Vapor Pressure: Ln P = (A/T) + B (39)

Unknowns are Q, F
o
, P, V
L
,
V
, T can be obtained by using the closed system of
equations of (35) to (39).












22

(8) Multicomponent Flash Drum
A liquid stream at high T & P is flashed into a drum, i.e., its pressure is reduced. This
sudden expansion is irreversible and occurs at constant enthalpy. If the drum pressure
is lower than the bubble point pressure of the feed at the feed temperature, some of the
liquid feed will vaporize.



Figure 7: Flash drum.




23

Adiabatic condition, i.e. no heat loss, is assumed.

Liquid density,
L
, is a known function of temperature and composition:


L
= f (x
j
, T) (48)

Vapor density,
v
, is a known function of temperature, composition and pressure. If the
perfect gas law can be used:

v
= P M
av v
/ R T (49)

Where M
av v
is the average molecular weight of the gas and it can be calculated from:

M
av v
= M
j
y
j
(50)

(A) Steady-state Model
The simplest model of this system is one that neglects dynamics completely. Pressure
is assumed constant, and the steady state total and component continuity equations and
steady state energy balance are used. Vapor and liquid phases are assumed to be in
equilibrium.

Total continuity (# of Equation = 1):

o
F
o
=
v
F
v
+
L
F
L
(51)
24

Component continuity (# of Equations = N
C
- 1):
(
o
F
o
/ M
av o
) x
o J
= (
v
F
v
/ M
av v
) y
J
+ (
L
F
L
/ M
av L
) x
J
(52)

Vapor-Liquid equilibrium (# of Equations = N
C
):
y
J
= f (x
J
, T, P) (53)

Energy equation (# of equation = 1):

o
F
o
h
o
=
v
F
v
H +
L
F
L
h (54)

Thermal properties (# of equations = 2):
h
o
= f (x
o J
, T
o
)
h = f (x
J
, T) (55)
H = f (y
J
, T, P)

Comments:
1. The average molecular weight M
av
are calculated from the mole fractions in the
appropriate stream (use equation 50).

2. The number of variables in the system is 9+2(N
C
1):
These are
v
, F
v
, M
av v
, y
1
, y
2
,, y
NC-1


L
, F
L
, M
av L
, x
1
, x
2
,, x
NC-1

T, h, H
25


3. Pressure P and all the feed properties are given.

4. Number of equations available are (2N
C
+7):
Total continuity (1 Equation), Component continuity (N
C
1 Equations), Energy
Equation, Vapor- Liquid equilibrium (N
C
Equations), Densities of vapor and liquid
(2 Equations), Thermal properties (2 Equations), average molecular weight (2
Equations).

5. The system is specified by the algebraic equations listed above to provide the
traditional steady state equilibrium-flash calculation.


Case Study for Multi-component Flash Drum

Consider F, V, and L to be the molar flow rates of feed, and vapor and liquid
products from the flash drum. Also consider z
if
, y
i
, and x
i
to be the mole fractions of
the feed, and vapor and liquid products.

Total material balance: F = V + L (1)
Energy balance: F h
F
+ Q = V H
V
+ L h
L
(2)

26

When Q = 0 (i.e. adiabatic flash) or Q 0 (i.e. non- adiabatic flash). Equation 2 can
be rearranged after eliminating L using equation 1 to give the fraction of feed that
has been vaporized. Take R = V/F

From equation 2, h
F
+ (Q /F) = (V/F) H
V
+ (F-V/F) h
L

Then h
F
+ (Q /F) = R H
V
+ (1-R) h
L
= R H
V
+h
L
R h
L

Then h
F
+ (Q /F) = R (H
V
h
L
) + h
L


R = (V/F) = [h
F
h
L
+ (Q/F)] / (H
V
- h
L
) (3)

Component material balance:
F z
iF
= V y
i
+ L x
i
(N
C
equations) (4)
F z
iF
= V y
i
+ (F V) x
i

z
iF
= (V/F) y
i
+ (1 R) x
i

x
i
= (z
iF
- R y
i
) / (1 R) (5)

Let k
i
= y
i
/x
i

Where k - factor: vapor liquid equilibrium constant (distribution coefficient)
k
i
= (mole fraction in vapor / mole fraction in liquid) at equilibrium.

Since k
i
= y
i
/x
i
and x
i
= (z
iF
- R y
i
) / (1 R)
Then y
i
/k
i
= (z
iF
- R y
i
) / (1 R)
27

By making some mathematical manipulation:
y
i
= k
i
z
iF
/ [1 + R (k
i
1)] (6)

Taking into consideration as well that:
x
i
= y
i
= 1.0 (7)

Given F, z
iF
, i=1 to N
C
, T
F
, and T & P of the flash drum

Required products flow rates and their compositions

Steps
1. Calculate p
v
for each component @ T & P
2. Calculate k
i
for each component
3. Guess R = V/F
4. Calculate y
i
for each component using equation 6
5. Check if y
i
= 1.0, if yes keep going otherwise go back to step 3
6. Calculate x
i
for each component using equation x
i
= y
i
/k
i

7. Calculate L & V from the correct value of R and F.




28


(B) Dynamics of Multicomponent Adiabatic Flash Drum
A more practical dynamic model can be developed if we ignore the dynamics of the
vapor phase. The vapor is assumed to be always in equilibrium with the liquid.
Therefore, the conservation equations are written for the liquid phase only.


Assume P
F
> P
D
, and use pressure control on the drum and level control for the liquid in the
drum.

For adiabatic flash (Q = 0).

L
=
L
(x
i
, T)

v
= 1/v = M
V
avg
P
D
/RT
where M
V
avg
=
i
Nc
M
i
y
i

h
F
= h
F
(z
Fi
, T
F
)
h
L
= h
L
(x
i
, T)
H
V
= H
V
(y
i
, T, P
D
)

Given: P
D
and feed properties F,
F
, T
F
(or h
F
), z
Fi
, i = 1, 2 Nc, and M
F
avg
.

Variables or unknowns of the system:

v
, F
v
, M
V
avg
, y
i
, i = 1, 2, Nc 3+Nc

L
, F
L
, M
L
avg
, x
i
, i = 1, 2, Nc 3+Nc
T, h
L
, H
V
3_____
2Nc + 9
29





Equations of the system: (Neglecting the mass of vapor inside the drum)

Total Material Balance:
d(V
L

L
)/dt =
F
F -
V
F
V
-
L
F
L
1

Component Material Balance:
d(V
L

L
x
i
/M
L
avg
)/dt =
F
F z
Fi
/M
F
avg
-
V
F
V
y
i
/M
V
avg
-
L
F
L
x
i
/ M
L
avg
Nc-1

Energy Balance:
d(V
L

L
h
L
)/dt =
F
F h
F
-
V
F
V
H
V
-
L
F
L
h
L
1

Vapor-Liquid Equilibrium Relationship:
y
i
= x
i

i
P
i
s
/ P
D
Nc
Enthalpy Relationships: 2
Density Relationships: 2
Molecular Weight Relationships: 2
Level Control: F
L
= F
L
(V
L
) 1
Pressure Control: F
V
= F

(P
D
) 1_____
2Nc + 9
30

(9) Ideal Binary Distillation Column
Distillation column is probably the most popular and important process studied in the
chemical engineering literature.


Distillation is used in many chemical processes for:
o Separating of feed streams
o Purification of final and intermediate product streams.

Most columns handle multicomponent feeds, but many can be approximated by binary
or pseudo-binary mixtures. The purpose of studying this simplified case is to introduce
the basic equations. Then a more realistic system will be modeled.
31


32

Assumptions:
o Binary system

o Constant relative volatility throughout the column

o Theoretical trays (100% efficiency), i.e., vapor leaving the tray is in equilibrium
with the liquid on the tray. So we can use the simple vapor-liquid equilibrium
relationship of:


1 ) 1 (

n
n
n
x
x
y

(77)
Where x
n
: liquid composition on the n
th
tray (mole fraction of the more volatile
component)
y
n
: vapor composition on the n
th
tray (mole fraction of mole volatile
component)
: relative volatility (i.e. vapor pressure of the more volatile component
divided by the vapor pressure of the less volatile component)

Description:
o Single feed stream is fed as saturated liquid (at its bubble point) onto the feed tray
N
F
.
33


o Feed flow rate F (mol/min) and composition z (mole fraction of the more volatile
component).

o The overhead vapor is totally condensed in a condenser and flows into the reflux
drum, whose holdup of liquid is M
D
(moles).

o The contents of the drum is assumed to be perfectly mixed with composition x
D
.

o The liquid in the drum is at its bubble point.

o Reflux is pumped back to the top tray N
T
of the column at a rate R.

o Overhead distillate product rate is D.

o Neglect any delay time (i.e. deadtime) in the vapor line from the top of the
column to the reflux drum and in the reflux line back to the top tray. In industrial
scale columns this is usually a good assumption, but not in small scale laboratory
columns.

o Notice that y
NT
is not equal to x
D
dynamically. They are equal only at steady state.

34

o At the base of the column, liquid bottom product is removed at a rate B with a
composition x
B
. Vapor boilup is generated in a thermosiphon reboiler at a rate V.

o Liquid circulates from the bottom of the column through the tubes in the vertical
tube-in-shell reboiler because of the smaller density of the vapor-liquid mixture in
the reboiler in the reboiler tubes.

o We will assume that the liquids in the reboiler and in the base of the column are
perfectly mixed together and have the same composition x
B
and total holdup M
B

(moles). The circulation rates through well-designed thermosiphon reboilers are
quite high, so this assumption is usually a good one.

o The composition of the vapor leaving the base of the column and entering tray 1
is y
B
. It is in equilibrium with the liquid with composition x
B
.

o The column contains a total N
T
of theoretical trays.

o The liquid holdup on each tray including the downcomer is M
n
. The liquid on
each tray is assumed to be perfectly mixed with composition x
n
.

o The holdup of the vapor is assumed to be negligible throughout the system
because the vapor density is much smaller than the liquid density.
35


o An important assumption to make is that of equimolal overflow. If the molar heats
of vaporization of the two components are about the same, whenever one mole of
vapor condenses, it vaporizes a mole of liquid. Heat loses and sensible heat
effects are assumed negligible. These assumptions mean that the vapor and liquid
rates through the stripping and rectifying sections will be constant under steady
state conditions. Therefore the operating lines on the familiar McCabe-Thiele
diagram are straight lines.

o However, we are interested here in dynamic conditions. The assumptions above
including negligible vapor holdup mean that the vapor rate through all trays of the
column is the same dynamically as well as steady state:

V = V
1
= V
2
= .= V
NT


o Remember these Vs are not necessarily constant with time. The vapor boilup can
be manipulated dynamically. The mathematical effect of assuming equimolal
overflow is that we do not need an energy equation for each tray. This is quite a
significant simplification.

o The liquid rates throughout the column will not be the same dynamically. They
will depend on the fluid mechanics of the tray. Often a simple Francis weir
36

formula relationship is used to relate the liquid holdup on the tray M
n
to the liquid
flow rate leaving the tray L
n
.

F
L
= 3.33 L
w
(h
ow
)

1.5
(78)

Where F
L
: liquid flow rate over weir (ft
3
/s)
L
w
: length of weir (ft)
h
ow
: height of liquid over weir (ft)

o More rigorous relationship can be obtained from the detailed tray hydraulic
equations to include the effects of vapor rate, densities, compositions, etc. We will
assume a simple functional relationship between liquid holdup and liquid rate as:

M
n
= f (L
n
) (79)

o Finally we neglect the dynamics of the condenser and reboiler because the
dynamic response of these heat exchangers is much faster than the response of the
column itself.




37

Calculations:

1- Variables Specified:
Feed rate, F, composition, z
F
, and enthalpy, h
F
.
Product purities required, x
B
and x
D


2- Other Parameters which must be selected:
Column pressure: P affects vapor-liquid equilibrium, VLE.
Reflux ratio: have to be calculated from R
min
.

3- Data Required:
VLE data: very important to have good thermodynamic data.
Physical properties: molecular weight, enthalpy and density for both vapor
and liquid.

4- Design Procedures:
a- Graphical (Binary systems) Methods:
Equimolal Overflow: use McCabe-Thiele x-y diagram.
Non-equimolal Overflow: use Ponchon-Savarit H-x-y diagram.

b- Computer: plate-to-plate calculations using equations for
Total material balance: 1 per Tray
38

Component material balances: Nc-1 per Tray
Energy balance: 1 per Tray
VLE relationship: 1 per Component per Tray
Enthalpies:
For liquids: h = f (x
i
, T) 1 per Tray
For vapors H = f (y
i
, T, P) 1 per Tray

5- Results should include
1. Total number of trays, N
T
, and height of the column after specifying tray
spacing.
2. Feed tray location, N
F
.
3. Liquid and vapor loads (rates): L and V.
4. Column diameter: vapor load limiting (the feed tray and/or the top tray has
the highest vapor load).
5. Reboiler duty, Q
R
.
6. Condenser duty, Q
C
.
7. Top and bottom temperatures and the temperature distribution in the
column.
8. Estimate the overall heat transfer coefficient, U, and then calculate the heat
transfer areas of the condenser and reboiler.
9. Design temperature and pressure of the column.
10. Calculate the column shell thickness.
39

With all these assumptions in mind, we are ready to write the equations
describing the system as follows:

1. Condenser & Reflux Drum
Total continuity:

Component continuity:

2. Top tray (@ N
T
)
Total continuity:

Component continuity:





40

3. Next to top (@ N
T
1)
Total continuity:

Component continuity:

4. n
th
tray
Total continuity:

Component continuity:








41

Feed tray (n =N
F
)
Total continuity:

Component continuity:


5. First tray (n =1)
Total continuity:

Component continuity:







42

6. Reboiler & column base
Total continuity:

Component continuity:



Comments:
1. Each tray and the column base have equilibrium equations (Equation 77).
2. Each tray has a hydraulic equation (Equation 79)
3. We also need two equations representing the level controllers on the column
base and reflux drum shown in the distillation diagram. These two equations
should be given as:
D = f
1
(M
D
)
B = f
2
(M
B
) (94)




43

4. Let us now examine the degrees of freedom of the system. The feed rate F and
composition z are given.

Number of variables:









44

Number of equations:


5. Therefore the system is underspecified by two equations. From a control
viewpoint this means that there are only two variables that can be controlled
(i.e. can be fixed).

The two variables that must be specified are reflux ratio R and vapor boilup V.
They can be held constant or they can be changed by two controllers to try to
hold some other two variables constant.


45

(10) Batch Distillation with Holdup
Batch distillation is frequently used for small volume products. One column can be
used to separate a multicomponent mixture instead of requiring (N
C
1) continuous
columns.

The energy consumption in batch distillation is usually higher than in continuous, but
with small volume.

The following Figure shows a typical batch distillation column. Fresh feed is charged
into the still pot and heated until it begins to boil. The vapor moves up the column and
is condensed in the condenser.

The condensate liquid runs into the reflux drum. When a liquid level has been
established in the column, reflux is pumped back to the top tray in the column.

The column is run on total reflux until the overhead distillate composition of the
lightest component x
D1
reaches its specification purity. Then a distillate product, which
is the lightest component, is withdrawn at some rate. Eventually, the amount of
component 1 in the still pot gets very low and the x
D1
purity of the distillate drops.

There is a period of time when the distillate contains too little of component 1 to be
used for the product and also too little of component 2 to be used for the next heavier
46

product. Therefore, a slop cut must be withdrawn until x
D2
builds up to its
specification. Then a second product is withdrawn. Thus multiple products can be
made from a single column.

Figure 14 Batch distillation

47


The process can run at varying pressures and reflux ratios during each of the product
and slop cuts.

Optimum design of the columns (i.e. diameter and number of trays) and optimum
operation are important in reducing batch times.

We assume the following:
o Theoretical trays
o Equimolal overflow
o Constant relative volatilities
o Tray liquid hold up is constant
o Reflux drum hold up is constant
o Vapor boilup rate is constant at V (moles/hr)
o The reflux drum, column trays, and still pot are all initially filled with
material of composition x
BOi


The total amount of material charged to the column is M
BO
(moles). This material can
be fresh feed with composition z
i
or a mixture of fresh feed and the slop cuts.

48

The composition in the still pot at the beginning of the batch is x
BOi
. The composition
in the still pot at any point in time is x
Bi
. The instantaneous holdup in the still pot is
M
B
.
The equations describing the batch distillation of a multicomponent mixture are given
below as:

1. Still pot:







2. Tray n:

49





3. Tray N
T
(top tray):




4. Reflux drum: